Sailing Performance In a Nonuniform Wind

Merci beaucoup

 

I thought it was less a question of sheer but more a question that different parts of the course had different general wind and current patterns.

Essentially saying that you want a different kind of boat depending on which part of the course you were on.

True this gives you some seriously skewed polars, but that's nothing new. Bill Buchan built a slightly assym. hull for the Star he used in Pusan.

 

I was not being clear in my statement. The explanation in Finnish text of this "Speedseminar" which I wrote in -92 is same as Ockam's. The area between the tangency points is a no-no zone, where you never want to sail. Either sail with the spinnaker and lower or with the genoa and higher than the course to the mark.

 

They were alleged to have used different foils, but in fact used the same foils on both sides and throughout the regatta.

However, they did place the ballast sail on one side.

 

C and C' are not better. The line C-C' lies inside E-E', so the average of the two tacks will be less.

The other tack is not irrelevant because you need to use the opposite tack to make up the cross-wind distance. As long as you have to head for a mark in the zone E-E', you need to spend some time on the unfavored tack. By footing on the unfavored tack (E vs C), you are gaining cross-track distance and spend less time on that tack. By pinching on the favored tack (E' vs C'), you are spending more time on the favored tack and incurring less cross-wind distance that has to be made up. The points of tangency of the convex hull show what the optimum amounts of footing and pinching are.

I don't see how it breaks down at the ends. If you wanted to sail in the direction of E', I don't think it would be faster to tack between C and C' because you'd lose more at C than you gain at C'.

 

If it's time to move on to part 2 of the question

Let's check the assumption. Why do you think The vmg will change? If you have a first set of polars keyed to one height and a second set of polars keyed to a different height; and both were prepared with the same wind shear in place - why would the vmg be different. The polars would be asymmetrical and different from each other, but I think they would be a transform of each other and the transform would be determined by the wind shear distribution and require that vmc would everywhere transform to the same value. Now if vmc is everywhere the same, what does that say about vmg? I think it will be the same value, but at an angle differing by the shear between the two points.

All of the above is intuition. Anyone care to puzzle it out?

 

Which is consistent with what you would do if you were sailing in wind that was stronger in one direction than the other: Foot for the pressure, pinch to stay in the pressure.

 

Baltic, that's a completely different question than what is being discussed here. So far, the conditions are uniform over the course, with a true wind shear, and presumably, a velocity gradient, although that hasn't entered into the discussion yet.

 

Well as I suggested above, I don't think there is a great difference between whether there is actually a velocity gradient or whether it is due to directional sheer, the effect is similar, on one tack you will be able to generate more power than on the other.

That is effectively the same thing as sailing in a wind that oscillates and where the puffs from one direction are lighter (or stronger) than the wind from the other direction. Why?

Because on the tack where you can sheet the lower section of the main to generate full power and twist the top enough to generate some power out of the sheered wind, you will get more power than on the tack where you have to choose between power down low or power up high

imagine a 30 deg backing sheer

on Port tack, you will have the boom at 30 AWA and the top eased by 15 deg of twist so it is sailing in 45deg of AWA (somewhat stalled but still generating power

On STB tack you will have the choice of sailing with the boom/lower section at 30 deg AWA and the top at 0deg AWA, or bearing off and sailing with the top at 30 AWA and the lower section completely stalled at 60AWA. (more realistically you will sail with an oversheeted main to hook the uppers and get say 20deg AWA and ease outhaul to get an effective 45 deg AWA down low.


So that Is effectively the same as having almost twice the wind force on Port than on Stb.

And in that sort of wind, you sail your highest CMG on port, and then foot on STB - just as you would if you were sailing in wind puffs of two different strengths



So essentially instead of having a rotated polar, what you have is a higher wind Polar on port and a lower wind polar on STB.


[edit] edited the first sentence for clarity (sorry I've been spending the day speaking technical French and my brain is bent sideways)

 

I came at it from a slightly different perspective. If you have a wind shear and a velocity gradient and are sailing to windward then -

1. The velocity gradient causes the moving boat's AWA at the sail's foot to be smaller than that at the head on both tacks.

2. The shear gradient causes AWA to be smaller at the foot one one tack, but larger on the other tack.

When you combine these effects and shape the sail for best results, you get a higher L/D when the effects are cancelling than when they are conspiring. Furthemore, the average L/D {(L/D)_port + (L/D)_stbd}/2 will be a bit less than the symmetrical L/D of the nonsheared wind.

 

Note in my previous post, I don't mean to suggest that all the aspects of an oscillatory wind are the same, since in an oscillatory wind you have amongst other issues, the question of when to tack during the oscillation to optimize your power and accel

What I was trying to suggest was a bit more of a clearly delineated analysis. I think the "heeled boat" analysis is inaccurate. Because it seeks to simulate a distortion of the curves, when what is really going on is that in a sheered wind, on one board you have more power than on the other

Just as if you were saling in a wind that has a radically different vector direction than a strong current

 

I don't think this is necessarily so, it all depends on what turns out to be the optimal power generation for the sail

 

Actually, it doesn't matter why the polars are asymmetrical. The polar itself contains all the information needed to optimize the straight-line performance, with the assumption that there's no loss due to tacking or gybing. The asymmetry could come from the wind, the waves, or the boat's configuration.

I'm coming to realize that the title of this thread is misleading. The wind is what it is. You can take any height as the reference height and reference the wind shear and gradient from there to get the wind at any level. From that, you can calculate the forces and moments on the boat, and ultimately the speed polar.

Similarly, the fundamental performance equation, Vb = Vt*sin(gamma - beta)/sin(beta), still holds, even though beta and gamma will be changing depending on the wind at the reference height. There is a range of combinations of gamma and beta that can result in the same Vb.

However, if you want to go to the next step and say Vmg = Vb*cos(gamma), now you are locking the reference wind direction. Now things get a bit sticky, because the Vmg can be asymmetrical for other reasons than the wind. You could have a perfectly uniform wind and still have an asymmetrical polar with Vmg that doesn't point into the wind. Vmg = Vb*cos(gamma) has an implicit assumption of a symmetrical polar, or at least one in which the Vmg is aligned with the true wind.

 

You are quite right it does not. I was focussing on the C E end of the line and previously tried a graphical method which showed a slight improvement for C over E-E' however having now done a full calculation E-E' is about 3.3% better than C for a course to a mark 65 degrees off to port from the true wind.
At the other end it is now obvious to me that because C' is beyond E', E' must be quicker. Sorry for raising a red herring.

 

Oops. I stuck my foot in it when I said "So far, the conditions are uniform". That wasn't directed at your title. I was speaking of uniform over the course. You were speaking nonuniform over height. I'm going to make an edit to my old post.