As a consequence of trying to explain why a 15 hp outboard is not too
effective in pushing a trawler into a 20 kt wind, I have received a number of
requests for the calculations supporting my statement that most of the power
is used in overcoming wind resistance. Rather than reply reply to each person
individually, I'll take the lazy person's way out and simply repost the
calculations to the entire TWL. Anyone considering a low power "get home"
system should understand that slogging upwind into a howling gale is not a
practical proposition.

My estimate of 12 engine hp per 100 sq. ft. of projected area to hold
position in a 20 kt. wind is based on the following calculations.

Wind drag on a boat is a function of the projected area at right angles to
the wind, the square of the wind speed, the density of the air, and the
dimensionless coefficient of drag which depends on the shape of the boat.
Drag coefficients have been determined from wind tunnel tests. Some
representative drag coefficients are:

Open parachute (or efficient spinnaker) = 2.0
Hollow hemisphere, concave to wind = 1.7
Flat rectangular plate = 1.28
Wires, cylinders, masts = 1.0
Cargo ship, wind dead ahead = .95
Fishing trawler, wind dead ahead = .9 to 1.05, depending on superstructure,
outriggers, etc.
Streamlined passenger vessel = .70
Recreational trawler = .70 to 1.0, depending on superstructure, masts,
outriggers, etc.
Sphere = .47
Hollow hemisphere, convex to wind = .38
Modern automobile = .26 to .35
Airplane = .09

Using Area in sq. ft., wind Velocity in knots, and the U.S. Standard
Atmosphere for air density, the equation for Drag in lbs. can be written as:

Drag = .00339 x Coefficient of Drag x Knots^2 x Area

In a 20 kt. wind, a boat with an area of 100 sq. ft. at right angles to the
wind with a drag coefficient of 1.0  will have 135.6 lbs. of wind pressure on
its surface. To determine the amount of power required to hold position,
let's calculate how much power it would take to move that 100 sq. ft. area
into still air at a speed of 20 kts. or 33.77 feet per second. The required
power comes to 4580 ft. lbs. per second or 8.33 hp.

But that's power applied directly directly to the boat hull with 100%
efficiency. In practice, most boat propellers operate at 50% efficiency or
less. Ignoring power transmission losses, the boat's engine would have to
generate 16.66 hp to hold position in a 20 kt. wind. Since the power required
varies directly as the coefficient of drag, a streamlined boat might hold
position with only about 12 hp. The force of the wind increases as the square
of its speed. The pressure of a 40 kt. wind is 4 times stronger than that of
a 20 kt. wind. Thus holding each 100 sq. ft. area in position in an 40 kt.
wind might require the boat's engine to produce between 52 to 75 hp. A
typical hurricane condition 80 kt. wind produces 4 times the pressure of a 40
kt. wind. To hold position in this strength wind would require an engine
output of between 186 to 267 hp. for each 100 sq. ft. of projected area.

Clearly the prudent strategy in extreme wind conditions is not to buck the
wind but to run before the wind streaming a drogue to avoid broaching.

Larry Z

Last spring we were bringing our 48 LRC back to Texas from Florida. As we
approached Biloxi in the Mississippi sound, we encountered a thunderstorm
microburst with wind gusts between 80 and 100 mph hour. This storm was short
in duration, but swamped several boats, including a couple small shrimp
boats. We were less than five miles to safe harbor. The storm came on us so
fast and was so contained in size we were not prepared. We sustained light
damage to our canvass but generally faired well. However the Hatteras LRC
has a great deal of wind resistance and as we tried to head to port during
the storm, she just ignored us and pointed her nose to the wind like a fine
sailboat. I don't know how much this had to do with design and how much
resulted from wind load, but the 252 horsepower I had would not redirect my
boat during the height of this storm.

Mike Finley
Hatteras 48LRC
Donna K
-----Original Message-----
From: trawler-world-list-admin@lists.samurai.com
[mailto:trawler-world-list-admin@lists.samurai.com]On Behalf Of
LRZeitlin@aol.com
Sent: Wednesday, December 11, 2002 1:58 PM
To: trawler-world-list@samurai.com
Subject: TWL: Power into wind

As a consequence of trying to explain why a 15 hp outboard is not too
effective in pushing a trawler into a 20 kt wind, I have received a number
of
requests for the calculations supporting my statement that most of the power
is used in overcoming wind resistance. Rather than reply reply to each
person
individually, I'll take the lazy person's way out and simply repost the