# Horsepower/Sail Area equivalence

Discussion in 'Sailboats' started by bob the builder, Jul 26, 2009.

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### bob the buildernovice

thanks guys,

for all the time you've taken.

(it'll take awhile to digest "at 15m/s and C = 1, the force is 15kgf = 13.8kgf)

(still prefer Kg Force - the POWER expended holding 1 Liter of milk to your mouth)

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### bob the buildernovice

just FYI,

The question i was actually asking was;

what wind speed on a stationary Sail gives a force on 1 m2 which = 1hp?

(however rick did interperate this more sensibly, and his explanation is in fact of more use to me : o )

so, both boats are stationary at a jetty. the boat has a huge propeller, no efficiency losses blah blah blah, and has 1hp motor, and is holding 75Kg 1 meter off the ground

sail force = sail area * pressure
=1m2 x 750 Newtons (yes? ~ 750 watts ~1hp)
=75Kg

so wind speed = 30 ms or 60 knots

is this correct?

(and i'm still digesting)

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### yipsterdesigner

sounds good, but thomas thirt jones would say "a 1000 square-foot sail in a 10 knot breeze gives about 1 hp

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### daiquiriEngineering and Design

Let me analyse your question item by item, as numbered above.

1)
As others have pointed out in "trying to show their feathers" (cute, what does it mean..? ), you are again comparing apples and grapes. A force is a force, not a power, and is measured in Newtons (N) or, with lots of precaution, in kilogram-force (kgf, also called kilopond ,kp). The first one is a SI unit, the latter two are not.
A power is a force multiplied by a velocity, to state it in a very simple and a bit restrictive way. It is measured in Watts (W) or, another common way, in horse-power (HP).

So, knowing this, your question (item 1 cited above) translates to:
- what wind speed on a stationary sail gives a force which is a power?
The answer is: no wind speed. A force is not a power and will never be.

A more correct question would be:
- what wind speed on a stationary sail gives a force which equals a bollard pull of a propeller driven by a 1 hp engine?

There we arrive to other issues related to the physics of a boat propulsion, as follows.

2)
So you have a 1 HP motor driving a huge prop, performing a bollard pull. If that is all that you know about your boat, no answer can be given again.
What is the shaft rpm at which your engine gives 1 HP? What are the prop's diameter and pitch? What is the transmission ratio of the drive?
This is the minimum info you need to know if you want to perform the calculation.

There is an approximate formula in Dave Gerr's "Propeller Handbook", taken from the work of Kenneth Barnaby, which states that:
Ts = 12 (SHP * D)^0.67
where:
Ts is a static thrust, in pounds
SHP is a shaft horse-power
D is a prop's diameter, in inches
Even with this simple formula you have to know at least the prop diameter. And the formula doesn't take into consideration the prop pitch, which is a big flaw.

That's the part which regards the powerboat at one end of the rope.

On the other end of the rope you have a sailboat. So you need to know what is the force acting on sails at some given wind speed. That force will hopefully equal the bollard pull of the powerboat. So:

3)
Please take a look at the sailforce diagrams I have included below. These graphs are valid for a bermuda (marconi) rig of aspect-ratio 5. Only for that one. Change the type of rig or the geometrical proportions and you will obtain some completely different numbers.

Graph n.1 - Propulsive-force coefficient (Cr) and Lateral-force coefficient (Ch) vs. apparent wind angle (Beta) and vs. heel angle (Phi)
These are non-dimensional coefficients which are used to calculate the force on sails through the equation:
You can see that both the propulsive and the lateral force are strongly dependant on Beta angle and on heel. In the case of bollard pull you are interested in total force coefficient (Ct), which is calculated as:
Ct = Square-root-of (Cr^2 + Ch^2).

Knowing the Ct you can calculate the total aerodynamic force acting on this particular sail as:
F = 0.5*rho*(Vapp^2)*S*Ct
Where:
- rho is the air desity
- Vapp is the apparent wind speed (which, for a bollard pull only, coincides with the actual wind speed)
- S is the sail area.

All units must be coherent, either all imperial or all SI.
Don't forget, change the sail type or a geometry and all the coefficients will vary strongly.

Now you maybe start to realize that the answer is not trivial at all, and cannot be resumed in one simple formula.
But that's not all. Let's see the next graph.

Graph n.2 - Lift-force coeficient (Cl) and Drag-force coefficient (Cd) vs. camber
Again, the graph is valid only for this particular sail.
I assume that you know what camber is, so let me point just one thing visible in the graph:
- increasing the camber (of the same sail) from 5% to 14% creates a 70%difference in maximum lift coefficient. The change of the total force follows accordingly.
So, knowing that one, things are starting to get really messy...

Graph n.3 - Lift-force coeficient (Cl) and Drag-force coefficient (Cd) vs. maximum camber position
Keeping the camber fixed at 7.5%, the position is varied from 1/3 from the mast to 2/3 from the mast. You can see that all the coefficients vary again, though not so drammatically as seen in previous graphs.

I didn't include some other graphs which indicate variations of force coefficients vs. vang tension, for example, or vs. different rig types. A big plenty of variables when you deal with sails.

All of this can give you the meaning of what I wrote in the post #23.
No unique, general answer can be given to your question. You need first to specify all of the variables involved and seen above, and then you can try to obtain an approximate numerical result.

As about your calculation in the item 3), I don't get where those numbers come from. Apart that, you are mixing units, powers, forces... So, the answer is - no, it is not correct. It would make you fail an exam in physics in any college.

P.S.
I will pass over your post #27. It is a drivel, I'm not interested in replying to it and fueling other flames.
I just want to point out that I have never partecipated to the discussion about catamaran beams which you have cited. So I don't recognize myself in none of your words, you must have mistaken me for someone else. Peace and love.

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Last edited: Aug 13, 2009
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### bob the buildernovice

thank you so much

i understand

and good on you for all the time.

mal

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### yipsterdesigner

to 9 hp reaching in 20 knot

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### hoytedowOld Woodbutcher

Very interesting thread. I'll be back to study this one for sure.

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### ecflyerJunior Member

Hi Bob,
Such high teck replies to a simple question. Yikes! I found this info from either a book or magazine article many years ago-- so memory is foggy!
0.015 hp/sqr ft of sail @ 7-10 kts wind speed
0.020 hp/sqr ft of sail @ 11-16 kts ''
0.040 hp/sqr ft of sail @ 17-21 kts "
0.070 hp/sqr ft of sail @ 22-27 kts "

I hope this is what you were looking for!
Have a Great Day!
Earl

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### ecflyerJunior Member

Hi Bob,
I should have added that drawing a simple graph and using projection, we find 9 sqr ft approx equals 1 hp @ 40 knots wind speed.

Have a Great Day!
Earl

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### CT 249Senior Member

Ummm, is 'power' really 'misused' in common talk? It seems like it's a 13th century word, as is 'force', so surely the way that millions of people have used the words for hundreds of years is not 'misuse'.

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### bob the buildernovice

thanks earl !,

(one small thing though - do you realise that there are 195 countries in the world, and you have used some strange gibberish which is apparently used in only one country on the entire planet?

also,
"drawing a simple graph, by projection etc
9 sqr ft approx equals 1 hp @ 40 knots wind speed.

whereas i used science!

Force = (Drag Coefficient of a 1 meter square surface in wind at sea level)/2 x the velocity squared (in meters per second)

F = force
p = Air density = 1
V = Wind Velocity
Cd = Coefficient of drag = 1.2
A = Area = 1

or
F=.6 x V squared

1HP= ~ 750 Newtons
so
75Kg = X meters per second

solving for X, i got 30.8 meters per second, or 60knots
(which was utterly wrong)

see?
you did it with gibberish AND projection and got the right answer,
whereas i,
being a pure math student, with 12 years of university, did it with science and maths, and got the wrong answer.

this shits me.

(just thought i'd share.)

the right answer is 21.21 m/s or 40 knots

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### BeauVrolykSailor

I hate to say it, but the fact that a lot of folks have used a word the wrong way for centuries doesn't make the mis-use correct. Also, which I suppose is relatively obvious, every technical field as strict definitions for terms and Physics and Naval Architecture are typical of this. These definitions have little to do with common usage.

Great example: When my teenaged son used to say: "Heavy Dad!" he did not mean that the sentence I had just spoken had mass. Just as now, he's older, when he says "That is SICK!" he does not mean something is infected with a disease.

We, in the "normal" world, use words in an extraordinarily sloppy way. Just listen to any random speech by the thankfully retired X-President George Bush the Second, his language is rife with examples. Again, this doesn't make any of it "right" or even "gooder English", to actually quote our X-President. It simply means that normal people get to be sloppy in their speech patterns while scientists have discovered that this leads to errors and have decided not to be as sloppy. (BTW, it leads to massive errors in Presidents too.)

BV

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### hoytedowOld Woodbutcher

And gay used to mean joyful or happy before it was hijacked and twisted into something else. What ex-president Bush has to do with this thread is beyond my comprehension. I enjoyed the previous posts regarding horsepower, however.

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### CT 249Senior Member

I've got no issue with specialists using their own definitions of words; as you say, we all do it.

However, surely that doesn't mean that the specialists can mock or take issue with those who don't always use the term in the 'right' way, which was what happened here. No big deal, but it just seemed uneccessary to carp at the OP for his use of words.

BTW, I'm willing to bet that just about every physicist uses words in a way that can be criticised by 'true experts' in lynguistics. My father-in-law is, by profession, a scholar in ancient Greek and Latin.... if you want to see pedantry in action, he's your man!

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### Dave GudemanSenior Member

prescriptivist vs. descriptivist

In linguistics and philosophy of language, there are two common approaches to what you mean when you talk about the "meaning" of a word. Prescriptivists believe that it is the job of linguists, English teachers, dictionary writers, etc. to prescribe the proper usage of words. In other words, words have a correct and proper meaning regardless of how any one uses them and it is possible for everyone in the world to use a word incorrectly.

By contrast, descriptivists say that words get their meaning from their common usage and from the way that they are commonly understood, so the job of linguists, English teachers and dictionary writers is to describe the actual usage of a word. In this view it is not possible for everyone to use a word wrongly because the way that everyone uses the word is by definition the right way to use it. In this view, if a dictionary definition of a word does not correctly describe the way that everyone uses the word, then it is the dictionary that is wrong, not everyone else.

In previous centuries, most people who studied language were prescriptivists. Nowadays most of them are descriptivists. The problem with prescriptivism is that it begs the question of exactly what it is that defines the meaning of a word. If a word is not defined by the way that it is used, exactly how is it defined? Are we to give random dictionary writers the power to tell us how to speak? What if one dictionary writer differs from another one? How would we know which one has the true and proper definition? What if a dictionary writer wanted to be funny so he defined "girl" as "a human male over the age of 70". Would we be forced to change our language to match the dictionary?

Of course that is absurd, and so the only reasonable view seems to be that the job of a dictionary writer is to describe speech. But then, of course, people who want to speak clearly and understandably will want to use words the same way that everyone else does, and so they will use the dictionary prescriptively but the prescriptive authority of the dictionary flows from it's accurate description of the pre-existing meaning of the word.

The point being that pedantic insistence on everyone using the "proper" technical definition of a word is not only annoying, it is also logically unfounded.