BA
Brian Allen
Tue, Aug 2, 2022 11:46 PM
Hi,
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
Any ideas?
Thanks,
Brian
Hi,
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
Any ideas?
Thanks,
Brian
JB
Jordan Brown
Wed, Aug 3, 2022 12:04 AM
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster in
complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that stretching
is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can chime in -
but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major (long) and minor
(short) axis dimensions; the other parts of the curve mathematically
flow from those two measurements.
You can make something else, but you can't make it using this technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2: https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3*u*(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls together.
Note that the four coordinates correspond to the coordinates of the
control points in a GUI drawing program: the start point, control point
associated with the start, the control point associated with the end,
and the end point. I have sometimes created OpenSCAD objects by tracing
an object in a drawing program and then reading off the coordinates of
the various control points.
Note that Bézier functions are not the only mathematical curves. (For
instance, they cannot exactly match an arc of a circle.) I'm sure there
are numerous other functions that also produce curves.
On 8/2/2022 4:46 PM, Brian Allen wrote:
> I created an oval using:
> inMm = 25.4;
> fullW = 35.5 * inMm;
> fullL = 54.5 * inMm;
> resize([fullW, fullL, 1])
> circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster in
complicated cases.
> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that stretching
is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can chime in -
but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major (long) and minor
(short) axis dimensions; the other parts of the curve mathematically
flow from those two measurements.
You can make something else, but you can't make it using this technique.
> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2: https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3*u*(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls together.
Note that the four coordinates correspond to the coordinates of the
control points in a GUI drawing program: the start point, control point
associated with the start, the control point associated with the end,
and the end point. I have sometimes created OpenSCAD objects by tracing
an object in a drawing program and then reading off the coordinates of
the various control points.
Note that Bézier functions are not the only mathematical curves. (For
instance, they cannot exactly match an arc of a circle.) I'm sure there
are numerous other functions that also produce curves.
JB
Jordan Brown
Wed, Aug 3, 2022 12:25 AM
More generally, if you can define a function that returns the values
around your curve, you can use list comprehension to generate a list of
points that you can then feed to polygon().
For instance, here's a sequence that will generate a unit circle:
// Define f to be a function that accepts a value between 0 and 1
// and returns a corresponding coordinate.
function f(i) = let(a = i*360) [ cos(a), sin(a) ];
points = [ for (i = [0:0.01:1]) f(i) ];
polygon(points);
Here's a sequence that will generate a race track:
// Define f to be a function that accepts a value between 0 and 1
// and returns a corresponding coordinate.
function f(i) =
i < 0.125
? let(a = i*2*360) [ cos(a) + 0.5, sin(a) ]
: i < 0.375
? [ (0.25 - i)*4, 1 ]
: i < 0.625
? let(a = (i-0.25)*2*360) [ cos(a) - 0.5, sin(a) ]
: i < 0.875
? [ (i - 0.75)*4, -1 ]
: let(a = (i-1)*2*360) [ cos(a) + 0.5, sin(a) ];
points = [ for (i = [0:0.01:1]) f(i) ];
polygon(points);
Not that there aren't easier ways to do both of those (circle, hull of
two circles), but the technique is general: it will work for any shape
that you can represent as a function of [0..1].
More generally, if you can define a function that returns the values
around your curve, you can use list comprehension to generate a list of
points that you can then feed to polygon().
For instance, here's a sequence that will generate a unit circle:
// Define f to be a function that accepts a value between 0 and 1
// and returns a corresponding coordinate.
function f(i) = let(a = i*360) [ cos(a), sin(a) ];
points = [ for (i = [0:0.01:1]) f(i) ];
polygon(points);
Here's a sequence that will generate a race track:
// Define f to be a function that accepts a value between 0 and 1
// and returns a corresponding coordinate.
function f(i) =
i < 0.125
? let(a = i*2*360) [ cos(a) + 0.5, sin(a) ]
: i < 0.375
? [ (0.25 - i)*4, 1 ]
: i < 0.625
? let(a = (i-0.25)*2*360) [ cos(a) - 0.5, sin(a) ]
: i < 0.875
? [ (i - 0.75)*4, -1 ]
: let(a = (i-1)*2*360) [ cos(a) + 0.5, sin(a) ];
points = [ for (i = [0:0.01:1]) f(i) ];
polygon(points);
Not that there aren't easier ways to do both of those (circle, hull of
two circles), but the technique is general: it will work for any shape
that you can represent as a function of [0..1].
MM
Michael Marx
Wed, Aug 3, 2022 12:45 AM
Or old school.
Scan the shape to PNG, load it via surface(). Or DXF/SVG via import().
Or
Plot your points on graph paper, for the top half of the curve.
Put the values in an array, calculate the negative side, make a polygon.
From: Jordan Brown [mailto:openscad@jordan.maileater.net]
Sent: Wed, 3 Aug 2022 10:26
To: OpenSCAD general discussion; Brian Allen
Subject: [OpenSCAD] Re: Transforming an Oval
More generally, if you can define a function that returns the values around your curve, you can use list comprehension to generate a list of points that you can then feed to polygon().
For instance, here's a sequence that will generate a unit circle:
// Define f to be a function that accepts a value between 0 and 1
// and returns a corresponding coordinate.
function f(i) = let(a = i*360) [ cos(a), sin(a) ];
points = [ for (i = [0:0.01:1]) f(i) ];
polygon(points);
Here's a sequence that will generate a race track:
// Define f to be a function that accepts a value between 0 and 1
// and returns a corresponding coordinate.
function f(i) =
i < 0.125
? let(a = i2360) [ cos(a) + 0.5, sin(a) ]
: i < 0.375
? [ (0.25 - i)*4, 1 ]
: i < 0.625
? let(a = (i-0.25)2360) [ cos(a) - 0.5, sin(a) ]
: i < 0.875
? [ (i - 0.75)*4, -1 ]
: let(a = (i-1)2360) [ cos(a) + 0.5, sin(a) ];
points = [ for (i = [0:0.01:1]) f(i) ];
polygon(points);
Not that there aren't easier ways to do both of those (circle, hull of two circles), but the technique is general: it will work for any shape that you can represent as a function of [0..1].
--
This email has been checked for viruses by AVG.
https://www.avg.com
Or old school.
Scan the shape to PNG, load it via surface(). Or DXF/SVG via import().
Or
Plot your points on graph paper, for the top half of the curve.
Put the values in an array, calculate the negative side, make a polygon.
_____
From: Jordan Brown [mailto:openscad@jordan.maileater.net]
Sent: Wed, 3 Aug 2022 10:26
To: OpenSCAD general discussion; Brian Allen
Subject: [OpenSCAD] Re: Transforming an Oval
More generally, if you can define a function that returns the values around your curve, you can use list comprehension to generate a list of points that you can then feed to polygon().
For instance, here's a sequence that will generate a unit circle:
// Define f to be a function that accepts a value between 0 and 1
// and returns a corresponding coordinate.
function f(i) = let(a = i*360) [ cos(a), sin(a) ];
points = [ for (i = [0:0.01:1]) f(i) ];
polygon(points);
Here's a sequence that will generate a race track:
// Define f to be a function that accepts a value between 0 and 1
// and returns a corresponding coordinate.
function f(i) =
i < 0.125
? let(a = i*2*360) [ cos(a) + 0.5, sin(a) ]
: i < 0.375
? [ (0.25 - i)*4, 1 ]
: i < 0.625
? let(a = (i-0.25)*2*360) [ cos(a) - 0.5, sin(a) ]
: i < 0.875
? [ (i - 0.75)*4, -1 ]
: let(a = (i-1)*2*360) [ cos(a) + 0.5, sin(a) ];
points = [ for (i = [0:0.01:1]) f(i) ];
polygon(points);
Not that there aren't easier ways to do both of those (circle, hull of two circles), but the technique is general: it will work for any shape that you can represent as a function of [0..1].
--
This email has been checked for viruses by AVG.
https://www.avg.com
BA
Brian Allen
Wed, Aug 3, 2022 1:19 AM
Thank you both. I really appreciate it. One of those solutions will work for my needs, I'm sure.
Brian
On Tue, Aug 2, 2022, at 6:45 PM, Michael Marx wrote:
Or old school.
Scan the shape to PNG, load it via surface(). Or DXF/SVG via import().
Or
Plot your points on graph paper, for the top half of the curve.
Put the values in an array, calculate the negative side, make a polygon.
From: Jordan Brown [mailto:openscad@jordan.maileater.net]
Sent: Wed, 3 Aug 2022 10:26
To: OpenSCAD general discussion; Brian Allen
Subject: [OpenSCAD] Re: Transforming an Oval
More generally, if you can define a function that returns the values around your curve, you can use list comprehension to generate a list of points that you can then feed to polygon().
For instance, here's a sequence that will generate a unit circle:
// Define f to be a function that accepts a value between 0 and 1
// and returns a corresponding coordinate.
function f(i) = let(a = i*360) [ cos(a), sin(a) ];
points = [ for (i = [0:0.01:1]) f(i) ];
polygon(points);
Here's a sequence that will generate a race track:
// Define f to be a function that accepts a value between 0 and 1
// and returns a corresponding coordinate.
function f(i) =
i < 0.125
? let(a = i2360) [ cos(a) + 0.5, sin(a) ]
: i < 0.375
? [ (0.25 - i)*4, 1 ]
: i < 0.625
? let(a = (i-0.25)2360) [ cos(a) - 0.5, sin(a) ]
: i < 0.875
? [ (i - 0.75)*4, -1 ]
: let(a = (i-1)2360) [ cos(a) + 0.5, sin(a) ];
points = [ for (i = [0:0.01:1]) f(i) ];
polygon(points);
Not that there aren't easier ways to do both of those (circle, hull of two circles), but the technique is general: it will work for any shape that you can represent as a function of [0..1].
Thank you both. I really appreciate it. One of those solutions will work for my needs, I'm sure.
Brian
On Tue, Aug 2, 2022, at 6:45 PM, Michael Marx wrote:
> Or old school.
>
> Scan the shape to PNG, load it via surface(). Or DXF/SVG via import().
> Or
> Plot your points on graph paper, for the top half of the curve.
> Put the values in an array, calculate the negative side, make a polygon.
>
>
> **From:** Jordan Brown [mailto:openscad@jordan.maileater.net]
> **Sent:** Wed, 3 Aug 2022 10:26
> **To:** OpenSCAD general discussion; Brian Allen
> **Subject:** [OpenSCAD] Re: Transforming an Oval
>
> More generally, if you can define a function that returns the values around your curve, you can use list comprehension to generate a list of points that you can then feed to polygon().
>
> For instance, here's a sequence that will generate a unit circle:
> // Define f to be a function that accepts a value between 0 and 1
> // and returns a corresponding coordinate.
> function f(i) = let(a = i*360) [ cos(a), sin(a) ];
>
> points = [ for (i = [0:0.01:1]) f(i) ];
> polygon(points);
> Here's a sequence that will generate a race track:
> // Define f to be a function that accepts a value between 0 and 1
> // and returns a corresponding coordinate.
> function f(i) =
> i < 0.125
> ? let(a = i*2*360) [ cos(a) + 0.5, sin(a) ]
> : i < 0.375
> ? [ (0.25 - i)*4, 1 ]
> : i < 0.625
> ? let(a = (i-0.25)*2*360) [ cos(a) - 0.5, sin(a) ]
> : i < 0.875
> ? [ (i - 0.75)*4, -1 ]
> : let(a = (i-1)*2*360) [ cos(a) + 0.5, sin(a) ];
>
> points = [ for (i = [0:0.01:1]) f(i) ];
> polygon(points);
> Not that there aren't easier ways to do both of those (circle, hull of two circles), but the technique is general: it will work for any shape that you can represent as a function of [0..1].
>
>
>
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Sanjeev Prabhakar
Wed, Aug 3, 2022 2:35 AM
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor axis "r1" and
"r2" respectively and with center "cp" and number of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, openscad@jordan.maileater.net
wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster in complicated
cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that stretching is
an ellipse.
I'm not 100% sure - maybe one of the real math wizards can chime in - but
I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major (long) and minor
(short) axis dimensions; the other parts of the curve mathematically flow
from those two measurements.
You can make something else, but you can't make it using this technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2: https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls together.
Note that the four coordinates correspond to the coordinates of the
control points in a GUI drawing program: the start point, control point
associated with the start, the control point associated with the end, and
the end point. I have sometimes created OpenSCAD objects by tracing an
object in a drawing program and then reading off the coordinates of the
various control points.
Note that Bézier functions are not the only mathematical curves. (For
instance, they cannot exactly match an arc of a circle.) I'm sure there
are numerous other functions that also produce curves.
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor axis "r1" and
"r2" respectively and with center "cp" and number of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <openscad@jordan.maileater.net>
wrote:
> On 8/2/2022 4:46 PM, Brian Allen wrote:
>
> I created an oval using:
> inMm = 25.4;
> fullW = 35.5 * inMm;
> fullL = 54.5 * inMm;
> resize([fullW, fullL, 1])
> circle(d=fullW);
>
>
> Tip: use scale() rather than resize(); it previews faster in complicated
> cases.
>
> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>
>
> You're taking a circle and stretching it; the result of that stretching is
> an ellipse.
>
> I'm not 100% sure - maybe one of the real math wizards can chime in - but
> I think the shape you are asking for would not be an ellipse.
>
> I believe that an ellipse is fully defined by its major (long) and minor
> (short) axis dimensions; the other parts of the curve mathematically flow
> from those two measurements.
>
> You can make something else, but you can't make it using this technique.
>
> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>
>
> You can use a Bézier function.
>
> Using BOSL2: https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>
> DIY, here's a function that I cobbled together some years ago:
>
> // Bezier functions from https://www.thingiverse.com/thing:8443
> // but that yielded either single points or a raft of triangles;
> // this yields a vector of points that you can then concatenate
> // with other pieces to form a single polygon.
> // If we were really clever, I think it would be possible to
> // automatically space the output points based on how linear
> // the curve is at that point. But right now I'm not that clever.
> function BEZ03(u) = pow((1-u), 3);
> function BEZ13(u) = 3*u*(pow((1-u),2));
> function BEZ23(u) = 3*(pow(u,2))*(1-u);
> function BEZ33(u) = pow(u,3);
>
> function PointAlongBez4(p0, p1, p2, p3, u) = [
> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>
> // p0 - start point
> // p1 - control point 1, line departs p0 headed this way
> // p2 - control point 2, line arrives at p3 from this way
> // p3 - end point
> // segs - number of segments
> function bez(p0, p1, p2, p3, segs) = [
> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
> ];
>
> It is that final function bez() that I actually call, e.g.
>
> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>
> often concatenating the results of several bez() calls together.
>
> Note that the four coordinates correspond to the coordinates of the
> control points in a GUI drawing program: the start point, control point
> associated with the start, the control point associated with the end, and
> the end point. I have sometimes created OpenSCAD objects by tracing an
> object in a drawing program and then reading off the coordinates of the
> various control points.
>
>
> Note that Bézier functions are not the only mathematical curves. (For
> instance, they cannot exactly match an arc of a circle.) I'm sure there
> are numerous other functions that also produce curves.
> _______________________________________________
> OpenSCAD mailing list
> To unsubscribe send an email to discuss-leave@lists.openscad.org
>
AM
Adrian Mariano
Wed, Aug 3, 2022 2:50 AM
Here's a challenge for you: draw an "ellipse" polygon with N segments that
are all the same length that circumscribes a perfect ellipse with given
semiaxes a and b. I tried to solve this problem and could not find a
solution. I can draw an inscribed N segment "ellipse" polygon with all
segments the same length---though it's not easy---but the circumscribed
case eludes me. I think of the inscribed case as, every vertex is located
on the ellipse and the circumscribed case, every segment is tangent to the
ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar sprabhakar2006@gmail.com
wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor axis "r1" and
"r2" respectively and with center "cp" and number of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, openscad@jordan.maileater.net
wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster in complicated
cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that stretching
is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can chime in - but
I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major (long) and minor
(short) axis dimensions; the other parts of the curve mathematically flow
from those two measurements.
You can make something else, but you can't make it using this technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2: https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls together.
Note that the four coordinates correspond to the coordinates of the
control points in a GUI drawing program: the start point, control point
associated with the start, the control point associated with the end, and
the end point. I have sometimes created OpenSCAD objects by tracing an
object in a drawing program and then reading off the coordinates of the
various control points.
Note that Bézier functions are not the only mathematical curves. (For
instance, they cannot exactly match an arc of a circle.) I'm sure there
are numerous other functions that also produce curves.
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org
Here's a challenge for you: draw an "ellipse" polygon with N segments that
are all the same length that circumscribes a perfect ellipse with given
semiaxes a and b. I tried to solve this problem and could not find a
solution. I can draw an *inscribed* N segment "ellipse" polygon with all
segments the same length---though it's not easy---but the circumscribed
case eludes me. I think of the inscribed case as, every vertex is located
on the ellipse and the circumscribed case, every segment is tangent to the
ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <sprabhakar2006@gmail.com>
wrote:
> In openscad drawing a ellipse is not too difficult
> Refer following code
>
> function to draw an ellipse with semi-major and semi-minor axis "r1" and
> "r2" respectively and with center "cp" and number of segment "s"
> /
> // example:
> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>
> function ellipse(r1,r2,cp,s=30)=
> let(
> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
> )sec;
>
> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <openscad@jordan.maileater.net>
> wrote:
>
>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>
>> I created an oval using:
>> inMm = 25.4;
>> fullW = 35.5 * inMm;
>> fullL = 54.5 * inMm;
>> resize([fullW, fullL, 1])
>> circle(d=fullW);
>>
>>
>> Tip: use scale() rather than resize(); it previews faster in complicated
>> cases.
>>
>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>
>>
>> You're taking a circle and stretching it; the result of that stretching
>> is an ellipse.
>>
>> I'm not 100% sure - maybe one of the real math wizards can chime in - but
>> I think the shape you are asking for would not be an ellipse.
>>
>> I believe that an ellipse is fully defined by its major (long) and minor
>> (short) axis dimensions; the other parts of the curve mathematically flow
>> from those two measurements.
>>
>> You can make something else, but you can't make it using this technique.
>>
>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>
>>
>> You can use a Bézier function.
>>
>> Using BOSL2: https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>
>> DIY, here's a function that I cobbled together some years ago:
>>
>> // Bezier functions from https://www.thingiverse.com/thing:8443
>> // but that yielded either single points or a raft of triangles;
>> // this yields a vector of points that you can then concatenate
>> // with other pieces to form a single polygon.
>> // If we were really clever, I think it would be possible to
>> // automatically space the output points based on how linear
>> // the curve is at that point. But right now I'm not that clever.
>> function BEZ03(u) = pow((1-u), 3);
>> function BEZ13(u) = 3*u*(pow((1-u),2));
>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>> function BEZ33(u) = pow(u,3);
>>
>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>
>> // p0 - start point
>> // p1 - control point 1, line departs p0 headed this way
>> // p2 - control point 2, line arrives at p3 from this way
>> // p3 - end point
>> // segs - number of segments
>> function bez(p0, p1, p2, p3, segs) = [
>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>> ];
>>
>> It is that final function bez() that I actually call, e.g.
>>
>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>
>> often concatenating the results of several bez() calls together.
>>
>> Note that the four coordinates correspond to the coordinates of the
>> control points in a GUI drawing program: the start point, control point
>> associated with the start, the control point associated with the end, and
>> the end point. I have sometimes created OpenSCAD objects by tracing an
>> object in a drawing program and then reading off the coordinates of the
>> various control points.
>>
>>
>> Note that Bézier functions are not the only mathematical curves. (For
>> instance, they cannot exactly match an arc of a circle.) I'm sure there
>> are numerous other functions that also produce curves.
>> _______________________________________________
>> OpenSCAD mailing list
>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>
> _______________________________________________
> OpenSCAD mailing list
> To unsubscribe send an email to discuss-leave@lists.openscad.org
>
GH
gene heskett
Wed, Aug 3, 2022 2:53 AM
On 8/2/22 19:50, Brian Allen wrote:
Hi,
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
Any ideas?
Thanks,
Brian
I'm not sure if it would meet your needs, but I made the elliptical
armature for a 50/1 harmonic drive by
using the scale module on a cylinder, then scaling it 1.05 in the x
direction, and .95 in the y direction.
It worked pretty good and printed in PETG lasted well too, but given the
motors to drive it, and
the heat in the flexible bearings I also printed, it was too slow to do
what I wanted to do. So I
bought a 5/1 worm, put a 3NM 3 phase motor on it and I was off to the
races. Its the B drive
on my 6040 milling machine.
Cheers, Gene Heskett.
"There are four boxes to be used in defense of liberty:
soap, ballot, jury, and ammo. Please use in that order."
-Ed Howdershelt (Author, 1940)
If we desire respect for the law, we must first make the law respectable.
On 8/2/22 19:50, Brian Allen wrote:
> Hi,
>
> I created an oval using:
> inMm = 25.4;
> fullW = 35.5 * inMm;
> fullL = 54.5 * inMm;
> resize([fullW, fullL, 1])
> circle(d=fullW);
>
> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>
> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>
> Any ideas?
>
> Thanks,
> Brian
> I'm not sure if it would meet your needs, but I made the elliptical
> armature for a 50/1 harmonic drive by
using the scale module on a cylinder, then scaling it 1.05 in the x
direction, and .95 in the y direction.
It worked pretty good and printed in PETG lasted well too, but given the
motors to drive it, and
the heat in the flexible bearings I also printed, it was too slow to do
what I wanted to do. So I
bought a 5/1 worm, put a 3NM 3 phase motor on it and I was off to the
races. Its the B drive
on my 6040 milling machine.
>
> _______________________________________________
> OpenSCAD mailing list
> To unsubscribe send an email to discuss-leave@lists.openscad.org
Cheers, Gene Heskett.
--
"There are four boxes to be used in defense of liberty:
soap, ballot, jury, and ammo. Please use in that order."
-Ed Howdershelt (Author, 1940)
If we desire respect for the law, we must first make the law respectable.
- Louis D. Brandeis
Genes Web page <http://geneslinuxbox.net:6309/>
FH
Father Horton
Wed, Aug 3, 2022 2:55 AM
Except by brute force (step through it with sufficiently small steps), it’s
not possible because there’s no closed form expression for the distance
between two points on a general ellipse (as opposed to special cases like a
circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano avm4@cornell.edu wrote:
Here's a challenge for you: draw an "ellipse" polygon with N segments
that are all the same length that circumscribes a perfect ellipse with
given semiaxes a and b. I tried to solve this problem and could not find a
solution. I can draw an inscribed N segment "ellipse" polygon with all
segments the same length---though it's not easy---but the circumscribed
case eludes me. I think of the inscribed case as, every vertex is located
on the ellipse and the circumscribed case, every segment is tangent to the
ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor axis "r1" and
"r2" respectively and with center "cp" and number of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, openscad@jordan.maileater.net
wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster in
complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that stretching
is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can chime in -
but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major (long) and minor
(short) axis dimensions; the other parts of the curve mathematically flow
from those two measurements.
You can make something else, but you can't make it using this technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls together.
Note that the four coordinates correspond to the coordinates of the
control points in a GUI drawing program: the start point, control point
associated with the start, the control point associated with the end, and
the end point. I have sometimes created OpenSCAD objects by tracing an
object in a drawing program and then reading off the coordinates of the
various control points.
Note that Bézier functions are not the only mathematical curves. (For
instance, they cannot exactly match an arc of a circle.) I'm sure there
are numerous other functions that also produce curves.
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org
Except by brute force (step through it with sufficiently small steps), it’s
not possible because there’s no closed form expression for the distance
between two points on a general ellipse (as opposed to special cases like a
circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <avm4@cornell.edu> wrote:
> Here's a challenge for you: draw an "ellipse" polygon with N segments
> that are all the same length that circumscribes a perfect ellipse with
> given semiaxes a and b. I tried to solve this problem and could not find a
> solution. I can draw an *inscribed* N segment "ellipse" polygon with all
> segments the same length---though it's not easy---but the circumscribed
> case eludes me. I think of the inscribed case as, every vertex is located
> on the ellipse and the circumscribed case, every segment is tangent to the
> ellipse
>
> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
> sprabhakar2006@gmail.com> wrote:
>
>> In openscad drawing a ellipse is not too difficult
>> Refer following code
>>
>> function to draw an ellipse with semi-major and semi-minor axis "r1" and
>> "r2" respectively and with center "cp" and number of segment "s"
>> /
>> // example:
>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>
>> function ellipse(r1,r2,cp,s=30)=
>> let(
>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>> )sec;
>>
>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <openscad@jordan.maileater.net>
>> wrote:
>>
>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>
>>> I created an oval using:
>>> inMm = 25.4;
>>> fullW = 35.5 * inMm;
>>> fullL = 54.5 * inMm;
>>> resize([fullW, fullL, 1])
>>> circle(d=fullW);
>>>
>>>
>>> Tip: use scale() rather than resize(); it previews faster in
>>> complicated cases.
>>>
>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>
>>>
>>> You're taking a circle and stretching it; the result of that stretching
>>> is an ellipse.
>>>
>>> I'm not 100% sure - maybe one of the real math wizards can chime in -
>>> but I think the shape you are asking for would not be an ellipse.
>>>
>>> I believe that an ellipse is fully defined by its major (long) and minor
>>> (short) axis dimensions; the other parts of the curve mathematically flow
>>> from those two measurements.
>>>
>>> You can make something else, but you can't make it using this technique.
>>>
>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>
>>>
>>> You can use a Bézier function.
>>>
>>> Using BOSL2:
>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>
>>> DIY, here's a function that I cobbled together some years ago:
>>>
>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>> // but that yielded either single points or a raft of triangles;
>>> // this yields a vector of points that you can then concatenate
>>> // with other pieces to form a single polygon.
>>> // If we were really clever, I think it would be possible to
>>> // automatically space the output points based on how linear
>>> // the curve is at that point. But right now I'm not that clever.
>>> function BEZ03(u) = pow((1-u), 3);
>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>> function BEZ33(u) = pow(u,3);
>>>
>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>
>>> // p0 - start point
>>> // p1 - control point 1, line departs p0 headed this way
>>> // p2 - control point 2, line arrives at p3 from this way
>>> // p3 - end point
>>> // segs - number of segments
>>> function bez(p0, p1, p2, p3, segs) = [
>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>> ];
>>>
>>> It is that final function bez() that I actually call, e.g.
>>>
>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>
>>> often concatenating the results of several bez() calls together.
>>>
>>> Note that the four coordinates correspond to the coordinates of the
>>> control points in a GUI drawing program: the start point, control point
>>> associated with the start, the control point associated with the end, and
>>> the end point. I have sometimes created OpenSCAD objects by tracing an
>>> object in a drawing program and then reading off the coordinates of the
>>> various control points.
>>>
>>>
>>> Note that Bézier functions are not the only mathematical curves. (For
>>> instance, they cannot exactly match an arc of a circle.) I'm sure there
>>> are numerous other functions that also produce curves.
>>> _______________________________________________
>>> OpenSCAD mailing list
>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>
>> _______________________________________________
>> OpenSCAD mailing list
>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>
> _______________________________________________
> OpenSCAD mailing list
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>
AM
Adrian Mariano
Wed, Aug 3, 2022 3:09 AM
Eh, there's no such thing as "brute force" in this situation. Step through
what with sufficiently small steps? Of course an iterative algorithm is
necessary. This is true for most interesting problems. No big deal. Show
me an iterative algorithm that converges to the circumscribing polygon.
When I tried to devise such an algorithm I couldn't find one that would
converge. Writing an algorithm that would converge to the inscribed
polygon seemed to be straight forward. Note also that computing arc
length of the ellipse doesn't really help here. You can do that by solving
an elliptical integral, or summing up small segments. But what's the
connection between arc length of the ellipse and a circumscribing polygon
with equal segment lengths? If there's a connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton fatherhorton@gmail.com
wrote:
Except by brute force (step through it with sufficiently small steps),
it’s not possible because there’s no closed form expression for the
distance between two points on a general ellipse (as opposed to special
cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano avm4@cornell.edu wrote:
Here's a challenge for you: draw an "ellipse" polygon with N segments
that are all the same length that circumscribes a perfect ellipse with
given semiaxes a and b. I tried to solve this problem and could not find a
solution. I can draw an inscribed N segment "ellipse" polygon with all
segments the same length---though it's not easy---but the circumscribed
case eludes me. I think of the inscribed case as, every vertex is located
on the ellipse and the circumscribed case, every segment is tangent to the
ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor axis "r1" and
"r2" respectively and with center "cp" and number of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster in
complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that stretching
is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can chime in -
but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major (long) and
minor (short) axis dimensions; the other parts of the curve mathematically
flow from those two measurements.
You can make something else, but you can't make it using this technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls together.
Note that the four coordinates correspond to the coordinates of the
control points in a GUI drawing program: the start point, control point
associated with the start, the control point associated with the end, and
the end point. I have sometimes created OpenSCAD objects by tracing an
object in a drawing program and then reading off the coordinates of the
various control points.
Note that Bézier functions are not the only mathematical curves. (For
instance, they cannot exactly match an arc of a circle.) I'm sure there
are numerous other functions that also produce curves.
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org
Eh, there's no such thing as "brute force" in this situation. Step through
*what* with sufficiently small steps? Of course an iterative algorithm is
necessary. This is true for most interesting problems. No big deal. Show
me an iterative algorithm that converges to the circumscribing polygon.
When I tried to devise such an algorithm I couldn't find one that would
converge. Writing an algorithm that would converge to the *inscribed*
polygon seemed to be straight forward. Note also that computing arc
length of the ellipse doesn't really help here. You can do that by solving
an elliptical integral, or summing up small segments. But what's the
connection between arc length of the ellipse and a circumscribing polygon
with equal segment lengths? If there's a connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton <fatherhorton@gmail.com>
wrote:
> Except by brute force (step through it with sufficiently small steps),
> it’s not possible because there’s no closed form expression for the
> distance between two points on a general ellipse (as opposed to special
> cases like a circle).
>
> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <avm4@cornell.edu> wrote:
>
>> Here's a challenge for you: draw an "ellipse" polygon with N segments
>> that are all the same length that circumscribes a perfect ellipse with
>> given semiaxes a and b. I tried to solve this problem and could not find a
>> solution. I can draw an *inscribed* N segment "ellipse" polygon with all
>> segments the same length---though it's not easy---but the circumscribed
>> case eludes me. I think of the inscribed case as, every vertex is located
>> on the ellipse and the circumscribed case, every segment is tangent to the
>> ellipse
>>
>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>> sprabhakar2006@gmail.com> wrote:
>>
>>> In openscad drawing a ellipse is not too difficult
>>> Refer following code
>>>
>>> function to draw an ellipse with semi-major and semi-minor axis "r1" and
>>> "r2" respectively and with center "cp" and number of segment "s"
>>> /
>>> // example:
>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>
>>> function ellipse(r1,r2,cp,s=30)=
>>> let(
>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>> )sec;
>>>
>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>> openscad@jordan.maileater.net> wrote:
>>>
>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>
>>>> I created an oval using:
>>>> inMm = 25.4;
>>>> fullW = 35.5 * inMm;
>>>> fullL = 54.5 * inMm;
>>>> resize([fullW, fullL, 1])
>>>> circle(d=fullW);
>>>>
>>>>
>>>> Tip: use scale() rather than resize(); it previews faster in
>>>> complicated cases.
>>>>
>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>
>>>>
>>>> You're taking a circle and stretching it; the result of that stretching
>>>> is an ellipse.
>>>>
>>>> I'm not 100% sure - maybe one of the real math wizards can chime in -
>>>> but I think the shape you are asking for would not be an ellipse.
>>>>
>>>> I believe that an ellipse is fully defined by its major (long) and
>>>> minor (short) axis dimensions; the other parts of the curve mathematically
>>>> flow from those two measurements.
>>>>
>>>> You can make something else, but you can't make it using this technique.
>>>>
>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>
>>>>
>>>> You can use a Bézier function.
>>>>
>>>> Using BOSL2:
>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>
>>>> DIY, here's a function that I cobbled together some years ago:
>>>>
>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>> // but that yielded either single points or a raft of triangles;
>>>> // this yields a vector of points that you can then concatenate
>>>> // with other pieces to form a single polygon.
>>>> // If we were really clever, I think it would be possible to
>>>> // automatically space the output points based on how linear
>>>> // the curve is at that point. But right now I'm not that clever.
>>>> function BEZ03(u) = pow((1-u), 3);
>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>> function BEZ33(u) = pow(u,3);
>>>>
>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>
>>>> // p0 - start point
>>>> // p1 - control point 1, line departs p0 headed this way
>>>> // p2 - control point 2, line arrives at p3 from this way
>>>> // p3 - end point
>>>> // segs - number of segments
>>>> function bez(p0, p1, p2, p3, segs) = [
>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>> ];
>>>>
>>>> It is that final function bez() that I actually call, e.g.
>>>>
>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>
>>>> often concatenating the results of several bez() calls together.
>>>>
>>>> Note that the four coordinates correspond to the coordinates of the
>>>> control points in a GUI drawing program: the start point, control point
>>>> associated with the start, the control point associated with the end, and
>>>> the end point. I have sometimes created OpenSCAD objects by tracing an
>>>> object in a drawing program and then reading off the coordinates of the
>>>> various control points.
>>>>
>>>>
>>>> Note that Bézier functions are not the only mathematical curves. (For
>>>> instance, they cannot exactly match an arc of a circle.) I'm sure there
>>>> are numerous other functions that also produce curves.
>>>> _______________________________________________
>>>> OpenSCAD mailing list
>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>
>>> _______________________________________________
>>> OpenSCAD mailing list
>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>
>> _______________________________________________
>> OpenSCAD mailing list
>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>
> _______________________________________________
> OpenSCAD mailing list
> To unsubscribe send an email to discuss-leave@lists.openscad.org
>
JB
Jordan Brown
Wed, Aug 3, 2022 3:11 AM
Isn't this a general ellipse for axes a and b?
module ellipse(a, b) {
scale([a/b, 1]) circle(b);
}
ellipse(10,20);
Of course, a function that generates the points is useful in other ways.
Isn't this a general ellipse for axes a and b?
module ellipse(a, b) {
scale([a/b, 1]) circle(b);
}
ellipse(10,20);
Of course, a function that generates the points is useful in other ways.
SP
Sanjeev Prabhakar
Wed, Aug 3, 2022 4:01 AM
I am quite occupied these days
Will give it a try this weekend
On Wed, 3 Aug, 2022, 8:21 am Adrian Mariano, avm4@cornell.edu wrote:
Here's a challenge for you: draw an "ellipse" polygon with N segments
that are all the same length that circumscribes a perfect ellipse with
given semiaxes a and b. I tried to solve this problem and could not find a
solution. I can draw an inscribed N segment "ellipse" polygon with all
segments the same length---though it's not easy---but the circumscribed
case eludes me. I think of the inscribed case as, every vertex is located
on the ellipse and the circumscribed case, every segment is tangent to the
ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor axis "r1" and
"r2" respectively and with center "cp" and number of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, openscad@jordan.maileater.net
wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster in
complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that stretching
is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can chime in -
but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major (long) and minor
(short) axis dimensions; the other parts of the curve mathematically flow
from those two measurements.
You can make something else, but you can't make it using this technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls together.
Note that the four coordinates correspond to the coordinates of the
control points in a GUI drawing program: the start point, control point
associated with the start, the control point associated with the end, and
the end point. I have sometimes created OpenSCAD objects by tracing an
object in a drawing program and then reading off the coordinates of the
various control points.
Note that Bézier functions are not the only mathematical curves. (For
instance, they cannot exactly match an arc of a circle.) I'm sure there
are numerous other functions that also produce curves.
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org
I am quite occupied these days
Will give it a try this weekend
On Wed, 3 Aug, 2022, 8:21 am Adrian Mariano, <avm4@cornell.edu> wrote:
> Here's a challenge for you: draw an "ellipse" polygon with N segments
> that are all the same length that circumscribes a perfect ellipse with
> given semiaxes a and b. I tried to solve this problem and could not find a
> solution. I can draw an *inscribed* N segment "ellipse" polygon with all
> segments the same length---though it's not easy---but the circumscribed
> case eludes me. I think of the inscribed case as, every vertex is located
> on the ellipse and the circumscribed case, every segment is tangent to the
> ellipse
>
> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
> sprabhakar2006@gmail.com> wrote:
>
>> In openscad drawing a ellipse is not too difficult
>> Refer following code
>>
>> function to draw an ellipse with semi-major and semi-minor axis "r1" and
>> "r2" respectively and with center "cp" and number of segment "s"
>> /
>> // example:
>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>
>> function ellipse(r1,r2,cp,s=30)=
>> let(
>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>> )sec;
>>
>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <openscad@jordan.maileater.net>
>> wrote:
>>
>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>
>>> I created an oval using:
>>> inMm = 25.4;
>>> fullW = 35.5 * inMm;
>>> fullL = 54.5 * inMm;
>>> resize([fullW, fullL, 1])
>>> circle(d=fullW);
>>>
>>>
>>> Tip: use scale() rather than resize(); it previews faster in
>>> complicated cases.
>>>
>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>
>>>
>>> You're taking a circle and stretching it; the result of that stretching
>>> is an ellipse.
>>>
>>> I'm not 100% sure - maybe one of the real math wizards can chime in -
>>> but I think the shape you are asking for would not be an ellipse.
>>>
>>> I believe that an ellipse is fully defined by its major (long) and minor
>>> (short) axis dimensions; the other parts of the curve mathematically flow
>>> from those two measurements.
>>>
>>> You can make something else, but you can't make it using this technique.
>>>
>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>
>>>
>>> You can use a Bézier function.
>>>
>>> Using BOSL2:
>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>
>>> DIY, here's a function that I cobbled together some years ago:
>>>
>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>> // but that yielded either single points or a raft of triangles;
>>> // this yields a vector of points that you can then concatenate
>>> // with other pieces to form a single polygon.
>>> // If we were really clever, I think it would be possible to
>>> // automatically space the output points based on how linear
>>> // the curve is at that point. But right now I'm not that clever.
>>> function BEZ03(u) = pow((1-u), 3);
>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>> function BEZ33(u) = pow(u,3);
>>>
>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>
>>> // p0 - start point
>>> // p1 - control point 1, line departs p0 headed this way
>>> // p2 - control point 2, line arrives at p3 from this way
>>> // p3 - end point
>>> // segs - number of segments
>>> function bez(p0, p1, p2, p3, segs) = [
>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>> ];
>>>
>>> It is that final function bez() that I actually call, e.g.
>>>
>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>
>>> often concatenating the results of several bez() calls together.
>>>
>>> Note that the four coordinates correspond to the coordinates of the
>>> control points in a GUI drawing program: the start point, control point
>>> associated with the start, the control point associated with the end, and
>>> the end point. I have sometimes created OpenSCAD objects by tracing an
>>> object in a drawing program and then reading off the coordinates of the
>>> various control points.
>>>
>>>
>>> Note that Bézier functions are not the only mathematical curves. (For
>>> instance, they cannot exactly match an arc of a circle.) I'm sure there
>>> are numerous other functions that also produce curves.
>>> _______________________________________________
>>> OpenSCAD mailing list
>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>
>> _______________________________________________
>> OpenSCAD mailing list
>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>
> _______________________________________________
> OpenSCAD mailing list
> To unsubscribe send an email to discuss-leave@lists.openscad.org
>
RW
Raymond West
Wed, Aug 3, 2022 11:47 AM
sort of interesting. I think whatever you do to a circular object using
scad basic operations, you end up with the same shape curve. For an
oval, you only need to match a quarter of the outline. (I had thought
that a slice through a tilted cone would give something different)
$fn=80;
projection()
translate([0,0,-20])
rotate([0,25,0])
cylinder(h=50,d1=50,d2=20);
scale([.905,1])
#circle(25);
for a one off, as others said, draw a polygon/whatever. I guess a more
interesting result could be by creating slices of a circle( as Jordan
mentioned) but also vary the length of each slice by some other
function. Make it smaller and offset can iron out the bumps if necessary.
Best wishes,
Ray
On 03/08/2022 00:46, Brian Allen wrote:
Hi,
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
Any ideas?
Thanks,
Brian
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org
sort of interesting. I think whatever you do to a circular object using
scad basic operations, you end up with the same shape curve. For an
oval, you only need to match a quarter of the outline. (I had thought
that a slice through a tilted cone would give something different)
$fn=80;
projection()
translate([0,0,-20])
rotate([0,25,0])
cylinder(h=50,d1=50,d2=20);
scale([.905,1])
#circle(25);
for a one off, as others said, draw a polygon/whatever. I guess a more
interesting result could be by creating slices of a circle( as Jordan
mentioned) but also vary the length of each slice by some other
function. Make it smaller and offset can iron out the bumps if necessary.
Best wishes,
Ray
On 03/08/2022 00:46, Brian Allen wrote:
> Hi,
>
> I created an oval using:
> inMm = 25.4;
> fullW = 35.5 * inMm;
> fullL = 54.5 * inMm;
> resize([fullW, fullL, 1])
> circle(d=fullW);
>
> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>
> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>
> Any ideas?
>
> Thanks,
> Brian
>
> _______________________________________________
> OpenSCAD mailing list
> To unsubscribe send an email to discuss-leave@lists.openscad.org
BC
Bob Carlson
Wed, Aug 3, 2022 8:07 PM
Wouldn’t it come pretty close to use you inscribed points as the tangent points, the find the intersections of the tangents?
-Bob
Tucson AZ
On Aug 2, 2022, at 19:50, Adrian Mariano avm4@cornell.edu wrote:
Here's a challenge for you: draw an "ellipse" polygon with N segments that are all the same length that circumscribes a perfect ellipse with given semiaxes a and b. I tried to solve this problem and could not find a solution. I can draw an inscribed N segment "ellipse" polygon with all segments the same length---though it's not easy---but the circumscribed case eludes me. I think of the inscribed case as, every vertex is located on the ellipse and the circumscribed case, every segment is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <sprabhakar2006@gmail.com mailto:sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor axis "r1" and "r2" respectively and with center "cp" and number of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <openscad@jordan.maileater.net mailto:openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster in complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can chime in - but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major (long) and minor (short) axis dimensions; the other parts of the curve mathematically flow from those two measurements.
You can make something else, but you can't make it using this technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2: https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443 https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls together.
Note that the four coordinates correspond to the coordinates of the control points in a GUI drawing program: the start point, control point associated with the start, the control point associated with the end, and the end point. I have sometimes created OpenSCAD objects by tracing an object in a drawing program and then reading off the coordinates of the various control points.
Note that Bézier functions are not the only mathematical curves. (For instance, they cannot exactly match an arc of a circle.) I'm sure there are numerous other functions that also produce curves.
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Wouldn’t it come pretty close to use you inscribed points as the tangent points, the find the intersections of the tangents?
-Bob
Tucson AZ
On Aug 2, 2022, at 19:50, Adrian Mariano <avm4@cornell.edu> wrote:
Here's a challenge for you: draw an "ellipse" polygon with N segments that are all the same length that circumscribes a perfect ellipse with given semiaxes a and b. I tried to solve this problem and could not find a solution. I can draw an *inscribed* N segment "ellipse" polygon with all segments the same length---though it's not easy---but the circumscribed case eludes me. I think of the inscribed case as, every vertex is located on the ellipse and the circumscribed case, every segment is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <sprabhakar2006@gmail.com <mailto:sprabhakar2006@gmail.com>> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor axis "r1" and "r2" respectively and with center "cp" and number of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <openscad@jordan.maileater.net <mailto:openscad@jordan.maileater.net>> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
> I created an oval using:
> inMm = 25.4;
> fullW = 35.5 * inMm;
> fullL = 54.5 * inMm;
> resize([fullW, fullL, 1])
> circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster in complicated cases.
> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can chime in - but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major (long) and minor (short) axis dimensions; the other parts of the curve mathematically flow from those two measurements.
You can make something else, but you can't make it using this technique.
> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2: https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves <https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves>
DIY, here's a function that I cobbled together some years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443 <https://www.thingiverse.com/thing:8443>
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3*u*(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls together.
Note that the four coordinates correspond to the coordinates of the control points in a GUI drawing program: the start point, control point associated with the start, the control point associated with the end, and the end point. I have sometimes created OpenSCAD objects by tracing an object in a drawing program and then reading off the coordinates of the various control points.
Note that Bézier functions are not the only mathematical curves. (For instance, they cannot exactly match an arc of a circle.) I'm sure there are numerous other functions that also produce curves.
_______________________________________________
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org <mailto:discuss-leave@lists.openscad.org>
_______________________________________________
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org <mailto:discuss-leave@lists.openscad.org>
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AM
Adrian Mariano
Wed, Aug 3, 2022 10:26 PM
No, using inscribed equal segment points as tangent points is a terrible
approximation of the equal segment circumscribed polygon. Here's an
example for N=6 where green is inscribed with equal lengths and red is the
derived circumscribed polygon and some edges are almost double the length
of others. And if the ellipse gets skinnier it gets even worse. Note
also that there are presumably two (symmetric) solutions to the problem,
one tangent at the right point of the ellipse and one with a vertex at the
right tip of the ellipse.
If I recall correctly, I tried to start with this and refine it to get a
good solution, but my approach doesn't converge.
On Wed, Aug 3, 2022 at 4:08 PM Bob Carlson bob@rjcarlson.com wrote:
Wouldn’t it come pretty close to use you inscribed points as the tangent
points, the find the intersections of the tangents?
-Bob
Tucson AZ
On Aug 2, 2022, at 19:50, Adrian Mariano avm4@cornell.edu wrote:
Here's a challenge for you: draw an "ellipse" polygon with N segments
that are all the same length that circumscribes a perfect ellipse with
given semiaxes a and b. I tried to solve this problem and could not find a
solution. I can draw an inscribed N segment "ellipse" polygon with all
segments the same length---though it's not easy---but the circumscribed
case eludes me. I think of the inscribed case as, every vertex is located
on the ellipse and the circumscribed case, every segment is tangent to the
ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor axis "r1" and
"r2" respectively and with center "cp" and number of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, openscad@jordan.maileater.net
wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster in
complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that stretching
is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can chime in -
but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major (long) and minor
(short) axis dimensions; the other parts of the curve mathematically flow
from those two measurements.
You can make something else, but you can't make it using this technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls together.
Note that the four coordinates correspond to the coordinates of the
control points in a GUI drawing program: the start point, control point
associated with the start, the control point associated with the end, and
the end point. I have sometimes created OpenSCAD objects by tracing an
object in a drawing program and then reading off the coordinates of the
various control points.
Note that Bézier functions are not the only mathematical curves. (For
instance, they cannot exactly match an arc of a circle.) I'm sure there
are numerous other functions that also produce curves.
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org
No, using inscribed equal segment points as tangent points is a terrible
approximation of the equal segment circumscribed polygon. Here's an
example for N=6 where green is inscribed with equal lengths and red is the
derived circumscribed polygon and some edges are almost double the length
of others. And if the ellipse gets skinnier it gets even worse. Note
also that there are presumably two (symmetric) solutions to the problem,
one tangent at the right point of the ellipse and one with a vertex at the
right tip of the ellipse.
If I recall correctly, I tried to start with this and refine it to get a
good solution, but my approach doesn't converge.
On Wed, Aug 3, 2022 at 4:08 PM Bob Carlson <bob@rjcarlson.com> wrote:
> Wouldn’t it come pretty close to use you inscribed points as the tangent
> points, the find the intersections of the tangents?
>
> -Bob
> Tucson AZ
>
>
>
> On Aug 2, 2022, at 19:50, Adrian Mariano <avm4@cornell.edu> wrote:
>
> Here's a challenge for you: draw an "ellipse" polygon with N segments
> that are all the same length that circumscribes a perfect ellipse with
> given semiaxes a and b. I tried to solve this problem and could not find a
> solution. I can draw an *inscribed* N segment "ellipse" polygon with all
> segments the same length---though it's not easy---but the circumscribed
> case eludes me. I think of the inscribed case as, every vertex is located
> on the ellipse and the circumscribed case, every segment is tangent to the
> ellipse
>
> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
> sprabhakar2006@gmail.com> wrote:
>
>> In openscad drawing a ellipse is not too difficult
>> Refer following code
>>
>> function to draw an ellipse with semi-major and semi-minor axis "r1" and
>> "r2" respectively and with center "cp" and number of segment "s"
>> /
>> // example:
>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>
>> function ellipse(r1,r2,cp,s=30)=
>> let(
>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>> )sec;
>>
>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <openscad@jordan.maileater.net>
>> wrote:
>>
>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>
>>> I created an oval using:
>>> inMm = 25.4;
>>> fullW = 35.5 * inMm;
>>> fullL = 54.5 * inMm;
>>> resize([fullW, fullL, 1])
>>> circle(d=fullW);
>>>
>>>
>>> Tip: use scale() rather than resize(); it previews faster in
>>> complicated cases.
>>>
>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>
>>>
>>> You're taking a circle and stretching it; the result of that stretching
>>> is an ellipse.
>>>
>>> I'm not 100% sure - maybe one of the real math wizards can chime in -
>>> but I think the shape you are asking for would not be an ellipse.
>>>
>>> I believe that an ellipse is fully defined by its major (long) and minor
>>> (short) axis dimensions; the other parts of the curve mathematically flow
>>> from those two measurements.
>>>
>>> You can make something else, but you can't make it using this technique.
>>>
>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>
>>>
>>> You can use a Bézier function.
>>>
>>> Using BOSL2:
>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>
>>> DIY, here's a function that I cobbled together some years ago:
>>>
>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>> // but that yielded either single points or a raft of triangles;
>>> // this yields a vector of points that you can then concatenate
>>> // with other pieces to form a single polygon.
>>> // If we were really clever, I think it would be possible to
>>> // automatically space the output points based on how linear
>>> // the curve is at that point. But right now I'm not that clever.
>>> function BEZ03(u) = pow((1-u), 3);
>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>> function BEZ33(u) = pow(u,3);
>>>
>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>
>>> // p0 - start point
>>> // p1 - control point 1, line departs p0 headed this way
>>> // p2 - control point 2, line arrives at p3 from this way
>>> // p3 - end point
>>> // segs - number of segments
>>> function bez(p0, p1, p2, p3, segs) = [
>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>> ];
>>>
>>> It is that final function bez() that I actually call, e.g.
>>>
>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>
>>> often concatenating the results of several bez() calls together.
>>>
>>> Note that the four coordinates correspond to the coordinates of the
>>> control points in a GUI drawing program: the start point, control point
>>> associated with the start, the control point associated with the end, and
>>> the end point. I have sometimes created OpenSCAD objects by tracing an
>>> object in a drawing program and then reading off the coordinates of the
>>> various control points.
>>>
>>>
>>> Note that Bézier functions are not the only mathematical curves. (For
>>> instance, they cannot exactly match an arc of a circle.) I'm sure there
>>> are numerous other functions that also produce curves.
>>> _______________________________________________
>>> OpenSCAD mailing list
>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>
>> _______________________________________________
>> OpenSCAD mailing list
>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>
> _______________________________________________
> OpenSCAD mailing list
> To unsubscribe send an email to discuss-leave@lists.openscad.org
>
> _______________________________________________
> OpenSCAD mailing list
> To unsubscribe send an email to discuss-leave@lists.openscad.org
>
FH
Father Horton
Wed, Aug 3, 2022 11:40 PM
I think our emails crossed. I was addressing how to approximate an ellipse
with constant-length segments in general, not how to solve your problem.
When I had to do something similar, I used polar coordinates and stepped up
the value of theta until the distance from the previous point was correct.
(In the constant-length version, I'd keep a running total of the maximum
desired length so that round-off errors don't accumulate.)
Figuring out the circumscribed polygon is, as you note, a greater
challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano avm4@cornell.edu wrote:
Eh, there's no such thing as "brute force" in this situation. Step
through what with sufficiently small steps? Of course an iterative
algorithm is necessary. This is true for most interesting problems. No
big deal. Show me an iterative algorithm that converges to the
circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton fatherhorton@gmail.com
wrote:
Except by brute force (step through it with sufficiently small steps),
it’s not possible because there’s no closed form expression for the
distance between two points on a general ellipse (as opposed to special
cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano avm4@cornell.edu wrote:
Here's a challenge for you: draw an "ellipse" polygon with N segments
that are all the same length that circumscribes a perfect ellipse with
given semiaxes a and b. I tried to solve this problem and could not find a
solution. I can draw an inscribed N segment "ellipse" polygon with all
segments the same length---though it's not easy---but the circumscribed
case eludes me. I think of the inscribed case as, every vertex is located
on the ellipse and the circumscribed case, every segment is tangent to the
ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor axis "r1"
and "r2" respectively and with center "cp" and number of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster in
complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that
stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can chime in -
but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major (long) and
minor (short) axis dimensions; the other parts of the curve mathematically
flow from those two measurements.
You can make something else, but you can't make it using this
technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls together.
Note that the four coordinates correspond to the coordinates of the
control points in a GUI drawing program: the start point, control point
associated with the start, the control point associated with the end, and
the end point. I have sometimes created OpenSCAD objects by tracing an
object in a drawing program and then reading off the coordinates of the
various control points.
Note that Bézier functions are not the only mathematical curves. (For
instance, they cannot exactly match an arc of a circle.) I'm sure there
are numerous other functions that also produce curves.
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org
I think our emails crossed. I was addressing how to approximate an ellipse
with constant-length segments in general, not how to solve your problem.
When I had to do something similar, I used polar coordinates and stepped up
the value of theta until the distance from the previous point was correct.
(In the constant-length version, I'd keep a running total of the maximum
desired length so that round-off errors don't accumulate.)
Figuring out the circumscribed polygon is, as you note, a greater
challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <avm4@cornell.edu> wrote:
> Eh, there's no such thing as "brute force" in this situation. Step
> through *what* with sufficiently small steps? Of course an iterative
> algorithm is necessary. This is true for most interesting problems. No
> big deal. Show me an iterative algorithm that converges to the
> circumscribing polygon. When I tried to devise such an algorithm I
> couldn't find one that would converge. Writing an algorithm that would
> converge to the *inscribed* polygon seemed to be straight forward. Note
> also that computing arc length of the ellipse doesn't really help here.
> You can do that by solving an elliptical integral, or summing up small
> segments. But what's the connection between arc length of the ellipse and
> a circumscribing polygon with equal segment lengths? If there's a
> connection, I don't see it.
>
> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <fatherhorton@gmail.com>
> wrote:
>
>> Except by brute force (step through it with sufficiently small steps),
>> it’s not possible because there’s no closed form expression for the
>> distance between two points on a general ellipse (as opposed to special
>> cases like a circle).
>>
>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <avm4@cornell.edu> wrote:
>>
>>> Here's a challenge for you: draw an "ellipse" polygon with N segments
>>> that are all the same length that circumscribes a perfect ellipse with
>>> given semiaxes a and b. I tried to solve this problem and could not find a
>>> solution. I can draw an *inscribed* N segment "ellipse" polygon with all
>>> segments the same length---though it's not easy---but the circumscribed
>>> case eludes me. I think of the inscribed case as, every vertex is located
>>> on the ellipse and the circumscribed case, every segment is tangent to the
>>> ellipse
>>>
>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>>> sprabhakar2006@gmail.com> wrote:
>>>
>>>> In openscad drawing a ellipse is not too difficult
>>>> Refer following code
>>>>
>>>> function to draw an ellipse with semi-major and semi-minor axis "r1"
>>>> and "r2" respectively and with center "cp" and number of segment "s"
>>>> /
>>>> // example:
>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>>
>>>> function ellipse(r1,r2,cp,s=30)=
>>>> let(
>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>>> )sec;
>>>>
>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>>> openscad@jordan.maileater.net> wrote:
>>>>
>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>>
>>>>> I created an oval using:
>>>>> inMm = 25.4;
>>>>> fullW = 35.5 * inMm;
>>>>> fullL = 54.5 * inMm;
>>>>> resize([fullW, fullL, 1])
>>>>> circle(d=fullW);
>>>>>
>>>>>
>>>>> Tip: use scale() rather than resize(); it previews faster in
>>>>> complicated cases.
>>>>>
>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>>
>>>>>
>>>>> You're taking a circle and stretching it; the result of that
>>>>> stretching is an ellipse.
>>>>>
>>>>> I'm not 100% sure - maybe one of the real math wizards can chime in -
>>>>> but I think the shape you are asking for would not be an ellipse.
>>>>>
>>>>> I believe that an ellipse is fully defined by its major (long) and
>>>>> minor (short) axis dimensions; the other parts of the curve mathematically
>>>>> flow from those two measurements.
>>>>>
>>>>> You can make something else, but you can't make it using this
>>>>> technique.
>>>>>
>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>>
>>>>>
>>>>> You can use a Bézier function.
>>>>>
>>>>> Using BOSL2:
>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>>
>>>>> DIY, here's a function that I cobbled together some years ago:
>>>>>
>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>>> // but that yielded either single points or a raft of triangles;
>>>>> // this yields a vector of points that you can then concatenate
>>>>> // with other pieces to form a single polygon.
>>>>> // If we were really clever, I think it would be possible to
>>>>> // automatically space the output points based on how linear
>>>>> // the curve is at that point. But right now I'm not that clever.
>>>>> function BEZ03(u) = pow((1-u), 3);
>>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>>> function BEZ33(u) = pow(u,3);
>>>>>
>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>>
>>>>> // p0 - start point
>>>>> // p1 - control point 1, line departs p0 headed this way
>>>>> // p2 - control point 2, line arrives at p3 from this way
>>>>> // p3 - end point
>>>>> // segs - number of segments
>>>>> function bez(p0, p1, p2, p3, segs) = [
>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>>> ];
>>>>>
>>>>> It is that final function bez() that I actually call, e.g.
>>>>>
>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>>
>>>>> often concatenating the results of several bez() calls together.
>>>>>
>>>>> Note that the four coordinates correspond to the coordinates of the
>>>>> control points in a GUI drawing program: the start point, control point
>>>>> associated with the start, the control point associated with the end, and
>>>>> the end point. I have sometimes created OpenSCAD objects by tracing an
>>>>> object in a drawing program and then reading off the coordinates of the
>>>>> various control points.
>>>>>
>>>>>
>>>>> Note that Bézier functions are not the only mathematical curves. (For
>>>>> instance, they cannot exactly match an arc of a circle.) I'm sure there
>>>>> are numerous other functions that also produce curves.
>>>>> _______________________________________________
>>>>> OpenSCAD mailing list
>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>
>>>> _______________________________________________
>>>> OpenSCAD mailing list
>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>
>>> _______________________________________________
>>> OpenSCAD mailing list
>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>
>> _______________________________________________
>> OpenSCAD mailing list
>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>
> _______________________________________________
> OpenSCAD mailing list
> To unsubscribe send an email to discuss-leave@lists.openscad.org
>
AM
Adrian Mariano
Thu, Aug 4, 2022 12:13 AM
I think you need to be a little careful about defining what you want to
do. If you want to approximate an ellipse with points that are distributed
uniformly over the ellipse, that could be done by approximating the arc
length of the ellipse to find the necessary points---e.g. advance distance
d along the ellipse at each step. But if you want to approximate an
ellipse with a polygon with equal length sides, that approach does not do
it, because the length of the segments will not be constant as the span
sections of the ellipse with differing curvature. And since the perimeter
of the desired polygon is unknown, you can't just step it out analogously.
The circle doesn't present this challenge due to having constant
curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton fatherhorton@gmail.com wrote:
I think our emails crossed. I was addressing how to approximate an ellipse
with constant-length segments in general, not how to solve your problem.
When I had to do something similar, I used polar coordinates and stepped up
the value of theta until the distance from the previous point was correct.
(In the constant-length version, I'd keep a running total of the maximum
desired length so that round-off errors don't accumulate.)
Figuring out the circumscribed polygon is, as you note, a greater
challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano avm4@cornell.edu wrote:
Eh, there's no such thing as "brute force" in this situation. Step
through what with sufficiently small steps? Of course an iterative
algorithm is necessary. This is true for most interesting problems. No
big deal. Show me an iterative algorithm that converges to the
circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton fatherhorton@gmail.com
wrote:
Except by brute force (step through it with sufficiently small steps),
it’s not possible because there’s no closed form expression for the
distance between two points on a general ellipse (as opposed to special
cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano avm4@cornell.edu wrote:
Here's a challenge for you: draw an "ellipse" polygon with N segments
that are all the same length that circumscribes a perfect ellipse with
given semiaxes a and b. I tried to solve this problem and could not find a
solution. I can draw an inscribed N segment "ellipse" polygon with all
segments the same length---though it's not easy---but the circumscribed
case eludes me. I think of the inscribed case as, every vertex is located
on the ellipse and the circumscribed case, every segment is tangent to the
ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor axis "r1"
and "r2" respectively and with center "cp" and number of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster in
complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that
stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can chime in -
but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major (long) and
minor (short) axis dimensions; the other parts of the curve mathematically
flow from those two measurements.
You can make something else, but you can't make it using this
technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls together.
Note that the four coordinates correspond to the coordinates of the
control points in a GUI drawing program: the start point, control point
associated with the start, the control point associated with the end, and
the end point. I have sometimes created OpenSCAD objects by tracing an
object in a drawing program and then reading off the coordinates of the
various control points.
Note that Bézier functions are not the only mathematical curves.
(For instance, they cannot exactly match an arc of a circle.) I'm sure
there are numerous other functions that also produce curves.
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org
I think you need to be a little careful about defining what you want to
do. If you want to approximate an ellipse with points that are distributed
uniformly over the ellipse, that could be done by approximating the arc
length of the ellipse to find the necessary points---e.g. advance distance
d along the ellipse at each step. But if you want to approximate an
ellipse with a polygon with equal length sides, that approach does not do
it, because the length of the segments will not be constant as the span
sections of the ellipse with differing curvature. And since the perimeter
of the desired polygon is unknown, you can't just step it out analogously.
The circle doesn't present this challenge due to having constant
curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton <fatherhorton@gmail.com> wrote:
> I think our emails crossed. I was addressing how to approximate an ellipse
> with constant-length segments in general, not how to solve your problem.
> When I had to do something similar, I used polar coordinates and stepped up
> the value of theta until the distance from the previous point was correct.
> (In the constant-length version, I'd keep a running total of the maximum
> desired length so that round-off errors don't accumulate.)
>
> Figuring out the circumscribed polygon is, as you note, a greater
> challenge. I can visualize how to do it for n = 4, but it doesn't
> extrapolate at all.
>
> On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <avm4@cornell.edu> wrote:
>
>> Eh, there's no such thing as "brute force" in this situation. Step
>> through *what* with sufficiently small steps? Of course an iterative
>> algorithm is necessary. This is true for most interesting problems. No
>> big deal. Show me an iterative algorithm that converges to the
>> circumscribing polygon. When I tried to devise such an algorithm I
>> couldn't find one that would converge. Writing an algorithm that would
>> converge to the *inscribed* polygon seemed to be straight forward. Note
>> also that computing arc length of the ellipse doesn't really help here.
>> You can do that by solving an elliptical integral, or summing up small
>> segments. But what's the connection between arc length of the ellipse and
>> a circumscribing polygon with equal segment lengths? If there's a
>> connection, I don't see it.
>>
>> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <fatherhorton@gmail.com>
>> wrote:
>>
>>> Except by brute force (step through it with sufficiently small steps),
>>> it’s not possible because there’s no closed form expression for the
>>> distance between two points on a general ellipse (as opposed to special
>>> cases like a circle).
>>>
>>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <avm4@cornell.edu> wrote:
>>>
>>>> Here's a challenge for you: draw an "ellipse" polygon with N segments
>>>> that are all the same length that circumscribes a perfect ellipse with
>>>> given semiaxes a and b. I tried to solve this problem and could not find a
>>>> solution. I can draw an *inscribed* N segment "ellipse" polygon with all
>>>> segments the same length---though it's not easy---but the circumscribed
>>>> case eludes me. I think of the inscribed case as, every vertex is located
>>>> on the ellipse and the circumscribed case, every segment is tangent to the
>>>> ellipse
>>>>
>>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>>>> sprabhakar2006@gmail.com> wrote:
>>>>
>>>>> In openscad drawing a ellipse is not too difficult
>>>>> Refer following code
>>>>>
>>>>> function to draw an ellipse with semi-major and semi-minor axis "r1"
>>>>> and "r2" respectively and with center "cp" and number of segment "s"
>>>>> /
>>>>> // example:
>>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>>>
>>>>> function ellipse(r1,r2,cp,s=30)=
>>>>> let(
>>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>>>> )sec;
>>>>>
>>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>>>> openscad@jordan.maileater.net> wrote:
>>>>>
>>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>>>
>>>>>> I created an oval using:
>>>>>> inMm = 25.4;
>>>>>> fullW = 35.5 * inMm;
>>>>>> fullL = 54.5 * inMm;
>>>>>> resize([fullW, fullL, 1])
>>>>>> circle(d=fullW);
>>>>>>
>>>>>>
>>>>>> Tip: use scale() rather than resize(); it previews faster in
>>>>>> complicated cases.
>>>>>>
>>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>>>
>>>>>>
>>>>>> You're taking a circle and stretching it; the result of that
>>>>>> stretching is an ellipse.
>>>>>>
>>>>>> I'm not 100% sure - maybe one of the real math wizards can chime in -
>>>>>> but I think the shape you are asking for would not be an ellipse.
>>>>>>
>>>>>> I believe that an ellipse is fully defined by its major (long) and
>>>>>> minor (short) axis dimensions; the other parts of the curve mathematically
>>>>>> flow from those two measurements.
>>>>>>
>>>>>> You can make something else, but you can't make it using this
>>>>>> technique.
>>>>>>
>>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>>>
>>>>>>
>>>>>> You can use a Bézier function.
>>>>>>
>>>>>> Using BOSL2:
>>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>>>
>>>>>> DIY, here's a function that I cobbled together some years ago:
>>>>>>
>>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>>>> // but that yielded either single points or a raft of triangles;
>>>>>> // this yields a vector of points that you can then concatenate
>>>>>> // with other pieces to form a single polygon.
>>>>>> // If we were really clever, I think it would be possible to
>>>>>> // automatically space the output points based on how linear
>>>>>> // the curve is at that point. But right now I'm not that clever.
>>>>>> function BEZ03(u) = pow((1-u), 3);
>>>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>>>> function BEZ33(u) = pow(u,3);
>>>>>>
>>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>>>
>>>>>> // p0 - start point
>>>>>> // p1 - control point 1, line departs p0 headed this way
>>>>>> // p2 - control point 2, line arrives at p3 from this way
>>>>>> // p3 - end point
>>>>>> // segs - number of segments
>>>>>> function bez(p0, p1, p2, p3, segs) = [
>>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>>>> ];
>>>>>>
>>>>>> It is that final function bez() that I actually call, e.g.
>>>>>>
>>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>>>
>>>>>> often concatenating the results of several bez() calls together.
>>>>>>
>>>>>> Note that the four coordinates correspond to the coordinates of the
>>>>>> control points in a GUI drawing program: the start point, control point
>>>>>> associated with the start, the control point associated with the end, and
>>>>>> the end point. I have sometimes created OpenSCAD objects by tracing an
>>>>>> object in a drawing program and then reading off the coordinates of the
>>>>>> various control points.
>>>>>>
>>>>>>
>>>>>> Note that Bézier functions are not the only mathematical curves.
>>>>>> (For instance, they cannot exactly match an arc of a circle.) I'm sure
>>>>>> there are numerous other functions that also produce curves.
>>>>>> _______________________________________________
>>>>>> OpenSCAD mailing list
>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>
>>>>> _______________________________________________
>>>>> OpenSCAD mailing list
>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>
>>>> _______________________________________________
>>>> OpenSCAD mailing list
>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>
>>> _______________________________________________
>>> OpenSCAD mailing list
>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>
>> _______________________________________________
>> OpenSCAD mailing list
>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>
> _______________________________________________
> OpenSCAD mailing list
> To unsubscribe send an email to discuss-leave@lists.openscad.org
>
SP
Sanjeev Prabhakar
Fri, Aug 5, 2022 4:43 PM
I think circumscribed polygon segments cannot be equal lengths.
inscribed ones may be approximately equal, but would be a very tough
problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano avm4@cornell.edu wrote:
I think you need to be a little careful about defining what you want to
do. If you want to approximate an ellipse with points that are distributed
uniformly over the ellipse, that could be done by approximating the arc
length of the ellipse to find the necessary points---e.g. advance distance
d along the ellipse at each step. But if you want to approximate an
ellipse with a polygon with equal length sides, that approach does not do
it, because the length of the segments will not be constant as the span
sections of the ellipse with differing curvature. And since the perimeter
of the desired polygon is unknown, you can't just step it out analogously.
The circle doesn't present this challenge due to having constant
curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton fatherhorton@gmail.com
wrote:
I think our emails crossed. I was addressing how to approximate an
ellipse with constant-length segments in general, not how to solve your
problem. When I had to do something similar, I used polar coordinates and
stepped up the value of theta until the distance from the previous point
was correct. (In the constant-length version, I'd keep a running total of
the maximum desired length so that round-off errors don't accumulate.)
Figuring out the circumscribed polygon is, as you note, a greater
challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano avm4@cornell.edu wrote:
Eh, there's no such thing as "brute force" in this situation. Step
through what with sufficiently small steps? Of course an iterative
algorithm is necessary. This is true for most interesting problems. No
big deal. Show me an iterative algorithm that converges to the
circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton fatherhorton@gmail.com
wrote:
Except by brute force (step through it with sufficiently small steps),
it’s not possible because there’s no closed form expression for the
distance between two points on a general ellipse (as opposed to special
cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano avm4@cornell.edu wrote:
Here's a challenge for you: draw an "ellipse" polygon with N segments
that are all the same length that circumscribes a perfect ellipse with
given semiaxes a and b. I tried to solve this problem and could not find a
solution. I can draw an inscribed N segment "ellipse" polygon with all
segments the same length---though it's not easy---but the circumscribed
case eludes me. I think of the inscribed case as, every vertex is located
on the ellipse and the circumscribed case, every segment is tangent to the
ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor axis "r1"
and "r2" respectively and with center "cp" and number of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster in
complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that
stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can chime in
- but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major (long) and
minor (short) axis dimensions; the other parts of the curve mathematically
flow from those two measurements.
You can make something else, but you can't make it using this
technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls together.
Note that the four coordinates correspond to the coordinates of the
control points in a GUI drawing program: the start point, control point
associated with the start, the control point associated with the end, and
the end point. I have sometimes created OpenSCAD objects by tracing an
object in a drawing program and then reading off the coordinates of the
various control points.
Note that Bézier functions are not the only mathematical curves.
(For instance, they cannot exactly match an arc of a circle.) I'm sure
there are numerous other functions that also produce curves.
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org
I think circumscribed polygon segments cannot be equal lengths.
inscribed ones may be approximately equal, but would be a very tough
problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <avm4@cornell.edu> wrote:
> I think you need to be a little careful about defining what you want to
> do. If you want to approximate an ellipse with points that are distributed
> uniformly over the ellipse, that could be done by approximating the arc
> length of the ellipse to find the necessary points---e.g. advance distance
> d along the ellipse at each step. But if you want to approximate an
> ellipse with a polygon with equal length sides, that approach does not do
> it, because the length of the segments will not be constant as the span
> sections of the ellipse with differing curvature. And since the perimeter
> of the desired polygon is unknown, you can't just step it out analogously.
> The circle doesn't present this challenge due to having constant
> curvature.
>
> On Wed, Aug 3, 2022 at 7:41 PM Father Horton <fatherhorton@gmail.com>
> wrote:
>
>> I think our emails crossed. I was addressing how to approximate an
>> ellipse with constant-length segments in general, not how to solve your
>> problem. When I had to do something similar, I used polar coordinates and
>> stepped up the value of theta until the distance from the previous point
>> was correct. (In the constant-length version, I'd keep a running total of
>> the maximum desired length so that round-off errors don't accumulate.)
>>
>> Figuring out the circumscribed polygon is, as you note, a greater
>> challenge. I can visualize how to do it for n = 4, but it doesn't
>> extrapolate at all.
>>
>> On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <avm4@cornell.edu> wrote:
>>
>>> Eh, there's no such thing as "brute force" in this situation. Step
>>> through *what* with sufficiently small steps? Of course an iterative
>>> algorithm is necessary. This is true for most interesting problems. No
>>> big deal. Show me an iterative algorithm that converges to the
>>> circumscribing polygon. When I tried to devise such an algorithm I
>>> couldn't find one that would converge. Writing an algorithm that would
>>> converge to the *inscribed* polygon seemed to be straight forward. Note
>>> also that computing arc length of the ellipse doesn't really help here.
>>> You can do that by solving an elliptical integral, or summing up small
>>> segments. But what's the connection between arc length of the ellipse and
>>> a circumscribing polygon with equal segment lengths? If there's a
>>> connection, I don't see it.
>>>
>>> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <fatherhorton@gmail.com>
>>> wrote:
>>>
>>>> Except by brute force (step through it with sufficiently small steps),
>>>> it’s not possible because there’s no closed form expression for the
>>>> distance between two points on a general ellipse (as opposed to special
>>>> cases like a circle).
>>>>
>>>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <avm4@cornell.edu> wrote:
>>>>
>>>>> Here's a challenge for you: draw an "ellipse" polygon with N segments
>>>>> that are all the same length that circumscribes a perfect ellipse with
>>>>> given semiaxes a and b. I tried to solve this problem and could not find a
>>>>> solution. I can draw an *inscribed* N segment "ellipse" polygon with all
>>>>> segments the same length---though it's not easy---but the circumscribed
>>>>> case eludes me. I think of the inscribed case as, every vertex is located
>>>>> on the ellipse and the circumscribed case, every segment is tangent to the
>>>>> ellipse
>>>>>
>>>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>
>>>>>> In openscad drawing a ellipse is not too difficult
>>>>>> Refer following code
>>>>>>
>>>>>> function to draw an ellipse with semi-major and semi-minor axis "r1"
>>>>>> and "r2" respectively and with center "cp" and number of segment "s"
>>>>>> /
>>>>>> // example:
>>>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>>>>
>>>>>> function ellipse(r1,r2,cp,s=30)=
>>>>>> let(
>>>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>>>>> )sec;
>>>>>>
>>>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>>>>> openscad@jordan.maileater.net> wrote:
>>>>>>
>>>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>>>>
>>>>>>> I created an oval using:
>>>>>>> inMm = 25.4;
>>>>>>> fullW = 35.5 * inMm;
>>>>>>> fullL = 54.5 * inMm;
>>>>>>> resize([fullW, fullL, 1])
>>>>>>> circle(d=fullW);
>>>>>>>
>>>>>>>
>>>>>>> Tip: use scale() rather than resize(); it previews faster in
>>>>>>> complicated cases.
>>>>>>>
>>>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>>>>
>>>>>>>
>>>>>>> You're taking a circle and stretching it; the result of that
>>>>>>> stretching is an ellipse.
>>>>>>>
>>>>>>> I'm not 100% sure - maybe one of the real math wizards can chime in
>>>>>>> - but I think the shape you are asking for would not be an ellipse.
>>>>>>>
>>>>>>> I believe that an ellipse is fully defined by its major (long) and
>>>>>>> minor (short) axis dimensions; the other parts of the curve mathematically
>>>>>>> flow from those two measurements.
>>>>>>>
>>>>>>> You can make something else, but you can't make it using this
>>>>>>> technique.
>>>>>>>
>>>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>>>>
>>>>>>>
>>>>>>> You can use a Bézier function.
>>>>>>>
>>>>>>> Using BOSL2:
>>>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>>>>
>>>>>>> DIY, here's a function that I cobbled together some years ago:
>>>>>>>
>>>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>>>>> // but that yielded either single points or a raft of triangles;
>>>>>>> // this yields a vector of points that you can then concatenate
>>>>>>> // with other pieces to form a single polygon.
>>>>>>> // If we were really clever, I think it would be possible to
>>>>>>> // automatically space the output points based on how linear
>>>>>>> // the curve is at that point. But right now I'm not that clever.
>>>>>>> function BEZ03(u) = pow((1-u), 3);
>>>>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>>>>> function BEZ33(u) = pow(u,3);
>>>>>>>
>>>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>>>>
>>>>>>> // p0 - start point
>>>>>>> // p1 - control point 1, line departs p0 headed this way
>>>>>>> // p2 - control point 2, line arrives at p3 from this way
>>>>>>> // p3 - end point
>>>>>>> // segs - number of segments
>>>>>>> function bez(p0, p1, p2, p3, segs) = [
>>>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>>>>> ];
>>>>>>>
>>>>>>> It is that final function bez() that I actually call, e.g.
>>>>>>>
>>>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>>>>
>>>>>>> often concatenating the results of several bez() calls together.
>>>>>>>
>>>>>>> Note that the four coordinates correspond to the coordinates of the
>>>>>>> control points in a GUI drawing program: the start point, control point
>>>>>>> associated with the start, the control point associated with the end, and
>>>>>>> the end point. I have sometimes created OpenSCAD objects by tracing an
>>>>>>> object in a drawing program and then reading off the coordinates of the
>>>>>>> various control points.
>>>>>>>
>>>>>>>
>>>>>>> Note that Bézier functions are not the only mathematical curves.
>>>>>>> (For instance, they cannot exactly match an arc of a circle.) I'm sure
>>>>>>> there are numerous other functions that also produce curves.
>>>>>>> _______________________________________________
>>>>>>> OpenSCAD mailing list
>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>
>>>>>> _______________________________________________
>>>>>> OpenSCAD mailing list
>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>
>>>>> _______________________________________________
>>>>> OpenSCAD mailing list
>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>
>>>> _______________________________________________
>>>> OpenSCAD mailing list
>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>
>>> _______________________________________________
>>> OpenSCAD mailing list
>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>
>> _______________________________________________
>> OpenSCAD mailing list
>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>
> _______________________________________________
> OpenSCAD mailing list
> To unsubscribe send an email to discuss-leave@lists.openscad.org
>
AM
Adrian Mariano
Fri, Aug 5, 2022 8:57 PM
You think that a circumscribing polygon with equal length segments does not
exist?? I'm skeptical of this claim.
For circumscribed, it's not so hard. You start with a uniform-in-angle
point distribution, compute the segment lengths, compare to the mean
segment length to get an error, and then compute an adjustment based on the
error. It converges quickly.
So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but surely good enough.
[image: image.png]
The code for this is in BOSL2 in shapes2d.scad. A similar algorithm should
exist for the circumscribed case, but as I said, my efforts to find it have
failed.
On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar sprabhakar2006@gmail.com
wrote:
I think circumscribed polygon segments cannot be equal lengths.
inscribed ones may be approximately equal, but would be a very tough
problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano avm4@cornell.edu wrote:
I think you need to be a little careful about defining what you want to
do. If you want to approximate an ellipse with points that are distributed
uniformly over the ellipse, that could be done by approximating the arc
length of the ellipse to find the necessary points---e.g. advance distance
d along the ellipse at each step. But if you want to approximate an
ellipse with a polygon with equal length sides, that approach does not do
it, because the length of the segments will not be constant as the span
sections of the ellipse with differing curvature. And since the perimeter
of the desired polygon is unknown, you can't just step it out analogously.
The circle doesn't present this challenge due to having constant
curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton fatherhorton@gmail.com
wrote:
I think our emails crossed. I was addressing how to approximate an
ellipse with constant-length segments in general, not how to solve your
problem. When I had to do something similar, I used polar coordinates and
stepped up the value of theta until the distance from the previous point
was correct. (In the constant-length version, I'd keep a running total of
the maximum desired length so that round-off errors don't accumulate.)
Figuring out the circumscribed polygon is, as you note, a greater
challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano avm4@cornell.edu wrote:
Eh, there's no such thing as "brute force" in this situation. Step
through what with sufficiently small steps? Of course an iterative
algorithm is necessary. This is true for most interesting problems. No
big deal. Show me an iterative algorithm that converges to the
circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton fatherhorton@gmail.com
wrote:
Except by brute force (step through it with sufficiently small steps),
it’s not possible because there’s no closed form expression for the
distance between two points on a general ellipse (as opposed to special
cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano avm4@cornell.edu
wrote:
Here's a challenge for you: draw an "ellipse" polygon with N
segments that are all the same length that circumscribes a perfect ellipse
with given semiaxes a and b. I tried to solve this problem and could not
find a solution. I can draw an inscribed N segment "ellipse" polygon
with all segments the same length---though it's not easy---but the
circumscribed case eludes me. I think of the inscribed case as, every
vertex is located on the ellipse and the circumscribed case, every segment
is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor axis "r1"
and "r2" respectively and with center "cp" and number of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster in
complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that
stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can chime in
- but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major (long) and
minor (short) axis dimensions; the other parts of the curve mathematically
flow from those two measurements.
You can make something else, but you can't make it using this
technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls together.
Note that the four coordinates correspond to the coordinates of the
control points in a GUI drawing program: the start point, control point
associated with the start, the control point associated with the end, and
the end point. I have sometimes created OpenSCAD objects by tracing an
object in a drawing program and then reading off the coordinates of the
various control points.
Note that Bézier functions are not the only mathematical curves.
(For instance, they cannot exactly match an arc of a circle.) I'm sure
there are numerous other functions that also produce curves.
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org
You think that a circumscribing polygon with equal length segments does not
exist?? I'm skeptical of this claim.
For circumscribed, it's not so hard. You start with a uniform-in-angle
point distribution, compute the segment lengths, compare to the mean
segment length to get an error, and then compute an adjustment based on the
error. It converges quickly.
So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but surely good enough.
[image: image.png]
The code for this is in BOSL2 in shapes2d.scad. A similar algorithm should
exist for the circumscribed case, but as I said, my efforts to find it have
failed.
On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <sprabhakar2006@gmail.com>
wrote:
> I think circumscribed polygon segments cannot be equal lengths.
> inscribed ones may be approximately equal, but would be a very tough
> problem to solve.
> I tried something.
>
>
>
> On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <avm4@cornell.edu> wrote:
>
>> I think you need to be a little careful about defining what you want to
>> do. If you want to approximate an ellipse with points that are distributed
>> uniformly over the ellipse, that could be done by approximating the arc
>> length of the ellipse to find the necessary points---e.g. advance distance
>> d along the ellipse at each step. But if you want to approximate an
>> ellipse with a polygon with equal length sides, that approach does not do
>> it, because the length of the segments will not be constant as the span
>> sections of the ellipse with differing curvature. And since the perimeter
>> of the desired polygon is unknown, you can't just step it out analogously.
>> The circle doesn't present this challenge due to having constant
>> curvature.
>>
>> On Wed, Aug 3, 2022 at 7:41 PM Father Horton <fatherhorton@gmail.com>
>> wrote:
>>
>>> I think our emails crossed. I was addressing how to approximate an
>>> ellipse with constant-length segments in general, not how to solve your
>>> problem. When I had to do something similar, I used polar coordinates and
>>> stepped up the value of theta until the distance from the previous point
>>> was correct. (In the constant-length version, I'd keep a running total of
>>> the maximum desired length so that round-off errors don't accumulate.)
>>>
>>> Figuring out the circumscribed polygon is, as you note, a greater
>>> challenge. I can visualize how to do it for n = 4, but it doesn't
>>> extrapolate at all.
>>>
>>> On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <avm4@cornell.edu> wrote:
>>>
>>>> Eh, there's no such thing as "brute force" in this situation. Step
>>>> through *what* with sufficiently small steps? Of course an iterative
>>>> algorithm is necessary. This is true for most interesting problems. No
>>>> big deal. Show me an iterative algorithm that converges to the
>>>> circumscribing polygon. When I tried to devise such an algorithm I
>>>> couldn't find one that would converge. Writing an algorithm that would
>>>> converge to the *inscribed* polygon seemed to be straight forward. Note
>>>> also that computing arc length of the ellipse doesn't really help here.
>>>> You can do that by solving an elliptical integral, or summing up small
>>>> segments. But what's the connection between arc length of the ellipse and
>>>> a circumscribing polygon with equal segment lengths? If there's a
>>>> connection, I don't see it.
>>>>
>>>> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <fatherhorton@gmail.com>
>>>> wrote:
>>>>
>>>>> Except by brute force (step through it with sufficiently small steps),
>>>>> it’s not possible because there’s no closed form expression for the
>>>>> distance between two points on a general ellipse (as opposed to special
>>>>> cases like a circle).
>>>>>
>>>>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <avm4@cornell.edu>
>>>>> wrote:
>>>>>
>>>>>> Here's a challenge for you: draw an "ellipse" polygon with N
>>>>>> segments that are all the same length that circumscribes a perfect ellipse
>>>>>> with given semiaxes a and b. I tried to solve this problem and could not
>>>>>> find a solution. I can draw an *inscribed* N segment "ellipse" polygon
>>>>>> with all segments the same length---though it's not easy---but the
>>>>>> circumscribed case eludes me. I think of the inscribed case as, every
>>>>>> vertex is located on the ellipse and the circumscribed case, every segment
>>>>>> is tangent to the ellipse
>>>>>>
>>>>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>
>>>>>>> In openscad drawing a ellipse is not too difficult
>>>>>>> Refer following code
>>>>>>>
>>>>>>> function to draw an ellipse with semi-major and semi-minor axis "r1"
>>>>>>> and "r2" respectively and with center "cp" and number of segment "s"
>>>>>>> /
>>>>>>> // example:
>>>>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>>>>>
>>>>>>> function ellipse(r1,r2,cp,s=30)=
>>>>>>> let(
>>>>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>>>>>> )sec;
>>>>>>>
>>>>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>>>>>> openscad@jordan.maileater.net> wrote:
>>>>>>>
>>>>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>>>>>
>>>>>>>> I created an oval using:
>>>>>>>> inMm = 25.4;
>>>>>>>> fullW = 35.5 * inMm;
>>>>>>>> fullL = 54.5 * inMm;
>>>>>>>> resize([fullW, fullL, 1])
>>>>>>>> circle(d=fullW);
>>>>>>>>
>>>>>>>>
>>>>>>>> Tip: use scale() rather than resize(); it previews faster in
>>>>>>>> complicated cases.
>>>>>>>>
>>>>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>>>>>
>>>>>>>>
>>>>>>>> You're taking a circle and stretching it; the result of that
>>>>>>>> stretching is an ellipse.
>>>>>>>>
>>>>>>>> I'm not 100% sure - maybe one of the real math wizards can chime in
>>>>>>>> - but I think the shape you are asking for would not be an ellipse.
>>>>>>>>
>>>>>>>> I believe that an ellipse is fully defined by its major (long) and
>>>>>>>> minor (short) axis dimensions; the other parts of the curve mathematically
>>>>>>>> flow from those two measurements.
>>>>>>>>
>>>>>>>> You can make something else, but you can't make it using this
>>>>>>>> technique.
>>>>>>>>
>>>>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>>>>>
>>>>>>>>
>>>>>>>> You can use a Bézier function.
>>>>>>>>
>>>>>>>> Using BOSL2:
>>>>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>>>>>
>>>>>>>> DIY, here's a function that I cobbled together some years ago:
>>>>>>>>
>>>>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>>>>>> // but that yielded either single points or a raft of triangles;
>>>>>>>> // this yields a vector of points that you can then concatenate
>>>>>>>> // with other pieces to form a single polygon.
>>>>>>>> // If we were really clever, I think it would be possible to
>>>>>>>> // automatically space the output points based on how linear
>>>>>>>> // the curve is at that point. But right now I'm not that clever.
>>>>>>>> function BEZ03(u) = pow((1-u), 3);
>>>>>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>>>>>> function BEZ33(u) = pow(u,3);
>>>>>>>>
>>>>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>>>>>
>>>>>>>> // p0 - start point
>>>>>>>> // p1 - control point 1, line departs p0 headed this way
>>>>>>>> // p2 - control point 2, line arrives at p3 from this way
>>>>>>>> // p3 - end point
>>>>>>>> // segs - number of segments
>>>>>>>> function bez(p0, p1, p2, p3, segs) = [
>>>>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>>>>>> ];
>>>>>>>>
>>>>>>>> It is that final function bez() that I actually call, e.g.
>>>>>>>>
>>>>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>>>>>
>>>>>>>> often concatenating the results of several bez() calls together.
>>>>>>>>
>>>>>>>> Note that the four coordinates correspond to the coordinates of the
>>>>>>>> control points in a GUI drawing program: the start point, control point
>>>>>>>> associated with the start, the control point associated with the end, and
>>>>>>>> the end point. I have sometimes created OpenSCAD objects by tracing an
>>>>>>>> object in a drawing program and then reading off the coordinates of the
>>>>>>>> various control points.
>>>>>>>>
>>>>>>>>
>>>>>>>> Note that Bézier functions are not the only mathematical curves.
>>>>>>>> (For instance, they cannot exactly match an arc of a circle.) I'm sure
>>>>>>>> there are numerous other functions that also produce curves.
>>>>>>>> _______________________________________________
>>>>>>>> OpenSCAD mailing list
>>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>>
>>>>>>> _______________________________________________
>>>>>>> OpenSCAD mailing list
>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>
>>>>>> _______________________________________________
>>>>>> OpenSCAD mailing list
>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>
>>>>> _______________________________________________
>>>>> OpenSCAD mailing list
>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>
>>>> _______________________________________________
>>>> OpenSCAD mailing list
>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>
>>> _______________________________________________
>>> OpenSCAD mailing list
>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>
>> _______________________________________________
>> OpenSCAD mailing list
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>>
> _______________________________________________
> OpenSCAD mailing list
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>
NH
nop head
Fri, Aug 5, 2022 10:32 PM
A long discussion but I have to ask why do you need an elliptical polygon
with equal tangential sides?
On Fri, 5 Aug 2022, 21:58 Adrian Mariano, avm4@cornell.edu wrote:
You think that a circumscribing polygon with equal length segments does
not exist?? I'm skeptical of this claim.
For circumscribed, it's not so hard. You start with a uniform-in-angle
point distribution, compute the segment lengths, compare to the mean
segment length to get an error, and then compute an adjustment based on the
error. It converges quickly.
So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but surely good enough.
[image: image.png]
The code for this is in BOSL2 in shapes2d.scad. A similar algorithm should
exist for the circumscribed case, but as I said, my efforts to find it have
failed.
On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
I think circumscribed polygon segments cannot be equal lengths.
inscribed ones may be approximately equal, but would be a very tough
problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano avm4@cornell.edu wrote:
I think you need to be a little careful about defining what you want to
do. If you want to approximate an ellipse with points that are distributed
uniformly over the ellipse, that could be done by approximating the arc
length of the ellipse to find the necessary points---e.g. advance distance
d along the ellipse at each step. But if you want to approximate an
ellipse with a polygon with equal length sides, that approach does not do
it, because the length of the segments will not be constant as the span
sections of the ellipse with differing curvature. And since the perimeter
of the desired polygon is unknown, you can't just step it out analogously.
The circle doesn't present this challenge due to having constant
curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton fatherhorton@gmail.com
wrote:
I think our emails crossed. I was addressing how to approximate an
ellipse with constant-length segments in general, not how to solve your
problem. When I had to do something similar, I used polar coordinates and
stepped up the value of theta until the distance from the previous point
was correct. (In the constant-length version, I'd keep a running total of
the maximum desired length so that round-off errors don't accumulate.)
Figuring out the circumscribed polygon is, as you note, a greater
challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano avm4@cornell.edu
wrote:
Eh, there's no such thing as "brute force" in this situation. Step
through what with sufficiently small steps? Of course an iterative
algorithm is necessary. This is true for most interesting problems. No
big deal. Show me an iterative algorithm that converges to the
circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton fatherhorton@gmail.com
wrote:
Except by brute force (step through it with sufficiently small
steps), it’s not possible because there’s no closed form expression for the
distance between two points on a general ellipse (as opposed to special
cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano avm4@cornell.edu
wrote:
Here's a challenge for you: draw an "ellipse" polygon with N
segments that are all the same length that circumscribes a perfect ellipse
with given semiaxes a and b. I tried to solve this problem and could not
find a solution. I can draw an inscribed N segment "ellipse" polygon
with all segments the same length---though it's not easy---but the
circumscribed case eludes me. I think of the inscribed case as, every
vertex is located on the ellipse and the circumscribed case, every segment
is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor axis
"r1" and "r2" respectively and with center "cp" and number of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster in
complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that
stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can chime
in - but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major (long) and
minor (short) axis dimensions; the other parts of the curve mathematically
flow from those two measurements.
You can make something else, but you can't make it using this
technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls together.
Note that the four coordinates correspond to the coordinates of
the control points in a GUI drawing program: the start point, control
point associated with the start, the control point associated with the end,
and the end point. I have sometimes created OpenSCAD objects by tracing an
object in a drawing program and then reading off the coordinates of the
various control points.
Note that Bézier functions are not the only mathematical curves.
(For instance, they cannot exactly match an arc of a circle.) I'm sure
there are numerous other functions that also produce curves.
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org
A long discussion but I have to ask why do you need an elliptical polygon
with equal tangential sides?
On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu> wrote:
> You think that a circumscribing polygon with equal length segments does
> not exist?? I'm skeptical of this claim.
>
> For circumscribed, it's not so hard. You start with a uniform-in-angle
> point distribution, compute the segment lengths, compare to the mean
> segment length to get an error, and then compute an adjustment based on the
> error. It converges quickly.
>
> So for example, I have segments lengths of:
>
> ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
> 5.57297]
>
> Maybe not exactly equal, since tolerance is 1e-9, but surely good enough.
> [image: image.png]
>
> The code for this is in BOSL2 in shapes2d.scad. A similar algorithm should
> exist for the circumscribed case, but as I said, my efforts to find it have
> failed.
>
> On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
> sprabhakar2006@gmail.com> wrote:
>
>> I think circumscribed polygon segments cannot be equal lengths.
>> inscribed ones may be approximately equal, but would be a very tough
>> problem to solve.
>> I tried something.
>>
>>
>>
>> On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <avm4@cornell.edu> wrote:
>>
>>> I think you need to be a little careful about defining what you want to
>>> do. If you want to approximate an ellipse with points that are distributed
>>> uniformly over the ellipse, that could be done by approximating the arc
>>> length of the ellipse to find the necessary points---e.g. advance distance
>>> d along the ellipse at each step. But if you want to approximate an
>>> ellipse with a polygon with equal length sides, that approach does not do
>>> it, because the length of the segments will not be constant as the span
>>> sections of the ellipse with differing curvature. And since the perimeter
>>> of the desired polygon is unknown, you can't just step it out analogously.
>>> The circle doesn't present this challenge due to having constant
>>> curvature.
>>>
>>> On Wed, Aug 3, 2022 at 7:41 PM Father Horton <fatherhorton@gmail.com>
>>> wrote:
>>>
>>>> I think our emails crossed. I was addressing how to approximate an
>>>> ellipse with constant-length segments in general, not how to solve your
>>>> problem. When I had to do something similar, I used polar coordinates and
>>>> stepped up the value of theta until the distance from the previous point
>>>> was correct. (In the constant-length version, I'd keep a running total of
>>>> the maximum desired length so that round-off errors don't accumulate.)
>>>>
>>>> Figuring out the circumscribed polygon is, as you note, a greater
>>>> challenge. I can visualize how to do it for n = 4, but it doesn't
>>>> extrapolate at all.
>>>>
>>>> On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <avm4@cornell.edu>
>>>> wrote:
>>>>
>>>>> Eh, there's no such thing as "brute force" in this situation. Step
>>>>> through *what* with sufficiently small steps? Of course an iterative
>>>>> algorithm is necessary. This is true for most interesting problems. No
>>>>> big deal. Show me an iterative algorithm that converges to the
>>>>> circumscribing polygon. When I tried to devise such an algorithm I
>>>>> couldn't find one that would converge. Writing an algorithm that would
>>>>> converge to the *inscribed* polygon seemed to be straight forward. Note
>>>>> also that computing arc length of the ellipse doesn't really help here.
>>>>> You can do that by solving an elliptical integral, or summing up small
>>>>> segments. But what's the connection between arc length of the ellipse and
>>>>> a circumscribing polygon with equal segment lengths? If there's a
>>>>> connection, I don't see it.
>>>>>
>>>>> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <fatherhorton@gmail.com>
>>>>> wrote:
>>>>>
>>>>>> Except by brute force (step through it with sufficiently small
>>>>>> steps), it’s not possible because there’s no closed form expression for the
>>>>>> distance between two points on a general ellipse (as opposed to special
>>>>>> cases like a circle).
>>>>>>
>>>>>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <avm4@cornell.edu>
>>>>>> wrote:
>>>>>>
>>>>>>> Here's a challenge for you: draw an "ellipse" polygon with N
>>>>>>> segments that are all the same length that circumscribes a perfect ellipse
>>>>>>> with given semiaxes a and b. I tried to solve this problem and could not
>>>>>>> find a solution. I can draw an *inscribed* N segment "ellipse" polygon
>>>>>>> with all segments the same length---though it's not easy---but the
>>>>>>> circumscribed case eludes me. I think of the inscribed case as, every
>>>>>>> vertex is located on the ellipse and the circumscribed case, every segment
>>>>>>> is tangent to the ellipse
>>>>>>>
>>>>>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>
>>>>>>>> In openscad drawing a ellipse is not too difficult
>>>>>>>> Refer following code
>>>>>>>>
>>>>>>>> function to draw an ellipse with semi-major and semi-minor axis
>>>>>>>> "r1" and "r2" respectively and with center "cp" and number of segment "s"
>>>>>>>> /
>>>>>>>> // example:
>>>>>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>>>>>>
>>>>>>>> function ellipse(r1,r2,cp,s=30)=
>>>>>>>> let(
>>>>>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>>>>>>> )sec;
>>>>>>>>
>>>>>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>>>>>>> openscad@jordan.maileater.net> wrote:
>>>>>>>>
>>>>>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>>>>>>
>>>>>>>>> I created an oval using:
>>>>>>>>> inMm = 25.4;
>>>>>>>>> fullW = 35.5 * inMm;
>>>>>>>>> fullL = 54.5 * inMm;
>>>>>>>>> resize([fullW, fullL, 1])
>>>>>>>>> circle(d=fullW);
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Tip: use scale() rather than resize(); it previews faster in
>>>>>>>>> complicated cases.
>>>>>>>>>
>>>>>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> You're taking a circle and stretching it; the result of that
>>>>>>>>> stretching is an ellipse.
>>>>>>>>>
>>>>>>>>> I'm not 100% sure - maybe one of the real math wizards can chime
>>>>>>>>> in - but I think the shape you are asking for would not be an ellipse.
>>>>>>>>>
>>>>>>>>> I believe that an ellipse is fully defined by its major (long) and
>>>>>>>>> minor (short) axis dimensions; the other parts of the curve mathematically
>>>>>>>>> flow from those two measurements.
>>>>>>>>>
>>>>>>>>> You can make something else, but you can't make it using this
>>>>>>>>> technique.
>>>>>>>>>
>>>>>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> You can use a Bézier function.
>>>>>>>>>
>>>>>>>>> Using BOSL2:
>>>>>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>>>>>>
>>>>>>>>> DIY, here's a function that I cobbled together some years ago:
>>>>>>>>>
>>>>>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>>>>>>> // but that yielded either single points or a raft of triangles;
>>>>>>>>> // this yields a vector of points that you can then concatenate
>>>>>>>>> // with other pieces to form a single polygon.
>>>>>>>>> // If we were really clever, I think it would be possible to
>>>>>>>>> // automatically space the output points based on how linear
>>>>>>>>> // the curve is at that point. But right now I'm not that clever.
>>>>>>>>> function BEZ03(u) = pow((1-u), 3);
>>>>>>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>>>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>>>>>>> function BEZ33(u) = pow(u,3);
>>>>>>>>>
>>>>>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>>>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>>>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>>>>>>
>>>>>>>>> // p0 - start point
>>>>>>>>> // p1 - control point 1, line departs p0 headed this way
>>>>>>>>> // p2 - control point 2, line arrives at p3 from this way
>>>>>>>>> // p3 - end point
>>>>>>>>> // segs - number of segments
>>>>>>>>> function bez(p0, p1, p2, p3, segs) = [
>>>>>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>>>>>>> ];
>>>>>>>>>
>>>>>>>>> It is that final function bez() that I actually call, e.g.
>>>>>>>>>
>>>>>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>>>>>>
>>>>>>>>> often concatenating the results of several bez() calls together.
>>>>>>>>>
>>>>>>>>> Note that the four coordinates correspond to the coordinates of
>>>>>>>>> the control points in a GUI drawing program: the start point, control
>>>>>>>>> point associated with the start, the control point associated with the end,
>>>>>>>>> and the end point. I have sometimes created OpenSCAD objects by tracing an
>>>>>>>>> object in a drawing program and then reading off the coordinates of the
>>>>>>>>> various control points.
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Note that Bézier functions are not the only mathematical curves.
>>>>>>>>> (For instance, they cannot exactly match an arc of a circle.) I'm sure
>>>>>>>>> there are numerous other functions that also produce curves.
>>>>>>>>> _______________________________________________
>>>>>>>>> OpenSCAD mailing list
>>>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>>>
>>>>>>>> _______________________________________________
>>>>>>>> OpenSCAD mailing list
>>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>>
>>>>>>> _______________________________________________
>>>>>>> OpenSCAD mailing list
>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>
>>>>>> _______________________________________________
>>>>>> OpenSCAD mailing list
>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>
>>>>> _______________________________________________
>>>>> OpenSCAD mailing list
>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>
>>>> _______________________________________________
>>>> OpenSCAD mailing list
>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>
>>> _______________________________________________
>>> OpenSCAD mailing list
>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
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>> _______________________________________________
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AM
Adrian Mariano
Fri, Aug 5, 2022 10:38 PM
I don't have a specific need for a circumscribing elliptical polygon with
equal sides. I was writing code to make ellipses and it seemed like equal
side approximations would be nice, so I solved the case for inscribed, but
couldn't solve the circumscribed case. So it's an unsolved puzzle I've got
that this discussion reminded me of.
On Fri, Aug 5, 2022 at 6:33 PM nop head nop.head@gmail.com wrote:
A long discussion but I have to ask why do you need an elliptical polygon
with equal tangential sides?
On Fri, 5 Aug 2022, 21:58 Adrian Mariano, avm4@cornell.edu wrote:
You think that a circumscribing polygon with equal length segments does
not exist?? I'm skeptical of this claim.
For circumscribed, it's not so hard. You start with a uniform-in-angle
point distribution, compute the segment lengths, compare to the mean
segment length to get an error, and then compute an adjustment based on the
error. It converges quickly.
So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but surely good enough.
[image: image.png]
The code for this is in BOSL2 in shapes2d.scad. A similar algorithm
should exist for the circumscribed case, but as I said, my efforts to find
it have failed.
On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
I think circumscribed polygon segments cannot be equal lengths.
inscribed ones may be approximately equal, but would be a very tough
problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano avm4@cornell.edu wrote:
I think you need to be a little careful about defining what you want to
do. If you want to approximate an ellipse with points that are distributed
uniformly over the ellipse, that could be done by approximating the arc
length of the ellipse to find the necessary points---e.g. advance distance
d along the ellipse at each step. But if you want to approximate an
ellipse with a polygon with equal length sides, that approach does not do
it, because the length of the segments will not be constant as the span
sections of the ellipse with differing curvature. And since the perimeter
of the desired polygon is unknown, you can't just step it out analogously.
The circle doesn't present this challenge due to having constant
curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton fatherhorton@gmail.com
wrote:
I think our emails crossed. I was addressing how to approximate an
ellipse with constant-length segments in general, not how to solve your
problem. When I had to do something similar, I used polar coordinates and
stepped up the value of theta until the distance from the previous point
was correct. (In the constant-length version, I'd keep a running total of
the maximum desired length so that round-off errors don't accumulate.)
Figuring out the circumscribed polygon is, as you note, a greater
challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano avm4@cornell.edu
wrote:
Eh, there's no such thing as "brute force" in this situation. Step
through what with sufficiently small steps? Of course an iterative
algorithm is necessary. This is true for most interesting problems. No
big deal. Show me an iterative algorithm that converges to the
circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton fatherhorton@gmail.com
wrote:
Except by brute force (step through it with sufficiently small
steps), it’s not possible because there’s no closed form expression for the
distance between two points on a general ellipse (as opposed to special
cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano avm4@cornell.edu
wrote:
Here's a challenge for you: draw an "ellipse" polygon with N
segments that are all the same length that circumscribes a perfect ellipse
with given semiaxes a and b. I tried to solve this problem and could not
find a solution. I can draw an inscribed N segment "ellipse" polygon
with all segments the same length---though it's not easy---but the
circumscribed case eludes me. I think of the inscribed case as, every
vertex is located on the ellipse and the circumscribed case, every segment
is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor axis
"r1" and "r2" respectively and with center "cp" and number of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster in
complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that
stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can chime
in - but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major (long)
and minor (short) axis dimensions; the other parts of the curve
mathematically flow from those two measurements.
You can make something else, but you can't make it using this
technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls together.
Note that the four coordinates correspond to the coordinates of
the control points in a GUI drawing program: the start point, control
point associated with the start, the control point associated with the end,
and the end point. I have sometimes created OpenSCAD objects by tracing an
object in a drawing program and then reading off the coordinates of the
various control points.
Note that Bézier functions are not the only mathematical curves.
(For instance, they cannot exactly match an arc of a circle.) I'm sure
there are numerous other functions that also produce curves.
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org
I don't have a specific need for a circumscribing elliptical polygon with
equal sides. I was writing code to make ellipses and it seemed like equal
side approximations would be nice, so I solved the case for inscribed, but
couldn't solve the circumscribed case. So it's an unsolved puzzle I've got
that this discussion reminded me of.
On Fri, Aug 5, 2022 at 6:33 PM nop head <nop.head@gmail.com> wrote:
> A long discussion but I have to ask why do you need an elliptical polygon
> with equal tangential sides?
>
> On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu> wrote:
>
>> You think that a circumscribing polygon with equal length segments does
>> not exist?? I'm skeptical of this claim.
>>
>> For circumscribed, it's not so hard. You start with a uniform-in-angle
>> point distribution, compute the segment lengths, compare to the mean
>> segment length to get an error, and then compute an adjustment based on the
>> error. It converges quickly.
>>
>> So for example, I have segments lengths of:
>>
>> ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
>> 5.57297]
>>
>> Maybe not exactly equal, since tolerance is 1e-9, but surely good enough.
>> [image: image.png]
>>
>> The code for this is in BOSL2 in shapes2d.scad. A similar algorithm
>> should exist for the circumscribed case, but as I said, my efforts to find
>> it have failed.
>>
>> On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
>> sprabhakar2006@gmail.com> wrote:
>>
>>> I think circumscribed polygon segments cannot be equal lengths.
>>> inscribed ones may be approximately equal, but would be a very tough
>>> problem to solve.
>>> I tried something.
>>>
>>>
>>>
>>> On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <avm4@cornell.edu> wrote:
>>>
>>>> I think you need to be a little careful about defining what you want to
>>>> do. If you want to approximate an ellipse with points that are distributed
>>>> uniformly over the ellipse, that could be done by approximating the arc
>>>> length of the ellipse to find the necessary points---e.g. advance distance
>>>> d along the ellipse at each step. But if you want to approximate an
>>>> ellipse with a polygon with equal length sides, that approach does not do
>>>> it, because the length of the segments will not be constant as the span
>>>> sections of the ellipse with differing curvature. And since the perimeter
>>>> of the desired polygon is unknown, you can't just step it out analogously.
>>>> The circle doesn't present this challenge due to having constant
>>>> curvature.
>>>>
>>>> On Wed, Aug 3, 2022 at 7:41 PM Father Horton <fatherhorton@gmail.com>
>>>> wrote:
>>>>
>>>>> I think our emails crossed. I was addressing how to approximate an
>>>>> ellipse with constant-length segments in general, not how to solve your
>>>>> problem. When I had to do something similar, I used polar coordinates and
>>>>> stepped up the value of theta until the distance from the previous point
>>>>> was correct. (In the constant-length version, I'd keep a running total of
>>>>> the maximum desired length so that round-off errors don't accumulate.)
>>>>>
>>>>> Figuring out the circumscribed polygon is, as you note, a greater
>>>>> challenge. I can visualize how to do it for n = 4, but it doesn't
>>>>> extrapolate at all.
>>>>>
>>>>> On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <avm4@cornell.edu>
>>>>> wrote:
>>>>>
>>>>>> Eh, there's no such thing as "brute force" in this situation. Step
>>>>>> through *what* with sufficiently small steps? Of course an iterative
>>>>>> algorithm is necessary. This is true for most interesting problems. No
>>>>>> big deal. Show me an iterative algorithm that converges to the
>>>>>> circumscribing polygon. When I tried to devise such an algorithm I
>>>>>> couldn't find one that would converge. Writing an algorithm that would
>>>>>> converge to the *inscribed* polygon seemed to be straight forward. Note
>>>>>> also that computing arc length of the ellipse doesn't really help here.
>>>>>> You can do that by solving an elliptical integral, or summing up small
>>>>>> segments. But what's the connection between arc length of the ellipse and
>>>>>> a circumscribing polygon with equal segment lengths? If there's a
>>>>>> connection, I don't see it.
>>>>>>
>>>>>> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <fatherhorton@gmail.com>
>>>>>> wrote:
>>>>>>
>>>>>>> Except by brute force (step through it with sufficiently small
>>>>>>> steps), it’s not possible because there’s no closed form expression for the
>>>>>>> distance between two points on a general ellipse (as opposed to special
>>>>>>> cases like a circle).
>>>>>>>
>>>>>>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <avm4@cornell.edu>
>>>>>>> wrote:
>>>>>>>
>>>>>>>> Here's a challenge for you: draw an "ellipse" polygon with N
>>>>>>>> segments that are all the same length that circumscribes a perfect ellipse
>>>>>>>> with given semiaxes a and b. I tried to solve this problem and could not
>>>>>>>> find a solution. I can draw an *inscribed* N segment "ellipse" polygon
>>>>>>>> with all segments the same length---though it's not easy---but the
>>>>>>>> circumscribed case eludes me. I think of the inscribed case as, every
>>>>>>>> vertex is located on the ellipse and the circumscribed case, every segment
>>>>>>>> is tangent to the ellipse
>>>>>>>>
>>>>>>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>
>>>>>>>>> In openscad drawing a ellipse is not too difficult
>>>>>>>>> Refer following code
>>>>>>>>>
>>>>>>>>> function to draw an ellipse with semi-major and semi-minor axis
>>>>>>>>> "r1" and "r2" respectively and with center "cp" and number of segment "s"
>>>>>>>>> /
>>>>>>>>> // example:
>>>>>>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>>>>>>>
>>>>>>>>> function ellipse(r1,r2,cp,s=30)=
>>>>>>>>> let(
>>>>>>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>>>>>>>> )sec;
>>>>>>>>>
>>>>>>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>>>>>>>> openscad@jordan.maileater.net> wrote:
>>>>>>>>>
>>>>>>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>>>>>>>
>>>>>>>>>> I created an oval using:
>>>>>>>>>> inMm = 25.4;
>>>>>>>>>> fullW = 35.5 * inMm;
>>>>>>>>>> fullL = 54.5 * inMm;
>>>>>>>>>> resize([fullW, fullL, 1])
>>>>>>>>>> circle(d=fullW);
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Tip: use scale() rather than resize(); it previews faster in
>>>>>>>>>> complicated cases.
>>>>>>>>>>
>>>>>>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> You're taking a circle and stretching it; the result of that
>>>>>>>>>> stretching is an ellipse.
>>>>>>>>>>
>>>>>>>>>> I'm not 100% sure - maybe one of the real math wizards can chime
>>>>>>>>>> in - but I think the shape you are asking for would not be an ellipse.
>>>>>>>>>>
>>>>>>>>>> I believe that an ellipse is fully defined by its major (long)
>>>>>>>>>> and minor (short) axis dimensions; the other parts of the curve
>>>>>>>>>> mathematically flow from those two measurements.
>>>>>>>>>>
>>>>>>>>>> You can make something else, but you can't make it using this
>>>>>>>>>> technique.
>>>>>>>>>>
>>>>>>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> You can use a Bézier function.
>>>>>>>>>>
>>>>>>>>>> Using BOSL2:
>>>>>>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>>>>>>>
>>>>>>>>>> DIY, here's a function that I cobbled together some years ago:
>>>>>>>>>>
>>>>>>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>>>>>>>> // but that yielded either single points or a raft of triangles;
>>>>>>>>>> // this yields a vector of points that you can then concatenate
>>>>>>>>>> // with other pieces to form a single polygon.
>>>>>>>>>> // If we were really clever, I think it would be possible to
>>>>>>>>>> // automatically space the output points based on how linear
>>>>>>>>>> // the curve is at that point. But right now I'm not that clever.
>>>>>>>>>> function BEZ03(u) = pow((1-u), 3);
>>>>>>>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>>>>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>>>>>>>> function BEZ33(u) = pow(u,3);
>>>>>>>>>>
>>>>>>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>>>>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>>>>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>>>>>>>
>>>>>>>>>> // p0 - start point
>>>>>>>>>> // p1 - control point 1, line departs p0 headed this way
>>>>>>>>>> // p2 - control point 2, line arrives at p3 from this way
>>>>>>>>>> // p3 - end point
>>>>>>>>>> // segs - number of segments
>>>>>>>>>> function bez(p0, p1, p2, p3, segs) = [
>>>>>>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>>>>>>>> ];
>>>>>>>>>>
>>>>>>>>>> It is that final function bez() that I actually call, e.g.
>>>>>>>>>>
>>>>>>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>>>>>>>
>>>>>>>>>> often concatenating the results of several bez() calls together.
>>>>>>>>>>
>>>>>>>>>> Note that the four coordinates correspond to the coordinates of
>>>>>>>>>> the control points in a GUI drawing program: the start point, control
>>>>>>>>>> point associated with the start, the control point associated with the end,
>>>>>>>>>> and the end point. I have sometimes created OpenSCAD objects by tracing an
>>>>>>>>>> object in a drawing program and then reading off the coordinates of the
>>>>>>>>>> various control points.
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Note that Bézier functions are not the only mathematical curves.
>>>>>>>>>> (For instance, they cannot exactly match an arc of a circle.) I'm sure
>>>>>>>>>> there are numerous other functions that also produce curves.
>>>>>>>>>> _______________________________________________
>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>>>>
>>>>>>>>> _______________________________________________
>>>>>>>>> OpenSCAD mailing list
>>>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>>>
>>>>>>>> _______________________________________________
>>>>>>>> OpenSCAD mailing list
>>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>>
>>>>>>> _______________________________________________
>>>>>>> OpenSCAD mailing list
>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>
>>>>>> _______________________________________________
>>>>>> OpenSCAD mailing list
>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>
>>>>> _______________________________________________
>>>>> OpenSCAD mailing list
>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>
>>>> _______________________________________________
>>>> OpenSCAD mailing list
>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>
>>> _______________________________________________
>>> OpenSCAD mailing list
>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>
>> _______________________________________________
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>>
> _______________________________________________
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>
BC
Bob Carlson
Fri, Aug 5, 2022 11:15 PM
Seems like I remember Nop Head talking about using circumscribed holes for screws and even rotating the polygons to produce a more even hole.
If an ellipse is the nominal shape, the using a circumscribed polygon for a hole and an inscribed polygon for a solid might make sense.
Btw, can someone confirm my intuition that a circle scaled in one dimension is not an ellipse?
-Bob
Tucson AZ
On Aug 5, 2022, at 15:38, Adrian Mariano avm4@cornell.edu wrote:
I don't have a specific need for a circumscribing elliptical polygon with equal sides. I was writing code to make ellipses and it seemed like equal side approximations would be nice, so I solved the case for inscribed, but couldn't solve the circumscribed case. So it's an unsolved puzzle I've got that this discussion reminded me of.
On Fri, Aug 5, 2022 at 6:33 PM nop head <nop.head@gmail.com mailto:nop.head@gmail.com> wrote:
A long discussion but I have to ask why do you need an elliptical polygon with equal tangential sides?
On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu mailto:avm4@cornell.edu> wrote:
You think that a circumscribing polygon with equal length segments does not exist?? I'm skeptical of this claim.
For circumscribed, it's not so hard. You start with a uniform-in-angle point distribution, compute the segment lengths, compare to the mean segment length to get an error, and then compute an adjustment based on the error. It converges quickly.
So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but surely good enough.
<image.png>
The code for this is in BOSL2 in shapes2d.scad. A similar algorithm should exist for the circumscribed case, but as I said, my efforts to find it have failed.
On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <sprabhakar2006@gmail.com mailto:sprabhakar2006@gmail.com> wrote:
I think circumscribed polygon segments cannot be equal lengths.
inscribed ones may be approximately equal, but would be a very tough problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <avm4@cornell.edu mailto:avm4@cornell.edu> wrote:
I think you need to be a little careful about defining what you want to do. If you want to approximate an ellipse with points that are distributed uniformly over the ellipse, that could be done by approximating the arc length of the ellipse to find the necessary points---e.g. advance distance d along the ellipse at each step. But if you want to approximate an ellipse with a polygon with equal length sides, that approach does not do it, because the length of the segments will not be constant as the span sections of the ellipse with differing curvature. And since the perimeter of the desired polygon is unknown, you can't just step it out analogously. The circle doesn't present this challenge due to having constant curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton <fatherhorton@gmail.com mailto:fatherhorton@gmail.com> wrote:
I think our emails crossed. I was addressing how to approximate an ellipse with constant-length segments in general, not how to solve your problem. When I had to do something similar, I used polar coordinates and stepped up the value of theta until the distance from the previous point was correct. (In the constant-length version, I'd keep a running total of the maximum desired length so that round-off errors don't accumulate.)
Figuring out the circumscribed polygon is, as you note, a greater challenge. I can visualize how to do it for n = 4, but it doesn't extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <avm4@cornell.edu mailto:avm4@cornell.edu> wrote:
Eh, there's no such thing as "brute force" in this situation. Step through what with sufficiently small steps? Of course an iterative algorithm is necessary. This is true for most interesting problems. No big deal. Show me an iterative algorithm that converges to the circumscribing polygon. When I tried to devise such an algorithm I couldn't find one that would converge. Writing an algorithm that would converge to the inscribed polygon seemed to be straight forward. Note also that computing arc length of the ellipse doesn't really help here. You can do that by solving an elliptical integral, or summing up small segments. But what's the connection between arc length of the ellipse and a circumscribing polygon with equal segment lengths? If there's a connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton <fatherhorton@gmail.com mailto:fatherhorton@gmail.com> wrote:
Except by brute force (step through it with sufficiently small steps), it’s not possible because there’s no closed form expression for the distance between two points on a general ellipse (as opposed to special cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <avm4@cornell.edu mailto:avm4@cornell.edu> wrote:
Here's a challenge for you: draw an "ellipse" polygon with N segments that are all the same length that circumscribes a perfect ellipse with given semiaxes a and b. I tried to solve this problem and could not find a solution. I can draw an inscribed N segment "ellipse" polygon with all segments the same length---though it's not easy---but the circumscribed case eludes me. I think of the inscribed case as, every vertex is located on the ellipse and the circumscribed case, every segment is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <sprabhakar2006@gmail.com mailto:sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor axis "r1" and "r2" respectively and with center "cp" and number of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <openscad@jordan.maileater.net mailto:openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster in complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can chime in - but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major (long) and minor (short) axis dimensions; the other parts of the curve mathematically flow from those two measurements.
You can make something else, but you can't make it using this technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2: https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443 https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls together.
Note that the four coordinates correspond to the coordinates of the control points in a GUI drawing program: the start point, control point associated with the start, the control point associated with the end, and the end point. I have sometimes created OpenSCAD objects by tracing an object in a drawing program and then reading off the coordinates of the various control points.
Note that Bézier functions are not the only mathematical curves. (For instance, they cannot exactly match an arc of a circle.) I'm sure there are numerous other functions that also produce curves.
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Seems like I remember Nop Head talking about using circumscribed holes for screws and even rotating the polygons to produce a more even hole.
If an ellipse is the nominal shape, the using a circumscribed polygon for a hole and an inscribed polygon for a solid might make sense.
Btw, can someone confirm my intuition that a circle scaled in one dimension is not an ellipse?
-Bob
Tucson AZ
On Aug 5, 2022, at 15:38, Adrian Mariano <avm4@cornell.edu> wrote:
I don't have a specific need for a circumscribing elliptical polygon with equal sides. I was writing code to make ellipses and it seemed like equal side approximations would be nice, so I solved the case for inscribed, but couldn't solve the circumscribed case. So it's an unsolved puzzle I've got that this discussion reminded me of.
On Fri, Aug 5, 2022 at 6:33 PM nop head <nop.head@gmail.com <mailto:nop.head@gmail.com>> wrote:
A long discussion but I have to ask why do you need an elliptical polygon with equal tangential sides?
On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu <mailto:avm4@cornell.edu>> wrote:
You think that a circumscribing polygon with equal length segments does not exist?? I'm skeptical of this claim.
For circumscribed, it's not so hard. You start with a uniform-in-angle point distribution, compute the segment lengths, compare to the mean segment length to get an error, and then compute an adjustment based on the error. It converges quickly.
So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but surely good enough.
<image.png>
The code for this is in BOSL2 in shapes2d.scad. A similar algorithm should exist for the circumscribed case, but as I said, my efforts to find it have failed.
On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <sprabhakar2006@gmail.com <mailto:sprabhakar2006@gmail.com>> wrote:
I think circumscribed polygon segments cannot be equal lengths.
inscribed ones may be approximately equal, but would be a very tough problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <avm4@cornell.edu <mailto:avm4@cornell.edu>> wrote:
I think you need to be a little careful about defining what you want to do. If you want to approximate an ellipse with points that are distributed uniformly over the ellipse, that could be done by approximating the arc length of the ellipse to find the necessary points---e.g. advance distance d along the ellipse at each step. But if you want to approximate an ellipse with a polygon with equal length sides, that approach does not do it, because the length of the segments will not be constant as the span sections of the ellipse with differing curvature. And since the perimeter of the desired polygon is unknown, you can't just step it out analogously. The circle doesn't present this challenge due to having constant curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton <fatherhorton@gmail.com <mailto:fatherhorton@gmail.com>> wrote:
I think our emails crossed. I was addressing how to approximate an ellipse with constant-length segments in general, not how to solve your problem. When I had to do something similar, I used polar coordinates and stepped up the value of theta until the distance from the previous point was correct. (In the constant-length version, I'd keep a running total of the maximum desired length so that round-off errors don't accumulate.)
Figuring out the circumscribed polygon is, as you note, a greater challenge. I can visualize how to do it for n = 4, but it doesn't extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <avm4@cornell.edu <mailto:avm4@cornell.edu>> wrote:
Eh, there's no such thing as "brute force" in this situation. Step through *what* with sufficiently small steps? Of course an iterative algorithm is necessary. This is true for most interesting problems. No big deal. Show me an iterative algorithm that converges to the circumscribing polygon. When I tried to devise such an algorithm I couldn't find one that would converge. Writing an algorithm that would converge to the *inscribed* polygon seemed to be straight forward. Note also that computing arc length of the ellipse doesn't really help here. You can do that by solving an elliptical integral, or summing up small segments. But what's the connection between arc length of the ellipse and a circumscribing polygon with equal segment lengths? If there's a connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton <fatherhorton@gmail.com <mailto:fatherhorton@gmail.com>> wrote:
Except by brute force (step through it with sufficiently small steps), it’s not possible because there’s no closed form expression for the distance between two points on a general ellipse (as opposed to special cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <avm4@cornell.edu <mailto:avm4@cornell.edu>> wrote:
Here's a challenge for you: draw an "ellipse" polygon with N segments that are all the same length that circumscribes a perfect ellipse with given semiaxes a and b. I tried to solve this problem and could not find a solution. I can draw an *inscribed* N segment "ellipse" polygon with all segments the same length---though it's not easy---but the circumscribed case eludes me. I think of the inscribed case as, every vertex is located on the ellipse and the circumscribed case, every segment is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <sprabhakar2006@gmail.com <mailto:sprabhakar2006@gmail.com>> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor axis "r1" and "r2" respectively and with center "cp" and number of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <openscad@jordan.maileater.net <mailto:openscad@jordan.maileater.net>> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
> I created an oval using:
> inMm = 25.4;
> fullW = 35.5 * inMm;
> fullL = 54.5 * inMm;
> resize([fullW, fullL, 1])
> circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster in complicated cases.
> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can chime in - but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major (long) and minor (short) axis dimensions; the other parts of the curve mathematically flow from those two measurements.
You can make something else, but you can't make it using this technique.
> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2: https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves <https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves>
DIY, here's a function that I cobbled together some years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443 <https://www.thingiverse.com/thing:8443>
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3*u*(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls together.
Note that the four coordinates correspond to the coordinates of the control points in a GUI drawing program: the start point, control point associated with the start, the control point associated with the end, and the end point. I have sometimes created OpenSCAD objects by tracing an object in a drawing program and then reading off the coordinates of the various control points.
Note that Bézier functions are not the only mathematical curves. (For instance, they cannot exactly match an arc of a circle.) I'm sure there are numerous other functions that also produce curves.
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AM
Adrian Mariano
Fri, Aug 5, 2022 11:19 PM
Your intuition is incorrect: a circle scaled in one direction IS an
ellipse. The motivation for being interested in circumscribing is to make
holes, though the constraint of equal length sides is probably not what
you'd want for that purpose. You can make a nonuniform circumscribing
polygon for an ellipse by simply making a polygon that circumscribes a
circle and scaling it.
On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson bob@rjcarlson.com wrote:
Seems like I remember Nop Head talking about using circumscribed holes for
screws and even rotating the polygons to produce a more even hole.
If an ellipse is the nominal shape, the using a circumscribed polygon for
a hole and an inscribed polygon for a solid might make sense.
Btw, can someone confirm my intuition that a circle scaled in one
dimension is not an ellipse?
-Bob
Tucson AZ
On Aug 5, 2022, at 15:38, Adrian Mariano avm4@cornell.edu wrote:
I don't have a specific need for a circumscribing elliptical polygon with
equal sides. I was writing code to make ellipses and it seemed like equal
side approximations would be nice, so I solved the case for inscribed, but
couldn't solve the circumscribed case. So it's an unsolved puzzle I've got
that this discussion reminded me of.
On Fri, Aug 5, 2022 at 6:33 PM nop head nop.head@gmail.com wrote:
A long discussion but I have to ask why do you need an elliptical polygon
with equal tangential sides?
On Fri, 5 Aug 2022, 21:58 Adrian Mariano, avm4@cornell.edu wrote:
You think that a circumscribing polygon with equal length segments does
not exist?? I'm skeptical of this claim.
For circumscribed, it's not so hard. You start with a uniform-in-angle
point distribution, compute the segment lengths, compare to the mean
segment length to get an error, and then compute an adjustment based on the
error. It converges quickly.
So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but surely good
enough.
<image.png>
The code for this is in BOSL2 in shapes2d.scad. A similar algorithm
should exist for the circumscribed case, but as I said, my efforts to find
it have failed.
On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
I think circumscribed polygon segments cannot be equal lengths.
inscribed ones may be approximately equal, but would be a very tough
problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano avm4@cornell.edu wrote:
I think you need to be a little careful about defining what you want
to do. If you want to approximate an ellipse with points that are
distributed uniformly over the ellipse, that could be done by approximating
the arc length of the ellipse to find the necessary points---e.g. advance
distance d along the ellipse at each step. But if you want to approximate
an ellipse with a polygon with equal length sides, that approach does not
do it, because the length of the segments will not be constant as the span
sections of the ellipse with differing curvature. And since the perimeter
of the desired polygon is unknown, you can't just step it out analogously.
The circle doesn't present this challenge due to having constant
curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton fatherhorton@gmail.com
wrote:
I think our emails crossed. I was addressing how to approximate an
ellipse with constant-length segments in general, not how to solve your
problem. When I had to do something similar, I used polar coordinates and
stepped up the value of theta until the distance from the previous point
was correct. (In the constant-length version, I'd keep a running total of
the maximum desired length so that round-off errors don't accumulate.)
Figuring out the circumscribed polygon is, as you note, a greater
challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano avm4@cornell.edu
wrote:
Eh, there's no such thing as "brute force" in this situation. Step
through what with sufficiently small steps? Of course an iterative
algorithm is necessary. This is true for most interesting problems. No
big deal. Show me an iterative algorithm that converges to the
circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
fatherhorton@gmail.com> wrote:
Except by brute force (step through it with sufficiently small
steps), it’s not possible because there’s no closed form expression for the
distance between two points on a general ellipse (as opposed to special
cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano avm4@cornell.edu
wrote:
Here's a challenge for you: draw an "ellipse" polygon with N
segments that are all the same length that circumscribes a perfect ellipse
with given semiaxes a and b. I tried to solve this problem and could not
find a solution. I can draw an inscribed N segment "ellipse" polygon
with all segments the same length---though it's not easy---but the
circumscribed case eludes me. I think of the inscribed case as, every
vertex is located on the ellipse and the circumscribed case, every segment
is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor axis
"r1" and "r2" respectively and with center "cp" and number of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster in
complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that
stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can chime
in - but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major (long)
and minor (short) axis dimensions; the other parts of the curve
mathematically flow from those two measurements.
You can make something else, but you can't make it using this
technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls together.
Note that the four coordinates correspond to the coordinates of
the control points in a GUI drawing program: the start point, control
point associated with the start, the control point associated with the end,
and the end point. I have sometimes created OpenSCAD objects by tracing an
object in a drawing program and then reading off the coordinates of the
various control points.
Note that Bézier functions are not the only mathematical
curves. (For instance, they cannot exactly match an arc of a circle.) I'm
sure there are numerous other functions that also produce curves.
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org
Your intuition is incorrect: a circle scaled in one direction *IS* an
ellipse. The motivation for being interested in circumscribing is to make
holes, though the constraint of equal length sides is probably not what
you'd want for that purpose. You can make a nonuniform circumscribing
polygon for an ellipse by simply making a polygon that circumscribes a
circle and scaling it.
On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson <bob@rjcarlson.com> wrote:
> Seems like I remember Nop Head talking about using circumscribed holes for
> screws and even rotating the polygons to produce a more even hole.
>
> If an ellipse is the nominal shape, the using a circumscribed polygon for
> a hole and an inscribed polygon for a solid might make sense.
>
> Btw, can someone confirm my intuition that a circle scaled in one
> dimension is not an ellipse?
>
> -Bob
> Tucson AZ
>
>
>
> On Aug 5, 2022, at 15:38, Adrian Mariano <avm4@cornell.edu> wrote:
>
> I don't have a specific need for a circumscribing elliptical polygon with
> equal sides. I was writing code to make ellipses and it seemed like equal
> side approximations would be nice, so I solved the case for inscribed, but
> couldn't solve the circumscribed case. So it's an unsolved puzzle I've got
> that this discussion reminded me of.
>
> On Fri, Aug 5, 2022 at 6:33 PM nop head <nop.head@gmail.com> wrote:
>
>> A long discussion but I have to ask why do you need an elliptical polygon
>> with equal tangential sides?
>>
>> On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu> wrote:
>>
>>> You think that a circumscribing polygon with equal length segments does
>>> not exist?? I'm skeptical of this claim.
>>>
>>> For circumscribed, it's not so hard. You start with a uniform-in-angle
>>> point distribution, compute the segment lengths, compare to the mean
>>> segment length to get an error, and then compute an adjustment based on the
>>> error. It converges quickly.
>>>
>>> So for example, I have segments lengths of:
>>> ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
>>> 5.57297]
>>>
>>> Maybe not exactly equal, since tolerance is 1e-9, but surely good
>>> enough.
>>> <image.png>
>>>
>>> The code for this is in BOSL2 in shapes2d.scad. A similar algorithm
>>> should exist for the circumscribed case, but as I said, my efforts to find
>>> it have failed.
>>>
>>> On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
>>> sprabhakar2006@gmail.com> wrote:
>>>
>>>> I think circumscribed polygon segments cannot be equal lengths.
>>>> inscribed ones may be approximately equal, but would be a very tough
>>>> problem to solve.
>>>> I tried something.
>>>>
>>>>
>>>>
>>>> On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <avm4@cornell.edu> wrote:
>>>>
>>>>> I think you need to be a little careful about defining what you want
>>>>> to do. If you want to approximate an ellipse with points that are
>>>>> distributed uniformly over the ellipse, that could be done by approximating
>>>>> the arc length of the ellipse to find the necessary points---e.g. advance
>>>>> distance d along the ellipse at each step. But if you want to approximate
>>>>> an ellipse with a polygon with equal length sides, that approach does not
>>>>> do it, because the length of the segments will not be constant as the span
>>>>> sections of the ellipse with differing curvature. And since the perimeter
>>>>> of the desired polygon is unknown, you can't just step it out analogously.
>>>>> The circle doesn't present this challenge due to having constant
>>>>> curvature.
>>>>>
>>>>> On Wed, Aug 3, 2022 at 7:41 PM Father Horton <fatherhorton@gmail.com>
>>>>> wrote:
>>>>>
>>>>>> I think our emails crossed. I was addressing how to approximate an
>>>>>> ellipse with constant-length segments in general, not how to solve your
>>>>>> problem. When I had to do something similar, I used polar coordinates and
>>>>>> stepped up the value of theta until the distance from the previous point
>>>>>> was correct. (In the constant-length version, I'd keep a running total of
>>>>>> the maximum desired length so that round-off errors don't accumulate.)
>>>>>>
>>>>>> Figuring out the circumscribed polygon is, as you note, a greater
>>>>>> challenge. I can visualize how to do it for n = 4, but it doesn't
>>>>>> extrapolate at all.
>>>>>>
>>>>>> On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <avm4@cornell.edu>
>>>>>> wrote:
>>>>>>
>>>>>>> Eh, there's no such thing as "brute force" in this situation. Step
>>>>>>> through *what* with sufficiently small steps? Of course an iterative
>>>>>>> algorithm is necessary. This is true for most interesting problems. No
>>>>>>> big deal. Show me an iterative algorithm that converges to the
>>>>>>> circumscribing polygon. When I tried to devise such an algorithm I
>>>>>>> couldn't find one that would converge. Writing an algorithm that would
>>>>>>> converge to the *inscribed* polygon seemed to be straight forward. Note
>>>>>>> also that computing arc length of the ellipse doesn't really help here.
>>>>>>> You can do that by solving an elliptical integral, or summing up small
>>>>>>> segments. But what's the connection between arc length of the ellipse and
>>>>>>> a circumscribing polygon with equal segment lengths? If there's a
>>>>>>> connection, I don't see it.
>>>>>>>
>>>>>>> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>
>>>>>>>> Except by brute force (step through it with sufficiently small
>>>>>>>> steps), it’s not possible because there’s no closed form expression for the
>>>>>>>> distance between two points on a general ellipse (as opposed to special
>>>>>>>> cases like a circle).
>>>>>>>>
>>>>>>>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <avm4@cornell.edu>
>>>>>>>> wrote:
>>>>>>>>
>>>>>>>>> Here's a challenge for you: draw an "ellipse" polygon with N
>>>>>>>>> segments that are all the same length that circumscribes a perfect ellipse
>>>>>>>>> with given semiaxes a and b. I tried to solve this problem and could not
>>>>>>>>> find a solution. I can draw an *inscribed* N segment "ellipse" polygon
>>>>>>>>> with all segments the same length---though it's not easy---but the
>>>>>>>>> circumscribed case eludes me. I think of the inscribed case as, every
>>>>>>>>> vertex is located on the ellipse and the circumscribed case, every segment
>>>>>>>>> is tangent to the ellipse
>>>>>>>>>
>>>>>>>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>
>>>>>>>>>> In openscad drawing a ellipse is not too difficult
>>>>>>>>>> Refer following code
>>>>>>>>>>
>>>>>>>>>> function to draw an ellipse with semi-major and semi-minor axis
>>>>>>>>>> "r1" and "r2" respectively and with center "cp" and number of segment "s"
>>>>>>>>>> /
>>>>>>>>>> // example:
>>>>>>>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>>>>>>>>
>>>>>>>>>> function ellipse(r1,r2,cp,s=30)=
>>>>>>>>>> let(
>>>>>>>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>>>>>>>>> )sec;
>>>>>>>>>>
>>>>>>>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>>>>>>>>> openscad@jordan.maileater.net> wrote:
>>>>>>>>>>
>>>>>>>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>>>>>>>>
>>>>>>>>>>> I created an oval using:
>>>>>>>>>>> inMm = 25.4;
>>>>>>>>>>> fullW = 35.5 * inMm;
>>>>>>>>>>> fullL = 54.5 * inMm;
>>>>>>>>>>> resize([fullW, fullL, 1])
>>>>>>>>>>> circle(d=fullW);
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> Tip: use scale() rather than resize(); it previews faster in
>>>>>>>>>>> complicated cases.
>>>>>>>>>>>
>>>>>>>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> You're taking a circle and stretching it; the result of that
>>>>>>>>>>> stretching is an ellipse.
>>>>>>>>>>>
>>>>>>>>>>> I'm not 100% sure - maybe one of the real math wizards can chime
>>>>>>>>>>> in - but I think the shape you are asking for would not be an ellipse.
>>>>>>>>>>>
>>>>>>>>>>> I believe that an ellipse is fully defined by its major (long)
>>>>>>>>>>> and minor (short) axis dimensions; the other parts of the curve
>>>>>>>>>>> mathematically flow from those two measurements.
>>>>>>>>>>>
>>>>>>>>>>> You can make something else, but you can't make it using this
>>>>>>>>>>> technique.
>>>>>>>>>>>
>>>>>>>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> You can use a Bézier function.
>>>>>>>>>>>
>>>>>>>>>>> Using BOSL2:
>>>>>>>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>>>>>>>>
>>>>>>>>>>> DIY, here's a function that I cobbled together some years ago:
>>>>>>>>>>>
>>>>>>>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>>>>>>>>> // but that yielded either single points or a raft of triangles;
>>>>>>>>>>> // this yields a vector of points that you can then concatenate
>>>>>>>>>>> // with other pieces to form a single polygon.
>>>>>>>>>>> // If we were really clever, I think it would be possible to
>>>>>>>>>>> // automatically space the output points based on how linear
>>>>>>>>>>> // the curve is at that point. But right now I'm not that clever.
>>>>>>>>>>> function BEZ03(u) = pow((1-u), 3);
>>>>>>>>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>>>>>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>>>>>>>>> function BEZ33(u) = pow(u,3);
>>>>>>>>>>>
>>>>>>>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>>>>>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>>>>>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>>>>>>>>
>>>>>>>>>>> // p0 - start point
>>>>>>>>>>> // p1 - control point 1, line departs p0 headed this way
>>>>>>>>>>> // p2 - control point 2, line arrives at p3 from this way
>>>>>>>>>>> // p3 - end point
>>>>>>>>>>> // segs - number of segments
>>>>>>>>>>> function bez(p0, p1, p2, p3, segs) = [
>>>>>>>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>>>>>>>>> ];
>>>>>>>>>>>
>>>>>>>>>>> It is that final function bez() that I actually call, e.g.
>>>>>>>>>>>
>>>>>>>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>>>>>>>>
>>>>>>>>>>> often concatenating the results of several bez() calls together.
>>>>>>>>>>>
>>>>>>>>>>> Note that the four coordinates correspond to the coordinates of
>>>>>>>>>>> the control points in a GUI drawing program: the start point, control
>>>>>>>>>>> point associated with the start, the control point associated with the end,
>>>>>>>>>>> and the end point. I have sometimes created OpenSCAD objects by tracing an
>>>>>>>>>>> object in a drawing program and then reading off the coordinates of the
>>>>>>>>>>> various control points.
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> Note that Bézier functions are not the only mathematical
>>>>>>>>>>> curves. (For instance, they cannot exactly match an arc of a circle.) I'm
>>>>>>>>>>> sure there are numerous other functions that also produce curves.
>>>>>>>>>>> _______________________________________________
>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>>>>>
>>>>>>>>>> _______________________________________________
>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>>>>
>>>>>>>>> _______________________________________________
>>>>>>>>> OpenSCAD mailing list
>>>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>>>
>>>>>>>> _______________________________________________
>>>>>>>> OpenSCAD mailing list
>>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>>
>>>>>>> _______________________________________________
>>>>>>> OpenSCAD mailing list
>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>
>>>>>> _______________________________________________
>>>>>> OpenSCAD mailing list
>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>
>>>>> _______________________________________________
>>>>> OpenSCAD mailing list
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>>>>>
>>>> _______________________________________________
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>
NH
nop head
Sat, Aug 6, 2022 12:49 AM
I have been thinking about trying to make the equivalent of polyholes for
cycloidal gears. The profile varies between convex and concave, so it makes
my brain hurt.
On Sat, 6 Aug 2022, 00:20 Adrian Mariano, avm4@cornell.edu wrote:
Your intuition is incorrect: a circle scaled in one direction IS an
ellipse. The motivation for being interested in circumscribing is to make
holes, though the constraint of equal length sides is probably not what
you'd want for that purpose. You can make a nonuniform circumscribing
polygon for an ellipse by simply making a polygon that circumscribes a
circle and scaling it.
On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson bob@rjcarlson.com wrote:
Seems like I remember Nop Head talking about using circumscribed holes
for screws and even rotating the polygons to produce a more even hole.
If an ellipse is the nominal shape, the using a circumscribed polygon for
a hole and an inscribed polygon for a solid might make sense.
Btw, can someone confirm my intuition that a circle scaled in one
dimension is not an ellipse?
-Bob
Tucson AZ
On Aug 5, 2022, at 15:38, Adrian Mariano avm4@cornell.edu wrote:
I don't have a specific need for a circumscribing elliptical polygon with
equal sides. I was writing code to make ellipses and it seemed like equal
side approximations would be nice, so I solved the case for inscribed, but
couldn't solve the circumscribed case. So it's an unsolved puzzle I've got
that this discussion reminded me of.
On Fri, Aug 5, 2022 at 6:33 PM nop head nop.head@gmail.com wrote:
A long discussion but I have to ask why do you need an elliptical
polygon with equal tangential sides?
On Fri, 5 Aug 2022, 21:58 Adrian Mariano, avm4@cornell.edu wrote:
You think that a circumscribing polygon with equal length segments does
not exist?? I'm skeptical of this claim.
For circumscribed, it's not so hard. You start with a uniform-in-angle
point distribution, compute the segment lengths, compare to the mean
segment length to get an error, and then compute an adjustment based on the
error. It converges quickly.
So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but surely good
enough.
<image.png>
The code for this is in BOSL2 in shapes2d.scad. A similar algorithm
should exist for the circumscribed case, but as I said, my efforts to find
it have failed.
On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
I think circumscribed polygon segments cannot be equal lengths.
inscribed ones may be approximately equal, but would be a very tough
problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano avm4@cornell.edu wrote:
I think you need to be a little careful about defining what you want
to do. If you want to approximate an ellipse with points that are
distributed uniformly over the ellipse, that could be done by approximating
the arc length of the ellipse to find the necessary points---e.g. advance
distance d along the ellipse at each step. But if you want to approximate
an ellipse with a polygon with equal length sides, that approach does not
do it, because the length of the segments will not be constant as the span
sections of the ellipse with differing curvature. And since the perimeter
of the desired polygon is unknown, you can't just step it out analogously.
The circle doesn't present this challenge due to having constant
curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton fatherhorton@gmail.com
wrote:
I think our emails crossed. I was addressing how to approximate an
ellipse with constant-length segments in general, not how to solve your
problem. When I had to do something similar, I used polar coordinates and
stepped up the value of theta until the distance from the previous point
was correct. (In the constant-length version, I'd keep a running total of
the maximum desired length so that round-off errors don't accumulate.)
Figuring out the circumscribed polygon is, as you note, a greater
challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano avm4@cornell.edu
wrote:
Eh, there's no such thing as "brute force" in this situation. Step
through what with sufficiently small steps? Of course an iterative
algorithm is necessary. This is true for most interesting problems. No
big deal. Show me an iterative algorithm that converges to the
circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
fatherhorton@gmail.com> wrote:
Except by brute force (step through it with sufficiently small
steps), it’s not possible because there’s no closed form expression for the
distance between two points on a general ellipse (as opposed to special
cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano avm4@cornell.edu
wrote:
Here's a challenge for you: draw an "ellipse" polygon with N
segments that are all the same length that circumscribes a perfect ellipse
with given semiaxes a and b. I tried to solve this problem and could not
find a solution. I can draw an inscribed N segment "ellipse" polygon
with all segments the same length---though it's not easy---but the
circumscribed case eludes me. I think of the inscribed case as, every
vertex is located on the ellipse and the circumscribed case, every segment
is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor axis
"r1" and "r2" respectively and with center "cp" and number of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster in
complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that
stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can
chime in - but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major (long)
and minor (short) axis dimensions; the other parts of the curve
mathematically flow from those two measurements.
You can make something else, but you can't make it using this
technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls together.
Note that the four coordinates correspond to the coordinates of
the control points in a GUI drawing program: the start point, control
point associated with the start, the control point associated with the end,
and the end point. I have sometimes created OpenSCAD objects by tracing an
object in a drawing program and then reading off the coordinates of the
various control points.
Note that Bézier functions are not the only mathematical
curves. (For instance, they cannot exactly match an arc of a circle.) I'm
sure there are numerous other functions that also produce curves.
OpenSCAD mailing list
To unsubscribe send an email to
discuss-leave@lists.openscad.org
I have been thinking about trying to make the equivalent of polyholes for
cycloidal gears. The profile varies between convex and concave, so it makes
my brain hurt.
On Sat, 6 Aug 2022, 00:20 Adrian Mariano, <avm4@cornell.edu> wrote:
> Your intuition is incorrect: a circle scaled in one direction *IS* an
> ellipse. The motivation for being interested in circumscribing is to make
> holes, though the constraint of equal length sides is probably not what
> you'd want for that purpose. You can make a nonuniform circumscribing
> polygon for an ellipse by simply making a polygon that circumscribes a
> circle and scaling it.
>
> On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson <bob@rjcarlson.com> wrote:
>
>> Seems like I remember Nop Head talking about using circumscribed holes
>> for screws and even rotating the polygons to produce a more even hole.
>>
>> If an ellipse is the nominal shape, the using a circumscribed polygon for
>> a hole and an inscribed polygon for a solid might make sense.
>>
>> Btw, can someone confirm my intuition that a circle scaled in one
>> dimension is not an ellipse?
>>
>> -Bob
>> Tucson AZ
>>
>>
>>
>> On Aug 5, 2022, at 15:38, Adrian Mariano <avm4@cornell.edu> wrote:
>>
>> I don't have a specific need for a circumscribing elliptical polygon with
>> equal sides. I was writing code to make ellipses and it seemed like equal
>> side approximations would be nice, so I solved the case for inscribed, but
>> couldn't solve the circumscribed case. So it's an unsolved puzzle I've got
>> that this discussion reminded me of.
>>
>> On Fri, Aug 5, 2022 at 6:33 PM nop head <nop.head@gmail.com> wrote:
>>
>>> A long discussion but I have to ask why do you need an elliptical
>>> polygon with equal tangential sides?
>>>
>>> On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu> wrote:
>>>
>>>> You think that a circumscribing polygon with equal length segments does
>>>> not exist?? I'm skeptical of this claim.
>>>>
>>>> For circumscribed, it's not so hard. You start with a uniform-in-angle
>>>> point distribution, compute the segment lengths, compare to the mean
>>>> segment length to get an error, and then compute an adjustment based on the
>>>> error. It converges quickly.
>>>>
>>>> So for example, I have segments lengths of:
>>>> ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
>>>> 5.57297]
>>>>
>>>> Maybe not exactly equal, since tolerance is 1e-9, but surely good
>>>> enough.
>>>> <image.png>
>>>>
>>>> The code for this is in BOSL2 in shapes2d.scad. A similar algorithm
>>>> should exist for the circumscribed case, but as I said, my efforts to find
>>>> it have failed.
>>>>
>>>> On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
>>>> sprabhakar2006@gmail.com> wrote:
>>>>
>>>>> I think circumscribed polygon segments cannot be equal lengths.
>>>>> inscribed ones may be approximately equal, but would be a very tough
>>>>> problem to solve.
>>>>> I tried something.
>>>>>
>>>>>
>>>>>
>>>>> On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <avm4@cornell.edu> wrote:
>>>>>
>>>>>> I think you need to be a little careful about defining what you want
>>>>>> to do. If you want to approximate an ellipse with points that are
>>>>>> distributed uniformly over the ellipse, that could be done by approximating
>>>>>> the arc length of the ellipse to find the necessary points---e.g. advance
>>>>>> distance d along the ellipse at each step. But if you want to approximate
>>>>>> an ellipse with a polygon with equal length sides, that approach does not
>>>>>> do it, because the length of the segments will not be constant as the span
>>>>>> sections of the ellipse with differing curvature. And since the perimeter
>>>>>> of the desired polygon is unknown, you can't just step it out analogously.
>>>>>> The circle doesn't present this challenge due to having constant
>>>>>> curvature.
>>>>>>
>>>>>> On Wed, Aug 3, 2022 at 7:41 PM Father Horton <fatherhorton@gmail.com>
>>>>>> wrote:
>>>>>>
>>>>>>> I think our emails crossed. I was addressing how to approximate an
>>>>>>> ellipse with constant-length segments in general, not how to solve your
>>>>>>> problem. When I had to do something similar, I used polar coordinates and
>>>>>>> stepped up the value of theta until the distance from the previous point
>>>>>>> was correct. (In the constant-length version, I'd keep a running total of
>>>>>>> the maximum desired length so that round-off errors don't accumulate.)
>>>>>>>
>>>>>>> Figuring out the circumscribed polygon is, as you note, a greater
>>>>>>> challenge. I can visualize how to do it for n = 4, but it doesn't
>>>>>>> extrapolate at all.
>>>>>>>
>>>>>>> On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <avm4@cornell.edu>
>>>>>>> wrote:
>>>>>>>
>>>>>>>> Eh, there's no such thing as "brute force" in this situation. Step
>>>>>>>> through *what* with sufficiently small steps? Of course an iterative
>>>>>>>> algorithm is necessary. This is true for most interesting problems. No
>>>>>>>> big deal. Show me an iterative algorithm that converges to the
>>>>>>>> circumscribing polygon. When I tried to devise such an algorithm I
>>>>>>>> couldn't find one that would converge. Writing an algorithm that would
>>>>>>>> converge to the *inscribed* polygon seemed to be straight forward. Note
>>>>>>>> also that computing arc length of the ellipse doesn't really help here.
>>>>>>>> You can do that by solving an elliptical integral, or summing up small
>>>>>>>> segments. But what's the connection between arc length of the ellipse and
>>>>>>>> a circumscribing polygon with equal segment lengths? If there's a
>>>>>>>> connection, I don't see it.
>>>>>>>>
>>>>>>>> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>
>>>>>>>>> Except by brute force (step through it with sufficiently small
>>>>>>>>> steps), it’s not possible because there’s no closed form expression for the
>>>>>>>>> distance between two points on a general ellipse (as opposed to special
>>>>>>>>> cases like a circle).
>>>>>>>>>
>>>>>>>>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <avm4@cornell.edu>
>>>>>>>>> wrote:
>>>>>>>>>
>>>>>>>>>> Here's a challenge for you: draw an "ellipse" polygon with N
>>>>>>>>>> segments that are all the same length that circumscribes a perfect ellipse
>>>>>>>>>> with given semiaxes a and b. I tried to solve this problem and could not
>>>>>>>>>> find a solution. I can draw an *inscribed* N segment "ellipse" polygon
>>>>>>>>>> with all segments the same length---though it's not easy---but the
>>>>>>>>>> circumscribed case eludes me. I think of the inscribed case as, every
>>>>>>>>>> vertex is located on the ellipse and the circumscribed case, every segment
>>>>>>>>>> is tangent to the ellipse
>>>>>>>>>>
>>>>>>>>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>
>>>>>>>>>>> In openscad drawing a ellipse is not too difficult
>>>>>>>>>>> Refer following code
>>>>>>>>>>>
>>>>>>>>>>> function to draw an ellipse with semi-major and semi-minor axis
>>>>>>>>>>> "r1" and "r2" respectively and with center "cp" and number of segment "s"
>>>>>>>>>>> /
>>>>>>>>>>> // example:
>>>>>>>>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>>>>>>>>>
>>>>>>>>>>> function ellipse(r1,r2,cp,s=30)=
>>>>>>>>>>> let(
>>>>>>>>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>>>>>>>>>> )sec;
>>>>>>>>>>>
>>>>>>>>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>>>>>>>>>> openscad@jordan.maileater.net> wrote:
>>>>>>>>>>>
>>>>>>>>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>>>>>>>>>
>>>>>>>>>>>> I created an oval using:
>>>>>>>>>>>> inMm = 25.4;
>>>>>>>>>>>> fullW = 35.5 * inMm;
>>>>>>>>>>>> fullL = 54.5 * inMm;
>>>>>>>>>>>> resize([fullW, fullL, 1])
>>>>>>>>>>>> circle(d=fullW);
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Tip: use scale() rather than resize(); it previews faster in
>>>>>>>>>>>> complicated cases.
>>>>>>>>>>>>
>>>>>>>>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> You're taking a circle and stretching it; the result of that
>>>>>>>>>>>> stretching is an ellipse.
>>>>>>>>>>>>
>>>>>>>>>>>> I'm not 100% sure - maybe one of the real math wizards can
>>>>>>>>>>>> chime in - but I think the shape you are asking for would not be an ellipse.
>>>>>>>>>>>>
>>>>>>>>>>>> I believe that an ellipse is fully defined by its major (long)
>>>>>>>>>>>> and minor (short) axis dimensions; the other parts of the curve
>>>>>>>>>>>> mathematically flow from those two measurements.
>>>>>>>>>>>>
>>>>>>>>>>>> You can make something else, but you can't make it using this
>>>>>>>>>>>> technique.
>>>>>>>>>>>>
>>>>>>>>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> You can use a Bézier function.
>>>>>>>>>>>>
>>>>>>>>>>>> Using BOSL2:
>>>>>>>>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>>>>>>>>>
>>>>>>>>>>>> DIY, here's a function that I cobbled together some years ago:
>>>>>>>>>>>>
>>>>>>>>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>>>>>>>>>> // but that yielded either single points or a raft of triangles;
>>>>>>>>>>>> // this yields a vector of points that you can then concatenate
>>>>>>>>>>>> // with other pieces to form a single polygon.
>>>>>>>>>>>> // If we were really clever, I think it would be possible to
>>>>>>>>>>>> // automatically space the output points based on how linear
>>>>>>>>>>>> // the curve is at that point. But right now I'm not that clever.
>>>>>>>>>>>> function BEZ03(u) = pow((1-u), 3);
>>>>>>>>>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>>>>>>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>>>>>>>>>> function BEZ33(u) = pow(u,3);
>>>>>>>>>>>>
>>>>>>>>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>>>>>>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>>>>>>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>>>>>>>>>
>>>>>>>>>>>> // p0 - start point
>>>>>>>>>>>> // p1 - control point 1, line departs p0 headed this way
>>>>>>>>>>>> // p2 - control point 2, line arrives at p3 from this way
>>>>>>>>>>>> // p3 - end point
>>>>>>>>>>>> // segs - number of segments
>>>>>>>>>>>> function bez(p0, p1, p2, p3, segs) = [
>>>>>>>>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>>>>>>>>>> ];
>>>>>>>>>>>>
>>>>>>>>>>>> It is that final function bez() that I actually call, e.g.
>>>>>>>>>>>>
>>>>>>>>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>>>>>>>>>
>>>>>>>>>>>> often concatenating the results of several bez() calls together.
>>>>>>>>>>>>
>>>>>>>>>>>> Note that the four coordinates correspond to the coordinates of
>>>>>>>>>>>> the control points in a GUI drawing program: the start point, control
>>>>>>>>>>>> point associated with the start, the control point associated with the end,
>>>>>>>>>>>> and the end point. I have sometimes created OpenSCAD objects by tracing an
>>>>>>>>>>>> object in a drawing program and then reading off the coordinates of the
>>>>>>>>>>>> various control points.
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Note that Bézier functions are not the only mathematical
>>>>>>>>>>>> curves. (For instance, they cannot exactly match an arc of a circle.) I'm
>>>>>>>>>>>> sure there are numerous other functions that also produce curves.
>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>
>>>>>>>>>>> _______________________________________________
>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>>>>>
>>>>>>>>>> _______________________________________________
>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>>>>
>>>>>>>>> _______________________________________________
>>>>>>>>> OpenSCAD mailing list
>>>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>>>
>>>>>>>> _______________________________________________
>>>>>>>> OpenSCAD mailing list
>>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>>
>>>>>>> _______________________________________________
>>>>>>> OpenSCAD mailing list
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>>>>>>>
>>>>>> _______________________________________________
>>>>>> OpenSCAD mailing list
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>>>>>>
>>>>> _______________________________________________
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>>>>>
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>>>>
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>
SP
Sanjeev Prabhakar
Sat, Aug 6, 2022 12:54 AM
Circumscribed means tangents from points on ellipse
How can the tangents from equally spaced arc lengths or equally spaced
angles be equal for an ellipse
You need to change the arc lengths dynamically and also accurately estimate
the segment length which fits the ellipse .
On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, avm4@cornell.edu wrote:
Your intuition is incorrect: a circle scaled in one direction IS an
ellipse. The motivation for being interested in circumscribing is to make
holes, though the constraint of equal length sides is probably not what
you'd want for that purpose. You can make a nonuniform circumscribing
polygon for an ellipse by simply making a polygon that circumscribes a
circle and scaling it.
On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson bob@rjcarlson.com wrote:
Seems like I remember Nop Head talking about using circumscribed holes
for screws and even rotating the polygons to produce a more even hole.
If an ellipse is the nominal shape, the using a circumscribed polygon for
a hole and an inscribed polygon for a solid might make sense.
Btw, can someone confirm my intuition that a circle scaled in one
dimension is not an ellipse?
-Bob
Tucson AZ
On Aug 5, 2022, at 15:38, Adrian Mariano avm4@cornell.edu wrote:
I don't have a specific need for a circumscribing elliptical polygon with
equal sides. I was writing code to make ellipses and it seemed like equal
side approximations would be nice, so I solved the case for inscribed, but
couldn't solve the circumscribed case. So it's an unsolved puzzle I've got
that this discussion reminded me of.
On Fri, Aug 5, 2022 at 6:33 PM nop head nop.head@gmail.com wrote:
A long discussion but I have to ask why do you need an elliptical
polygon with equal tangential sides?
On Fri, 5 Aug 2022, 21:58 Adrian Mariano, avm4@cornell.edu wrote:
You think that a circumscribing polygon with equal length segments does
not exist?? I'm skeptical of this claim.
For circumscribed, it's not so hard. You start with a uniform-in-angle
point distribution, compute the segment lengths, compare to the mean
segment length to get an error, and then compute an adjustment based on the
error. It converges quickly.
So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but surely good
enough.
<image.png>
The code for this is in BOSL2 in shapes2d.scad. A similar algorithm
should exist for the circumscribed case, but as I said, my efforts to find
it have failed.
On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
I think circumscribed polygon segments cannot be equal lengths.
inscribed ones may be approximately equal, but would be a very tough
problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano avm4@cornell.edu wrote:
I think you need to be a little careful about defining what you want
to do. If you want to approximate an ellipse with points that are
distributed uniformly over the ellipse, that could be done by approximating
the arc length of the ellipse to find the necessary points---e.g. advance
distance d along the ellipse at each step. But if you want to approximate
an ellipse with a polygon with equal length sides, that approach does not
do it, because the length of the segments will not be constant as the span
sections of the ellipse with differing curvature. And since the perimeter
of the desired polygon is unknown, you can't just step it out analogously.
The circle doesn't present this challenge due to having constant
curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton fatherhorton@gmail.com
wrote:
I think our emails crossed. I was addressing how to approximate an
ellipse with constant-length segments in general, not how to solve your
problem. When I had to do something similar, I used polar coordinates and
stepped up the value of theta until the distance from the previous point
was correct. (In the constant-length version, I'd keep a running total of
the maximum desired length so that round-off errors don't accumulate.)
Figuring out the circumscribed polygon is, as you note, a greater
challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano avm4@cornell.edu
wrote:
Eh, there's no such thing as "brute force" in this situation. Step
through what with sufficiently small steps? Of course an iterative
algorithm is necessary. This is true for most interesting problems. No
big deal. Show me an iterative algorithm that converges to the
circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
fatherhorton@gmail.com> wrote:
Except by brute force (step through it with sufficiently small
steps), it’s not possible because there’s no closed form expression for the
distance between two points on a general ellipse (as opposed to special
cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano avm4@cornell.edu
wrote:
Here's a challenge for you: draw an "ellipse" polygon with N
segments that are all the same length that circumscribes a perfect ellipse
with given semiaxes a and b. I tried to solve this problem and could not
find a solution. I can draw an inscribed N segment "ellipse" polygon
with all segments the same length---though it's not easy---but the
circumscribed case eludes me. I think of the inscribed case as, every
vertex is located on the ellipse and the circumscribed case, every segment
is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor axis
"r1" and "r2" respectively and with center "cp" and number of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster in
complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that
stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can
chime in - but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major (long)
and minor (short) axis dimensions; the other parts of the curve
mathematically flow from those two measurements.
You can make something else, but you can't make it using this
technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls together.
Note that the four coordinates correspond to the coordinates of
the control points in a GUI drawing program: the start point, control
point associated with the start, the control point associated with the end,
and the end point. I have sometimes created OpenSCAD objects by tracing an
object in a drawing program and then reading off the coordinates of the
various control points.
Note that Bézier functions are not the only mathematical
curves. (For instance, they cannot exactly match an arc of a circle.) I'm
sure there are numerous other functions that also produce curves.
OpenSCAD mailing list
To unsubscribe send an email to
discuss-leave@lists.openscad.org
Circumscribed means tangents from points on ellipse
How can the tangents from equally spaced arc lengths or equally spaced
angles be equal for an ellipse
You need to change the arc lengths dynamically and also accurately estimate
the segment length which fits the ellipse .
On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, <avm4@cornell.edu> wrote:
> Your intuition is incorrect: a circle scaled in one direction *IS* an
> ellipse. The motivation for being interested in circumscribing is to make
> holes, though the constraint of equal length sides is probably not what
> you'd want for that purpose. You can make a nonuniform circumscribing
> polygon for an ellipse by simply making a polygon that circumscribes a
> circle and scaling it.
>
> On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson <bob@rjcarlson.com> wrote:
>
>> Seems like I remember Nop Head talking about using circumscribed holes
>> for screws and even rotating the polygons to produce a more even hole.
>>
>> If an ellipse is the nominal shape, the using a circumscribed polygon for
>> a hole and an inscribed polygon for a solid might make sense.
>>
>> Btw, can someone confirm my intuition that a circle scaled in one
>> dimension is not an ellipse?
>>
>> -Bob
>> Tucson AZ
>>
>>
>>
>> On Aug 5, 2022, at 15:38, Adrian Mariano <avm4@cornell.edu> wrote:
>>
>> I don't have a specific need for a circumscribing elliptical polygon with
>> equal sides. I was writing code to make ellipses and it seemed like equal
>> side approximations would be nice, so I solved the case for inscribed, but
>> couldn't solve the circumscribed case. So it's an unsolved puzzle I've got
>> that this discussion reminded me of.
>>
>> On Fri, Aug 5, 2022 at 6:33 PM nop head <nop.head@gmail.com> wrote:
>>
>>> A long discussion but I have to ask why do you need an elliptical
>>> polygon with equal tangential sides?
>>>
>>> On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu> wrote:
>>>
>>>> You think that a circumscribing polygon with equal length segments does
>>>> not exist?? I'm skeptical of this claim.
>>>>
>>>> For circumscribed, it's not so hard. You start with a uniform-in-angle
>>>> point distribution, compute the segment lengths, compare to the mean
>>>> segment length to get an error, and then compute an adjustment based on the
>>>> error. It converges quickly.
>>>>
>>>> So for example, I have segments lengths of:
>>>> ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
>>>> 5.57297]
>>>>
>>>> Maybe not exactly equal, since tolerance is 1e-9, but surely good
>>>> enough.
>>>> <image.png>
>>>>
>>>> The code for this is in BOSL2 in shapes2d.scad. A similar algorithm
>>>> should exist for the circumscribed case, but as I said, my efforts to find
>>>> it have failed.
>>>>
>>>> On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
>>>> sprabhakar2006@gmail.com> wrote:
>>>>
>>>>> I think circumscribed polygon segments cannot be equal lengths.
>>>>> inscribed ones may be approximately equal, but would be a very tough
>>>>> problem to solve.
>>>>> I tried something.
>>>>>
>>>>>
>>>>>
>>>>> On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <avm4@cornell.edu> wrote:
>>>>>
>>>>>> I think you need to be a little careful about defining what you want
>>>>>> to do. If you want to approximate an ellipse with points that are
>>>>>> distributed uniformly over the ellipse, that could be done by approximating
>>>>>> the arc length of the ellipse to find the necessary points---e.g. advance
>>>>>> distance d along the ellipse at each step. But if you want to approximate
>>>>>> an ellipse with a polygon with equal length sides, that approach does not
>>>>>> do it, because the length of the segments will not be constant as the span
>>>>>> sections of the ellipse with differing curvature. And since the perimeter
>>>>>> of the desired polygon is unknown, you can't just step it out analogously.
>>>>>> The circle doesn't present this challenge due to having constant
>>>>>> curvature.
>>>>>>
>>>>>> On Wed, Aug 3, 2022 at 7:41 PM Father Horton <fatherhorton@gmail.com>
>>>>>> wrote:
>>>>>>
>>>>>>> I think our emails crossed. I was addressing how to approximate an
>>>>>>> ellipse with constant-length segments in general, not how to solve your
>>>>>>> problem. When I had to do something similar, I used polar coordinates and
>>>>>>> stepped up the value of theta until the distance from the previous point
>>>>>>> was correct. (In the constant-length version, I'd keep a running total of
>>>>>>> the maximum desired length so that round-off errors don't accumulate.)
>>>>>>>
>>>>>>> Figuring out the circumscribed polygon is, as you note, a greater
>>>>>>> challenge. I can visualize how to do it for n = 4, but it doesn't
>>>>>>> extrapolate at all.
>>>>>>>
>>>>>>> On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <avm4@cornell.edu>
>>>>>>> wrote:
>>>>>>>
>>>>>>>> Eh, there's no such thing as "brute force" in this situation. Step
>>>>>>>> through *what* with sufficiently small steps? Of course an iterative
>>>>>>>> algorithm is necessary. This is true for most interesting problems. No
>>>>>>>> big deal. Show me an iterative algorithm that converges to the
>>>>>>>> circumscribing polygon. When I tried to devise such an algorithm I
>>>>>>>> couldn't find one that would converge. Writing an algorithm that would
>>>>>>>> converge to the *inscribed* polygon seemed to be straight forward. Note
>>>>>>>> also that computing arc length of the ellipse doesn't really help here.
>>>>>>>> You can do that by solving an elliptical integral, or summing up small
>>>>>>>> segments. But what's the connection between arc length of the ellipse and
>>>>>>>> a circumscribing polygon with equal segment lengths? If there's a
>>>>>>>> connection, I don't see it.
>>>>>>>>
>>>>>>>> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>
>>>>>>>>> Except by brute force (step through it with sufficiently small
>>>>>>>>> steps), it’s not possible because there’s no closed form expression for the
>>>>>>>>> distance between two points on a general ellipse (as opposed to special
>>>>>>>>> cases like a circle).
>>>>>>>>>
>>>>>>>>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <avm4@cornell.edu>
>>>>>>>>> wrote:
>>>>>>>>>
>>>>>>>>>> Here's a challenge for you: draw an "ellipse" polygon with N
>>>>>>>>>> segments that are all the same length that circumscribes a perfect ellipse
>>>>>>>>>> with given semiaxes a and b. I tried to solve this problem and could not
>>>>>>>>>> find a solution. I can draw an *inscribed* N segment "ellipse" polygon
>>>>>>>>>> with all segments the same length---though it's not easy---but the
>>>>>>>>>> circumscribed case eludes me. I think of the inscribed case as, every
>>>>>>>>>> vertex is located on the ellipse and the circumscribed case, every segment
>>>>>>>>>> is tangent to the ellipse
>>>>>>>>>>
>>>>>>>>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>
>>>>>>>>>>> In openscad drawing a ellipse is not too difficult
>>>>>>>>>>> Refer following code
>>>>>>>>>>>
>>>>>>>>>>> function to draw an ellipse with semi-major and semi-minor axis
>>>>>>>>>>> "r1" and "r2" respectively and with center "cp" and number of segment "s"
>>>>>>>>>>> /
>>>>>>>>>>> // example:
>>>>>>>>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>>>>>>>>>
>>>>>>>>>>> function ellipse(r1,r2,cp,s=30)=
>>>>>>>>>>> let(
>>>>>>>>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>>>>>>>>>> )sec;
>>>>>>>>>>>
>>>>>>>>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>>>>>>>>>> openscad@jordan.maileater.net> wrote:
>>>>>>>>>>>
>>>>>>>>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>>>>>>>>>
>>>>>>>>>>>> I created an oval using:
>>>>>>>>>>>> inMm = 25.4;
>>>>>>>>>>>> fullW = 35.5 * inMm;
>>>>>>>>>>>> fullL = 54.5 * inMm;
>>>>>>>>>>>> resize([fullW, fullL, 1])
>>>>>>>>>>>> circle(d=fullW);
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Tip: use scale() rather than resize(); it previews faster in
>>>>>>>>>>>> complicated cases.
>>>>>>>>>>>>
>>>>>>>>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> You're taking a circle and stretching it; the result of that
>>>>>>>>>>>> stretching is an ellipse.
>>>>>>>>>>>>
>>>>>>>>>>>> I'm not 100% sure - maybe one of the real math wizards can
>>>>>>>>>>>> chime in - but I think the shape you are asking for would not be an ellipse.
>>>>>>>>>>>>
>>>>>>>>>>>> I believe that an ellipse is fully defined by its major (long)
>>>>>>>>>>>> and minor (short) axis dimensions; the other parts of the curve
>>>>>>>>>>>> mathematically flow from those two measurements.
>>>>>>>>>>>>
>>>>>>>>>>>> You can make something else, but you can't make it using this
>>>>>>>>>>>> technique.
>>>>>>>>>>>>
>>>>>>>>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> You can use a Bézier function.
>>>>>>>>>>>>
>>>>>>>>>>>> Using BOSL2:
>>>>>>>>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>>>>>>>>>
>>>>>>>>>>>> DIY, here's a function that I cobbled together some years ago:
>>>>>>>>>>>>
>>>>>>>>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>>>>>>>>>> // but that yielded either single points or a raft of triangles;
>>>>>>>>>>>> // this yields a vector of points that you can then concatenate
>>>>>>>>>>>> // with other pieces to form a single polygon.
>>>>>>>>>>>> // If we were really clever, I think it would be possible to
>>>>>>>>>>>> // automatically space the output points based on how linear
>>>>>>>>>>>> // the curve is at that point. But right now I'm not that clever.
>>>>>>>>>>>> function BEZ03(u) = pow((1-u), 3);
>>>>>>>>>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>>>>>>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>>>>>>>>>> function BEZ33(u) = pow(u,3);
>>>>>>>>>>>>
>>>>>>>>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>>>>>>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>>>>>>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>>>>>>>>>
>>>>>>>>>>>> // p0 - start point
>>>>>>>>>>>> // p1 - control point 1, line departs p0 headed this way
>>>>>>>>>>>> // p2 - control point 2, line arrives at p3 from this way
>>>>>>>>>>>> // p3 - end point
>>>>>>>>>>>> // segs - number of segments
>>>>>>>>>>>> function bez(p0, p1, p2, p3, segs) = [
>>>>>>>>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>>>>>>>>>> ];
>>>>>>>>>>>>
>>>>>>>>>>>> It is that final function bez() that I actually call, e.g.
>>>>>>>>>>>>
>>>>>>>>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>>>>>>>>>
>>>>>>>>>>>> often concatenating the results of several bez() calls together.
>>>>>>>>>>>>
>>>>>>>>>>>> Note that the four coordinates correspond to the coordinates of
>>>>>>>>>>>> the control points in a GUI drawing program: the start point, control
>>>>>>>>>>>> point associated with the start, the control point associated with the end,
>>>>>>>>>>>> and the end point. I have sometimes created OpenSCAD objects by tracing an
>>>>>>>>>>>> object in a drawing program and then reading off the coordinates of the
>>>>>>>>>>>> various control points.
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Note that Bézier functions are not the only mathematical
>>>>>>>>>>>> curves. (For instance, they cannot exactly match an arc of a circle.) I'm
>>>>>>>>>>>> sure there are numerous other functions that also produce curves.
>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>
>>>>>>>>>>> _______________________________________________
>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>>>>>
>>>>>>>>>> _______________________________________________
>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>>>>
>>>>>>>>> _______________________________________________
>>>>>>>>> OpenSCAD mailing list
>>>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>>>
>>>>>>>> _______________________________________________
>>>>>>>> OpenSCAD mailing list
>>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>>
>>>>>>> _______________________________________________
>>>>>>> OpenSCAD mailing list
>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>
>>>>>> _______________________________________________
>>>>>> OpenSCAD mailing list
>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>
>>>>> _______________________________________________
>>>>> OpenSCAD mailing list
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>>>>>
>>>> _______________________________________________
>>>> OpenSCAD mailing list
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AM
Adrian Mariano
Sat, Aug 6, 2022 4:43 AM
Circumscribed means tangents to points on the ellipse. But there are no
arc lengths (equal or unequal) in this problem, nor equally spaced angles.
The angles of each segment will be different. But the lengths of the
segments are not the same as the arc lengths, which is why arc length does
not matter. Here's a solution to the N=6 polygon sort of inspired by the
idea Jordan suggested, where I adjust the first segment until the second
segment has the right length. The rest of the solution is then obtained
using symmetry, which means there is only a single unknown that can simply
be solved for. But this doesn't generalize to higher N.
[image: image.png]
include<BOSL2/std.scad>
// ellipse semiaxes
// If a is much smaller than b it will fail
a=5;
b=2;
// Find point on ellipse at specified x coordinate
function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
// Find tangent line to ellipse at point (u,w)
// and determine the x value where that line has
// given y value.
function findx(a,b,u,w,y) =
let(
A = -ub^2/w/a^2,
B = w+(ub)^2/w/a^2
)
(y-B)/A;
// First segment if tangent point is at x=u
function first_seg(a,b,u) =
let(
w = ellipse_pt(a,b,u),
x1 = findx(a,b,u,w,0),
x2 = findx(a,b,u,w,b)
)
[[x1,0],[x2,b]];
err = function (u) let(pt = first_seg(a,b,u))
2*pt[1].x - norm(pt[0]-pt[1]);
x = root_find(err,.2a,.9a);
first = first_seg(a,b,x);
top = concat(first,reverse(xflip(first)));
poly = concat(top, reverse(yflip(list_tail(top))));
stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
color("red")stroke(poly,width=.05,closed=true);
// Verify equality of segment lengths
echo(path_segment_lengths(poly,closed=true));
On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar sprabhakar2006@gmail.com
wrote:
Circumscribed means tangents from points on ellipse
How can the tangents from equally spaced arc lengths or equally spaced
angles be equal for an ellipse
You need to change the arc lengths dynamically and also accurately
estimate the segment length which fits the ellipse .
On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, avm4@cornell.edu wrote:
Your intuition is incorrect: a circle scaled in one direction IS an
ellipse. The motivation for being interested in circumscribing is to make
holes, though the constraint of equal length sides is probably not what
you'd want for that purpose. You can make a nonuniform circumscribing
polygon for an ellipse by simply making a polygon that circumscribes a
circle and scaling it.
On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson bob@rjcarlson.com wrote:
Seems like I remember Nop Head talking about using circumscribed holes
for screws and even rotating the polygons to produce a more even hole.
If an ellipse is the nominal shape, the using a circumscribed polygon
for a hole and an inscribed polygon for a solid might make sense.
Btw, can someone confirm my intuition that a circle scaled in one
dimension is not an ellipse?
-Bob
Tucson AZ
On Aug 5, 2022, at 15:38, Adrian Mariano avm4@cornell.edu wrote:
I don't have a specific need for a circumscribing elliptical polygon
with equal sides. I was writing code to make ellipses and it seemed like
equal side approximations would be nice, so I solved the case for
inscribed, but couldn't solve the circumscribed case. So it's an unsolved
puzzle I've got that this discussion reminded me of.
On Fri, Aug 5, 2022 at 6:33 PM nop head nop.head@gmail.com wrote:
A long discussion but I have to ask why do you need an elliptical
polygon with equal tangential sides?
On Fri, 5 Aug 2022, 21:58 Adrian Mariano, avm4@cornell.edu wrote:
You think that a circumscribing polygon with equal length segments
does not exist?? I'm skeptical of this claim.
For circumscribed, it's not so hard. You start with a
uniform-in-angle point distribution, compute the segment lengths, compare
to the mean segment length to get an error, and then compute an adjustment
based on the error. It converges quickly.
So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but surely good
enough.
<image.png>
The code for this is in BOSL2 in shapes2d.scad. A similar algorithm
should exist for the circumscribed case, but as I said, my efforts to find
it have failed.
On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
I think circumscribed polygon segments cannot be equal lengths.
inscribed ones may be approximately equal, but would be a very tough
problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano avm4@cornell.edu wrote:
I think you need to be a little careful about defining what you want
to do. If you want to approximate an ellipse with points that are
distributed uniformly over the ellipse, that could be done by approximating
the arc length of the ellipse to find the necessary points---e.g. advance
distance d along the ellipse at each step. But if you want to approximate
an ellipse with a polygon with equal length sides, that approach does not
do it, because the length of the segments will not be constant as the span
sections of the ellipse with differing curvature. And since the perimeter
of the desired polygon is unknown, you can't just step it out analogously.
The circle doesn't present this challenge due to having constant
curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton fatherhorton@gmail.com
wrote:
I think our emails crossed. I was addressing how to approximate an
ellipse with constant-length segments in general, not how to solve your
problem. When I had to do something similar, I used polar coordinates and
stepped up the value of theta until the distance from the previous point
was correct. (In the constant-length version, I'd keep a running total of
the maximum desired length so that round-off errors don't accumulate.)
Figuring out the circumscribed polygon is, as you note, a greater
challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano avm4@cornell.edu
wrote:
Eh, there's no such thing as "brute force" in this situation.
Step through what with sufficiently small steps? Of course an iterative
algorithm is necessary. This is true for most interesting problems. No
big deal. Show me an iterative algorithm that converges to the
circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
fatherhorton@gmail.com> wrote:
Except by brute force (step through it with sufficiently small
steps), it’s not possible because there’s no closed form expression for the
distance between two points on a general ellipse (as opposed to special
cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano avm4@cornell.edu
wrote:
Here's a challenge for you: draw an "ellipse" polygon with N
segments that are all the same length that circumscribes a perfect ellipse
with given semiaxes a and b. I tried to solve this problem and could not
find a solution. I can draw an inscribed N segment "ellipse" polygon
with all segments the same length---though it's not easy---but the
circumscribed case eludes me. I think of the inscribed case as, every
vertex is located on the ellipse and the circumscribed case, every segment
is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor axis
"r1" and "r2" respectively and with center "cp" and number of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster in
complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that
stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can
chime in - but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major (long)
and minor (short) axis dimensions; the other parts of the curve
mathematically flow from those two measurements.
You can make something else, but you can't make it using this
technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls
together.
Note that the four coordinates correspond to the coordinates
of the control points in a GUI drawing program: the start point, control
point associated with the start, the control point associated with the end,
and the end point. I have sometimes created OpenSCAD objects by tracing an
object in a drawing program and then reading off the coordinates of the
various control points.
Note that Bézier functions are not the only mathematical
curves. (For instance, they cannot exactly match an arc of a circle.) I'm
sure there are numerous other functions that also produce curves.
OpenSCAD mailing list
To unsubscribe send an email to
discuss-leave@lists.openscad.org
Circumscribed means tangents to points on the ellipse. But there are no
arc lengths (equal or unequal) in this problem, nor equally spaced angles.
The angles of each segment will be different. But the lengths of the
segments are not the same as the arc lengths, which is why arc length does
not matter. Here's a solution to the N=6 polygon sort of inspired by the
idea Jordan suggested, where I adjust the first segment until the second
segment has the right length. The rest of the solution is then obtained
using symmetry, which means there is only a single unknown that can simply
be solved for. But this doesn't generalize to higher N.
[image: image.png]
include<BOSL2/std.scad>
// ellipse semiaxes
// If a is much smaller than b it will fail
a=5;
b=2;
// Find point on ellipse at specified x coordinate
function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
// Find tangent line to ellipse at point (u,w)
// and determine the x value where that line has
// given y value.
function findx(a,b,u,w,y) =
let(
A = -u*b^2/w/a^2,
B = w+(u*b)^2/w/a^2
)
(y-B)/A;
// First segment if tangent point is at x=u
function first_seg(a,b,u) =
let(
w = ellipse_pt(a,b,u),
x1 = findx(a,b,u,w,0),
x2 = findx(a,b,u,w,b)
)
[[x1,0],[x2,b]];
err = function (u) let(pt = first_seg(a,b,u))
2*pt[1].x - norm(pt[0]-pt[1]);
x = root_find(err,.2*a,.9*a);
first = first_seg(a,b,x);
top = concat(first,reverse(xflip(first)));
poly = concat(top, reverse(yflip(list_tail(top))));
stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
color("red")stroke(poly,width=.05,closed=true);
// Verify equality of segment lengths
echo(path_segment_lengths(poly,closed=true));
On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <sprabhakar2006@gmail.com>
wrote:
> Circumscribed means tangents from points on ellipse
> How can the tangents from equally spaced arc lengths or equally spaced
> angles be equal for an ellipse
> You need to change the arc lengths dynamically and also accurately
> estimate the segment length which fits the ellipse .
>
>
>
> On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, <avm4@cornell.edu> wrote:
>
>> Your intuition is incorrect: a circle scaled in one direction *IS* an
>> ellipse. The motivation for being interested in circumscribing is to make
>> holes, though the constraint of equal length sides is probably not what
>> you'd want for that purpose. You can make a nonuniform circumscribing
>> polygon for an ellipse by simply making a polygon that circumscribes a
>> circle and scaling it.
>>
>> On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson <bob@rjcarlson.com> wrote:
>>
>>> Seems like I remember Nop Head talking about using circumscribed holes
>>> for screws and even rotating the polygons to produce a more even hole.
>>>
>>> If an ellipse is the nominal shape, the using a circumscribed polygon
>>> for a hole and an inscribed polygon for a solid might make sense.
>>>
>>> Btw, can someone confirm my intuition that a circle scaled in one
>>> dimension is not an ellipse?
>>>
>>> -Bob
>>> Tucson AZ
>>>
>>>
>>>
>>> On Aug 5, 2022, at 15:38, Adrian Mariano <avm4@cornell.edu> wrote:
>>>
>>> I don't have a specific need for a circumscribing elliptical polygon
>>> with equal sides. I was writing code to make ellipses and it seemed like
>>> equal side approximations would be nice, so I solved the case for
>>> inscribed, but couldn't solve the circumscribed case. So it's an unsolved
>>> puzzle I've got that this discussion reminded me of.
>>>
>>> On Fri, Aug 5, 2022 at 6:33 PM nop head <nop.head@gmail.com> wrote:
>>>
>>>> A long discussion but I have to ask why do you need an elliptical
>>>> polygon with equal tangential sides?
>>>>
>>>> On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu> wrote:
>>>>
>>>>> You think that a circumscribing polygon with equal length segments
>>>>> does not exist?? I'm skeptical of this claim.
>>>>>
>>>>> For circumscribed, it's not so hard. You start with a
>>>>> uniform-in-angle point distribution, compute the segment lengths, compare
>>>>> to the mean segment length to get an error, and then compute an adjustment
>>>>> based on the error. It converges quickly.
>>>>>
>>>>> So for example, I have segments lengths of:
>>>>> ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
>>>>> 5.57297]
>>>>>
>>>>> Maybe not exactly equal, since tolerance is 1e-9, but surely good
>>>>> enough.
>>>>> <image.png>
>>>>>
>>>>> The code for this is in BOSL2 in shapes2d.scad. A similar algorithm
>>>>> should exist for the circumscribed case, but as I said, my efforts to find
>>>>> it have failed.
>>>>>
>>>>> On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>
>>>>>> I think circumscribed polygon segments cannot be equal lengths.
>>>>>> inscribed ones may be approximately equal, but would be a very tough
>>>>>> problem to solve.
>>>>>> I tried something.
>>>>>>
>>>>>>
>>>>>>
>>>>>> On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <avm4@cornell.edu> wrote:
>>>>>>
>>>>>>> I think you need to be a little careful about defining what you want
>>>>>>> to do. If you want to approximate an ellipse with points that are
>>>>>>> distributed uniformly over the ellipse, that could be done by approximating
>>>>>>> the arc length of the ellipse to find the necessary points---e.g. advance
>>>>>>> distance d along the ellipse at each step. But if you want to approximate
>>>>>>> an ellipse with a polygon with equal length sides, that approach does not
>>>>>>> do it, because the length of the segments will not be constant as the span
>>>>>>> sections of the ellipse with differing curvature. And since the perimeter
>>>>>>> of the desired polygon is unknown, you can't just step it out analogously.
>>>>>>> The circle doesn't present this challenge due to having constant
>>>>>>> curvature.
>>>>>>>
>>>>>>> On Wed, Aug 3, 2022 at 7:41 PM Father Horton <fatherhorton@gmail.com>
>>>>>>> wrote:
>>>>>>>
>>>>>>>> I think our emails crossed. I was addressing how to approximate an
>>>>>>>> ellipse with constant-length segments in general, not how to solve your
>>>>>>>> problem. When I had to do something similar, I used polar coordinates and
>>>>>>>> stepped up the value of theta until the distance from the previous point
>>>>>>>> was correct. (In the constant-length version, I'd keep a running total of
>>>>>>>> the maximum desired length so that round-off errors don't accumulate.)
>>>>>>>>
>>>>>>>> Figuring out the circumscribed polygon is, as you note, a greater
>>>>>>>> challenge. I can visualize how to do it for n = 4, but it doesn't
>>>>>>>> extrapolate at all.
>>>>>>>>
>>>>>>>> On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <avm4@cornell.edu>
>>>>>>>> wrote:
>>>>>>>>
>>>>>>>>> Eh, there's no such thing as "brute force" in this situation.
>>>>>>>>> Step through *what* with sufficiently small steps? Of course an iterative
>>>>>>>>> algorithm is necessary. This is true for most interesting problems. No
>>>>>>>>> big deal. Show me an iterative algorithm that converges to the
>>>>>>>>> circumscribing polygon. When I tried to devise such an algorithm I
>>>>>>>>> couldn't find one that would converge. Writing an algorithm that would
>>>>>>>>> converge to the *inscribed* polygon seemed to be straight forward. Note
>>>>>>>>> also that computing arc length of the ellipse doesn't really help here.
>>>>>>>>> You can do that by solving an elliptical integral, or summing up small
>>>>>>>>> segments. But what's the connection between arc length of the ellipse and
>>>>>>>>> a circumscribing polygon with equal segment lengths? If there's a
>>>>>>>>> connection, I don't see it.
>>>>>>>>>
>>>>>>>>> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>
>>>>>>>>>> Except by brute force (step through it with sufficiently small
>>>>>>>>>> steps), it’s not possible because there’s no closed form expression for the
>>>>>>>>>> distance between two points on a general ellipse (as opposed to special
>>>>>>>>>> cases like a circle).
>>>>>>>>>>
>>>>>>>>>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <avm4@cornell.edu>
>>>>>>>>>> wrote:
>>>>>>>>>>
>>>>>>>>>>> Here's a challenge for you: draw an "ellipse" polygon with N
>>>>>>>>>>> segments that are all the same length that circumscribes a perfect ellipse
>>>>>>>>>>> with given semiaxes a and b. I tried to solve this problem and could not
>>>>>>>>>>> find a solution. I can draw an *inscribed* N segment "ellipse" polygon
>>>>>>>>>>> with all segments the same length---though it's not easy---but the
>>>>>>>>>>> circumscribed case eludes me. I think of the inscribed case as, every
>>>>>>>>>>> vertex is located on the ellipse and the circumscribed case, every segment
>>>>>>>>>>> is tangent to the ellipse
>>>>>>>>>>>
>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>>
>>>>>>>>>>>> In openscad drawing a ellipse is not too difficult
>>>>>>>>>>>> Refer following code
>>>>>>>>>>>>
>>>>>>>>>>>> function to draw an ellipse with semi-major and semi-minor axis
>>>>>>>>>>>> "r1" and "r2" respectively and with center "cp" and number of segment "s"
>>>>>>>>>>>> /
>>>>>>>>>>>> // example:
>>>>>>>>>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>>>>>>>>>>
>>>>>>>>>>>> function ellipse(r1,r2,cp,s=30)=
>>>>>>>>>>>> let(
>>>>>>>>>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>>>>>>>>>>> )sec;
>>>>>>>>>>>>
>>>>>>>>>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>>>>>>>>>>> openscad@jordan.maileater.net> wrote:
>>>>>>>>>>>>
>>>>>>>>>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>> I created an oval using:
>>>>>>>>>>>>> inMm = 25.4;
>>>>>>>>>>>>> fullW = 35.5 * inMm;
>>>>>>>>>>>>> fullL = 54.5 * inMm;
>>>>>>>>>>>>> resize([fullW, fullL, 1])
>>>>>>>>>>>>> circle(d=fullW);
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> Tip: use scale() rather than resize(); it previews faster in
>>>>>>>>>>>>> complicated cases.
>>>>>>>>>>>>>
>>>>>>>>>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> You're taking a circle and stretching it; the result of that
>>>>>>>>>>>>> stretching is an ellipse.
>>>>>>>>>>>>>
>>>>>>>>>>>>> I'm not 100% sure - maybe one of the real math wizards can
>>>>>>>>>>>>> chime in - but I think the shape you are asking for would not be an ellipse.
>>>>>>>>>>>>>
>>>>>>>>>>>>> I believe that an ellipse is fully defined by its major (long)
>>>>>>>>>>>>> and minor (short) axis dimensions; the other parts of the curve
>>>>>>>>>>>>> mathematically flow from those two measurements.
>>>>>>>>>>>>>
>>>>>>>>>>>>> You can make something else, but you can't make it using this
>>>>>>>>>>>>> technique.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> You can use a Bézier function.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Using BOSL2:
>>>>>>>>>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>>>>>>>>>>
>>>>>>>>>>>>> DIY, here's a function that I cobbled together some years ago:
>>>>>>>>>>>>>
>>>>>>>>>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>>>>>>>>>>> // but that yielded either single points or a raft of triangles;
>>>>>>>>>>>>> // this yields a vector of points that you can then concatenate
>>>>>>>>>>>>> // with other pieces to form a single polygon.
>>>>>>>>>>>>> // If we were really clever, I think it would be possible to
>>>>>>>>>>>>> // automatically space the output points based on how linear
>>>>>>>>>>>>> // the curve is at that point. But right now I'm not that clever.
>>>>>>>>>>>>> function BEZ03(u) = pow((1-u), 3);
>>>>>>>>>>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>>>>>>>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>>>>>>>>>>> function BEZ33(u) = pow(u,3);
>>>>>>>>>>>>>
>>>>>>>>>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>>>>>>>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>>>>>>>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>>>>>>>>>>
>>>>>>>>>>>>> // p0 - start point
>>>>>>>>>>>>> // p1 - control point 1, line departs p0 headed this way
>>>>>>>>>>>>> // p2 - control point 2, line arrives at p3 from this way
>>>>>>>>>>>>> // p3 - end point
>>>>>>>>>>>>> // segs - number of segments
>>>>>>>>>>>>> function bez(p0, p1, p2, p3, segs) = [
>>>>>>>>>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>>>>>>>>>>> ];
>>>>>>>>>>>>>
>>>>>>>>>>>>> It is that final function bez() that I actually call, e.g.
>>>>>>>>>>>>>
>>>>>>>>>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>>>>>>>>>>
>>>>>>>>>>>>> often concatenating the results of several bez() calls
>>>>>>>>>>>>> together.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Note that the four coordinates correspond to the coordinates
>>>>>>>>>>>>> of the control points in a GUI drawing program: the start point, control
>>>>>>>>>>>>> point associated with the start, the control point associated with the end,
>>>>>>>>>>>>> and the end point. I have sometimes created OpenSCAD objects by tracing an
>>>>>>>>>>>>> object in a drawing program and then reading off the coordinates of the
>>>>>>>>>>>>> various control points.
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> Note that Bézier functions are not the only mathematical
>>>>>>>>>>>>> curves. (For instance, they cannot exactly match an arc of a circle.) I'm
>>>>>>>>>>>>> sure there are numerous other functions that also produce curves.
>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>
>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>
>>>>>>>>>>> _______________________________________________
>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>>>>>
>>>>>>>>>> _______________________________________________
>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>>>>
>>>>>>>>> _______________________________________________
>>>>>>>>> OpenSCAD mailing list
>>>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>>>
>>>>>>>> _______________________________________________
>>>>>>>> OpenSCAD mailing list
>>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>>
>>>>>>> _______________________________________________
>>>>>>> OpenSCAD mailing list
>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>
>>>>>> _______________________________________________
>>>>>> OpenSCAD mailing list
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>>>>>>
>>>>> _______________________________________________
>>>>> OpenSCAD mailing list
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>>>>>
>>>> _______________________________________________
>>>> OpenSCAD mailing list
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>>>>
>>> _______________________________________________
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DM
Douglas Miller
Sat, Aug 6, 2022 11:49 AM
That is correct: the shape Brian wants is not an ellipse, but (probably)
a hyperellipse.
The standard equation of an ellipse is (x/a)^2 + (y/b)^2 = 1 where a, b > 0
By contrast, the standard equation of a hyperellipse is |x/a|^n +
|y/b|^n = 1 where a, b > 0 and n > 2
This can be generated by this function
function f(i,a,b,n) =
let(theta=i*360)
[a * cos(theta), sign(180-theta) * b *
pow((1-pow(abs(cos(theta)),n)),1/n) ];
For example:
points = [ for(i=[0:0.01:1]) f(i,12,7,2.5) ]; // semi-major axis = 12,
semi-minor axis = 7, n = 2.5
polygon(points);
I imagine that Brian will want to use a value of n that is only slightly
greater than 2. A good starting value for experimentation appears to be
n = 2.25.
On 8/2/2022 8:04 PM, Jordan Brown wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
[...]
You're taking a circle and stretching it; the result of that
stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can chime in -
but I think the shape you are asking for would not be an ellipse.
That is correct: the shape Brian wants is not an ellipse, but (probably)
a hyperellipse.
The standard equation of an ellipse is (x/a)^2 + (y/b)^2 = 1 where a, b > 0
By contrast, the standard equation of a hyperellipse is |x/a|^n +
|y/b|^n = 1 where a, b > 0 and n > 2
This can be generated by this function
function f(i,a,b,n) =
let(theta=i*360)
[a * cos(theta), sign(180-theta) * b *
pow((1-pow(abs(cos(theta)),n)),1/n) ];
For example:
points = [ for(i=[0:0.01:1]) f(i,12,7,2.5) ]; // semi-major axis = 12,
semi-minor axis = 7, n = 2.5
polygon(points);
I imagine that Brian will want to use a value of n that is only slightly
greater than 2. A good starting value for experimentation appears to be
n = 2.25.
On 8/2/2022 8:04 PM, Jordan Brown wrote:
> On 8/2/2022 4:46 PM, Brian Allen wrote:
>> I created an oval using:
>> inMm = 25.4;
>> fullW = 35.5 * inMm;
>> fullL = 54.5 * inMm;
>> resize([fullW, fullL, 1])
>> circle(d=fullW);
> [...]
> You're taking a circle and stretching it; the result of that
> stretching is an ellipse.
>
> I'm not 100% sure - maybe one of the real math wizards can chime in -
> but I think the shape you are asking for would not be an ellipse.
RP
Ronaldo Persiano
Sun, Aug 7, 2022 8:28 PM
Adrian, why to use a (either inscribed or circumscribed) polygon with equal
side as an approximation of an ellipsis? It doesn't seem to be a good
approximation of it. I get a better approximation by taking an angle
uniform sample of the curve in the parameter space.
[image: Uniformly_sampled_ellipsis.png]
Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano avm4@cornell.edu
escreveu:
Circumscribed means tangents to points on the ellipse. But there are no
arc lengths (equal or unequal) in this problem, nor equally spaced angles.
The angles of each segment will be different. But the lengths of the
segments are not the same as the arc lengths, which is why arc length does
not matter. Here's a solution to the N=6 polygon sort of inspired by the
idea Jordan suggested, where I adjust the first segment until the second
segment has the right length. The rest of the solution is then obtained
using symmetry, which means there is only a single unknown that can simply
be solved for. But this doesn't generalize to higher N.
[image: image.png]
include<BOSL2/std.scad>
// ellipse semiaxes
// If a is much smaller than b it will fail
a=5;
b=2;
// Find point on ellipse at specified x coordinate
function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
// Find tangent line to ellipse at point (u,w)
// and determine the x value where that line has
// given y value.
function findx(a,b,u,w,y) =
let(
A = -ub^2/w/a^2,
B = w+(ub)^2/w/a^2
)
(y-B)/A;
// First segment if tangent point is at x=u
function first_seg(a,b,u) =
let(
w = ellipse_pt(a,b,u),
x1 = findx(a,b,u,w,0),
x2 = findx(a,b,u,w,b)
)
[[x1,0],[x2,b]];
err = function (u) let(pt = first_seg(a,b,u))
2*pt[1].x - norm(pt[0]-pt[1]);
x = root_find(err,.2a,.9a);
first = first_seg(a,b,x);
top = concat(first,reverse(xflip(first)));
poly = concat(top, reverse(yflip(list_tail(top))));
stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
color("red")stroke(poly,width=.05,closed=true);
// Verify equality of segment lengths
echo(path_segment_lengths(poly,closed=true));
On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar sprabhakar2006@gmail.com
wrote:
Circumscribed means tangents from points on ellipse
How can the tangents from equally spaced arc lengths or equally spaced
angles be equal for an ellipse
You need to change the arc lengths dynamically and also accurately
estimate the segment length which fits the ellipse .
On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, avm4@cornell.edu wrote:
Your intuition is incorrect: a circle scaled in one direction IS an
ellipse. The motivation for being interested in circumscribing is to make
holes, though the constraint of equal length sides is probably not what
you'd want for that purpose. You can make a nonuniform circumscribing
polygon for an ellipse by simply making a polygon that circumscribes a
circle and scaling it.
On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson bob@rjcarlson.com wrote:
Seems like I remember Nop Head talking about using circumscribed holes
for screws and even rotating the polygons to produce a more even hole.
If an ellipse is the nominal shape, the using a circumscribed polygon
for a hole and an inscribed polygon for a solid might make sense.
Btw, can someone confirm my intuition that a circle scaled in one
dimension is not an ellipse?
-Bob
Tucson AZ
On Aug 5, 2022, at 15:38, Adrian Mariano avm4@cornell.edu wrote:
I don't have a specific need for a circumscribing elliptical polygon
with equal sides. I was writing code to make ellipses and it seemed like
equal side approximations would be nice, so I solved the case for
inscribed, but couldn't solve the circumscribed case. So it's an unsolved
puzzle I've got that this discussion reminded me of.
On Fri, Aug 5, 2022 at 6:33 PM nop head nop.head@gmail.com wrote:
A long discussion but I have to ask why do you need an elliptical
polygon with equal tangential sides?
On Fri, 5 Aug 2022, 21:58 Adrian Mariano, avm4@cornell.edu wrote:
You think that a circumscribing polygon with equal length segments
does not exist?? I'm skeptical of this claim.
For circumscribed, it's not so hard. You start with a
uniform-in-angle point distribution, compute the segment lengths, compare
to the mean segment length to get an error, and then compute an adjustment
based on the error. It converges quickly.
So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but surely good
enough.
<image.png>
The code for this is in BOSL2 in shapes2d.scad. A similar algorithm
should exist for the circumscribed case, but as I said, my efforts to find
it have failed.
On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
I think circumscribed polygon segments cannot be equal lengths.
inscribed ones may be approximately equal, but would be a very tough
problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano avm4@cornell.edu
wrote:
I think you need to be a little careful about defining what you
want to do. If you want to approximate an ellipse with points that are
distributed uniformly over the ellipse, that could be done by approximating
the arc length of the ellipse to find the necessary points---e.g. advance
distance d along the ellipse at each step. But if you want to approximate
an ellipse with a polygon with equal length sides, that approach does not
do it, because the length of the segments will not be constant as the span
sections of the ellipse with differing curvature. And since the perimeter
of the desired polygon is unknown, you can't just step it out analogously.
The circle doesn't present this challenge due to having constant
curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
fatherhorton@gmail.com> wrote:
I think our emails crossed. I was addressing how to approximate an
ellipse with constant-length segments in general, not how to solve your
problem. When I had to do something similar, I used polar coordinates and
stepped up the value of theta until the distance from the previous point
was correct. (In the constant-length version, I'd keep a running total of
the maximum desired length so that round-off errors don't accumulate.)
Figuring out the circumscribed polygon is, as you note, a greater
challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano avm4@cornell.edu
wrote:
Eh, there's no such thing as "brute force" in this situation.
Step through what with sufficiently small steps? Of course an iterative
algorithm is necessary. This is true for most interesting problems. No
big deal. Show me an iterative algorithm that converges to the
circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
fatherhorton@gmail.com> wrote:
Except by brute force (step through it with sufficiently small
steps), it’s not possible because there’s no closed form expression for the
distance between two points on a general ellipse (as opposed to special
cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano avm4@cornell.edu
wrote:
Here's a challenge for you: draw an "ellipse" polygon with N
segments that are all the same length that circumscribes a perfect ellipse
with given semiaxes a and b. I tried to solve this problem and could not
find a solution. I can draw an inscribed N segment "ellipse" polygon
with all segments the same length---though it's not easy---but the
circumscribed case eludes me. I think of the inscribed case as, every
vertex is located on the ellipse and the circumscribed case, every segment
is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor
axis "r1" and "r2" respectively and with center "cp" and number of segment
"s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster in
complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that
stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can
chime in - but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major
(long) and minor (short) axis dimensions; the other parts of the curve
mathematically flow from those two measurements.
You can make something else, but you can't make it using this
technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls
together.
Note that the four coordinates correspond to the coordinates
of the control points in a GUI drawing program: the start point, control
point associated with the start, the control point associated with the end,
and the end point. I have sometimes created OpenSCAD objects by tracing an
object in a drawing program and then reading off the coordinates of the
various control points.
Note that Bézier functions are not the only mathematical
curves. (For instance, they cannot exactly match an arc of a circle.) I'm
sure there are numerous other functions that also produce curves.
OpenSCAD mailing list
To unsubscribe send an email to
discuss-leave@lists.openscad.org
Adrian, why to use a (either inscribed or circumscribed) polygon with equal
side as an approximation of an ellipsis? It doesn't seem to be a good
approximation of it. I get a better approximation by taking an angle
uniform sample of the curve in the parameter space.
[image: Uniformly_sampled_ellipsis.png]
Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano <avm4@cornell.edu>
escreveu:
> Circumscribed means tangents to points on the ellipse. But there are no
> arc lengths (equal or unequal) in this problem, nor equally spaced angles.
> The angles of each segment will be different. But the lengths of the
> segments are not the same as the arc lengths, which is why arc length does
> not matter. Here's a solution to the N=6 polygon sort of inspired by the
> idea Jordan suggested, where I adjust the first segment until the second
> segment has the right length. The rest of the solution is then obtained
> using symmetry, which means there is only a single unknown that can simply
> be solved for. But this doesn't generalize to higher N.
>
> [image: image.png]
> include<BOSL2/std.scad>
>
> // ellipse semiaxes
> // If a is much smaller than b it will fail
> a=5;
> b=2;
>
> // Find point on ellipse at specified x coordinate
> function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
>
>
> // Find tangent line to ellipse at point (u,w)
> // and determine the x value where that line has
> // given y value.
> function findx(a,b,u,w,y) =
> let(
> A = -u*b^2/w/a^2,
> B = w+(u*b)^2/w/a^2
> )
> (y-B)/A;
>
>
> // First segment if tangent point is at x=u
> function first_seg(a,b,u) =
> let(
> w = ellipse_pt(a,b,u),
> x1 = findx(a,b,u,w,0),
> x2 = findx(a,b,u,w,b)
> )
> [[x1,0],[x2,b]];
>
>
> err = function (u) let(pt = first_seg(a,b,u))
> 2*pt[1].x - norm(pt[0]-pt[1]);
>
> x = root_find(err,.2*a,.9*a);
> first = first_seg(a,b,x);
> top = concat(first,reverse(xflip(first)));
> poly = concat(top, reverse(yflip(list_tail(top))));
>
> stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
> color("red")stroke(poly,width=.05,closed=true);
>
> // Verify equality of segment lengths
> echo(path_segment_lengths(poly,closed=true));
>
>
>
>
>
> On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <sprabhakar2006@gmail.com>
> wrote:
>
>> Circumscribed means tangents from points on ellipse
>> How can the tangents from equally spaced arc lengths or equally spaced
>> angles be equal for an ellipse
>> You need to change the arc lengths dynamically and also accurately
>> estimate the segment length which fits the ellipse .
>>
>>
>>
>> On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, <avm4@cornell.edu> wrote:
>>
>>> Your intuition is incorrect: a circle scaled in one direction *IS* an
>>> ellipse. The motivation for being interested in circumscribing is to make
>>> holes, though the constraint of equal length sides is probably not what
>>> you'd want for that purpose. You can make a nonuniform circumscribing
>>> polygon for an ellipse by simply making a polygon that circumscribes a
>>> circle and scaling it.
>>>
>>> On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson <bob@rjcarlson.com> wrote:
>>>
>>>> Seems like I remember Nop Head talking about using circumscribed holes
>>>> for screws and even rotating the polygons to produce a more even hole.
>>>>
>>>> If an ellipse is the nominal shape, the using a circumscribed polygon
>>>> for a hole and an inscribed polygon for a solid might make sense.
>>>>
>>>> Btw, can someone confirm my intuition that a circle scaled in one
>>>> dimension is not an ellipse?
>>>>
>>>> -Bob
>>>> Tucson AZ
>>>>
>>>>
>>>>
>>>> On Aug 5, 2022, at 15:38, Adrian Mariano <avm4@cornell.edu> wrote:
>>>>
>>>> I don't have a specific need for a circumscribing elliptical polygon
>>>> with equal sides. I was writing code to make ellipses and it seemed like
>>>> equal side approximations would be nice, so I solved the case for
>>>> inscribed, but couldn't solve the circumscribed case. So it's an unsolved
>>>> puzzle I've got that this discussion reminded me of.
>>>>
>>>> On Fri, Aug 5, 2022 at 6:33 PM nop head <nop.head@gmail.com> wrote:
>>>>
>>>>> A long discussion but I have to ask why do you need an elliptical
>>>>> polygon with equal tangential sides?
>>>>>
>>>>> On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu> wrote:
>>>>>
>>>>>> You think that a circumscribing polygon with equal length segments
>>>>>> does not exist?? I'm skeptical of this claim.
>>>>>>
>>>>>> For circumscribed, it's not so hard. You start with a
>>>>>> uniform-in-angle point distribution, compute the segment lengths, compare
>>>>>> to the mean segment length to get an error, and then compute an adjustment
>>>>>> based on the error. It converges quickly.
>>>>>>
>>>>>> So for example, I have segments lengths of:
>>>>>> ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
>>>>>> 5.57297]
>>>>>>
>>>>>> Maybe not exactly equal, since tolerance is 1e-9, but surely good
>>>>>> enough.
>>>>>> <image.png>
>>>>>>
>>>>>> The code for this is in BOSL2 in shapes2d.scad. A similar algorithm
>>>>>> should exist for the circumscribed case, but as I said, my efforts to find
>>>>>> it have failed.
>>>>>>
>>>>>> On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>
>>>>>>> I think circumscribed polygon segments cannot be equal lengths.
>>>>>>> inscribed ones may be approximately equal, but would be a very tough
>>>>>>> problem to solve.
>>>>>>> I tried something.
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <avm4@cornell.edu>
>>>>>>> wrote:
>>>>>>>
>>>>>>>> I think you need to be a little careful about defining what you
>>>>>>>> want to do. If you want to approximate an ellipse with points that are
>>>>>>>> distributed uniformly over the ellipse, that could be done by approximating
>>>>>>>> the arc length of the ellipse to find the necessary points---e.g. advance
>>>>>>>> distance d along the ellipse at each step. But if you want to approximate
>>>>>>>> an ellipse with a polygon with equal length sides, that approach does not
>>>>>>>> do it, because the length of the segments will not be constant as the span
>>>>>>>> sections of the ellipse with differing curvature. And since the perimeter
>>>>>>>> of the desired polygon is unknown, you can't just step it out analogously.
>>>>>>>> The circle doesn't present this challenge due to having constant
>>>>>>>> curvature.
>>>>>>>>
>>>>>>>> On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>
>>>>>>>>> I think our emails crossed. I was addressing how to approximate an
>>>>>>>>> ellipse with constant-length segments in general, not how to solve your
>>>>>>>>> problem. When I had to do something similar, I used polar coordinates and
>>>>>>>>> stepped up the value of theta until the distance from the previous point
>>>>>>>>> was correct. (In the constant-length version, I'd keep a running total of
>>>>>>>>> the maximum desired length so that round-off errors don't accumulate.)
>>>>>>>>>
>>>>>>>>> Figuring out the circumscribed polygon is, as you note, a greater
>>>>>>>>> challenge. I can visualize how to do it for n = 4, but it doesn't
>>>>>>>>> extrapolate at all.
>>>>>>>>>
>>>>>>>>> On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <avm4@cornell.edu>
>>>>>>>>> wrote:
>>>>>>>>>
>>>>>>>>>> Eh, there's no such thing as "brute force" in this situation.
>>>>>>>>>> Step through *what* with sufficiently small steps? Of course an iterative
>>>>>>>>>> algorithm is necessary. This is true for most interesting problems. No
>>>>>>>>>> big deal. Show me an iterative algorithm that converges to the
>>>>>>>>>> circumscribing polygon. When I tried to devise such an algorithm I
>>>>>>>>>> couldn't find one that would converge. Writing an algorithm that would
>>>>>>>>>> converge to the *inscribed* polygon seemed to be straight forward. Note
>>>>>>>>>> also that computing arc length of the ellipse doesn't really help here.
>>>>>>>>>> You can do that by solving an elliptical integral, or summing up small
>>>>>>>>>> segments. But what's the connection between arc length of the ellipse and
>>>>>>>>>> a circumscribing polygon with equal segment lengths? If there's a
>>>>>>>>>> connection, I don't see it.
>>>>>>>>>>
>>>>>>>>>> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>
>>>>>>>>>>> Except by brute force (step through it with sufficiently small
>>>>>>>>>>> steps), it’s not possible because there’s no closed form expression for the
>>>>>>>>>>> distance between two points on a general ellipse (as opposed to special
>>>>>>>>>>> cases like a circle).
>>>>>>>>>>>
>>>>>>>>>>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <avm4@cornell.edu>
>>>>>>>>>>> wrote:
>>>>>>>>>>>
>>>>>>>>>>>> Here's a challenge for you: draw an "ellipse" polygon with N
>>>>>>>>>>>> segments that are all the same length that circumscribes a perfect ellipse
>>>>>>>>>>>> with given semiaxes a and b. I tried to solve this problem and could not
>>>>>>>>>>>> find a solution. I can draw an *inscribed* N segment "ellipse" polygon
>>>>>>>>>>>> with all segments the same length---though it's not easy---but the
>>>>>>>>>>>> circumscribed case eludes me. I think of the inscribed case as, every
>>>>>>>>>>>> vertex is located on the ellipse and the circumscribed case, every segment
>>>>>>>>>>>> is tangent to the ellipse
>>>>>>>>>>>>
>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>>>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>>>
>>>>>>>>>>>>> In openscad drawing a ellipse is not too difficult
>>>>>>>>>>>>> Refer following code
>>>>>>>>>>>>>
>>>>>>>>>>>>> function to draw an ellipse with semi-major and semi-minor
>>>>>>>>>>>>> axis "r1" and "r2" respectively and with center "cp" and number of segment
>>>>>>>>>>>>> "s"
>>>>>>>>>>>>> /
>>>>>>>>>>>>> // example:
>>>>>>>>>>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>>>>>>>>>>>
>>>>>>>>>>>>> function ellipse(r1,r2,cp,s=30)=
>>>>>>>>>>>>> let(
>>>>>>>>>>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>>>>>>>>>>>> )sec;
>>>>>>>>>>>>>
>>>>>>>>>>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>>>>>>>>>>>> openscad@jordan.maileater.net> wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> I created an oval using:
>>>>>>>>>>>>>> inMm = 25.4;
>>>>>>>>>>>>>> fullW = 35.5 * inMm;
>>>>>>>>>>>>>> fullL = 54.5 * inMm;
>>>>>>>>>>>>>> resize([fullW, fullL, 1])
>>>>>>>>>>>>>> circle(d=fullW);
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Tip: use scale() rather than resize(); it previews faster in
>>>>>>>>>>>>>> complicated cases.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> You're taking a circle and stretching it; the result of that
>>>>>>>>>>>>>> stretching is an ellipse.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> I'm not 100% sure - maybe one of the real math wizards can
>>>>>>>>>>>>>> chime in - but I think the shape you are asking for would not be an ellipse.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> I believe that an ellipse is fully defined by its major
>>>>>>>>>>>>>> (long) and minor (short) axis dimensions; the other parts of the curve
>>>>>>>>>>>>>> mathematically flow from those two measurements.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> You can make something else, but you can't make it using this
>>>>>>>>>>>>>> technique.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> You can use a Bézier function.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Using BOSL2:
>>>>>>>>>>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> DIY, here's a function that I cobbled together some years ago:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>>>>>>>>>>>> // but that yielded either single points or a raft of triangles;
>>>>>>>>>>>>>> // this yields a vector of points that you can then concatenate
>>>>>>>>>>>>>> // with other pieces to form a single polygon.
>>>>>>>>>>>>>> // If we were really clever, I think it would be possible to
>>>>>>>>>>>>>> // automatically space the output points based on how linear
>>>>>>>>>>>>>> // the curve is at that point. But right now I'm not that clever.
>>>>>>>>>>>>>> function BEZ03(u) = pow((1-u), 3);
>>>>>>>>>>>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>>>>>>>>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>>>>>>>>>>>> function BEZ33(u) = pow(u,3);
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>>>>>>>>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>>>>>>>>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> // p0 - start point
>>>>>>>>>>>>>> // p1 - control point 1, line departs p0 headed this way
>>>>>>>>>>>>>> // p2 - control point 2, line arrives at p3 from this way
>>>>>>>>>>>>>> // p3 - end point
>>>>>>>>>>>>>> // segs - number of segments
>>>>>>>>>>>>>> function bez(p0, p1, p2, p3, segs) = [
>>>>>>>>>>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>>>>>>>>>>>> ];
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> It is that final function bez() that I actually call, e.g.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> often concatenating the results of several bez() calls
>>>>>>>>>>>>>> together.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Note that the four coordinates correspond to the coordinates
>>>>>>>>>>>>>> of the control points in a GUI drawing program: the start point, control
>>>>>>>>>>>>>> point associated with the start, the control point associated with the end,
>>>>>>>>>>>>>> and the end point. I have sometimes created OpenSCAD objects by tracing an
>>>>>>>>>>>>>> object in a drawing program and then reading off the coordinates of the
>>>>>>>>>>>>>> various control points.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Note that Bézier functions are not the only mathematical
>>>>>>>>>>>>>> curves. (For instance, they cannot exactly match an arc of a circle.) I'm
>>>>>>>>>>>>>> sure there are numerous other functions that also produce curves.
>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>
>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>> OpenSCAD mailing list
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>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>
>>>>>>>>>>>> _______________________________________________
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>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>
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>>>>>>>>>>>
>>>>>>>>>> _______________________________________________
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>>>>>>>>>>
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>>>>>>>>>
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>
FH
Father Horton
Sun, Aug 7, 2022 9:32 PM
How about this to construct N equal length segments around an ellipse with
axes a and b:
- Select an initial segment length estimate (e.g., Ramanujan's formula /
N) = L
- Construct a half-segment from (0, b) to (0, L/2).
- Find a line tangent to the ellipse and passing through (0, L/2).
- Find the point on the tangent line that is L away from (0, L/2).
- Repeat (3) substituting the new point for (0, L/2). until you have N
segments.
If everything worked perfectly, the last point found would be (0, -L/2). If
there's a gap, increase L and try again. If there's overlap, decrease L and
try again.
On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano rcmpersiano@gmail.com
wrote:
Adrian, why to use a (either inscribed or circumscribed) polygon with
equal side as an approximation of an ellipsis? It doesn't seem to be a good
approximation of it. I get a better approximation by taking an angle
uniform sample of the curve in the parameter space.
[image: Uniformly_sampled_ellipsis.png]
Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano avm4@cornell.edu
escreveu:
Circumscribed means tangents to points on the ellipse. But there are no
arc lengths (equal or unequal) in this problem, nor equally spaced angles.
The angles of each segment will be different. But the lengths of the
segments are not the same as the arc lengths, which is why arc length does
not matter. Here's a solution to the N=6 polygon sort of inspired by the
idea Jordan suggested, where I adjust the first segment until the second
segment has the right length. The rest of the solution is then obtained
using symmetry, which means there is only a single unknown that can simply
be solved for. But this doesn't generalize to higher N.
[image: image.png]
include<BOSL2/std.scad>
// ellipse semiaxes
// If a is much smaller than b it will fail
a=5;
b=2;
// Find point on ellipse at specified x coordinate
function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
// Find tangent line to ellipse at point (u,w)
// and determine the x value where that line has
// given y value.
function findx(a,b,u,w,y) =
let(
A = -ub^2/w/a^2,
B = w+(ub)^2/w/a^2
)
(y-B)/A;
// First segment if tangent point is at x=u
function first_seg(a,b,u) =
let(
w = ellipse_pt(a,b,u),
x1 = findx(a,b,u,w,0),
x2 = findx(a,b,u,w,b)
)
[[x1,0],[x2,b]];
err = function (u) let(pt = first_seg(a,b,u))
2*pt[1].x - norm(pt[0]-pt[1]);
x = root_find(err,.2a,.9a);
first = first_seg(a,b,x);
top = concat(first,reverse(xflip(first)));
poly = concat(top, reverse(yflip(list_tail(top))));
stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
color("red")stroke(poly,width=.05,closed=true);
// Verify equality of segment lengths
echo(path_segment_lengths(poly,closed=true));
On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
Circumscribed means tangents from points on ellipse
How can the tangents from equally spaced arc lengths or equally spaced
angles be equal for an ellipse
You need to change the arc lengths dynamically and also accurately
estimate the segment length which fits the ellipse .
On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, avm4@cornell.edu wrote:
Your intuition is incorrect: a circle scaled in one direction IS an
ellipse. The motivation for being interested in circumscribing is to make
holes, though the constraint of equal length sides is probably not what
you'd want for that purpose. You can make a nonuniform circumscribing
polygon for an ellipse by simply making a polygon that circumscribes a
circle and scaling it.
On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson bob@rjcarlson.com wrote:
Seems like I remember Nop Head talking about using circumscribed holes
for screws and even rotating the polygons to produce a more even hole.
If an ellipse is the nominal shape, the using a circumscribed polygon
for a hole and an inscribed polygon for a solid might make sense.
Btw, can someone confirm my intuition that a circle scaled in one
dimension is not an ellipse?
-Bob
Tucson AZ
On Aug 5, 2022, at 15:38, Adrian Mariano avm4@cornell.edu wrote:
I don't have a specific need for a circumscribing elliptical polygon
with equal sides. I was writing code to make ellipses and it seemed like
equal side approximations would be nice, so I solved the case for
inscribed, but couldn't solve the circumscribed case. So it's an unsolved
puzzle I've got that this discussion reminded me of.
On Fri, Aug 5, 2022 at 6:33 PM nop head nop.head@gmail.com wrote:
A long discussion but I have to ask why do you need an elliptical
polygon with equal tangential sides?
On Fri, 5 Aug 2022, 21:58 Adrian Mariano, avm4@cornell.edu wrote:
You think that a circumscribing polygon with equal length segments
does not exist?? I'm skeptical of this claim.
For circumscribed, it's not so hard. You start with a
uniform-in-angle point distribution, compute the segment lengths, compare
to the mean segment length to get an error, and then compute an adjustment
based on the error. It converges quickly.
So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
5.57297, 5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but surely good
enough.
<image.png>
The code for this is in BOSL2 in shapes2d.scad. A similar algorithm
should exist for the circumscribed case, but as I said, my efforts to find
it have failed.
On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
I think circumscribed polygon segments cannot be equal lengths.
inscribed ones may be approximately equal, but would be a very
tough problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano avm4@cornell.edu
wrote:
I think you need to be a little careful about defining what you
want to do. If you want to approximate an ellipse with points that are
distributed uniformly over the ellipse, that could be done by approximating
the arc length of the ellipse to find the necessary points---e.g. advance
distance d along the ellipse at each step. But if you want to approximate
an ellipse with a polygon with equal length sides, that approach does not
do it, because the length of the segments will not be constant as the span
sections of the ellipse with differing curvature. And since the perimeter
of the desired polygon is unknown, you can't just step it out analogously.
The circle doesn't present this challenge due to having constant
curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
fatherhorton@gmail.com> wrote:
I think our emails crossed. I was addressing how to approximate
an ellipse with constant-length segments in general, not how to solve your
problem. When I had to do something similar, I used polar coordinates and
stepped up the value of theta until the distance from the previous point
was correct. (In the constant-length version, I'd keep a running total of
the maximum desired length so that round-off errors don't accumulate.)
Figuring out the circumscribed polygon is, as you note, a greater
challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano avm4@cornell.edu
wrote:
Eh, there's no such thing as "brute force" in this situation.
Step through what with sufficiently small steps? Of course an iterative
algorithm is necessary. This is true for most interesting problems. No
big deal. Show me an iterative algorithm that converges to the
circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
fatherhorton@gmail.com> wrote:
Except by brute force (step through it with sufficiently small
steps), it’s not possible because there’s no closed form expression for the
distance between two points on a general ellipse (as opposed to special
cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano avm4@cornell.edu
wrote:
Here's a challenge for you: draw an "ellipse" polygon with N
segments that are all the same length that circumscribes a perfect ellipse
with given semiaxes a and b. I tried to solve this problem and could not
find a solution. I can draw an inscribed N segment "ellipse" polygon
with all segments the same length---though it's not easy---but the
circumscribed case eludes me. I think of the inscribed case as, every
vertex is located on the ellipse and the circumscribed case, every segment
is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor
axis "r1" and "r2" respectively and with center "cp" and number of segment
"s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster
in complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that
stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can
chime in - but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major
(long) and minor (short) axis dimensions; the other parts of the curve
mathematically flow from those two measurements.
You can make something else, but you can't make it using
this technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years
ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls
together.
Note that the four coordinates correspond to the coordinates
of the control points in a GUI drawing program: the start point, control
point associated with the start, the control point associated with the end,
and the end point. I have sometimes created OpenSCAD objects by tracing an
object in a drawing program and then reading off the coordinates of the
various control points.
Note that Bézier functions are not the only mathematical
curves. (For instance, they cannot exactly match an arc of a circle.) I'm
sure there are numerous other functions that also produce curves.
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discuss-leave@lists.openscad.org
How about this to construct N equal length segments around an ellipse with
axes a and b:
1) Select an initial segment length estimate (e.g., Ramanujan's formula /
N) = L
2) Construct a half-segment from (0, b) to (0, L/2).
3) Find a line tangent to the ellipse and passing through (0, L/2).
3) Find the point on the tangent line that is L away from (0, L/2).
4) Repeat (3) substituting the new point for (0, L/2). until you have N
segments.
If everything worked perfectly, the last point found would be (0, -L/2). If
there's a gap, increase L and try again. If there's overlap, decrease L and
try again.
On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano <rcmpersiano@gmail.com>
wrote:
> Adrian, why to use a (either inscribed or circumscribed) polygon with
> equal side as an approximation of an ellipsis? It doesn't seem to be a good
> approximation of it. I get a better approximation by taking an angle
> uniform sample of the curve in the parameter space.
>
> [image: Uniformly_sampled_ellipsis.png]
>
>
> Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano <avm4@cornell.edu>
> escreveu:
>
>> Circumscribed means tangents to points on the ellipse. But there are no
>> arc lengths (equal or unequal) in this problem, nor equally spaced angles.
>> The angles of each segment will be different. But the lengths of the
>> segments are not the same as the arc lengths, which is why arc length does
>> not matter. Here's a solution to the N=6 polygon sort of inspired by the
>> idea Jordan suggested, where I adjust the first segment until the second
>> segment has the right length. The rest of the solution is then obtained
>> using symmetry, which means there is only a single unknown that can simply
>> be solved for. But this doesn't generalize to higher N.
>>
>> [image: image.png]
>> include<BOSL2/std.scad>
>>
>> // ellipse semiaxes
>> // If a is much smaller than b it will fail
>> a=5;
>> b=2;
>>
>> // Find point on ellipse at specified x coordinate
>> function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
>>
>>
>> // Find tangent line to ellipse at point (u,w)
>> // and determine the x value where that line has
>> // given y value.
>> function findx(a,b,u,w,y) =
>> let(
>> A = -u*b^2/w/a^2,
>> B = w+(u*b)^2/w/a^2
>> )
>> (y-B)/A;
>>
>>
>> // First segment if tangent point is at x=u
>> function first_seg(a,b,u) =
>> let(
>> w = ellipse_pt(a,b,u),
>> x1 = findx(a,b,u,w,0),
>> x2 = findx(a,b,u,w,b)
>> )
>> [[x1,0],[x2,b]];
>>
>>
>> err = function (u) let(pt = first_seg(a,b,u))
>> 2*pt[1].x - norm(pt[0]-pt[1]);
>>
>> x = root_find(err,.2*a,.9*a);
>> first = first_seg(a,b,x);
>> top = concat(first,reverse(xflip(first)));
>> poly = concat(top, reverse(yflip(list_tail(top))));
>>
>> stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
>> color("red")stroke(poly,width=.05,closed=true);
>>
>> // Verify equality of segment lengths
>> echo(path_segment_lengths(poly,closed=true));
>>
>>
>>
>>
>>
>> On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
>> sprabhakar2006@gmail.com> wrote:
>>
>>> Circumscribed means tangents from points on ellipse
>>> How can the tangents from equally spaced arc lengths or equally spaced
>>> angles be equal for an ellipse
>>> You need to change the arc lengths dynamically and also accurately
>>> estimate the segment length which fits the ellipse .
>>>
>>>
>>>
>>> On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, <avm4@cornell.edu> wrote:
>>>
>>>> Your intuition is incorrect: a circle scaled in one direction *IS* an
>>>> ellipse. The motivation for being interested in circumscribing is to make
>>>> holes, though the constraint of equal length sides is probably not what
>>>> you'd want for that purpose. You can make a nonuniform circumscribing
>>>> polygon for an ellipse by simply making a polygon that circumscribes a
>>>> circle and scaling it.
>>>>
>>>> On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson <bob@rjcarlson.com> wrote:
>>>>
>>>>> Seems like I remember Nop Head talking about using circumscribed holes
>>>>> for screws and even rotating the polygons to produce a more even hole.
>>>>>
>>>>> If an ellipse is the nominal shape, the using a circumscribed polygon
>>>>> for a hole and an inscribed polygon for a solid might make sense.
>>>>>
>>>>> Btw, can someone confirm my intuition that a circle scaled in one
>>>>> dimension is not an ellipse?
>>>>>
>>>>> -Bob
>>>>> Tucson AZ
>>>>>
>>>>>
>>>>>
>>>>> On Aug 5, 2022, at 15:38, Adrian Mariano <avm4@cornell.edu> wrote:
>>>>>
>>>>> I don't have a specific need for a circumscribing elliptical polygon
>>>>> with equal sides. I was writing code to make ellipses and it seemed like
>>>>> equal side approximations would be nice, so I solved the case for
>>>>> inscribed, but couldn't solve the circumscribed case. So it's an unsolved
>>>>> puzzle I've got that this discussion reminded me of.
>>>>>
>>>>> On Fri, Aug 5, 2022 at 6:33 PM nop head <nop.head@gmail.com> wrote:
>>>>>
>>>>>> A long discussion but I have to ask why do you need an elliptical
>>>>>> polygon with equal tangential sides?
>>>>>>
>>>>>> On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu> wrote:
>>>>>>
>>>>>>> You think that a circumscribing polygon with equal length segments
>>>>>>> does not exist?? I'm skeptical of this claim.
>>>>>>>
>>>>>>> For circumscribed, it's not so hard. You start with a
>>>>>>> uniform-in-angle point distribution, compute the segment lengths, compare
>>>>>>> to the mean segment length to get an error, and then compute an adjustment
>>>>>>> based on the error. It converges quickly.
>>>>>>>
>>>>>>> So for example, I have segments lengths of:
>>>>>>> ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
>>>>>>> 5.57297, 5.57297]
>>>>>>>
>>>>>>> Maybe not exactly equal, since tolerance is 1e-9, but surely good
>>>>>>> enough.
>>>>>>> <image.png>
>>>>>>>
>>>>>>> The code for this is in BOSL2 in shapes2d.scad. A similar algorithm
>>>>>>> should exist for the circumscribed case, but as I said, my efforts to find
>>>>>>> it have failed.
>>>>>>>
>>>>>>> On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>
>>>>>>>> I think circumscribed polygon segments cannot be equal lengths.
>>>>>>>> inscribed ones may be approximately equal, but would be a very
>>>>>>>> tough problem to solve.
>>>>>>>> I tried something.
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>> On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <avm4@cornell.edu>
>>>>>>>> wrote:
>>>>>>>>
>>>>>>>>> I think you need to be a little careful about defining what you
>>>>>>>>> want to do. If you want to approximate an ellipse with points that are
>>>>>>>>> distributed uniformly over the ellipse, that could be done by approximating
>>>>>>>>> the arc length of the ellipse to find the necessary points---e.g. advance
>>>>>>>>> distance d along the ellipse at each step. But if you want to approximate
>>>>>>>>> an ellipse with a polygon with equal length sides, that approach does not
>>>>>>>>> do it, because the length of the segments will not be constant as the span
>>>>>>>>> sections of the ellipse with differing curvature. And since the perimeter
>>>>>>>>> of the desired polygon is unknown, you can't just step it out analogously.
>>>>>>>>> The circle doesn't present this challenge due to having constant
>>>>>>>>> curvature.
>>>>>>>>>
>>>>>>>>> On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>
>>>>>>>>>> I think our emails crossed. I was addressing how to approximate
>>>>>>>>>> an ellipse with constant-length segments in general, not how to solve your
>>>>>>>>>> problem. When I had to do something similar, I used polar coordinates and
>>>>>>>>>> stepped up the value of theta until the distance from the previous point
>>>>>>>>>> was correct. (In the constant-length version, I'd keep a running total of
>>>>>>>>>> the maximum desired length so that round-off errors don't accumulate.)
>>>>>>>>>>
>>>>>>>>>> Figuring out the circumscribed polygon is, as you note, a greater
>>>>>>>>>> challenge. I can visualize how to do it for n = 4, but it doesn't
>>>>>>>>>> extrapolate at all.
>>>>>>>>>>
>>>>>>>>>> On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <avm4@cornell.edu>
>>>>>>>>>> wrote:
>>>>>>>>>>
>>>>>>>>>>> Eh, there's no such thing as "brute force" in this situation.
>>>>>>>>>>> Step through *what* with sufficiently small steps? Of course an iterative
>>>>>>>>>>> algorithm is necessary. This is true for most interesting problems. No
>>>>>>>>>>> big deal. Show me an iterative algorithm that converges to the
>>>>>>>>>>> circumscribing polygon. When I tried to devise such an algorithm I
>>>>>>>>>>> couldn't find one that would converge. Writing an algorithm that would
>>>>>>>>>>> converge to the *inscribed* polygon seemed to be straight forward. Note
>>>>>>>>>>> also that computing arc length of the ellipse doesn't really help here.
>>>>>>>>>>> You can do that by solving an elliptical integral, or summing up small
>>>>>>>>>>> segments. But what's the connection between arc length of the ellipse and
>>>>>>>>>>> a circumscribing polygon with equal segment lengths? If there's a
>>>>>>>>>>> connection, I don't see it.
>>>>>>>>>>>
>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
>>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>>
>>>>>>>>>>>> Except by brute force (step through it with sufficiently small
>>>>>>>>>>>> steps), it’s not possible because there’s no closed form expression for the
>>>>>>>>>>>> distance between two points on a general ellipse (as opposed to special
>>>>>>>>>>>> cases like a circle).
>>>>>>>>>>>>
>>>>>>>>>>>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <avm4@cornell.edu>
>>>>>>>>>>>> wrote:
>>>>>>>>>>>>
>>>>>>>>>>>>> Here's a challenge for you: draw an "ellipse" polygon with N
>>>>>>>>>>>>> segments that are all the same length that circumscribes a perfect ellipse
>>>>>>>>>>>>> with given semiaxes a and b. I tried to solve this problem and could not
>>>>>>>>>>>>> find a solution. I can draw an *inscribed* N segment "ellipse" polygon
>>>>>>>>>>>>> with all segments the same length---though it's not easy---but the
>>>>>>>>>>>>> circumscribed case eludes me. I think of the inscribed case as, every
>>>>>>>>>>>>> vertex is located on the ellipse and the circumscribed case, every segment
>>>>>>>>>>>>> is tangent to the ellipse
>>>>>>>>>>>>>
>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>>>>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> In openscad drawing a ellipse is not too difficult
>>>>>>>>>>>>>> Refer following code
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> function to draw an ellipse with semi-major and semi-minor
>>>>>>>>>>>>>> axis "r1" and "r2" respectively and with center "cp" and number of segment
>>>>>>>>>>>>>> "s"
>>>>>>>>>>>>>> /
>>>>>>>>>>>>>> // example:
>>>>>>>>>>>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> function ellipse(r1,r2,cp,s=30)=
>>>>>>>>>>>>>> let(
>>>>>>>>>>>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>>>>>>>>>>>>> )sec;
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>>>>>>>>>>>>> openscad@jordan.maileater.net> wrote:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> I created an oval using:
>>>>>>>>>>>>>>> inMm = 25.4;
>>>>>>>>>>>>>>> fullW = 35.5 * inMm;
>>>>>>>>>>>>>>> fullL = 54.5 * inMm;
>>>>>>>>>>>>>>> resize([fullW, fullL, 1])
>>>>>>>>>>>>>>> circle(d=fullW);
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Tip: use scale() rather than resize(); it previews faster
>>>>>>>>>>>>>>> in complicated cases.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> You're taking a circle and stretching it; the result of that
>>>>>>>>>>>>>>> stretching is an ellipse.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> I'm not 100% sure - maybe one of the real math wizards can
>>>>>>>>>>>>>>> chime in - but I think the shape you are asking for would not be an ellipse.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> I believe that an ellipse is fully defined by its major
>>>>>>>>>>>>>>> (long) and minor (short) axis dimensions; the other parts of the curve
>>>>>>>>>>>>>>> mathematically flow from those two measurements.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> You can make something else, but you can't make it using
>>>>>>>>>>>>>>> this technique.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> You can use a Bézier function.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Using BOSL2:
>>>>>>>>>>>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> DIY, here's a function that I cobbled together some years
>>>>>>>>>>>>>>> ago:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>>>>>>>>>>>>> // but that yielded either single points or a raft of triangles;
>>>>>>>>>>>>>>> // this yields a vector of points that you can then concatenate
>>>>>>>>>>>>>>> // with other pieces to form a single polygon.
>>>>>>>>>>>>>>> // If we were really clever, I think it would be possible to
>>>>>>>>>>>>>>> // automatically space the output points based on how linear
>>>>>>>>>>>>>>> // the curve is at that point. But right now I'm not that clever.
>>>>>>>>>>>>>>> function BEZ03(u) = pow((1-u), 3);
>>>>>>>>>>>>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>>>>>>>>>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>>>>>>>>>>>>> function BEZ33(u) = pow(u,3);
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>>>>>>>>>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>>>>>>>>>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> // p0 - start point
>>>>>>>>>>>>>>> // p1 - control point 1, line departs p0 headed this way
>>>>>>>>>>>>>>> // p2 - control point 2, line arrives at p3 from this way
>>>>>>>>>>>>>>> // p3 - end point
>>>>>>>>>>>>>>> // segs - number of segments
>>>>>>>>>>>>>>> function bez(p0, p1, p2, p3, segs) = [
>>>>>>>>>>>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>>>>>>>>>>>>> ];
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> It is that final function bez() that I actually call, e.g.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> often concatenating the results of several bez() calls
>>>>>>>>>>>>>>> together.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Note that the four coordinates correspond to the coordinates
>>>>>>>>>>>>>>> of the control points in a GUI drawing program: the start point, control
>>>>>>>>>>>>>>> point associated with the start, the control point associated with the end,
>>>>>>>>>>>>>>> and the end point. I have sometimes created OpenSCAD objects by tracing an
>>>>>>>>>>>>>>> object in a drawing program and then reading off the coordinates of the
>>>>>>>>>>>>>>> various control points.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Note that Bézier functions are not the only mathematical
>>>>>>>>>>>>>>> curves. (For instance, they cannot exactly match an arc of a circle.) I'm
>>>>>>>>>>>>>>> sure there are numerous other functions that also produce curves.
>>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>> OpenSCAD mailing list
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>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>
>>>>>>>>>>>>> _______________________________________________
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>>>>>>>>>>>>>
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AM
Adrian Mariano
Mon, Aug 8, 2022 1:14 AM
I think I thought it might be nice to be able to produce ellipse-like
polygons with equal side length. I experimented with a few things and
ended up concluding that just stretching out the points on a circle
produces a pretty good result if your goal is to approximate an ellipse
with a polygon that has unequal sides. I wrote code to sample an ellipse
uniformly in arc length and the result is a poor approximation of the
ellipse, with too many points at the flat side and not enough at the tips.
On Sun, Aug 7, 2022 at 4:30 PM Ronaldo Persiano rcmpersiano@gmail.com
wrote:
Adrian, why to use a (either inscribed or circumscribed) polygon with
equal side as an approximation of an ellipsis? It doesn't seem to be a good
approximation of it. I get a better approximation by taking an angle
uniform sample of the curve in the parameter space.
[image: Uniformly_sampled_ellipsis.png]
Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano avm4@cornell.edu
escreveu:
Circumscribed means tangents to points on the ellipse. But there are no
arc lengths (equal or unequal) in this problem, nor equally spaced angles.
The angles of each segment will be different. But the lengths of the
segments are not the same as the arc lengths, which is why arc length does
not matter. Here's a solution to the N=6 polygon sort of inspired by the
idea Jordan suggested, where I adjust the first segment until the second
segment has the right length. The rest of the solution is then obtained
using symmetry, which means there is only a single unknown that can simply
be solved for. But this doesn't generalize to higher N.
[image: image.png]
include<BOSL2/std.scad>
// ellipse semiaxes
// If a is much smaller than b it will fail
a=5;
b=2;
// Find point on ellipse at specified x coordinate
function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
// Find tangent line to ellipse at point (u,w)
// and determine the x value where that line has
// given y value.
function findx(a,b,u,w,y) =
let(
A = -ub^2/w/a^2,
B = w+(ub)^2/w/a^2
)
(y-B)/A;
// First segment if tangent point is at x=u
function first_seg(a,b,u) =
let(
w = ellipse_pt(a,b,u),
x1 = findx(a,b,u,w,0),
x2 = findx(a,b,u,w,b)
)
[[x1,0],[x2,b]];
err = function (u) let(pt = first_seg(a,b,u))
2*pt[1].x - norm(pt[0]-pt[1]);
x = root_find(err,.2a,.9a);
first = first_seg(a,b,x);
top = concat(first,reverse(xflip(first)));
poly = concat(top, reverse(yflip(list_tail(top))));
stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
color("red")stroke(poly,width=.05,closed=true);
// Verify equality of segment lengths
echo(path_segment_lengths(poly,closed=true));
On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
Circumscribed means tangents from points on ellipse
How can the tangents from equally spaced arc lengths or equally spaced
angles be equal for an ellipse
You need to change the arc lengths dynamically and also accurately
estimate the segment length which fits the ellipse .
On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, avm4@cornell.edu wrote:
Your intuition is incorrect: a circle scaled in one direction IS an
ellipse. The motivation for being interested in circumscribing is to make
holes, though the constraint of equal length sides is probably not what
you'd want for that purpose. You can make a nonuniform circumscribing
polygon for an ellipse by simply making a polygon that circumscribes a
circle and scaling it.
On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson bob@rjcarlson.com wrote:
Seems like I remember Nop Head talking about using circumscribed holes
for screws and even rotating the polygons to produce a more even hole.
If an ellipse is the nominal shape, the using a circumscribed polygon
for a hole and an inscribed polygon for a solid might make sense.
Btw, can someone confirm my intuition that a circle scaled in one
dimension is not an ellipse?
-Bob
Tucson AZ
On Aug 5, 2022, at 15:38, Adrian Mariano avm4@cornell.edu wrote:
I don't have a specific need for a circumscribing elliptical polygon
with equal sides. I was writing code to make ellipses and it seemed like
equal side approximations would be nice, so I solved the case for
inscribed, but couldn't solve the circumscribed case. So it's an unsolved
puzzle I've got that this discussion reminded me of.
On Fri, Aug 5, 2022 at 6:33 PM nop head nop.head@gmail.com wrote:
A long discussion but I have to ask why do you need an elliptical
polygon with equal tangential sides?
On Fri, 5 Aug 2022, 21:58 Adrian Mariano, avm4@cornell.edu wrote:
You think that a circumscribing polygon with equal length segments
does not exist?? I'm skeptical of this claim.
For circumscribed, it's not so hard. You start with a
uniform-in-angle point distribution, compute the segment lengths, compare
to the mean segment length to get an error, and then compute an adjustment
based on the error. It converges quickly.
So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
5.57297, 5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but surely good
enough.
<image.png>
The code for this is in BOSL2 in shapes2d.scad. A similar algorithm
should exist for the circumscribed case, but as I said, my efforts to find
it have failed.
On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
I think circumscribed polygon segments cannot be equal lengths.
inscribed ones may be approximately equal, but would be a very
tough problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano avm4@cornell.edu
wrote:
I think you need to be a little careful about defining what you
want to do. If you want to approximate an ellipse with points that are
distributed uniformly over the ellipse, that could be done by approximating
the arc length of the ellipse to find the necessary points---e.g. advance
distance d along the ellipse at each step. But if you want to approximate
an ellipse with a polygon with equal length sides, that approach does not
do it, because the length of the segments will not be constant as the span
sections of the ellipse with differing curvature. And since the perimeter
of the desired polygon is unknown, you can't just step it out analogously.
The circle doesn't present this challenge due to having constant
curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
fatherhorton@gmail.com> wrote:
I think our emails crossed. I was addressing how to approximate
an ellipse with constant-length segments in general, not how to solve your
problem. When I had to do something similar, I used polar coordinates and
stepped up the value of theta until the distance from the previous point
was correct. (In the constant-length version, I'd keep a running total of
the maximum desired length so that round-off errors don't accumulate.)
Figuring out the circumscribed polygon is, as you note, a greater
challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano avm4@cornell.edu
wrote:
Eh, there's no such thing as "brute force" in this situation.
Step through what with sufficiently small steps? Of course an iterative
algorithm is necessary. This is true for most interesting problems. No
big deal. Show me an iterative algorithm that converges to the
circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
fatherhorton@gmail.com> wrote:
Except by brute force (step through it with sufficiently small
steps), it’s not possible because there’s no closed form expression for the
distance between two points on a general ellipse (as opposed to special
cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano avm4@cornell.edu
wrote:
Here's a challenge for you: draw an "ellipse" polygon with N
segments that are all the same length that circumscribes a perfect ellipse
with given semiaxes a and b. I tried to solve this problem and could not
find a solution. I can draw an inscribed N segment "ellipse" polygon
with all segments the same length---though it's not easy---but the
circumscribed case eludes me. I think of the inscribed case as, every
vertex is located on the ellipse and the circumscribed case, every segment
is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor
axis "r1" and "r2" respectively and with center "cp" and number of segment
"s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster
in complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of that
stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can
chime in - but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major
(long) and minor (short) axis dimensions; the other parts of the curve
mathematically flow from those two measurements.
You can make something else, but you can't make it using
this technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years
ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls
together.
Note that the four coordinates correspond to the coordinates
of the control points in a GUI drawing program: the start point, control
point associated with the start, the control point associated with the end,
and the end point. I have sometimes created OpenSCAD objects by tracing an
object in a drawing program and then reading off the coordinates of the
various control points.
Note that Bézier functions are not the only mathematical
curves. (For instance, they cannot exactly match an arc of a circle.) I'm
sure there are numerous other functions that also produce curves.
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To unsubscribe send an email to
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I think I thought it might be nice to be able to produce ellipse-like
polygons with equal side length. I experimented with a few things and
ended up concluding that just stretching out the points on a circle
produces a pretty good result if your goal is to approximate an ellipse
with a polygon that has unequal sides. I wrote code to sample an ellipse
uniformly in arc length and the result is a poor approximation of the
ellipse, with too many points at the flat side and not enough at the tips.
On Sun, Aug 7, 2022 at 4:30 PM Ronaldo Persiano <rcmpersiano@gmail.com>
wrote:
> Adrian, why to use a (either inscribed or circumscribed) polygon with
> equal side as an approximation of an ellipsis? It doesn't seem to be a good
> approximation of it. I get a better approximation by taking an angle
> uniform sample of the curve in the parameter space.
>
> [image: Uniformly_sampled_ellipsis.png]
>
>
> Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano <avm4@cornell.edu>
> escreveu:
>
>> Circumscribed means tangents to points on the ellipse. But there are no
>> arc lengths (equal or unequal) in this problem, nor equally spaced angles.
>> The angles of each segment will be different. But the lengths of the
>> segments are not the same as the arc lengths, which is why arc length does
>> not matter. Here's a solution to the N=6 polygon sort of inspired by the
>> idea Jordan suggested, where I adjust the first segment until the second
>> segment has the right length. The rest of the solution is then obtained
>> using symmetry, which means there is only a single unknown that can simply
>> be solved for. But this doesn't generalize to higher N.
>>
>> [image: image.png]
>> include<BOSL2/std.scad>
>>
>> // ellipse semiaxes
>> // If a is much smaller than b it will fail
>> a=5;
>> b=2;
>>
>> // Find point on ellipse at specified x coordinate
>> function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
>>
>>
>> // Find tangent line to ellipse at point (u,w)
>> // and determine the x value where that line has
>> // given y value.
>> function findx(a,b,u,w,y) =
>> let(
>> A = -u*b^2/w/a^2,
>> B = w+(u*b)^2/w/a^2
>> )
>> (y-B)/A;
>>
>>
>> // First segment if tangent point is at x=u
>> function first_seg(a,b,u) =
>> let(
>> w = ellipse_pt(a,b,u),
>> x1 = findx(a,b,u,w,0),
>> x2 = findx(a,b,u,w,b)
>> )
>> [[x1,0],[x2,b]];
>>
>>
>> err = function (u) let(pt = first_seg(a,b,u))
>> 2*pt[1].x - norm(pt[0]-pt[1]);
>>
>> x = root_find(err,.2*a,.9*a);
>> first = first_seg(a,b,x);
>> top = concat(first,reverse(xflip(first)));
>> poly = concat(top, reverse(yflip(list_tail(top))));
>>
>> stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
>> color("red")stroke(poly,width=.05,closed=true);
>>
>> // Verify equality of segment lengths
>> echo(path_segment_lengths(poly,closed=true));
>>
>>
>>
>>
>>
>> On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
>> sprabhakar2006@gmail.com> wrote:
>>
>>> Circumscribed means tangents from points on ellipse
>>> How can the tangents from equally spaced arc lengths or equally spaced
>>> angles be equal for an ellipse
>>> You need to change the arc lengths dynamically and also accurately
>>> estimate the segment length which fits the ellipse .
>>>
>>>
>>>
>>> On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, <avm4@cornell.edu> wrote:
>>>
>>>> Your intuition is incorrect: a circle scaled in one direction *IS* an
>>>> ellipse. The motivation for being interested in circumscribing is to make
>>>> holes, though the constraint of equal length sides is probably not what
>>>> you'd want for that purpose. You can make a nonuniform circumscribing
>>>> polygon for an ellipse by simply making a polygon that circumscribes a
>>>> circle and scaling it.
>>>>
>>>> On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson <bob@rjcarlson.com> wrote:
>>>>
>>>>> Seems like I remember Nop Head talking about using circumscribed holes
>>>>> for screws and even rotating the polygons to produce a more even hole.
>>>>>
>>>>> If an ellipse is the nominal shape, the using a circumscribed polygon
>>>>> for a hole and an inscribed polygon for a solid might make sense.
>>>>>
>>>>> Btw, can someone confirm my intuition that a circle scaled in one
>>>>> dimension is not an ellipse?
>>>>>
>>>>> -Bob
>>>>> Tucson AZ
>>>>>
>>>>>
>>>>>
>>>>> On Aug 5, 2022, at 15:38, Adrian Mariano <avm4@cornell.edu> wrote:
>>>>>
>>>>> I don't have a specific need for a circumscribing elliptical polygon
>>>>> with equal sides. I was writing code to make ellipses and it seemed like
>>>>> equal side approximations would be nice, so I solved the case for
>>>>> inscribed, but couldn't solve the circumscribed case. So it's an unsolved
>>>>> puzzle I've got that this discussion reminded me of.
>>>>>
>>>>> On Fri, Aug 5, 2022 at 6:33 PM nop head <nop.head@gmail.com> wrote:
>>>>>
>>>>>> A long discussion but I have to ask why do you need an elliptical
>>>>>> polygon with equal tangential sides?
>>>>>>
>>>>>> On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu> wrote:
>>>>>>
>>>>>>> You think that a circumscribing polygon with equal length segments
>>>>>>> does not exist?? I'm skeptical of this claim.
>>>>>>>
>>>>>>> For circumscribed, it's not so hard. You start with a
>>>>>>> uniform-in-angle point distribution, compute the segment lengths, compare
>>>>>>> to the mean segment length to get an error, and then compute an adjustment
>>>>>>> based on the error. It converges quickly.
>>>>>>>
>>>>>>> So for example, I have segments lengths of:
>>>>>>> ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
>>>>>>> 5.57297, 5.57297]
>>>>>>>
>>>>>>> Maybe not exactly equal, since tolerance is 1e-9, but surely good
>>>>>>> enough.
>>>>>>> <image.png>
>>>>>>>
>>>>>>> The code for this is in BOSL2 in shapes2d.scad. A similar algorithm
>>>>>>> should exist for the circumscribed case, but as I said, my efforts to find
>>>>>>> it have failed.
>>>>>>>
>>>>>>> On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>
>>>>>>>> I think circumscribed polygon segments cannot be equal lengths.
>>>>>>>> inscribed ones may be approximately equal, but would be a very
>>>>>>>> tough problem to solve.
>>>>>>>> I tried something.
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>> On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <avm4@cornell.edu>
>>>>>>>> wrote:
>>>>>>>>
>>>>>>>>> I think you need to be a little careful about defining what you
>>>>>>>>> want to do. If you want to approximate an ellipse with points that are
>>>>>>>>> distributed uniformly over the ellipse, that could be done by approximating
>>>>>>>>> the arc length of the ellipse to find the necessary points---e.g. advance
>>>>>>>>> distance d along the ellipse at each step. But if you want to approximate
>>>>>>>>> an ellipse with a polygon with equal length sides, that approach does not
>>>>>>>>> do it, because the length of the segments will not be constant as the span
>>>>>>>>> sections of the ellipse with differing curvature. And since the perimeter
>>>>>>>>> of the desired polygon is unknown, you can't just step it out analogously.
>>>>>>>>> The circle doesn't present this challenge due to having constant
>>>>>>>>> curvature.
>>>>>>>>>
>>>>>>>>> On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>
>>>>>>>>>> I think our emails crossed. I was addressing how to approximate
>>>>>>>>>> an ellipse with constant-length segments in general, not how to solve your
>>>>>>>>>> problem. When I had to do something similar, I used polar coordinates and
>>>>>>>>>> stepped up the value of theta until the distance from the previous point
>>>>>>>>>> was correct. (In the constant-length version, I'd keep a running total of
>>>>>>>>>> the maximum desired length so that round-off errors don't accumulate.)
>>>>>>>>>>
>>>>>>>>>> Figuring out the circumscribed polygon is, as you note, a greater
>>>>>>>>>> challenge. I can visualize how to do it for n = 4, but it doesn't
>>>>>>>>>> extrapolate at all.
>>>>>>>>>>
>>>>>>>>>> On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <avm4@cornell.edu>
>>>>>>>>>> wrote:
>>>>>>>>>>
>>>>>>>>>>> Eh, there's no such thing as "brute force" in this situation.
>>>>>>>>>>> Step through *what* with sufficiently small steps? Of course an iterative
>>>>>>>>>>> algorithm is necessary. This is true for most interesting problems. No
>>>>>>>>>>> big deal. Show me an iterative algorithm that converges to the
>>>>>>>>>>> circumscribing polygon. When I tried to devise such an algorithm I
>>>>>>>>>>> couldn't find one that would converge. Writing an algorithm that would
>>>>>>>>>>> converge to the *inscribed* polygon seemed to be straight forward. Note
>>>>>>>>>>> also that computing arc length of the ellipse doesn't really help here.
>>>>>>>>>>> You can do that by solving an elliptical integral, or summing up small
>>>>>>>>>>> segments. But what's the connection between arc length of the ellipse and
>>>>>>>>>>> a circumscribing polygon with equal segment lengths? If there's a
>>>>>>>>>>> connection, I don't see it.
>>>>>>>>>>>
>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
>>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>>
>>>>>>>>>>>> Except by brute force (step through it with sufficiently small
>>>>>>>>>>>> steps), it’s not possible because there’s no closed form expression for the
>>>>>>>>>>>> distance between two points on a general ellipse (as opposed to special
>>>>>>>>>>>> cases like a circle).
>>>>>>>>>>>>
>>>>>>>>>>>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <avm4@cornell.edu>
>>>>>>>>>>>> wrote:
>>>>>>>>>>>>
>>>>>>>>>>>>> Here's a challenge for you: draw an "ellipse" polygon with N
>>>>>>>>>>>>> segments that are all the same length that circumscribes a perfect ellipse
>>>>>>>>>>>>> with given semiaxes a and b. I tried to solve this problem and could not
>>>>>>>>>>>>> find a solution. I can draw an *inscribed* N segment "ellipse" polygon
>>>>>>>>>>>>> with all segments the same length---though it's not easy---but the
>>>>>>>>>>>>> circumscribed case eludes me. I think of the inscribed case as, every
>>>>>>>>>>>>> vertex is located on the ellipse and the circumscribed case, every segment
>>>>>>>>>>>>> is tangent to the ellipse
>>>>>>>>>>>>>
>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>>>>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> In openscad drawing a ellipse is not too difficult
>>>>>>>>>>>>>> Refer following code
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> function to draw an ellipse with semi-major and semi-minor
>>>>>>>>>>>>>> axis "r1" and "r2" respectively and with center "cp" and number of segment
>>>>>>>>>>>>>> "s"
>>>>>>>>>>>>>> /
>>>>>>>>>>>>>> // example:
>>>>>>>>>>>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> function ellipse(r1,r2,cp,s=30)=
>>>>>>>>>>>>>> let(
>>>>>>>>>>>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>>>>>>>>>>>>> )sec;
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>>>>>>>>>>>>> openscad@jordan.maileater.net> wrote:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> I created an oval using:
>>>>>>>>>>>>>>> inMm = 25.4;
>>>>>>>>>>>>>>> fullW = 35.5 * inMm;
>>>>>>>>>>>>>>> fullL = 54.5 * inMm;
>>>>>>>>>>>>>>> resize([fullW, fullL, 1])
>>>>>>>>>>>>>>> circle(d=fullW);
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Tip: use scale() rather than resize(); it previews faster
>>>>>>>>>>>>>>> in complicated cases.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> You're taking a circle and stretching it; the result of that
>>>>>>>>>>>>>>> stretching is an ellipse.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> I'm not 100% sure - maybe one of the real math wizards can
>>>>>>>>>>>>>>> chime in - but I think the shape you are asking for would not be an ellipse.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> I believe that an ellipse is fully defined by its major
>>>>>>>>>>>>>>> (long) and minor (short) axis dimensions; the other parts of the curve
>>>>>>>>>>>>>>> mathematically flow from those two measurements.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> You can make something else, but you can't make it using
>>>>>>>>>>>>>>> this technique.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> You can use a Bézier function.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Using BOSL2:
>>>>>>>>>>>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> DIY, here's a function that I cobbled together some years
>>>>>>>>>>>>>>> ago:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>>>>>>>>>>>>> // but that yielded either single points or a raft of triangles;
>>>>>>>>>>>>>>> // this yields a vector of points that you can then concatenate
>>>>>>>>>>>>>>> // with other pieces to form a single polygon.
>>>>>>>>>>>>>>> // If we were really clever, I think it would be possible to
>>>>>>>>>>>>>>> // automatically space the output points based on how linear
>>>>>>>>>>>>>>> // the curve is at that point. But right now I'm not that clever.
>>>>>>>>>>>>>>> function BEZ03(u) = pow((1-u), 3);
>>>>>>>>>>>>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>>>>>>>>>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>>>>>>>>>>>>> function BEZ33(u) = pow(u,3);
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>>>>>>>>>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>>>>>>>>>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> // p0 - start point
>>>>>>>>>>>>>>> // p1 - control point 1, line departs p0 headed this way
>>>>>>>>>>>>>>> // p2 - control point 2, line arrives at p3 from this way
>>>>>>>>>>>>>>> // p3 - end point
>>>>>>>>>>>>>>> // segs - number of segments
>>>>>>>>>>>>>>> function bez(p0, p1, p2, p3, segs) = [
>>>>>>>>>>>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>>>>>>>>>>>>> ];
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> It is that final function bez() that I actually call, e.g.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> often concatenating the results of several bez() calls
>>>>>>>>>>>>>>> together.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Note that the four coordinates correspond to the coordinates
>>>>>>>>>>>>>>> of the control points in a GUI drawing program: the start point, control
>>>>>>>>>>>>>>> point associated with the start, the control point associated with the end,
>>>>>>>>>>>>>>> and the end point. I have sometimes created OpenSCAD objects by tracing an
>>>>>>>>>>>>>>> object in a drawing program and then reading off the coordinates of the
>>>>>>>>>>>>>>> various control points.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Note that Bézier functions are not the only mathematical
>>>>>>>>>>>>>>> curves. (For instance, they cannot exactly match an arc of a circle.) I'm
>>>>>>>>>>>>>>> sure there are numerous other functions that also produce curves.
>>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>
>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>> OpenSCAD mailing list
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>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>
>>>>>>>>>>>> _______________________________________________
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>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>
>>>>>>>>>>> _______________________________________________
>>>>>>>>>>> OpenSCAD mailing list
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>>>>>>>>>>>
>>>>>>>>>> _______________________________________________
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>>>>>>>>>>
>>>>>>>>> _______________________________________________
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>>>>>>>>>
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>>>>>>>>
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AM
Adrian Mariano
Mon, Aug 8, 2022 1:32 AM
Father Horton,
I'm having trouble understanding your proposed method. A segment from
(0,b) to (0,L/2) might be inside the ellipse depending on N, and then there
won't be a tangent.
It also sounds like your idea is somewhat similar to Jordan's idea:
basically start at one side and work around the ellipse and see if it
works. If not, make a correction. The problem with this idea is that
there are two parameters for the first segment, so how do you make a
correction? (The two parameters are the location of the endpoint of the
first segment and its length.)
On Sun, Aug 7, 2022 at 5:33 PM Father Horton fatherhorton@gmail.com wrote:
How about this to construct N equal length segments around an ellipse with
axes a and b:
- Select an initial segment length estimate (e.g., Ramanujan's formula /
N) = L
- Construct a half-segment from (0, b) to (0, L/2).
- Find a line tangent to the ellipse and passing through (0, L/2).
- Find the point on the tangent line that is L away from (0, L/2).
- Repeat (3) substituting the new point for (0, L/2). until you have N
segments.
If everything worked perfectly, the last point found would be (0, -L/2).
If there's a gap, increase L and try again. If there's overlap, decrease L
and try again.
On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano rcmpersiano@gmail.com
wrote:
Adrian, why to use a (either inscribed or circumscribed) polygon with
equal side as an approximation of an ellipsis? It doesn't seem to be a good
approximation of it. I get a better approximation by taking an angle
uniform sample of the curve in the parameter space.
[image: Uniformly_sampled_ellipsis.png]
Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano avm4@cornell.edu
escreveu:
Circumscribed means tangents to points on the ellipse. But there are no
arc lengths (equal or unequal) in this problem, nor equally spaced angles.
The angles of each segment will be different. But the lengths of the
segments are not the same as the arc lengths, which is why arc length does
not matter. Here's a solution to the N=6 polygon sort of inspired by the
idea Jordan suggested, where I adjust the first segment until the second
segment has the right length. The rest of the solution is then obtained
using symmetry, which means there is only a single unknown that can simply
be solved for. But this doesn't generalize to higher N.
[image: image.png]
include<BOSL2/std.scad>
// ellipse semiaxes
// If a is much smaller than b it will fail
a=5;
b=2;
// Find point on ellipse at specified x coordinate
function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
// Find tangent line to ellipse at point (u,w)
// and determine the x value where that line has
// given y value.
function findx(a,b,u,w,y) =
let(
A = -ub^2/w/a^2,
B = w+(ub)^2/w/a^2
)
(y-B)/A;
// First segment if tangent point is at x=u
function first_seg(a,b,u) =
let(
w = ellipse_pt(a,b,u),
x1 = findx(a,b,u,w,0),
x2 = findx(a,b,u,w,b)
)
[[x1,0],[x2,b]];
err = function (u) let(pt = first_seg(a,b,u))
2*pt[1].x - norm(pt[0]-pt[1]);
x = root_find(err,.2a,.9a);
first = first_seg(a,b,x);
top = concat(first,reverse(xflip(first)));
poly = concat(top, reverse(yflip(list_tail(top))));
stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
color("red")stroke(poly,width=.05,closed=true);
// Verify equality of segment lengths
echo(path_segment_lengths(poly,closed=true));
On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
Circumscribed means tangents from points on ellipse
How can the tangents from equally spaced arc lengths or equally spaced
angles be equal for an ellipse
You need to change the arc lengths dynamically and also accurately
estimate the segment length which fits the ellipse .
On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, avm4@cornell.edu wrote:
Your intuition is incorrect: a circle scaled in one direction IS an
ellipse. The motivation for being interested in circumscribing is to make
holes, though the constraint of equal length sides is probably not what
you'd want for that purpose. You can make a nonuniform circumscribing
polygon for an ellipse by simply making a polygon that circumscribes a
circle and scaling it.
On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson bob@rjcarlson.com wrote:
Seems like I remember Nop Head talking about using circumscribed
holes for screws and even rotating the polygons to produce a more even
hole.
If an ellipse is the nominal shape, the using a circumscribed polygon
for a hole and an inscribed polygon for a solid might make sense.
Btw, can someone confirm my intuition that a circle scaled in one
dimension is not an ellipse?
-Bob
Tucson AZ
On Aug 5, 2022, at 15:38, Adrian Mariano avm4@cornell.edu wrote:
I don't have a specific need for a circumscribing elliptical polygon
with equal sides. I was writing code to make ellipses and it seemed like
equal side approximations would be nice, so I solved the case for
inscribed, but couldn't solve the circumscribed case. So it's an unsolved
puzzle I've got that this discussion reminded me of.
On Fri, Aug 5, 2022 at 6:33 PM nop head nop.head@gmail.com wrote:
A long discussion but I have to ask why do you need an elliptical
polygon with equal tangential sides?
On Fri, 5 Aug 2022, 21:58 Adrian Mariano, avm4@cornell.edu wrote:
You think that a circumscribing polygon with equal length segments
does not exist?? I'm skeptical of this claim.
For circumscribed, it's not so hard. You start with a
uniform-in-angle point distribution, compute the segment lengths, compare
to the mean segment length to get an error, and then compute an adjustment
based on the error. It converges quickly.
So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
5.57297, 5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but surely good
enough.
<image.png>
The code for this is in BOSL2 in shapes2d.scad. A similar algorithm
should exist for the circumscribed case, but as I said, my efforts to find
it have failed.
On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
I think circumscribed polygon segments cannot be equal lengths.
inscribed ones may be approximately equal, but would be a very
tough problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano avm4@cornell.edu
wrote:
I think you need to be a little careful about defining what you
want to do. If you want to approximate an ellipse with points that are
distributed uniformly over the ellipse, that could be done by approximating
the arc length of the ellipse to find the necessary points---e.g. advance
distance d along the ellipse at each step. But if you want to approximate
an ellipse with a polygon with equal length sides, that approach does not
do it, because the length of the segments will not be constant as the span
sections of the ellipse with differing curvature. And since the perimeter
of the desired polygon is unknown, you can't just step it out analogously.
The circle doesn't present this challenge due to having constant
curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
fatherhorton@gmail.com> wrote:
I think our emails crossed. I was addressing how to approximate
an ellipse with constant-length segments in general, not how to solve your
problem. When I had to do something similar, I used polar coordinates and
stepped up the value of theta until the distance from the previous point
was correct. (In the constant-length version, I'd keep a running total of
the maximum desired length so that round-off errors don't accumulate.)
Figuring out the circumscribed polygon is, as you note, a
greater challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano avm4@cornell.edu
wrote:
Eh, there's no such thing as "brute force" in this situation.
Step through what with sufficiently small steps? Of course an iterative
algorithm is necessary. This is true for most interesting problems. No
big deal. Show me an iterative algorithm that converges to the
circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
fatherhorton@gmail.com> wrote:
Except by brute force (step through it with sufficiently small
steps), it’s not possible because there’s no closed form expression for the
distance between two points on a general ellipse (as opposed to special
cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <
avm4@cornell.edu> wrote:
Here's a challenge for you: draw an "ellipse" polygon with N
segments that are all the same length that circumscribes a perfect ellipse
with given semiaxes a and b. I tried to solve this problem and could not
find a solution. I can draw an inscribed N segment "ellipse" polygon
with all segments the same length---though it's not easy---but the
circumscribed case eludes me. I think of the inscribed case as, every
vertex is located on the ellipse and the circumscribed case, every segment
is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor
axis "r1" and "r2" respectively and with center "cp" and number of segment
"s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster
in complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of
that stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can
chime in - but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major
(long) and minor (short) axis dimensions; the other parts of the curve
mathematically flow from those two measurements.
You can make something else, but you can't make it using
this technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years
ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls
together.
Note that the four coordinates correspond to the
coordinates of the control points in a GUI drawing program: the start
point, control point associated with the start, the control point
associated with the end, and the end point. I have sometimes created
OpenSCAD objects by tracing an object in a drawing program and then reading
off the coordinates of the various control points.
Note that Bézier functions are not the only mathematical
curves. (For instance, they cannot exactly match an arc of a circle.) I'm
sure there are numerous other functions that also produce curves.
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Father Horton,
I'm having trouble understanding your proposed method. A segment from
(0,b) to (0,L/2) might be inside the ellipse depending on N, and then there
won't be a tangent.
It also sounds like your idea is somewhat similar to Jordan's idea:
basically start at one side and work around the ellipse and see if it
works. If not, make a correction. The problem with this idea is that
there are two parameters for the first segment, so how do you make a
correction? (The two parameters are the location of the endpoint of the
first segment and its length.)
On Sun, Aug 7, 2022 at 5:33 PM Father Horton <fatherhorton@gmail.com> wrote:
> How about this to construct N equal length segments around an ellipse with
> axes a and b:
>
> 1) Select an initial segment length estimate (e.g., Ramanujan's formula /
> N) = L
> 2) Construct a half-segment from (0, b) to (0, L/2).
> 3) Find a line tangent to the ellipse and passing through (0, L/2).
> 3) Find the point on the tangent line that is L away from (0, L/2).
> 4) Repeat (3) substituting the new point for (0, L/2). until you have N
> segments.
>
> If everything worked perfectly, the last point found would be (0, -L/2).
> If there's a gap, increase L and try again. If there's overlap, decrease L
> and try again.
>
> On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano <rcmpersiano@gmail.com>
> wrote:
>
>> Adrian, why to use a (either inscribed or circumscribed) polygon with
>> equal side as an approximation of an ellipsis? It doesn't seem to be a good
>> approximation of it. I get a better approximation by taking an angle
>> uniform sample of the curve in the parameter space.
>>
>> [image: Uniformly_sampled_ellipsis.png]
>>
>>
>> Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano <avm4@cornell.edu>
>> escreveu:
>>
>>> Circumscribed means tangents to points on the ellipse. But there are no
>>> arc lengths (equal or unequal) in this problem, nor equally spaced angles.
>>> The angles of each segment will be different. But the lengths of the
>>> segments are not the same as the arc lengths, which is why arc length does
>>> not matter. Here's a solution to the N=6 polygon sort of inspired by the
>>> idea Jordan suggested, where I adjust the first segment until the second
>>> segment has the right length. The rest of the solution is then obtained
>>> using symmetry, which means there is only a single unknown that can simply
>>> be solved for. But this doesn't generalize to higher N.
>>>
>>> [image: image.png]
>>> include<BOSL2/std.scad>
>>>
>>> // ellipse semiaxes
>>> // If a is much smaller than b it will fail
>>> a=5;
>>> b=2;
>>>
>>> // Find point on ellipse at specified x coordinate
>>> function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
>>>
>>>
>>> // Find tangent line to ellipse at point (u,w)
>>> // and determine the x value where that line has
>>> // given y value.
>>> function findx(a,b,u,w,y) =
>>> let(
>>> A = -u*b^2/w/a^2,
>>> B = w+(u*b)^2/w/a^2
>>> )
>>> (y-B)/A;
>>>
>>>
>>> // First segment if tangent point is at x=u
>>> function first_seg(a,b,u) =
>>> let(
>>> w = ellipse_pt(a,b,u),
>>> x1 = findx(a,b,u,w,0),
>>> x2 = findx(a,b,u,w,b)
>>> )
>>> [[x1,0],[x2,b]];
>>>
>>>
>>> err = function (u) let(pt = first_seg(a,b,u))
>>> 2*pt[1].x - norm(pt[0]-pt[1]);
>>>
>>> x = root_find(err,.2*a,.9*a);
>>> first = first_seg(a,b,x);
>>> top = concat(first,reverse(xflip(first)));
>>> poly = concat(top, reverse(yflip(list_tail(top))));
>>>
>>> stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
>>> color("red")stroke(poly,width=.05,closed=true);
>>>
>>> // Verify equality of segment lengths
>>> echo(path_segment_lengths(poly,closed=true));
>>>
>>>
>>>
>>>
>>>
>>> On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
>>> sprabhakar2006@gmail.com> wrote:
>>>
>>>> Circumscribed means tangents from points on ellipse
>>>> How can the tangents from equally spaced arc lengths or equally spaced
>>>> angles be equal for an ellipse
>>>> You need to change the arc lengths dynamically and also accurately
>>>> estimate the segment length which fits the ellipse .
>>>>
>>>>
>>>>
>>>> On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, <avm4@cornell.edu> wrote:
>>>>
>>>>> Your intuition is incorrect: a circle scaled in one direction *IS* an
>>>>> ellipse. The motivation for being interested in circumscribing is to make
>>>>> holes, though the constraint of equal length sides is probably not what
>>>>> you'd want for that purpose. You can make a nonuniform circumscribing
>>>>> polygon for an ellipse by simply making a polygon that circumscribes a
>>>>> circle and scaling it.
>>>>>
>>>>> On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson <bob@rjcarlson.com> wrote:
>>>>>
>>>>>> Seems like I remember Nop Head talking about using circumscribed
>>>>>> holes for screws and even rotating the polygons to produce a more even
>>>>>> hole.
>>>>>>
>>>>>> If an ellipse is the nominal shape, the using a circumscribed polygon
>>>>>> for a hole and an inscribed polygon for a solid might make sense.
>>>>>>
>>>>>> Btw, can someone confirm my intuition that a circle scaled in one
>>>>>> dimension is not an ellipse?
>>>>>>
>>>>>> -Bob
>>>>>> Tucson AZ
>>>>>>
>>>>>>
>>>>>>
>>>>>> On Aug 5, 2022, at 15:38, Adrian Mariano <avm4@cornell.edu> wrote:
>>>>>>
>>>>>> I don't have a specific need for a circumscribing elliptical polygon
>>>>>> with equal sides. I was writing code to make ellipses and it seemed like
>>>>>> equal side approximations would be nice, so I solved the case for
>>>>>> inscribed, but couldn't solve the circumscribed case. So it's an unsolved
>>>>>> puzzle I've got that this discussion reminded me of.
>>>>>>
>>>>>> On Fri, Aug 5, 2022 at 6:33 PM nop head <nop.head@gmail.com> wrote:
>>>>>>
>>>>>>> A long discussion but I have to ask why do you need an elliptical
>>>>>>> polygon with equal tangential sides?
>>>>>>>
>>>>>>> On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu> wrote:
>>>>>>>
>>>>>>>> You think that a circumscribing polygon with equal length segments
>>>>>>>> does not exist?? I'm skeptical of this claim.
>>>>>>>>
>>>>>>>> For circumscribed, it's not so hard. You start with a
>>>>>>>> uniform-in-angle point distribution, compute the segment lengths, compare
>>>>>>>> to the mean segment length to get an error, and then compute an adjustment
>>>>>>>> based on the error. It converges quickly.
>>>>>>>>
>>>>>>>> So for example, I have segments lengths of:
>>>>>>>> ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
>>>>>>>> 5.57297, 5.57297]
>>>>>>>>
>>>>>>>> Maybe not exactly equal, since tolerance is 1e-9, but surely good
>>>>>>>> enough.
>>>>>>>> <image.png>
>>>>>>>>
>>>>>>>> The code for this is in BOSL2 in shapes2d.scad. A similar algorithm
>>>>>>>> should exist for the circumscribed case, but as I said, my efforts to find
>>>>>>>> it have failed.
>>>>>>>>
>>>>>>>> On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>
>>>>>>>>> I think circumscribed polygon segments cannot be equal lengths.
>>>>>>>>> inscribed ones may be approximately equal, but would be a very
>>>>>>>>> tough problem to solve.
>>>>>>>>> I tried something.
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <avm4@cornell.edu>
>>>>>>>>> wrote:
>>>>>>>>>
>>>>>>>>>> I think you need to be a little careful about defining what you
>>>>>>>>>> want to do. If you want to approximate an ellipse with points that are
>>>>>>>>>> distributed uniformly over the ellipse, that could be done by approximating
>>>>>>>>>> the arc length of the ellipse to find the necessary points---e.g. advance
>>>>>>>>>> distance d along the ellipse at each step. But if you want to approximate
>>>>>>>>>> an ellipse with a polygon with equal length sides, that approach does not
>>>>>>>>>> do it, because the length of the segments will not be constant as the span
>>>>>>>>>> sections of the ellipse with differing curvature. And since the perimeter
>>>>>>>>>> of the desired polygon is unknown, you can't just step it out analogously.
>>>>>>>>>> The circle doesn't present this challenge due to having constant
>>>>>>>>>> curvature.
>>>>>>>>>>
>>>>>>>>>> On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>
>>>>>>>>>>> I think our emails crossed. I was addressing how to approximate
>>>>>>>>>>> an ellipse with constant-length segments in general, not how to solve your
>>>>>>>>>>> problem. When I had to do something similar, I used polar coordinates and
>>>>>>>>>>> stepped up the value of theta until the distance from the previous point
>>>>>>>>>>> was correct. (In the constant-length version, I'd keep a running total of
>>>>>>>>>>> the maximum desired length so that round-off errors don't accumulate.)
>>>>>>>>>>>
>>>>>>>>>>> Figuring out the circumscribed polygon is, as you note, a
>>>>>>>>>>> greater challenge. I can visualize how to do it for n = 4, but it doesn't
>>>>>>>>>>> extrapolate at all.
>>>>>>>>>>>
>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <avm4@cornell.edu>
>>>>>>>>>>> wrote:
>>>>>>>>>>>
>>>>>>>>>>>> Eh, there's no such thing as "brute force" in this situation.
>>>>>>>>>>>> Step through *what* with sufficiently small steps? Of course an iterative
>>>>>>>>>>>> algorithm is necessary. This is true for most interesting problems. No
>>>>>>>>>>>> big deal. Show me an iterative algorithm that converges to the
>>>>>>>>>>>> circumscribing polygon. When I tried to devise such an algorithm I
>>>>>>>>>>>> couldn't find one that would converge. Writing an algorithm that would
>>>>>>>>>>>> converge to the *inscribed* polygon seemed to be straight forward. Note
>>>>>>>>>>>> also that computing arc length of the ellipse doesn't really help here.
>>>>>>>>>>>> You can do that by solving an elliptical integral, or summing up small
>>>>>>>>>>>> segments. But what's the connection between arc length of the ellipse and
>>>>>>>>>>>> a circumscribing polygon with equal segment lengths? If there's a
>>>>>>>>>>>> connection, I don't see it.
>>>>>>>>>>>>
>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
>>>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>>>
>>>>>>>>>>>>> Except by brute force (step through it with sufficiently small
>>>>>>>>>>>>> steps), it’s not possible because there’s no closed form expression for the
>>>>>>>>>>>>> distance between two points on a general ellipse (as opposed to special
>>>>>>>>>>>>> cases like a circle).
>>>>>>>>>>>>>
>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <
>>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> Here's a challenge for you: draw an "ellipse" polygon with N
>>>>>>>>>>>>>> segments that are all the same length that circumscribes a perfect ellipse
>>>>>>>>>>>>>> with given semiaxes a and b. I tried to solve this problem and could not
>>>>>>>>>>>>>> find a solution. I can draw an *inscribed* N segment "ellipse" polygon
>>>>>>>>>>>>>> with all segments the same length---though it's not easy---but the
>>>>>>>>>>>>>> circumscribed case eludes me. I think of the inscribed case as, every
>>>>>>>>>>>>>> vertex is located on the ellipse and the circumscribed case, every segment
>>>>>>>>>>>>>> is tangent to the ellipse
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>>>>>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> In openscad drawing a ellipse is not too difficult
>>>>>>>>>>>>>>> Refer following code
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> function to draw an ellipse with semi-major and semi-minor
>>>>>>>>>>>>>>> axis "r1" and "r2" respectively and with center "cp" and number of segment
>>>>>>>>>>>>>>> "s"
>>>>>>>>>>>>>>> /
>>>>>>>>>>>>>>> // example:
>>>>>>>>>>>>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> function ellipse(r1,r2,cp,s=30)=
>>>>>>>>>>>>>>> let(
>>>>>>>>>>>>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>>>>>>>>>>>>>> )sec;
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>>>>>>>>>>>>>> openscad@jordan.maileater.net> wrote:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> I created an oval using:
>>>>>>>>>>>>>>>> inMm = 25.4;
>>>>>>>>>>>>>>>> fullW = 35.5 * inMm;
>>>>>>>>>>>>>>>> fullL = 54.5 * inMm;
>>>>>>>>>>>>>>>> resize([fullW, fullL, 1])
>>>>>>>>>>>>>>>> circle(d=fullW);
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Tip: use scale() rather than resize(); it previews faster
>>>>>>>>>>>>>>>> in complicated cases.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> You're taking a circle and stretching it; the result of
>>>>>>>>>>>>>>>> that stretching is an ellipse.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> I'm not 100% sure - maybe one of the real math wizards can
>>>>>>>>>>>>>>>> chime in - but I think the shape you are asking for would not be an ellipse.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> I believe that an ellipse is fully defined by its major
>>>>>>>>>>>>>>>> (long) and minor (short) axis dimensions; the other parts of the curve
>>>>>>>>>>>>>>>> mathematically flow from those two measurements.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> You can make something else, but you can't make it using
>>>>>>>>>>>>>>>> this technique.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> You can use a Bézier function.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Using BOSL2:
>>>>>>>>>>>>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> DIY, here's a function that I cobbled together some years
>>>>>>>>>>>>>>>> ago:
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>>>>>>>>>>>>>> // but that yielded either single points or a raft of triangles;
>>>>>>>>>>>>>>>> // this yields a vector of points that you can then concatenate
>>>>>>>>>>>>>>>> // with other pieces to form a single polygon.
>>>>>>>>>>>>>>>> // If we were really clever, I think it would be possible to
>>>>>>>>>>>>>>>> // automatically space the output points based on how linear
>>>>>>>>>>>>>>>> // the curve is at that point. But right now I'm not that clever.
>>>>>>>>>>>>>>>> function BEZ03(u) = pow((1-u), 3);
>>>>>>>>>>>>>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>>>>>>>>>>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>>>>>>>>>>>>>> function BEZ33(u) = pow(u,3);
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>>>>>>>>>>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>>>>>>>>>>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> // p0 - start point
>>>>>>>>>>>>>>>> // p1 - control point 1, line departs p0 headed this way
>>>>>>>>>>>>>>>> // p2 - control point 2, line arrives at p3 from this way
>>>>>>>>>>>>>>>> // p3 - end point
>>>>>>>>>>>>>>>> // segs - number of segments
>>>>>>>>>>>>>>>> function bez(p0, p1, p2, p3, segs) = [
>>>>>>>>>>>>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>>>>>>>>>>>>>> ];
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> It is that final function bez() that I actually call, e.g.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> often concatenating the results of several bez() calls
>>>>>>>>>>>>>>>> together.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Note that the four coordinates correspond to the
>>>>>>>>>>>>>>>> coordinates of the control points in a GUI drawing program: the start
>>>>>>>>>>>>>>>> point, control point associated with the start, the control point
>>>>>>>>>>>>>>>> associated with the end, and the end point. I have sometimes created
>>>>>>>>>>>>>>>> OpenSCAD objects by tracing an object in a drawing program and then reading
>>>>>>>>>>>>>>>> off the coordinates of the various control points.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Note that Bézier functions are not the only mathematical
>>>>>>>>>>>>>>>> curves. (For instance, they cannot exactly match an arc of a circle.) I'm
>>>>>>>>>>>>>>>> sure there are numerous other functions that also produce curves.
>>>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>
>>>>>>>>>>>>> _______________________________________________
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>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>
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>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>
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>>>>>>>>>>>
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>>>>>>>>>>
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>>>>>>>>>
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>>>>>>>>
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>>>>>>>
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>>>>>>
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FH
Father Horton
Mon, Aug 8, 2022 1:52 AM
I did that wrong. It's not (0, b) to (0, L/2). It's (0, b) to (L/2, b).
That's always outside the ellipse.
I will look at the second problem as soon as I figure out how to calculate
a tangent line to an ellipse containing a point not on the ellipse. Google
is not helping much.
On Sun, Aug 7, 2022 at 8:32 PM Adrian Mariano avm4@cornell.edu wrote:
Father Horton,
I'm having trouble understanding your proposed method. A segment from
(0,b) to (0,L/2) might be inside the ellipse depending on N, and then there
won't be a tangent.
It also sounds like your idea is somewhat similar to Jordan's idea:
basically start at one side and work around the ellipse and see if it
works. If not, make a correction. The problem with this idea is that
there are two parameters for the first segment, so how do you make a
correction? (The two parameters are the location of the endpoint of the
first segment and its length.)
On Sun, Aug 7, 2022 at 5:33 PM Father Horton fatherhorton@gmail.com
wrote:
How about this to construct N equal length segments around an ellipse
with axes a and b:
- Select an initial segment length estimate (e.g., Ramanujan's formula /
N) = L
- Construct a half-segment from (0, b) to (0, L/2).
- Find a line tangent to the ellipse and passing through (0, L/2).
- Find the point on the tangent line that is L away from (0, L/2).
- Repeat (3) substituting the new point for (0, L/2). until you have N
segments.
If everything worked perfectly, the last point found would be (0, -L/2).
If there's a gap, increase L and try again. If there's overlap, decrease L
and try again.
On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano rcmpersiano@gmail.com
wrote:
Adrian, why to use a (either inscribed or circumscribed) polygon with
equal side as an approximation of an ellipsis? It doesn't seem to be a good
approximation of it. I get a better approximation by taking an angle
uniform sample of the curve in the parameter space.
[image: Uniformly_sampled_ellipsis.png]
Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano avm4@cornell.edu
escreveu:
Circumscribed means tangents to points on the ellipse. But there are
no arc lengths (equal or unequal) in this problem, nor equally spaced
angles. The angles of each segment will be different. But the lengths of
the segments are not the same as the arc lengths, which is why arc length
does not matter. Here's a solution to the N=6 polygon sort of inspired by
the idea Jordan suggested, where I adjust the first segment until the
second segment has the right length. The rest of the solution is then
obtained using symmetry, which means there is only a single unknown that
can simply be solved for. But this doesn't generalize to higher N.
[image: image.png]
include<BOSL2/std.scad>
// ellipse semiaxes
// If a is much smaller than b it will fail
a=5;
b=2;
// Find point on ellipse at specified x coordinate
function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
// Find tangent line to ellipse at point (u,w)
// and determine the x value where that line has
// given y value.
function findx(a,b,u,w,y) =
let(
A = -ub^2/w/a^2,
B = w+(ub)^2/w/a^2
)
(y-B)/A;
// First segment if tangent point is at x=u
function first_seg(a,b,u) =
let(
w = ellipse_pt(a,b,u),
x1 = findx(a,b,u,w,0),
x2 = findx(a,b,u,w,b)
)
[[x1,0],[x2,b]];
err = function (u) let(pt = first_seg(a,b,u))
2*pt[1].x - norm(pt[0]-pt[1]);
x = root_find(err,.2a,.9a);
first = first_seg(a,b,x);
top = concat(first,reverse(xflip(first)));
poly = concat(top, reverse(yflip(list_tail(top))));
stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
color("red")stroke(poly,width=.05,closed=true);
// Verify equality of segment lengths
echo(path_segment_lengths(poly,closed=true));
On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
Circumscribed means tangents from points on ellipse
How can the tangents from equally spaced arc lengths or equally spaced
angles be equal for an ellipse
You need to change the arc lengths dynamically and also accurately
estimate the segment length which fits the ellipse .
On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, avm4@cornell.edu wrote:
Your intuition is incorrect: a circle scaled in one direction IS
an ellipse. The motivation for being interested in circumscribing is to
make holes, though the constraint of equal length sides is probably not
what you'd want for that purpose. You can make a nonuniform circumscribing
polygon for an ellipse by simply making a polygon that circumscribes a
circle and scaling it.
On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson bob@rjcarlson.com wrote:
Seems like I remember Nop Head talking about using circumscribed
holes for screws and even rotating the polygons to produce a more even
hole.
If an ellipse is the nominal shape, the using a circumscribed
polygon for a hole and an inscribed polygon for a solid might make sense.
Btw, can someone confirm my intuition that a circle scaled in one
dimension is not an ellipse?
-Bob
Tucson AZ
On Aug 5, 2022, at 15:38, Adrian Mariano avm4@cornell.edu wrote:
I don't have a specific need for a circumscribing elliptical polygon
with equal sides. I was writing code to make ellipses and it seemed like
equal side approximations would be nice, so I solved the case for
inscribed, but couldn't solve the circumscribed case. So it's an unsolved
puzzle I've got that this discussion reminded me of.
On Fri, Aug 5, 2022 at 6:33 PM nop head nop.head@gmail.com wrote:
A long discussion but I have to ask why do you need an elliptical
polygon with equal tangential sides?
On Fri, 5 Aug 2022, 21:58 Adrian Mariano, avm4@cornell.edu wrote:
You think that a circumscribing polygon with equal length segments
does not exist?? I'm skeptical of this claim.
For circumscribed, it's not so hard. You start with a
uniform-in-angle point distribution, compute the segment lengths, compare
to the mean segment length to get an error, and then compute an adjustment
based on the error. It converges quickly.
So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
5.57297, 5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but surely good
enough.
<image.png>
The code for this is in BOSL2 in shapes2d.scad. A similar
algorithm should exist for the circumscribed case, but as I said, my
efforts to find it have failed.
On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
I think circumscribed polygon segments cannot be equal lengths.
inscribed ones may be approximately equal, but would be a very
tough problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano avm4@cornell.edu
wrote:
I think you need to be a little careful about defining what you
want to do. If you want to approximate an ellipse with points that are
distributed uniformly over the ellipse, that could be done by approximating
the arc length of the ellipse to find the necessary points---e.g. advance
distance d along the ellipse at each step. But if you want to approximate
an ellipse with a polygon with equal length sides, that approach does not
do it, because the length of the segments will not be constant as the span
sections of the ellipse with differing curvature. And since the perimeter
of the desired polygon is unknown, you can't just step it out analogously.
The circle doesn't present this challenge due to having constant
curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
fatherhorton@gmail.com> wrote:
I think our emails crossed. I was addressing how to approximate
an ellipse with constant-length segments in general, not how to solve your
problem. When I had to do something similar, I used polar coordinates and
stepped up the value of theta until the distance from the previous point
was correct. (In the constant-length version, I'd keep a running total of
the maximum desired length so that round-off errors don't accumulate.)
Figuring out the circumscribed polygon is, as you note, a
greater challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <
avm4@cornell.edu> wrote:
Eh, there's no such thing as "brute force" in this situation.
Step through what with sufficiently small steps? Of course an iterative
algorithm is necessary. This is true for most interesting problems. No
big deal. Show me an iterative algorithm that converges to the
circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
fatherhorton@gmail.com> wrote:
Except by brute force (step through it with sufficiently
small steps), it’s not possible because there’s no closed form expression
for the distance between two points on a general ellipse (as opposed to
special cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <
avm4@cornell.edu> wrote:
Here's a challenge for you: draw an "ellipse" polygon with
N segments that are all the same length that circumscribes a perfect
ellipse with given semiaxes a and b. I tried to solve this problem and
could not find a solution. I can draw an inscribed N segment "ellipse"
polygon with all segments the same length---though it's not easy---but the
circumscribed case eludes me. I think of the inscribed case as, every
vertex is located on the ellipse and the circumscribed case, every segment
is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor
axis "r1" and "r2" respectively and with center "cp" and number of segment
"s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster
in complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of
that stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can
chime in - but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major
(long) and minor (short) axis dimensions; the other parts of the curve
mathematically flow from those two measurements.
You can make something else, but you can't make it using
this technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years
ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls
together.
Note that the four coordinates correspond to the
coordinates of the control points in a GUI drawing program: the start
point, control point associated with the start, the control point
associated with the end, and the end point. I have sometimes created
OpenSCAD objects by tracing an object in a drawing program and then reading
off the coordinates of the various control points.
Note that Bézier functions are not the only mathematical
curves. (For instance, they cannot exactly match an arc of a circle.) I'm
sure there are numerous other functions that also produce curves.
OpenSCAD mailing list
To unsubscribe send an email to
discuss-leave@lists.openscad.org
I did that wrong. It's not (0, b) to (0, L/2). It's (0, b) to (L/2, b).
That's always outside the ellipse.
I will look at the second problem as soon as I figure out how to calculate
a tangent line to an ellipse containing a point not on the ellipse. Google
is not helping much.
On Sun, Aug 7, 2022 at 8:32 PM Adrian Mariano <avm4@cornell.edu> wrote:
> Father Horton,
>
> I'm having trouble understanding your proposed method. A segment from
> (0,b) to (0,L/2) might be inside the ellipse depending on N, and then there
> won't be a tangent.
>
> It also sounds like your idea is somewhat similar to Jordan's idea:
> basically start at one side and work around the ellipse and see if it
> works. If not, make a correction. The problem with this idea is that
> there are two parameters for the first segment, so how do you make a
> correction? (The two parameters are the location of the endpoint of the
> first segment and its length.)
>
> On Sun, Aug 7, 2022 at 5:33 PM Father Horton <fatherhorton@gmail.com>
> wrote:
>
>> How about this to construct N equal length segments around an ellipse
>> with axes a and b:
>>
>> 1) Select an initial segment length estimate (e.g., Ramanujan's formula /
>> N) = L
>> 2) Construct a half-segment from (0, b) to (0, L/2).
>> 3) Find a line tangent to the ellipse and passing through (0, L/2).
>> 3) Find the point on the tangent line that is L away from (0, L/2).
>> 4) Repeat (3) substituting the new point for (0, L/2). until you have N
>> segments.
>>
>> If everything worked perfectly, the last point found would be (0, -L/2).
>> If there's a gap, increase L and try again. If there's overlap, decrease L
>> and try again.
>>
>> On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano <rcmpersiano@gmail.com>
>> wrote:
>>
>>> Adrian, why to use a (either inscribed or circumscribed) polygon with
>>> equal side as an approximation of an ellipsis? It doesn't seem to be a good
>>> approximation of it. I get a better approximation by taking an angle
>>> uniform sample of the curve in the parameter space.
>>>
>>> [image: Uniformly_sampled_ellipsis.png]
>>>
>>>
>>> Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano <avm4@cornell.edu>
>>> escreveu:
>>>
>>>> Circumscribed means tangents to points on the ellipse. But there are
>>>> no arc lengths (equal or unequal) in this problem, nor equally spaced
>>>> angles. The angles of each segment will be different. But the lengths of
>>>> the segments are not the same as the arc lengths, which is why arc length
>>>> does not matter. Here's a solution to the N=6 polygon sort of inspired by
>>>> the idea Jordan suggested, where I adjust the first segment until the
>>>> second segment has the right length. The rest of the solution is then
>>>> obtained using symmetry, which means there is only a single unknown that
>>>> can simply be solved for. But this doesn't generalize to higher N.
>>>>
>>>> [image: image.png]
>>>> include<BOSL2/std.scad>
>>>>
>>>> // ellipse semiaxes
>>>> // If a is much smaller than b it will fail
>>>> a=5;
>>>> b=2;
>>>>
>>>> // Find point on ellipse at specified x coordinate
>>>> function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
>>>>
>>>>
>>>> // Find tangent line to ellipse at point (u,w)
>>>> // and determine the x value where that line has
>>>> // given y value.
>>>> function findx(a,b,u,w,y) =
>>>> let(
>>>> A = -u*b^2/w/a^2,
>>>> B = w+(u*b)^2/w/a^2
>>>> )
>>>> (y-B)/A;
>>>>
>>>>
>>>> // First segment if tangent point is at x=u
>>>> function first_seg(a,b,u) =
>>>> let(
>>>> w = ellipse_pt(a,b,u),
>>>> x1 = findx(a,b,u,w,0),
>>>> x2 = findx(a,b,u,w,b)
>>>> )
>>>> [[x1,0],[x2,b]];
>>>>
>>>>
>>>> err = function (u) let(pt = first_seg(a,b,u))
>>>> 2*pt[1].x - norm(pt[0]-pt[1]);
>>>>
>>>> x = root_find(err,.2*a,.9*a);
>>>> first = first_seg(a,b,x);
>>>> top = concat(first,reverse(xflip(first)));
>>>> poly = concat(top, reverse(yflip(list_tail(top))));
>>>>
>>>> stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
>>>> color("red")stroke(poly,width=.05,closed=true);
>>>>
>>>> // Verify equality of segment lengths
>>>> echo(path_segment_lengths(poly,closed=true));
>>>>
>>>>
>>>>
>>>>
>>>>
>>>> On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
>>>> sprabhakar2006@gmail.com> wrote:
>>>>
>>>>> Circumscribed means tangents from points on ellipse
>>>>> How can the tangents from equally spaced arc lengths or equally spaced
>>>>> angles be equal for an ellipse
>>>>> You need to change the arc lengths dynamically and also accurately
>>>>> estimate the segment length which fits the ellipse .
>>>>>
>>>>>
>>>>>
>>>>> On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, <avm4@cornell.edu> wrote:
>>>>>
>>>>>> Your intuition is incorrect: a circle scaled in one direction *IS*
>>>>>> an ellipse. The motivation for being interested in circumscribing is to
>>>>>> make holes, though the constraint of equal length sides is probably not
>>>>>> what you'd want for that purpose. You can make a nonuniform circumscribing
>>>>>> polygon for an ellipse by simply making a polygon that circumscribes a
>>>>>> circle and scaling it.
>>>>>>
>>>>>> On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson <bob@rjcarlson.com> wrote:
>>>>>>
>>>>>>> Seems like I remember Nop Head talking about using circumscribed
>>>>>>> holes for screws and even rotating the polygons to produce a more even
>>>>>>> hole.
>>>>>>>
>>>>>>> If an ellipse is the nominal shape, the using a circumscribed
>>>>>>> polygon for a hole and an inscribed polygon for a solid might make sense.
>>>>>>>
>>>>>>> Btw, can someone confirm my intuition that a circle scaled in one
>>>>>>> dimension is not an ellipse?
>>>>>>>
>>>>>>> -Bob
>>>>>>> Tucson AZ
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> On Aug 5, 2022, at 15:38, Adrian Mariano <avm4@cornell.edu> wrote:
>>>>>>>
>>>>>>> I don't have a specific need for a circumscribing elliptical polygon
>>>>>>> with equal sides. I was writing code to make ellipses and it seemed like
>>>>>>> equal side approximations would be nice, so I solved the case for
>>>>>>> inscribed, but couldn't solve the circumscribed case. So it's an unsolved
>>>>>>> puzzle I've got that this discussion reminded me of.
>>>>>>>
>>>>>>> On Fri, Aug 5, 2022 at 6:33 PM nop head <nop.head@gmail.com> wrote:
>>>>>>>
>>>>>>>> A long discussion but I have to ask why do you need an elliptical
>>>>>>>> polygon with equal tangential sides?
>>>>>>>>
>>>>>>>> On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu> wrote:
>>>>>>>>
>>>>>>>>> You think that a circumscribing polygon with equal length segments
>>>>>>>>> does not exist?? I'm skeptical of this claim.
>>>>>>>>>
>>>>>>>>> For circumscribed, it's not so hard. You start with a
>>>>>>>>> uniform-in-angle point distribution, compute the segment lengths, compare
>>>>>>>>> to the mean segment length to get an error, and then compute an adjustment
>>>>>>>>> based on the error. It converges quickly.
>>>>>>>>>
>>>>>>>>> So for example, I have segments lengths of:
>>>>>>>>> ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
>>>>>>>>> 5.57297, 5.57297]
>>>>>>>>>
>>>>>>>>> Maybe not exactly equal, since tolerance is 1e-9, but surely good
>>>>>>>>> enough.
>>>>>>>>> <image.png>
>>>>>>>>>
>>>>>>>>> The code for this is in BOSL2 in shapes2d.scad. A similar
>>>>>>>>> algorithm should exist for the circumscribed case, but as I said, my
>>>>>>>>> efforts to find it have failed.
>>>>>>>>>
>>>>>>>>> On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>
>>>>>>>>>> I think circumscribed polygon segments cannot be equal lengths.
>>>>>>>>>> inscribed ones may be approximately equal, but would be a very
>>>>>>>>>> tough problem to solve.
>>>>>>>>>> I tried something.
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <avm4@cornell.edu>
>>>>>>>>>> wrote:
>>>>>>>>>>
>>>>>>>>>>> I think you need to be a little careful about defining what you
>>>>>>>>>>> want to do. If you want to approximate an ellipse with points that are
>>>>>>>>>>> distributed uniformly over the ellipse, that could be done by approximating
>>>>>>>>>>> the arc length of the ellipse to find the necessary points---e.g. advance
>>>>>>>>>>> distance d along the ellipse at each step. But if you want to approximate
>>>>>>>>>>> an ellipse with a polygon with equal length sides, that approach does not
>>>>>>>>>>> do it, because the length of the segments will not be constant as the span
>>>>>>>>>>> sections of the ellipse with differing curvature. And since the perimeter
>>>>>>>>>>> of the desired polygon is unknown, you can't just step it out analogously.
>>>>>>>>>>> The circle doesn't present this challenge due to having constant
>>>>>>>>>>> curvature.
>>>>>>>>>>>
>>>>>>>>>>> On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
>>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>>
>>>>>>>>>>>> I think our emails crossed. I was addressing how to approximate
>>>>>>>>>>>> an ellipse with constant-length segments in general, not how to solve your
>>>>>>>>>>>> problem. When I had to do something similar, I used polar coordinates and
>>>>>>>>>>>> stepped up the value of theta until the distance from the previous point
>>>>>>>>>>>> was correct. (In the constant-length version, I'd keep a running total of
>>>>>>>>>>>> the maximum desired length so that round-off errors don't accumulate.)
>>>>>>>>>>>>
>>>>>>>>>>>> Figuring out the circumscribed polygon is, as you note, a
>>>>>>>>>>>> greater challenge. I can visualize how to do it for n = 4, but it doesn't
>>>>>>>>>>>> extrapolate at all.
>>>>>>>>>>>>
>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <
>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>
>>>>>>>>>>>>> Eh, there's no such thing as "brute force" in this situation.
>>>>>>>>>>>>> Step through *what* with sufficiently small steps? Of course an iterative
>>>>>>>>>>>>> algorithm is necessary. This is true for most interesting problems. No
>>>>>>>>>>>>> big deal. Show me an iterative algorithm that converges to the
>>>>>>>>>>>>> circumscribing polygon. When I tried to devise such an algorithm I
>>>>>>>>>>>>> couldn't find one that would converge. Writing an algorithm that would
>>>>>>>>>>>>> converge to the *inscribed* polygon seemed to be straight forward. Note
>>>>>>>>>>>>> also that computing arc length of the ellipse doesn't really help here.
>>>>>>>>>>>>> You can do that by solving an elliptical integral, or summing up small
>>>>>>>>>>>>> segments. But what's the connection between arc length of the ellipse and
>>>>>>>>>>>>> a circumscribing polygon with equal segment lengths? If there's a
>>>>>>>>>>>>> connection, I don't see it.
>>>>>>>>>>>>>
>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
>>>>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> Except by brute force (step through it with sufficiently
>>>>>>>>>>>>>> small steps), it’s not possible because there’s no closed form expression
>>>>>>>>>>>>>> for the distance between two points on a general ellipse (as opposed to
>>>>>>>>>>>>>> special cases like a circle).
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <
>>>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Here's a challenge for you: draw an "ellipse" polygon with
>>>>>>>>>>>>>>> N segments that are all the same length that circumscribes a perfect
>>>>>>>>>>>>>>> ellipse with given semiaxes a and b. I tried to solve this problem and
>>>>>>>>>>>>>>> could not find a solution. I can draw an *inscribed* N segment "ellipse"
>>>>>>>>>>>>>>> polygon with all segments the same length---though it's not easy---but the
>>>>>>>>>>>>>>> circumscribed case eludes me. I think of the inscribed case as, every
>>>>>>>>>>>>>>> vertex is located on the ellipse and the circumscribed case, every segment
>>>>>>>>>>>>>>> is tangent to the ellipse
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>>>>>>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> In openscad drawing a ellipse is not too difficult
>>>>>>>>>>>>>>>> Refer following code
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> function to draw an ellipse with semi-major and semi-minor
>>>>>>>>>>>>>>>> axis "r1" and "r2" respectively and with center "cp" and number of segment
>>>>>>>>>>>>>>>> "s"
>>>>>>>>>>>>>>>> /
>>>>>>>>>>>>>>>> // example:
>>>>>>>>>>>>>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> function ellipse(r1,r2,cp,s=30)=
>>>>>>>>>>>>>>>> let(
>>>>>>>>>>>>>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>>>>>>>>>>>>>>> )sec;
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>>>>>>>>>>>>>>> openscad@jordan.maileater.net> wrote:
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> I created an oval using:
>>>>>>>>>>>>>>>>> inMm = 25.4;
>>>>>>>>>>>>>>>>> fullW = 35.5 * inMm;
>>>>>>>>>>>>>>>>> fullL = 54.5 * inMm;
>>>>>>>>>>>>>>>>> resize([fullW, fullL, 1])
>>>>>>>>>>>>>>>>> circle(d=fullW);
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Tip: use scale() rather than resize(); it previews faster
>>>>>>>>>>>>>>>>> in complicated cases.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> You're taking a circle and stretching it; the result of
>>>>>>>>>>>>>>>>> that stretching is an ellipse.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> I'm not 100% sure - maybe one of the real math wizards can
>>>>>>>>>>>>>>>>> chime in - but I think the shape you are asking for would not be an ellipse.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> I believe that an ellipse is fully defined by its major
>>>>>>>>>>>>>>>>> (long) and minor (short) axis dimensions; the other parts of the curve
>>>>>>>>>>>>>>>>> mathematically flow from those two measurements.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> You can make something else, but you can't make it using
>>>>>>>>>>>>>>>>> this technique.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> You can use a Bézier function.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Using BOSL2:
>>>>>>>>>>>>>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> DIY, here's a function that I cobbled together some years
>>>>>>>>>>>>>>>>> ago:
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>>>>>>>>>>>>>>> // but that yielded either single points or a raft of triangles;
>>>>>>>>>>>>>>>>> // this yields a vector of points that you can then concatenate
>>>>>>>>>>>>>>>>> // with other pieces to form a single polygon.
>>>>>>>>>>>>>>>>> // If we were really clever, I think it would be possible to
>>>>>>>>>>>>>>>>> // automatically space the output points based on how linear
>>>>>>>>>>>>>>>>> // the curve is at that point. But right now I'm not that clever.
>>>>>>>>>>>>>>>>> function BEZ03(u) = pow((1-u), 3);
>>>>>>>>>>>>>>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>>>>>>>>>>>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>>>>>>>>>>>>>>> function BEZ33(u) = pow(u,3);
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>>>>>>>>>>>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>>>>>>>>>>>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> // p0 - start point
>>>>>>>>>>>>>>>>> // p1 - control point 1, line departs p0 headed this way
>>>>>>>>>>>>>>>>> // p2 - control point 2, line arrives at p3 from this way
>>>>>>>>>>>>>>>>> // p3 - end point
>>>>>>>>>>>>>>>>> // segs - number of segments
>>>>>>>>>>>>>>>>> function bez(p0, p1, p2, p3, segs) = [
>>>>>>>>>>>>>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>>>>>>>>>>>>>>> ];
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> It is that final function bez() that I actually call, e.g.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> often concatenating the results of several bez() calls
>>>>>>>>>>>>>>>>> together.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Note that the four coordinates correspond to the
>>>>>>>>>>>>>>>>> coordinates of the control points in a GUI drawing program: the start
>>>>>>>>>>>>>>>>> point, control point associated with the start, the control point
>>>>>>>>>>>>>>>>> associated with the end, and the end point. I have sometimes created
>>>>>>>>>>>>>>>>> OpenSCAD objects by tracing an object in a drawing program and then reading
>>>>>>>>>>>>>>>>> off the coordinates of the various control points.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Note that Bézier functions are not the only mathematical
>>>>>>>>>>>>>>>>> curves. (For instance, they cannot exactly match an arc of a circle.) I'm
>>>>>>>>>>>>>>>>> sure there are numerous other functions that also produce curves.
>>>>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>
>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>
>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>
>>>>>>>>>>> _______________________________________________
>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>>>>>>>>>>>
>>>>>>>>>> _______________________________________________
>>>>>>>>>> OpenSCAD mailing list
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>>>>>>>>>>
>>>>>>>>> _______________________________________________
>>>>>>>>> OpenSCAD mailing list
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>>>>>>>>>
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>>>>>>>> OpenSCAD mailing list
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>>>>>>>
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>>>>>>>
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>>>>>>
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>>>>>
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>
SP
Sanjeev Prabhakar
Mon, Aug 8, 2022 2:24 AM
If you think of ellipse as a circle which is inclined at an angle e.g. a
radius 10 circle inclined at 60 degrees will produce an ellipse with 10 and
5 semi axis.
In such a case no matter how you choose your circumscribed polygon, the
first tangent will be a vertical line from point [a,0] and when rotated it
will be of length lcos 60 (where l is the height of point from [a,0]).
Based on this height a circumscribed circle may be choose and equal lengths
inscribed polygon with lengths 2l*cos 60, might work.
I don't know if the points will be same numbers in such a case.
I haven't tried this
Regards
Sanjeev
On Mon, 8 Aug, 2022, 7:04 am Adrian Mariano, avm4@cornell.edu wrote:
Father Horton,
I'm having trouble understanding your proposed method. A segment from
(0,b) to (0,L/2) might be inside the ellipse depending on N, and then there
won't be a tangent.
It also sounds like your idea is somewhat similar to Jordan's idea:
basically start at one side and work around the ellipse and see if it
works. If not, make a correction. The problem with this idea is that
there are two parameters for the first segment, so how do you make a
correction? (The two parameters are the location of the endpoint of the
first segment and its length.)
On Sun, Aug 7, 2022 at 5:33 PM Father Horton fatherhorton@gmail.com
wrote:
How about this to construct N equal length segments around an ellipse
with axes a and b:
- Select an initial segment length estimate (e.g., Ramanujan's formula /
N) = L
- Construct a half-segment from (0, b) to (0, L/2).
- Find a line tangent to the ellipse and passing through (0, L/2).
- Find the point on the tangent line that is L away from (0, L/2).
- Repeat (3) substituting the new point for (0, L/2). until you have N
segments.
If everything worked perfectly, the last point found would be (0, -L/2).
If there's a gap, increase L and try again. If there's overlap, decrease L
and try again.
On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano rcmpersiano@gmail.com
wrote:
Adrian, why to use a (either inscribed or circumscribed) polygon with
equal side as an approximation of an ellipsis? It doesn't seem to be a good
approximation of it. I get a better approximation by taking an angle
uniform sample of the curve in the parameter space.
[image: Uniformly_sampled_ellipsis.png]
Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano avm4@cornell.edu
escreveu:
Circumscribed means tangents to points on the ellipse. But there are
no arc lengths (equal or unequal) in this problem, nor equally spaced
angles. The angles of each segment will be different. But the lengths of
the segments are not the same as the arc lengths, which is why arc length
does not matter. Here's a solution to the N=6 polygon sort of inspired by
the idea Jordan suggested, where I adjust the first segment until the
second segment has the right length. The rest of the solution is then
obtained using symmetry, which means there is only a single unknown that
can simply be solved for. But this doesn't generalize to higher N.
[image: image.png]
include<BOSL2/std.scad>
// ellipse semiaxes
// If a is much smaller than b it will fail
a=5;
b=2;
// Find point on ellipse at specified x coordinate
function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
// Find tangent line to ellipse at point (u,w)
// and determine the x value where that line has
// given y value.
function findx(a,b,u,w,y) =
let(
A = -ub^2/w/a^2,
B = w+(ub)^2/w/a^2
)
(y-B)/A;
// First segment if tangent point is at x=u
function first_seg(a,b,u) =
let(
w = ellipse_pt(a,b,u),
x1 = findx(a,b,u,w,0),
x2 = findx(a,b,u,w,b)
)
[[x1,0],[x2,b]];
err = function (u) let(pt = first_seg(a,b,u))
2*pt[1].x - norm(pt[0]-pt[1]);
x = root_find(err,.2a,.9a);
first = first_seg(a,b,x);
top = concat(first,reverse(xflip(first)));
poly = concat(top, reverse(yflip(list_tail(top))));
stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
color("red")stroke(poly,width=.05,closed=true);
// Verify equality of segment lengths
echo(path_segment_lengths(poly,closed=true));
On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
Circumscribed means tangents from points on ellipse
How can the tangents from equally spaced arc lengths or equally spaced
angles be equal for an ellipse
You need to change the arc lengths dynamically and also accurately
estimate the segment length which fits the ellipse .
On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, avm4@cornell.edu wrote:
Your intuition is incorrect: a circle scaled in one direction IS
an ellipse. The motivation for being interested in circumscribing is to
make holes, though the constraint of equal length sides is probably not
what you'd want for that purpose. You can make a nonuniform circumscribing
polygon for an ellipse by simply making a polygon that circumscribes a
circle and scaling it.
On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson bob@rjcarlson.com wrote:
Seems like I remember Nop Head talking about using circumscribed
holes for screws and even rotating the polygons to produce a more even
hole.
If an ellipse is the nominal shape, the using a circumscribed
polygon for a hole and an inscribed polygon for a solid might make sense.
Btw, can someone confirm my intuition that a circle scaled in one
dimension is not an ellipse?
-Bob
Tucson AZ
On Aug 5, 2022, at 15:38, Adrian Mariano avm4@cornell.edu wrote:
I don't have a specific need for a circumscribing elliptical polygon
with equal sides. I was writing code to make ellipses and it seemed like
equal side approximations would be nice, so I solved the case for
inscribed, but couldn't solve the circumscribed case. So it's an unsolved
puzzle I've got that this discussion reminded me of.
On Fri, Aug 5, 2022 at 6:33 PM nop head nop.head@gmail.com wrote:
A long discussion but I have to ask why do you need an elliptical
polygon with equal tangential sides?
On Fri, 5 Aug 2022, 21:58 Adrian Mariano, avm4@cornell.edu wrote:
You think that a circumscribing polygon with equal length segments
does not exist?? I'm skeptical of this claim.
For circumscribed, it's not so hard. You start with a
uniform-in-angle point distribution, compute the segment lengths, compare
to the mean segment length to get an error, and then compute an adjustment
based on the error. It converges quickly.
So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
5.57297, 5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but surely good
enough.
<image.png>
The code for this is in BOSL2 in shapes2d.scad. A similar
algorithm should exist for the circumscribed case, but as I said, my
efforts to find it have failed.
On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
I think circumscribed polygon segments cannot be equal lengths.
inscribed ones may be approximately equal, but would be a very
tough problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano avm4@cornell.edu
wrote:
I think you need to be a little careful about defining what you
want to do. If you want to approximate an ellipse with points that are
distributed uniformly over the ellipse, that could be done by approximating
the arc length of the ellipse to find the necessary points---e.g. advance
distance d along the ellipse at each step. But if you want to approximate
an ellipse with a polygon with equal length sides, that approach does not
do it, because the length of the segments will not be constant as the span
sections of the ellipse with differing curvature. And since the perimeter
of the desired polygon is unknown, you can't just step it out analogously.
The circle doesn't present this challenge due to having constant
curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
fatherhorton@gmail.com> wrote:
I think our emails crossed. I was addressing how to approximate
an ellipse with constant-length segments in general, not how to solve your
problem. When I had to do something similar, I used polar coordinates and
stepped up the value of theta until the distance from the previous point
was correct. (In the constant-length version, I'd keep a running total of
the maximum desired length so that round-off errors don't accumulate.)
Figuring out the circumscribed polygon is, as you note, a
greater challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <
avm4@cornell.edu> wrote:
Eh, there's no such thing as "brute force" in this situation.
Step through what with sufficiently small steps? Of course an iterative
algorithm is necessary. This is true for most interesting problems. No
big deal. Show me an iterative algorithm that converges to the
circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
fatherhorton@gmail.com> wrote:
Except by brute force (step through it with sufficiently
small steps), it’s not possible because there’s no closed form expression
for the distance between two points on a general ellipse (as opposed to
special cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <
avm4@cornell.edu> wrote:
Here's a challenge for you: draw an "ellipse" polygon with
N segments that are all the same length that circumscribes a perfect
ellipse with given semiaxes a and b. I tried to solve this problem and
could not find a solution. I can draw an inscribed N segment "ellipse"
polygon with all segments the same length---though it's not easy---but the
circumscribed case eludes me. I think of the inscribed case as, every
vertex is located on the ellipse and the circumscribed case, every segment
is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor
axis "r1" and "r2" respectively and with center "cp" and number of segment
"s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews faster
in complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of
that stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can
chime in - but I think the shape you are asking for would not be an ellipse.
I believe that an ellipse is fully defined by its major
(long) and minor (short) axis dimensions; the other parts of the curve
mathematically flow from those two measurements.
You can make something else, but you can't make it using
this technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years
ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls
together.
Note that the four coordinates correspond to the
coordinates of the control points in a GUI drawing program: the start
point, control point associated with the start, the control point
associated with the end, and the end point. I have sometimes created
OpenSCAD objects by tracing an object in a drawing program and then reading
off the coordinates of the various control points.
Note that Bézier functions are not the only mathematical
curves. (For instance, they cannot exactly match an arc of a circle.) I'm
sure there are numerous other functions that also produce curves.
OpenSCAD mailing list
To unsubscribe send an email to
discuss-leave@lists.openscad.org
If you think of ellipse as a circle which is inclined at an angle e.g. a
radius 10 circle inclined at 60 degrees will produce an ellipse with 10 and
5 semi axis.
In such a case no matter how you choose your circumscribed polygon, the
first tangent will be a vertical line from point [a,0] and when rotated it
will be of length l*cos 60 (where l is the height of point from [a,0]).
Based on this height a circumscribed circle may be choose and equal lengths
inscribed polygon with lengths 2*l*cos 60, might work.
I don't know if the points will be same numbers in such a case.
I haven't tried this
Regards
Sanjeev
On Mon, 8 Aug, 2022, 7:04 am Adrian Mariano, <avm4@cornell.edu> wrote:
> Father Horton,
>
> I'm having trouble understanding your proposed method. A segment from
> (0,b) to (0,L/2) might be inside the ellipse depending on N, and then there
> won't be a tangent.
>
> It also sounds like your idea is somewhat similar to Jordan's idea:
> basically start at one side and work around the ellipse and see if it
> works. If not, make a correction. The problem with this idea is that
> there are two parameters for the first segment, so how do you make a
> correction? (The two parameters are the location of the endpoint of the
> first segment and its length.)
>
> On Sun, Aug 7, 2022 at 5:33 PM Father Horton <fatherhorton@gmail.com>
> wrote:
>
>> How about this to construct N equal length segments around an ellipse
>> with axes a and b:
>>
>> 1) Select an initial segment length estimate (e.g., Ramanujan's formula /
>> N) = L
>> 2) Construct a half-segment from (0, b) to (0, L/2).
>> 3) Find a line tangent to the ellipse and passing through (0, L/2).
>> 3) Find the point on the tangent line that is L away from (0, L/2).
>> 4) Repeat (3) substituting the new point for (0, L/2). until you have N
>> segments.
>>
>> If everything worked perfectly, the last point found would be (0, -L/2).
>> If there's a gap, increase L and try again. If there's overlap, decrease L
>> and try again.
>>
>> On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano <rcmpersiano@gmail.com>
>> wrote:
>>
>>> Adrian, why to use a (either inscribed or circumscribed) polygon with
>>> equal side as an approximation of an ellipsis? It doesn't seem to be a good
>>> approximation of it. I get a better approximation by taking an angle
>>> uniform sample of the curve in the parameter space.
>>>
>>> [image: Uniformly_sampled_ellipsis.png]
>>>
>>>
>>> Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano <avm4@cornell.edu>
>>> escreveu:
>>>
>>>> Circumscribed means tangents to points on the ellipse. But there are
>>>> no arc lengths (equal or unequal) in this problem, nor equally spaced
>>>> angles. The angles of each segment will be different. But the lengths of
>>>> the segments are not the same as the arc lengths, which is why arc length
>>>> does not matter. Here's a solution to the N=6 polygon sort of inspired by
>>>> the idea Jordan suggested, where I adjust the first segment until the
>>>> second segment has the right length. The rest of the solution is then
>>>> obtained using symmetry, which means there is only a single unknown that
>>>> can simply be solved for. But this doesn't generalize to higher N.
>>>>
>>>> [image: image.png]
>>>> include<BOSL2/std.scad>
>>>>
>>>> // ellipse semiaxes
>>>> // If a is much smaller than b it will fail
>>>> a=5;
>>>> b=2;
>>>>
>>>> // Find point on ellipse at specified x coordinate
>>>> function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
>>>>
>>>>
>>>> // Find tangent line to ellipse at point (u,w)
>>>> // and determine the x value where that line has
>>>> // given y value.
>>>> function findx(a,b,u,w,y) =
>>>> let(
>>>> A = -u*b^2/w/a^2,
>>>> B = w+(u*b)^2/w/a^2
>>>> )
>>>> (y-B)/A;
>>>>
>>>>
>>>> // First segment if tangent point is at x=u
>>>> function first_seg(a,b,u) =
>>>> let(
>>>> w = ellipse_pt(a,b,u),
>>>> x1 = findx(a,b,u,w,0),
>>>> x2 = findx(a,b,u,w,b)
>>>> )
>>>> [[x1,0],[x2,b]];
>>>>
>>>>
>>>> err = function (u) let(pt = first_seg(a,b,u))
>>>> 2*pt[1].x - norm(pt[0]-pt[1]);
>>>>
>>>> x = root_find(err,.2*a,.9*a);
>>>> first = first_seg(a,b,x);
>>>> top = concat(first,reverse(xflip(first)));
>>>> poly = concat(top, reverse(yflip(list_tail(top))));
>>>>
>>>> stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
>>>> color("red")stroke(poly,width=.05,closed=true);
>>>>
>>>> // Verify equality of segment lengths
>>>> echo(path_segment_lengths(poly,closed=true));
>>>>
>>>>
>>>>
>>>>
>>>>
>>>> On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
>>>> sprabhakar2006@gmail.com> wrote:
>>>>
>>>>> Circumscribed means tangents from points on ellipse
>>>>> How can the tangents from equally spaced arc lengths or equally spaced
>>>>> angles be equal for an ellipse
>>>>> You need to change the arc lengths dynamically and also accurately
>>>>> estimate the segment length which fits the ellipse .
>>>>>
>>>>>
>>>>>
>>>>> On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, <avm4@cornell.edu> wrote:
>>>>>
>>>>>> Your intuition is incorrect: a circle scaled in one direction *IS*
>>>>>> an ellipse. The motivation for being interested in circumscribing is to
>>>>>> make holes, though the constraint of equal length sides is probably not
>>>>>> what you'd want for that purpose. You can make a nonuniform circumscribing
>>>>>> polygon for an ellipse by simply making a polygon that circumscribes a
>>>>>> circle and scaling it.
>>>>>>
>>>>>> On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson <bob@rjcarlson.com> wrote:
>>>>>>
>>>>>>> Seems like I remember Nop Head talking about using circumscribed
>>>>>>> holes for screws and even rotating the polygons to produce a more even
>>>>>>> hole.
>>>>>>>
>>>>>>> If an ellipse is the nominal shape, the using a circumscribed
>>>>>>> polygon for a hole and an inscribed polygon for a solid might make sense.
>>>>>>>
>>>>>>> Btw, can someone confirm my intuition that a circle scaled in one
>>>>>>> dimension is not an ellipse?
>>>>>>>
>>>>>>> -Bob
>>>>>>> Tucson AZ
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> On Aug 5, 2022, at 15:38, Adrian Mariano <avm4@cornell.edu> wrote:
>>>>>>>
>>>>>>> I don't have a specific need for a circumscribing elliptical polygon
>>>>>>> with equal sides. I was writing code to make ellipses and it seemed like
>>>>>>> equal side approximations would be nice, so I solved the case for
>>>>>>> inscribed, but couldn't solve the circumscribed case. So it's an unsolved
>>>>>>> puzzle I've got that this discussion reminded me of.
>>>>>>>
>>>>>>> On Fri, Aug 5, 2022 at 6:33 PM nop head <nop.head@gmail.com> wrote:
>>>>>>>
>>>>>>>> A long discussion but I have to ask why do you need an elliptical
>>>>>>>> polygon with equal tangential sides?
>>>>>>>>
>>>>>>>> On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu> wrote:
>>>>>>>>
>>>>>>>>> You think that a circumscribing polygon with equal length segments
>>>>>>>>> does not exist?? I'm skeptical of this claim.
>>>>>>>>>
>>>>>>>>> For circumscribed, it's not so hard. You start with a
>>>>>>>>> uniform-in-angle point distribution, compute the segment lengths, compare
>>>>>>>>> to the mean segment length to get an error, and then compute an adjustment
>>>>>>>>> based on the error. It converges quickly.
>>>>>>>>>
>>>>>>>>> So for example, I have segments lengths of:
>>>>>>>>> ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
>>>>>>>>> 5.57297, 5.57297]
>>>>>>>>>
>>>>>>>>> Maybe not exactly equal, since tolerance is 1e-9, but surely good
>>>>>>>>> enough.
>>>>>>>>> <image.png>
>>>>>>>>>
>>>>>>>>> The code for this is in BOSL2 in shapes2d.scad. A similar
>>>>>>>>> algorithm should exist for the circumscribed case, but as I said, my
>>>>>>>>> efforts to find it have failed.
>>>>>>>>>
>>>>>>>>> On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>
>>>>>>>>>> I think circumscribed polygon segments cannot be equal lengths.
>>>>>>>>>> inscribed ones may be approximately equal, but would be a very
>>>>>>>>>> tough problem to solve.
>>>>>>>>>> I tried something.
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <avm4@cornell.edu>
>>>>>>>>>> wrote:
>>>>>>>>>>
>>>>>>>>>>> I think you need to be a little careful about defining what you
>>>>>>>>>>> want to do. If you want to approximate an ellipse with points that are
>>>>>>>>>>> distributed uniformly over the ellipse, that could be done by approximating
>>>>>>>>>>> the arc length of the ellipse to find the necessary points---e.g. advance
>>>>>>>>>>> distance d along the ellipse at each step. But if you want to approximate
>>>>>>>>>>> an ellipse with a polygon with equal length sides, that approach does not
>>>>>>>>>>> do it, because the length of the segments will not be constant as the span
>>>>>>>>>>> sections of the ellipse with differing curvature. And since the perimeter
>>>>>>>>>>> of the desired polygon is unknown, you can't just step it out analogously.
>>>>>>>>>>> The circle doesn't present this challenge due to having constant
>>>>>>>>>>> curvature.
>>>>>>>>>>>
>>>>>>>>>>> On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
>>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>>
>>>>>>>>>>>> I think our emails crossed. I was addressing how to approximate
>>>>>>>>>>>> an ellipse with constant-length segments in general, not how to solve your
>>>>>>>>>>>> problem. When I had to do something similar, I used polar coordinates and
>>>>>>>>>>>> stepped up the value of theta until the distance from the previous point
>>>>>>>>>>>> was correct. (In the constant-length version, I'd keep a running total of
>>>>>>>>>>>> the maximum desired length so that round-off errors don't accumulate.)
>>>>>>>>>>>>
>>>>>>>>>>>> Figuring out the circumscribed polygon is, as you note, a
>>>>>>>>>>>> greater challenge. I can visualize how to do it for n = 4, but it doesn't
>>>>>>>>>>>> extrapolate at all.
>>>>>>>>>>>>
>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <
>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>
>>>>>>>>>>>>> Eh, there's no such thing as "brute force" in this situation.
>>>>>>>>>>>>> Step through *what* with sufficiently small steps? Of course an iterative
>>>>>>>>>>>>> algorithm is necessary. This is true for most interesting problems. No
>>>>>>>>>>>>> big deal. Show me an iterative algorithm that converges to the
>>>>>>>>>>>>> circumscribing polygon. When I tried to devise such an algorithm I
>>>>>>>>>>>>> couldn't find one that would converge. Writing an algorithm that would
>>>>>>>>>>>>> converge to the *inscribed* polygon seemed to be straight forward. Note
>>>>>>>>>>>>> also that computing arc length of the ellipse doesn't really help here.
>>>>>>>>>>>>> You can do that by solving an elliptical integral, or summing up small
>>>>>>>>>>>>> segments. But what's the connection between arc length of the ellipse and
>>>>>>>>>>>>> a circumscribing polygon with equal segment lengths? If there's a
>>>>>>>>>>>>> connection, I don't see it.
>>>>>>>>>>>>>
>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
>>>>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> Except by brute force (step through it with sufficiently
>>>>>>>>>>>>>> small steps), it’s not possible because there’s no closed form expression
>>>>>>>>>>>>>> for the distance between two points on a general ellipse (as opposed to
>>>>>>>>>>>>>> special cases like a circle).
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <
>>>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Here's a challenge for you: draw an "ellipse" polygon with
>>>>>>>>>>>>>>> N segments that are all the same length that circumscribes a perfect
>>>>>>>>>>>>>>> ellipse with given semiaxes a and b. I tried to solve this problem and
>>>>>>>>>>>>>>> could not find a solution. I can draw an *inscribed* N segment "ellipse"
>>>>>>>>>>>>>>> polygon with all segments the same length---though it's not easy---but the
>>>>>>>>>>>>>>> circumscribed case eludes me. I think of the inscribed case as, every
>>>>>>>>>>>>>>> vertex is located on the ellipse and the circumscribed case, every segment
>>>>>>>>>>>>>>> is tangent to the ellipse
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>>>>>>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> In openscad drawing a ellipse is not too difficult
>>>>>>>>>>>>>>>> Refer following code
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> function to draw an ellipse with semi-major and semi-minor
>>>>>>>>>>>>>>>> axis "r1" and "r2" respectively and with center "cp" and number of segment
>>>>>>>>>>>>>>>> "s"
>>>>>>>>>>>>>>>> /
>>>>>>>>>>>>>>>> // example:
>>>>>>>>>>>>>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> function ellipse(r1,r2,cp,s=30)=
>>>>>>>>>>>>>>>> let(
>>>>>>>>>>>>>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>>>>>>>>>>>>>>> )sec;
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>>>>>>>>>>>>>>> openscad@jordan.maileater.net> wrote:
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> I created an oval using:
>>>>>>>>>>>>>>>>> inMm = 25.4;
>>>>>>>>>>>>>>>>> fullW = 35.5 * inMm;
>>>>>>>>>>>>>>>>> fullL = 54.5 * inMm;
>>>>>>>>>>>>>>>>> resize([fullW, fullL, 1])
>>>>>>>>>>>>>>>>> circle(d=fullW);
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Tip: use scale() rather than resize(); it previews faster
>>>>>>>>>>>>>>>>> in complicated cases.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> You're taking a circle and stretching it; the result of
>>>>>>>>>>>>>>>>> that stretching is an ellipse.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> I'm not 100% sure - maybe one of the real math wizards can
>>>>>>>>>>>>>>>>> chime in - but I think the shape you are asking for would not be an ellipse.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> I believe that an ellipse is fully defined by its major
>>>>>>>>>>>>>>>>> (long) and minor (short) axis dimensions; the other parts of the curve
>>>>>>>>>>>>>>>>> mathematically flow from those two measurements.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> You can make something else, but you can't make it using
>>>>>>>>>>>>>>>>> this technique.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> You can use a Bézier function.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Using BOSL2:
>>>>>>>>>>>>>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> DIY, here's a function that I cobbled together some years
>>>>>>>>>>>>>>>>> ago:
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>>>>>>>>>>>>>>> // but that yielded either single points or a raft of triangles;
>>>>>>>>>>>>>>>>> // this yields a vector of points that you can then concatenate
>>>>>>>>>>>>>>>>> // with other pieces to form a single polygon.
>>>>>>>>>>>>>>>>> // If we were really clever, I think it would be possible to
>>>>>>>>>>>>>>>>> // automatically space the output points based on how linear
>>>>>>>>>>>>>>>>> // the curve is at that point. But right now I'm not that clever.
>>>>>>>>>>>>>>>>> function BEZ03(u) = pow((1-u), 3);
>>>>>>>>>>>>>>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>>>>>>>>>>>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>>>>>>>>>>>>>>> function BEZ33(u) = pow(u,3);
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>>>>>>>>>>>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>>>>>>>>>>>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> // p0 - start point
>>>>>>>>>>>>>>>>> // p1 - control point 1, line departs p0 headed this way
>>>>>>>>>>>>>>>>> // p2 - control point 2, line arrives at p3 from this way
>>>>>>>>>>>>>>>>> // p3 - end point
>>>>>>>>>>>>>>>>> // segs - number of segments
>>>>>>>>>>>>>>>>> function bez(p0, p1, p2, p3, segs) = [
>>>>>>>>>>>>>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>>>>>>>>>>>>>>> ];
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> It is that final function bez() that I actually call, e.g.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> often concatenating the results of several bez() calls
>>>>>>>>>>>>>>>>> together.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Note that the four coordinates correspond to the
>>>>>>>>>>>>>>>>> coordinates of the control points in a GUI drawing program: the start
>>>>>>>>>>>>>>>>> point, control point associated with the start, the control point
>>>>>>>>>>>>>>>>> associated with the end, and the end point. I have sometimes created
>>>>>>>>>>>>>>>>> OpenSCAD objects by tracing an object in a drawing program and then reading
>>>>>>>>>>>>>>>>> off the coordinates of the various control points.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Note that Bézier functions are not the only mathematical
>>>>>>>>>>>>>>>>> curves. (For instance, they cannot exactly match an arc of a circle.) I'm
>>>>>>>>>>>>>>>>> sure there are numerous other functions that also produce curves.
>>>>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> _______________________________________________
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>>>>>>>>>>>>>>>>
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>>>>>>>>>>>>>>>
>>>>>>>>>>>>>> _______________________________________________
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>>>>>>>>>>>>> _______________________________________________
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DM
Douglas Miller
Mon, Aug 8, 2022 2:32 AM
That is correct: the shape Brian wants is not an ellipse, but (probably)
a hyperellipse.
The standard equation of an ellipse is (x/a)^2 + (y/b)^2 = 1 where a, b > 0
By contrast, the standard equation of a hyperellipse is |x/a|^n +
|y/b|^n = 1 where a, b > 0 and n > 2
This can be generated by this function
function f(i,a,b,n) =
let(theta=i*360)
[a * cos(theta), sign(180-theta) * b *
pow((1-pow(abs(cos(theta)),n)),1/n) ];
For example:
points = [ for(i=[0:0.01:1]) f(i,12,7,2.5) ]; // semi-major axis = 12,
semi-minor axis = 7, n = 2.5
polygon(points);
I imagine that Brian will want to use a value of n that is only slightly
greater than 2. A good starting value for experimentation appears to be
n = 2.25.
On 8/2/2022 8:04 PM, Jordan Brown wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
[...]
You're taking a circle and stretching it; the result of that
stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can chime in -
but I think the shape you are asking for would not be an ellipse.
That is correct: the shape Brian wants is not an ellipse, but (probably)
a hyperellipse.
The standard equation of an ellipse is (x/a)^2 + (y/b)^2 = 1 where a, b > 0
By contrast, the standard equation of a hyperellipse is |x/a|^n +
|y/b|^n = 1 where a, b > 0 and n > 2
This can be generated by this function
function f(i,a,b,n) =
let(theta=i*360)
[a * cos(theta), sign(180-theta) * b *
pow((1-pow(abs(cos(theta)),n)),1/n) ];
For example:
points = [ for(i=[0:0.01:1]) f(i,12,7,2.5) ]; // semi-major axis = 12,
semi-minor axis = 7, n = 2.5
polygon(points);
I imagine that Brian will want to use a value of n that is only slightly
greater than 2. A good starting value for experimentation appears to be
n = 2.25.
On 8/2/2022 8:04 PM, Jordan Brown wrote:
> On 8/2/2022 4:46 PM, Brian Allen wrote:
>> I created an oval using:
>> inMm = 25.4;
>> fullW = 35.5 * inMm;
>> fullL = 54.5 * inMm;
>> resize([fullW, fullL, 1])
>> circle(d=fullW);
> [...]
> You're taking a circle and stretching it; the result of that
> stretching is an ellipse.
>
> I'm not 100% sure - maybe one of the real math wizards can chime in -
> but I think the shape you are asking for would not be an ellipse.
AM
Adrian Mariano
Mon, Aug 8, 2022 2:41 AM
Now I understand your idea. I think it should be possible to make that
work for the case of N=4k+2. It does seem like there are some
complications, like you try to extend the segment and the length is closer
than the tangent, so it's impossible. Also things may be complicated if
the algorithm tries to go around the tip of the ellipse (that is, if the
path is too long).
For finding a tangent to an ellipse, transform the problem to the circle,
solve, and then transform back. In BOSL2:
function ellipse_tangents(a,b,pt) =
let(
r=a,
scaledpt = yscale(a/b,pt),
tangents = circle_point_tangents(r,[0,0],scaledpt)
)
yscale(b/a,tangents);
On Sun, Aug 7, 2022 at 9:54 PM Father Horton fatherhorton@gmail.com wrote:
I did that wrong. It's not (0, b) to (0, L/2). It's (0, b) to (L/2, b).
That's always outside the ellipse.
I will look at the second problem as soon as I figure out how to calculate
a tangent line to an ellipse containing a point not on the ellipse. Google
is not helping much.
On Sun, Aug 7, 2022 at 8:32 PM Adrian Mariano avm4@cornell.edu wrote:
Father Horton,
I'm having trouble understanding your proposed method. A segment from
(0,b) to (0,L/2) might be inside the ellipse depending on N, and then there
won't be a tangent.
It also sounds like your idea is somewhat similar to Jordan's idea:
basically start at one side and work around the ellipse and see if it
works. If not, make a correction. The problem with this idea is that
there are two parameters for the first segment, so how do you make a
correction? (The two parameters are the location of the endpoint of the
first segment and its length.)
On Sun, Aug 7, 2022 at 5:33 PM Father Horton fatherhorton@gmail.com
wrote:
How about this to construct N equal length segments around an ellipse
with axes a and b:
- Select an initial segment length estimate (e.g., Ramanujan's formula
/ N) = L
- Construct a half-segment from (0, b) to (0, L/2).
- Find a line tangent to the ellipse and passing through (0, L/2).
- Find the point on the tangent line that is L away from (0, L/2).
- Repeat (3) substituting the new point for (0, L/2). until you have N
segments.
If everything worked perfectly, the last point found would be (0, -L/2).
If there's a gap, increase L and try again. If there's overlap, decrease L
and try again.
On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano rcmpersiano@gmail.com
wrote:
Adrian, why to use a (either inscribed or circumscribed) polygon with
equal side as an approximation of an ellipsis? It doesn't seem to be a good
approximation of it. I get a better approximation by taking an angle
uniform sample of the curve in the parameter space.
[image: Uniformly_sampled_ellipsis.png]
Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano avm4@cornell.edu
escreveu:
Circumscribed means tangents to points on the ellipse. But there are
no arc lengths (equal or unequal) in this problem, nor equally spaced
angles. The angles of each segment will be different. But the lengths of
the segments are not the same as the arc lengths, which is why arc length
does not matter. Here's a solution to the N=6 polygon sort of inspired by
the idea Jordan suggested, where I adjust the first segment until the
second segment has the right length. The rest of the solution is then
obtained using symmetry, which means there is only a single unknown that
can simply be solved for. But this doesn't generalize to higher N.
[image: image.png]
include<BOSL2/std.scad>
// ellipse semiaxes
// If a is much smaller than b it will fail
a=5;
b=2;
// Find point on ellipse at specified x coordinate
function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
// Find tangent line to ellipse at point (u,w)
// and determine the x value where that line has
// given y value.
function findx(a,b,u,w,y) =
let(
A = -ub^2/w/a^2,
B = w+(ub)^2/w/a^2
)
(y-B)/A;
// First segment if tangent point is at x=u
function first_seg(a,b,u) =
let(
w = ellipse_pt(a,b,u),
x1 = findx(a,b,u,w,0),
x2 = findx(a,b,u,w,b)
)
[[x1,0],[x2,b]];
err = function (u) let(pt = first_seg(a,b,u))
2*pt[1].x - norm(pt[0]-pt[1]);
x = root_find(err,.2a,.9a);
first = first_seg(a,b,x);
top = concat(first,reverse(xflip(first)));
poly = concat(top, reverse(yflip(list_tail(top))));
stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
color("red")stroke(poly,width=.05,closed=true);
// Verify equality of segment lengths
echo(path_segment_lengths(poly,closed=true));
On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
Circumscribed means tangents from points on ellipse
How can the tangents from equally spaced arc lengths or equally
spaced angles be equal for an ellipse
You need to change the arc lengths dynamically and also accurately
estimate the segment length which fits the ellipse .
On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, avm4@cornell.edu
wrote:
Your intuition is incorrect: a circle scaled in one direction IS
an ellipse. The motivation for being interested in circumscribing is to
make holes, though the constraint of equal length sides is probably not
what you'd want for that purpose. You can make a nonuniform circumscribing
polygon for an ellipse by simply making a polygon that circumscribes a
circle and scaling it.
On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson bob@rjcarlson.com
wrote:
Seems like I remember Nop Head talking about using circumscribed
holes for screws and even rotating the polygons to produce a more even
hole.
If an ellipse is the nominal shape, the using a circumscribed
polygon for a hole and an inscribed polygon for a solid might make sense.
Btw, can someone confirm my intuition that a circle scaled in one
dimension is not an ellipse?
-Bob
Tucson AZ
On Aug 5, 2022, at 15:38, Adrian Mariano avm4@cornell.edu wrote:
I don't have a specific need for a circumscribing elliptical
polygon with equal sides. I was writing code to make ellipses and it
seemed like equal side approximations would be nice, so I solved the case
for inscribed, but couldn't solve the circumscribed case. So it's an
unsolved puzzle I've got that this discussion reminded me of.
On Fri, Aug 5, 2022 at 6:33 PM nop head nop.head@gmail.com wrote:
A long discussion but I have to ask why do you need an elliptical
polygon with equal tangential sides?
On Fri, 5 Aug 2022, 21:58 Adrian Mariano, avm4@cornell.edu
wrote:
You think that a circumscribing polygon with equal length
segments does not exist?? I'm skeptical of this claim.
For circumscribed, it's not so hard. You start with a
uniform-in-angle point distribution, compute the segment lengths, compare
to the mean segment length to get an error, and then compute an adjustment
based on the error. It converges quickly.
So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
5.57297, 5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but surely good
enough.
<image.png>
The code for this is in BOSL2 in shapes2d.scad. A similar
algorithm should exist for the circumscribed case, but as I said, my
efforts to find it have failed.
On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
I think circumscribed polygon segments cannot be equal lengths.
inscribed ones may be approximately equal, but would be a very
tough problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano avm4@cornell.edu
wrote:
I think you need to be a little careful about defining what you
want to do. If you want to approximate an ellipse with points that are
distributed uniformly over the ellipse, that could be done by approximating
the arc length of the ellipse to find the necessary points---e.g. advance
distance d along the ellipse at each step. But if you want to approximate
an ellipse with a polygon with equal length sides, that approach does not
do it, because the length of the segments will not be constant as the span
sections of the ellipse with differing curvature. And since the perimeter
of the desired polygon is unknown, you can't just step it out analogously.
The circle doesn't present this challenge due to having constant
curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
fatherhorton@gmail.com> wrote:
I think our emails crossed. I was addressing how to
approximate an ellipse with constant-length segments in general, not how to
solve your problem. When I had to do something similar, I used polar
coordinates and stepped up the value of theta until the distance from the
previous point was correct. (In the constant-length version, I'd keep a
running total of the maximum desired length so that round-off errors don't
accumulate.)
Figuring out the circumscribed polygon is, as you note, a
greater challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <
avm4@cornell.edu> wrote:
Eh, there's no such thing as "brute force" in this
situation. Step through what with sufficiently small steps? Of course
an iterative algorithm is necessary. This is true for most interesting
problems. No big deal. Show me an iterative algorithm that converges to
the circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
fatherhorton@gmail.com> wrote:
Except by brute force (step through it with sufficiently
small steps), it’s not possible because there’s no closed form expression
for the distance between two points on a general ellipse (as opposed to
special cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <
avm4@cornell.edu> wrote:
Here's a challenge for you: draw an "ellipse" polygon with
N segments that are all the same length that circumscribes a perfect
ellipse with given semiaxes a and b. I tried to solve this problem and
could not find a solution. I can draw an inscribed N segment "ellipse"
polygon with all segments the same length---though it's not easy---but the
circumscribed case eludes me. I think of the inscribed case as, every
vertex is located on the ellipse and the circumscribed case, every segment
is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and semi-minor
axis "r1" and "r2" respectively and with center "cp" and number of segment
"s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews
faster in complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of
that stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards
can chime in - but I think the shape you are asking for would not be an
ellipse.
I believe that an ellipse is fully defined by its major
(long) and minor (short) axis dimensions; the other parts of the curve
mathematically flow from those two measurements.
You can make something else, but you can't make it using
this technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some years
ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls
together.
Note that the four coordinates correspond to the
coordinates of the control points in a GUI drawing program: the start
point, control point associated with the start, the control point
associated with the end, and the end point. I have sometimes created
OpenSCAD objects by tracing an object in a drawing program and then reading
off the coordinates of the various control points.
Note that Bézier functions are not the only mathematical
curves. (For instance, they cannot exactly match an arc of a circle.) I'm
sure there are numerous other functions that also produce curves.
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Now I understand your idea. I think it should be possible to make that
work for the case of N=4k+2. It does seem like there are some
complications, like you try to extend the segment and the length is closer
than the tangent, so it's impossible. Also things may be complicated if
the algorithm tries to go around the tip of the ellipse (that is, if the
path is too long).
For finding a tangent to an ellipse, transform the problem to the circle,
solve, and then transform back. In BOSL2:
function ellipse_tangents(a,b,pt) =
let(
r=a,
scaledpt = yscale(a/b,pt),
tangents = circle_point_tangents(r,[0,0],scaledpt)
)
yscale(b/a,tangents);
On Sun, Aug 7, 2022 at 9:54 PM Father Horton <fatherhorton@gmail.com> wrote:
> I did that wrong. It's not (0, b) to (0, L/2). It's (0, b) to (L/2, b).
> That's always outside the ellipse.
>
> I will look at the second problem as soon as I figure out how to calculate
> a tangent line to an ellipse containing a point not on the ellipse. Google
> is not helping much.
>
> On Sun, Aug 7, 2022 at 8:32 PM Adrian Mariano <avm4@cornell.edu> wrote:
>
>> Father Horton,
>>
>> I'm having trouble understanding your proposed method. A segment from
>> (0,b) to (0,L/2) might be inside the ellipse depending on N, and then there
>> won't be a tangent.
>>
>> It also sounds like your idea is somewhat similar to Jordan's idea:
>> basically start at one side and work around the ellipse and see if it
>> works. If not, make a correction. The problem with this idea is that
>> there are two parameters for the first segment, so how do you make a
>> correction? (The two parameters are the location of the endpoint of the
>> first segment and its length.)
>>
>> On Sun, Aug 7, 2022 at 5:33 PM Father Horton <fatherhorton@gmail.com>
>> wrote:
>>
>>> How about this to construct N equal length segments around an ellipse
>>> with axes a and b:
>>>
>>> 1) Select an initial segment length estimate (e.g., Ramanujan's formula
>>> / N) = L
>>> 2) Construct a half-segment from (0, b) to (0, L/2).
>>> 3) Find a line tangent to the ellipse and passing through (0, L/2).
>>> 3) Find the point on the tangent line that is L away from (0, L/2).
>>> 4) Repeat (3) substituting the new point for (0, L/2). until you have N
>>> segments.
>>>
>>> If everything worked perfectly, the last point found would be (0, -L/2).
>>> If there's a gap, increase L and try again. If there's overlap, decrease L
>>> and try again.
>>>
>>> On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano <rcmpersiano@gmail.com>
>>> wrote:
>>>
>>>> Adrian, why to use a (either inscribed or circumscribed) polygon with
>>>> equal side as an approximation of an ellipsis? It doesn't seem to be a good
>>>> approximation of it. I get a better approximation by taking an angle
>>>> uniform sample of the curve in the parameter space.
>>>>
>>>> [image: Uniformly_sampled_ellipsis.png]
>>>>
>>>>
>>>> Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano <avm4@cornell.edu>
>>>> escreveu:
>>>>
>>>>> Circumscribed means tangents to points on the ellipse. But there are
>>>>> no arc lengths (equal or unequal) in this problem, nor equally spaced
>>>>> angles. The angles of each segment will be different. But the lengths of
>>>>> the segments are not the same as the arc lengths, which is why arc length
>>>>> does not matter. Here's a solution to the N=6 polygon sort of inspired by
>>>>> the idea Jordan suggested, where I adjust the first segment until the
>>>>> second segment has the right length. The rest of the solution is then
>>>>> obtained using symmetry, which means there is only a single unknown that
>>>>> can simply be solved for. But this doesn't generalize to higher N.
>>>>>
>>>>> [image: image.png]
>>>>> include<BOSL2/std.scad>
>>>>>
>>>>> // ellipse semiaxes
>>>>> // If a is much smaller than b it will fail
>>>>> a=5;
>>>>> b=2;
>>>>>
>>>>> // Find point on ellipse at specified x coordinate
>>>>> function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
>>>>>
>>>>>
>>>>> // Find tangent line to ellipse at point (u,w)
>>>>> // and determine the x value where that line has
>>>>> // given y value.
>>>>> function findx(a,b,u,w,y) =
>>>>> let(
>>>>> A = -u*b^2/w/a^2,
>>>>> B = w+(u*b)^2/w/a^2
>>>>> )
>>>>> (y-B)/A;
>>>>>
>>>>>
>>>>> // First segment if tangent point is at x=u
>>>>> function first_seg(a,b,u) =
>>>>> let(
>>>>> w = ellipse_pt(a,b,u),
>>>>> x1 = findx(a,b,u,w,0),
>>>>> x2 = findx(a,b,u,w,b)
>>>>> )
>>>>> [[x1,0],[x2,b]];
>>>>>
>>>>>
>>>>> err = function (u) let(pt = first_seg(a,b,u))
>>>>> 2*pt[1].x - norm(pt[0]-pt[1]);
>>>>>
>>>>> x = root_find(err,.2*a,.9*a);
>>>>> first = first_seg(a,b,x);
>>>>> top = concat(first,reverse(xflip(first)));
>>>>> poly = concat(top, reverse(yflip(list_tail(top))));
>>>>>
>>>>> stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
>>>>> color("red")stroke(poly,width=.05,closed=true);
>>>>>
>>>>> // Verify equality of segment lengths
>>>>> echo(path_segment_lengths(poly,closed=true));
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>> On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>
>>>>>> Circumscribed means tangents from points on ellipse
>>>>>> How can the tangents from equally spaced arc lengths or equally
>>>>>> spaced angles be equal for an ellipse
>>>>>> You need to change the arc lengths dynamically and also accurately
>>>>>> estimate the segment length which fits the ellipse .
>>>>>>
>>>>>>
>>>>>>
>>>>>> On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, <avm4@cornell.edu>
>>>>>> wrote:
>>>>>>
>>>>>>> Your intuition is incorrect: a circle scaled in one direction *IS*
>>>>>>> an ellipse. The motivation for being interested in circumscribing is to
>>>>>>> make holes, though the constraint of equal length sides is probably not
>>>>>>> what you'd want for that purpose. You can make a nonuniform circumscribing
>>>>>>> polygon for an ellipse by simply making a polygon that circumscribes a
>>>>>>> circle and scaling it.
>>>>>>>
>>>>>>> On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson <bob@rjcarlson.com>
>>>>>>> wrote:
>>>>>>>
>>>>>>>> Seems like I remember Nop Head talking about using circumscribed
>>>>>>>> holes for screws and even rotating the polygons to produce a more even
>>>>>>>> hole.
>>>>>>>>
>>>>>>>> If an ellipse is the nominal shape, the using a circumscribed
>>>>>>>> polygon for a hole and an inscribed polygon for a solid might make sense.
>>>>>>>>
>>>>>>>> Btw, can someone confirm my intuition that a circle scaled in one
>>>>>>>> dimension is not an ellipse?
>>>>>>>>
>>>>>>>> -Bob
>>>>>>>> Tucson AZ
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>> On Aug 5, 2022, at 15:38, Adrian Mariano <avm4@cornell.edu> wrote:
>>>>>>>>
>>>>>>>> I don't have a specific need for a circumscribing elliptical
>>>>>>>> polygon with equal sides. I was writing code to make ellipses and it
>>>>>>>> seemed like equal side approximations would be nice, so I solved the case
>>>>>>>> for inscribed, but couldn't solve the circumscribed case. So it's an
>>>>>>>> unsolved puzzle I've got that this discussion reminded me of.
>>>>>>>>
>>>>>>>> On Fri, Aug 5, 2022 at 6:33 PM nop head <nop.head@gmail.com> wrote:
>>>>>>>>
>>>>>>>>> A long discussion but I have to ask why do you need an elliptical
>>>>>>>>> polygon with equal tangential sides?
>>>>>>>>>
>>>>>>>>> On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu>
>>>>>>>>> wrote:
>>>>>>>>>
>>>>>>>>>> You think that a circumscribing polygon with equal length
>>>>>>>>>> segments does not exist?? I'm skeptical of this claim.
>>>>>>>>>>
>>>>>>>>>> For circumscribed, it's not so hard. You start with a
>>>>>>>>>> uniform-in-angle point distribution, compute the segment lengths, compare
>>>>>>>>>> to the mean segment length to get an error, and then compute an adjustment
>>>>>>>>>> based on the error. It converges quickly.
>>>>>>>>>>
>>>>>>>>>> So for example, I have segments lengths of:
>>>>>>>>>> ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
>>>>>>>>>> 5.57297, 5.57297]
>>>>>>>>>>
>>>>>>>>>> Maybe not exactly equal, since tolerance is 1e-9, but surely good
>>>>>>>>>> enough.
>>>>>>>>>> <image.png>
>>>>>>>>>>
>>>>>>>>>> The code for this is in BOSL2 in shapes2d.scad. A similar
>>>>>>>>>> algorithm should exist for the circumscribed case, but as I said, my
>>>>>>>>>> efforts to find it have failed.
>>>>>>>>>>
>>>>>>>>>> On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>
>>>>>>>>>>> I think circumscribed polygon segments cannot be equal lengths.
>>>>>>>>>>> inscribed ones may be approximately equal, but would be a very
>>>>>>>>>>> tough problem to solve.
>>>>>>>>>>> I tried something.
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <avm4@cornell.edu>
>>>>>>>>>>> wrote:
>>>>>>>>>>>
>>>>>>>>>>>> I think you need to be a little careful about defining what you
>>>>>>>>>>>> want to do. If you want to approximate an ellipse with points that are
>>>>>>>>>>>> distributed uniformly over the ellipse, that could be done by approximating
>>>>>>>>>>>> the arc length of the ellipse to find the necessary points---e.g. advance
>>>>>>>>>>>> distance d along the ellipse at each step. But if you want to approximate
>>>>>>>>>>>> an ellipse with a polygon with equal length sides, that approach does not
>>>>>>>>>>>> do it, because the length of the segments will not be constant as the span
>>>>>>>>>>>> sections of the ellipse with differing curvature. And since the perimeter
>>>>>>>>>>>> of the desired polygon is unknown, you can't just step it out analogously.
>>>>>>>>>>>> The circle doesn't present this challenge due to having constant
>>>>>>>>>>>> curvature.
>>>>>>>>>>>>
>>>>>>>>>>>> On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
>>>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>>>
>>>>>>>>>>>>> I think our emails crossed. I was addressing how to
>>>>>>>>>>>>> approximate an ellipse with constant-length segments in general, not how to
>>>>>>>>>>>>> solve your problem. When I had to do something similar, I used polar
>>>>>>>>>>>>> coordinates and stepped up the value of theta until the distance from the
>>>>>>>>>>>>> previous point was correct. (In the constant-length version, I'd keep a
>>>>>>>>>>>>> running total of the maximum desired length so that round-off errors don't
>>>>>>>>>>>>> accumulate.)
>>>>>>>>>>>>>
>>>>>>>>>>>>> Figuring out the circumscribed polygon is, as you note, a
>>>>>>>>>>>>> greater challenge. I can visualize how to do it for n = 4, but it doesn't
>>>>>>>>>>>>> extrapolate at all.
>>>>>>>>>>>>>
>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <
>>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> Eh, there's no such thing as "brute force" in this
>>>>>>>>>>>>>> situation. Step through *what* with sufficiently small steps? Of course
>>>>>>>>>>>>>> an iterative algorithm is necessary. This is true for most interesting
>>>>>>>>>>>>>> problems. No big deal. Show me an iterative algorithm that converges to
>>>>>>>>>>>>>> the circumscribing polygon. When I tried to devise such an algorithm I
>>>>>>>>>>>>>> couldn't find one that would converge. Writing an algorithm that would
>>>>>>>>>>>>>> converge to the *inscribed* polygon seemed to be straight forward. Note
>>>>>>>>>>>>>> also that computing arc length of the ellipse doesn't really help here.
>>>>>>>>>>>>>> You can do that by solving an elliptical integral, or summing up small
>>>>>>>>>>>>>> segments. But what's the connection between arc length of the ellipse and
>>>>>>>>>>>>>> a circumscribing polygon with equal segment lengths? If there's a
>>>>>>>>>>>>>> connection, I don't see it.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
>>>>>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Except by brute force (step through it with sufficiently
>>>>>>>>>>>>>>> small steps), it’s not possible because there’s no closed form expression
>>>>>>>>>>>>>>> for the distance between two points on a general ellipse (as opposed to
>>>>>>>>>>>>>>> special cases like a circle).
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <
>>>>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Here's a challenge for you: draw an "ellipse" polygon with
>>>>>>>>>>>>>>>> N segments that are all the same length that circumscribes a perfect
>>>>>>>>>>>>>>>> ellipse with given semiaxes a and b. I tried to solve this problem and
>>>>>>>>>>>>>>>> could not find a solution. I can draw an *inscribed* N segment "ellipse"
>>>>>>>>>>>>>>>> polygon with all segments the same length---though it's not easy---but the
>>>>>>>>>>>>>>>> circumscribed case eludes me. I think of the inscribed case as, every
>>>>>>>>>>>>>>>> vertex is located on the ellipse and the circumscribed case, every segment
>>>>>>>>>>>>>>>> is tangent to the ellipse
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>>>>>>>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> In openscad drawing a ellipse is not too difficult
>>>>>>>>>>>>>>>>> Refer following code
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> function to draw an ellipse with semi-major and semi-minor
>>>>>>>>>>>>>>>>> axis "r1" and "r2" respectively and with center "cp" and number of segment
>>>>>>>>>>>>>>>>> "s"
>>>>>>>>>>>>>>>>> /
>>>>>>>>>>>>>>>>> // example:
>>>>>>>>>>>>>>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> function ellipse(r1,r2,cp,s=30)=
>>>>>>>>>>>>>>>>> let(
>>>>>>>>>>>>>>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>>>>>>>>>>>>>>>> )sec;
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>>>>>>>>>>>>>>>> openscad@jordan.maileater.net> wrote:
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> I created an oval using:
>>>>>>>>>>>>>>>>>> inMm = 25.4;
>>>>>>>>>>>>>>>>>> fullW = 35.5 * inMm;
>>>>>>>>>>>>>>>>>> fullL = 54.5 * inMm;
>>>>>>>>>>>>>>>>>> resize([fullW, fullL, 1])
>>>>>>>>>>>>>>>>>> circle(d=fullW);
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> Tip: use scale() rather than resize(); it previews
>>>>>>>>>>>>>>>>>> faster in complicated cases.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> You're taking a circle and stretching it; the result of
>>>>>>>>>>>>>>>>>> that stretching is an ellipse.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> I'm not 100% sure - maybe one of the real math wizards
>>>>>>>>>>>>>>>>>> can chime in - but I think the shape you are asking for would not be an
>>>>>>>>>>>>>>>>>> ellipse.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> I believe that an ellipse is fully defined by its major
>>>>>>>>>>>>>>>>>> (long) and minor (short) axis dimensions; the other parts of the curve
>>>>>>>>>>>>>>>>>> mathematically flow from those two measurements.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> You can make something else, but you can't make it using
>>>>>>>>>>>>>>>>>> this technique.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> You can use a Bézier function.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> Using BOSL2:
>>>>>>>>>>>>>>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> DIY, here's a function that I cobbled together some years
>>>>>>>>>>>>>>>>>> ago:
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>>>>>>>>>>>>>>>> // but that yielded either single points or a raft of triangles;
>>>>>>>>>>>>>>>>>> // this yields a vector of points that you can then concatenate
>>>>>>>>>>>>>>>>>> // with other pieces to form a single polygon.
>>>>>>>>>>>>>>>>>> // If we were really clever, I think it would be possible to
>>>>>>>>>>>>>>>>>> // automatically space the output points based on how linear
>>>>>>>>>>>>>>>>>> // the curve is at that point. But right now I'm not that clever.
>>>>>>>>>>>>>>>>>> function BEZ03(u) = pow((1-u), 3);
>>>>>>>>>>>>>>>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>>>>>>>>>>>>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>>>>>>>>>>>>>>>> function BEZ33(u) = pow(u,3);
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>>>>>>>>>>>>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>>>>>>>>>>>>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> // p0 - start point
>>>>>>>>>>>>>>>>>> // p1 - control point 1, line departs p0 headed this way
>>>>>>>>>>>>>>>>>> // p2 - control point 2, line arrives at p3 from this way
>>>>>>>>>>>>>>>>>> // p3 - end point
>>>>>>>>>>>>>>>>>> // segs - number of segments
>>>>>>>>>>>>>>>>>> function bez(p0, p1, p2, p3, segs) = [
>>>>>>>>>>>>>>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>>>>>>>>>>>>>>>> ];
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> It is that final function bez() that I actually call, e.g.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> often concatenating the results of several bez() calls
>>>>>>>>>>>>>>>>>> together.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> Note that the four coordinates correspond to the
>>>>>>>>>>>>>>>>>> coordinates of the control points in a GUI drawing program: the start
>>>>>>>>>>>>>>>>>> point, control point associated with the start, the control point
>>>>>>>>>>>>>>>>>> associated with the end, and the end point. I have sometimes created
>>>>>>>>>>>>>>>>>> OpenSCAD objects by tracing an object in a drawing program and then reading
>>>>>>>>>>>>>>>>>> off the coordinates of the various control points.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> Note that Bézier functions are not the only mathematical
>>>>>>>>>>>>>>>>>> curves. (For instance, they cannot exactly match an arc of a circle.) I'm
>>>>>>>>>>>>>>>>>> sure there are numerous other functions that also produce curves.
>>>>>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> _______________________________________________
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>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> _______________________________________________
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>>>>>>>>>>>
>>>>>>>>>> _______________________________________________
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>>>>>>>>>>
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>>>>>>>>>
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>
FH
Father Horton
Mon, Aug 8, 2022 2:43 AM
Google showed me that method when I came up with the right terms. But it's
too late at night for me to try to implement it. Maybe tomorrow.
On Sun, Aug 7, 2022 at 9:41 PM Adrian Mariano avm4@cornell.edu wrote:
Now I understand your idea. I think it should be possible to make that
work for the case of N=4k+2. It does seem like there are some
complications, like you try to extend the segment and the length is closer
than the tangent, so it's impossible. Also things may be complicated if
the algorithm tries to go around the tip of the ellipse (that is, if the
path is too long).
For finding a tangent to an ellipse, transform the problem to the circle,
solve, and then transform back. In BOSL2:
function ellipse_tangents(a,b,pt) =
let(
r=a,
scaledpt = yscale(a/b,pt),
tangents = circle_point_tangents(r,[0,0],scaledpt)
)
yscale(b/a,tangents);
On Sun, Aug 7, 2022 at 9:54 PM Father Horton fatherhorton@gmail.com
wrote:
I did that wrong. It's not (0, b) to (0, L/2). It's (0, b) to (L/2, b).
That's always outside the ellipse.
I will look at the second problem as soon as I figure out how to
calculate a tangent line to an ellipse containing a point not on the
ellipse. Google is not helping much.
On Sun, Aug 7, 2022 at 8:32 PM Adrian Mariano avm4@cornell.edu wrote:
Father Horton,
I'm having trouble understanding your proposed method. A segment from
(0,b) to (0,L/2) might be inside the ellipse depending on N, and then there
won't be a tangent.
It also sounds like your idea is somewhat similar to Jordan's idea:
basically start at one side and work around the ellipse and see if it
works. If not, make a correction. The problem with this idea is that
there are two parameters for the first segment, so how do you make a
correction? (The two parameters are the location of the endpoint of the
first segment and its length.)
On Sun, Aug 7, 2022 at 5:33 PM Father Horton fatherhorton@gmail.com
wrote:
How about this to construct N equal length segments around an ellipse
with axes a and b:
- Select an initial segment length estimate (e.g., Ramanujan's formula
/ N) = L
- Construct a half-segment from (0, b) to (0, L/2).
- Find a line tangent to the ellipse and passing through (0, L/2).
- Find the point on the tangent line that is L away from (0, L/2).
- Repeat (3) substituting the new point for (0, L/2). until you have N
segments.
If everything worked perfectly, the last point found would be (0,
-L/2). If there's a gap, increase L and try again. If there's overlap,
decrease L and try again.
On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano rcmpersiano@gmail.com
wrote:
Adrian, why to use a (either inscribed or circumscribed) polygon with
equal side as an approximation of an ellipsis? It doesn't seem to be a good
approximation of it. I get a better approximation by taking an angle
uniform sample of the curve in the parameter space.
[image: Uniformly_sampled_ellipsis.png]
Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano avm4@cornell.edu
escreveu:
Circumscribed means tangents to points on the ellipse. But there are
no arc lengths (equal or unequal) in this problem, nor equally spaced
angles. The angles of each segment will be different. But the lengths of
the segments are not the same as the arc lengths, which is why arc length
does not matter. Here's a solution to the N=6 polygon sort of inspired by
the idea Jordan suggested, where I adjust the first segment until the
second segment has the right length. The rest of the solution is then
obtained using symmetry, which means there is only a single unknown that
can simply be solved for. But this doesn't generalize to higher N.
[image: image.png]
include<BOSL2/std.scad>
// ellipse semiaxes
// If a is much smaller than b it will fail
a=5;
b=2;
// Find point on ellipse at specified x coordinate
function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
// Find tangent line to ellipse at point (u,w)
// and determine the x value where that line has
// given y value.
function findx(a,b,u,w,y) =
let(
A = -ub^2/w/a^2,
B = w+(ub)^2/w/a^2
)
(y-B)/A;
// First segment if tangent point is at x=u
function first_seg(a,b,u) =
let(
w = ellipse_pt(a,b,u),
x1 = findx(a,b,u,w,0),
x2 = findx(a,b,u,w,b)
)
[[x1,0],[x2,b]];
err = function (u) let(pt = first_seg(a,b,u))
2*pt[1].x - norm(pt[0]-pt[1]);
x = root_find(err,.2a,.9a);
first = first_seg(a,b,x);
top = concat(first,reverse(xflip(first)));
poly = concat(top, reverse(yflip(list_tail(top))));
stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
color("red")stroke(poly,width=.05,closed=true);
// Verify equality of segment lengths
echo(path_segment_lengths(poly,closed=true));
On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
Circumscribed means tangents from points on ellipse
How can the tangents from equally spaced arc lengths or equally
spaced angles be equal for an ellipse
You need to change the arc lengths dynamically and also accurately
estimate the segment length which fits the ellipse .
On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, avm4@cornell.edu
wrote:
Your intuition is incorrect: a circle scaled in one direction IS
an ellipse. The motivation for being interested in circumscribing is to
make holes, though the constraint of equal length sides is probably not
what you'd want for that purpose. You can make a nonuniform circumscribing
polygon for an ellipse by simply making a polygon that circumscribes a
circle and scaling it.
On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson bob@rjcarlson.com
wrote:
Seems like I remember Nop Head talking about using circumscribed
holes for screws and even rotating the polygons to produce a more even
hole.
If an ellipse is the nominal shape, the using a circumscribed
polygon for a hole and an inscribed polygon for a solid might make sense.
Btw, can someone confirm my intuition that a circle scaled in one
dimension is not an ellipse?
-Bob
Tucson AZ
On Aug 5, 2022, at 15:38, Adrian Mariano avm4@cornell.edu wrote:
I don't have a specific need for a circumscribing elliptical
polygon with equal sides. I was writing code to make ellipses and it
seemed like equal side approximations would be nice, so I solved the case
for inscribed, but couldn't solve the circumscribed case. So it's an
unsolved puzzle I've got that this discussion reminded me of.
On Fri, Aug 5, 2022 at 6:33 PM nop head nop.head@gmail.com
wrote:
A long discussion but I have to ask why do you need an elliptical
polygon with equal tangential sides?
On Fri, 5 Aug 2022, 21:58 Adrian Mariano, avm4@cornell.edu
wrote:
You think that a circumscribing polygon with equal length
segments does not exist?? I'm skeptical of this claim.
For circumscribed, it's not so hard. You start with a
uniform-in-angle point distribution, compute the segment lengths, compare
to the mean segment length to get an error, and then compute an adjustment
based on the error. It converges quickly.
So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
5.57297, 5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but surely
good enough.
<image.png>
The code for this is in BOSL2 in shapes2d.scad. A similar
algorithm should exist for the circumscribed case, but as I said, my
efforts to find it have failed.
On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
I think circumscribed polygon segments cannot be equal lengths.
inscribed ones may be approximately equal, but would be a very
tough problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano avm4@cornell.edu
wrote:
I think you need to be a little careful about defining what
you want to do. If you want to approximate an ellipse with points that are
distributed uniformly over the ellipse, that could be done by approximating
the arc length of the ellipse to find the necessary points---e.g. advance
distance d along the ellipse at each step. But if you want to approximate
an ellipse with a polygon with equal length sides, that approach does not
do it, because the length of the segments will not be constant as the span
sections of the ellipse with differing curvature. And since the perimeter
of the desired polygon is unknown, you can't just step it out analogously.
The circle doesn't present this challenge due to having constant
curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
fatherhorton@gmail.com> wrote:
I think our emails crossed. I was addressing how to
approximate an ellipse with constant-length segments in general, not how to
solve your problem. When I had to do something similar, I used polar
coordinates and stepped up the value of theta until the distance from the
previous point was correct. (In the constant-length version, I'd keep a
running total of the maximum desired length so that round-off errors don't
accumulate.)
Figuring out the circumscribed polygon is, as you note, a
greater challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <
avm4@cornell.edu> wrote:
Eh, there's no such thing as "brute force" in this
situation. Step through what with sufficiently small steps? Of course
an iterative algorithm is necessary. This is true for most interesting
problems. No big deal. Show me an iterative algorithm that converges to
the circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
fatherhorton@gmail.com> wrote:
Except by brute force (step through it with sufficiently
small steps), it’s not possible because there’s no closed form expression
for the distance between two points on a general ellipse (as opposed to
special cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <
avm4@cornell.edu> wrote:
Here's a challenge for you: draw an "ellipse" polygon
with N segments that are all the same length that circumscribes a perfect
ellipse with given semiaxes a and b. I tried to solve this problem and
could not find a solution. I can draw an inscribed N segment "ellipse"
polygon with all segments the same length---though it's not easy---but the
circumscribed case eludes me. I think of the inscribed case as, every
vertex is located on the ellipse and the circumscribed case, every segment
is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and
semi-minor axis "r1" and "r2" respectively and with center "cp" and number
of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews
faster in complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of
that stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards
can chime in - but I think the shape you are asking for would not be an
ellipse.
I believe that an ellipse is fully defined by its major
(long) and minor (short) axis dimensions; the other parts of the curve
mathematically flow from those two measurements.
You can make something else, but you can't make it using
this technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some
years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call,
e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls
together.
Note that the four coordinates correspond to the
coordinates of the control points in a GUI drawing program: the start
point, control point associated with the start, the control point
associated with the end, and the end point. I have sometimes created
OpenSCAD objects by tracing an object in a drawing program and then reading
off the coordinates of the various control points.
Note that Bézier functions are not the only mathematical
curves. (For instance, they cannot exactly match an arc of a circle.) I'm
sure there are numerous other functions that also produce curves.
OpenSCAD mailing list
To unsubscribe send an email to
discuss-leave@lists.openscad.org
Google showed me that method when I came up with the right terms. But it's
too late at night for me to try to implement it. Maybe tomorrow.
On Sun, Aug 7, 2022 at 9:41 PM Adrian Mariano <avm4@cornell.edu> wrote:
> Now I understand your idea. I think it should be possible to make that
> work for the case of N=4k+2. It does seem like there are some
> complications, like you try to extend the segment and the length is closer
> than the tangent, so it's impossible. Also things may be complicated if
> the algorithm tries to go around the tip of the ellipse (that is, if the
> path is too long).
>
> For finding a tangent to an ellipse, transform the problem to the circle,
> solve, and then transform back. In BOSL2:
>
> function ellipse_tangents(a,b,pt) =
> let(
> r=a,
> scaledpt = yscale(a/b,pt),
> tangents = circle_point_tangents(r,[0,0],scaledpt)
> )
> yscale(b/a,tangents);
>
> On Sun, Aug 7, 2022 at 9:54 PM Father Horton <fatherhorton@gmail.com>
> wrote:
>
>> I did that wrong. It's not (0, b) to (0, L/2). It's (0, b) to (L/2, b).
>> That's always outside the ellipse.
>>
>> I will look at the second problem as soon as I figure out how to
>> calculate a tangent line to an ellipse containing a point not on the
>> ellipse. Google is not helping much.
>>
>> On Sun, Aug 7, 2022 at 8:32 PM Adrian Mariano <avm4@cornell.edu> wrote:
>>
>>> Father Horton,
>>>
>>> I'm having trouble understanding your proposed method. A segment from
>>> (0,b) to (0,L/2) might be inside the ellipse depending on N, and then there
>>> won't be a tangent.
>>>
>>> It also sounds like your idea is somewhat similar to Jordan's idea:
>>> basically start at one side and work around the ellipse and see if it
>>> works. If not, make a correction. The problem with this idea is that
>>> there are two parameters for the first segment, so how do you make a
>>> correction? (The two parameters are the location of the endpoint of the
>>> first segment and its length.)
>>>
>>> On Sun, Aug 7, 2022 at 5:33 PM Father Horton <fatherhorton@gmail.com>
>>> wrote:
>>>
>>>> How about this to construct N equal length segments around an ellipse
>>>> with axes a and b:
>>>>
>>>> 1) Select an initial segment length estimate (e.g., Ramanujan's formula
>>>> / N) = L
>>>> 2) Construct a half-segment from (0, b) to (0, L/2).
>>>> 3) Find a line tangent to the ellipse and passing through (0, L/2).
>>>> 3) Find the point on the tangent line that is L away from (0, L/2).
>>>> 4) Repeat (3) substituting the new point for (0, L/2). until you have N
>>>> segments.
>>>>
>>>> If everything worked perfectly, the last point found would be (0,
>>>> -L/2). If there's a gap, increase L and try again. If there's overlap,
>>>> decrease L and try again.
>>>>
>>>> On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano <rcmpersiano@gmail.com>
>>>> wrote:
>>>>
>>>>> Adrian, why to use a (either inscribed or circumscribed) polygon with
>>>>> equal side as an approximation of an ellipsis? It doesn't seem to be a good
>>>>> approximation of it. I get a better approximation by taking an angle
>>>>> uniform sample of the curve in the parameter space.
>>>>>
>>>>> [image: Uniformly_sampled_ellipsis.png]
>>>>>
>>>>>
>>>>> Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano <avm4@cornell.edu>
>>>>> escreveu:
>>>>>
>>>>>> Circumscribed means tangents to points on the ellipse. But there are
>>>>>> no arc lengths (equal or unequal) in this problem, nor equally spaced
>>>>>> angles. The angles of each segment will be different. But the lengths of
>>>>>> the segments are not the same as the arc lengths, which is why arc length
>>>>>> does not matter. Here's a solution to the N=6 polygon sort of inspired by
>>>>>> the idea Jordan suggested, where I adjust the first segment until the
>>>>>> second segment has the right length. The rest of the solution is then
>>>>>> obtained using symmetry, which means there is only a single unknown that
>>>>>> can simply be solved for. But this doesn't generalize to higher N.
>>>>>>
>>>>>> [image: image.png]
>>>>>> include<BOSL2/std.scad>
>>>>>>
>>>>>> // ellipse semiaxes
>>>>>> // If a is much smaller than b it will fail
>>>>>> a=5;
>>>>>> b=2;
>>>>>>
>>>>>> // Find point on ellipse at specified x coordinate
>>>>>> function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
>>>>>>
>>>>>>
>>>>>> // Find tangent line to ellipse at point (u,w)
>>>>>> // and determine the x value where that line has
>>>>>> // given y value.
>>>>>> function findx(a,b,u,w,y) =
>>>>>> let(
>>>>>> A = -u*b^2/w/a^2,
>>>>>> B = w+(u*b)^2/w/a^2
>>>>>> )
>>>>>> (y-B)/A;
>>>>>>
>>>>>>
>>>>>> // First segment if tangent point is at x=u
>>>>>> function first_seg(a,b,u) =
>>>>>> let(
>>>>>> w = ellipse_pt(a,b,u),
>>>>>> x1 = findx(a,b,u,w,0),
>>>>>> x2 = findx(a,b,u,w,b)
>>>>>> )
>>>>>> [[x1,0],[x2,b]];
>>>>>>
>>>>>>
>>>>>> err = function (u) let(pt = first_seg(a,b,u))
>>>>>> 2*pt[1].x - norm(pt[0]-pt[1]);
>>>>>>
>>>>>> x = root_find(err,.2*a,.9*a);
>>>>>> first = first_seg(a,b,x);
>>>>>> top = concat(first,reverse(xflip(first)));
>>>>>> poly = concat(top, reverse(yflip(list_tail(top))));
>>>>>>
>>>>>> stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
>>>>>> color("red")stroke(poly,width=.05,closed=true);
>>>>>>
>>>>>> // Verify equality of segment lengths
>>>>>> echo(path_segment_lengths(poly,closed=true));
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>> On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>
>>>>>>> Circumscribed means tangents from points on ellipse
>>>>>>> How can the tangents from equally spaced arc lengths or equally
>>>>>>> spaced angles be equal for an ellipse
>>>>>>> You need to change the arc lengths dynamically and also accurately
>>>>>>> estimate the segment length which fits the ellipse .
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, <avm4@cornell.edu>
>>>>>>> wrote:
>>>>>>>
>>>>>>>> Your intuition is incorrect: a circle scaled in one direction *IS*
>>>>>>>> an ellipse. The motivation for being interested in circumscribing is to
>>>>>>>> make holes, though the constraint of equal length sides is probably not
>>>>>>>> what you'd want for that purpose. You can make a nonuniform circumscribing
>>>>>>>> polygon for an ellipse by simply making a polygon that circumscribes a
>>>>>>>> circle and scaling it.
>>>>>>>>
>>>>>>>> On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson <bob@rjcarlson.com>
>>>>>>>> wrote:
>>>>>>>>
>>>>>>>>> Seems like I remember Nop Head talking about using circumscribed
>>>>>>>>> holes for screws and even rotating the polygons to produce a more even
>>>>>>>>> hole.
>>>>>>>>>
>>>>>>>>> If an ellipse is the nominal shape, the using a circumscribed
>>>>>>>>> polygon for a hole and an inscribed polygon for a solid might make sense.
>>>>>>>>>
>>>>>>>>> Btw, can someone confirm my intuition that a circle scaled in one
>>>>>>>>> dimension is not an ellipse?
>>>>>>>>>
>>>>>>>>> -Bob
>>>>>>>>> Tucson AZ
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> On Aug 5, 2022, at 15:38, Adrian Mariano <avm4@cornell.edu> wrote:
>>>>>>>>>
>>>>>>>>> I don't have a specific need for a circumscribing elliptical
>>>>>>>>> polygon with equal sides. I was writing code to make ellipses and it
>>>>>>>>> seemed like equal side approximations would be nice, so I solved the case
>>>>>>>>> for inscribed, but couldn't solve the circumscribed case. So it's an
>>>>>>>>> unsolved puzzle I've got that this discussion reminded me of.
>>>>>>>>>
>>>>>>>>> On Fri, Aug 5, 2022 at 6:33 PM nop head <nop.head@gmail.com>
>>>>>>>>> wrote:
>>>>>>>>>
>>>>>>>>>> A long discussion but I have to ask why do you need an elliptical
>>>>>>>>>> polygon with equal tangential sides?
>>>>>>>>>>
>>>>>>>>>> On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu>
>>>>>>>>>> wrote:
>>>>>>>>>>
>>>>>>>>>>> You think that a circumscribing polygon with equal length
>>>>>>>>>>> segments does not exist?? I'm skeptical of this claim.
>>>>>>>>>>>
>>>>>>>>>>> For circumscribed, it's not so hard. You start with a
>>>>>>>>>>> uniform-in-angle point distribution, compute the segment lengths, compare
>>>>>>>>>>> to the mean segment length to get an error, and then compute an adjustment
>>>>>>>>>>> based on the error. It converges quickly.
>>>>>>>>>>>
>>>>>>>>>>> So for example, I have segments lengths of:
>>>>>>>>>>> ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
>>>>>>>>>>> 5.57297, 5.57297]
>>>>>>>>>>>
>>>>>>>>>>> Maybe not exactly equal, since tolerance is 1e-9, but surely
>>>>>>>>>>> good enough.
>>>>>>>>>>> <image.png>
>>>>>>>>>>>
>>>>>>>>>>> The code for this is in BOSL2 in shapes2d.scad. A similar
>>>>>>>>>>> algorithm should exist for the circumscribed case, but as I said, my
>>>>>>>>>>> efforts to find it have failed.
>>>>>>>>>>>
>>>>>>>>>>> On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
>>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>>
>>>>>>>>>>>> I think circumscribed polygon segments cannot be equal lengths.
>>>>>>>>>>>> inscribed ones may be approximately equal, but would be a very
>>>>>>>>>>>> tough problem to solve.
>>>>>>>>>>>> I tried something.
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <avm4@cornell.edu>
>>>>>>>>>>>> wrote:
>>>>>>>>>>>>
>>>>>>>>>>>>> I think you need to be a little careful about defining what
>>>>>>>>>>>>> you want to do. If you want to approximate an ellipse with points that are
>>>>>>>>>>>>> distributed uniformly over the ellipse, that could be done by approximating
>>>>>>>>>>>>> the arc length of the ellipse to find the necessary points---e.g. advance
>>>>>>>>>>>>> distance d along the ellipse at each step. But if you want to approximate
>>>>>>>>>>>>> an ellipse with a polygon with equal length sides, that approach does not
>>>>>>>>>>>>> do it, because the length of the segments will not be constant as the span
>>>>>>>>>>>>> sections of the ellipse with differing curvature. And since the perimeter
>>>>>>>>>>>>> of the desired polygon is unknown, you can't just step it out analogously.
>>>>>>>>>>>>> The circle doesn't present this challenge due to having constant
>>>>>>>>>>>>> curvature.
>>>>>>>>>>>>>
>>>>>>>>>>>>> On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
>>>>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> I think our emails crossed. I was addressing how to
>>>>>>>>>>>>>> approximate an ellipse with constant-length segments in general, not how to
>>>>>>>>>>>>>> solve your problem. When I had to do something similar, I used polar
>>>>>>>>>>>>>> coordinates and stepped up the value of theta until the distance from the
>>>>>>>>>>>>>> previous point was correct. (In the constant-length version, I'd keep a
>>>>>>>>>>>>>> running total of the maximum desired length so that round-off errors don't
>>>>>>>>>>>>>> accumulate.)
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Figuring out the circumscribed polygon is, as you note, a
>>>>>>>>>>>>>> greater challenge. I can visualize how to do it for n = 4, but it doesn't
>>>>>>>>>>>>>> extrapolate at all.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <
>>>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Eh, there's no such thing as "brute force" in this
>>>>>>>>>>>>>>> situation. Step through *what* with sufficiently small steps? Of course
>>>>>>>>>>>>>>> an iterative algorithm is necessary. This is true for most interesting
>>>>>>>>>>>>>>> problems. No big deal. Show me an iterative algorithm that converges to
>>>>>>>>>>>>>>> the circumscribing polygon. When I tried to devise such an algorithm I
>>>>>>>>>>>>>>> couldn't find one that would converge. Writing an algorithm that would
>>>>>>>>>>>>>>> converge to the *inscribed* polygon seemed to be straight forward. Note
>>>>>>>>>>>>>>> also that computing arc length of the ellipse doesn't really help here.
>>>>>>>>>>>>>>> You can do that by solving an elliptical integral, or summing up small
>>>>>>>>>>>>>>> segments. But what's the connection between arc length of the ellipse and
>>>>>>>>>>>>>>> a circumscribing polygon with equal segment lengths? If there's a
>>>>>>>>>>>>>>> connection, I don't see it.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
>>>>>>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Except by brute force (step through it with sufficiently
>>>>>>>>>>>>>>>> small steps), it’s not possible because there’s no closed form expression
>>>>>>>>>>>>>>>> for the distance between two points on a general ellipse (as opposed to
>>>>>>>>>>>>>>>> special cases like a circle).
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <
>>>>>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Here's a challenge for you: draw an "ellipse" polygon
>>>>>>>>>>>>>>>>> with N segments that are all the same length that circumscribes a perfect
>>>>>>>>>>>>>>>>> ellipse with given semiaxes a and b. I tried to solve this problem and
>>>>>>>>>>>>>>>>> could not find a solution. I can draw an *inscribed* N segment "ellipse"
>>>>>>>>>>>>>>>>> polygon with all segments the same length---though it's not easy---but the
>>>>>>>>>>>>>>>>> circumscribed case eludes me. I think of the inscribed case as, every
>>>>>>>>>>>>>>>>> vertex is located on the ellipse and the circumscribed case, every segment
>>>>>>>>>>>>>>>>> is tangent to the ellipse
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>>>>>>>>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> In openscad drawing a ellipse is not too difficult
>>>>>>>>>>>>>>>>>> Refer following code
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> function to draw an ellipse with semi-major and
>>>>>>>>>>>>>>>>>> semi-minor axis "r1" and "r2" respectively and with center "cp" and number
>>>>>>>>>>>>>>>>>> of segment "s"
>>>>>>>>>>>>>>>>>> /
>>>>>>>>>>>>>>>>>> // example:
>>>>>>>>>>>>>>>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> function ellipse(r1,r2,cp,s=30)=
>>>>>>>>>>>>>>>>>> let(
>>>>>>>>>>>>>>>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>>>>>>>>>>>>>>>>> )sec;
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>>>>>>>>>>>>>>>>> openscad@jordan.maileater.net> wrote:
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> I created an oval using:
>>>>>>>>>>>>>>>>>>> inMm = 25.4;
>>>>>>>>>>>>>>>>>>> fullW = 35.5 * inMm;
>>>>>>>>>>>>>>>>>>> fullL = 54.5 * inMm;
>>>>>>>>>>>>>>>>>>> resize([fullW, fullL, 1])
>>>>>>>>>>>>>>>>>>> circle(d=fullW);
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> Tip: use scale() rather than resize(); it previews
>>>>>>>>>>>>>>>>>>> faster in complicated cases.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> You're taking a circle and stretching it; the result of
>>>>>>>>>>>>>>>>>>> that stretching is an ellipse.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> I'm not 100% sure - maybe one of the real math wizards
>>>>>>>>>>>>>>>>>>> can chime in - but I think the shape you are asking for would not be an
>>>>>>>>>>>>>>>>>>> ellipse.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> I believe that an ellipse is fully defined by its major
>>>>>>>>>>>>>>>>>>> (long) and minor (short) axis dimensions; the other parts of the curve
>>>>>>>>>>>>>>>>>>> mathematically flow from those two measurements.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> You can make something else, but you can't make it using
>>>>>>>>>>>>>>>>>>> this technique.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> You can use a Bézier function.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> Using BOSL2:
>>>>>>>>>>>>>>>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> DIY, here's a function that I cobbled together some
>>>>>>>>>>>>>>>>>>> years ago:
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>>>>>>>>>>>>>>>>> // but that yielded either single points or a raft of triangles;
>>>>>>>>>>>>>>>>>>> // this yields a vector of points that you can then concatenate
>>>>>>>>>>>>>>>>>>> // with other pieces to form a single polygon.
>>>>>>>>>>>>>>>>>>> // If we were really clever, I think it would be possible to
>>>>>>>>>>>>>>>>>>> // automatically space the output points based on how linear
>>>>>>>>>>>>>>>>>>> // the curve is at that point. But right now I'm not that clever.
>>>>>>>>>>>>>>>>>>> function BEZ03(u) = pow((1-u), 3);
>>>>>>>>>>>>>>>>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>>>>>>>>>>>>>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>>>>>>>>>>>>>>>>> function BEZ33(u) = pow(u,3);
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>>>>>>>>>>>>>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>>>>>>>>>>>>>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> // p0 - start point
>>>>>>>>>>>>>>>>>>> // p1 - control point 1, line departs p0 headed this way
>>>>>>>>>>>>>>>>>>> // p2 - control point 2, line arrives at p3 from this way
>>>>>>>>>>>>>>>>>>> // p3 - end point
>>>>>>>>>>>>>>>>>>> // segs - number of segments
>>>>>>>>>>>>>>>>>>> function bez(p0, p1, p2, p3, segs) = [
>>>>>>>>>>>>>>>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>>>>>>>>>>>>>>>>> ];
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> It is that final function bez() that I actually call,
>>>>>>>>>>>>>>>>>>> e.g.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> often concatenating the results of several bez() calls
>>>>>>>>>>>>>>>>>>> together.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> Note that the four coordinates correspond to the
>>>>>>>>>>>>>>>>>>> coordinates of the control points in a GUI drawing program: the start
>>>>>>>>>>>>>>>>>>> point, control point associated with the start, the control point
>>>>>>>>>>>>>>>>>>> associated with the end, and the end point. I have sometimes created
>>>>>>>>>>>>>>>>>>> OpenSCAD objects by tracing an object in a drawing program and then reading
>>>>>>>>>>>>>>>>>>> off the coordinates of the various control points.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> Note that Bézier functions are not the only mathematical
>>>>>>>>>>>>>>>>>>> curves. (For instance, they cannot exactly match an arc of a circle.) I'm
>>>>>>>>>>>>>>>>>>> sure there are numerous other functions that also produce curves.
>>>>>>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> _______________________________________________
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>>>>>>>>>>>>>>>>> _______________________________________________
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>
FH
Father Horton
Mon, Aug 8, 2022 5:41 PM
I'm not sure why there should be a restriction on N; I think it's general.
I did a spreadsheet that looks like it's doing the right thing in the first
quadrant, but it (or I) gets lost when I have to select the right value for
the sign of the square roots involved as the process enters other quadrants.
On Sun, Aug 7, 2022 at 9:43 PM Father Horton fatherhorton@gmail.com wrote:
Google showed me that method when I came up with the right terms. But it's
too late at night for me to try to implement it. Maybe tomorrow.
On Sun, Aug 7, 2022 at 9:41 PM Adrian Mariano avm4@cornell.edu wrote:
Now I understand your idea. I think it should be possible to make that
work for the case of N=4k+2. It does seem like there are some
complications, like you try to extend the segment and the length is closer
than the tangent, so it's impossible. Also things may be complicated if
the algorithm tries to go around the tip of the ellipse (that is, if the
path is too long).
For finding a tangent to an ellipse, transform the problem to the circle,
solve, and then transform back. In BOSL2:
function ellipse_tangents(a,b,pt) =
let(
r=a,
scaledpt = yscale(a/b,pt),
tangents = circle_point_tangents(r,[0,0],scaledpt)
)
yscale(b/a,tangents);
On Sun, Aug 7, 2022 at 9:54 PM Father Horton fatherhorton@gmail.com
wrote:
I did that wrong. It's not (0, b) to (0, L/2). It's (0, b) to (L/2, b).
That's always outside the ellipse.
I will look at the second problem as soon as I figure out how to
calculate a tangent line to an ellipse containing a point not on the
ellipse. Google is not helping much.
On Sun, Aug 7, 2022 at 8:32 PM Adrian Mariano avm4@cornell.edu wrote:
Father Horton,
I'm having trouble understanding your proposed method. A segment from
(0,b) to (0,L/2) might be inside the ellipse depending on N, and then there
won't be a tangent.
It also sounds like your idea is somewhat similar to Jordan's idea:
basically start at one side and work around the ellipse and see if it
works. If not, make a correction. The problem with this idea is that
there are two parameters for the first segment, so how do you make a
correction? (The two parameters are the location of the endpoint of the
first segment and its length.)
On Sun, Aug 7, 2022 at 5:33 PM Father Horton fatherhorton@gmail.com
wrote:
How about this to construct N equal length segments around an ellipse
with axes a and b:
- Select an initial segment length estimate (e.g., Ramanujan's
formula / N) = L
- Construct a half-segment from (0, b) to (0, L/2).
- Find a line tangent to the ellipse and passing through (0, L/2).
- Find the point on the tangent line that is L away from (0, L/2).
- Repeat (3) substituting the new point for (0, L/2). until you have
N segments.
If everything worked perfectly, the last point found would be (0,
-L/2). If there's a gap, increase L and try again. If there's overlap,
decrease L and try again.
On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano rcmpersiano@gmail.com
wrote:
Adrian, why to use a (either inscribed or circumscribed) polygon with
equal side as an approximation of an ellipsis? It doesn't seem to be a good
approximation of it. I get a better approximation by taking an angle
uniform sample of the curve in the parameter space.
[image: Uniformly_sampled_ellipsis.png]
Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano avm4@cornell.edu
escreveu:
Circumscribed means tangents to points on the ellipse. But there
are no arc lengths (equal or unequal) in this problem, nor equally spaced
angles. The angles of each segment will be different. But the lengths of
the segments are not the same as the arc lengths, which is why arc length
does not matter. Here's a solution to the N=6 polygon sort of inspired by
the idea Jordan suggested, where I adjust the first segment until the
second segment has the right length. The rest of the solution is then
obtained using symmetry, which means there is only a single unknown that
can simply be solved for. But this doesn't generalize to higher N.
[image: image.png]
include<BOSL2/std.scad>
// ellipse semiaxes
// If a is much smaller than b it will fail
a=5;
b=2;
// Find point on ellipse at specified x coordinate
function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
// Find tangent line to ellipse at point (u,w)
// and determine the x value where that line has
// given y value.
function findx(a,b,u,w,y) =
let(
A = -ub^2/w/a^2,
B = w+(ub)^2/w/a^2
)
(y-B)/A;
// First segment if tangent point is at x=u
function first_seg(a,b,u) =
let(
w = ellipse_pt(a,b,u),
x1 = findx(a,b,u,w,0),
x2 = findx(a,b,u,w,b)
)
[[x1,0],[x2,b]];
err = function (u) let(pt = first_seg(a,b,u))
2*pt[1].x - norm(pt[0]-pt[1]);
x = root_find(err,.2a,.9a);
first = first_seg(a,b,x);
top = concat(first,reverse(xflip(first)));
poly = concat(top, reverse(yflip(list_tail(top))));
stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
color("red")stroke(poly,width=.05,closed=true);
// Verify equality of segment lengths
echo(path_segment_lengths(poly,closed=true));
On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
Circumscribed means tangents from points on ellipse
How can the tangents from equally spaced arc lengths or equally
spaced angles be equal for an ellipse
You need to change the arc lengths dynamically and also accurately
estimate the segment length which fits the ellipse .
On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, avm4@cornell.edu
wrote:
Your intuition is incorrect: a circle scaled in one direction IS
an ellipse. The motivation for being interested in circumscribing is to
make holes, though the constraint of equal length sides is probably not
what you'd want for that purpose. You can make a nonuniform circumscribing
polygon for an ellipse by simply making a polygon that circumscribes a
circle and scaling it.
On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson bob@rjcarlson.com
wrote:
Seems like I remember Nop Head talking about using circumscribed
holes for screws and even rotating the polygons to produce a more even
hole.
If an ellipse is the nominal shape, the using a circumscribed
polygon for a hole and an inscribed polygon for a solid might make sense.
Btw, can someone confirm my intuition that a circle scaled in one
dimension is not an ellipse?
-Bob
Tucson AZ
On Aug 5, 2022, at 15:38, Adrian Mariano avm4@cornell.edu
wrote:
I don't have a specific need for a circumscribing elliptical
polygon with equal sides. I was writing code to make ellipses and it
seemed like equal side approximations would be nice, so I solved the case
for inscribed, but couldn't solve the circumscribed case. So it's an
unsolved puzzle I've got that this discussion reminded me of.
On Fri, Aug 5, 2022 at 6:33 PM nop head nop.head@gmail.com
wrote:
A long discussion but I have to ask why do you need an
elliptical polygon with equal tangential sides?
On Fri, 5 Aug 2022, 21:58 Adrian Mariano, avm4@cornell.edu
wrote:
You think that a circumscribing polygon with equal length
segments does not exist?? I'm skeptical of this claim.
For circumscribed, it's not so hard. You start with a
uniform-in-angle point distribution, compute the segment lengths, compare
to the mean segment length to get an error, and then compute an adjustment
based on the error. It converges quickly.
So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
5.57297, 5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but surely
good enough.
<image.png>
The code for this is in BOSL2 in shapes2d.scad. A similar
algorithm should exist for the circumscribed case, but as I said, my
efforts to find it have failed.
On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
I think circumscribed polygon segments cannot be equal lengths.
inscribed ones may be approximately equal, but would be a very
tough problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano avm4@cornell.edu
wrote:
I think you need to be a little careful about defining what
you want to do. If you want to approximate an ellipse with points that are
distributed uniformly over the ellipse, that could be done by approximating
the arc length of the ellipse to find the necessary points---e.g. advance
distance d along the ellipse at each step. But if you want to approximate
an ellipse with a polygon with equal length sides, that approach does not
do it, because the length of the segments will not be constant as the span
sections of the ellipse with differing curvature. And since the perimeter
of the desired polygon is unknown, you can't just step it out analogously.
The circle doesn't present this challenge due to having constant
curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
fatherhorton@gmail.com> wrote:
I think our emails crossed. I was addressing how to
approximate an ellipse with constant-length segments in general, not how to
solve your problem. When I had to do something similar, I used polar
coordinates and stepped up the value of theta until the distance from the
previous point was correct. (In the constant-length version, I'd keep a
running total of the maximum desired length so that round-off errors don't
accumulate.)
Figuring out the circumscribed polygon is, as you note, a
greater challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <
avm4@cornell.edu> wrote:
Eh, there's no such thing as "brute force" in this
situation. Step through what with sufficiently small steps? Of course
an iterative algorithm is necessary. This is true for most interesting
problems. No big deal. Show me an iterative algorithm that converges to
the circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
fatherhorton@gmail.com> wrote:
Except by brute force (step through it with sufficiently
small steps), it’s not possible because there’s no closed form expression
for the distance between two points on a general ellipse (as opposed to
special cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <
avm4@cornell.edu> wrote:
Here's a challenge for you: draw an "ellipse" polygon
with N segments that are all the same length that circumscribes a perfect
ellipse with given semiaxes a and b. I tried to solve this problem and
could not find a solution. I can draw an inscribed N segment "ellipse"
polygon with all segments the same length---though it's not easy---but the
circumscribed case eludes me. I think of the inscribed case as, every
vertex is located on the ellipse and the circumscribed case, every segment
is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and
semi-minor axis "r1" and "r2" respectively and with center "cp" and number
of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews
faster in complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result of
that stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards
can chime in - but I think the shape you are asking for would not be an
ellipse.
I believe that an ellipse is fully defined by its major
(long) and minor (short) axis dimensions; the other parts of the curve
mathematically flow from those two measurements.
You can make something else, but you can't make it
using this technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some
years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call,
e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls
together.
Note that the four coordinates correspond to the
coordinates of the control points in a GUI drawing program: the start
point, control point associated with the start, the control point
associated with the end, and the end point. I have sometimes created
OpenSCAD objects by tracing an object in a drawing program and then reading
off the coordinates of the various control points.
Note that Bézier functions are not the only
mathematical curves. (For instance, they cannot exactly match an arc of a
circle.) I'm sure there are numerous other functions that also produce
curves.
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I'm not sure why there should be a restriction on N; I think it's general.
I did a spreadsheet that looks like it's doing the right thing in the first
quadrant, but it (or I) gets lost when I have to select the right value for
the sign of the square roots involved as the process enters other quadrants.
On Sun, Aug 7, 2022 at 9:43 PM Father Horton <fatherhorton@gmail.com> wrote:
> Google showed me that method when I came up with the right terms. But it's
> too late at night for me to try to implement it. Maybe tomorrow.
>
> On Sun, Aug 7, 2022 at 9:41 PM Adrian Mariano <avm4@cornell.edu> wrote:
>
>> Now I understand your idea. I think it should be possible to make that
>> work for the case of N=4k+2. It does seem like there are some
>> complications, like you try to extend the segment and the length is closer
>> than the tangent, so it's impossible. Also things may be complicated if
>> the algorithm tries to go around the tip of the ellipse (that is, if the
>> path is too long).
>>
>> For finding a tangent to an ellipse, transform the problem to the circle,
>> solve, and then transform back. In BOSL2:
>>
>> function ellipse_tangents(a,b,pt) =
>> let(
>> r=a,
>> scaledpt = yscale(a/b,pt),
>> tangents = circle_point_tangents(r,[0,0],scaledpt)
>> )
>> yscale(b/a,tangents);
>>
>> On Sun, Aug 7, 2022 at 9:54 PM Father Horton <fatherhorton@gmail.com>
>> wrote:
>>
>>> I did that wrong. It's not (0, b) to (0, L/2). It's (0, b) to (L/2, b).
>>> That's always outside the ellipse.
>>>
>>> I will look at the second problem as soon as I figure out how to
>>> calculate a tangent line to an ellipse containing a point not on the
>>> ellipse. Google is not helping much.
>>>
>>> On Sun, Aug 7, 2022 at 8:32 PM Adrian Mariano <avm4@cornell.edu> wrote:
>>>
>>>> Father Horton,
>>>>
>>>> I'm having trouble understanding your proposed method. A segment from
>>>> (0,b) to (0,L/2) might be inside the ellipse depending on N, and then there
>>>> won't be a tangent.
>>>>
>>>> It also sounds like your idea is somewhat similar to Jordan's idea:
>>>> basically start at one side and work around the ellipse and see if it
>>>> works. If not, make a correction. The problem with this idea is that
>>>> there are two parameters for the first segment, so how do you make a
>>>> correction? (The two parameters are the location of the endpoint of the
>>>> first segment and its length.)
>>>>
>>>> On Sun, Aug 7, 2022 at 5:33 PM Father Horton <fatherhorton@gmail.com>
>>>> wrote:
>>>>
>>>>> How about this to construct N equal length segments around an ellipse
>>>>> with axes a and b:
>>>>>
>>>>> 1) Select an initial segment length estimate (e.g., Ramanujan's
>>>>> formula / N) = L
>>>>> 2) Construct a half-segment from (0, b) to (0, L/2).
>>>>> 3) Find a line tangent to the ellipse and passing through (0, L/2).
>>>>> 3) Find the point on the tangent line that is L away from (0, L/2).
>>>>> 4) Repeat (3) substituting the new point for (0, L/2). until you have
>>>>> N segments.
>>>>>
>>>>> If everything worked perfectly, the last point found would be (0,
>>>>> -L/2). If there's a gap, increase L and try again. If there's overlap,
>>>>> decrease L and try again.
>>>>>
>>>>> On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano <rcmpersiano@gmail.com>
>>>>> wrote:
>>>>>
>>>>>> Adrian, why to use a (either inscribed or circumscribed) polygon with
>>>>>> equal side as an approximation of an ellipsis? It doesn't seem to be a good
>>>>>> approximation of it. I get a better approximation by taking an angle
>>>>>> uniform sample of the curve in the parameter space.
>>>>>>
>>>>>> [image: Uniformly_sampled_ellipsis.png]
>>>>>>
>>>>>>
>>>>>> Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano <avm4@cornell.edu>
>>>>>> escreveu:
>>>>>>
>>>>>>> Circumscribed means tangents to points on the ellipse. But there
>>>>>>> are no arc lengths (equal or unequal) in this problem, nor equally spaced
>>>>>>> angles. The angles of each segment will be different. But the lengths of
>>>>>>> the segments are not the same as the arc lengths, which is why arc length
>>>>>>> does not matter. Here's a solution to the N=6 polygon sort of inspired by
>>>>>>> the idea Jordan suggested, where I adjust the first segment until the
>>>>>>> second segment has the right length. The rest of the solution is then
>>>>>>> obtained using symmetry, which means there is only a single unknown that
>>>>>>> can simply be solved for. But this doesn't generalize to higher N.
>>>>>>>
>>>>>>> [image: image.png]
>>>>>>> include<BOSL2/std.scad>
>>>>>>>
>>>>>>> // ellipse semiaxes
>>>>>>> // If a is much smaller than b it will fail
>>>>>>> a=5;
>>>>>>> b=2;
>>>>>>>
>>>>>>> // Find point on ellipse at specified x coordinate
>>>>>>> function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
>>>>>>>
>>>>>>>
>>>>>>> // Find tangent line to ellipse at point (u,w)
>>>>>>> // and determine the x value where that line has
>>>>>>> // given y value.
>>>>>>> function findx(a,b,u,w,y) =
>>>>>>> let(
>>>>>>> A = -u*b^2/w/a^2,
>>>>>>> B = w+(u*b)^2/w/a^2
>>>>>>> )
>>>>>>> (y-B)/A;
>>>>>>>
>>>>>>>
>>>>>>> // First segment if tangent point is at x=u
>>>>>>> function first_seg(a,b,u) =
>>>>>>> let(
>>>>>>> w = ellipse_pt(a,b,u),
>>>>>>> x1 = findx(a,b,u,w,0),
>>>>>>> x2 = findx(a,b,u,w,b)
>>>>>>> )
>>>>>>> [[x1,0],[x2,b]];
>>>>>>>
>>>>>>>
>>>>>>> err = function (u) let(pt = first_seg(a,b,u))
>>>>>>> 2*pt[1].x - norm(pt[0]-pt[1]);
>>>>>>>
>>>>>>> x = root_find(err,.2*a,.9*a);
>>>>>>> first = first_seg(a,b,x);
>>>>>>> top = concat(first,reverse(xflip(first)));
>>>>>>> poly = concat(top, reverse(yflip(list_tail(top))));
>>>>>>>
>>>>>>> stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
>>>>>>> color("red")stroke(poly,width=.05,closed=true);
>>>>>>>
>>>>>>> // Verify equality of segment lengths
>>>>>>> echo(path_segment_lengths(poly,closed=true));
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>
>>>>>>>> Circumscribed means tangents from points on ellipse
>>>>>>>> How can the tangents from equally spaced arc lengths or equally
>>>>>>>> spaced angles be equal for an ellipse
>>>>>>>> You need to change the arc lengths dynamically and also accurately
>>>>>>>> estimate the segment length which fits the ellipse .
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>> On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, <avm4@cornell.edu>
>>>>>>>> wrote:
>>>>>>>>
>>>>>>>>> Your intuition is incorrect: a circle scaled in one direction *IS*
>>>>>>>>> an ellipse. The motivation for being interested in circumscribing is to
>>>>>>>>> make holes, though the constraint of equal length sides is probably not
>>>>>>>>> what you'd want for that purpose. You can make a nonuniform circumscribing
>>>>>>>>> polygon for an ellipse by simply making a polygon that circumscribes a
>>>>>>>>> circle and scaling it.
>>>>>>>>>
>>>>>>>>> On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson <bob@rjcarlson.com>
>>>>>>>>> wrote:
>>>>>>>>>
>>>>>>>>>> Seems like I remember Nop Head talking about using circumscribed
>>>>>>>>>> holes for screws and even rotating the polygons to produce a more even
>>>>>>>>>> hole.
>>>>>>>>>>
>>>>>>>>>> If an ellipse is the nominal shape, the using a circumscribed
>>>>>>>>>> polygon for a hole and an inscribed polygon for a solid might make sense.
>>>>>>>>>>
>>>>>>>>>> Btw, can someone confirm my intuition that a circle scaled in one
>>>>>>>>>> dimension is not an ellipse?
>>>>>>>>>>
>>>>>>>>>> -Bob
>>>>>>>>>> Tucson AZ
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> On Aug 5, 2022, at 15:38, Adrian Mariano <avm4@cornell.edu>
>>>>>>>>>> wrote:
>>>>>>>>>>
>>>>>>>>>> I don't have a specific need for a circumscribing elliptical
>>>>>>>>>> polygon with equal sides. I was writing code to make ellipses and it
>>>>>>>>>> seemed like equal side approximations would be nice, so I solved the case
>>>>>>>>>> for inscribed, but couldn't solve the circumscribed case. So it's an
>>>>>>>>>> unsolved puzzle I've got that this discussion reminded me of.
>>>>>>>>>>
>>>>>>>>>> On Fri, Aug 5, 2022 at 6:33 PM nop head <nop.head@gmail.com>
>>>>>>>>>> wrote:
>>>>>>>>>>
>>>>>>>>>>> A long discussion but I have to ask why do you need an
>>>>>>>>>>> elliptical polygon with equal tangential sides?
>>>>>>>>>>>
>>>>>>>>>>> On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu>
>>>>>>>>>>> wrote:
>>>>>>>>>>>
>>>>>>>>>>>> You think that a circumscribing polygon with equal length
>>>>>>>>>>>> segments does not exist?? I'm skeptical of this claim.
>>>>>>>>>>>>
>>>>>>>>>>>> For circumscribed, it's not so hard. You start with a
>>>>>>>>>>>> uniform-in-angle point distribution, compute the segment lengths, compare
>>>>>>>>>>>> to the mean segment length to get an error, and then compute an adjustment
>>>>>>>>>>>> based on the error. It converges quickly.
>>>>>>>>>>>>
>>>>>>>>>>>> So for example, I have segments lengths of:
>>>>>>>>>>>> ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
>>>>>>>>>>>> 5.57297, 5.57297]
>>>>>>>>>>>>
>>>>>>>>>>>> Maybe not exactly equal, since tolerance is 1e-9, but surely
>>>>>>>>>>>> good enough.
>>>>>>>>>>>> <image.png>
>>>>>>>>>>>>
>>>>>>>>>>>> The code for this is in BOSL2 in shapes2d.scad. A similar
>>>>>>>>>>>> algorithm should exist for the circumscribed case, but as I said, my
>>>>>>>>>>>> efforts to find it have failed.
>>>>>>>>>>>>
>>>>>>>>>>>> On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
>>>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>>>
>>>>>>>>>>>>> I think circumscribed polygon segments cannot be equal lengths.
>>>>>>>>>>>>> inscribed ones may be approximately equal, but would be a very
>>>>>>>>>>>>> tough problem to solve.
>>>>>>>>>>>>> I tried something.
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <avm4@cornell.edu>
>>>>>>>>>>>>> wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> I think you need to be a little careful about defining what
>>>>>>>>>>>>>> you want to do. If you want to approximate an ellipse with points that are
>>>>>>>>>>>>>> distributed uniformly over the ellipse, that could be done by approximating
>>>>>>>>>>>>>> the arc length of the ellipse to find the necessary points---e.g. advance
>>>>>>>>>>>>>> distance d along the ellipse at each step. But if you want to approximate
>>>>>>>>>>>>>> an ellipse with a polygon with equal length sides, that approach does not
>>>>>>>>>>>>>> do it, because the length of the segments will not be constant as the span
>>>>>>>>>>>>>> sections of the ellipse with differing curvature. And since the perimeter
>>>>>>>>>>>>>> of the desired polygon is unknown, you can't just step it out analogously.
>>>>>>>>>>>>>> The circle doesn't present this challenge due to having constant
>>>>>>>>>>>>>> curvature.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
>>>>>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> I think our emails crossed. I was addressing how to
>>>>>>>>>>>>>>> approximate an ellipse with constant-length segments in general, not how to
>>>>>>>>>>>>>>> solve your problem. When I had to do something similar, I used polar
>>>>>>>>>>>>>>> coordinates and stepped up the value of theta until the distance from the
>>>>>>>>>>>>>>> previous point was correct. (In the constant-length version, I'd keep a
>>>>>>>>>>>>>>> running total of the maximum desired length so that round-off errors don't
>>>>>>>>>>>>>>> accumulate.)
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Figuring out the circumscribed polygon is, as you note, a
>>>>>>>>>>>>>>> greater challenge. I can visualize how to do it for n = 4, but it doesn't
>>>>>>>>>>>>>>> extrapolate at all.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <
>>>>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Eh, there's no such thing as "brute force" in this
>>>>>>>>>>>>>>>> situation. Step through *what* with sufficiently small steps? Of course
>>>>>>>>>>>>>>>> an iterative algorithm is necessary. This is true for most interesting
>>>>>>>>>>>>>>>> problems. No big deal. Show me an iterative algorithm that converges to
>>>>>>>>>>>>>>>> the circumscribing polygon. When I tried to devise such an algorithm I
>>>>>>>>>>>>>>>> couldn't find one that would converge. Writing an algorithm that would
>>>>>>>>>>>>>>>> converge to the *inscribed* polygon seemed to be straight forward. Note
>>>>>>>>>>>>>>>> also that computing arc length of the ellipse doesn't really help here.
>>>>>>>>>>>>>>>> You can do that by solving an elliptical integral, or summing up small
>>>>>>>>>>>>>>>> segments. But what's the connection between arc length of the ellipse and
>>>>>>>>>>>>>>>> a circumscribing polygon with equal segment lengths? If there's a
>>>>>>>>>>>>>>>> connection, I don't see it.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
>>>>>>>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Except by brute force (step through it with sufficiently
>>>>>>>>>>>>>>>>> small steps), it’s not possible because there’s no closed form expression
>>>>>>>>>>>>>>>>> for the distance between two points on a general ellipse (as opposed to
>>>>>>>>>>>>>>>>> special cases like a circle).
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <
>>>>>>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> Here's a challenge for you: draw an "ellipse" polygon
>>>>>>>>>>>>>>>>>> with N segments that are all the same length that circumscribes a perfect
>>>>>>>>>>>>>>>>>> ellipse with given semiaxes a and b. I tried to solve this problem and
>>>>>>>>>>>>>>>>>> could not find a solution. I can draw an *inscribed* N segment "ellipse"
>>>>>>>>>>>>>>>>>> polygon with all segments the same length---though it's not easy---but the
>>>>>>>>>>>>>>>>>> circumscribed case eludes me. I think of the inscribed case as, every
>>>>>>>>>>>>>>>>>> vertex is located on the ellipse and the circumscribed case, every segment
>>>>>>>>>>>>>>>>>> is tangent to the ellipse
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>>>>>>>>>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> In openscad drawing a ellipse is not too difficult
>>>>>>>>>>>>>>>>>>> Refer following code
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> function to draw an ellipse with semi-major and
>>>>>>>>>>>>>>>>>>> semi-minor axis "r1" and "r2" respectively and with center "cp" and number
>>>>>>>>>>>>>>>>>>> of segment "s"
>>>>>>>>>>>>>>>>>>> /
>>>>>>>>>>>>>>>>>>> // example:
>>>>>>>>>>>>>>>>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> function ellipse(r1,r2,cp,s=30)=
>>>>>>>>>>>>>>>>>>> let(
>>>>>>>>>>>>>>>>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>>>>>>>>>>>>>>>>>> )sec;
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>>>>>>>>>>>>>>>>>> openscad@jordan.maileater.net> wrote:
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> I created an oval using:
>>>>>>>>>>>>>>>>>>>> inMm = 25.4;
>>>>>>>>>>>>>>>>>>>> fullW = 35.5 * inMm;
>>>>>>>>>>>>>>>>>>>> fullL = 54.5 * inMm;
>>>>>>>>>>>>>>>>>>>> resize([fullW, fullL, 1])
>>>>>>>>>>>>>>>>>>>> circle(d=fullW);
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> Tip: use scale() rather than resize(); it previews
>>>>>>>>>>>>>>>>>>>> faster in complicated cases.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> You're taking a circle and stretching it; the result of
>>>>>>>>>>>>>>>>>>>> that stretching is an ellipse.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> I'm not 100% sure - maybe one of the real math wizards
>>>>>>>>>>>>>>>>>>>> can chime in - but I think the shape you are asking for would not be an
>>>>>>>>>>>>>>>>>>>> ellipse.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> I believe that an ellipse is fully defined by its major
>>>>>>>>>>>>>>>>>>>> (long) and minor (short) axis dimensions; the other parts of the curve
>>>>>>>>>>>>>>>>>>>> mathematically flow from those two measurements.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> You can make something else, but you can't make it
>>>>>>>>>>>>>>>>>>>> using this technique.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> You can use a Bézier function.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> Using BOSL2:
>>>>>>>>>>>>>>>>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> DIY, here's a function that I cobbled together some
>>>>>>>>>>>>>>>>>>>> years ago:
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>>>>>>>>>>>>>>>>>> // but that yielded either single points or a raft of triangles;
>>>>>>>>>>>>>>>>>>>> // this yields a vector of points that you can then concatenate
>>>>>>>>>>>>>>>>>>>> // with other pieces to form a single polygon.
>>>>>>>>>>>>>>>>>>>> // If we were really clever, I think it would be possible to
>>>>>>>>>>>>>>>>>>>> // automatically space the output points based on how linear
>>>>>>>>>>>>>>>>>>>> // the curve is at that point. But right now I'm not that clever.
>>>>>>>>>>>>>>>>>>>> function BEZ03(u) = pow((1-u), 3);
>>>>>>>>>>>>>>>>>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>>>>>>>>>>>>>>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>>>>>>>>>>>>>>>>>> function BEZ33(u) = pow(u,3);
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>>>>>>>>>>>>>>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>>>>>>>>>>>>>>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> // p0 - start point
>>>>>>>>>>>>>>>>>>>> // p1 - control point 1, line departs p0 headed this way
>>>>>>>>>>>>>>>>>>>> // p2 - control point 2, line arrives at p3 from this way
>>>>>>>>>>>>>>>>>>>> // p3 - end point
>>>>>>>>>>>>>>>>>>>> // segs - number of segments
>>>>>>>>>>>>>>>>>>>> function bez(p0, p1, p2, p3, segs) = [
>>>>>>>>>>>>>>>>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>>>>>>>>>>>>>>>>>> ];
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> It is that final function bez() that I actually call,
>>>>>>>>>>>>>>>>>>>> e.g.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> often concatenating the results of several bez() calls
>>>>>>>>>>>>>>>>>>>> together.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> Note that the four coordinates correspond to the
>>>>>>>>>>>>>>>>>>>> coordinates of the control points in a GUI drawing program: the start
>>>>>>>>>>>>>>>>>>>> point, control point associated with the start, the control point
>>>>>>>>>>>>>>>>>>>> associated with the end, and the end point. I have sometimes created
>>>>>>>>>>>>>>>>>>>> OpenSCAD objects by tracing an object in a drawing program and then reading
>>>>>>>>>>>>>>>>>>>> off the coordinates of the various control points.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> Note that Bézier functions are not the only
>>>>>>>>>>>>>>>>>>>> mathematical curves. (For instance, they cannot exactly match an arc of a
>>>>>>>>>>>>>>>>>>>> circle.) I'm sure there are numerous other functions that also produce
>>>>>>>>>>>>>>>>>>>> curves.
>>>>>>>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>>>>>>> OpenSCAD mailing list
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>>>>>>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> _______________________________________________
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>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> _______________________________________________
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>>>>>>>>>>>
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>>>>>>>>>>
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>>>>>>>>>>
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>
AM
Adrian Mariano
Mon, Aug 8, 2022 6:30 PM
The restriction arises because of the end condition. There are 4 cases
depending on what happens at the four ends of the ellipse (the points where
it intersects the coordinate axes). The four possibilities are at each end
either a corner or a flat tangent at the end. Your approach as I
understand it begins with a flat and constructs a quarter of the polygon,
ending with a point. Such a polygon must have 4k+2 sides. I do not see
how the end condition can be formulated to end on a flat. Also I do not
see how it can start on a point. If n is odd you can start at one end and
go half way around and end on a point. But if n is divisible by 4 you can
have 4 flats or four points and the method cannot produce either of them.
Have I missed something?
On Mon, Aug 8, 2022, 13:42 Father Horton fatherhorton@gmail.com wrote:
I'm not sure why there should be a restriction on N; I think it's general.
I did a spreadsheet that looks like it's doing the right thing in the first
quadrant, but it (or I) gets lost when I have to select the right value for
the sign of the square roots involved as the process enters other quadrants.
On Sun, Aug 7, 2022 at 9:43 PM Father Horton fatherhorton@gmail.com
wrote:
Google showed me that method when I came up with the right terms. But
it's too late at night for me to try to implement it. Maybe tomorrow.
On Sun, Aug 7, 2022 at 9:41 PM Adrian Mariano avm4@cornell.edu wrote:
Now I understand your idea. I think it should be possible to make that
work for the case of N=4k+2. It does seem like there are some
complications, like you try to extend the segment and the length is closer
than the tangent, so it's impossible. Also things may be complicated if
the algorithm tries to go around the tip of the ellipse (that is, if the
path is too long).
For finding a tangent to an ellipse, transform the problem to the
circle, solve, and then transform back. In BOSL2:
function ellipse_tangents(a,b,pt) =
let(
r=a,
scaledpt = yscale(a/b,pt),
tangents = circle_point_tangents(r,[0,0],scaledpt)
)
yscale(b/a,tangents);
On Sun, Aug 7, 2022 at 9:54 PM Father Horton fatherhorton@gmail.com
wrote:
I did that wrong. It's not (0, b) to (0, L/2). It's (0, b) to (L/2, b).
That's always outside the ellipse.
I will look at the second problem as soon as I figure out how to
calculate a tangent line to an ellipse containing a point not on the
ellipse. Google is not helping much.
On Sun, Aug 7, 2022 at 8:32 PM Adrian Mariano avm4@cornell.edu wrote:
Father Horton,
I'm having trouble understanding your proposed method. A segment from
(0,b) to (0,L/2) might be inside the ellipse depending on N, and then there
won't be a tangent.
It also sounds like your idea is somewhat similar to Jordan's idea:
basically start at one side and work around the ellipse and see if it
works. If not, make a correction. The problem with this idea is that
there are two parameters for the first segment, so how do you make a
correction? (The two parameters are the location of the endpoint of the
first segment and its length.)
On Sun, Aug 7, 2022 at 5:33 PM Father Horton fatherhorton@gmail.com
wrote:
How about this to construct N equal length segments around an ellipse
with axes a and b:
- Select an initial segment length estimate (e.g., Ramanujan's
formula / N) = L
- Construct a half-segment from (0, b) to (0, L/2).
- Find a line tangent to the ellipse and passing through (0, L/2).
- Find the point on the tangent line that is L away from (0, L/2).
- Repeat (3) substituting the new point for (0, L/2). until you have
N segments.
If everything worked perfectly, the last point found would be (0,
-L/2). If there's a gap, increase L and try again. If there's overlap,
decrease L and try again.
On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano <
rcmpersiano@gmail.com> wrote:
Adrian, why to use a (either inscribed or circumscribed) polygon
with equal side as an approximation of an ellipsis? It doesn't seem to be a
good approximation of it. I get a better approximation by taking an angle
uniform sample of the curve in the parameter space.
[image: Uniformly_sampled_ellipsis.png]
Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano <
avm4@cornell.edu> escreveu:
Circumscribed means tangents to points on the ellipse. But there
are no arc lengths (equal or unequal) in this problem, nor equally spaced
angles. The angles of each segment will be different. But the lengths of
the segments are not the same as the arc lengths, which is why arc length
does not matter. Here's a solution to the N=6 polygon sort of inspired by
the idea Jordan suggested, where I adjust the first segment until the
second segment has the right length. The rest of the solution is then
obtained using symmetry, which means there is only a single unknown that
can simply be solved for. But this doesn't generalize to higher N.
[image: image.png]
include<BOSL2/std.scad>
// ellipse semiaxes
// If a is much smaller than b it will fail
a=5;
b=2;
// Find point on ellipse at specified x coordinate
function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
// Find tangent line to ellipse at point (u,w)
// and determine the x value where that line has
// given y value.
function findx(a,b,u,w,y) =
let(
A = -ub^2/w/a^2,
B = w+(ub)^2/w/a^2
)
(y-B)/A;
// First segment if tangent point is at x=u
function first_seg(a,b,u) =
let(
w = ellipse_pt(a,b,u),
x1 = findx(a,b,u,w,0),
x2 = findx(a,b,u,w,b)
)
[[x1,0],[x2,b]];
err = function (u) let(pt = first_seg(a,b,u))
2*pt[1].x - norm(pt[0]-pt[1]);
x = root_find(err,.2a,.9a);
first = first_seg(a,b,x);
top = concat(first,reverse(xflip(first)));
poly = concat(top, reverse(yflip(list_tail(top))));
stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
color("red")stroke(poly,width=.05,closed=true);
// Verify equality of segment lengths
echo(path_segment_lengths(poly,closed=true));
On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
Circumscribed means tangents from points on ellipse
How can the tangents from equally spaced arc lengths or equally
spaced angles be equal for an ellipse
You need to change the arc lengths dynamically and also accurately
estimate the segment length which fits the ellipse .
On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, avm4@cornell.edu
wrote:
Your intuition is incorrect: a circle scaled in one direction
IS an ellipse. The motivation for being interested in
circumscribing is to make holes, though the constraint of equal length
sides is probably not what you'd want for that purpose. You can make a
nonuniform circumscribing polygon for an ellipse by simply making a polygon
that circumscribes a circle and scaling it.
On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson bob@rjcarlson.com
wrote:
Seems like I remember Nop Head talking about using circumscribed
holes for screws and even rotating the polygons to produce a more even
hole.
If an ellipse is the nominal shape, the using a circumscribed
polygon for a hole and an inscribed polygon for a solid might make sense.
Btw, can someone confirm my intuition that a circle scaled in
one dimension is not an ellipse?
-Bob
Tucson AZ
On Aug 5, 2022, at 15:38, Adrian Mariano avm4@cornell.edu
wrote:
I don't have a specific need for a circumscribing elliptical
polygon with equal sides. I was writing code to make ellipses and it
seemed like equal side approximations would be nice, so I solved the case
for inscribed, but couldn't solve the circumscribed case. So it's an
unsolved puzzle I've got that this discussion reminded me of.
On Fri, Aug 5, 2022 at 6:33 PM nop head nop.head@gmail.com
wrote:
A long discussion but I have to ask why do you need an
elliptical polygon with equal tangential sides?
On Fri, 5 Aug 2022, 21:58 Adrian Mariano, avm4@cornell.edu
wrote:
You think that a circumscribing polygon with equal length
segments does not exist?? I'm skeptical of this claim.
For circumscribed, it's not so hard. You start with a
uniform-in-angle point distribution, compute the segment lengths, compare
to the mean segment length to get an error, and then compute an adjustment
based on the error. It converges quickly.
So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
5.57297, 5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but surely
good enough.
<image.png>
The code for this is in BOSL2 in shapes2d.scad. A similar
algorithm should exist for the circumscribed case, but as I said, my
efforts to find it have failed.
On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
I think circumscribed polygon segments cannot be equal
lengths.
inscribed ones may be approximately equal, but would be a
very tough problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano avm4@cornell.edu
wrote:
I think you need to be a little careful about defining what
you want to do. If you want to approximate an ellipse with points that are
distributed uniformly over the ellipse, that could be done by approximating
the arc length of the ellipse to find the necessary points---e.g. advance
distance d along the ellipse at each step. But if you want to approximate
an ellipse with a polygon with equal length sides, that approach does not
do it, because the length of the segments will not be constant as the span
sections of the ellipse with differing curvature. And since the perimeter
of the desired polygon is unknown, you can't just step it out analogously.
The circle doesn't present this challenge due to having constant
curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
fatherhorton@gmail.com> wrote:
I think our emails crossed. I was addressing how to
approximate an ellipse with constant-length segments in general, not how to
solve your problem. When I had to do something similar, I used polar
coordinates and stepped up the value of theta until the distance from the
previous point was correct. (In the constant-length version, I'd keep a
running total of the maximum desired length so that round-off errors don't
accumulate.)
Figuring out the circumscribed polygon is, as you note, a
greater challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <
avm4@cornell.edu> wrote:
Eh, there's no such thing as "brute force" in this
situation. Step through what with sufficiently small steps? Of course
an iterative algorithm is necessary. This is true for most interesting
problems. No big deal. Show me an iterative algorithm that converges to
the circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
fatherhorton@gmail.com> wrote:
Except by brute force (step through it with sufficiently
small steps), it’s not possible because there’s no closed form expression
for the distance between two points on a general ellipse (as opposed to
special cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <
avm4@cornell.edu> wrote:
Here's a challenge for you: draw an "ellipse" polygon
with N segments that are all the same length that circumscribes a perfect
ellipse with given semiaxes a and b. I tried to solve this problem and
could not find a solution. I can draw an inscribed N segment "ellipse"
polygon with all segments the same length---though it's not easy---but the
circumscribed case eludes me. I think of the inscribed case as, every
vertex is located on the ellipse and the circumscribed case, every segment
is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and
semi-minor axis "r1" and "r2" respectively and with center "cp" and number
of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews
faster in complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result
of that stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards
can chime in - but I think the shape you are asking for would not be an
ellipse.
I believe that an ellipse is fully defined by its
major (long) and minor (short) axis dimensions; the other parts of the
curve mathematically flow from those two measurements.
You can make something else, but you can't make it
using this technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some
years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call,
e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez() calls
together.
Note that the four coordinates correspond to the
coordinates of the control points in a GUI drawing program: the start
point, control point associated with the start, the control point
associated with the end, and the end point. I have sometimes created
OpenSCAD objects by tracing an object in a drawing program and then reading
off the coordinates of the various control points.
Note that Bézier functions are not the only
mathematical curves. (For instance, they cannot exactly match an arc of a
circle.) I'm sure there are numerous other functions that also produce
curves.
OpenSCAD mailing list
To unsubscribe send an email to
discuss-leave@lists.openscad.org
The restriction arises because of the end condition. There are 4 cases
depending on what happens at the four ends of the ellipse (the points where
it intersects the coordinate axes). The four possibilities are at each end
either a corner or a flat tangent at the end. Your approach as I
understand it begins with a flat and constructs a quarter of the polygon,
ending with a point. Such a polygon must have 4k+2 sides. I do not see
how the end condition can be formulated to end on a flat. Also I do not
see how it can start on a point. If n is odd you can start at one end and
go half way around and end on a point. But if n is divisible by 4 you can
have 4 flats or four points and the method cannot produce either of them.
Have I missed something?
On Mon, Aug 8, 2022, 13:42 Father Horton <fatherhorton@gmail.com> wrote:
> I'm not sure why there should be a restriction on N; I think it's general.
> I did a spreadsheet that looks like it's doing the right thing in the first
> quadrant, but it (or I) gets lost when I have to select the right value for
> the sign of the square roots involved as the process enters other quadrants.
>
> On Sun, Aug 7, 2022 at 9:43 PM Father Horton <fatherhorton@gmail.com>
> wrote:
>
>> Google showed me that method when I came up with the right terms. But
>> it's too late at night for me to try to implement it. Maybe tomorrow.
>>
>> On Sun, Aug 7, 2022 at 9:41 PM Adrian Mariano <avm4@cornell.edu> wrote:
>>
>>> Now I understand your idea. I think it should be possible to make that
>>> work for the case of N=4k+2. It does seem like there are some
>>> complications, like you try to extend the segment and the length is closer
>>> than the tangent, so it's impossible. Also things may be complicated if
>>> the algorithm tries to go around the tip of the ellipse (that is, if the
>>> path is too long).
>>>
>>> For finding a tangent to an ellipse, transform the problem to the
>>> circle, solve, and then transform back. In BOSL2:
>>>
>>> function ellipse_tangents(a,b,pt) =
>>> let(
>>> r=a,
>>> scaledpt = yscale(a/b,pt),
>>> tangents = circle_point_tangents(r,[0,0],scaledpt)
>>> )
>>> yscale(b/a,tangents);
>>>
>>> On Sun, Aug 7, 2022 at 9:54 PM Father Horton <fatherhorton@gmail.com>
>>> wrote:
>>>
>>>> I did that wrong. It's not (0, b) to (0, L/2). It's (0, b) to (L/2, b).
>>>> That's always outside the ellipse.
>>>>
>>>> I will look at the second problem as soon as I figure out how to
>>>> calculate a tangent line to an ellipse containing a point not on the
>>>> ellipse. Google is not helping much.
>>>>
>>>> On Sun, Aug 7, 2022 at 8:32 PM Adrian Mariano <avm4@cornell.edu> wrote:
>>>>
>>>>> Father Horton,
>>>>>
>>>>> I'm having trouble understanding your proposed method. A segment from
>>>>> (0,b) to (0,L/2) might be inside the ellipse depending on N, and then there
>>>>> won't be a tangent.
>>>>>
>>>>> It also sounds like your idea is somewhat similar to Jordan's idea:
>>>>> basically start at one side and work around the ellipse and see if it
>>>>> works. If not, make a correction. The problem with this idea is that
>>>>> there are two parameters for the first segment, so how do you make a
>>>>> correction? (The two parameters are the location of the endpoint of the
>>>>> first segment and its length.)
>>>>>
>>>>> On Sun, Aug 7, 2022 at 5:33 PM Father Horton <fatherhorton@gmail.com>
>>>>> wrote:
>>>>>
>>>>>> How about this to construct N equal length segments around an ellipse
>>>>>> with axes a and b:
>>>>>>
>>>>>> 1) Select an initial segment length estimate (e.g., Ramanujan's
>>>>>> formula / N) = L
>>>>>> 2) Construct a half-segment from (0, b) to (0, L/2).
>>>>>> 3) Find a line tangent to the ellipse and passing through (0, L/2).
>>>>>> 3) Find the point on the tangent line that is L away from (0, L/2).
>>>>>> 4) Repeat (3) substituting the new point for (0, L/2). until you have
>>>>>> N segments.
>>>>>>
>>>>>> If everything worked perfectly, the last point found would be (0,
>>>>>> -L/2). If there's a gap, increase L and try again. If there's overlap,
>>>>>> decrease L and try again.
>>>>>>
>>>>>> On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano <
>>>>>> rcmpersiano@gmail.com> wrote:
>>>>>>
>>>>>>> Adrian, why to use a (either inscribed or circumscribed) polygon
>>>>>>> with equal side as an approximation of an ellipsis? It doesn't seem to be a
>>>>>>> good approximation of it. I get a better approximation by taking an angle
>>>>>>> uniform sample of the curve in the parameter space.
>>>>>>>
>>>>>>> [image: Uniformly_sampled_ellipsis.png]
>>>>>>>
>>>>>>>
>>>>>>> Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano <
>>>>>>> avm4@cornell.edu> escreveu:
>>>>>>>
>>>>>>>> Circumscribed means tangents to points on the ellipse. But there
>>>>>>>> are no arc lengths (equal or unequal) in this problem, nor equally spaced
>>>>>>>> angles. The angles of each segment will be different. But the lengths of
>>>>>>>> the segments are not the same as the arc lengths, which is why arc length
>>>>>>>> does not matter. Here's a solution to the N=6 polygon sort of inspired by
>>>>>>>> the idea Jordan suggested, where I adjust the first segment until the
>>>>>>>> second segment has the right length. The rest of the solution is then
>>>>>>>> obtained using symmetry, which means there is only a single unknown that
>>>>>>>> can simply be solved for. But this doesn't generalize to higher N.
>>>>>>>>
>>>>>>>> [image: image.png]
>>>>>>>> include<BOSL2/std.scad>
>>>>>>>>
>>>>>>>> // ellipse semiaxes
>>>>>>>> // If a is much smaller than b it will fail
>>>>>>>> a=5;
>>>>>>>> b=2;
>>>>>>>>
>>>>>>>> // Find point on ellipse at specified x coordinate
>>>>>>>> function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
>>>>>>>>
>>>>>>>>
>>>>>>>> // Find tangent line to ellipse at point (u,w)
>>>>>>>> // and determine the x value where that line has
>>>>>>>> // given y value.
>>>>>>>> function findx(a,b,u,w,y) =
>>>>>>>> let(
>>>>>>>> A = -u*b^2/w/a^2,
>>>>>>>> B = w+(u*b)^2/w/a^2
>>>>>>>> )
>>>>>>>> (y-B)/A;
>>>>>>>>
>>>>>>>>
>>>>>>>> // First segment if tangent point is at x=u
>>>>>>>> function first_seg(a,b,u) =
>>>>>>>> let(
>>>>>>>> w = ellipse_pt(a,b,u),
>>>>>>>> x1 = findx(a,b,u,w,0),
>>>>>>>> x2 = findx(a,b,u,w,b)
>>>>>>>> )
>>>>>>>> [[x1,0],[x2,b]];
>>>>>>>>
>>>>>>>>
>>>>>>>> err = function (u) let(pt = first_seg(a,b,u))
>>>>>>>> 2*pt[1].x - norm(pt[0]-pt[1]);
>>>>>>>>
>>>>>>>> x = root_find(err,.2*a,.9*a);
>>>>>>>> first = first_seg(a,b,x);
>>>>>>>> top = concat(first,reverse(xflip(first)));
>>>>>>>> poly = concat(top, reverse(yflip(list_tail(top))));
>>>>>>>>
>>>>>>>> stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
>>>>>>>> color("red")stroke(poly,width=.05,closed=true);
>>>>>>>>
>>>>>>>> // Verify equality of segment lengths
>>>>>>>> echo(path_segment_lengths(poly,closed=true));
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>> On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>
>>>>>>>>> Circumscribed means tangents from points on ellipse
>>>>>>>>> How can the tangents from equally spaced arc lengths or equally
>>>>>>>>> spaced angles be equal for an ellipse
>>>>>>>>> You need to change the arc lengths dynamically and also accurately
>>>>>>>>> estimate the segment length which fits the ellipse .
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, <avm4@cornell.edu>
>>>>>>>>> wrote:
>>>>>>>>>
>>>>>>>>>> Your intuition is incorrect: a circle scaled in one direction
>>>>>>>>>> *IS* an ellipse. The motivation for being interested in
>>>>>>>>>> circumscribing is to make holes, though the constraint of equal length
>>>>>>>>>> sides is probably not what you'd want for that purpose. You can make a
>>>>>>>>>> nonuniform circumscribing polygon for an ellipse by simply making a polygon
>>>>>>>>>> that circumscribes a circle and scaling it.
>>>>>>>>>>
>>>>>>>>>> On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson <bob@rjcarlson.com>
>>>>>>>>>> wrote:
>>>>>>>>>>
>>>>>>>>>>> Seems like I remember Nop Head talking about using circumscribed
>>>>>>>>>>> holes for screws and even rotating the polygons to produce a more even
>>>>>>>>>>> hole.
>>>>>>>>>>>
>>>>>>>>>>> If an ellipse is the nominal shape, the using a circumscribed
>>>>>>>>>>> polygon for a hole and an inscribed polygon for a solid might make sense.
>>>>>>>>>>>
>>>>>>>>>>> Btw, can someone confirm my intuition that a circle scaled in
>>>>>>>>>>> one dimension is not an ellipse?
>>>>>>>>>>>
>>>>>>>>>>> -Bob
>>>>>>>>>>> Tucson AZ
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> On Aug 5, 2022, at 15:38, Adrian Mariano <avm4@cornell.edu>
>>>>>>>>>>> wrote:
>>>>>>>>>>>
>>>>>>>>>>> I don't have a specific need for a circumscribing elliptical
>>>>>>>>>>> polygon with equal sides. I was writing code to make ellipses and it
>>>>>>>>>>> seemed like equal side approximations would be nice, so I solved the case
>>>>>>>>>>> for inscribed, but couldn't solve the circumscribed case. So it's an
>>>>>>>>>>> unsolved puzzle I've got that this discussion reminded me of.
>>>>>>>>>>>
>>>>>>>>>>> On Fri, Aug 5, 2022 at 6:33 PM nop head <nop.head@gmail.com>
>>>>>>>>>>> wrote:
>>>>>>>>>>>
>>>>>>>>>>>> A long discussion but I have to ask why do you need an
>>>>>>>>>>>> elliptical polygon with equal tangential sides?
>>>>>>>>>>>>
>>>>>>>>>>>> On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu>
>>>>>>>>>>>> wrote:
>>>>>>>>>>>>
>>>>>>>>>>>>> You think that a circumscribing polygon with equal length
>>>>>>>>>>>>> segments does not exist?? I'm skeptical of this claim.
>>>>>>>>>>>>>
>>>>>>>>>>>>> For circumscribed, it's not so hard. You start with a
>>>>>>>>>>>>> uniform-in-angle point distribution, compute the segment lengths, compare
>>>>>>>>>>>>> to the mean segment length to get an error, and then compute an adjustment
>>>>>>>>>>>>> based on the error. It converges quickly.
>>>>>>>>>>>>>
>>>>>>>>>>>>> So for example, I have segments lengths of:
>>>>>>>>>>>>> ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
>>>>>>>>>>>>> 5.57297, 5.57297]
>>>>>>>>>>>>>
>>>>>>>>>>>>> Maybe not exactly equal, since tolerance is 1e-9, but surely
>>>>>>>>>>>>> good enough.
>>>>>>>>>>>>> <image.png>
>>>>>>>>>>>>>
>>>>>>>>>>>>> The code for this is in BOSL2 in shapes2d.scad. A similar
>>>>>>>>>>>>> algorithm should exist for the circumscribed case, but as I said, my
>>>>>>>>>>>>> efforts to find it have failed.
>>>>>>>>>>>>>
>>>>>>>>>>>>> On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
>>>>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> I think circumscribed polygon segments cannot be equal
>>>>>>>>>>>>>> lengths.
>>>>>>>>>>>>>> inscribed ones may be approximately equal, but would be a
>>>>>>>>>>>>>> very tough problem to solve.
>>>>>>>>>>>>>> I tried something.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <avm4@cornell.edu>
>>>>>>>>>>>>>> wrote:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> I think you need to be a little careful about defining what
>>>>>>>>>>>>>>> you want to do. If you want to approximate an ellipse with points that are
>>>>>>>>>>>>>>> distributed uniformly over the ellipse, that could be done by approximating
>>>>>>>>>>>>>>> the arc length of the ellipse to find the necessary points---e.g. advance
>>>>>>>>>>>>>>> distance d along the ellipse at each step. But if you want to approximate
>>>>>>>>>>>>>>> an ellipse with a polygon with equal length sides, that approach does not
>>>>>>>>>>>>>>> do it, because the length of the segments will not be constant as the span
>>>>>>>>>>>>>>> sections of the ellipse with differing curvature. And since the perimeter
>>>>>>>>>>>>>>> of the desired polygon is unknown, you can't just step it out analogously.
>>>>>>>>>>>>>>> The circle doesn't present this challenge due to having constant
>>>>>>>>>>>>>>> curvature.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
>>>>>>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> I think our emails crossed. I was addressing how to
>>>>>>>>>>>>>>>> approximate an ellipse with constant-length segments in general, not how to
>>>>>>>>>>>>>>>> solve your problem. When I had to do something similar, I used polar
>>>>>>>>>>>>>>>> coordinates and stepped up the value of theta until the distance from the
>>>>>>>>>>>>>>>> previous point was correct. (In the constant-length version, I'd keep a
>>>>>>>>>>>>>>>> running total of the maximum desired length so that round-off errors don't
>>>>>>>>>>>>>>>> accumulate.)
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Figuring out the circumscribed polygon is, as you note, a
>>>>>>>>>>>>>>>> greater challenge. I can visualize how to do it for n = 4, but it doesn't
>>>>>>>>>>>>>>>> extrapolate at all.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <
>>>>>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Eh, there's no such thing as "brute force" in this
>>>>>>>>>>>>>>>>> situation. Step through *what* with sufficiently small steps? Of course
>>>>>>>>>>>>>>>>> an iterative algorithm is necessary. This is true for most interesting
>>>>>>>>>>>>>>>>> problems. No big deal. Show me an iterative algorithm that converges to
>>>>>>>>>>>>>>>>> the circumscribing polygon. When I tried to devise such an algorithm I
>>>>>>>>>>>>>>>>> couldn't find one that would converge. Writing an algorithm that would
>>>>>>>>>>>>>>>>> converge to the *inscribed* polygon seemed to be straight forward. Note
>>>>>>>>>>>>>>>>> also that computing arc length of the ellipse doesn't really help here.
>>>>>>>>>>>>>>>>> You can do that by solving an elliptical integral, or summing up small
>>>>>>>>>>>>>>>>> segments. But what's the connection between arc length of the ellipse and
>>>>>>>>>>>>>>>>> a circumscribing polygon with equal segment lengths? If there's a
>>>>>>>>>>>>>>>>> connection, I don't see it.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
>>>>>>>>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> Except by brute force (step through it with sufficiently
>>>>>>>>>>>>>>>>>> small steps), it’s not possible because there’s no closed form expression
>>>>>>>>>>>>>>>>>> for the distance between two points on a general ellipse (as opposed to
>>>>>>>>>>>>>>>>>> special cases like a circle).
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <
>>>>>>>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> Here's a challenge for you: draw an "ellipse" polygon
>>>>>>>>>>>>>>>>>>> with N segments that are all the same length that circumscribes a perfect
>>>>>>>>>>>>>>>>>>> ellipse with given semiaxes a and b. I tried to solve this problem and
>>>>>>>>>>>>>>>>>>> could not find a solution. I can draw an *inscribed* N segment "ellipse"
>>>>>>>>>>>>>>>>>>> polygon with all segments the same length---though it's not easy---but the
>>>>>>>>>>>>>>>>>>> circumscribed case eludes me. I think of the inscribed case as, every
>>>>>>>>>>>>>>>>>>> vertex is located on the ellipse and the circumscribed case, every segment
>>>>>>>>>>>>>>>>>>> is tangent to the ellipse
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>>>>>>>>>>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> In openscad drawing a ellipse is not too difficult
>>>>>>>>>>>>>>>>>>>> Refer following code
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> function to draw an ellipse with semi-major and
>>>>>>>>>>>>>>>>>>>> semi-minor axis "r1" and "r2" respectively and with center "cp" and number
>>>>>>>>>>>>>>>>>>>> of segment "s"
>>>>>>>>>>>>>>>>>>>> /
>>>>>>>>>>>>>>>>>>>> // example:
>>>>>>>>>>>>>>>>>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> function ellipse(r1,r2,cp,s=30)=
>>>>>>>>>>>>>>>>>>>> let(
>>>>>>>>>>>>>>>>>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>>>>>>>>>>>>>>>>>>> )sec;
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>>>>>>>>>>>>>>>>>>> openscad@jordan.maileater.net> wrote:
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> I created an oval using:
>>>>>>>>>>>>>>>>>>>>> inMm = 25.4;
>>>>>>>>>>>>>>>>>>>>> fullW = 35.5 * inMm;
>>>>>>>>>>>>>>>>>>>>> fullL = 54.5 * inMm;
>>>>>>>>>>>>>>>>>>>>> resize([fullW, fullL, 1])
>>>>>>>>>>>>>>>>>>>>> circle(d=fullW);
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> Tip: use scale() rather than resize(); it previews
>>>>>>>>>>>>>>>>>>>>> faster in complicated cases.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> You're taking a circle and stretching it; the result
>>>>>>>>>>>>>>>>>>>>> of that stretching is an ellipse.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> I'm not 100% sure - maybe one of the real math wizards
>>>>>>>>>>>>>>>>>>>>> can chime in - but I think the shape you are asking for would not be an
>>>>>>>>>>>>>>>>>>>>> ellipse.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> I believe that an ellipse is fully defined by its
>>>>>>>>>>>>>>>>>>>>> major (long) and minor (short) axis dimensions; the other parts of the
>>>>>>>>>>>>>>>>>>>>> curve mathematically flow from those two measurements.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> You can make something else, but you can't make it
>>>>>>>>>>>>>>>>>>>>> using this technique.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> You can use a Bézier function.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> Using BOSL2:
>>>>>>>>>>>>>>>>>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> DIY, here's a function that I cobbled together some
>>>>>>>>>>>>>>>>>>>>> years ago:
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>>>>>>>>>>>>>>>>>>> // but that yielded either single points or a raft of triangles;
>>>>>>>>>>>>>>>>>>>>> // this yields a vector of points that you can then concatenate
>>>>>>>>>>>>>>>>>>>>> // with other pieces to form a single polygon.
>>>>>>>>>>>>>>>>>>>>> // If we were really clever, I think it would be possible to
>>>>>>>>>>>>>>>>>>>>> // automatically space the output points based on how linear
>>>>>>>>>>>>>>>>>>>>> // the curve is at that point. But right now I'm not that clever.
>>>>>>>>>>>>>>>>>>>>> function BEZ03(u) = pow((1-u), 3);
>>>>>>>>>>>>>>>>>>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>>>>>>>>>>>>>>>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>>>>>>>>>>>>>>>>>>> function BEZ33(u) = pow(u,3);
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>>>>>>>>>>>>>>>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>>>>>>>>>>>>>>>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> // p0 - start point
>>>>>>>>>>>>>>>>>>>>> // p1 - control point 1, line departs p0 headed this way
>>>>>>>>>>>>>>>>>>>>> // p2 - control point 2, line arrives at p3 from this way
>>>>>>>>>>>>>>>>>>>>> // p3 - end point
>>>>>>>>>>>>>>>>>>>>> // segs - number of segments
>>>>>>>>>>>>>>>>>>>>> function bez(p0, p1, p2, p3, segs) = [
>>>>>>>>>>>>>>>>>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>>>>>>>>>>>>>>>>>>> ];
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> It is that final function bez() that I actually call,
>>>>>>>>>>>>>>>>>>>>> e.g.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> often concatenating the results of several bez() calls
>>>>>>>>>>>>>>>>>>>>> together.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> Note that the four coordinates correspond to the
>>>>>>>>>>>>>>>>>>>>> coordinates of the control points in a GUI drawing program: the start
>>>>>>>>>>>>>>>>>>>>> point, control point associated with the start, the control point
>>>>>>>>>>>>>>>>>>>>> associated with the end, and the end point. I have sometimes created
>>>>>>>>>>>>>>>>>>>>> OpenSCAD objects by tracing an object in a drawing program and then reading
>>>>>>>>>>>>>>>>>>>>> off the coordinates of the various control points.
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> Note that Bézier functions are not the only
>>>>>>>>>>>>>>>>>>>>> mathematical curves. (For instance, they cannot exactly match an arc of a
>>>>>>>>>>>>>>>>>>>>> circle.) I'm sure there are numerous other functions that also produce
>>>>>>>>>>>>>>>>>>>>> curves.
>>>>>>>>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> _______________________________________________
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>>>>>>>>>>>>>>>>>>> _______________________________________________
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>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> _______________________________________________
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>
FH
Father Horton
Mon, Aug 8, 2022 7:35 PM
Here's a Google sheet that shows what I have in mind.
https://docs.google.com/spreadsheets/d/13YOJKkWWfcgZNKVn_nlBtQZ_bbxNvN6SdVv1bVWLDP8/edit?usp=sharing
I have to adjust the signs on the square roots manually, so I didn't
iterate until I got equal lengths, but this is the idea. Blue points are
end points of the circumscribing segments and red points are the tangent
points on the ellipse. I have other things to do at the moment, but I may
revisit later to get the signs to work out right and then see if it will
iterate to an acceptable result.
On Mon, Aug 8, 2022 at 1:31 PM Adrian Mariano avm4@cornell.edu wrote:
The restriction arises because of the end condition. There are 4 cases
depending on what happens at the four ends of the ellipse (the points where
it intersects the coordinate axes). The four possibilities are at each end
either a corner or a flat tangent at the end. Your approach as I
understand it begins with a flat and constructs a quarter of the polygon,
ending with a point. Such a polygon must have 4k+2 sides. I do not see
how the end condition can be formulated to end on a flat. Also I do not
see how it can start on a point. If n is odd you can start at one end and
go half way around and end on a point. But if n is divisible by 4 you can
have 4 flats or four points and the method cannot produce either of them.
Have I missed something?
On Mon, Aug 8, 2022, 13:42 Father Horton fatherhorton@gmail.com wrote:
I'm not sure why there should be a restriction on N; I think it's
general. I did a spreadsheet that looks like it's doing the right thing in
the first quadrant, but it (or I) gets lost when I have to select the right
value for the sign of the square roots involved as the process enters other
quadrants.
On Sun, Aug 7, 2022 at 9:43 PM Father Horton fatherhorton@gmail.com
wrote:
Google showed me that method when I came up with the right terms. But
it's too late at night for me to try to implement it. Maybe tomorrow.
On Sun, Aug 7, 2022 at 9:41 PM Adrian Mariano avm4@cornell.edu wrote:
Now I understand your idea. I think it should be possible to make that
work for the case of N=4k+2. It does seem like there are some
complications, like you try to extend the segment and the length is closer
than the tangent, so it's impossible. Also things may be complicated if
the algorithm tries to go around the tip of the ellipse (that is, if the
path is too long).
For finding a tangent to an ellipse, transform the problem to the
circle, solve, and then transform back. In BOSL2:
function ellipse_tangents(a,b,pt) =
let(
r=a,
scaledpt = yscale(a/b,pt),
tangents = circle_point_tangents(r,[0,0],scaledpt)
)
yscale(b/a,tangents);
On Sun, Aug 7, 2022 at 9:54 PM Father Horton fatherhorton@gmail.com
wrote:
I did that wrong. It's not (0, b) to (0, L/2). It's (0, b) to (L/2,
b). That's always outside the ellipse.
I will look at the second problem as soon as I figure out how to
calculate a tangent line to an ellipse containing a point not on the
ellipse. Google is not helping much.
On Sun, Aug 7, 2022 at 8:32 PM Adrian Mariano avm4@cornell.edu
wrote:
Father Horton,
I'm having trouble understanding your proposed method. A segment
from (0,b) to (0,L/2) might be inside the ellipse depending on N, and then
there won't be a tangent.
It also sounds like your idea is somewhat similar to Jordan's idea:
basically start at one side and work around the ellipse and see if it
works. If not, make a correction. The problem with this idea is that
there are two parameters for the first segment, so how do you make a
correction? (The two parameters are the location of the endpoint of the
first segment and its length.)
On Sun, Aug 7, 2022 at 5:33 PM Father Horton fatherhorton@gmail.com
wrote:
How about this to construct N equal length segments around an
ellipse with axes a and b:
- Select an initial segment length estimate (e.g., Ramanujan's
formula / N) = L
- Construct a half-segment from (0, b) to (0, L/2).
- Find a line tangent to the ellipse and passing through (0, L/2).
- Find the point on the tangent line that is L away from (0, L/2).
- Repeat (3) substituting the new point for (0, L/2). until you
have N segments.
If everything worked perfectly, the last point found would be (0,
-L/2). If there's a gap, increase L and try again. If there's overlap,
decrease L and try again.
On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano <
rcmpersiano@gmail.com> wrote:
Adrian, why to use a (either inscribed or circumscribed) polygon
with equal side as an approximation of an ellipsis? It doesn't seem to be a
good approximation of it. I get a better approximation by taking an angle
uniform sample of the curve in the parameter space.
[image: Uniformly_sampled_ellipsis.png]
Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano <
avm4@cornell.edu> escreveu:
Circumscribed means tangents to points on the ellipse. But there
are no arc lengths (equal or unequal) in this problem, nor equally spaced
angles. The angles of each segment will be different. But the lengths of
the segments are not the same as the arc lengths, which is why arc length
does not matter. Here's a solution to the N=6 polygon sort of inspired by
the idea Jordan suggested, where I adjust the first segment until the
second segment has the right length. The rest of the solution is then
obtained using symmetry, which means there is only a single unknown that
can simply be solved for. But this doesn't generalize to higher N.
[image: image.png]
include<BOSL2/std.scad>
// ellipse semiaxes
// If a is much smaller than b it will fail
a=5;
b=2;
// Find point on ellipse at specified x coordinate
function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
// Find tangent line to ellipse at point (u,w)
// and determine the x value where that line has
// given y value.
function findx(a,b,u,w,y) =
let(
A = -ub^2/w/a^2,
B = w+(ub)^2/w/a^2
)
(y-B)/A;
// First segment if tangent point is at x=u
function first_seg(a,b,u) =
let(
w = ellipse_pt(a,b,u),
x1 = findx(a,b,u,w,0),
x2 = findx(a,b,u,w,b)
)
[[x1,0],[x2,b]];
err = function (u) let(pt = first_seg(a,b,u))
2*pt[1].x - norm(pt[0]-pt[1]);
x = root_find(err,.2a,.9a);
first = first_seg(a,b,x);
top = concat(first,reverse(xflip(first)));
poly = concat(top, reverse(yflip(list_tail(top))));
stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
color("red")stroke(poly,width=.05,closed=true);
// Verify equality of segment lengths
echo(path_segment_lengths(poly,closed=true));
On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
Circumscribed means tangents from points on ellipse
How can the tangents from equally spaced arc lengths or equally
spaced angles be equal for an ellipse
You need to change the arc lengths dynamically and also
accurately estimate the segment length which fits the ellipse .
On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, avm4@cornell.edu
wrote:
Your intuition is incorrect: a circle scaled in one direction
IS an ellipse. The motivation for being interested in
circumscribing is to make holes, though the constraint of equal length
sides is probably not what you'd want for that purpose. You can make a
nonuniform circumscribing polygon for an ellipse by simply making a polygon
that circumscribes a circle and scaling it.
On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson bob@rjcarlson.com
wrote:
Seems like I remember Nop Head talking about using
circumscribed holes for screws and even rotating the polygons to produce a
more even hole.
If an ellipse is the nominal shape, the using a circumscribed
polygon for a hole and an inscribed polygon for a solid might make sense.
Btw, can someone confirm my intuition that a circle scaled in
one dimension is not an ellipse?
-Bob
Tucson AZ
On Aug 5, 2022, at 15:38, Adrian Mariano avm4@cornell.edu
wrote:
I don't have a specific need for a circumscribing elliptical
polygon with equal sides. I was writing code to make ellipses and it
seemed like equal side approximations would be nice, so I solved the case
for inscribed, but couldn't solve the circumscribed case. So it's an
unsolved puzzle I've got that this discussion reminded me of.
On Fri, Aug 5, 2022 at 6:33 PM nop head nop.head@gmail.com
wrote:
A long discussion but I have to ask why do you need an
elliptical polygon with equal tangential sides?
On Fri, 5 Aug 2022, 21:58 Adrian Mariano, avm4@cornell.edu
wrote:
You think that a circumscribing polygon with equal length
segments does not exist?? I'm skeptical of this claim.
For circumscribed, it's not so hard. You start with a
uniform-in-angle point distribution, compute the segment lengths, compare
to the mean segment length to get an error, and then compute an adjustment
based on the error. It converges quickly.
So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
5.57297, 5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but surely
good enough.
<image.png>
The code for this is in BOSL2 in shapes2d.scad. A similar
algorithm should exist for the circumscribed case, but as I said, my
efforts to find it have failed.
On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
I think circumscribed polygon segments cannot be equal
lengths.
inscribed ones may be approximately equal, but would be a
very tough problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <
avm4@cornell.edu> wrote:
I think you need to be a little careful about defining what
you want to do. If you want to approximate an ellipse with points that are
distributed uniformly over the ellipse, that could be done by approximating
the arc length of the ellipse to find the necessary points---e.g. advance
distance d along the ellipse at each step. But if you want to approximate
an ellipse with a polygon with equal length sides, that approach does not
do it, because the length of the segments will not be constant as the span
sections of the ellipse with differing curvature. And since the perimeter
of the desired polygon is unknown, you can't just step it out analogously.
The circle doesn't present this challenge due to having constant
curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
fatherhorton@gmail.com> wrote:
I think our emails crossed. I was addressing how to
approximate an ellipse with constant-length segments in general, not how to
solve your problem. When I had to do something similar, I used polar
coordinates and stepped up the value of theta until the distance from the
previous point was correct. (In the constant-length version, I'd keep a
running total of the maximum desired length so that round-off errors don't
accumulate.)
Figuring out the circumscribed polygon is, as you note, a
greater challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <
avm4@cornell.edu> wrote:
Eh, there's no such thing as "brute force" in this
situation. Step through what with sufficiently small steps? Of course
an iterative algorithm is necessary. This is true for most interesting
problems. No big deal. Show me an iterative algorithm that converges to
the circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
fatherhorton@gmail.com> wrote:
Except by brute force (step through it with sufficiently
small steps), it’s not possible because there’s no closed form expression
for the distance between two points on a general ellipse (as opposed to
special cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <
avm4@cornell.edu> wrote:
Here's a challenge for you: draw an "ellipse" polygon
with N segments that are all the same length that circumscribes a perfect
ellipse with given semiaxes a and b. I tried to solve this problem and
could not find a solution. I can draw an inscribed N segment "ellipse"
polygon with all segments the same length---though it's not easy---but the
circumscribed case eludes me. I think of the inscribed case as, every
vertex is located on the ellipse and the circumscribed case, every segment
is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and
semi-minor axis "r1" and "r2" respectively and with center "cp" and number
of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews
faster in complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result
of that stretching is an ellipse.
I'm not 100% sure - maybe one of the real math
wizards can chime in - but I think the shape you are asking for would not
be an ellipse.
I believe that an ellipse is fully defined by its
major (long) and minor (short) axis dimensions; the other parts of the
curve mathematically flow from those two measurements.
You can make something else, but you can't make it
using this technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some
years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually call,
e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez()
calls together.
Note that the four coordinates correspond to the
coordinates of the control points in a GUI drawing program: the start
point, control point associated with the start, the control point
associated with the end, and the end point. I have sometimes created
OpenSCAD objects by tracing an object in a drawing program and then reading
off the coordinates of the various control points.
Note that Bézier functions are not the only
mathematical curves. (For instance, they cannot exactly match an arc of a
circle.) I'm sure there are numerous other functions that also produce
curves.
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Here's a Google sheet that shows what I have in mind.
https://docs.google.com/spreadsheets/d/13YOJKkWWfcgZNKVn_nlBtQZ_bbxNvN6SdVv1bVWLDP8/edit?usp=sharing
I have to adjust the signs on the square roots manually, so I didn't
iterate until I got equal lengths, but this is the idea. Blue points are
end points of the circumscribing segments and red points are the tangent
points on the ellipse. I have other things to do at the moment, but I may
revisit later to get the signs to work out right and then see if it will
iterate to an acceptable result.
On Mon, Aug 8, 2022 at 1:31 PM Adrian Mariano <avm4@cornell.edu> wrote:
> The restriction arises because of the end condition. There are 4 cases
> depending on what happens at the four ends of the ellipse (the points where
> it intersects the coordinate axes). The four possibilities are at each end
> either a corner or a flat tangent at the end. Your approach as I
> understand it begins with a flat and constructs a quarter of the polygon,
> ending with a point. Such a polygon must have 4k+2 sides. I do not see
> how the end condition can be formulated to end on a flat. Also I do not
> see how it can start on a point. If n is odd you can start at one end and
> go half way around and end on a point. But if n is divisible by 4 you can
> have 4 flats or four points and the method cannot produce either of them.
> Have I missed something?
>
>
> On Mon, Aug 8, 2022, 13:42 Father Horton <fatherhorton@gmail.com> wrote:
>
>> I'm not sure why there should be a restriction on N; I think it's
>> general. I did a spreadsheet that looks like it's doing the right thing in
>> the first quadrant, but it (or I) gets lost when I have to select the right
>> value for the sign of the square roots involved as the process enters other
>> quadrants.
>>
>> On Sun, Aug 7, 2022 at 9:43 PM Father Horton <fatherhorton@gmail.com>
>> wrote:
>>
>>> Google showed me that method when I came up with the right terms. But
>>> it's too late at night for me to try to implement it. Maybe tomorrow.
>>>
>>> On Sun, Aug 7, 2022 at 9:41 PM Adrian Mariano <avm4@cornell.edu> wrote:
>>>
>>>> Now I understand your idea. I think it should be possible to make that
>>>> work for the case of N=4k+2. It does seem like there are some
>>>> complications, like you try to extend the segment and the length is closer
>>>> than the tangent, so it's impossible. Also things may be complicated if
>>>> the algorithm tries to go around the tip of the ellipse (that is, if the
>>>> path is too long).
>>>>
>>>> For finding a tangent to an ellipse, transform the problem to the
>>>> circle, solve, and then transform back. In BOSL2:
>>>>
>>>> function ellipse_tangents(a,b,pt) =
>>>> let(
>>>> r=a,
>>>> scaledpt = yscale(a/b,pt),
>>>> tangents = circle_point_tangents(r,[0,0],scaledpt)
>>>> )
>>>> yscale(b/a,tangents);
>>>>
>>>> On Sun, Aug 7, 2022 at 9:54 PM Father Horton <fatherhorton@gmail.com>
>>>> wrote:
>>>>
>>>>> I did that wrong. It's not (0, b) to (0, L/2). It's (0, b) to (L/2,
>>>>> b). That's always outside the ellipse.
>>>>>
>>>>> I will look at the second problem as soon as I figure out how to
>>>>> calculate a tangent line to an ellipse containing a point not on the
>>>>> ellipse. Google is not helping much.
>>>>>
>>>>> On Sun, Aug 7, 2022 at 8:32 PM Adrian Mariano <avm4@cornell.edu>
>>>>> wrote:
>>>>>
>>>>>> Father Horton,
>>>>>>
>>>>>> I'm having trouble understanding your proposed method. A segment
>>>>>> from (0,b) to (0,L/2) might be inside the ellipse depending on N, and then
>>>>>> there won't be a tangent.
>>>>>>
>>>>>> It also sounds like your idea is somewhat similar to Jordan's idea:
>>>>>> basically start at one side and work around the ellipse and see if it
>>>>>> works. If not, make a correction. The problem with this idea is that
>>>>>> there are two parameters for the first segment, so how do you make a
>>>>>> correction? (The two parameters are the location of the endpoint of the
>>>>>> first segment and its length.)
>>>>>>
>>>>>> On Sun, Aug 7, 2022 at 5:33 PM Father Horton <fatherhorton@gmail.com>
>>>>>> wrote:
>>>>>>
>>>>>>> How about this to construct N equal length segments around an
>>>>>>> ellipse with axes a and b:
>>>>>>>
>>>>>>> 1) Select an initial segment length estimate (e.g., Ramanujan's
>>>>>>> formula / N) = L
>>>>>>> 2) Construct a half-segment from (0, b) to (0, L/2).
>>>>>>> 3) Find a line tangent to the ellipse and passing through (0, L/2).
>>>>>>> 3) Find the point on the tangent line that is L away from (0, L/2).
>>>>>>> 4) Repeat (3) substituting the new point for (0, L/2). until you
>>>>>>> have N segments.
>>>>>>>
>>>>>>> If everything worked perfectly, the last point found would be (0,
>>>>>>> -L/2). If there's a gap, increase L and try again. If there's overlap,
>>>>>>> decrease L and try again.
>>>>>>>
>>>>>>> On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano <
>>>>>>> rcmpersiano@gmail.com> wrote:
>>>>>>>
>>>>>>>> Adrian, why to use a (either inscribed or circumscribed) polygon
>>>>>>>> with equal side as an approximation of an ellipsis? It doesn't seem to be a
>>>>>>>> good approximation of it. I get a better approximation by taking an angle
>>>>>>>> uniform sample of the curve in the parameter space.
>>>>>>>>
>>>>>>>> [image: Uniformly_sampled_ellipsis.png]
>>>>>>>>
>>>>>>>>
>>>>>>>> Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano <
>>>>>>>> avm4@cornell.edu> escreveu:
>>>>>>>>
>>>>>>>>> Circumscribed means tangents to points on the ellipse. But there
>>>>>>>>> are no arc lengths (equal or unequal) in this problem, nor equally spaced
>>>>>>>>> angles. The angles of each segment will be different. But the lengths of
>>>>>>>>> the segments are not the same as the arc lengths, which is why arc length
>>>>>>>>> does not matter. Here's a solution to the N=6 polygon sort of inspired by
>>>>>>>>> the idea Jordan suggested, where I adjust the first segment until the
>>>>>>>>> second segment has the right length. The rest of the solution is then
>>>>>>>>> obtained using symmetry, which means there is only a single unknown that
>>>>>>>>> can simply be solved for. But this doesn't generalize to higher N.
>>>>>>>>>
>>>>>>>>> [image: image.png]
>>>>>>>>> include<BOSL2/std.scad>
>>>>>>>>>
>>>>>>>>> // ellipse semiaxes
>>>>>>>>> // If a is much smaller than b it will fail
>>>>>>>>> a=5;
>>>>>>>>> b=2;
>>>>>>>>>
>>>>>>>>> // Find point on ellipse at specified x coordinate
>>>>>>>>> function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> // Find tangent line to ellipse at point (u,w)
>>>>>>>>> // and determine the x value where that line has
>>>>>>>>> // given y value.
>>>>>>>>> function findx(a,b,u,w,y) =
>>>>>>>>> let(
>>>>>>>>> A = -u*b^2/w/a^2,
>>>>>>>>> B = w+(u*b)^2/w/a^2
>>>>>>>>> )
>>>>>>>>> (y-B)/A;
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> // First segment if tangent point is at x=u
>>>>>>>>> function first_seg(a,b,u) =
>>>>>>>>> let(
>>>>>>>>> w = ellipse_pt(a,b,u),
>>>>>>>>> x1 = findx(a,b,u,w,0),
>>>>>>>>> x2 = findx(a,b,u,w,b)
>>>>>>>>> )
>>>>>>>>> [[x1,0],[x2,b]];
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> err = function (u) let(pt = first_seg(a,b,u))
>>>>>>>>> 2*pt[1].x - norm(pt[0]-pt[1]);
>>>>>>>>>
>>>>>>>>> x = root_find(err,.2*a,.9*a);
>>>>>>>>> first = first_seg(a,b,x);
>>>>>>>>> top = concat(first,reverse(xflip(first)));
>>>>>>>>> poly = concat(top, reverse(yflip(list_tail(top))));
>>>>>>>>>
>>>>>>>>> stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
>>>>>>>>> color("red")stroke(poly,width=.05,closed=true);
>>>>>>>>>
>>>>>>>>> // Verify equality of segment lengths
>>>>>>>>> echo(path_segment_lengths(poly,closed=true));
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>
>>>>>>>>>> Circumscribed means tangents from points on ellipse
>>>>>>>>>> How can the tangents from equally spaced arc lengths or equally
>>>>>>>>>> spaced angles be equal for an ellipse
>>>>>>>>>> You need to change the arc lengths dynamically and also
>>>>>>>>>> accurately estimate the segment length which fits the ellipse .
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, <avm4@cornell.edu>
>>>>>>>>>> wrote:
>>>>>>>>>>
>>>>>>>>>>> Your intuition is incorrect: a circle scaled in one direction
>>>>>>>>>>> *IS* an ellipse. The motivation for being interested in
>>>>>>>>>>> circumscribing is to make holes, though the constraint of equal length
>>>>>>>>>>> sides is probably not what you'd want for that purpose. You can make a
>>>>>>>>>>> nonuniform circumscribing polygon for an ellipse by simply making a polygon
>>>>>>>>>>> that circumscribes a circle and scaling it.
>>>>>>>>>>>
>>>>>>>>>>> On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson <bob@rjcarlson.com>
>>>>>>>>>>> wrote:
>>>>>>>>>>>
>>>>>>>>>>>> Seems like I remember Nop Head talking about using
>>>>>>>>>>>> circumscribed holes for screws and even rotating the polygons to produce a
>>>>>>>>>>>> more even hole.
>>>>>>>>>>>>
>>>>>>>>>>>> If an ellipse is the nominal shape, the using a circumscribed
>>>>>>>>>>>> polygon for a hole and an inscribed polygon for a solid might make sense.
>>>>>>>>>>>>
>>>>>>>>>>>> Btw, can someone confirm my intuition that a circle scaled in
>>>>>>>>>>>> one dimension is not an ellipse?
>>>>>>>>>>>>
>>>>>>>>>>>> -Bob
>>>>>>>>>>>> Tucson AZ
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> On Aug 5, 2022, at 15:38, Adrian Mariano <avm4@cornell.edu>
>>>>>>>>>>>> wrote:
>>>>>>>>>>>>
>>>>>>>>>>>> I don't have a specific need for a circumscribing elliptical
>>>>>>>>>>>> polygon with equal sides. I was writing code to make ellipses and it
>>>>>>>>>>>> seemed like equal side approximations would be nice, so I solved the case
>>>>>>>>>>>> for inscribed, but couldn't solve the circumscribed case. So it's an
>>>>>>>>>>>> unsolved puzzle I've got that this discussion reminded me of.
>>>>>>>>>>>>
>>>>>>>>>>>> On Fri, Aug 5, 2022 at 6:33 PM nop head <nop.head@gmail.com>
>>>>>>>>>>>> wrote:
>>>>>>>>>>>>
>>>>>>>>>>>>> A long discussion but I have to ask why do you need an
>>>>>>>>>>>>> elliptical polygon with equal tangential sides?
>>>>>>>>>>>>>
>>>>>>>>>>>>> On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu>
>>>>>>>>>>>>> wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> You think that a circumscribing polygon with equal length
>>>>>>>>>>>>>> segments does not exist?? I'm skeptical of this claim.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> For circumscribed, it's not so hard. You start with a
>>>>>>>>>>>>>> uniform-in-angle point distribution, compute the segment lengths, compare
>>>>>>>>>>>>>> to the mean segment length to get an error, and then compute an adjustment
>>>>>>>>>>>>>> based on the error. It converges quickly.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> So for example, I have segments lengths of:
>>>>>>>>>>>>>> ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
>>>>>>>>>>>>>> 5.57297, 5.57297]
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Maybe not exactly equal, since tolerance is 1e-9, but surely
>>>>>>>>>>>>>> good enough.
>>>>>>>>>>>>>> <image.png>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> The code for this is in BOSL2 in shapes2d.scad. A similar
>>>>>>>>>>>>>> algorithm should exist for the circumscribed case, but as I said, my
>>>>>>>>>>>>>> efforts to find it have failed.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
>>>>>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> I think circumscribed polygon segments cannot be equal
>>>>>>>>>>>>>>> lengths.
>>>>>>>>>>>>>>> inscribed ones may be approximately equal, but would be a
>>>>>>>>>>>>>>> very tough problem to solve.
>>>>>>>>>>>>>>> I tried something.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <
>>>>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> I think you need to be a little careful about defining what
>>>>>>>>>>>>>>>> you want to do. If you want to approximate an ellipse with points that are
>>>>>>>>>>>>>>>> distributed uniformly over the ellipse, that could be done by approximating
>>>>>>>>>>>>>>>> the arc length of the ellipse to find the necessary points---e.g. advance
>>>>>>>>>>>>>>>> distance d along the ellipse at each step. But if you want to approximate
>>>>>>>>>>>>>>>> an ellipse with a polygon with equal length sides, that approach does not
>>>>>>>>>>>>>>>> do it, because the length of the segments will not be constant as the span
>>>>>>>>>>>>>>>> sections of the ellipse with differing curvature. And since the perimeter
>>>>>>>>>>>>>>>> of the desired polygon is unknown, you can't just step it out analogously.
>>>>>>>>>>>>>>>> The circle doesn't present this challenge due to having constant
>>>>>>>>>>>>>>>> curvature.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
>>>>>>>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> I think our emails crossed. I was addressing how to
>>>>>>>>>>>>>>>>> approximate an ellipse with constant-length segments in general, not how to
>>>>>>>>>>>>>>>>> solve your problem. When I had to do something similar, I used polar
>>>>>>>>>>>>>>>>> coordinates and stepped up the value of theta until the distance from the
>>>>>>>>>>>>>>>>> previous point was correct. (In the constant-length version, I'd keep a
>>>>>>>>>>>>>>>>> running total of the maximum desired length so that round-off errors don't
>>>>>>>>>>>>>>>>> accumulate.)
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Figuring out the circumscribed polygon is, as you note, a
>>>>>>>>>>>>>>>>> greater challenge. I can visualize how to do it for n = 4, but it doesn't
>>>>>>>>>>>>>>>>> extrapolate at all.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <
>>>>>>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> Eh, there's no such thing as "brute force" in this
>>>>>>>>>>>>>>>>>> situation. Step through *what* with sufficiently small steps? Of course
>>>>>>>>>>>>>>>>>> an iterative algorithm is necessary. This is true for most interesting
>>>>>>>>>>>>>>>>>> problems. No big deal. Show me an iterative algorithm that converges to
>>>>>>>>>>>>>>>>>> the circumscribing polygon. When I tried to devise such an algorithm I
>>>>>>>>>>>>>>>>>> couldn't find one that would converge. Writing an algorithm that would
>>>>>>>>>>>>>>>>>> converge to the *inscribed* polygon seemed to be straight forward. Note
>>>>>>>>>>>>>>>>>> also that computing arc length of the ellipse doesn't really help here.
>>>>>>>>>>>>>>>>>> You can do that by solving an elliptical integral, or summing up small
>>>>>>>>>>>>>>>>>> segments. But what's the connection between arc length of the ellipse and
>>>>>>>>>>>>>>>>>> a circumscribing polygon with equal segment lengths? If there's a
>>>>>>>>>>>>>>>>>> connection, I don't see it.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
>>>>>>>>>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> Except by brute force (step through it with sufficiently
>>>>>>>>>>>>>>>>>>> small steps), it’s not possible because there’s no closed form expression
>>>>>>>>>>>>>>>>>>> for the distance between two points on a general ellipse (as opposed to
>>>>>>>>>>>>>>>>>>> special cases like a circle).
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <
>>>>>>>>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> Here's a challenge for you: draw an "ellipse" polygon
>>>>>>>>>>>>>>>>>>>> with N segments that are all the same length that circumscribes a perfect
>>>>>>>>>>>>>>>>>>>> ellipse with given semiaxes a and b. I tried to solve this problem and
>>>>>>>>>>>>>>>>>>>> could not find a solution. I can draw an *inscribed* N segment "ellipse"
>>>>>>>>>>>>>>>>>>>> polygon with all segments the same length---though it's not easy---but the
>>>>>>>>>>>>>>>>>>>> circumscribed case eludes me. I think of the inscribed case as, every
>>>>>>>>>>>>>>>>>>>> vertex is located on the ellipse and the circumscribed case, every segment
>>>>>>>>>>>>>>>>>>>> is tangent to the ellipse
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>>>>>>>>>>>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> In openscad drawing a ellipse is not too difficult
>>>>>>>>>>>>>>>>>>>>> Refer following code
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> function to draw an ellipse with semi-major and
>>>>>>>>>>>>>>>>>>>>> semi-minor axis "r1" and "r2" respectively and with center "cp" and number
>>>>>>>>>>>>>>>>>>>>> of segment "s"
>>>>>>>>>>>>>>>>>>>>> /
>>>>>>>>>>>>>>>>>>>>> // example:
>>>>>>>>>>>>>>>>>>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> function ellipse(r1,r2,cp,s=30)=
>>>>>>>>>>>>>>>>>>>>> let(
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>>>>>>>>>>>>>>>>>>>> )sec;
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>>>>>>>>>>>>>>>>>>>> openscad@jordan.maileater.net> wrote:
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> I created an oval using:
>>>>>>>>>>>>>>>>>>>>>> inMm = 25.4;
>>>>>>>>>>>>>>>>>>>>>> fullW = 35.5 * inMm;
>>>>>>>>>>>>>>>>>>>>>> fullL = 54.5 * inMm;
>>>>>>>>>>>>>>>>>>>>>> resize([fullW, fullL, 1])
>>>>>>>>>>>>>>>>>>>>>> circle(d=fullW);
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> Tip: use scale() rather than resize(); it previews
>>>>>>>>>>>>>>>>>>>>>> faster in complicated cases.
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> You're taking a circle and stretching it; the result
>>>>>>>>>>>>>>>>>>>>>> of that stretching is an ellipse.
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> I'm not 100% sure - maybe one of the real math
>>>>>>>>>>>>>>>>>>>>>> wizards can chime in - but I think the shape you are asking for would not
>>>>>>>>>>>>>>>>>>>>>> be an ellipse.
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> I believe that an ellipse is fully defined by its
>>>>>>>>>>>>>>>>>>>>>> major (long) and minor (short) axis dimensions; the other parts of the
>>>>>>>>>>>>>>>>>>>>>> curve mathematically flow from those two measurements.
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> You can make something else, but you can't make it
>>>>>>>>>>>>>>>>>>>>>> using this technique.
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> You can use a Bézier function.
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> Using BOSL2:
>>>>>>>>>>>>>>>>>>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> DIY, here's a function that I cobbled together some
>>>>>>>>>>>>>>>>>>>>>> years ago:
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>>>>>>>>>>>>>>>>>>>> // but that yielded either single points or a raft of triangles;
>>>>>>>>>>>>>>>>>>>>>> // this yields a vector of points that you can then concatenate
>>>>>>>>>>>>>>>>>>>>>> // with other pieces to form a single polygon.
>>>>>>>>>>>>>>>>>>>>>> // If we were really clever, I think it would be possible to
>>>>>>>>>>>>>>>>>>>>>> // automatically space the output points based on how linear
>>>>>>>>>>>>>>>>>>>>>> // the curve is at that point. But right now I'm not that clever.
>>>>>>>>>>>>>>>>>>>>>> function BEZ03(u) = pow((1-u), 3);
>>>>>>>>>>>>>>>>>>>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>>>>>>>>>>>>>>>>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>>>>>>>>>>>>>>>>>>>> function BEZ33(u) = pow(u,3);
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>>>>>>>>>>>>>>>>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>>>>>>>>>>>>>>>>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> // p0 - start point
>>>>>>>>>>>>>>>>>>>>>> // p1 - control point 1, line departs p0 headed this way
>>>>>>>>>>>>>>>>>>>>>> // p2 - control point 2, line arrives at p3 from this way
>>>>>>>>>>>>>>>>>>>>>> // p3 - end point
>>>>>>>>>>>>>>>>>>>>>> // segs - number of segments
>>>>>>>>>>>>>>>>>>>>>> function bez(p0, p1, p2, p3, segs) = [
>>>>>>>>>>>>>>>>>>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>>>>>>>>>>>>>>>>>>>> ];
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> It is that final function bez() that I actually call,
>>>>>>>>>>>>>>>>>>>>>> e.g.
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> often concatenating the results of several bez()
>>>>>>>>>>>>>>>>>>>>>> calls together.
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> Note that the four coordinates correspond to the
>>>>>>>>>>>>>>>>>>>>>> coordinates of the control points in a GUI drawing program: the start
>>>>>>>>>>>>>>>>>>>>>> point, control point associated with the start, the control point
>>>>>>>>>>>>>>>>>>>>>> associated with the end, and the end point. I have sometimes created
>>>>>>>>>>>>>>>>>>>>>> OpenSCAD objects by tracing an object in a drawing program and then reading
>>>>>>>>>>>>>>>>>>>>>> off the coordinates of the various control points.
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> Note that Bézier functions are not the only
>>>>>>>>>>>>>>>>>>>>>> mathematical curves. (For instance, they cannot exactly match an arc of a
>>>>>>>>>>>>>>>>>>>>>> circle.) I'm sure there are numerous other functions that also produce
>>>>>>>>>>>>>>>>>>>>>> curves.
>>>>>>>>>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> _______________________________________________
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>
FH
Father Horton
Mon, Aug 8, 2022 8:11 PM
My end condition is that the last point calculated is (-L/2, b), but as I
think about it, it does make sense that N has to be a multiple of 2,
because the whole thing has to be symmetrical around the y-axis.
On Mon, Aug 8, 2022 at 2:35 PM Father Horton fatherhorton@gmail.com wrote:
Here's a Google sheet that shows what I have in mind.
https://docs.google.com/spreadsheets/d/13YOJKkWWfcgZNKVn_nlBtQZ_bbxNvN6SdVv1bVWLDP8/edit?usp=sharing
I have to adjust the signs on the square roots manually, so I didn't
iterate until I got equal lengths, but this is the idea. Blue points are
end points of the circumscribing segments and red points are the tangent
points on the ellipse. I have other things to do at the moment, but I may
revisit later to get the signs to work out right and then see if it will
iterate to an acceptable result.
On Mon, Aug 8, 2022 at 1:31 PM Adrian Mariano avm4@cornell.edu wrote:
The restriction arises because of the end condition. There are 4 cases
depending on what happens at the four ends of the ellipse (the points where
it intersects the coordinate axes). The four possibilities are at each end
either a corner or a flat tangent at the end. Your approach as I
understand it begins with a flat and constructs a quarter of the polygon,
ending with a point. Such a polygon must have 4k+2 sides. I do not see
how the end condition can be formulated to end on a flat. Also I do not
see how it can start on a point. If n is odd you can start at one end and
go half way around and end on a point. But if n is divisible by 4 you can
have 4 flats or four points and the method cannot produce either of them.
Have I missed something?
On Mon, Aug 8, 2022, 13:42 Father Horton fatherhorton@gmail.com wrote:
I'm not sure why there should be a restriction on N; I think it's
general. I did a spreadsheet that looks like it's doing the right thing in
the first quadrant, but it (or I) gets lost when I have to select the right
value for the sign of the square roots involved as the process enters other
quadrants.
On Sun, Aug 7, 2022 at 9:43 PM Father Horton fatherhorton@gmail.com
wrote:
Google showed me that method when I came up with the right terms. But
it's too late at night for me to try to implement it. Maybe tomorrow.
On Sun, Aug 7, 2022 at 9:41 PM Adrian Mariano avm4@cornell.edu wrote:
Now I understand your idea. I think it should be possible to make
that work for the case of N=4k+2. It does seem like there are some
complications, like you try to extend the segment and the length is closer
than the tangent, so it's impossible. Also things may be complicated if
the algorithm tries to go around the tip of the ellipse (that is, if the
path is too long).
For finding a tangent to an ellipse, transform the problem to the
circle, solve, and then transform back. In BOSL2:
function ellipse_tangents(a,b,pt) =
let(
r=a,
scaledpt = yscale(a/b,pt),
tangents = circle_point_tangents(r,[0,0],scaledpt)
)
yscale(b/a,tangents);
On Sun, Aug 7, 2022 at 9:54 PM Father Horton fatherhorton@gmail.com
wrote:
I did that wrong. It's not (0, b) to (0, L/2). It's (0, b) to (L/2,
b). That's always outside the ellipse.
I will look at the second problem as soon as I figure out how to
calculate a tangent line to an ellipse containing a point not on the
ellipse. Google is not helping much.
On Sun, Aug 7, 2022 at 8:32 PM Adrian Mariano avm4@cornell.edu
wrote:
Father Horton,
I'm having trouble understanding your proposed method. A segment
from (0,b) to (0,L/2) might be inside the ellipse depending on N, and then
there won't be a tangent.
It also sounds like your idea is somewhat similar to Jordan's idea:
basically start at one side and work around the ellipse and see if it
works. If not, make a correction. The problem with this idea is that
there are two parameters for the first segment, so how do you make a
correction? (The two parameters are the location of the endpoint of the
first segment and its length.)
On Sun, Aug 7, 2022 at 5:33 PM Father Horton fatherhorton@gmail.com
wrote:
How about this to construct N equal length segments around an
ellipse with axes a and b:
- Select an initial segment length estimate (e.g., Ramanujan's
formula / N) = L
- Construct a half-segment from (0, b) to (0, L/2).
- Find a line tangent to the ellipse and passing through (0, L/2).
- Find the point on the tangent line that is L away from (0, L/2).
- Repeat (3) substituting the new point for (0, L/2). until you
have N segments.
If everything worked perfectly, the last point found would be (0,
-L/2). If there's a gap, increase L and try again. If there's overlap,
decrease L and try again.
On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano <
rcmpersiano@gmail.com> wrote:
Adrian, why to use a (either inscribed or circumscribed) polygon
with equal side as an approximation of an ellipsis? It doesn't seem to be a
good approximation of it. I get a better approximation by taking an angle
uniform sample of the curve in the parameter space.
[image: Uniformly_sampled_ellipsis.png]
Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano <
avm4@cornell.edu> escreveu:
Circumscribed means tangents to points on the ellipse. But there
are no arc lengths (equal or unequal) in this problem, nor equally spaced
angles. The angles of each segment will be different. But the lengths of
the segments are not the same as the arc lengths, which is why arc length
does not matter. Here's a solution to the N=6 polygon sort of inspired by
the idea Jordan suggested, where I adjust the first segment until the
second segment has the right length. The rest of the solution is then
obtained using symmetry, which means there is only a single unknown that
can simply be solved for. But this doesn't generalize to higher N.
[image: image.png]
include<BOSL2/std.scad>
// ellipse semiaxes
// If a is much smaller than b it will fail
a=5;
b=2;
// Find point on ellipse at specified x coordinate
function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
// Find tangent line to ellipse at point (u,w)
// and determine the x value where that line has
// given y value.
function findx(a,b,u,w,y) =
let(
A = -ub^2/w/a^2,
B = w+(ub)^2/w/a^2
)
(y-B)/A;
// First segment if tangent point is at x=u
function first_seg(a,b,u) =
let(
w = ellipse_pt(a,b,u),
x1 = findx(a,b,u,w,0),
x2 = findx(a,b,u,w,b)
)
[[x1,0],[x2,b]];
err = function (u) let(pt = first_seg(a,b,u))
2*pt[1].x - norm(pt[0]-pt[1]);
x = root_find(err,.2a,.9a);
first = first_seg(a,b,x);
top = concat(first,reverse(xflip(first)));
poly = concat(top, reverse(yflip(list_tail(top))));
stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
color("red")stroke(poly,width=.05,closed=true);
// Verify equality of segment lengths
echo(path_segment_lengths(poly,closed=true));
On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
Circumscribed means tangents from points on ellipse
How can the tangents from equally spaced arc lengths or equally
spaced angles be equal for an ellipse
You need to change the arc lengths dynamically and also
accurately estimate the segment length which fits the ellipse .
On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, avm4@cornell.edu
wrote:
Your intuition is incorrect: a circle scaled in one direction
IS an ellipse. The motivation for being interested in
circumscribing is to make holes, though the constraint of equal length
sides is probably not what you'd want for that purpose. You can make a
nonuniform circumscribing polygon for an ellipse by simply making a polygon
that circumscribes a circle and scaling it.
On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson bob@rjcarlson.com
wrote:
Seems like I remember Nop Head talking about using
circumscribed holes for screws and even rotating the polygons to produce a
more even hole.
If an ellipse is the nominal shape, the using a circumscribed
polygon for a hole and an inscribed polygon for a solid might make sense.
Btw, can someone confirm my intuition that a circle scaled in
one dimension is not an ellipse?
-Bob
Tucson AZ
On Aug 5, 2022, at 15:38, Adrian Mariano avm4@cornell.edu
wrote:
I don't have a specific need for a circumscribing elliptical
polygon with equal sides. I was writing code to make ellipses and it
seemed like equal side approximations would be nice, so I solved the case
for inscribed, but couldn't solve the circumscribed case. So it's an
unsolved puzzle I've got that this discussion reminded me of.
On Fri, Aug 5, 2022 at 6:33 PM nop head nop.head@gmail.com
wrote:
A long discussion but I have to ask why do you need an
elliptical polygon with equal tangential sides?
On Fri, 5 Aug 2022, 21:58 Adrian Mariano, avm4@cornell.edu
wrote:
You think that a circumscribing polygon with equal length
segments does not exist?? I'm skeptical of this claim.
For circumscribed, it's not so hard. You start with a
uniform-in-angle point distribution, compute the segment lengths, compare
to the mean segment length to get an error, and then compute an adjustment
based on the error. It converges quickly.
So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
5.57297, 5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but surely
good enough.
<image.png>
The code for this is in BOSL2 in shapes2d.scad. A similar
algorithm should exist for the circumscribed case, but as I said, my
efforts to find it have failed.
On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
I think circumscribed polygon segments cannot be equal
lengths.
inscribed ones may be approximately equal, but would be a
very tough problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <
avm4@cornell.edu> wrote:
I think you need to be a little careful about defining
what you want to do. If you want to approximate an ellipse with points
that are distributed uniformly over the ellipse, that could be done by
approximating the arc length of the ellipse to find the necessary
points---e.g. advance distance d along the ellipse at each step. But if
you want to approximate an ellipse with a polygon with equal length sides,
that approach does not do it, because the length of the segments will not
be constant as the span sections of the ellipse with differing curvature.
And since the perimeter of the desired polygon is unknown, you can't just
step it out analogously. The circle doesn't present this challenge due to
having constant curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
fatherhorton@gmail.com> wrote:
I think our emails crossed. I was addressing how to
approximate an ellipse with constant-length segments in general, not how to
solve your problem. When I had to do something similar, I used polar
coordinates and stepped up the value of theta until the distance from the
previous point was correct. (In the constant-length version, I'd keep a
running total of the maximum desired length so that round-off errors don't
accumulate.)
Figuring out the circumscribed polygon is, as you note, a
greater challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <
avm4@cornell.edu> wrote:
Eh, there's no such thing as "brute force" in this
situation. Step through what with sufficiently small steps? Of course
an iterative algorithm is necessary. This is true for most interesting
problems. No big deal. Show me an iterative algorithm that converges to
the circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
fatherhorton@gmail.com> wrote:
Except by brute force (step through it with
sufficiently small steps), it’s not possible because there’s no closed form
expression for the distance between two points on a general ellipse (as
opposed to special cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <
avm4@cornell.edu> wrote:
Here's a challenge for you: draw an "ellipse" polygon
with N segments that are all the same length that circumscribes a perfect
ellipse with given semiaxes a and b. I tried to solve this problem and
could not find a solution. I can draw an inscribed N segment "ellipse"
polygon with all segments the same length---though it's not easy---but the
circumscribed case eludes me. I think of the inscribed case as, every
vertex is located on the ellipse and the circumscribed case, every segment
is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and
semi-minor axis "r1" and "r2" respectively and with center "cp" and number
of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews
faster in complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the result
of that stretching is an ellipse.
I'm not 100% sure - maybe one of the real math
wizards can chime in - but I think the shape you are asking for would not
be an ellipse.
I believe that an ellipse is fully defined by its
major (long) and minor (short) axis dimensions; the other parts of the
curve mathematically flow from those two measurements.
You can make something else, but you can't make it
using this technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some
years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually
call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez()
calls together.
Note that the four coordinates correspond to the
coordinates of the control points in a GUI drawing program: the start
point, control point associated with the start, the control point
associated with the end, and the end point. I have sometimes created
OpenSCAD objects by tracing an object in a drawing program and then reading
off the coordinates of the various control points.
Note that Bézier functions are not the only
mathematical curves. (For instance, they cannot exactly match an arc of a
circle.) I'm sure there are numerous other functions that also produce
curves.
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My end condition is that the last point calculated is (-L/2, b), but as I
think about it, it does make sense that N has to be a multiple of 2,
because the whole thing has to be symmetrical around the y-axis.
On Mon, Aug 8, 2022 at 2:35 PM Father Horton <fatherhorton@gmail.com> wrote:
> Here's a Google sheet that shows what I have in mind.
> https://docs.google.com/spreadsheets/d/13YOJKkWWfcgZNKVn_nlBtQZ_bbxNvN6SdVv1bVWLDP8/edit?usp=sharing
>
> I have to adjust the signs on the square roots manually, so I didn't
> iterate until I got equal lengths, but this is the idea. Blue points are
> end points of the circumscribing segments and red points are the tangent
> points on the ellipse. I have other things to do at the moment, but I may
> revisit later to get the signs to work out right and then see if it will
> iterate to an acceptable result.
>
> On Mon, Aug 8, 2022 at 1:31 PM Adrian Mariano <avm4@cornell.edu> wrote:
>
>> The restriction arises because of the end condition. There are 4 cases
>> depending on what happens at the four ends of the ellipse (the points where
>> it intersects the coordinate axes). The four possibilities are at each end
>> either a corner or a flat tangent at the end. Your approach as I
>> understand it begins with a flat and constructs a quarter of the polygon,
>> ending with a point. Such a polygon must have 4k+2 sides. I do not see
>> how the end condition can be formulated to end on a flat. Also I do not
>> see how it can start on a point. If n is odd you can start at one end and
>> go half way around and end on a point. But if n is divisible by 4 you can
>> have 4 flats or four points and the method cannot produce either of them.
>> Have I missed something?
>>
>>
>> On Mon, Aug 8, 2022, 13:42 Father Horton <fatherhorton@gmail.com> wrote:
>>
>>> I'm not sure why there should be a restriction on N; I think it's
>>> general. I did a spreadsheet that looks like it's doing the right thing in
>>> the first quadrant, but it (or I) gets lost when I have to select the right
>>> value for the sign of the square roots involved as the process enters other
>>> quadrants.
>>>
>>> On Sun, Aug 7, 2022 at 9:43 PM Father Horton <fatherhorton@gmail.com>
>>> wrote:
>>>
>>>> Google showed me that method when I came up with the right terms. But
>>>> it's too late at night for me to try to implement it. Maybe tomorrow.
>>>>
>>>> On Sun, Aug 7, 2022 at 9:41 PM Adrian Mariano <avm4@cornell.edu> wrote:
>>>>
>>>>> Now I understand your idea. I think it should be possible to make
>>>>> that work for the case of N=4k+2. It does seem like there are some
>>>>> complications, like you try to extend the segment and the length is closer
>>>>> than the tangent, so it's impossible. Also things may be complicated if
>>>>> the algorithm tries to go around the tip of the ellipse (that is, if the
>>>>> path is too long).
>>>>>
>>>>> For finding a tangent to an ellipse, transform the problem to the
>>>>> circle, solve, and then transform back. In BOSL2:
>>>>>
>>>>> function ellipse_tangents(a,b,pt) =
>>>>> let(
>>>>> r=a,
>>>>> scaledpt = yscale(a/b,pt),
>>>>> tangents = circle_point_tangents(r,[0,0],scaledpt)
>>>>> )
>>>>> yscale(b/a,tangents);
>>>>>
>>>>> On Sun, Aug 7, 2022 at 9:54 PM Father Horton <fatherhorton@gmail.com>
>>>>> wrote:
>>>>>
>>>>>> I did that wrong. It's not (0, b) to (0, L/2). It's (0, b) to (L/2,
>>>>>> b). That's always outside the ellipse.
>>>>>>
>>>>>> I will look at the second problem as soon as I figure out how to
>>>>>> calculate a tangent line to an ellipse containing a point not on the
>>>>>> ellipse. Google is not helping much.
>>>>>>
>>>>>> On Sun, Aug 7, 2022 at 8:32 PM Adrian Mariano <avm4@cornell.edu>
>>>>>> wrote:
>>>>>>
>>>>>>> Father Horton,
>>>>>>>
>>>>>>> I'm having trouble understanding your proposed method. A segment
>>>>>>> from (0,b) to (0,L/2) might be inside the ellipse depending on N, and then
>>>>>>> there won't be a tangent.
>>>>>>>
>>>>>>> It also sounds like your idea is somewhat similar to Jordan's idea:
>>>>>>> basically start at one side and work around the ellipse and see if it
>>>>>>> works. If not, make a correction. The problem with this idea is that
>>>>>>> there are two parameters for the first segment, so how do you make a
>>>>>>> correction? (The two parameters are the location of the endpoint of the
>>>>>>> first segment and its length.)
>>>>>>>
>>>>>>> On Sun, Aug 7, 2022 at 5:33 PM Father Horton <fatherhorton@gmail.com>
>>>>>>> wrote:
>>>>>>>
>>>>>>>> How about this to construct N equal length segments around an
>>>>>>>> ellipse with axes a and b:
>>>>>>>>
>>>>>>>> 1) Select an initial segment length estimate (e.g., Ramanujan's
>>>>>>>> formula / N) = L
>>>>>>>> 2) Construct a half-segment from (0, b) to (0, L/2).
>>>>>>>> 3) Find a line tangent to the ellipse and passing through (0, L/2).
>>>>>>>> 3) Find the point on the tangent line that is L away from (0, L/2).
>>>>>>>> 4) Repeat (3) substituting the new point for (0, L/2). until you
>>>>>>>> have N segments.
>>>>>>>>
>>>>>>>> If everything worked perfectly, the last point found would be (0,
>>>>>>>> -L/2). If there's a gap, increase L and try again. If there's overlap,
>>>>>>>> decrease L and try again.
>>>>>>>>
>>>>>>>> On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano <
>>>>>>>> rcmpersiano@gmail.com> wrote:
>>>>>>>>
>>>>>>>>> Adrian, why to use a (either inscribed or circumscribed) polygon
>>>>>>>>> with equal side as an approximation of an ellipsis? It doesn't seem to be a
>>>>>>>>> good approximation of it. I get a better approximation by taking an angle
>>>>>>>>> uniform sample of the curve in the parameter space.
>>>>>>>>>
>>>>>>>>> [image: Uniformly_sampled_ellipsis.png]
>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano <
>>>>>>>>> avm4@cornell.edu> escreveu:
>>>>>>>>>
>>>>>>>>>> Circumscribed means tangents to points on the ellipse. But there
>>>>>>>>>> are no arc lengths (equal or unequal) in this problem, nor equally spaced
>>>>>>>>>> angles. The angles of each segment will be different. But the lengths of
>>>>>>>>>> the segments are not the same as the arc lengths, which is why arc length
>>>>>>>>>> does not matter. Here's a solution to the N=6 polygon sort of inspired by
>>>>>>>>>> the idea Jordan suggested, where I adjust the first segment until the
>>>>>>>>>> second segment has the right length. The rest of the solution is then
>>>>>>>>>> obtained using symmetry, which means there is only a single unknown that
>>>>>>>>>> can simply be solved for. But this doesn't generalize to higher N.
>>>>>>>>>>
>>>>>>>>>> [image: image.png]
>>>>>>>>>> include<BOSL2/std.scad>
>>>>>>>>>>
>>>>>>>>>> // ellipse semiaxes
>>>>>>>>>> // If a is much smaller than b it will fail
>>>>>>>>>> a=5;
>>>>>>>>>> b=2;
>>>>>>>>>>
>>>>>>>>>> // Find point on ellipse at specified x coordinate
>>>>>>>>>> function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> // Find tangent line to ellipse at point (u,w)
>>>>>>>>>> // and determine the x value where that line has
>>>>>>>>>> // given y value.
>>>>>>>>>> function findx(a,b,u,w,y) =
>>>>>>>>>> let(
>>>>>>>>>> A = -u*b^2/w/a^2,
>>>>>>>>>> B = w+(u*b)^2/w/a^2
>>>>>>>>>> )
>>>>>>>>>> (y-B)/A;
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> // First segment if tangent point is at x=u
>>>>>>>>>> function first_seg(a,b,u) =
>>>>>>>>>> let(
>>>>>>>>>> w = ellipse_pt(a,b,u),
>>>>>>>>>> x1 = findx(a,b,u,w,0),
>>>>>>>>>> x2 = findx(a,b,u,w,b)
>>>>>>>>>> )
>>>>>>>>>> [[x1,0],[x2,b]];
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> err = function (u) let(pt = first_seg(a,b,u))
>>>>>>>>>> 2*pt[1].x - norm(pt[0]-pt[1]);
>>>>>>>>>>
>>>>>>>>>> x = root_find(err,.2*a,.9*a);
>>>>>>>>>> first = first_seg(a,b,x);
>>>>>>>>>> top = concat(first,reverse(xflip(first)));
>>>>>>>>>> poly = concat(top, reverse(yflip(list_tail(top))));
>>>>>>>>>>
>>>>>>>>>> stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
>>>>>>>>>> color("red")stroke(poly,width=.05,closed=true);
>>>>>>>>>>
>>>>>>>>>> // Verify equality of segment lengths
>>>>>>>>>> echo(path_segment_lengths(poly,closed=true));
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>
>>>>>>>>>>> Circumscribed means tangents from points on ellipse
>>>>>>>>>>> How can the tangents from equally spaced arc lengths or equally
>>>>>>>>>>> spaced angles be equal for an ellipse
>>>>>>>>>>> You need to change the arc lengths dynamically and also
>>>>>>>>>>> accurately estimate the segment length which fits the ellipse .
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, <avm4@cornell.edu>
>>>>>>>>>>> wrote:
>>>>>>>>>>>
>>>>>>>>>>>> Your intuition is incorrect: a circle scaled in one direction
>>>>>>>>>>>> *IS* an ellipse. The motivation for being interested in
>>>>>>>>>>>> circumscribing is to make holes, though the constraint of equal length
>>>>>>>>>>>> sides is probably not what you'd want for that purpose. You can make a
>>>>>>>>>>>> nonuniform circumscribing polygon for an ellipse by simply making a polygon
>>>>>>>>>>>> that circumscribes a circle and scaling it.
>>>>>>>>>>>>
>>>>>>>>>>>> On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson <bob@rjcarlson.com>
>>>>>>>>>>>> wrote:
>>>>>>>>>>>>
>>>>>>>>>>>>> Seems like I remember Nop Head talking about using
>>>>>>>>>>>>> circumscribed holes for screws and even rotating the polygons to produce a
>>>>>>>>>>>>> more even hole.
>>>>>>>>>>>>>
>>>>>>>>>>>>> If an ellipse is the nominal shape, the using a circumscribed
>>>>>>>>>>>>> polygon for a hole and an inscribed polygon for a solid might make sense.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Btw, can someone confirm my intuition that a circle scaled in
>>>>>>>>>>>>> one dimension is not an ellipse?
>>>>>>>>>>>>>
>>>>>>>>>>>>> -Bob
>>>>>>>>>>>>> Tucson AZ
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> On Aug 5, 2022, at 15:38, Adrian Mariano <avm4@cornell.edu>
>>>>>>>>>>>>> wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>> I don't have a specific need for a circumscribing elliptical
>>>>>>>>>>>>> polygon with equal sides. I was writing code to make ellipses and it
>>>>>>>>>>>>> seemed like equal side approximations would be nice, so I solved the case
>>>>>>>>>>>>> for inscribed, but couldn't solve the circumscribed case. So it's an
>>>>>>>>>>>>> unsolved puzzle I've got that this discussion reminded me of.
>>>>>>>>>>>>>
>>>>>>>>>>>>> On Fri, Aug 5, 2022 at 6:33 PM nop head <nop.head@gmail.com>
>>>>>>>>>>>>> wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> A long discussion but I have to ask why do you need an
>>>>>>>>>>>>>> elliptical polygon with equal tangential sides?
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu>
>>>>>>>>>>>>>> wrote:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> You think that a circumscribing polygon with equal length
>>>>>>>>>>>>>>> segments does not exist?? I'm skeptical of this claim.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> For circumscribed, it's not so hard. You start with a
>>>>>>>>>>>>>>> uniform-in-angle point distribution, compute the segment lengths, compare
>>>>>>>>>>>>>>> to the mean segment length to get an error, and then compute an adjustment
>>>>>>>>>>>>>>> based on the error. It converges quickly.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> So for example, I have segments lengths of:
>>>>>>>>>>>>>>> ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
>>>>>>>>>>>>>>> 5.57297, 5.57297]
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Maybe not exactly equal, since tolerance is 1e-9, but surely
>>>>>>>>>>>>>>> good enough.
>>>>>>>>>>>>>>> <image.png>
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> The code for this is in BOSL2 in shapes2d.scad. A similar
>>>>>>>>>>>>>>> algorithm should exist for the circumscribed case, but as I said, my
>>>>>>>>>>>>>>> efforts to find it have failed.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
>>>>>>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> I think circumscribed polygon segments cannot be equal
>>>>>>>>>>>>>>>> lengths.
>>>>>>>>>>>>>>>> inscribed ones may be approximately equal, but would be a
>>>>>>>>>>>>>>>> very tough problem to solve.
>>>>>>>>>>>>>>>> I tried something.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <
>>>>>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> I think you need to be a little careful about defining
>>>>>>>>>>>>>>>>> what you want to do. If you want to approximate an ellipse with points
>>>>>>>>>>>>>>>>> that are distributed uniformly over the ellipse, that could be done by
>>>>>>>>>>>>>>>>> approximating the arc length of the ellipse to find the necessary
>>>>>>>>>>>>>>>>> points---e.g. advance distance d along the ellipse at each step. But if
>>>>>>>>>>>>>>>>> you want to approximate an ellipse with a polygon with equal length sides,
>>>>>>>>>>>>>>>>> that approach does not do it, because the length of the segments will not
>>>>>>>>>>>>>>>>> be constant as the span sections of the ellipse with differing curvature.
>>>>>>>>>>>>>>>>> And since the perimeter of the desired polygon is unknown, you can't just
>>>>>>>>>>>>>>>>> step it out analogously. The circle doesn't present this challenge due to
>>>>>>>>>>>>>>>>> having constant curvature.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
>>>>>>>>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> I think our emails crossed. I was addressing how to
>>>>>>>>>>>>>>>>>> approximate an ellipse with constant-length segments in general, not how to
>>>>>>>>>>>>>>>>>> solve your problem. When I had to do something similar, I used polar
>>>>>>>>>>>>>>>>>> coordinates and stepped up the value of theta until the distance from the
>>>>>>>>>>>>>>>>>> previous point was correct. (In the constant-length version, I'd keep a
>>>>>>>>>>>>>>>>>> running total of the maximum desired length so that round-off errors don't
>>>>>>>>>>>>>>>>>> accumulate.)
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> Figuring out the circumscribed polygon is, as you note, a
>>>>>>>>>>>>>>>>>> greater challenge. I can visualize how to do it for n = 4, but it doesn't
>>>>>>>>>>>>>>>>>> extrapolate at all.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <
>>>>>>>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> Eh, there's no such thing as "brute force" in this
>>>>>>>>>>>>>>>>>>> situation. Step through *what* with sufficiently small steps? Of course
>>>>>>>>>>>>>>>>>>> an iterative algorithm is necessary. This is true for most interesting
>>>>>>>>>>>>>>>>>>> problems. No big deal. Show me an iterative algorithm that converges to
>>>>>>>>>>>>>>>>>>> the circumscribing polygon. When I tried to devise such an algorithm I
>>>>>>>>>>>>>>>>>>> couldn't find one that would converge. Writing an algorithm that would
>>>>>>>>>>>>>>>>>>> converge to the *inscribed* polygon seemed to be straight forward. Note
>>>>>>>>>>>>>>>>>>> also that computing arc length of the ellipse doesn't really help here.
>>>>>>>>>>>>>>>>>>> You can do that by solving an elliptical integral, or summing up small
>>>>>>>>>>>>>>>>>>> segments. But what's the connection between arc length of the ellipse and
>>>>>>>>>>>>>>>>>>> a circumscribing polygon with equal segment lengths? If there's a
>>>>>>>>>>>>>>>>>>> connection, I don't see it.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
>>>>>>>>>>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> Except by brute force (step through it with
>>>>>>>>>>>>>>>>>>>> sufficiently small steps), it’s not possible because there’s no closed form
>>>>>>>>>>>>>>>>>>>> expression for the distance between two points on a general ellipse (as
>>>>>>>>>>>>>>>>>>>> opposed to special cases like a circle).
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <
>>>>>>>>>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> Here's a challenge for you: draw an "ellipse" polygon
>>>>>>>>>>>>>>>>>>>>> with N segments that are all the same length that circumscribes a perfect
>>>>>>>>>>>>>>>>>>>>> ellipse with given semiaxes a and b. I tried to solve this problem and
>>>>>>>>>>>>>>>>>>>>> could not find a solution. I can draw an *inscribed* N segment "ellipse"
>>>>>>>>>>>>>>>>>>>>> polygon with all segments the same length---though it's not easy---but the
>>>>>>>>>>>>>>>>>>>>> circumscribed case eludes me. I think of the inscribed case as, every
>>>>>>>>>>>>>>>>>>>>> vertex is located on the ellipse and the circumscribed case, every segment
>>>>>>>>>>>>>>>>>>>>> is tangent to the ellipse
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>>>>>>>>>>>>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> In openscad drawing a ellipse is not too difficult
>>>>>>>>>>>>>>>>>>>>>> Refer following code
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> function to draw an ellipse with semi-major and
>>>>>>>>>>>>>>>>>>>>>> semi-minor axis "r1" and "r2" respectively and with center "cp" and number
>>>>>>>>>>>>>>>>>>>>>> of segment "s"
>>>>>>>>>>>>>>>>>>>>>> /
>>>>>>>>>>>>>>>>>>>>>> // example:
>>>>>>>>>>>>>>>>>>>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> function ellipse(r1,r2,cp,s=30)=
>>>>>>>>>>>>>>>>>>>>>> let(
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>>>>>>>>>>>>>>>>>>>>> )sec;
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>>>>>>>>>>>>>>>>>>>>> openscad@jordan.maileater.net> wrote:
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> I created an oval using:
>>>>>>>>>>>>>>>>>>>>>>> inMm = 25.4;
>>>>>>>>>>>>>>>>>>>>>>> fullW = 35.5 * inMm;
>>>>>>>>>>>>>>>>>>>>>>> fullL = 54.5 * inMm;
>>>>>>>>>>>>>>>>>>>>>>> resize([fullW, fullL, 1])
>>>>>>>>>>>>>>>>>>>>>>> circle(d=fullW);
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> Tip: use scale() rather than resize(); it previews
>>>>>>>>>>>>>>>>>>>>>>> faster in complicated cases.
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> You're taking a circle and stretching it; the result
>>>>>>>>>>>>>>>>>>>>>>> of that stretching is an ellipse.
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> I'm not 100% sure - maybe one of the real math
>>>>>>>>>>>>>>>>>>>>>>> wizards can chime in - but I think the shape you are asking for would not
>>>>>>>>>>>>>>>>>>>>>>> be an ellipse.
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> I believe that an ellipse is fully defined by its
>>>>>>>>>>>>>>>>>>>>>>> major (long) and minor (short) axis dimensions; the other parts of the
>>>>>>>>>>>>>>>>>>>>>>> curve mathematically flow from those two measurements.
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> You can make something else, but you can't make it
>>>>>>>>>>>>>>>>>>>>>>> using this technique.
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> You can use a Bézier function.
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> Using BOSL2:
>>>>>>>>>>>>>>>>>>>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> DIY, here's a function that I cobbled together some
>>>>>>>>>>>>>>>>>>>>>>> years ago:
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>>>>>>>>>>>>>>>>>>>>> // but that yielded either single points or a raft of triangles;
>>>>>>>>>>>>>>>>>>>>>>> // this yields a vector of points that you can then concatenate
>>>>>>>>>>>>>>>>>>>>>>> // with other pieces to form a single polygon.
>>>>>>>>>>>>>>>>>>>>>>> // If we were really clever, I think it would be possible to
>>>>>>>>>>>>>>>>>>>>>>> // automatically space the output points based on how linear
>>>>>>>>>>>>>>>>>>>>>>> // the curve is at that point. But right now I'm not that clever.
>>>>>>>>>>>>>>>>>>>>>>> function BEZ03(u) = pow((1-u), 3);
>>>>>>>>>>>>>>>>>>>>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>>>>>>>>>>>>>>>>>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>>>>>>>>>>>>>>>>>>>>> function BEZ33(u) = pow(u,3);
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>>>>>>>>>>>>>>>>>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>>>>>>>>>>>>>>>>>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> // p0 - start point
>>>>>>>>>>>>>>>>>>>>>>> // p1 - control point 1, line departs p0 headed this way
>>>>>>>>>>>>>>>>>>>>>>> // p2 - control point 2, line arrives at p3 from this way
>>>>>>>>>>>>>>>>>>>>>>> // p3 - end point
>>>>>>>>>>>>>>>>>>>>>>> // segs - number of segments
>>>>>>>>>>>>>>>>>>>>>>> function bez(p0, p1, p2, p3, segs) = [
>>>>>>>>>>>>>>>>>>>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>>>>>>>>>>>>>>>>>>>>> ];
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> It is that final function bez() that I actually
>>>>>>>>>>>>>>>>>>>>>>> call, e.g.
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> often concatenating the results of several bez()
>>>>>>>>>>>>>>>>>>>>>>> calls together.
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> Note that the four coordinates correspond to the
>>>>>>>>>>>>>>>>>>>>>>> coordinates of the control points in a GUI drawing program: the start
>>>>>>>>>>>>>>>>>>>>>>> point, control point associated with the start, the control point
>>>>>>>>>>>>>>>>>>>>>>> associated with the end, and the end point. I have sometimes created
>>>>>>>>>>>>>>>>>>>>>>> OpenSCAD objects by tracing an object in a drawing program and then reading
>>>>>>>>>>>>>>>>>>>>>>> off the coordinates of the various control points.
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> Note that Bézier functions are not the only
>>>>>>>>>>>>>>>>>>>>>>> mathematical curves. (For instance, they cannot exactly match an arc of a
>>>>>>>>>>>>>>>>>>>>>>> circle.) I'm sure there are numerous other functions that also produce
>>>>>>>>>>>>>>>>>>>>>>> curves.
>>>>>>>>>>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>>>>>>>>>>
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MM
Michael Möller
Mon, Aug 8, 2022 8:22 PM
y and x if divisible by 4, right?
Michael, fra mobilen
man. 8. aug. 2022 22.12 skrev Father Horton fatherhorton@gmail.com:
My end condition is that the last point calculated is (-L/2, b), but as I
think about it, it does make sense that N has to be a multiple of 2,
because the whole thing has to be symmetrical around the y-axis.
On Mon, Aug 8, 2022 at 2:35 PM Father Horton fatherhorton@gmail.com
wrote:
Here's a Google sheet that shows what I have in mind.
https://docs.google.com/spreadsheets/d/13YOJKkWWfcgZNKVn_nlBtQZ_bbxNvN6SdVv1bVWLDP8/edit?usp=sharing
I have to adjust the signs on the square roots manually, so I didn't
iterate until I got equal lengths, but this is the idea. Blue points are
end points of the circumscribing segments and red points are the tangent
points on the ellipse. I have other things to do at the moment, but I may
revisit later to get the signs to work out right and then see if it will
iterate to an acceptable result.
On Mon, Aug 8, 2022 at 1:31 PM Adrian Mariano avm4@cornell.edu wrote:
The restriction arises because of the end condition. There are 4 cases
depending on what happens at the four ends of the ellipse (the points where
it intersects the coordinate axes). The four possibilities are at each end
either a corner or a flat tangent at the end. Your approach as I
understand it begins with a flat and constructs a quarter of the polygon,
ending with a point. Such a polygon must have 4k+2 sides. I do not see
how the end condition can be formulated to end on a flat. Also I do not
see how it can start on a point. If n is odd you can start at one end and
go half way around and end on a point. But if n is divisible by 4 you can
have 4 flats or four points and the method cannot produce either of them.
Have I missed something?
On Mon, Aug 8, 2022, 13:42 Father Horton fatherhorton@gmail.com wrote:
I'm not sure why there should be a restriction on N; I think it's
general. I did a spreadsheet that looks like it's doing the right thing in
the first quadrant, but it (or I) gets lost when I have to select the right
value for the sign of the square roots involved as the process enters other
quadrants.
On Sun, Aug 7, 2022 at 9:43 PM Father Horton fatherhorton@gmail.com
wrote:
Google showed me that method when I came up with the right terms. But
it's too late at night for me to try to implement it. Maybe tomorrow.
On Sun, Aug 7, 2022 at 9:41 PM Adrian Mariano avm4@cornell.edu
wrote:
Now I understand your idea. I think it should be possible to make
that work for the case of N=4k+2. It does seem like there are some
complications, like you try to extend the segment and the length is closer
than the tangent, so it's impossible. Also things may be complicated if
the algorithm tries to go around the tip of the ellipse (that is, if the
path is too long).
For finding a tangent to an ellipse, transform the problem to the
circle, solve, and then transform back. In BOSL2:
function ellipse_tangents(a,b,pt) =
let(
r=a,
scaledpt = yscale(a/b,pt),
tangents = circle_point_tangents(r,[0,0],scaledpt)
)
yscale(b/a,tangents);
On Sun, Aug 7, 2022 at 9:54 PM Father Horton fatherhorton@gmail.com
wrote:
I did that wrong. It's not (0, b) to (0, L/2). It's (0, b) to (L/2,
b). That's always outside the ellipse.
I will look at the second problem as soon as I figure out how to
calculate a tangent line to an ellipse containing a point not on the
ellipse. Google is not helping much.
On Sun, Aug 7, 2022 at 8:32 PM Adrian Mariano avm4@cornell.edu
wrote:
Father Horton,
I'm having trouble understanding your proposed method. A segment
from (0,b) to (0,L/2) might be inside the ellipse depending on N, and then
there won't be a tangent.
It also sounds like your idea is somewhat similar to Jordan's idea:
basically start at one side and work around the ellipse and see if it
works. If not, make a correction. The problem with this idea is that
there are two parameters for the first segment, so how do you make a
correction? (The two parameters are the location of the endpoint of the
first segment and its length.)
On Sun, Aug 7, 2022 at 5:33 PM Father Horton <
fatherhorton@gmail.com> wrote:
How about this to construct N equal length segments around an
ellipse with axes a and b:
- Select an initial segment length estimate (e.g., Ramanujan's
formula / N) = L
- Construct a half-segment from (0, b) to (0, L/2).
- Find a line tangent to the ellipse and passing through (0, L/2).
- Find the point on the tangent line that is L away from (0, L/2).
- Repeat (3) substituting the new point for (0, L/2). until you
have N segments.
If everything worked perfectly, the last point found would be (0,
-L/2). If there's a gap, increase L and try again. If there's overlap,
decrease L and try again.
On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano <
rcmpersiano@gmail.com> wrote:
Adrian, why to use a (either inscribed or circumscribed) polygon
with equal side as an approximation of an ellipsis? It doesn't seem to be a
good approximation of it. I get a better approximation by taking an angle
uniform sample of the curve in the parameter space.
[image: Uniformly_sampled_ellipsis.png]
Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano <
avm4@cornell.edu> escreveu:
Circumscribed means tangents to points on the ellipse. But
there are no arc lengths (equal or unequal) in this problem, nor equally
spaced angles. The angles of each segment will be different. But the
lengths of the segments are not the same as the arc lengths, which is why
arc length does not matter. Here's a solution to the N=6 polygon sort of
inspired by the idea Jordan suggested, where I adjust the first segment
until the second segment has the right length. The rest of the solution is
then obtained using symmetry, which means there is only a single unknown
that can simply be solved for. But this doesn't generalize to higher N.
[image: image.png]
include<BOSL2/std.scad>
// ellipse semiaxes
// If a is much smaller than b it will fail
a=5;
b=2;
// Find point on ellipse at specified x coordinate
function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
// Find tangent line to ellipse at point (u,w)
// and determine the x value where that line has
// given y value.
function findx(a,b,u,w,y) =
let(
A = -ub^2/w/a^2,
B = w+(ub)^2/w/a^2
)
(y-B)/A;
// First segment if tangent point is at x=u
function first_seg(a,b,u) =
let(
w = ellipse_pt(a,b,u),
x1 = findx(a,b,u,w,0),
x2 = findx(a,b,u,w,b)
)
[[x1,0],[x2,b]];
err = function (u) let(pt = first_seg(a,b,u))
2*pt[1].x - norm(pt[0]-pt[1]);
x = root_find(err,.2a,.9a);
first = first_seg(a,b,x);
top = concat(first,reverse(xflip(first)));
poly = concat(top, reverse(yflip(list_tail(top))));
stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
color("red")stroke(poly,width=.05,closed=true);
// Verify equality of segment lengths
echo(path_segment_lengths(poly,closed=true));
On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
Circumscribed means tangents from points on ellipse
How can the tangents from equally spaced arc lengths or equally
spaced angles be equal for an ellipse
You need to change the arc lengths dynamically and also
accurately estimate the segment length which fits the ellipse .
On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, avm4@cornell.edu
wrote:
Your intuition is incorrect: a circle scaled in one direction
IS an ellipse. The motivation for being interested in
circumscribing is to make holes, though the constraint of equal length
sides is probably not what you'd want for that purpose. You can make a
nonuniform circumscribing polygon for an ellipse by simply making a polygon
that circumscribes a circle and scaling it.
On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson bob@rjcarlson.com
wrote:
Seems like I remember Nop Head talking about using
circumscribed holes for screws and even rotating the polygons to produce a
more even hole.
If an ellipse is the nominal shape, the using a circumscribed
polygon for a hole and an inscribed polygon for a solid might make sense.
Btw, can someone confirm my intuition that a circle scaled in
one dimension is not an ellipse?
-Bob
Tucson AZ
On Aug 5, 2022, at 15:38, Adrian Mariano avm4@cornell.edu
wrote:
I don't have a specific need for a circumscribing elliptical
polygon with equal sides. I was writing code to make ellipses and it
seemed like equal side approximations would be nice, so I solved the case
for inscribed, but couldn't solve the circumscribed case. So it's an
unsolved puzzle I've got that this discussion reminded me of.
On Fri, Aug 5, 2022 at 6:33 PM nop head nop.head@gmail.com
wrote:
A long discussion but I have to ask why do you need an
elliptical polygon with equal tangential sides?
On Fri, 5 Aug 2022, 21:58 Adrian Mariano, avm4@cornell.edu
wrote:
You think that a circumscribing polygon with equal length
segments does not exist?? I'm skeptical of this claim.
For circumscribed, it's not so hard. You start with a
uniform-in-angle point distribution, compute the segment lengths, compare
to the mean segment length to get an error, and then compute an adjustment
based on the error. It converges quickly.
So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
5.57297, 5.57297, 5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but
surely good enough.
<image.png>
The code for this is in BOSL2 in shapes2d.scad. A similar
algorithm should exist for the circumscribed case, but as I said, my
efforts to find it have failed.
On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
I think circumscribed polygon segments cannot be equal
lengths.
inscribed ones may be approximately equal, but would be a
very tough problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <
avm4@cornell.edu> wrote:
I think you need to be a little careful about defining
what you want to do. If you want to approximate an ellipse with points
that are distributed uniformly over the ellipse, that could be done by
approximating the arc length of the ellipse to find the necessary
points---e.g. advance distance d along the ellipse at each step. But if
you want to approximate an ellipse with a polygon with equal length sides,
that approach does not do it, because the length of the segments will not
be constant as the span sections of the ellipse with differing curvature.
And since the perimeter of the desired polygon is unknown, you can't just
step it out analogously. The circle doesn't present this challenge due to
having constant curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
fatherhorton@gmail.com> wrote:
I think our emails crossed. I was addressing how to
approximate an ellipse with constant-length segments in general, not how to
solve your problem. When I had to do something similar, I used polar
coordinates and stepped up the value of theta until the distance from the
previous point was correct. (In the constant-length version, I'd keep a
running total of the maximum desired length so that round-off errors don't
accumulate.)
Figuring out the circumscribed polygon is, as you note,
a greater challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <
avm4@cornell.edu> wrote:
Eh, there's no such thing as "brute force" in this
situation. Step through what with sufficiently small steps? Of course
an iterative algorithm is necessary. This is true for most interesting
problems. No big deal. Show me an iterative algorithm that converges to
the circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
fatherhorton@gmail.com> wrote:
Except by brute force (step through it with
sufficiently small steps), it’s not possible because there’s no closed form
expression for the distance between two points on a general ellipse (as
opposed to special cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <
avm4@cornell.edu> wrote:
Here's a challenge for you: draw an "ellipse"
polygon with N segments that are all the same length that circumscribes a
perfect ellipse with given semiaxes a and b. I tried to solve this problem
and could not find a solution. I can draw an inscribed N segment
"ellipse" polygon with all segments the same length---though it's not
easy---but the circumscribed case eludes me. I think of the inscribed
case as, every vertex is located on the ellipse and the circumscribed case,
every segment is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and
semi-minor axis "r1" and "r2" respectively and with center "cp" and number
of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews
faster in complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the
result of that stretching is an ellipse.
I'm not 100% sure - maybe one of the real math
wizards can chime in - but I think the shape you are asking for would not
be an ellipse.
I believe that an ellipse is fully defined by its
major (long) and minor (short) axis dimensions; the other parts of the
curve mathematically flow from those two measurements.
You can make something else, but you can't make it
using this technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some
years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually
call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez()
calls together.
Note that the four coordinates correspond to the
coordinates of the control points in a GUI drawing program: the start
point, control point associated with the start, the control point
associated with the end, and the end point. I have sometimes created
OpenSCAD objects by tracing an object in a drawing program and then reading
off the coordinates of the various control points.
Note that Bézier functions are not the only
mathematical curves. (For instance, they cannot exactly match an arc of a
circle.) I'm sure there are numerous other functions that also produce
curves.
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To unsubscribe send an email to
discuss-leave@lists.openscad.org
y and x if divisible by 4, right?
Michael, fra mobilen
man. 8. aug. 2022 22.12 skrev Father Horton <fatherhorton@gmail.com>:
> My end condition is that the last point calculated is (-L/2, b), but as I
> think about it, it does make sense that N has to be a multiple of 2,
> because the whole thing has to be symmetrical around the y-axis.
>
> On Mon, Aug 8, 2022 at 2:35 PM Father Horton <fatherhorton@gmail.com>
> wrote:
>
>> Here's a Google sheet that shows what I have in mind.
>> https://docs.google.com/spreadsheets/d/13YOJKkWWfcgZNKVn_nlBtQZ_bbxNvN6SdVv1bVWLDP8/edit?usp=sharing
>>
>> I have to adjust the signs on the square roots manually, so I didn't
>> iterate until I got equal lengths, but this is the idea. Blue points are
>> end points of the circumscribing segments and red points are the tangent
>> points on the ellipse. I have other things to do at the moment, but I may
>> revisit later to get the signs to work out right and then see if it will
>> iterate to an acceptable result.
>>
>> On Mon, Aug 8, 2022 at 1:31 PM Adrian Mariano <avm4@cornell.edu> wrote:
>>
>>> The restriction arises because of the end condition. There are 4 cases
>>> depending on what happens at the four ends of the ellipse (the points where
>>> it intersects the coordinate axes). The four possibilities are at each end
>>> either a corner or a flat tangent at the end. Your approach as I
>>> understand it begins with a flat and constructs a quarter of the polygon,
>>> ending with a point. Such a polygon must have 4k+2 sides. I do not see
>>> how the end condition can be formulated to end on a flat. Also I do not
>>> see how it can start on a point. If n is odd you can start at one end and
>>> go half way around and end on a point. But if n is divisible by 4 you can
>>> have 4 flats or four points and the method cannot produce either of them.
>>> Have I missed something?
>>>
>>>
>>> On Mon, Aug 8, 2022, 13:42 Father Horton <fatherhorton@gmail.com> wrote:
>>>
>>>> I'm not sure why there should be a restriction on N; I think it's
>>>> general. I did a spreadsheet that looks like it's doing the right thing in
>>>> the first quadrant, but it (or I) gets lost when I have to select the right
>>>> value for the sign of the square roots involved as the process enters other
>>>> quadrants.
>>>>
>>>> On Sun, Aug 7, 2022 at 9:43 PM Father Horton <fatherhorton@gmail.com>
>>>> wrote:
>>>>
>>>>> Google showed me that method when I came up with the right terms. But
>>>>> it's too late at night for me to try to implement it. Maybe tomorrow.
>>>>>
>>>>> On Sun, Aug 7, 2022 at 9:41 PM Adrian Mariano <avm4@cornell.edu>
>>>>> wrote:
>>>>>
>>>>>> Now I understand your idea. I think it should be possible to make
>>>>>> that work for the case of N=4k+2. It does seem like there are some
>>>>>> complications, like you try to extend the segment and the length is closer
>>>>>> than the tangent, so it's impossible. Also things may be complicated if
>>>>>> the algorithm tries to go around the tip of the ellipse (that is, if the
>>>>>> path is too long).
>>>>>>
>>>>>> For finding a tangent to an ellipse, transform the problem to the
>>>>>> circle, solve, and then transform back. In BOSL2:
>>>>>>
>>>>>> function ellipse_tangents(a,b,pt) =
>>>>>> let(
>>>>>> r=a,
>>>>>> scaledpt = yscale(a/b,pt),
>>>>>> tangents = circle_point_tangents(r,[0,0],scaledpt)
>>>>>> )
>>>>>> yscale(b/a,tangents);
>>>>>>
>>>>>> On Sun, Aug 7, 2022 at 9:54 PM Father Horton <fatherhorton@gmail.com>
>>>>>> wrote:
>>>>>>
>>>>>>> I did that wrong. It's not (0, b) to (0, L/2). It's (0, b) to (L/2,
>>>>>>> b). That's always outside the ellipse.
>>>>>>>
>>>>>>> I will look at the second problem as soon as I figure out how to
>>>>>>> calculate a tangent line to an ellipse containing a point not on the
>>>>>>> ellipse. Google is not helping much.
>>>>>>>
>>>>>>> On Sun, Aug 7, 2022 at 8:32 PM Adrian Mariano <avm4@cornell.edu>
>>>>>>> wrote:
>>>>>>>
>>>>>>>> Father Horton,
>>>>>>>>
>>>>>>>> I'm having trouble understanding your proposed method. A segment
>>>>>>>> from (0,b) to (0,L/2) might be inside the ellipse depending on N, and then
>>>>>>>> there won't be a tangent.
>>>>>>>>
>>>>>>>> It also sounds like your idea is somewhat similar to Jordan's idea:
>>>>>>>> basically start at one side and work around the ellipse and see if it
>>>>>>>> works. If not, make a correction. The problem with this idea is that
>>>>>>>> there are two parameters for the first segment, so how do you make a
>>>>>>>> correction? (The two parameters are the location of the endpoint of the
>>>>>>>> first segment and its length.)
>>>>>>>>
>>>>>>>> On Sun, Aug 7, 2022 at 5:33 PM Father Horton <
>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>
>>>>>>>>> How about this to construct N equal length segments around an
>>>>>>>>> ellipse with axes a and b:
>>>>>>>>>
>>>>>>>>> 1) Select an initial segment length estimate (e.g., Ramanujan's
>>>>>>>>> formula / N) = L
>>>>>>>>> 2) Construct a half-segment from (0, b) to (0, L/2).
>>>>>>>>> 3) Find a line tangent to the ellipse and passing through (0, L/2).
>>>>>>>>> 3) Find the point on the tangent line that is L away from (0, L/2).
>>>>>>>>> 4) Repeat (3) substituting the new point for (0, L/2). until you
>>>>>>>>> have N segments.
>>>>>>>>>
>>>>>>>>> If everything worked perfectly, the last point found would be (0,
>>>>>>>>> -L/2). If there's a gap, increase L and try again. If there's overlap,
>>>>>>>>> decrease L and try again.
>>>>>>>>>
>>>>>>>>> On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano <
>>>>>>>>> rcmpersiano@gmail.com> wrote:
>>>>>>>>>
>>>>>>>>>> Adrian, why to use a (either inscribed or circumscribed) polygon
>>>>>>>>>> with equal side as an approximation of an ellipsis? It doesn't seem to be a
>>>>>>>>>> good approximation of it. I get a better approximation by taking an angle
>>>>>>>>>> uniform sample of the curve in the parameter space.
>>>>>>>>>>
>>>>>>>>>> [image: Uniformly_sampled_ellipsis.png]
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano <
>>>>>>>>>> avm4@cornell.edu> escreveu:
>>>>>>>>>>
>>>>>>>>>>> Circumscribed means tangents to points on the ellipse. But
>>>>>>>>>>> there are no arc lengths (equal or unequal) in this problem, nor equally
>>>>>>>>>>> spaced angles. The angles of each segment will be different. But the
>>>>>>>>>>> lengths of the segments are not the same as the arc lengths, which is why
>>>>>>>>>>> arc length does not matter. Here's a solution to the N=6 polygon sort of
>>>>>>>>>>> inspired by the idea Jordan suggested, where I adjust the first segment
>>>>>>>>>>> until the second segment has the right length. The rest of the solution is
>>>>>>>>>>> then obtained using symmetry, which means there is only a single unknown
>>>>>>>>>>> that can simply be solved for. But this doesn't generalize to higher N.
>>>>>>>>>>>
>>>>>>>>>>> [image: image.png]
>>>>>>>>>>> include<BOSL2/std.scad>
>>>>>>>>>>>
>>>>>>>>>>> // ellipse semiaxes
>>>>>>>>>>> // If a is much smaller than b it will fail
>>>>>>>>>>> a=5;
>>>>>>>>>>> b=2;
>>>>>>>>>>>
>>>>>>>>>>> // Find point on ellipse at specified x coordinate
>>>>>>>>>>> function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> // Find tangent line to ellipse at point (u,w)
>>>>>>>>>>> // and determine the x value where that line has
>>>>>>>>>>> // given y value.
>>>>>>>>>>> function findx(a,b,u,w,y) =
>>>>>>>>>>> let(
>>>>>>>>>>> A = -u*b^2/w/a^2,
>>>>>>>>>>> B = w+(u*b)^2/w/a^2
>>>>>>>>>>> )
>>>>>>>>>>> (y-B)/A;
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> // First segment if tangent point is at x=u
>>>>>>>>>>> function first_seg(a,b,u) =
>>>>>>>>>>> let(
>>>>>>>>>>> w = ellipse_pt(a,b,u),
>>>>>>>>>>> x1 = findx(a,b,u,w,0),
>>>>>>>>>>> x2 = findx(a,b,u,w,b)
>>>>>>>>>>> )
>>>>>>>>>>> [[x1,0],[x2,b]];
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> err = function (u) let(pt = first_seg(a,b,u))
>>>>>>>>>>> 2*pt[1].x - norm(pt[0]-pt[1]);
>>>>>>>>>>>
>>>>>>>>>>> x = root_find(err,.2*a,.9*a);
>>>>>>>>>>> first = first_seg(a,b,x);
>>>>>>>>>>> top = concat(first,reverse(xflip(first)));
>>>>>>>>>>> poly = concat(top, reverse(yflip(list_tail(top))));
>>>>>>>>>>>
>>>>>>>>>>> stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
>>>>>>>>>>> color("red")stroke(poly,width=.05,closed=true);
>>>>>>>>>>>
>>>>>>>>>>> // Verify equality of segment lengths
>>>>>>>>>>> echo(path_segment_lengths(poly,closed=true));
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
>>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>>
>>>>>>>>>>>> Circumscribed means tangents from points on ellipse
>>>>>>>>>>>> How can the tangents from equally spaced arc lengths or equally
>>>>>>>>>>>> spaced angles be equal for an ellipse
>>>>>>>>>>>> You need to change the arc lengths dynamically and also
>>>>>>>>>>>> accurately estimate the segment length which fits the ellipse .
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, <avm4@cornell.edu>
>>>>>>>>>>>> wrote:
>>>>>>>>>>>>
>>>>>>>>>>>>> Your intuition is incorrect: a circle scaled in one direction
>>>>>>>>>>>>> *IS* an ellipse. The motivation for being interested in
>>>>>>>>>>>>> circumscribing is to make holes, though the constraint of equal length
>>>>>>>>>>>>> sides is probably not what you'd want for that purpose. You can make a
>>>>>>>>>>>>> nonuniform circumscribing polygon for an ellipse by simply making a polygon
>>>>>>>>>>>>> that circumscribes a circle and scaling it.
>>>>>>>>>>>>>
>>>>>>>>>>>>> On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson <bob@rjcarlson.com>
>>>>>>>>>>>>> wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> Seems like I remember Nop Head talking about using
>>>>>>>>>>>>>> circumscribed holes for screws and even rotating the polygons to produce a
>>>>>>>>>>>>>> more even hole.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> If an ellipse is the nominal shape, the using a circumscribed
>>>>>>>>>>>>>> polygon for a hole and an inscribed polygon for a solid might make sense.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Btw, can someone confirm my intuition that a circle scaled in
>>>>>>>>>>>>>> one dimension is not an ellipse?
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> -Bob
>>>>>>>>>>>>>> Tucson AZ
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> On Aug 5, 2022, at 15:38, Adrian Mariano <avm4@cornell.edu>
>>>>>>>>>>>>>> wrote:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> I don't have a specific need for a circumscribing elliptical
>>>>>>>>>>>>>> polygon with equal sides. I was writing code to make ellipses and it
>>>>>>>>>>>>>> seemed like equal side approximations would be nice, so I solved the case
>>>>>>>>>>>>>> for inscribed, but couldn't solve the circumscribed case. So it's an
>>>>>>>>>>>>>> unsolved puzzle I've got that this discussion reminded me of.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> On Fri, Aug 5, 2022 at 6:33 PM nop head <nop.head@gmail.com>
>>>>>>>>>>>>>> wrote:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> A long discussion but I have to ask why do you need an
>>>>>>>>>>>>>>> elliptical polygon with equal tangential sides?
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu>
>>>>>>>>>>>>>>> wrote:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> You think that a circumscribing polygon with equal length
>>>>>>>>>>>>>>>> segments does not exist?? I'm skeptical of this claim.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> For circumscribed, it's not so hard. You start with a
>>>>>>>>>>>>>>>> uniform-in-angle point distribution, compute the segment lengths, compare
>>>>>>>>>>>>>>>> to the mean segment length to get an error, and then compute an adjustment
>>>>>>>>>>>>>>>> based on the error. It converges quickly.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> So for example, I have segments lengths of:
>>>>>>>>>>>>>>>> ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
>>>>>>>>>>>>>>>> 5.57297, 5.57297, 5.57297]
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Maybe not exactly equal, since tolerance is 1e-9, but
>>>>>>>>>>>>>>>> surely good enough.
>>>>>>>>>>>>>>>> <image.png>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> The code for this is in BOSL2 in shapes2d.scad. A similar
>>>>>>>>>>>>>>>> algorithm should exist for the circumscribed case, but as I said, my
>>>>>>>>>>>>>>>> efforts to find it have failed.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
>>>>>>>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> I think circumscribed polygon segments cannot be equal
>>>>>>>>>>>>>>>>> lengths.
>>>>>>>>>>>>>>>>> inscribed ones may be approximately equal, but would be a
>>>>>>>>>>>>>>>>> very tough problem to solve.
>>>>>>>>>>>>>>>>> I tried something.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <
>>>>>>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> I think you need to be a little careful about defining
>>>>>>>>>>>>>>>>>> what you want to do. If you want to approximate an ellipse with points
>>>>>>>>>>>>>>>>>> that are distributed uniformly over the ellipse, that could be done by
>>>>>>>>>>>>>>>>>> approximating the arc length of the ellipse to find the necessary
>>>>>>>>>>>>>>>>>> points---e.g. advance distance d along the ellipse at each step. But if
>>>>>>>>>>>>>>>>>> you want to approximate an ellipse with a polygon with equal length sides,
>>>>>>>>>>>>>>>>>> that approach does not do it, because the length of the segments will not
>>>>>>>>>>>>>>>>>> be constant as the span sections of the ellipse with differing curvature.
>>>>>>>>>>>>>>>>>> And since the perimeter of the desired polygon is unknown, you can't just
>>>>>>>>>>>>>>>>>> step it out analogously. The circle doesn't present this challenge due to
>>>>>>>>>>>>>>>>>> having constant curvature.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
>>>>>>>>>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> I think our emails crossed. I was addressing how to
>>>>>>>>>>>>>>>>>>> approximate an ellipse with constant-length segments in general, not how to
>>>>>>>>>>>>>>>>>>> solve your problem. When I had to do something similar, I used polar
>>>>>>>>>>>>>>>>>>> coordinates and stepped up the value of theta until the distance from the
>>>>>>>>>>>>>>>>>>> previous point was correct. (In the constant-length version, I'd keep a
>>>>>>>>>>>>>>>>>>> running total of the maximum desired length so that round-off errors don't
>>>>>>>>>>>>>>>>>>> accumulate.)
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> Figuring out the circumscribed polygon is, as you note,
>>>>>>>>>>>>>>>>>>> a greater challenge. I can visualize how to do it for n = 4, but it doesn't
>>>>>>>>>>>>>>>>>>> extrapolate at all.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <
>>>>>>>>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> Eh, there's no such thing as "brute force" in this
>>>>>>>>>>>>>>>>>>>> situation. Step through *what* with sufficiently small steps? Of course
>>>>>>>>>>>>>>>>>>>> an iterative algorithm is necessary. This is true for most interesting
>>>>>>>>>>>>>>>>>>>> problems. No big deal. Show me an iterative algorithm that converges to
>>>>>>>>>>>>>>>>>>>> the circumscribing polygon. When I tried to devise such an algorithm I
>>>>>>>>>>>>>>>>>>>> couldn't find one that would converge. Writing an algorithm that would
>>>>>>>>>>>>>>>>>>>> converge to the *inscribed* polygon seemed to be straight forward. Note
>>>>>>>>>>>>>>>>>>>> also that computing arc length of the ellipse doesn't really help here.
>>>>>>>>>>>>>>>>>>>> You can do that by solving an elliptical integral, or summing up small
>>>>>>>>>>>>>>>>>>>> segments. But what's the connection between arc length of the ellipse and
>>>>>>>>>>>>>>>>>>>> a circumscribing polygon with equal segment lengths? If there's a
>>>>>>>>>>>>>>>>>>>> connection, I don't see it.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
>>>>>>>>>>>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> Except by brute force (step through it with
>>>>>>>>>>>>>>>>>>>>> sufficiently small steps), it’s not possible because there’s no closed form
>>>>>>>>>>>>>>>>>>>>> expression for the distance between two points on a general ellipse (as
>>>>>>>>>>>>>>>>>>>>> opposed to special cases like a circle).
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <
>>>>>>>>>>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> Here's a challenge for you: draw an "ellipse"
>>>>>>>>>>>>>>>>>>>>>> polygon with N segments that are all the same length that circumscribes a
>>>>>>>>>>>>>>>>>>>>>> perfect ellipse with given semiaxes a and b. I tried to solve this problem
>>>>>>>>>>>>>>>>>>>>>> and could not find a solution. I can draw an *inscribed* N segment
>>>>>>>>>>>>>>>>>>>>>> "ellipse" polygon with all segments the same length---though it's not
>>>>>>>>>>>>>>>>>>>>>> easy---but the circumscribed case eludes me. I think of the inscribed
>>>>>>>>>>>>>>>>>>>>>> case as, every vertex is located on the ellipse and the circumscribed case,
>>>>>>>>>>>>>>>>>>>>>> every segment is tangent to the ellipse
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>>>>>>>>>>>>>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> In openscad drawing a ellipse is not too difficult
>>>>>>>>>>>>>>>>>>>>>>> Refer following code
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> function to draw an ellipse with semi-major and
>>>>>>>>>>>>>>>>>>>>>>> semi-minor axis "r1" and "r2" respectively and with center "cp" and number
>>>>>>>>>>>>>>>>>>>>>>> of segment "s"
>>>>>>>>>>>>>>>>>>>>>>> /
>>>>>>>>>>>>>>>>>>>>>>> // example:
>>>>>>>>>>>>>>>>>>>>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> function ellipse(r1,r2,cp,s=30)=
>>>>>>>>>>>>>>>>>>>>>>> let(
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>>>>>>>>>>>>>>>>>>>>>> )sec;
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>>>>>>>>>>>>>>>>>>>>>> openscad@jordan.maileater.net> wrote:
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> I created an oval using:
>>>>>>>>>>>>>>>>>>>>>>>> inMm = 25.4;
>>>>>>>>>>>>>>>>>>>>>>>> fullW = 35.5 * inMm;
>>>>>>>>>>>>>>>>>>>>>>>> fullL = 54.5 * inMm;
>>>>>>>>>>>>>>>>>>>>>>>> resize([fullW, fullL, 1])
>>>>>>>>>>>>>>>>>>>>>>>> circle(d=fullW);
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> Tip: use scale() rather than resize(); it previews
>>>>>>>>>>>>>>>>>>>>>>>> faster in complicated cases.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> You're taking a circle and stretching it; the
>>>>>>>>>>>>>>>>>>>>>>>> result of that stretching is an ellipse.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> I'm not 100% sure - maybe one of the real math
>>>>>>>>>>>>>>>>>>>>>>>> wizards can chime in - but I think the shape you are asking for would not
>>>>>>>>>>>>>>>>>>>>>>>> be an ellipse.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> I believe that an ellipse is fully defined by its
>>>>>>>>>>>>>>>>>>>>>>>> major (long) and minor (short) axis dimensions; the other parts of the
>>>>>>>>>>>>>>>>>>>>>>>> curve mathematically flow from those two measurements.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> You can make something else, but you can't make it
>>>>>>>>>>>>>>>>>>>>>>>> using this technique.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> You can use a Bézier function.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> Using BOSL2:
>>>>>>>>>>>>>>>>>>>>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> DIY, here's a function that I cobbled together some
>>>>>>>>>>>>>>>>>>>>>>>> years ago:
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>>>>>>>>>>>>>>>>>>>>>> // but that yielded either single points or a raft of triangles;
>>>>>>>>>>>>>>>>>>>>>>>> // this yields a vector of points that you can then concatenate
>>>>>>>>>>>>>>>>>>>>>>>> // with other pieces to form a single polygon.
>>>>>>>>>>>>>>>>>>>>>>>> // If we were really clever, I think it would be possible to
>>>>>>>>>>>>>>>>>>>>>>>> // automatically space the output points based on how linear
>>>>>>>>>>>>>>>>>>>>>>>> // the curve is at that point. But right now I'm not that clever.
>>>>>>>>>>>>>>>>>>>>>>>> function BEZ03(u) = pow((1-u), 3);
>>>>>>>>>>>>>>>>>>>>>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>>>>>>>>>>>>>>>>>>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>>>>>>>>>>>>>>>>>>>>>> function BEZ33(u) = pow(u,3);
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>>>>>>>>>>>>>>>>>>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>>>>>>>>>>>>>>>>>>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> // p0 - start point
>>>>>>>>>>>>>>>>>>>>>>>> // p1 - control point 1, line departs p0 headed this way
>>>>>>>>>>>>>>>>>>>>>>>> // p2 - control point 2, line arrives at p3 from this way
>>>>>>>>>>>>>>>>>>>>>>>> // p3 - end point
>>>>>>>>>>>>>>>>>>>>>>>> // segs - number of segments
>>>>>>>>>>>>>>>>>>>>>>>> function bez(p0, p1, p2, p3, segs) = [
>>>>>>>>>>>>>>>>>>>>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>>>>>>>>>>>>>>>>>>>>>> ];
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> It is that final function bez() that I actually
>>>>>>>>>>>>>>>>>>>>>>>> call, e.g.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> often concatenating the results of several bez()
>>>>>>>>>>>>>>>>>>>>>>>> calls together.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> Note that the four coordinates correspond to the
>>>>>>>>>>>>>>>>>>>>>>>> coordinates of the control points in a GUI drawing program: the start
>>>>>>>>>>>>>>>>>>>>>>>> point, control point associated with the start, the control point
>>>>>>>>>>>>>>>>>>>>>>>> associated with the end, and the end point. I have sometimes created
>>>>>>>>>>>>>>>>>>>>>>>> OpenSCAD objects by tracing an object in a drawing program and then reading
>>>>>>>>>>>>>>>>>>>>>>>> off the coordinates of the various control points.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> Note that Bézier functions are not the only
>>>>>>>>>>>>>>>>>>>>>>>> mathematical curves. (For instance, they cannot exactly match an arc of a
>>>>>>>>>>>>>>>>>>>>>>>> circle.) I'm sure there are numerous other functions that also produce
>>>>>>>>>>>>>>>>>>>>>>>> curves.
>>>>>>>>>>>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> _______________________________________________
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>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> _______________________________________________
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>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> _______________________________________________
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>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> _______________________________________________
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>>>>>>>>>>>>>>
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>
FH
Father Horton
Mon, Aug 8, 2022 8:29 PM
y and x if divisible by 4, right?
Right, which would make calculation easier since I wouldn't have to deal
with those nasty quadrant changes.
On Mon, Aug 8, 2022 at 3:23 PM Michael Möller <private2michael@gmail.com>
wrote:
> y and x if divisible by 4, right?
>
Right, which would make calculation easier since I wouldn't have to deal
with those nasty quadrant changes.
FH
Father Horton
Mon, Aug 8, 2022 8:39 PM
The spreadsheet now has a working 12-sided example (length picked by trial
and error).
The spreadsheet now has a working 12-sided example (length picked by trial
and error).
>
AM
Adrian Mariano
Mon, Aug 8, 2022 8:40 PM
So I hate spreadsheets and it's too hard to decipher what you've done by
looking at the spread sheet. I actually coded this up myself in OpenSCAD
And if there's a way to handle the N divisible by 4 case, I am not seeing
it. For the N=4k+2 case you start at [b,L/2] and work your way around
clockwise and place the segments to make a quarter-ellipse polygon.
Assuming you're length estimate isn't terrible you'll place all the
segments but end up slightly above/below the X axis. Based on how big the
vertical error is, you update the length and iterate to get a solution.
That produces a result that looks like this once you rotate and reflect to
get a full ellipse instead of just a 1/4 ellipse:
[image: image.png]
Notice how the polygon has a flat segment on top but is pointy on the
right?
If N is odd then you start at [-a,L/2] and work your way clockwise around
the ellipse, and then end with the same constraint. In this case there is
less symmetry and you have to build a half-ellipse.
[image: image.png]
I made the ellipse fatter because otherwise the side segments appear
coplanar. So here we have started on the left with a "flat" segment and
ended a half-ellipse on the right at a corner.
There are two other cases that arise for N divisible by 4 where we start
and end on a point or a flat. Starting on a point is problematic because
you have an extra parameter of how far away from the ellipse to put the
point. And ending on a flat is problematic because you have a target that
is two-dimensional, e.g. [a,L/2] and it's no longer apparent how to modify
the length to guarantee an improvement. If you and up [.1,.1] away from
the target...what should you do? Presumably this could be handled by
using a multi-dimensional optimization solver, but I don't have one of
those for OpenSCAD and it does seem kind of like overkill.
Here's the code:
include<BOSL2/std.scad>
function ellipse_tangents(a,b,pt) =
let(
r=a,
scaledpt = yscale(a/b,pt),
tangents = circle_point_tangents(r,[0,0],scaledpt)
)
yscale(b/a,tangents);
function ellipse_next_pt(a,b,pt,L) =
let(
tanglist = ellipse_tangents(a,b,pt),
//f=echo(pt=pt,tanglist=tanglist),
tang = tanglist[0].x>pt.x ? tanglist[0]
: len(tanglist)==2 && tanglist[1].x>pt.x ? tanglist[1]
: assert(false,"cannot find acceptable tangent")
)
assert(norm(tang-pt)<L,"Tangent is too far away")
L*unit(tang-pt)+pt;
function ellipse_path(a,b,L,N,path=[]) =
len(path)==0 ? ellipse_path(a,b,L,N,[[L/2,b]])
: len(path)==N ? path
: last(path).y<=0 ? path
: last(path).x>=a ? path
: ellipse_path(a,b,L,N,
[each path, ellipse_next_pt(a,b,last(path),L)]);
function ellipse_circ(a,b) = PI*(3*(a+b) - sqrt(10ab+3*(a^2+b^2)));
function cellipse(a,b,N) =
assert(N%4==2, "Must have N of form 4k+2")
let(
errfunc = function(L) let(path=(ellipse_path(a,b,L,(N+2)/4)))
-last(path).y+
((N+2)/4 - len(path)) * b ,
lengthguess = ellipse_circ(a,b)/N,
L = root_find(errfunc, 0.95lengthguess, 1.5lengthguess),
path = ellipse_path(a,b,L,(N+2)/4),
right = concat(path, yflip(list_tail(reverse(path))))
)
concat(right, reverse(xflip(right)));
function cellipse_odd(a,b,N) =
assert(N%2==1, "Must have N of form 2k+1")
let(
errfunc = function(L)
let(path=ellipse_path(a,b,L,(N+1)/2,[[-a,L/2]]))
-last(path).y+
((N+1)/2 - len(path)) * b ,
lengthguess = ellipse_circ(a,b)/N,
L = root_find(errfunc, 0.95lengthguess, 1.5lengthguess),
path = ellipse_path(a,b,L,(N+1)/2,[[-a,L/2]])
)
concat(path, list_tail(reverse(yflip(path))));
a=10;
b=6;
N = 7;
p = N%2==1 ? cellipse_odd(a,b,N)
: cellipse(a,b,N);
echo(length = len(p));
ellipse([a,b],$fn=164);
color("red")stroke([p],width=.05,dots=true,dots_width=4);
On Mon, Aug 8, 2022 at 4:12 PM Father Horton fatherhorton@gmail.com wrote:
My end condition is that the last point calculated is (-L/2, b), but as I
think about it, it does make sense that N has to be a multiple of 2,
because the whole thing has to be symmetrical around the y-axis.
On Mon, Aug 8, 2022 at 2:35 PM Father Horton fatherhorton@gmail.com
wrote:
Here's a Google sheet that shows what I have in mind.
https://docs.google.com/spreadsheets/d/13YOJKkWWfcgZNKVn_nlBtQZ_bbxNvN6SdVv1bVWLDP8/edit?usp=sharing
I have to adjust the signs on the square roots manually, so I didn't
iterate until I got equal lengths, but this is the idea. Blue points are
end points of the circumscribing segments and red points are the tangent
points on the ellipse. I have other things to do at the moment, but I may
revisit later to get the signs to work out right and then see if it will
iterate to an acceptable result.
On Mon, Aug 8, 2022 at 1:31 PM Adrian Mariano avm4@cornell.edu wrote:
The restriction arises because of the end condition. There are 4 cases
depending on what happens at the four ends of the ellipse (the points where
it intersects the coordinate axes). The four possibilities are at each end
either a corner or a flat tangent at the end. Your approach as I
understand it begins with a flat and constructs a quarter of the polygon,
ending with a point. Such a polygon must have 4k+2 sides. I do not see
how the end condition can be formulated to end on a flat. Also I do not
see how it can start on a point. If n is odd you can start at one end and
go half way around and end on a point. But if n is divisible by 4 you can
have 4 flats or four points and the method cannot produce either of them.
Have I missed something?
On Mon, Aug 8, 2022, 13:42 Father Horton fatherhorton@gmail.com wrote:
I'm not sure why there should be a restriction on N; I think it's
general. I did a spreadsheet that looks like it's doing the right thing in
the first quadrant, but it (or I) gets lost when I have to select the right
value for the sign of the square roots involved as the process enters other
quadrants.
On Sun, Aug 7, 2022 at 9:43 PM Father Horton fatherhorton@gmail.com
wrote:
Google showed me that method when I came up with the right terms. But
it's too late at night for me to try to implement it. Maybe tomorrow.
On Sun, Aug 7, 2022 at 9:41 PM Adrian Mariano avm4@cornell.edu
wrote:
Now I understand your idea. I think it should be possible to make
that work for the case of N=4k+2. It does seem like there are some
complications, like you try to extend the segment and the length is closer
than the tangent, so it's impossible. Also things may be complicated if
the algorithm tries to go around the tip of the ellipse (that is, if the
path is too long).
For finding a tangent to an ellipse, transform the problem to the
circle, solve, and then transform back. In BOSL2:
function ellipse_tangents(a,b,pt) =
let(
r=a,
scaledpt = yscale(a/b,pt),
tangents = circle_point_tangents(r,[0,0],scaledpt)
)
yscale(b/a,tangents);
On Sun, Aug 7, 2022 at 9:54 PM Father Horton fatherhorton@gmail.com
wrote:
I did that wrong. It's not (0, b) to (0, L/2). It's (0, b) to (L/2,
b). That's always outside the ellipse.
I will look at the second problem as soon as I figure out how to
calculate a tangent line to an ellipse containing a point not on the
ellipse. Google is not helping much.
On Sun, Aug 7, 2022 at 8:32 PM Adrian Mariano avm4@cornell.edu
wrote:
Father Horton,
I'm having trouble understanding your proposed method. A segment
from (0,b) to (0,L/2) might be inside the ellipse depending on N, and then
there won't be a tangent.
It also sounds like your idea is somewhat similar to Jordan's idea:
basically start at one side and work around the ellipse and see if it
works. If not, make a correction. The problem with this idea is that
there are two parameters for the first segment, so how do you make a
correction? (The two parameters are the location of the endpoint of the
first segment and its length.)
On Sun, Aug 7, 2022 at 5:33 PM Father Horton <
fatherhorton@gmail.com> wrote:
How about this to construct N equal length segments around an
ellipse with axes a and b:
- Select an initial segment length estimate (e.g., Ramanujan's
formula / N) = L
- Construct a half-segment from (0, b) to (0, L/2).
- Find a line tangent to the ellipse and passing through (0, L/2).
- Find the point on the tangent line that is L away from (0, L/2).
- Repeat (3) substituting the new point for (0, L/2). until you
have N segments.
If everything worked perfectly, the last point found would be (0,
-L/2). If there's a gap, increase L and try again. If there's overlap,
decrease L and try again.
On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano <
rcmpersiano@gmail.com> wrote:
Adrian, why to use a (either inscribed or circumscribed) polygon
with equal side as an approximation of an ellipsis? It doesn't seem to be a
good approximation of it. I get a better approximation by taking an angle
uniform sample of the curve in the parameter space.
[image: Uniformly_sampled_ellipsis.png]
Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano <
avm4@cornell.edu> escreveu:
Circumscribed means tangents to points on the ellipse. But
there are no arc lengths (equal or unequal) in this problem, nor equally
spaced angles. The angles of each segment will be different. But the
lengths of the segments are not the same as the arc lengths, which is why
arc length does not matter. Here's a solution to the N=6 polygon sort of
inspired by the idea Jordan suggested, where I adjust the first segment
until the second segment has the right length. The rest of the solution is
then obtained using symmetry, which means there is only a single unknown
that can simply be solved for. But this doesn't generalize to higher N.
[image: image.png]
include<BOSL2/std.scad>
// ellipse semiaxes
// If a is much smaller than b it will fail
a=5;
b=2;
// Find point on ellipse at specified x coordinate
function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
// Find tangent line to ellipse at point (u,w)
// and determine the x value where that line has
// given y value.
function findx(a,b,u,w,y) =
let(
A = -ub^2/w/a^2,
B = w+(ub)^2/w/a^2
)
(y-B)/A;
// First segment if tangent point is at x=u
function first_seg(a,b,u) =
let(
w = ellipse_pt(a,b,u),
x1 = findx(a,b,u,w,0),
x2 = findx(a,b,u,w,b)
)
[[x1,0],[x2,b]];
err = function (u) let(pt = first_seg(a,b,u))
2*pt[1].x - norm(pt[0]-pt[1]);
x = root_find(err,.2a,.9a);
first = first_seg(a,b,x);
top = concat(first,reverse(xflip(first)));
poly = concat(top, reverse(yflip(list_tail(top))));
stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
color("red")stroke(poly,width=.05,closed=true);
// Verify equality of segment lengths
echo(path_segment_lengths(poly,closed=true));
On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
Circumscribed means tangents from points on ellipse
How can the tangents from equally spaced arc lengths or equally
spaced angles be equal for an ellipse
You need to change the arc lengths dynamically and also
accurately estimate the segment length which fits the ellipse .
On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, avm4@cornell.edu
wrote:
Your intuition is incorrect: a circle scaled in one direction
IS an ellipse. The motivation for being interested in
circumscribing is to make holes, though the constraint of equal length
sides is probably not what you'd want for that purpose. You can make a
nonuniform circumscribing polygon for an ellipse by simply making a polygon
that circumscribes a circle and scaling it.
On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson bob@rjcarlson.com
wrote:
Seems like I remember Nop Head talking about using
circumscribed holes for screws and even rotating the polygons to produce a
more even hole.
If an ellipse is the nominal shape, the using a circumscribed
polygon for a hole and an inscribed polygon for a solid might make sense.
Btw, can someone confirm my intuition that a circle scaled in
one dimension is not an ellipse?
-Bob
Tucson AZ
On Aug 5, 2022, at 15:38, Adrian Mariano avm4@cornell.edu
wrote:
I don't have a specific need for a circumscribing elliptical
polygon with equal sides. I was writing code to make ellipses and it
seemed like equal side approximations would be nice, so I solved the case
for inscribed, but couldn't solve the circumscribed case. So it's an
unsolved puzzle I've got that this discussion reminded me of.
On Fri, Aug 5, 2022 at 6:33 PM nop head nop.head@gmail.com
wrote:
A long discussion but I have to ask why do you need an
elliptical polygon with equal tangential sides?
On Fri, 5 Aug 2022, 21:58 Adrian Mariano, avm4@cornell.edu
wrote:
You think that a circumscribing polygon with equal length
segments does not exist?? I'm skeptical of this claim.
For circumscribed, it's not so hard. You start with a
uniform-in-angle point distribution, compute the segment lengths, compare
to the mean segment length to get an error, and then compute an adjustment
based on the error. It converges quickly.
So for example, I have segments lengths of:
ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
5.57297, 5.57297, 5.57297]
Maybe not exactly equal, since tolerance is 1e-9, but
surely good enough.
<image.png>
The code for this is in BOSL2 in shapes2d.scad. A similar
algorithm should exist for the circumscribed case, but as I said, my
efforts to find it have failed.
On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
I think circumscribed polygon segments cannot be equal
lengths.
inscribed ones may be approximately equal, but would be a
very tough problem to solve.
I tried something.
On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <
avm4@cornell.edu> wrote:
I think you need to be a little careful about defining
what you want to do. If you want to approximate an ellipse with points
that are distributed uniformly over the ellipse, that could be done by
approximating the arc length of the ellipse to find the necessary
points---e.g. advance distance d along the ellipse at each step. But if
you want to approximate an ellipse with a polygon with equal length sides,
that approach does not do it, because the length of the segments will not
be constant as the span sections of the ellipse with differing curvature.
And since the perimeter of the desired polygon is unknown, you can't just
step it out analogously. The circle doesn't present this challenge due to
having constant curvature.
On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
fatherhorton@gmail.com> wrote:
I think our emails crossed. I was addressing how to
approximate an ellipse with constant-length segments in general, not how to
solve your problem. When I had to do something similar, I used polar
coordinates and stepped up the value of theta until the distance from the
previous point was correct. (In the constant-length version, I'd keep a
running total of the maximum desired length so that round-off errors don't
accumulate.)
Figuring out the circumscribed polygon is, as you note,
a greater challenge. I can visualize how to do it for n = 4, but it doesn't
extrapolate at all.
On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <
avm4@cornell.edu> wrote:
Eh, there's no such thing as "brute force" in this
situation. Step through what with sufficiently small steps? Of course
an iterative algorithm is necessary. This is true for most interesting
problems. No big deal. Show me an iterative algorithm that converges to
the circumscribing polygon. When I tried to devise such an algorithm I
couldn't find one that would converge. Writing an algorithm that would
converge to the inscribed polygon seemed to be straight forward. Note
also that computing arc length of the ellipse doesn't really help here.
You can do that by solving an elliptical integral, or summing up small
segments. But what's the connection between arc length of the ellipse and
a circumscribing polygon with equal segment lengths? If there's a
connection, I don't see it.
On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
fatherhorton@gmail.com> wrote:
Except by brute force (step through it with
sufficiently small steps), it’s not possible because there’s no closed form
expression for the distance between two points on a general ellipse (as
opposed to special cases like a circle).
On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <
avm4@cornell.edu> wrote:
Here's a challenge for you: draw an "ellipse"
polygon with N segments that are all the same length that circumscribes a
perfect ellipse with given semiaxes a and b. I tried to solve this problem
and could not find a solution. I can draw an inscribed N segment
"ellipse" polygon with all segments the same length---though it's not
easy---but the circumscribed case eludes me. I think of the inscribed
case as, every vertex is located on the ellipse and the circumscribed case,
every segment is tangent to the ellipse
On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
sprabhakar2006@gmail.com> wrote:
In openscad drawing a ellipse is not too difficult
Refer following code
function to draw an ellipse with semi-major and
semi-minor axis "r1" and "r2" respectively and with center "cp" and number
of segment "s"
/
// example:
// sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
function ellipse(r1,r2,cp,s=30)=
let(
sec=[for(i=[0:360/s:360-360/s])cp+[r1cos(i),r2sin(i)]]
)sec;
On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
openscad@jordan.maileater.net> wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
Tip: use scale() rather than resize(); it previews
faster in complicated cases.
It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
You're taking a circle and stretching it; the
result of that stretching is an ellipse.
I'm not 100% sure - maybe one of the real math
wizards can chime in - but I think the shape you are asking for would not
be an ellipse.
I believe that an ellipse is fully defined by its
major (long) and minor (short) axis dimensions; the other parts of the
curve mathematically flow from those two measurements.
You can make something else, but you can't make it
using this technique.
Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
You can use a Bézier function.
Using BOSL2:
https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
DIY, here's a function that I cobbled together some
years ago:
// Bezier functions from https://www.thingiverse.com/thing:8443
// but that yielded either single points or a raft of triangles;
// this yields a vector of points that you can then concatenate
// with other pieces to form a single polygon.
// If we were really clever, I think it would be possible to
// automatically space the output points based on how linear
// the curve is at that point. But right now I'm not that clever.
function BEZ03(u) = pow((1-u), 3);
function BEZ13(u) = 3u(pow((1-u),2));
function BEZ23(u) = 3*(pow(u,2))*(1-u);
function BEZ33(u) = pow(u,3);
function PointAlongBez4(p0, p1, p2, p3, u) = [
BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
// p0 - start point
// p1 - control point 1, line departs p0 headed this way
// p2 - control point 2, line arrives at p3 from this way
// p3 - end point
// segs - number of segments
function bez(p0, p1, p2, p3, segs) = [
for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
];
It is that final function bez() that I actually
call, e.g.
polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
often concatenating the results of several bez()
calls together.
Note that the four coordinates correspond to the
coordinates of the control points in a GUI drawing program: the start
point, control point associated with the start, the control point
associated with the end, and the end point. I have sometimes created
OpenSCAD objects by tracing an object in a drawing program and then reading
off the coordinates of the various control points.
Note that Bézier functions are not the only
mathematical curves. (For instance, they cannot exactly match an arc of a
circle.) I'm sure there are numerous other functions that also produce
curves.
OpenSCAD mailing list
To unsubscribe send an email to
discuss-leave@lists.openscad.org
So I hate spreadsheets and it's too hard to decipher what you've done by
looking at the spread sheet. I actually coded this up myself in OpenSCAD
And if there's a way to handle the N divisible by 4 case, I am not seeing
it. For the N=4k+2 case you start at [b,L/2] and work your way around
clockwise and place the segments to make a quarter-ellipse polygon.
Assuming you're length estimate isn't terrible you'll place all the
segments but end up slightly above/below the X axis. Based on how big the
vertical error is, you update the length and iterate to get a solution.
That produces a result that looks like this once you rotate and reflect to
get a full ellipse instead of just a 1/4 ellipse:
[image: image.png]
Notice how the polygon has a flat segment on top but is pointy on the
right?
If N is odd then you start at [-a,L/2] and work your way clockwise around
the ellipse, and then end with the same constraint. In this case there is
less symmetry and you have to build a half-ellipse.
[image: image.png]
I made the ellipse fatter because otherwise the side segments appear
coplanar. So here we have started on the left with a "flat" segment and
ended a half-ellipse on the right at a corner.
There are two other cases that arise for N divisible by 4 where we start
and end on a point or a flat. Starting on a point is problematic because
you have an extra parameter of how far away from the ellipse to put the
point. And ending on a flat is problematic because you have a target that
is two-dimensional, e.g. [a,L/2] and it's no longer apparent how to modify
the length to guarantee an improvement. If you and up [.1,.1] away from
the target...what should you do? Presumably this *could* be handled by
using a multi-dimensional optimization solver, but I don't have one of
those for OpenSCAD and it does seem kind of like overkill.
Here's the code:
include<BOSL2/std.scad>
function ellipse_tangents(a,b,pt) =
let(
r=a,
scaledpt = yscale(a/b,pt),
tangents = circle_point_tangents(r,[0,0],scaledpt)
)
yscale(b/a,tangents);
function ellipse_next_pt(a,b,pt,L) =
let(
tanglist = ellipse_tangents(a,b,pt),
//f=echo(pt=pt,tanglist=tanglist),
tang = tanglist[0].x>pt.x ? tanglist[0]
: len(tanglist)==2 && tanglist[1].x>pt.x ? tanglist[1]
: assert(false,"cannot find acceptable tangent")
)
assert(norm(tang-pt)<L,"Tangent is too far away")
L*unit(tang-pt)+pt;
function ellipse_path(a,b,L,N,path=[]) =
len(path)==0 ? ellipse_path(a,b,L,N,[[L/2,b]])
: len(path)==N ? path
: last(path).y<=0 ? path
: last(path).x>=a ? path
: ellipse_path(a,b,L,N,
[each path, ellipse_next_pt(a,b,last(path),L)]);
function ellipse_circ(a,b) = PI*(3*(a+b) - sqrt(10*a*b+3*(a^2+b^2)));
function cellipse(a,b,N) =
assert(N%4==2, "Must have N of form 4k+2")
let(
errfunc = function(L) let(path=(ellipse_path(a,b,L,(N+2)/4)))
-last(path).y+
((N+2)/4 - len(path)) * b ,
lengthguess = ellipse_circ(a,b)/N,
L = root_find(errfunc, 0.95*lengthguess, 1.5*lengthguess),
path = ellipse_path(a,b,L,(N+2)/4),
right = concat(path, yflip(list_tail(reverse(path))))
)
concat(right, reverse(xflip(right)));
function cellipse_odd(a,b,N) =
assert(N%2==1, "Must have N of form 2k+1")
let(
errfunc = function(L)
let(path=ellipse_path(a,b,L,(N+1)/2,[[-a,L/2]]))
-last(path).y+
((N+1)/2 - len(path)) * b ,
lengthguess = ellipse_circ(a,b)/N,
L = root_find(errfunc, 0.95*lengthguess, 1.5*lengthguess),
path = ellipse_path(a,b,L,(N+1)/2,[[-a,L/2]])
)
concat(path, list_tail(reverse(yflip(path))));
a=10;
b=6;
N = 7;
p = N%2==1 ? cellipse_odd(a,b,N)
: cellipse(a,b,N);
echo(length = len(p));
ellipse([a,b],$fn=164);
color("red")stroke([p],width=.05,dots=true,dots_width=4);
On Mon, Aug 8, 2022 at 4:12 PM Father Horton <fatherhorton@gmail.com> wrote:
> My end condition is that the last point calculated is (-L/2, b), but as I
> think about it, it does make sense that N has to be a multiple of 2,
> because the whole thing has to be symmetrical around the y-axis.
>
> On Mon, Aug 8, 2022 at 2:35 PM Father Horton <fatherhorton@gmail.com>
> wrote:
>
>> Here's a Google sheet that shows what I have in mind.
>> https://docs.google.com/spreadsheets/d/13YOJKkWWfcgZNKVn_nlBtQZ_bbxNvN6SdVv1bVWLDP8/edit?usp=sharing
>>
>> I have to adjust the signs on the square roots manually, so I didn't
>> iterate until I got equal lengths, but this is the idea. Blue points are
>> end points of the circumscribing segments and red points are the tangent
>> points on the ellipse. I have other things to do at the moment, but I may
>> revisit later to get the signs to work out right and then see if it will
>> iterate to an acceptable result.
>>
>> On Mon, Aug 8, 2022 at 1:31 PM Adrian Mariano <avm4@cornell.edu> wrote:
>>
>>> The restriction arises because of the end condition. There are 4 cases
>>> depending on what happens at the four ends of the ellipse (the points where
>>> it intersects the coordinate axes). The four possibilities are at each end
>>> either a corner or a flat tangent at the end. Your approach as I
>>> understand it begins with a flat and constructs a quarter of the polygon,
>>> ending with a point. Such a polygon must have 4k+2 sides. I do not see
>>> how the end condition can be formulated to end on a flat. Also I do not
>>> see how it can start on a point. If n is odd you can start at one end and
>>> go half way around and end on a point. But if n is divisible by 4 you can
>>> have 4 flats or four points and the method cannot produce either of them.
>>> Have I missed something?
>>>
>>>
>>> On Mon, Aug 8, 2022, 13:42 Father Horton <fatherhorton@gmail.com> wrote:
>>>
>>>> I'm not sure why there should be a restriction on N; I think it's
>>>> general. I did a spreadsheet that looks like it's doing the right thing in
>>>> the first quadrant, but it (or I) gets lost when I have to select the right
>>>> value for the sign of the square roots involved as the process enters other
>>>> quadrants.
>>>>
>>>> On Sun, Aug 7, 2022 at 9:43 PM Father Horton <fatherhorton@gmail.com>
>>>> wrote:
>>>>
>>>>> Google showed me that method when I came up with the right terms. But
>>>>> it's too late at night for me to try to implement it. Maybe tomorrow.
>>>>>
>>>>> On Sun, Aug 7, 2022 at 9:41 PM Adrian Mariano <avm4@cornell.edu>
>>>>> wrote:
>>>>>
>>>>>> Now I understand your idea. I think it should be possible to make
>>>>>> that work for the case of N=4k+2. It does seem like there are some
>>>>>> complications, like you try to extend the segment and the length is closer
>>>>>> than the tangent, so it's impossible. Also things may be complicated if
>>>>>> the algorithm tries to go around the tip of the ellipse (that is, if the
>>>>>> path is too long).
>>>>>>
>>>>>> For finding a tangent to an ellipse, transform the problem to the
>>>>>> circle, solve, and then transform back. In BOSL2:
>>>>>>
>>>>>> function ellipse_tangents(a,b,pt) =
>>>>>> let(
>>>>>> r=a,
>>>>>> scaledpt = yscale(a/b,pt),
>>>>>> tangents = circle_point_tangents(r,[0,0],scaledpt)
>>>>>> )
>>>>>> yscale(b/a,tangents);
>>>>>>
>>>>>> On Sun, Aug 7, 2022 at 9:54 PM Father Horton <fatherhorton@gmail.com>
>>>>>> wrote:
>>>>>>
>>>>>>> I did that wrong. It's not (0, b) to (0, L/2). It's (0, b) to (L/2,
>>>>>>> b). That's always outside the ellipse.
>>>>>>>
>>>>>>> I will look at the second problem as soon as I figure out how to
>>>>>>> calculate a tangent line to an ellipse containing a point not on the
>>>>>>> ellipse. Google is not helping much.
>>>>>>>
>>>>>>> On Sun, Aug 7, 2022 at 8:32 PM Adrian Mariano <avm4@cornell.edu>
>>>>>>> wrote:
>>>>>>>
>>>>>>>> Father Horton,
>>>>>>>>
>>>>>>>> I'm having trouble understanding your proposed method. A segment
>>>>>>>> from (0,b) to (0,L/2) might be inside the ellipse depending on N, and then
>>>>>>>> there won't be a tangent.
>>>>>>>>
>>>>>>>> It also sounds like your idea is somewhat similar to Jordan's idea:
>>>>>>>> basically start at one side and work around the ellipse and see if it
>>>>>>>> works. If not, make a correction. The problem with this idea is that
>>>>>>>> there are two parameters for the first segment, so how do you make a
>>>>>>>> correction? (The two parameters are the location of the endpoint of the
>>>>>>>> first segment and its length.)
>>>>>>>>
>>>>>>>> On Sun, Aug 7, 2022 at 5:33 PM Father Horton <
>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>
>>>>>>>>> How about this to construct N equal length segments around an
>>>>>>>>> ellipse with axes a and b:
>>>>>>>>>
>>>>>>>>> 1) Select an initial segment length estimate (e.g., Ramanujan's
>>>>>>>>> formula / N) = L
>>>>>>>>> 2) Construct a half-segment from (0, b) to (0, L/2).
>>>>>>>>> 3) Find a line tangent to the ellipse and passing through (0, L/2).
>>>>>>>>> 3) Find the point on the tangent line that is L away from (0, L/2).
>>>>>>>>> 4) Repeat (3) substituting the new point for (0, L/2). until you
>>>>>>>>> have N segments.
>>>>>>>>>
>>>>>>>>> If everything worked perfectly, the last point found would be (0,
>>>>>>>>> -L/2). If there's a gap, increase L and try again. If there's overlap,
>>>>>>>>> decrease L and try again.
>>>>>>>>>
>>>>>>>>> On Sun, Aug 7, 2022 at 3:29 PM Ronaldo Persiano <
>>>>>>>>> rcmpersiano@gmail.com> wrote:
>>>>>>>>>
>>>>>>>>>> Adrian, why to use a (either inscribed or circumscribed) polygon
>>>>>>>>>> with equal side as an approximation of an ellipsis? It doesn't seem to be a
>>>>>>>>>> good approximation of it. I get a better approximation by taking an angle
>>>>>>>>>> uniform sample of the curve in the parameter space.
>>>>>>>>>>
>>>>>>>>>> [image: Uniformly_sampled_ellipsis.png]
>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Em sáb., 6 de ago. de 2022 às 01:44, Adrian Mariano <
>>>>>>>>>> avm4@cornell.edu> escreveu:
>>>>>>>>>>
>>>>>>>>>>> Circumscribed means tangents to points on the ellipse. But
>>>>>>>>>>> there are no arc lengths (equal or unequal) in this problem, nor equally
>>>>>>>>>>> spaced angles. The angles of each segment will be different. But the
>>>>>>>>>>> lengths of the segments are not the same as the arc lengths, which is why
>>>>>>>>>>> arc length does not matter. Here's a solution to the N=6 polygon sort of
>>>>>>>>>>> inspired by the idea Jordan suggested, where I adjust the first segment
>>>>>>>>>>> until the second segment has the right length. The rest of the solution is
>>>>>>>>>>> then obtained using symmetry, which means there is only a single unknown
>>>>>>>>>>> that can simply be solved for. But this doesn't generalize to higher N.
>>>>>>>>>>>
>>>>>>>>>>> [image: image.png]
>>>>>>>>>>> include<BOSL2/std.scad>
>>>>>>>>>>>
>>>>>>>>>>> // ellipse semiaxes
>>>>>>>>>>> // If a is much smaller than b it will fail
>>>>>>>>>>> a=5;
>>>>>>>>>>> b=2;
>>>>>>>>>>>
>>>>>>>>>>> // Find point on ellipse at specified x coordinate
>>>>>>>>>>> function ellipse_pt(a,b,x) = sqrt(b^2*(1-x^2/a^2));
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> // Find tangent line to ellipse at point (u,w)
>>>>>>>>>>> // and determine the x value where that line has
>>>>>>>>>>> // given y value.
>>>>>>>>>>> function findx(a,b,u,w,y) =
>>>>>>>>>>> let(
>>>>>>>>>>> A = -u*b^2/w/a^2,
>>>>>>>>>>> B = w+(u*b)^2/w/a^2
>>>>>>>>>>> )
>>>>>>>>>>> (y-B)/A;
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> // First segment if tangent point is at x=u
>>>>>>>>>>> function first_seg(a,b,u) =
>>>>>>>>>>> let(
>>>>>>>>>>> w = ellipse_pt(a,b,u),
>>>>>>>>>>> x1 = findx(a,b,u,w,0),
>>>>>>>>>>> x2 = findx(a,b,u,w,b)
>>>>>>>>>>> )
>>>>>>>>>>> [[x1,0],[x2,b]];
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> err = function (u) let(pt = first_seg(a,b,u))
>>>>>>>>>>> 2*pt[1].x - norm(pt[0]-pt[1]);
>>>>>>>>>>>
>>>>>>>>>>> x = root_find(err,.2*a,.9*a);
>>>>>>>>>>> first = first_seg(a,b,x);
>>>>>>>>>>> top = concat(first,reverse(xflip(first)));
>>>>>>>>>>> poly = concat(top, reverse(yflip(list_tail(top))));
>>>>>>>>>>>
>>>>>>>>>>> stroke(ellipse([a,b]),width=.05,$fn=128,closed=true);
>>>>>>>>>>> color("red")stroke(poly,width=.05,closed=true);
>>>>>>>>>>>
>>>>>>>>>>> // Verify equality of segment lengths
>>>>>>>>>>> echo(path_segment_lengths(poly,closed=true));
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> On Fri, Aug 5, 2022 at 8:55 PM Sanjeev Prabhakar <
>>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>>
>>>>>>>>>>>> Circumscribed means tangents from points on ellipse
>>>>>>>>>>>> How can the tangents from equally spaced arc lengths or equally
>>>>>>>>>>>> spaced angles be equal for an ellipse
>>>>>>>>>>>> You need to change the arc lengths dynamically and also
>>>>>>>>>>>> accurately estimate the segment length which fits the ellipse .
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> On Sat, 6 Aug, 2022, 4:51 am Adrian Mariano, <avm4@cornell.edu>
>>>>>>>>>>>> wrote:
>>>>>>>>>>>>
>>>>>>>>>>>>> Your intuition is incorrect: a circle scaled in one direction
>>>>>>>>>>>>> *IS* an ellipse. The motivation for being interested in
>>>>>>>>>>>>> circumscribing is to make holes, though the constraint of equal length
>>>>>>>>>>>>> sides is probably not what you'd want for that purpose. You can make a
>>>>>>>>>>>>> nonuniform circumscribing polygon for an ellipse by simply making a polygon
>>>>>>>>>>>>> that circumscribes a circle and scaling it.
>>>>>>>>>>>>>
>>>>>>>>>>>>> On Fri, Aug 5, 2022 at 7:16 PM Bob Carlson <bob@rjcarlson.com>
>>>>>>>>>>>>> wrote:
>>>>>>>>>>>>>
>>>>>>>>>>>>>> Seems like I remember Nop Head talking about using
>>>>>>>>>>>>>> circumscribed holes for screws and even rotating the polygons to produce a
>>>>>>>>>>>>>> more even hole.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> If an ellipse is the nominal shape, the using a circumscribed
>>>>>>>>>>>>>> polygon for a hole and an inscribed polygon for a solid might make sense.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Btw, can someone confirm my intuition that a circle scaled in
>>>>>>>>>>>>>> one dimension is not an ellipse?
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> -Bob
>>>>>>>>>>>>>> Tucson AZ
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> On Aug 5, 2022, at 15:38, Adrian Mariano <avm4@cornell.edu>
>>>>>>>>>>>>>> wrote:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> I don't have a specific need for a circumscribing elliptical
>>>>>>>>>>>>>> polygon with equal sides. I was writing code to make ellipses and it
>>>>>>>>>>>>>> seemed like equal side approximations would be nice, so I solved the case
>>>>>>>>>>>>>> for inscribed, but couldn't solve the circumscribed case. So it's an
>>>>>>>>>>>>>> unsolved puzzle I've got that this discussion reminded me of.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> On Fri, Aug 5, 2022 at 6:33 PM nop head <nop.head@gmail.com>
>>>>>>>>>>>>>> wrote:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> A long discussion but I have to ask why do you need an
>>>>>>>>>>>>>>> elliptical polygon with equal tangential sides?
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> On Fri, 5 Aug 2022, 21:58 Adrian Mariano, <avm4@cornell.edu>
>>>>>>>>>>>>>>> wrote:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> You think that a circumscribing polygon with equal length
>>>>>>>>>>>>>>>> segments does not exist?? I'm skeptical of this claim.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> For circumscribed, it's not so hard. You start with a
>>>>>>>>>>>>>>>> uniform-in-angle point distribution, compute the segment lengths, compare
>>>>>>>>>>>>>>>> to the mean segment length to get an error, and then compute an adjustment
>>>>>>>>>>>>>>>> based on the error. It converges quickly.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> So for example, I have segments lengths of:
>>>>>>>>>>>>>>>> ECHO: [5.57297, 5.57297, 5.57297, 5.57297, 5.57297,
>>>>>>>>>>>>>>>> 5.57297, 5.57297, 5.57297]
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> Maybe not exactly equal, since tolerance is 1e-9, but
>>>>>>>>>>>>>>>> surely good enough.
>>>>>>>>>>>>>>>> <image.png>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> The code for this is in BOSL2 in shapes2d.scad. A similar
>>>>>>>>>>>>>>>> algorithm should exist for the circumscribed case, but as I said, my
>>>>>>>>>>>>>>>> efforts to find it have failed.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> On Fri, Aug 5, 2022 at 12:44 PM Sanjeev Prabhakar <
>>>>>>>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> I think circumscribed polygon segments cannot be equal
>>>>>>>>>>>>>>>>> lengths.
>>>>>>>>>>>>>>>>> inscribed ones may be approximately equal, but would be a
>>>>>>>>>>>>>>>>> very tough problem to solve.
>>>>>>>>>>>>>>>>> I tried something.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> On Thu, 4 Aug 2022 at 05:44, Adrian Mariano <
>>>>>>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> I think you need to be a little careful about defining
>>>>>>>>>>>>>>>>>> what you want to do. If you want to approximate an ellipse with points
>>>>>>>>>>>>>>>>>> that are distributed uniformly over the ellipse, that could be done by
>>>>>>>>>>>>>>>>>> approximating the arc length of the ellipse to find the necessary
>>>>>>>>>>>>>>>>>> points---e.g. advance distance d along the ellipse at each step. But if
>>>>>>>>>>>>>>>>>> you want to approximate an ellipse with a polygon with equal length sides,
>>>>>>>>>>>>>>>>>> that approach does not do it, because the length of the segments will not
>>>>>>>>>>>>>>>>>> be constant as the span sections of the ellipse with differing curvature.
>>>>>>>>>>>>>>>>>> And since the perimeter of the desired polygon is unknown, you can't just
>>>>>>>>>>>>>>>>>> step it out analogously. The circle doesn't present this challenge due to
>>>>>>>>>>>>>>>>>> having constant curvature.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>> On Wed, Aug 3, 2022 at 7:41 PM Father Horton <
>>>>>>>>>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> I think our emails crossed. I was addressing how to
>>>>>>>>>>>>>>>>>>> approximate an ellipse with constant-length segments in general, not how to
>>>>>>>>>>>>>>>>>>> solve your problem. When I had to do something similar, I used polar
>>>>>>>>>>>>>>>>>>> coordinates and stepped up the value of theta until the distance from the
>>>>>>>>>>>>>>>>>>> previous point was correct. (In the constant-length version, I'd keep a
>>>>>>>>>>>>>>>>>>> running total of the maximum desired length so that round-off errors don't
>>>>>>>>>>>>>>>>>>> accumulate.)
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> Figuring out the circumscribed polygon is, as you note,
>>>>>>>>>>>>>>>>>>> a greater challenge. I can visualize how to do it for n = 4, but it doesn't
>>>>>>>>>>>>>>>>>>> extrapolate at all.
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:10 PM Adrian Mariano <
>>>>>>>>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> Eh, there's no such thing as "brute force" in this
>>>>>>>>>>>>>>>>>>>> situation. Step through *what* with sufficiently small steps? Of course
>>>>>>>>>>>>>>>>>>>> an iterative algorithm is necessary. This is true for most interesting
>>>>>>>>>>>>>>>>>>>> problems. No big deal. Show me an iterative algorithm that converges to
>>>>>>>>>>>>>>>>>>>> the circumscribing polygon. When I tried to devise such an algorithm I
>>>>>>>>>>>>>>>>>>>> couldn't find one that would converge. Writing an algorithm that would
>>>>>>>>>>>>>>>>>>>> converge to the *inscribed* polygon seemed to be straight forward. Note
>>>>>>>>>>>>>>>>>>>> also that computing arc length of the ellipse doesn't really help here.
>>>>>>>>>>>>>>>>>>>> You can do that by solving an elliptical integral, or summing up small
>>>>>>>>>>>>>>>>>>>> segments. But what's the connection between arc length of the ellipse and
>>>>>>>>>>>>>>>>>>>> a circumscribing polygon with equal segment lengths? If there's a
>>>>>>>>>>>>>>>>>>>> connection, I don't see it.
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:56 PM Father Horton <
>>>>>>>>>>>>>>>>>>>> fatherhorton@gmail.com> wrote:
>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> Except by brute force (step through it with
>>>>>>>>>>>>>>>>>>>>> sufficiently small steps), it’s not possible because there’s no closed form
>>>>>>>>>>>>>>>>>>>>> expression for the distance between two points on a general ellipse (as
>>>>>>>>>>>>>>>>>>>>> opposed to special cases like a circle).
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 9:50 PM Adrian Mariano <
>>>>>>>>>>>>>>>>>>>>> avm4@cornell.edu> wrote:
>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> Here's a challenge for you: draw an "ellipse"
>>>>>>>>>>>>>>>>>>>>>> polygon with N segments that are all the same length that circumscribes a
>>>>>>>>>>>>>>>>>>>>>> perfect ellipse with given semiaxes a and b. I tried to solve this problem
>>>>>>>>>>>>>>>>>>>>>> and could not find a solution. I can draw an *inscribed* N segment
>>>>>>>>>>>>>>>>>>>>>> "ellipse" polygon with all segments the same length---though it's not
>>>>>>>>>>>>>>>>>>>>>> easy---but the circumscribed case eludes me. I think of the inscribed
>>>>>>>>>>>>>>>>>>>>>> case as, every vertex is located on the ellipse and the circumscribed case,
>>>>>>>>>>>>>>>>>>>>>> every segment is tangent to the ellipse
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> On Tue, Aug 2, 2022 at 10:36 PM Sanjeev Prabhakar <
>>>>>>>>>>>>>>>>>>>>>> sprabhakar2006@gmail.com> wrote:
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> In openscad drawing a ellipse is not too difficult
>>>>>>>>>>>>>>>>>>>>>>> Refer following code
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> function to draw an ellipse with semi-major and
>>>>>>>>>>>>>>>>>>>>>>> semi-minor axis "r1" and "r2" respectively and with center "cp" and number
>>>>>>>>>>>>>>>>>>>>>>> of segment "s"
>>>>>>>>>>>>>>>>>>>>>>> /
>>>>>>>>>>>>>>>>>>>>>>> // example:
>>>>>>>>>>>>>>>>>>>>>>> // sec=ellipse(r1=5,r2=3,cp=[2,3],s=30);
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> function ellipse(r1,r2,cp,s=30)=
>>>>>>>>>>>>>>>>>>>>>>> let(
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> sec=[for(i=[0:360/s:360-360/s])cp+[r1*cos(i),r2*sin(i)]]
>>>>>>>>>>>>>>>>>>>>>>> )sec;
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> On Wed, 3 Aug, 2022, 5:35 am Jordan Brown, <
>>>>>>>>>>>>>>>>>>>>>>> openscad@jordan.maileater.net> wrote:
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> On 8/2/2022 4:46 PM, Brian Allen wrote:
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> I created an oval using:
>>>>>>>>>>>>>>>>>>>>>>>> inMm = 25.4;
>>>>>>>>>>>>>>>>>>>>>>>> fullW = 35.5 * inMm;
>>>>>>>>>>>>>>>>>>>>>>>> fullL = 54.5 * inMm;
>>>>>>>>>>>>>>>>>>>>>>>> resize([fullW, fullL, 1])
>>>>>>>>>>>>>>>>>>>>>>>> circle(d=fullW);
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> Tip: use scale() rather than resize(); it previews
>>>>>>>>>>>>>>>>>>>>>>>> faster in complicated cases.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> It's the right size in the middle (35.5" wide at 50% of the length), but 25% along the length it's not quite as wide as I need it (it's 31.5" and I need it about 32"). If you can see the attached image I'm trying to make the yellow oval match the red "sanity check" lines. I'm trying to match the size and shape of an existing oval in the real world.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> You're taking a circle and stretching it; the
>>>>>>>>>>>>>>>>>>>>>>>> result of that stretching is an ellipse.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> I'm not 100% sure - maybe one of the real math
>>>>>>>>>>>>>>>>>>>>>>>> wizards can chime in - but I think the shape you are asking for would not
>>>>>>>>>>>>>>>>>>>>>>>> be an ellipse.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> I believe that an ellipse is fully defined by its
>>>>>>>>>>>>>>>>>>>>>>>> major (long) and minor (short) axis dimensions; the other parts of the
>>>>>>>>>>>>>>>>>>>>>>>> curve mathematically flow from those two measurements.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> You can make something else, but you can't make it
>>>>>>>>>>>>>>>>>>>>>>>> using this technique.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> Is there a way to stretch this in OpenSCAD? In a GUI vector graphics program I'd use a bézier curve and drag the handle to get 1/4 of it just right then mirror it in X and Y to get the oval I needed.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> You can use a Bézier function.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> Using BOSL2:
>>>>>>>>>>>>>>>>>>>>>>>> https://github.com/revarbat/BOSL2/wiki/Topics#bezier-curves
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> DIY, here's a function that I cobbled together some
>>>>>>>>>>>>>>>>>>>>>>>> years ago:
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> // Bezier functions from https://www.thingiverse.com/thing:8443
>>>>>>>>>>>>>>>>>>>>>>>> // but that yielded either single points or a raft of triangles;
>>>>>>>>>>>>>>>>>>>>>>>> // this yields a vector of points that you can then concatenate
>>>>>>>>>>>>>>>>>>>>>>>> // with other pieces to form a single polygon.
>>>>>>>>>>>>>>>>>>>>>>>> // If we were really clever, I think it would be possible to
>>>>>>>>>>>>>>>>>>>>>>>> // automatically space the output points based on how linear
>>>>>>>>>>>>>>>>>>>>>>>> // the curve is at that point. But right now I'm not that clever.
>>>>>>>>>>>>>>>>>>>>>>>> function BEZ03(u) = pow((1-u), 3);
>>>>>>>>>>>>>>>>>>>>>>>> function BEZ13(u) = 3*u*(pow((1-u),2));
>>>>>>>>>>>>>>>>>>>>>>>> function BEZ23(u) = 3*(pow(u,2))*(1-u);
>>>>>>>>>>>>>>>>>>>>>>>> function BEZ33(u) = pow(u,3);
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> function PointAlongBez4(p0, p1, p2, p3, u) = [
>>>>>>>>>>>>>>>>>>>>>>>> BEZ03(u)*p0[0]+BEZ13(u)*p1[0]+BEZ23(u)*p2[0]+BEZ33(u)*p3[0],
>>>>>>>>>>>>>>>>>>>>>>>> BEZ03(u)*p0[1]+BEZ13(u)*p1[1]+BEZ23(u)*p2[1]+BEZ33(u)*p3[1]];
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> // p0 - start point
>>>>>>>>>>>>>>>>>>>>>>>> // p1 - control point 1, line departs p0 headed this way
>>>>>>>>>>>>>>>>>>>>>>>> // p2 - control point 2, line arrives at p3 from this way
>>>>>>>>>>>>>>>>>>>>>>>> // p3 - end point
>>>>>>>>>>>>>>>>>>>>>>>> // segs - number of segments
>>>>>>>>>>>>>>>>>>>>>>>> function bez(p0, p1, p2, p3, segs) = [
>>>>>>>>>>>>>>>>>>>>>>>> for (i = [0:segs]) PointAlongBez4(p0, p1, p2, p3, i/segs)
>>>>>>>>>>>>>>>>>>>>>>>> ];
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> It is that final function bez() that I actually
>>>>>>>>>>>>>>>>>>>>>>>> call, e.g.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> polygon(bez([0,0], [0,4], [5,1], [7,0], 20));
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> often concatenating the results of several bez()
>>>>>>>>>>>>>>>>>>>>>>>> calls together.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> Note that the four coordinates correspond to the
>>>>>>>>>>>>>>>>>>>>>>>> coordinates of the control points in a GUI drawing program: the start
>>>>>>>>>>>>>>>>>>>>>>>> point, control point associated with the start, the control point
>>>>>>>>>>>>>>>>>>>>>>>> associated with the end, and the end point. I have sometimes created
>>>>>>>>>>>>>>>>>>>>>>>> OpenSCAD objects by tracing an object in a drawing program and then reading
>>>>>>>>>>>>>>>>>>>>>>>> off the coordinates of the various control points.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>> Note that Bézier functions are not the only
>>>>>>>>>>>>>>>>>>>>>>>> mathematical curves. (For instance, they cannot exactly match an arc of a
>>>>>>>>>>>>>>>>>>>>>>>> circle.) I'm sure there are numerous other functions that also produce
>>>>>>>>>>>>>>>>>>>>>>>> curves.
>>>>>>>>>>>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>> _______________________________________________
>>>>>>>>>>>>>>>>>>>>>>> OpenSCAD mailing list
>>>>>>>>>>>>>>>>>>>>>>> To unsubscribe send an email to
>>>>>>>>>>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>> _______________________________________________
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>>>>>>>>>>>>>>>>>>>>>> discuss-leave@lists.openscad.org
>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>> _______________________________________________
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>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>> _______________________________________________
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>>>>>>>>>>>>>>>>>>> _______________________________________________
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AM
Adrian Mariano
Mon, Aug 8, 2022 8:41 PM
It's never necessary to do more than half the ellipse, which I think
makes handling the multiplicity of tangents easier.
On Mon, Aug 8, 2022 at 4:30 PM Father Horton fatherhorton@gmail.com wrote:
On Mon, Aug 8, 2022 at 3:23 PM Michael Möller private2michael@gmail.com wrote:
y and x if divisible by 4, right?
Right, which would make calculation easier since I wouldn't have to deal with those nasty quadrant changes.
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org
It's never necessary to do more than half the ellipse, which I think
makes handling the multiplicity of tangents easier.
On Mon, Aug 8, 2022 at 4:30 PM Father Horton <fatherhorton@gmail.com> wrote:
>
>
>
> On Mon, Aug 8, 2022 at 3:23 PM Michael Möller <private2michael@gmail.com> wrote:
>>
>> y and x if divisible by 4, right?
>
>
> Right, which would make calculation easier since I wouldn't have to deal with those nasty quadrant changes.
> _______________________________________________
> OpenSCAD mailing list
> To unsubscribe send an email to discuss-leave@lists.openscad.org
FH
Father Horton
Mon, Aug 8, 2022 8:46 PM
Here's a Python version that gets the first quadrant right:
import math
major = 2
minor = 1
length = 1.03
px = length / 2
py = minor
for i in range(1, 3):
c = px / major
d = py / minor
print(f"Adjusted point off circle is ({c:.3f},{d:.3f})")
a = (2 * c + math.sqrt(4 * c * c - 4 * (d * d + c * c) * (1 - d * d)))
/ (2 * (d * d + c * c))
b = math.sqrt(1 - a * a)
print(f"Point on circle is ({a:.3f}, {b:.3f})")
ex = a * major
ey = b * minor
print(f"Point on ellipse is ({ex:.3f}, {ey:.3f})")
m = (py - ey) / (px - ex)
xhat = math.sqrt(length * length / (1 + m * m))
yhat = m * xhat
px = px + xhat
py = py + yhat
print(f"New point off circle is ({px:.3f}, {py:.3f})")
On Mon, Aug 8, 2022 at 3:41 PM Adrian Mariano avm4@cornell.edu wrote:
It's never necessary to do more than half the ellipse, which I think
makes handling the multiplicity of tangents easier.
On Mon, Aug 8, 2022 at 4:30 PM Father Horton fatherhorton@gmail.com
wrote:
y and x if divisible by 4, right?
Right, which would make calculation easier since I wouldn't have to deal
with those nasty quadrant changes.
Here's a Python version that gets the first quadrant right:
import math
major = 2
minor = 1
length = 1.03
px = length / 2
py = minor
for i in range(1, 3):
c = px / major
d = py / minor
print(f"Adjusted point off circle is ({c:.3f},{d:.3f})")
a = (2 * c + math.sqrt(4 * c * c - 4 * (d * d + c * c) * (1 - d * d)))
/ (2 * (d * d + c * c))
b = math.sqrt(1 - a * a)
print(f"Point on circle is ({a:.3f}, {b:.3f})")
ex = a * major
ey = b * minor
print(f"Point on ellipse is ({ex:.3f}, {ey:.3f})")
m = (py - ey) / (px - ex)
xhat = math.sqrt(length * length / (1 + m * m))
yhat = m * xhat
px = px + xhat
py = py + yhat
print(f"New point off circle is ({px:.3f}, {py:.3f})")
On Mon, Aug 8, 2022 at 3:41 PM Adrian Mariano <avm4@cornell.edu> wrote:
> It's never necessary to do more than half the ellipse, which I think
> makes handling the multiplicity of tangents easier.
>
> On Mon, Aug 8, 2022 at 4:30 PM Father Horton <fatherhorton@gmail.com>
> wrote:
> >
> >
> >
> > On Mon, Aug 8, 2022 at 3:23 PM Michael Möller <private2michael@gmail.com>
> wrote:
> >>
> >> y and x if divisible by 4, right?
> >
> >
> > Right, which would make calculation easier since I wouldn't have to deal
> with those nasty quadrant changes.
> > _______________________________________________
> > OpenSCAD mailing list
> > To unsubscribe send an email to discuss-leave@lists.openscad.org
> _______________________________________________
> OpenSCAD mailing list
> To unsubscribe send an email to discuss-leave@lists.openscad.org
>
AM
Adrian Mariano
Mon, Aug 8, 2022 9:28 PM
So that code just works around the ellipse, making 3 steps at the
fixed length, right? The hard part isn't doing that---it's being able
to automatically update so that you get the correct answer. What
happens if you try to do this manual process but for an ellipse that
starts with a full segment whose endpoint is [0,b+extra] for some
extra corner offset distance? Or one that starts the same but ends at
[b,L/2]. I haven't actually tried running these processes to see how
it behaves.
On Mon, Aug 8, 2022 at 4:47 PM Father Horton fatherhorton@gmail.com wrote:
Here's a Python version that gets the first quadrant right:
import math
major = 2
minor = 1
length = 1.03
px = length / 2
py = minor
for i in range(1, 3):
c = px / major
d = py / minor
print(f"Adjusted point off circle is ({c:.3f},{d:.3f})")
a = (2 * c + math.sqrt(4 * c * c - 4 * (d * d + c * c) * (1 - d * d))) / (2 * (d * d + c * c))
b = math.sqrt(1 - a * a)
print(f"Point on circle is ({a:.3f}, {b:.3f})")
ex = a * major
ey = b * minor
print(f"Point on ellipse is ({ex:.3f}, {ey:.3f})")
m = (py - ey) / (px - ex)
xhat = math.sqrt(length * length / (1 + m * m))
yhat = m * xhat
px = px + xhat
py = py + yhat
print(f"New point off circle is ({px:.3f}, {py:.3f})")
On Mon, Aug 8, 2022 at 3:41 PM Adrian Mariano avm4@cornell.edu wrote:
It's never necessary to do more than half the ellipse, which I think
makes handling the multiplicity of tangents easier.
On Mon, Aug 8, 2022 at 4:30 PM Father Horton fatherhorton@gmail.com wrote:
On Mon, Aug 8, 2022 at 3:23 PM Michael Möller private2michael@gmail.com wrote:
y and x if divisible by 4, right?
Right, which would make calculation easier since I wouldn't have to deal with those nasty quadrant changes.
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So that code just works around the ellipse, making 3 steps at the
fixed length, right? The hard part isn't doing that---it's being able
to automatically update so that you get the correct answer. What
happens if you try to do this manual process but for an ellipse that
starts with a full segment whose endpoint is [0,b+extra] for some
extra corner offset distance? Or one that starts the same but ends at
[b,L/2]. I haven't actually tried running these processes to see how
it behaves.
On Mon, Aug 8, 2022 at 4:47 PM Father Horton <fatherhorton@gmail.com> wrote:
>
> Here's a Python version that gets the first quadrant right:
>
> import math
>
> major = 2
> minor = 1
>
> length = 1.03
>
> px = length / 2
> py = minor
>
> for i in range(1, 3):
> c = px / major
> d = py / minor
> print(f"Adjusted point off circle is ({c:.3f},{d:.3f})")
>
>
> a = (2 * c + math.sqrt(4 * c * c - 4 * (d * d + c * c) * (1 - d * d))) / (2 * (d * d + c * c))
> b = math.sqrt(1 - a * a)
>
> print(f"Point on circle is ({a:.3f}, {b:.3f})")
>
> ex = a * major
> ey = b * minor
> print(f"Point on ellipse is ({ex:.3f}, {ey:.3f})")
>
> m = (py - ey) / (px - ex)
> xhat = math.sqrt(length * length / (1 + m * m))
> yhat = m * xhat
> px = px + xhat
> py = py + yhat
>
> print(f"New point off circle is ({px:.3f}, {py:.3f})")
>
>
> On Mon, Aug 8, 2022 at 3:41 PM Adrian Mariano <avm4@cornell.edu> wrote:
>>
>> It's never necessary to do more than half the ellipse, which I think
>> makes handling the multiplicity of tangents easier.
>>
>> On Mon, Aug 8, 2022 at 4:30 PM Father Horton <fatherhorton@gmail.com> wrote:
>> >
>> >
>> >
>> > On Mon, Aug 8, 2022 at 3:23 PM Michael Möller <private2michael@gmail.com> wrote:
>> >>
>> >> y and x if divisible by 4, right?
>> >
>> >
>> > Right, which would make calculation easier since I wouldn't have to deal with those nasty quadrant changes.
>> > _______________________________________________
>> > OpenSCAD mailing list
>> > To unsubscribe send an email to discuss-leave@lists.openscad.org
>> _______________________________________________
>> OpenSCAD mailing list
>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>
> _______________________________________________
> OpenSCAD mailing list
> To unsubscribe send an email to discuss-leave@lists.openscad.org
FH
Father Horton
Mon, Aug 8, 2022 9:35 PM
You change L until it works. I'd start with an estimate (Ramanujan's?) of
the perimeter of the ellipse and divide by N to get an estimate for L. Step
around and if L is too short, increase it by say 10%, and if it's too long,
decrease by 10%. Once you have the right distance bracketed, do a binary
search.
On Mon, Aug 8, 2022 at 4:29 PM Adrian Mariano avm4@cornell.edu wrote:
So that code just works around the ellipse, making 3 steps at the
fixed length, right? The hard part isn't doing that---it's being able
to automatically update so that you get the correct answer. What
happens if you try to do this manual process but for an ellipse that
starts with a full segment whose endpoint is [0,b+extra] for some
extra corner offset distance? Or one that starts the same but ends at
[b,L/2]. I haven't actually tried running these processes to see how
it behaves.
On Mon, Aug 8, 2022 at 4:47 PM Father Horton fatherhorton@gmail.com
wrote:
Here's a Python version that gets the first quadrant right:
import math
major = 2
minor = 1
length = 1.03
px = length / 2
py = minor
for i in range(1, 3):
c = px / major
d = py / minor
print(f"Adjusted point off circle is ({c:.3f},{d:.3f})")
a = (2 * c + math.sqrt(4 * c * c - 4 * (d * d + c * c) * (1 - d *
d))) / (2 * (d * d + c * c))
b = math.sqrt(1 - a * a)
print(f"Point on circle is ({a:.3f}, {b:.3f})")
ex = a * major
ey = b * minor
print(f"Point on ellipse is ({ex:.3f}, {ey:.3f})")
m = (py - ey) / (px - ex)
xhat = math.sqrt(length * length / (1 + m * m))
yhat = m * xhat
px = px + xhat
py = py + yhat
print(f"New point off circle is ({px:.3f}, {py:.3f})")
On Mon, Aug 8, 2022 at 3:41 PM Adrian Mariano avm4@cornell.edu wrote:
It's never necessary to do more than half the ellipse, which I think
makes handling the multiplicity of tangents easier.
On Mon, Aug 8, 2022 at 4:30 PM Father Horton fatherhorton@gmail.com
On Mon, Aug 8, 2022 at 3:23 PM Michael Möller <
y and x if divisible by 4, right?
Right, which would make calculation easier since I wouldn't have to
deal with those nasty quadrant changes.
You change L until it works. I'd start with an estimate (Ramanujan's?) of
the perimeter of the ellipse and divide by N to get an estimate for L. Step
around and if L is too short, increase it by say 10%, and if it's too long,
decrease by 10%. Once you have the right distance bracketed, do a binary
search.
On Mon, Aug 8, 2022 at 4:29 PM Adrian Mariano <avm4@cornell.edu> wrote:
> So that code just works around the ellipse, making 3 steps at the
> fixed length, right? The hard part isn't doing that---it's being able
> to automatically update so that you get the correct answer. What
> happens if you try to do this manual process but for an ellipse that
> starts with a full segment whose endpoint is [0,b+extra] for some
> extra corner offset distance? Or one that starts the same but ends at
> [b,L/2]. I haven't actually tried running these processes to see how
> it behaves.
>
> On Mon, Aug 8, 2022 at 4:47 PM Father Horton <fatherhorton@gmail.com>
> wrote:
> >
> > Here's a Python version that gets the first quadrant right:
> >
> > import math
> >
> > major = 2
> > minor = 1
> >
> > length = 1.03
> >
> > px = length / 2
> > py = minor
> >
> > for i in range(1, 3):
> > c = px / major
> > d = py / minor
> > print(f"Adjusted point off circle is ({c:.3f},{d:.3f})")
> >
> >
> > a = (2 * c + math.sqrt(4 * c * c - 4 * (d * d + c * c) * (1 - d *
> d))) / (2 * (d * d + c * c))
> > b = math.sqrt(1 - a * a)
> >
> > print(f"Point on circle is ({a:.3f}, {b:.3f})")
> >
> > ex = a * major
> > ey = b * minor
> > print(f"Point on ellipse is ({ex:.3f}, {ey:.3f})")
> >
> > m = (py - ey) / (px - ex)
> > xhat = math.sqrt(length * length / (1 + m * m))
> > yhat = m * xhat
> > px = px + xhat
> > py = py + yhat
> >
> > print(f"New point off circle is ({px:.3f}, {py:.3f})")
> >
> >
> > On Mon, Aug 8, 2022 at 3:41 PM Adrian Mariano <avm4@cornell.edu> wrote:
> >>
> >> It's never necessary to do more than half the ellipse, which I think
> >> makes handling the multiplicity of tangents easier.
> >>
> >> On Mon, Aug 8, 2022 at 4:30 PM Father Horton <fatherhorton@gmail.com>
> wrote:
> >> >
> >> >
> >> >
> >> > On Mon, Aug 8, 2022 at 3:23 PM Michael Möller <
> private2michael@gmail.com> wrote:
> >> >>
> >> >> y and x if divisible by 4, right?
> >> >
> >> >
> >> > Right, which would make calculation easier since I wouldn't have to
> deal with those nasty quadrant changes.
> >> > _______________________________________________
> >> > OpenSCAD mailing list
> >> > To unsubscribe send an email to discuss-leave@lists.openscad.org
> >> _______________________________________________
> >> OpenSCAD mailing list
> >> To unsubscribe send an email to discuss-leave@lists.openscad.org
> >
> > _______________________________________________
> > OpenSCAD mailing list
> > To unsubscribe send an email to discuss-leave@lists.openscad.org
> _______________________________________________
> OpenSCAD mailing list
> To unsubscribe send an email to discuss-leave@lists.openscad.org
>
AM
Adrian Mariano
Mon, Aug 8, 2022 10:18 PM
Did you see my post earlier that actually implements the entire
end-to-end process, including estimating L
When starting at a point there are TWO parameters, L and "extra", the
distance of the point away from the ellipse. If you have extra wrong
you will presumably never get a solution no matter how much you fiddle
with L. This could be addressed by hitting with a 2d optimization
method, but as I said, that seems overcomplex and I don't have such a
code implemented in OpenSCAD.
Actually I was just thinking a bit more about the case of starting a
quadrant and ending on a flat. That means start at [b,L/2] and end on
[a,L/2]. I tweaked the code to end with y=L/2 and it fails. That is,
the resulting point it finds doesn't have x=b. But there doesn't seem
to be another degree of freedom. It would therefore appear that the
case of four flat ellipse ends is generally impossible. Have I missed
something?
On Mon, Aug 8, 2022 at 5:37 PM Father Horton fatherhorton@gmail.com wrote:
You change L until it works. I'd start with an estimate (Ramanujan's?) of the perimeter of the ellipse and divide by N to get an estimate for L. Step around and if L is too short, increase it by say 10%, and if it's too long, decrease by 10%. Once you have the right distance bracketed, do a binary search.
On Mon, Aug 8, 2022 at 4:29 PM Adrian Mariano avm4@cornell.edu wrote:
So that code just works around the ellipse, making 3 steps at the
fixed length, right? The hard part isn't doing that---it's being able
to automatically update so that you get the correct answer. What
happens if you try to do this manual process but for an ellipse that
starts with a full segment whose endpoint is [0,b+extra] for some
extra corner offset distance? Or one that starts the same but ends at
[b,L/2]. I haven't actually tried running these processes to see how
it behaves.
On Mon, Aug 8, 2022 at 4:47 PM Father Horton fatherhorton@gmail.com wrote:
Here's a Python version that gets the first quadrant right:
import math
major = 2
minor = 1
length = 1.03
px = length / 2
py = minor
for i in range(1, 3):
c = px / major
d = py / minor
print(f"Adjusted point off circle is ({c:.3f},{d:.3f})")
a = (2 * c + math.sqrt(4 * c * c - 4 * (d * d + c * c) * (1 - d * d))) / (2 * (d * d + c * c))
b = math.sqrt(1 - a * a)
print(f"Point on circle is ({a:.3f}, {b:.3f})")
ex = a * major
ey = b * minor
print(f"Point on ellipse is ({ex:.3f}, {ey:.3f})")
m = (py - ey) / (px - ex)
xhat = math.sqrt(length * length / (1 + m * m))
yhat = m * xhat
px = px + xhat
py = py + yhat
print(f"New point off circle is ({px:.3f}, {py:.3f})")
On Mon, Aug 8, 2022 at 3:41 PM Adrian Mariano avm4@cornell.edu wrote:
It's never necessary to do more than half the ellipse, which I think
makes handling the multiplicity of tangents easier.
On Mon, Aug 8, 2022 at 4:30 PM Father Horton fatherhorton@gmail.com wrote:
On Mon, Aug 8, 2022 at 3:23 PM Michael Möller private2michael@gmail.com wrote:
y and x if divisible by 4, right?
Right, which would make calculation easier since I wouldn't have to deal with those nasty quadrant changes.
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To unsubscribe send an email to discuss-leave@lists.openscad.org
OpenSCAD mailing list
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OpenSCAD mailing list
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Did you see my post earlier that actually implements the entire
end-to-end process, including estimating L
When starting at a point there are TWO parameters, L and "extra", the
distance of the point away from the ellipse. If you have extra wrong
you will presumably never get a solution no matter how much you fiddle
with L. This could be addressed by hitting with a 2d optimization
method, but as I said, that seems overcomplex and I don't have such a
code implemented in OpenSCAD.
Actually I was just thinking a bit more about the case of starting a
quadrant and ending on a flat. That means start at [b,L/2] and end on
[a,L/2]. I tweaked the code to end with y=L/2 and it fails. That is,
the resulting point it finds doesn't have x=b. But there doesn't seem
to be another degree of freedom. It would therefore appear that the
case of four flat ellipse ends is generally impossible. Have I missed
something?
On Mon, Aug 8, 2022 at 5:37 PM Father Horton <fatherhorton@gmail.com> wrote:
>
> You change L until it works. I'd start with an estimate (Ramanujan's?) of the perimeter of the ellipse and divide by N to get an estimate for L. Step around and if L is too short, increase it by say 10%, and if it's too long, decrease by 10%. Once you have the right distance bracketed, do a binary search.
>
> On Mon, Aug 8, 2022 at 4:29 PM Adrian Mariano <avm4@cornell.edu> wrote:
>>
>> So that code just works around the ellipse, making 3 steps at the
>> fixed length, right? The hard part isn't doing that---it's being able
>> to automatically update so that you get the correct answer. What
>> happens if you try to do this manual process but for an ellipse that
>> starts with a full segment whose endpoint is [0,b+extra] for some
>> extra corner offset distance? Or one that starts the same but ends at
>> [b,L/2]. I haven't actually tried running these processes to see how
>> it behaves.
>>
>> On Mon, Aug 8, 2022 at 4:47 PM Father Horton <fatherhorton@gmail.com> wrote:
>> >
>> > Here's a Python version that gets the first quadrant right:
>> >
>> > import math
>> >
>> > major = 2
>> > minor = 1
>> >
>> > length = 1.03
>> >
>> > px = length / 2
>> > py = minor
>> >
>> > for i in range(1, 3):
>> > c = px / major
>> > d = py / minor
>> > print(f"Adjusted point off circle is ({c:.3f},{d:.3f})")
>> >
>> >
>> > a = (2 * c + math.sqrt(4 * c * c - 4 * (d * d + c * c) * (1 - d * d))) / (2 * (d * d + c * c))
>> > b = math.sqrt(1 - a * a)
>> >
>> > print(f"Point on circle is ({a:.3f}, {b:.3f})")
>> >
>> > ex = a * major
>> > ey = b * minor
>> > print(f"Point on ellipse is ({ex:.3f}, {ey:.3f})")
>> >
>> > m = (py - ey) / (px - ex)
>> > xhat = math.sqrt(length * length / (1 + m * m))
>> > yhat = m * xhat
>> > px = px + xhat
>> > py = py + yhat
>> >
>> > print(f"New point off circle is ({px:.3f}, {py:.3f})")
>> >
>> >
>> > On Mon, Aug 8, 2022 at 3:41 PM Adrian Mariano <avm4@cornell.edu> wrote:
>> >>
>> >> It's never necessary to do more than half the ellipse, which I think
>> >> makes handling the multiplicity of tangents easier.
>> >>
>> >> On Mon, Aug 8, 2022 at 4:30 PM Father Horton <fatherhorton@gmail.com> wrote:
>> >> >
>> >> >
>> >> >
>> >> > On Mon, Aug 8, 2022 at 3:23 PM Michael Möller <private2michael@gmail.com> wrote:
>> >> >>
>> >> >> y and x if divisible by 4, right?
>> >> >
>> >> >
>> >> > Right, which would make calculation easier since I wouldn't have to deal with those nasty quadrant changes.
>> >> > _______________________________________________
>> >> > OpenSCAD mailing list
>> >> > To unsubscribe send an email to discuss-leave@lists.openscad.org
>> >> _______________________________________________
>> >> OpenSCAD mailing list
>> >> To unsubscribe send an email to discuss-leave@lists.openscad.org
>> >
>> > _______________________________________________
>> > OpenSCAD mailing list
>> > To unsubscribe send an email to discuss-leave@lists.openscad.org
>> _______________________________________________
>> OpenSCAD mailing list
>> To unsubscribe send an email to discuss-leave@lists.openscad.org
>
> _______________________________________________
> OpenSCAD mailing list
> To unsubscribe send an email to discuss-leave@lists.openscad.org
BA
Brian Allen
Tue, Aug 9, 2022 1:14 AM
Thank you Douglas! That was super helpful.
Boy, I really opened a can of worms with this question! But I've enjoyed all of the discussion (the 3% of it that wasn't over my head). Thanks everyone!
Brian
On Sat, Aug 6, 2022, at 5:49 AM, Douglas Miller wrote:
That is correct: the shape Brian wants is not an ellipse, but (probably)
a hyperellipse.
The standard equation of an ellipse is (x/a)^2 + (y/b)^2 = 1 where a, b > 0
By contrast, the standard equation of a hyperellipse is |x/a|^n +
|y/b|^n = 1 where a, b > 0 and n > 2
This can be generated by this function
function f(i,a,b,n) =
let(theta=i*360)
[a * cos(theta), sign(180-theta) * b *
pow((1-pow(abs(cos(theta)),n)),1/n) ];
For example:
points = [ for(i=[0:0.01:1]) f(i,12,7,2.5) ]; // semi-major axis = 12,
semi-minor axis = 7, n = 2.5
polygon(points);
I imagine that Brian will want to use a value of n that is only slightly
greater than 2. A good starting value for experimentation appears to be
n = 2.25.
On 8/2/2022 8:04 PM, Jordan Brown wrote:
On 8/2/2022 4:46 PM, Brian Allen wrote:
I created an oval using:
inMm = 25.4;
fullW = 35.5 * inMm;
fullL = 54.5 * inMm;
resize([fullW, fullL, 1])
circle(d=fullW);
[...]
You're taking a circle and stretching it; the result of that
stretching is an ellipse.
I'm not 100% sure - maybe one of the real math wizards can chime in -
but I think the shape you are asking for would not be an ellipse.
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To unsubscribe send an email to discuss-leave@lists.openscad.org
Thank you Douglas! That was super helpful.
Boy, I really opened a can of worms with this question! But I've enjoyed all of the discussion (the 3% of it that wasn't over my head). Thanks everyone!
Brian
On Sat, Aug 6, 2022, at 5:49 AM, Douglas Miller wrote:
> That is correct: the shape Brian wants is not an ellipse, but (probably)
> a hyperellipse.
>
> The standard equation of an ellipse is (x/a)^2 + (y/b)^2 = 1 where a, b > 0
>
> By contrast, the standard equation of a hyperellipse is |x/a|^n +
> |y/b|^n = 1 where a, b > 0 and n > 2
>
> This can be generated by this function
>
> function f(i,a,b,n) =
> let(theta=i*360)
> [a * cos(theta), sign(180-theta) * b *
> pow((1-pow(abs(cos(theta)),n)),1/n) ];
>
> For example:
>
> points = [ for(i=[0:0.01:1]) f(i,12,7,2.5) ]; // semi-major axis = 12,
> semi-minor axis = 7, n = 2.5
> polygon(points);
>
> I imagine that Brian will want to use a value of n that is only slightly
> greater than 2. A good starting value for experimentation appears to be
> n = 2.25.
>
> On 8/2/2022 8:04 PM, Jordan Brown wrote:
> > On 8/2/2022 4:46 PM, Brian Allen wrote:
> >> I created an oval using:
> >> inMm = 25.4;
> >> fullW = 35.5 * inMm;
> >> fullL = 54.5 * inMm;
> >> resize([fullW, fullL, 1])
> >> circle(d=fullW);
> > [...]
> > You're taking a circle and stretching it; the result of that
> > stretching is an ellipse.
> >
> > I'm not 100% sure - maybe one of the real math wizards can chime in -
> > but I think the shape you are asking for would not be an ellipse.
>
>
> _______________________________________________
> OpenSCAD mailing list
> To unsubscribe send an email to discuss-leave@lists.openscad.org
>
