Wind Force calculations

[Breakin Wind]

Greetings all,

I have a couple topics I'm wanting/needing to address, but before I can go there I need to understand how wind force is calculated against a sail surface. I'm pretending for the moment that wind force is actually a real term, please feel free to correct me if that is wrong.

For the purposes of discussion, can we say a sail (or flat surface) is 100 sq ft in area and there is a 10 kt wind (or mph if that works easier). For this basic question let's assume the sail is flat and perpendicular to the wind, and not moving or tilted.

I'll leave the questions as that and see where it goes. Can anyone help?

Thanks, Scott

[jeadstx]

I actually wrote a paper on the "aerodynamics of sails" for a physics class in college about 40 years ago. Got a good grade on it as I recall. That said, I don't remember most of it. I do know that a sail is not flat, but rather it's shape is that of an air foil which creates lift much the way an airplane wing does for certain angles of sailing. When the wind is behind there are different dynamics going on. The current Americas Cup boats actually use a rigid wing sail that looks very much like an airplane wing with controll surfaces. Others on this forum can probably explain all the dynamics of the forces that interact on sails.

John

[GreenLake]

Scott,

an actual calculations of the forces on the sails based on the wind speed and angle is not a trivial matter. However, if you like to run some estimates, there are special purpose calculators on the web, for example http://www.wb-sails.fi/Portals/209338/news/SailPowerCalc/SailPowerCalc.htm.

Why don't you enter the relevant data for the DS for various wind speeds and report what you've learned?

Note that the calculator gives two different forces, because we are usually interested not in the total force, but two separate components of it. One is the driving force, being the component that aligns with the center axis of the boat, and the other one is of course the heeling force, which acts at right angles to the driving force.

From geometry of the sails you can further estimate the mythical point, center of effort, upon which these forces act. The height of the Center of Effort, times the heeling force would then give your heeling moment - which is in equilibrium with your righting moment (created by having ballast, you, hang over the side).

Sometimes it's interesting to know where the fore-and-aft position of the center of effort is, because it needs to be balanced to another mythical point, the center of lateral resistance, which is the equivalent of the center of effort for the underwater foils (CB and rudder with some contribution from the hull). If the CE is far behind the CLR the boat will strongly try to round up, and you would be fighting "weather helm".

OK, back to you, curious where your questions will lead you.

[GreenLake]

John beat me to it:

.. a sail is not flat, but rather it's shape is that of an air foil which creates lift much the way an airplane wing does for certain angles of sailing. When the wind is behind there are different dynamics going on. The current Americas Cup boats actually use a rigid wing sail that looks very much like an airplane wing with control surfaces.

For a foil, the forces are given as lift and drag, that is, the force components across and in line with the wind. Because the wind, going upwind, is at an angle with the direction the boat sails in, you use a different breakdown of the total force to analyze the effect the sails have on the boat's motion (as I just described in my answer to Scott). What the calculator programs do is that they look up published Lift/Drag ratios for the given sail configuration, assuming perfect trim, and re-analyze that for the wind angle given.

The force on the foil, incidentally, scales with something like the square of the wind speed. Double the speed is four times the force.

Hiking out only increases your righting moment linearly, leaning out twice as far gives you, at best, double your contribution to the righting moment. That's why, when a certain point is reached, very quickly the only available option is to depower the sail...

[Salty Dog]

I was taught when I first started sailing that the object was to make your sail a hard air foil like an airplane wing when beating to weather acting in conjunction w/ the center board for lift and forward motion. Make them soft and as much like a parashute when running with no CB at all.the same principal as the dinamics of a tumble weed. Pionts of sail in between are combinations of the two.That makes sence to me on a basic level and add all the sailtrim tweaks to that. with that in mind it seems that the sail has two different jobs to do, and it would depend on which sinario as to which calculation to use.

[Breakin Wind]

Thanks Guys, but I'm trying to understand something more basic, then I'll work out from there. I asked about a sail because it's a sailing forum and thought that would relate, but everybody's taking the question too deeply right now. I'll get to those other parts later, I'm just trying to understand a single principal if anyone is able to help.

A flat surface (not a sail then, a piece of 4 x 8 Sheetrock) floating (so no friction) just off the ground, with a 10 mph wind blowing on it squarely onto the broad face. I'm trying to understand how much force (lbs per square inch?) is being exerted on the face of the sheet rock. If I put a bathroom scale vertically on a post behind the Sheetrock, it should register zero with no wind. With a wind blowing against the Sheetrock, the scale will register something greater than zero (in that case I supposed the mass of the sheet rock will count for something because it's no longer stationary), but you get the drift?

Air has mass, and I googled the weight of air, and found it to be 14.7 lbs/sq/in at sea level (which actually I now question... shouldn't that be a cubic inch?). Not-withstanding, I can confirm (based on my bad experiences of trying to carry Sheetrock on a windy day) that air in motion exerts some portion of that mass/weight on a surface, but I don't know how to get to that number.

I assume it's some function of the mass of the air relative to the speed it's moving vs. the area it's impacting and flowing around, but I'm clearly not a fluid or aero dynamic person. Probably proved that quite clearly here.

Any ideas on this?
Thanks again - Scott

[Salty Dog]

Breaking Wind
I see what your saying. Like air would have a facter to convert velocity to psi and water would have a different factor to convert velocity to psi. i'm not sure where to find that I only chimed in because I like talking about boats I am not an engineer or any thing only from a laymans perspective.

Thanks Calvin

[Alan]

Scott,

It's 14.7 pounds per square inch, not cubic, because it's measured by the pressure on a flat surface. The column of air creating that pressure, one inch square, goes up into the atmosphere.

That's how I imagined it in high school science class, anyway.

Engineers, does this explanation work?

[Salty Dog]

I found this

The pressure developed by wind on a fixed structure is a similar problem. I chose to use data from the 3rd web site as an example. A 50 mph wind on a large highway sign develops 6.4 pounds per square foot. Pressure varies as the square of the velocity. So if the wind velocity on that sign were 60 mph the pressure would be
6.4 lbs/ft^2 * 60^2/50^2 = 9.2 lbs/ft^2
To convert to PSI:
9.2 lbs/ft^2 * (1 ft)^2/(12 inch)^2

I don't know if it will help or not maybe you can play around with it. from this I got 50 mph wind against 4X8 sheet of plywood = 32 ft2 X 6.4 = 204.8 lbs total force 294.4 lbs @ 60mph

[GreenLake]

From high school chemistry we remember that one mol of a gas takes up 22.4 liters at sea level (one mol contains as many molecules of a substance as given by Avogadro's number). The molecular weight of a gas molecule is the sum of the atomic weights. So for oxygen (O2) you get 2*16 or 32, for Nitrogen 28 etc, for an average of 29g per mol for air.

We can estimate how many mols would pass the sheetrock. If the latter was 100% efficient, it would stop all the air, and it would give up all its kinetic energy. We could calculate what that force would be. Keeping things in metric for a while, 100g of air is a little less than 3.5 mols, or 67.2 liters, let's round down to 66.6...

On a 10cm x 10cm surface (4"x4" or 16 sq inch) 66.6 liters make a column of 6.66m. So, if we have a wind of 6.66m / sec, we are stopping 100g of air every second from a speed of 13 some knots (there are approximately 2 knots for each m/sec, a nice and handy conversion).

Newton gives us the next two relations:

Kinetic energy = 1/2 * mass * square velocity: 50g * 6.66^2 = 2.218 kg * m^2 / sec^2

Force * distance = energy, so we divide the energy to by the distance (6.66m) to get the force required to decelerate all that air, and we get a force of 50g * 6.66 m / sec^2 or 32 Newtons.

(At first, it looks like the force is linear in the speed, but remember, when we double the speed, the speed in the formula doubles and in one second, two columns of air now pass the sheetrock, doubling the mass. So the force goes up with the square of the velocity in this estimate, which is what we were expecting.)

At 42 knots, we have four times the speed (and we are getting close to 50mph or 48.3), so let's compare this to force for the full sheet. From our results above, for each 4" x 4" square we get about 32 x 16 or 512 Newtons, and we can fit 12 x 24 of these into a 4' x 8', or 288, which gives us nearly 15o kN or something like 330 lbf.

That's higher than Calvin's estimate, but remember, the sheetrock isn't 100% effective in actually stopping the air, so we should expect a realistic number to be less. One usually uses a coefficient to express the effectiveness of a shape in resisting wind. A coefficient of approximately .6 is actually not too far off from realistic values for a flat plate, and with that we get 198lbf at 48.3mph. I'd say that's in pretty close agreement with Calvin's numbers.

Now, this estimate "works" because flat plate resistance is well approximated by this kind of Newtonian physics (if you fudge it with a coefficient). Now the forces on a sail, when trimmed for upwind (or say a roof over which the wind passes) are harder to relate to the loss of kinetic energy in the air column, but, ultimately, that's where the energy comes from. Downwind of another sailboat you have 'dirty' air, which is air that has both less energy left (slowed down) as well as having disturbed flow (so your sail is less efficient getting energy out of it).