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  #41  
Old 04-04-2012
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Re: Why is beam reach (or near to it) the fastest point of sail?

Quote:
Originally Posted by AdamLein View Post
I'd suggest that you can accelerate (gain kinetic energy) whenever the apparent wind is suitable (i.e. coming from a certain range of directions, and of a sufficient speed). If there's a 5 kt current, for example, setting due west, and your boat is pointed north and at rest relative to the current, she will experience a 5 kt beam wind that she can sail on. So, there's no true wind, but acceleration is still possible.
No, you seem to be getting confused about relative versus true wind. If the water is stationary and the wind is 5 knots then the true wind is 5 knots. If the water is moving at 5 knots and the wind is stationary but the boat is moving with the current then the true wind is still 5 knots.

True wind is with regard to the water (because that is what the boat is sailing in).
Relative wind is with regard to the boat.
These two are not the same.

If you want to include current then we have true water speed and relative water speed. Your boat's hull only sees relative water speed. Obviously with regard to travel then true water speed is important.

Quote:
True wind is important for considering sailing in places your boat isn't in. It's also important for considering the implications of wind against current. But I'm confident that it is irrelevant for figuring out how much energy your boat can get from the wind (irrelevant in the sense that apparent wind tells you everything that you need to know, whereas true wind doesn't).
Let's say the wind is blowing out of the north at 10 mph.
Now, let's say you are on a beam reach traveling east at 10 mph.
The apparent wind is 14.14 mph out of the north east.
So, lets make some naive assumptions.

When we started we couldn't sail directly north because that would have been into the wind. But now with the relative wind out of the NE we should be able to turn straight north and have the wind at 45 degrees and keep sailing, right?

No. That would clearly be impossible since the true wind doesn't change. If we swung to port and were traveling at 10 mph north then the relative wind would now be 20 mph out of the north and we would have no thrust whatsoever in our direction of travel. Clearly we would coast to a halt.

To be honest I find it somewhat amusing that you would suggest that energy is related to relative wind rather than true wind. That would be a pretty obvious violation of conservation of energy. It doesn't matter what frame of reference you are in, you can't get energy for free. Therefore the total available energy cannot increase beyond what it is before you start moving. If you actually could increase the available energy with increasing relative wind you would have a perpetual motion machine.

Perhaps you are confusing energy with dynamic forces. Dynamic forces do indeed increase with relative wind. In fact, they increase with the square of relative wind.
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Old 04-04-2012
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Re: Why is beam reach (or near to it) the fastest point of sail?

Quote:
Originally Posted by brehm62 View Post
No, you seem to be getting confused about relative versus true wind. If the water is stationary and the wind is 5 knots then the true wind is 5 knots. If the water is moving at 5 knots and the wind is stationary but the boat is moving with the current then the true wind is still 5 knots.

True wind is with regard to the water (because that is what the boat is sailing in).
Relative wind is with regard to the boat.
These two are not the same.
I guess I could be completely wrong. I'm pretty sure that true wind is relative to the land, and apparent wind is relative to the boat.

Quote:
Let's say the wind is blowing out of the north at 10 mph.
Now, let's say you are on a beam reach traveling east at 10 mph.
The apparent wind is 14.14 mph out of the north east.
So, lets make some naive assumptions.

When we started we couldn't sail directly north because that would have been into the wind. But now with the relative wind out of the NE we should be able to turn straight north and have the wind at 45 degrees and keep sailing, right?

No. That would clearly be impossible since the true wind doesn't change.
You're arguing something I never said. I never suggested that apparent wind would remain constant if the boat alters course or speed. Such a claim never came up in this thread, by me or anybody else.

Quote:
To be honest I find it somewhat amusing that you would suggest that energy is related to relative wind rather than true wind. That would be a pretty obvious violation of conservation of energy.
I'm glad you're amused. It's a pretty pedantic point, and I stand by my words. If you want to argue it, maybe let's do it in another thread without further sidetracking this one. Since you agree that force is relative, just pretend I said "force" in that paragraph.
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Old 04-04-2012
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Re: Why is beam reach (or near to it) the fastest point of sail?

Quote:
Originally Posted by AdamLein View Post
Wow, I should have checked in an hour ago.

Okay, so, brief comments on a variety of responses so far:

Luffing when wind "in line": I'm pretty sure the wind is "in line" with the luff of my sail when the telltales stream evenly aft on both sides. The sail isn't luffing because it's curved. The curve is maintained by the angled pull of the sheet.
I said the boom/sail. Again the sail is only curved because wind is pushing it more to one side than the other. This is because the boom is at an angle to the wind. If the boom is inline with the wind the sails must luff.

Quote:
Lift: perpendicular to freestream flow, not to the foil. Minor point; you could break down the force of the wind into any components you like, but only one is referred to as "lift" in the textbooks.

Acceleration on the lee side: my understanding is that faster flow separates sooner, which greatly increases drag, and that therefore you actually want to decelerate flow on the leeward side (while still keeping it faster than the windward flow), and that this is in fact how the slot effect works.

No, that's patently false. Lift is given by a integral of forces over the surface of the sail. It's not especially useful for internet forum discussions, but it's there.
That's not a quantification it's a trivial definition. Of course sum (integral) of the perpendicular forces over the sail tells you the lift. That doesn't help quantify the strength of this force relative to any other. It just defines how we'd find it.

Quote:
So yes, your "propeller pitch" model has an easy quantification --- which, as I keep complaining, is horribly wrong, because it predicts infinite speeds. Oh, except that you add drag, which you don't quantify. So your model either predicts infinite speeds, or is not quantitative, depending on which features you include :P
Obviously the force is related to wind speed and sail size as well as how efficiently the sail is positioned. The force is not infinite but the mechanics of how the force is generated allow speed to theoretically hit infinity. A friction-less boat could reach infinite speed. Are you saying it can't? I think the "propeller pitch" or wind deflection model is the easiest way to explain this. Lift remains for most an imprecise term that doesn't contain the boundaries for what a sailboat can do.

For example why can't a sailboat sail into the wind? My explaination is that a boat sails when wind hits the sails and get's deflected. Well when it's pointed into the wind there is no wind to deflect and this is totally clear on a simple diagram.

On the other hand lift is less concrete. Why can't the sail generate lift when the wind is flowing straight over it when a boat is in irons someone might ask. Even you had trouble allowing that a sail will luff in this situation. However I'll admit this is my own opinion of the relative simplicity of these two models.

Quote:
But, you do sail off apparent wind. If the apparent wind is in a direction that permits sailing, that is. Same goes for true wind.
I don't deny this but I don't think it's an effective way of explaining it. Once you allow for the fact that you're sailing off aparent wind, "A sailboat makes it own wind" is a phrase I've seen, you open yourself up to the kinds of mistakes that brehm62 points out. It becomes all to easy to get confused.
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Re: Why is beam reach (or near to it) the fastest point of sail?

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Originally Posted by asdf38 View Post
That's not a quantification it's a trivial definition. Of course sum (integral) of the perpendicular forces over the sail tells you the lift. That doesn't help quantify the strength of this force relative to any other. It just defines how we'd find it.
Right, so, having found it, and having found the others, now we know their relative strengths? I don't get your complaint.

Quote:
The force is not infinite but the mechanics of how the force is generated allow speed to theoretically hit infinity.
I was worried for a minute that I might be misunderstanding you; I'm glad this isn't the case.

Quote:
A friction-less boat could reach infinite speed. Are you saying it can't?
Nope! I'm saying that the drive-generating mechanism will never generate infinite force. A frictionless boat under the impetus of a constant force would indeed accelerate forever.

To go back to "the mechanics of how the force is generated allow speed to theoretically hit infinity". To me this means you are saying, "some configuration of the system will produce an infinite force", which is obviously wrong. A frictionless boat will never experience infinite force due to interaction with the wind.

Quote:
Lift remains for most an imprecise term that doesn't contain the boundaries for what a sailboat can do.
Gah!!!

Quote:
For example why can't a sailboat sail into the wind?
Because there would be no way to produce a lift which has a forward component.

Quote:
Why can't the sail generate lift when the wind is flowing straight over it when a boat is in irons someone might ask.
Sails can generate lift when the boat is in irons. Don't you know how to get out of irons? The lift won't have a forward component, so the boat will not go forward, but it will turn.

Quote:
Even you had trouble allowing that a sail will luff in this situation./quote]

I have no trouble allowing that there are situations when the sail will luff. I agree that the main will luff if the wind is parallel to the boom.



I don't deny this but I don't think it's an effective way of explaining it. Once you allow for the fact that you're sailing off aparent wind, "A sailboat makes it own wind" is a phrase I've seen, you open yourself up to the kinds of mistakes that brehm62 points out. It becomes all to easy to get confused.
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  #45  
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Re: Why is beam reach (or near to it) the fastest point of sail?

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Originally Posted by brehm62 View Post
No, this isn't related to relative wind. It's related to keel efficiency. A less efficient keel will reach maximum hull speed on a slight broad reach rather than a beam reach (because there is a bit less load on the keel).
Thanks for responding. It seems that everyone else is intent on beating each other senseless with the points they score. Thanks also to those who pointed out that in my iPhone brainlessness, I switched forward and backwards. As you say, the apparent wind moves forward as you travel faster.

However, still interested in whether polars would tell you anything more if based on apparent rather than true wind. It would make it easier to conceptualize what the masthead is telling me.
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Re: Why is beam reach (or near to it) the fastest point of sail?

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Originally Posted by AdamLein View Post
I guess I could be completely wrong. I'm pretty sure that true wind is relative to the land, and apparent wind is relative to the boat.
That's a third frame of reference which is used for navigating by dead reckoning. If we were talking about navigation then you would be correct because for those you reference wind or current from a fixed point.
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Why is beam reach (or near to it) the fastest point of sail?

Lot of nonsense here. Beam reach has least drift. Upwind get resistance and is actually slowest point. Downwind with spinnaker can allow you to exceed wind speed and even put uld boats planning. Just go out and see first hand. bumble bees actually do fly.
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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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Re: Why is beam reach (or near to it) the fastest point of sail?

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Originally Posted by reed1v View Post
Downwind with spinnaker can allow you to exceed wind speed
Yes, if you are using an asymmetric spinnaker on a broad run. If you are using a symmetric spinnaker running dead downwind then, no.

Of course, if the wind is not constant then whenever the wind drops you would still briefly be moving faster than the wind.
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Re: Why is beam reach (or near to it) the fastest point of sail?

Brehm62 :

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


Well explained that man!
A word of caution.
My High School maths teacher woulde be jumping up and down like a hairdresser at THREE horrible sins within your explanation....

1 : Never, but never put units in the middle of equations. Write them above and below the equations if you wish but never within the equation.

2 : The unit you use for force should be the lbf ... do not forget the "f" after it.

3 : Worst sin of all, NEVER pluralise a unit. Units have, by definition, no plural. If you put an "s" after a unit, it denotes "second", so let's have none of this "lbs" stuff ever again.


Apart from three sins, it was a good explanation.

The thread has become rather confrontational. For me, I just put the sails up, choose a course, and trim the sails until they just stop flapping. My ship is fastest when the relative wind is ahead of beam.

One of the beauties of sailing, methinks. I am fastest when the wind is slightly against me.

Now there is a pleasant thought.

On my beloved Loch Ness, the wind is either right on my nose, or right on the back of my neck, as the land masses either side channel the wind right at me, or behind me. Unless I choose to sail across the Loch, that is. It's only about a mile wide.
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