Challenge: Pounds of Thrust ?

[Hotel L]

Ok all you engineers out there....I have a question for you.

How do you compute the lbs of thrust a given sailboat generates at specific wind speeds (with the understanding sails are trimmed properly)?

[David M]

Pounds of thrust being generated by the sails is going to vary with point of sail, angle of inclination and wind velocity. I don't know if this can be calculated. Does anyone know?

I think you would need a test tank so that you know what force needs to be applied to the hull to achieve what speeds and at what heeling angles. As a boat heels the hull form becomes less efficient especially because the rudder has to be turned off center line in order to maintain a straight course through the water.

There are formulas for boat propellers that roughly convert horsepower to thrust. So if you knew the horsepower being applied to a propeller you could convert that into thrust using the formula Power = Force X Velocity times a coefficient for for the fact that it is not perfectly efficient conversion.

[Tellie]

Originally Posted by David M

Pounds of thrust being generated by the sails is going to vary with point of sail, angle of inclination and wind velocity. I don't know if this can be calculated. Does anyone know?

Not likely. In a perfect world where all the stars align and the wind, waves, current, hull conditions, trim, and the many variables of the condition of sails are perfect and constant...perhaps. If you happen to know where those conditions can be found please let me know. I've been searching for those cruising grounds all my life.

[DRS]

Odd .... I thought sails created lift not thrust. The egines on a plane create thrust the wings create lift. Sails work like the wings not the engines. Am I wrong ?

[David M]

Yes there are other factors. I was keeping to the significant basics.

Sails apply lift which is a force. Thrust is a force.

Right?

[Adelie]

The hard answer that will require you to look up some values and do some math is to use the formula:

F=C*A*(ro/2)(v^2)

where :
C -force coefficient
A -sail area
ro -air density
v -air speed across the sail

that I found at sail thrust formula - Boat Design Forums

[Blue Stocking]

Not the answer to your question, but I tell non-sailor guests that to obtain the same 7 knots under sail, requires 34 hp from my engine, under power.

[Hotel L]

Originally Posted by Adelie

The hard answer that will require you to look up some values and do some math is to use the formula:

F=C*A*(ro/2)(v^2)

where :
C -force coefficient
A -sail area
ro -air density
v -air speed across the sail

that I found at sail thrust formula - Boat Design Forums

Wow...my head is hurting. Great url by the way. I got everything figured out but the force coefficient.

I assumed a 80 degree day w/ 1180 MB and a dew point of 60 degrees, 12 kts of wind off the nose @ 120.26 degrees (puts it at a perfect 90) for a apparent of 13.89 kts. Island packet 35 has a sail area of 640 sq ft. But with all of the www I can't find a useful formula for C.

F=C*640*.54485*192.9321

All this and I still don't know what F actually will be....lbs I assume.

[Hotel L]

Ok...before someone corrects me I used way to high of an air pressure. Just short of the record ever recorded. My bad... Back to the calculator.

[Morgan3820]

Dave Gerr does just such a calculation in his book 'The Nature of Boats". An outstanding book by the way.

[Adelie]

Originally Posted by Hotel L

Ok...before someone corrects me I used way to high of an air pressure. Just short of the record ever recorded. My bad... Back to the calculator.

Air - 59oF - 0.0765 lbm/ft3
from Dynamic Pressure

Make sure all the units are SAE:
V is ft/s
For a run the C will be about 2
On a reach or close hauled ?


Or if you want to short cut all the research
Sail Wind Load = SA * ( WS ) 2 * 0.00431

SA: Sail Area in Square Feet.
WS: Wind Speed in Knots
from Wind Load Calculator

[sycankiz]

This formula calculates the drag force of the wind, for a given wind velocity, on a certain surface. Whether you chose to use this force to move your sailboat forward or to tear your sails is another issue though...

The metric version gives the result in Newtons, when you insert air density in kg/m3 (it is almost 1.2 under Standard Temperature and Pressure), wind velocity in m/sec, sail or surface area in m2.

The drag coefficient is the resistance of the unique shape of the object under wind effect. It defines how aerodynamic a certain shape or design is. If I recall right, for most cars we drive, it is 0.25 - 0.30.

If you want to calculate the drag force, say, on a rectangular plate, where the wind is blowing directly (i.e. 90 degrees) onto it, you should be taking the drag coefficient as 1. This means the force of the wind is not lessened at all by the shape of the effected object, i.e. the rectangular plate.

I don't see why the drag coefficient should be taken 2 in the above post. 2 means the effective force is doubled. And why should it be doubled unless the object in effect also blows back with the same velocity and two forces collide ?

For winds effecting the surfaces (i.e. sail surface) at angles other than 90 degrees, you can multiply the wind velocity with the cosine of the angle of attack, to find the effective velocity.

As for the drag coefficient of the boats, I'm not sure if they are being wind-tunnel tested during design or production but that should be the place to get those figures.

If you use your boat's sail area in the formula, you will get the actual amounts of forces your sails/rig are subject to when the wind is blowing at 90 degrees to the boat and sails.

You can't, in my opinion find the force at other angle of attacks, because although the effective velocity at an angle to the surface can be calculated as referred above, the drag coefficient remains unknown at different angles. At 90 degrees we can take it as 1, but we can't know this at say 45 degrees. At 45 degrees the boat shape is different, the coefficient is slightly reduced, at 0 degrees it is almost like an arrow, thus further reducing the coefficient. However, imo, estimations will be within acceptable limits.

Also you will see that; when wind increases from 15 knots to 18 knots; which is a %20 increase; the effective force will increase %44. If the wind is doubled from 15 to 30 knots, the force would increase 4 times.

A 30 knot wind blowing at 90 degrees on to a sail area of 26 m2, will be applying a total force of 3714 Newton, which is the equivalent of the force of the gravity on a 379 kg object on our lovely planet. (It means someone is pulling your sails/mast with 379 kg weight on the opposite direction the wind is blowing) On a 40 knot blow, this force is 6605 Newton, roughly 673 kg.

All this from long back, so I might have some errors here and there. Please feel free to correct.

[capngeo]

Thank God for you engineer types! Me? All I want to know is what speed I can make to figure an ETA (to sit and enjoy a cocktail under a coconut tree with a pretty girl in a hula skirt and lei)........... ***SIGH*** Nevermind, carry-on