Horsepower/Sail Area equivalence

[bob the builder]

good question,
at what windspeed does 1 meter square sail give 1 horsepower?


cheers,
mal

[Guest625101138]

Sails produce a force.

Power is force times velocity. So you cannot equate the force developed by a sail unless you know the speed of the boat.

The sail lift force is given by:
Lift Force = 0.5 x rho x V^2 x Cl x Area (Perpendicular to apparent wind)
Drag Force = 0.5 c rho a V^2 x Cd x Area (In line with the apparent wind)

V is the apparent wind velocity.

You have to do the vector sum of the forces. This will depend on the point of sailing.

The maximum value for Cl with a sloop rig having sails with aspect ratio of 4 will be about 2.5. The Cd for this rig is around 0.5 at maximum lift.

If you look past all the white noise on this thread you will see some images that give insight (same problem as your original thread):
Sail forces and sail plan ?

So working in metric. A boat doing 5m/s with apparent wind of 10m/s on a broad reach where the Cl is in line with travel, the force for every square meter in a sloop rig will be around:

Force = 0.5 x 1.2 x 10^2 x 2.5 = 150N

Power = Force x Velocity = 150 x 5 = 750W (Say 1HP)

[Manie B]

Nice one
thanks

[Guest625101138]

Originally Posted by Manie B

Nice one Rick
thanks

Manie
Could be useful to give an example or two of one of the boats you have been involved with.

Nothing like testing the maths.

I covered 27nm in 3 hours on a broad reach in a 3t deep keel yacht a long time ago. Takes a lot of power to do that.

I once saw a fellow lose a toe in a spinnaker sheet on a 39 footer. The sheet felt no pain as the spinnaker unloaded. A lot of tension to do that.

Rick W

[Manie B]

My current project the "Microcruiser" is going to be the really good one to test the maths, it is the culmination of my experiences and approach within my current knowledge and capabilities

My maths skills are nowhere near yours, so i appreciate your input

this microcruiser is a down sized Transat mini, i really wanted to do this.
The hull was modeled in freeship which also gave me the developed plates and the data as calculated in freeship. I did all the drawings in Turbo CAD16. Point is that i can answer design and drawing questions with accurate dimensions. I will also take good readings once on the water and then we can work backwards to determine what the maths says - this will eventually help me a lot, for future builds.

[Guest625101138]

Manie
I am looking forward to the data. Will be an interesting exercise.

[bob the builder]

1hp per meter,

thanks rick!.



got another question . . .


(purely as a matter of interest,
i can't follow you.
not even past the first line.)
Power is force times velocity?

whats the word that means static power then?
ie if the same sail boat is tied to a jetty?
(with one meter of sail and pulling 75Kg on a scale tied to a bollard)


if a weightlifter is perfectly steady holding 75 Kg above his head, he would dispute you saying he is using no power. so how much power is he using rick? if he's not moving that 75 Kg weight even a millimeter?


isn't 750 N = 75Kg?
so if the same boat has 30 m2 sail, why isn't there 30x75Kg force on the sail?
(please don't use equations. use sock puppets and song please.)

(also, don't answer unless you have the time. or if it's too exasperating.

[ancient kayaker]

I did the calc for the downwind case:

Power = sail force * boat speed
sail force = sail area * pressure
working in metric, the sail area = 1
pressure = ½ * (density of air) * (apparent wind speed)^2 * (shape factor)
The density of air is about 1.25 kg/m3
The maximum shape factor occures when the air is stopped dead but that never happens as some wind flows around the object: lets assume an efficient sail that achieves a shape factor of 0.5 (a typical sail is probably half of that)
wind velocity = V
boat speed = S
For the downwind case: apparent wind speed = V - S

Putting that all together:
power = 0.3125 * S * (V-S)^2 = 0.3125 (V^2*S - 2VS^2 + S^3)
differentiating wrt S:
dP/dS = 0.3125 (V^2 - 4VS + 3S^2)
solving for max power:
3S^2 - 4VS + V^2 = 0
or:
S = V/3

Thus:
power = 0.3125 * S * (V-S)^2 = 0.3125 *(V/3)*(2V/3)^2 = 0.046 * V^3 = 746 (1 hp)
or:
V = cuberoot(746/0.046) = 25.3 m/s
Using the wind pressures in the Beaufort scale yields 20 m/s

It would be interesting to see at what point off the wind the power would maximize ...

[marshmat]

whats the word that means static power then?
ie if the same sail boat is tied to a jetty?
(with one meter of sail and pulling 75Kg on a scale tied to a bollard)

if a weightlifter is perfectly steady holding 75 Kg above his head, he would dispute you saying he is using no power. so how much power is he using rick? if he's not moving that 75 Kg weight even a millimeter?

By definition:
Work = force * distance
Power = rate of doing work
= force * distance / time
= force * velocity

If you tie your sailboat to a bollard and let her pull, she's exerting a force, but neither the dock nor the boat is moving. Tugboat folks call this "bollard pull", ie. the maximum force that can be exerted at zero velocity.

Your weightlifter may grunt and complain, but he is only doing work while he lifts the barbell- when he is holding it there, he is doing no work on it. He is, however, producing a force; the energy his body is using to produce this force is all being dissipated as heat and no mechanical work is being done.

isn't 750 N = 75Kg?

NO!
The newton is a unit of force.
The kilogram is a unit of mass.
They are not interchangeable- there is no such thing as a "kilogram-force"; the use of that term is a holdover from the old days when some engineers didn't understand the difference and used pounds for both.
Now, an object with a mass of "x" kilograms will have a weight (the force exerted on it by gravity- not the same thing as mass) of "x" times "g", where g is the acceleration due to gravity, 9.8 m/s^2 at sea level. Your "x" kg block will weigh less on a mountaintop, and less still on the moon, but its mass will still be "x".
The force on your sails is measured in newtons. If you're getting 150 N per square metre (ie, 150 Pa) of wind pressure on your sails, and you have 10 square metres of sail, your driving force is 1500 N. If that makes you go at 5 metres per second, well, power = force * velocity, so 1500 N * 5 m/s = 7500 watts.
(recall, 1 horsepower = approx. 746 watts)

[bob the builder]

thanks marshmat

so a normal person would say the tug boat is exerting 1 horsepower thrust on the bollard, 75Kg held in the air.
and an engineer would say, "Sir, engineers use the same words to speak an entirely different language, and in our language you are speaking gibberish. the force exerted on the bollard is 750 Newtons, which is entirely different from 75Kg thrust"

even though a 75Kg pull on a rope on a bollard is exactly 750 Newtons pull on a rope on a bollard

is this right?

mal


(9.8=10 746=750 7.46=7.5 etc)

[marshmat]

Hi Bob,

Using some nice rounding-off to powers of ten, yes, a 75 kg mass hanging on the end of a rope will put a tension of 750 N on the rope.

The tugboat captain might brag about "My boat can do a bollard pull of 3,000 pounds". He'll then go on to say "That dinghy there weighs 200 pounds".

What is omitted here, and what is really the source of most of the confusion, is that in imperial measure, there are three units with the name "pound". There is a pound-force, a pound avoirdupois, and a (rarely used) pound troy.

Our captain, when he brags about his bollard pull, is talking in pounds-force. When talking about his dinghy, he's talking in pounds avoirdupois.

Because at sea level on Earth, something that has a mass of one pound avoirdupois has a weight of one pound-force, the two are often treated as interchangeable. So we usually just say "pound" for both force and mass.

In the SI system, different names are used for force and mass units, so that it is always clear which quantity is being discussed. One can mess around with nonexistent units like "kilogram-force" if he so chooses, but by far the easiest way is to just think of it in the proper units to start with:
Newtons for "how much force, tension, thrust, etc."
Kilograms for "how much material is that object made of"

[ancient kayaker]

I redid my calc in post #8 which had an error; the original wind speed was suspiciously low.

[PAR]

In force 4 winds, 50 sq. ft. is about 1 HP.

[Guest625101138]

Originally Posted by bob the builder

.......
if a weightlifter is perfectly steady holding 75 Kg above his head, he would dispute you saying he is using no power. so how much power is he using rick? if he's not moving that 75 Kg weight even a millimeter?


.........

The weightlifter is not exerting power just holding the thing. I could replace him with a lump of wood - it would move way less than a mm. How much power can a piece of wood produce.

You do not talk about the power of the frame of your house because it holds heavy tiles on the roof. Power is a function of force and how fast its point of application is travelling.

The weightlifter had to produce power to lift the weight but once the weight is in a steady position he is no longer producing power.

A tug boat doing a bollard pull achieves it by moving a large volume of water so it must develop power to move the water. However the dock is not being moved (hopefully) so there is no power applied to the dock, only force.

Rick W

[daiquiri]

It is basically about the misuse of the word "power" in common talk. It is too often wrongly used as a synonym of "force", hence all these doubts.