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post #48 of Old 04-05-2012
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Re: Why is beam reach (or near to it) the fastest point of sail?

Let me see if I can explain this.
Wind Power = 1/2 * Rho * Velocity^3 * Area

Rho is the density of air which is about 0.0807 lbs / cubic ft.
If we are using cubic feet then we need Velocity in feet also.
Let's say the wind is blowing 6 knots.
That is about 6.87 mph or 10.08 feet / sec.
Let's just round that down to 10 ft/sec.
Let's say we have 200 square feet of sail area.

Okay: 1/2 * 0.0807 lbs / ft^3 * (10 ft/sec)^3 * 200 ft^2
0.04035 lbs / ft^3 * 1000 ft^3/sec^3 * 200 ft^2
8,070 lbs (ft^3 / ft^3) * ft^2/sec^3
8,070 lbs ft^2/sec^3
Now we have to divide by gravity which is 32 ft/sec^2
8,070 lbs ft^2/sec^3 / 32 ft/sec^2
252.1875 lbs (ft^2/ft) /(sec^3 *sec^2)
252.1875 lbs ft /sec
1 Horsepower is 550 ft lbs /sec so this is a little less than half a horsepower
0.46 HP
This is all we have to power the boat; there isn't any other source. Will we get more than this if the boat is moving? No. This is all there is.

But ... (someone might protest) HP doesn't move the boat, force does. Well, this is true. So, would it be possible to get unlimited force from a fixed amount of HP? No. The amount of force available is dependent on the speed of the boat. Let's say we sail 10 ft per second perpendicular to the wind.

252.1875 lbs ft/sec / 10 ft/sec
= 25.21875 lbs of force
If we sail faster we get less force, not more.
This is all boilerplate physics and engineering. There is no way to get around this.

Could we go faster than 10 ft per second? Yes, if the total hull drag is less than 25 lbs of force at that speed then we'll have thrust left over to go faster. Could we go faster than 10 ft per second downwind? Clearly not because at 10 ft per second there would be no relative wind so force would be zero. Okay, so why can't we use relative wind to sail faster upwind than when reaching?

When we are reaching, 100% of the lift force of the sail is in the direction we want to go. So, under normal circumstances this would give us maximum speed.

When we angle upwind then only part of the lift force is in the direction we are traveling. But what about relative wind? We keep hearing how the sails generate force in regard to relative wind, not true wind. Now, being quite familiar with physics I know that it makes no difference if we view wind relative to the boat, a fixed point, or to the water. So, let's try relative wind.

The simplest way to illustrate this is to just make assumptions. Let's assume that the wind is blowing out of the north at 10 ft per sec. Let's assume that we can travel in any direction at the same 10 ft per sec. Now, I'm sure at this point someone is screaming that we can't just make that assumption. Actually we can. The way you normally do this process is you make an initial assumption and see what happens. Then, if you wanted a good model you would take the result and use it as the second starting point and repeat this over and over until the results changed very little. This is iteration. However, in this case, I don't think we'll have to iterate this over and over to see the truth. So, let's start with our assumptions and see what happens.

I'm going to take a constant vector for the wind of 10 ft per second blowing south and this is -90 degrees. East is 0 degrees, and North is 90 degrees. We aren't going use west but if we did it would work the same as east. Now, if the boat could travel at a constant velocity of the same 10 ft per second in any direction, what would the relative wind be at each point?

Angle of travel : Rel Wind velocity
90 : 20 (north, directly into the wind)
80 : 19.92
70 : 19.7
60 : 19.32
50 : 18.79
40 : 18.13
30 : 17.32
20 : 16.38
10 : 15.32
0 : 14.14 (east, perpendicular to the wind)
-10 : 12.86
-20 : 11.47
-30 : 10
-40 : 8.45
-50 : 6.84
-60 : 5.18
-70 : 3.47
-80 : 1.74
-90 : 0 (south, directly downwind)

We see exactly what we would expect: the velocity is highest when pointing directly into the wind and lowest when running dead down wind. If you are still having problems with the notion that the boat can travel at this speed in any direction then just imagine that we are starting from a tow or by using the motor.

Now with the boat speed and course fixed what could we actually get out of the sails? What we care about here is that lift (which gives us thrust) is perpendicular to the relative wind. And, if the relative wind is not perpendicular to our direction of travel then what we care about is that part of the wind vector that is perpendicular. So, what do we see?

Angle of travel : Relative size of lift component
90 : 0 (north, into the wind)
80 : 0.09
70 : 0.17
60 : 0.26
50 : 0.34
40 : 0.42
30 : 0.5
20 : 0.57
10 : 0.64
0 : 0.71 (east, perpendicular to the wind)
-10 : 0.77
-20 : 0.82
-30 : 0.87
-40 : 0.91
-50 : 0.94
-60 : 0.97
-70 : 0.98
-80 : 0.996
-90 : 1 (south, directly downwind)

Here we can see that as we head into the wind that the actual part of the wind that produces lift gets smaller and smaller. If you prefer percents just multiply by 100. However, to see what the total is we need to multiply this by the relative wind speed.

Angle of travel : Forward thrust from sails
90 : 0 (north, into the wind)
80 : 1.74
70 : 3.42
60 : 5
50 : 6.43
40 : 7.66
30 : 8.66
20 : 9.4
10 : 9.85
0 : 10 (east, perpendicular to the wind)
-10 : 9.85
-20 : 9.4
-30 : 8.66
-40 : 7.66
-50 : 6.43
-60 : 5
-70 : 3.42
-80 : 1.74
-90 : 0 (south, directly downwind)

Here we can clearly see that thrust is at a maximum near 0 degrees which is a track perpendicular to the wind (or a beam reach). For the sticklers here, this is a linear comparison. The actual force due to wind velocity is squared so the curve would be even sharper. However, squaring the numbers won't change the fact that it is largest near the middle or perpendicular to the wind.

Does this explain why we can sail fastest on a beam reach?

Note: I can show more intermediate steps if necessary. You get the combined wind velocity by using the Law of Cosines as the angle sweeps from 0 to 180 degrees. You get the vector angle by using the fact that for all triangles the interior angles always add up to 180 degrees. And, for an equilateral triangle the remaining two angles are the same so each one is half of 180 minus the sweep angle. Then you subtract the wind angle from the boat's track angle to get the relative wind angle. The lift is 90 degrees to this angle. Then you use cosine to get the portion that is in the same direction as the direction of travel. Then you multiply this by the relative wind velocity. Again, if you wanted to be more precise you would square the velocity first. However, this is negated in practice since drag is also squared.

Last edited by brehm62; 04-05-2012 at 02:34 AM.
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