Hi Gene,
You don't need to know the maths involved here, there are functions
written to do all the maths and you need to just know how to use them.
Regards
On Tue, 25 Jul, 2023, 7:42 am gene heskett, gheskett@shentel.net wrote:
On 7/24/23 20:49, Sanjeev Prabhakar wrote:
Haha, thanks
retirement is nearby also
While I as I look to my 89th in a couple months will be retired for 23
years. I am a C.E.T. with an 8th grade education, and don't have the
math skills to do what you are doing. Rather precocious electronically
before I was a teenager, my 8th grade algebra teacher thought he was a
George Carlin and far more interested in what was in the girls panties
than he was in actually teaching the algebra class he was being paid to
do. So I quit school and went to work fixing what was then a brand new
tech, TV's at 14 yo. You would not be far wrong to say I grew up with a
hot soldering iron in one hand and a scope probe in the other by the
time I hit 17. But real math? Nope... What little I can do was taught
to me by a TI SR51 or a log-log-duplex Pickett slipstick. What you are
doing might as well be black magic to me.
But if you have an electronics problem, I can probably help.
[...]
"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.
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On Mon, Jul 24, 2023 at 11:36:35PM +0530, Sanjeev Prabhakar wrote:
The approach explained in my earlier message will not work in such cases.
Isn't fillet about the same as:
minkovski with a sphere of the fillet radius, Then invert, again
minkovski with that sphere and then invert again.
That would at least give a defined result for my "difficult case".
Roger.
--
** R.E.Wolff@BitWizard.nl ** https://www.BitWizard.nl/ ** +31-15-2049110 **
** Delftechpark 11 2628 XJ Delft, The Netherlands. KVK: 27239233 **
f equals m times a. When your f is steady, and your m is going down
your a is going up. -- Chris Hadfield about flying up the space shuttle.
I have not used minkowski method in openscad, maybe someone else can comment
On Tue, 25 Jul 2023 at 14:34, Rogier Wolff R.E.Wolff@bitwizard.nl wrote:
On Mon, Jul 24, 2023 at 11:36:35PM +0530, Sanjeev Prabhakar wrote:
The approach explained in my earlier message will not work in such cases.
Isn't fillet about the same as:
minkovski with a sphere of the fillet radius, Then invert, again
minkovski with that sphere and then invert again.
That would at least give a defined result for my "difficult case".
Roger.
--
** R.E.Wolff@BitWizard.nl ** https://www.BitWizard.nl/ ** +31-15-2049110
**
** Delftechpark 11 2628 XJ Delft, The Netherlands. KVK: 27239233 **
f equals m times a. When your f is steady, and your m is going down
your a is going up. -- Chris Hadfield about flying up the space shuttle.
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To unsubscribe send an email to discuss-leave@lists.openscad.org
Three minkowskis of any non-trivial shape takes forever. This example takes
more than 5 minutes:
[image: round.png]
On Tue, 25 Jul 2023 at 15:56, Sanjeev Prabhakar sprabhakar2006@gmail.com
wrote:
I have not used minkowski method in openscad, maybe someone else can
comment
On Tue, 25 Jul 2023 at 14:34, Rogier Wolff R.E.Wolff@bitwizard.nl wrote:
On Mon, Jul 24, 2023 at 11:36:35PM +0530, Sanjeev Prabhakar wrote:
The approach explained in my earlier message will not work in such
cases.
Isn't fillet about the same as:
minkovski with a sphere of the fillet radius, Then invert, again
minkovski with that sphere and then invert again.
That would at least give a defined result for my "difficult case".
Roger.
--
** R.E.Wolff@BitWizard.nl ** https://www.BitWizard.nl/ ** +31-15-2049110
**
** Delftechpark 11 2628 XJ Delft, The Netherlands. KVK: 27239233
**
f equals m times a. When your f is steady, and your m is going down
your a is going up. -- Chris Hadfield about flying up the space shuttle.
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To unsubscribe send an email to discuss-leave@lists.openscad.org
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On 25.07.23 17:18, nop head wrote:
Three minkowskis of any non-trivial shape takes forever. This example takes more than 5 minutes.
What's the code for this example?
ciao,
Torsten.
It is the round_3D test from NopSCADLib. I have distilled it out to this:
$fa = 6;
$fs = 0.25;
inf = 1e10; // very big
eps = 1/128; // small fudge factor to stop CSG barfing on coincident
faces.
layer_height = 0.25;
teardrop_angle = 45;
function r2sides(r) = $fn ? $fn : ceil(max(min(360/ $fa, r * 2 * PI / $fs),
5)); //! Replicates the OpenSCAD logic to calculate the number of sides
from the radius
function r2sides4n(r) = floor((r2sides(r) + 3) / 4) * 4;
//! Round up the number of sides to a multiple of 4 to ensure
points land on all axes
module circle4n(r, d = undef) { //! Circle with
multiple of 4 vertices
R = is_undef(d) ? r : d / 2;
circle(R, $fn = r2sides4n(R));
}
module extrude_if(h, center = true) //! Extrudes 2D object
to 3D when h
is nonzero, otherwise leaves it 2D
if(h)
linear_extrude(h, center = center, convexity = 5) // 3D
children();
else
children(); // 2D
module teardrop(h, r, center = true, truncate = true, chamfer = 0,
chamfer_both_ends = true, plus = false) { //! For making horizontal holes
that don't need support material, set truncate = false
to make
traditional RepRap teardrops that don't even need bridging
module teardrop_2d(r, truncate) {
er = layer_height / 2 - eps; // Extrusion edge radius
R = plus ? r + er : r; // Corrected radius
offset = plus ? -er : 0; // Offset inwards
hull()
for(side = [0 : 1])
mirror([side, 0, 0])
intersection() {
hull()
translate([offset, 0]) {
circle4n(R);
if (teardrop_angle > 0) {
x = R * tan(teardrop_angle / 2);
if(truncate)
translate([0, R / 2])
square([2 * x, R], center =
true);
else
polygon([[0, 0], [eps, 0], [0, R +
x * tan(min(teardrop_angle, 90 - eps))]]);
}
}
translate([0, -2 * R])
square([R, 4 * R]);
}
}
render(convexity = 5)
extrude_if(h, center)
teardrop_2d(r, truncate);
}
module offset_3D(r, chamfer_base = false) { //! Offset 3D shape by
specified radius r
, positive or negative.
module ball(r)
if(chamfer_base)
rotate_extrude()
intersection() {
rotate(180)
teardrop(0, r);
translate([0, -r])
square([r, 2 * r]);
}
else
sphere(r);
if(r > 0)
minkowski() {
children();
ball(r);
}
else
if(r < 0)
render() difference() {
cube(inf / 2, center = true);
minkowski() {
difference() {
cube(inf, center = true);
children();
}
ball(-r);
}
}
else
children();
}
module round_3D(r, ir = undef, or = undef, chamfer_base = false) { //!
Round a 3D child single radius or separate inside and outside radii
IR = is_undef(ir) ? r : ir;
OR = is_undef(or) ? r : or;
offset_3D(OR, chamfer_base)
offset_3D(-OR -IR, chamfer_base)
offset_3D(IR, chamfer_base)
children();
}
module shape()
difference() {
square(40, center = true);
square([20, 20], center = true);
}
module rounds() {
translate([50, 0])
round_3D(or = 4, ir = 2, chamfer_base = true, $fn = 16)
linear_extrude(40, center = true)
shape();
}
rounds();
On Tue, 25 Jul 2023 at 16:37, Torsten Paul Torsten.Paul@gmx.de wrote:
On 25.07.23 17:18, nop head wrote:
Three minkowskis of any non-trivial shape takes forever. This example
takes more than 5 minutes.
What's the code for this example?
ciao,
Torsten.
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here is another approach for filleting such type of intersection, it was
suggested by someone earlier, I did some modification and it is super
efficient.
it took 0.4 sec to render.
But it works only for convex intersections.
fillet here is 5 mm
[image: Screenshot 2023-07-26 at 6.33.44 AM.png]
On Tue, 25 Jul 2023 at 20:25, Sanjeev Prabhakar sprabhakar2006@gmail.com
wrote:
I have not used minkowski method in openscad, maybe someone else can
comment
On Tue, 25 Jul 2023 at 14:34, Rogier Wolff R.E.Wolff@bitwizard.nl wrote:
On Mon, Jul 24, 2023 at 11:36:35PM +0530, Sanjeev Prabhakar wrote:
The approach explained in my earlier message will not work in such
cases.
Isn't fillet about the same as:
minkovski with a sphere of the fillet radius, Then invert, again
minkovski with that sphere and then invert again.
That would at least give a defined result for my "difficult case".
Roger.
--
** R.E.Wolff@BitWizard.nl ** https://www.BitWizard.nl/ ** +31-15-2049110
**
** Delftechpark 11 2628 XJ Delft, The Netherlands. KVK: 27239233
**
f equals m times a. When your f is steady, and your m is going down
your a is going up. -- Chris Hadfield about flying up the space shuttle.
OpenSCAD mailing list
To unsubscribe send an email to discuss-leave@lists.openscad.org
On Wed, 2023-07-26 at 06:35 +0530, Sanjeev Prabhakar wrote:
here is another approach for filleting such type of intersection, it
was suggested by someone earlier, I did some modification and it is
super efficient.
it took 0.4 sec to render.
On my machine...
F5: Total rendering time: 0:00:00.789
F6: Total rendering time: 0:00:00.290
Ubuntu MATE,
OpenSCAD version 2023.02.15.ai13654
But it works only for convex intersections.
fillet here is 5 mm
Screenshot 2023-07-26 at 6.33.44 AM.png
On Tue, 25 Jul 2023 at 20:25, Sanjeev Prabhakar
sprabhakar2006@gmail.com wrote:
I have not used minkowski method in openscad, maybe someone else
can comment
On Tue, 25 Jul 2023 at 14:34, Rogier Wolff R.E.Wolff@bitwizard.nl
wrote:
On Mon, Jul 24, 2023 at 11:36:35PM +0530, Sanjeev Prabhakar
wrote:
The approach explained in my earlier message will not work in
such cases.
Isn't fillet about the same as:
minkovski with a sphere of the fillet radius, Then invert, again
minkovski with that sphere and then invert again.
That would at least give a defined result for my "difficult
case".
Roger.
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To unsubscribe send an email to discuss-leave@lists.openscad.org