# Sail Area vs Keel Area - Stability Calculation

Discussion in 'Hydrodynamics and Aerodynamics' started by Dabrownone, May 28, 2012.

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### DabrownoneJunior Member

Hi, this is my first post.
I'm a mathematician and not a naval architect or an engineer, so this may actually be really obvious. However, I was fooling around with some equations and arrived at an interesting conclusion (at least to me ).
I'll do this from first principles:
A mass of fluid, m, strikes a flat plate of area A. The volume V of the fluid is given as V = A * dh, where dh is an arbitrarily small distance.
From this the following can be derived;
v = dh/dt, where v is the speed of the fluid.
p = m * (dh/dt), dp/dt = F = m * (dh/(d^2)t)
F = F * (V/V) = (m/V) * (dh/dt) * (dV/dt) = d * (dh/dt) * A * (dh/dt), where d is the density of the fluid.
Thus Force on the flat surface is F = A*d*v^2
I may be missing some constants, but i'll be applying this to a static equation, so I'll leave these out for now. I'm ignoring the effect of the fluid at the edges of the plate
To set up the effect of keel area to sail area, i'll set up the following model:
Imagine a flat rectangular plate with a small cylinder at a point somewhere towards the middle of plate, that is as wide as the plate and divides the plate into two rectangles with areas A1 and A2. For simplicity, have A1 be a perfect square.
A1 = (2*r1)^2, where r1 is the distance from the cylinder to halfway up to the top of the plate, making A1 a square with length 2*r1
A2 = (2*r1)(2*r2), where r2, the width, is half the distance from the cylinder to the bottom of the plate, and it has length r1 (this would be a lot simpler if there was a diagram; perhaps i'll attach one)

Place the plate so it is sitting vertically in water on the cylinder, so that A1 is above the water (sail area) and A2 is below (keel area). The question is, is it possible for the plate to remain in static equilibrium when hit with wind perpendicular to the plate given the damping provided by the keel and the difference in the densities of the two fluids.

To start, i'll apply the Force equation derived above:
A1*(v1^2)*d1 = A2*(v2^2)*d2, where v2 is the speed that the keel area is moving at it's center.
Say there was no damping, and the wind caused A1 to move with velocity v1, rotating the plate by an angle of d@, for a total distance of r1*d@ radians. A2, being on the same solid body, would also rotate by an angle of d@, for total distance r2*d@
Thus,
r1*(d@/dt) = v1; r2*(d@/dt) = v2
v2/r2 = v1/r1 => v2 = (v1*r2)/r1

Substitute this into the force equation
A1*(v1^2)*d1 = A2*(v2^2)*d2 = A2*(((v1*r2)/r1)^2)*d2

Substitute areas A1, A2

((2*r1)^2)*(v1^2)*d1 = (4*r1*r2)*(((v1*r2)/r1)^2)*d2

Solve for r1

4*(r1^2) = (d * (4*r1*r2)*(((v1*r2)/r1)^2))/(v1^2), where d is d2/d1, the proportional density of water to air

r1^2 = (d*(r2^3))/r1
r1^3 = d*r2^3,
Take the cube root
r1 = d^(1/3) * r2

Therefore, if r1 is equal to the third root of the proportional density times r2, the plate will stay in static equilibrium.

substituting this into the area equations gives:
let dd = d^(1/3)

A1 = (2*r1)^2 = 4*((dd)*r2)^2
A2 = (2*r1)(2*r2) = 4*(dd)*(r2)^2

A1 = dd * A1 = (d^(1/3)) * A2

Therefore, if the A1 (sail area) is equal to the third root of the proportional density times A2 (the keel area), then the plate will not rotate with the wind.
The density of both air and water vary somewhat with temperature, but I believe that taking d^(1/3) = 9 +/- 1 will be accurate for most temperatures. What is of note especially is that the wind speed, v1, dropped out as we solved for r1, which means that this should hold for any wind speed
So basically, if total sail area is ~ 9 times the keel area (or less), the boat in the model (the cylinder) will not heel, no matter how strong the wind is. This makes sense intuitively because the density of air, and therefore the momentum, is so low compared to that of water, that the damping force will always be greater than the wind force.
I realize that without diagrams and whatnot, this explanation may be as clear as mud. I'm open for discussion though; as I mentioned before though, i'm neither an engineer nor a naval architect, so don't flame me too hard please

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### daiquiriEngineering and Design

Ok, not too hard.

I just have to point out two small issues with your analysis:
1) The mathematical model of the sailboat you have used to derive your equations is as far as it can get from how a real sailboat works.
2) The mathematical model which you have used of aero/hydrodynamic forces on a flat plate is not correct. It agrees only from the dimensional point of view - for the rest it is disconnected from the reality of fluid dynamics.

Usually, when one undertakes a task of solving a particular problem, the first step is to perform a research on the current state of the art, to see whether someone else might have already solved the problem. You'll find out that the stuff you are trying to do has already been done in 99.9% of cases.

Since you are a matematician, you won't find difficult the math used in the following course of aerodynamics (for example): http://ocw.mit.edu/courses/aeronautics-and-astronautics/16-100-aerodynamics-fall-2005/lecture-notes/ . Once you have assimilated these basics of fluid dynamics, you'll be able to do more in-depth research, analysis and calculations. Then, perhaps, you may hope to arrive to some physically relevant and brand new discovery in this field.

Cheers

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### DabrownoneJunior Member

The goal wasn't really to model an actual sailboat, but explore one specific aspect of a boat. The model uses approximations because i'm not really concerned with making precise predictions; the point was to understand whether there existed a point where the force applied by the wind on the 'sails' (essentially everything above the water) with the damping afforded by the 'keel' would cancel each other. I agree that the model is in no way correct, in that you wouldn't actually build a boat like this (although I should build a model and see what happens; something this simple to build may even be within my grasp). My argument is basically that the side force of the keel can damp the rotation (heeling) of the boat, and could therefore be used to define a point where the damping will cancel out the side force generated by the wind.

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### daiquiriEngineering and Design

In order to have moments of side forces created by the keel and by the sail cancel out each other, they would have to act in the same direction. But, as is well-known to anyone interested in the theory and practice of sailing, that is not how a sailboat works. In realty the two forces act in the opposite directions, and that's what makes the sailboat move in the straight course:

(Pic taken from: http://ffden-2.phys.uaf.edu/211_fall2002.web.dir/Edwin_Monin/web 2002nov25 Folder/slide5[1].html)

If the forces above and below water were acting both in the same direction, there would be nothing to stop the boat from drifting sideways at an increasing rate.

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### PetrosSenior Member

Dabrownone,

I am an engineer, and I have built and sailed small boats since I was young. Your model ignores the lift to drag ratio of both the sail and the keel, and appears to misunderstand how the sail drives the boat forward. The reason you need a keel is not to counter the heeling moment of sail, but to counter the "slip" or side force the sail puts on the hull, so the boat moves forward rather than side ways on the water.

You need to study the mechanics of how a sail boat moves forward before you can put any meaningful equations to paper.

Fluid mechanics are not intuitively obvious, so first you have to understand how the sail interacts with both the moving air and the hull, and how the keel interacts with the water, and the hull. There have been boats on the water since before the dawn of recorded history, but it was only in the last few hundred years could they put a sail on a hull and do anything but go down wind with it. And modern designs were not developed until after WW1, when the science of fluid mechanics were well enough understood to make viable aircraft.

No matter the size of keel or sail, you will get heeling moment on it, the heeling moment is counteracted either by the crew (of a small boat) leaning over the side, or on a large keel boat, buy the weight of the keel. The optimum keel to sail ratio will vary depending on wind speed, hull speed, angle to the wind, etc. There is not "optimum" ratio, only for certain conditions, so you compromise the best overall performance. Larger sail or keel than you need creates more drag, smaller keel means you can not point as high into the wind (high keel loads with lower relative hull speed). So, you make a guess at which condition is more critical and just live with the less than optimum condition.

These optimin conditions, btw, also are affected by the foil shape in the water, and the size/shape of the sail.

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### DCockeySenior Member

Aero-Hydrodynamics and the Preformance of Sailing Yachts by Fabio Fossati is a good reference for someone who wants to learn about how sailboats work and is comfortable with vector geometry and calculus.

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### DabrownoneJunior Member

I think my explanation wasn't clear enough. This model has nothing to do with driving the boat forward. It is a static situation and only deals with a rotating plate, so the angular force as the sail is forced to rotate (heel). In fact, it may be easier to just imagine that the plate is locked into place, and only able to rotate around the cylinder, and does not move forward at all (or back, or sideways etc).
The wind would apply a force on the upper plate, which causes the plate to rotate with a specific amount of leverage. As the bottom plate (keel) is attached, it too will attempt to move with the upper plate. The point of the calculation was to show that the rotational moment on the upper plate could be sufficiently damped by the drag induced by the lower plate being forced to move through the water. Does the fact that the density of water is so much greater than the density of air not suggest to anyone that this is possible?
This is normally true, but is by no means an absolute. Are you telling me that a boat with bare poles in a 2 knot wind would heel if not for the weight of the keel or the crew? Both the keel weight and crew weight offer a counter-rotational moment that balances the boat, but is not the only manner in which rotation can be balanced.

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### BATAANSenior Member

This is a form of mental self-abuse, with no apparent (yet) practical use since it is all theory of simple forms and no application, and seems like a prodigious use of valuable time that could be spent building boats and sailing to see how it all really works.

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### StumbleSenior Member

Dab,

The 70' sailboat I raced on for a while with a 15' deep keep. Would absolutely heel in any type of cross breeze when tied to the dock (and thus under bare poles). Not much to be sure, but electronic heel indicators are sensitive things.

I am not a mathematician, but as I understands hat you are saying is if you apply a force to the mast of some magnitude. And a keel were of sufficient size, then the boat would not heel at all.

Not to be an ***, but newtons laws don't allow for that. When a force is applied the force must be transmitted somewhere. On boats that is to heeling rotation, or I guess heat. Assuming the body is perfectly rigid, and the impact is 100% efficient at transmitting the force so no heat is generated, that force has to impact the velocity of the rigid body in some since. Though if the force applied is small relative to the size of the body, of course it will result in minimal rotation.

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### DCockeySenior Member

The movement of the keel, rudder and hull in the water will dampen the motion of the boat. But that is dampen as in slow the motion, not prevent it. Other than buoyancy, a fluid does not cause a force/moment on an object unless there is relative motion between the fluid and the object. So for your example with the upper and lower plates on a cylinder which is locked into place and free to only rotate the effect of the lower plate in the still water will be to slow the motion of the cylinder. However the equilibrium heel angle due to air moving over the upper plate will be the same with or without the water in place other than buoyancy effects.

Yes, the boat will heel until there is an offset between the gravitational forces and buoyancy forces which causes a corresponding moment in the opposite direction. Fundamental hydrostatics. That offset can be caused by the boat heeling and/or the crew moving.

To echo what Stumble said there are two fundamental equations which cannot be ignored:
Sum of forces = linear acceleration X mass
Sum of moments = rotational acceleration X moment of inertia

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### DCockeySenior Member

Keep in mind that the "theory" of the original poster has nothing to do with generally accepted theory of hydrodynamics or sailing dynamics, nor with reality.

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### DabrownoneJunior Member

Ok, i'll say it as clear as possible; this isn't a dynamical equation, its a static one. Additionally, I need to amend my model to equate the torque generated by the two parts of the plate.
Therefore,
T = r x F,
Applying this to the results gives

r1^4 = d * r2^4

r1 = d^(1/4) * r2

This was actually what I had derived originally, but couldn't remember how I got it.
So this would indicate that if r1 ~ 5 * r2, then the torque generated would be balanced. While this isn't a 'real' sailboat, it indicates a relationship which is 'real'.
In regards to the 70' sailboat; most I have seen have a deep, thin keel with a very heavy bulb on the end. Considering that windage from rigging, poles, the hull above the waterline etc contributes to torque, and that the hull is designed to heel to a certain degree, and the action of the waves (which will apply a far greater torque as water is much more dense), it isn't surprising that most boats like this would heel in all conditions.
So why is this practical? Consider the other means of countering torque due to wind. These are form stability (which works to a point) and dynamic stability due to the righting action of the boats CE (which nowadays is usually down in the keel). Both of these have hard limits on their righting ability. For evidence of this, see the stability curves of any boat. Some are designed to right even with the mast less than parallel to the waves, and some would topple well before that. It indicates that as wind speed picks up, the force (and therefore torque) would increase quadratically (v^2 in the force equation), and would eventually overpower the ships ability to stabilize itself. The stability relationship I found is independent of windspeed, and will therefore dampen rotation (apply negative torque) all the time. A relationship of 5:1 sail to keel area would be hard to accomplish with modern boats, considering that everything above the water acts as 'sail' area, whereas only surfaces that dampen rotation act as 'keel' area.
Speaking of which, the 'keel' area could be any surface that resists rotation; for instance both keels in a bilge keels would have one side that resists, three keels would have three etc. As well, a flat bottom, or a chine that has been rotated below parallel with the water, would all actively resist rotating as well, providing a counter-torque to the above water action.
Ultimately, the action of the waves are the real danger, as they can toss you around far more readily than the wind, but here again the keel would at least resist this motion and slow it down, which is helpful.

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### DCockeySenior Member

"The stability relationship I found is independent of windspeed, and will therefore dampen rotation (apply negative torque) all the time."

"Speaking of which, the 'keel' area could be any surface that resists rotation; for instance both keels in a bilge keels would have one side that resists, three keels would have three etc. As well, a flat bottom, or a chine that has been rotated below parallel with the water, would all actively resist rotating as well, providing a counter-torque to the above water action."

For the water/"keel" interaction to generate a force (other than buoyancy which is not included in your equations) there must be relative motion between the keel an water.

Do you disagree?

If so then what is the mechanism which causes the force?

If not then what is the cause of the relative motion?

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### daiquiriEngineering and Design

Ok, let's put it this way - how would you apply your theory to a real boat? What kind of device should be added to a boat, which would work according to the principles you have exposed here?

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### DCockeySenior Member

Had another look at your original post, and your derivation includes a speed for the water past the keel. You then solve for the water speed in terms of the air speed on the basis of force balance. I bolded the relevant parts of your post.

I'm not able to completely follow your derivation, possibly because of the limitations of expressing differential equations as text though it also possible there are some errors.

My guess is the closest phyical version of what you are trying to model is a system with a "sail" in wind and a "keel" in a water current moving in the same direction as the wind but at a different speed. The sail and keel are connected and the combination is able to pivot about a fixed point at the division between the air and water. What you are arguring is along the lines of for any wind speed there will be a current speed which will cause a moment on the keel which will counterbalance the heeling moment of the wind on the sail.

Last edited: May 29, 2012
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