# Sailing Performance In a Nonuniform Wind

Discussion in 'Hydrodynamics and Aerodynamics' started by tspeer, Nov 13, 2013.

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### tspeerSenior Member

One question that I've been pondering for a long time is, "What is the meaning of velocity made good in a sheared wind?"

There are two parts to this issue. The first is how to interpret an asymmetric speed polar. The second is what to do when the direction of the true wind depends on the height at which it is measured? It doesn't seem right that you get a different answer for Vmg depending on what height you happen to choose to measure the wind.

Take the asymmetric speed polar first. Let's say that the true wind comes from the same direction at all heights (no shear). And for simplicity, let's say that the yacht is symmetrical except for having been ballasted to one side. One might get a polar like that shown in the first attachment below as the ballast is shifted from one side to the other.

When sailing to windward, the best courses to sail on the two tacks are where a straight line is tangent to the lobes of the polar, forming the convex hull. This is shown by the dashed line for the orange curve. It's best to pinch slightly on the favored tack and foot slightly on the bad tack, because the only reason to sail on the bad tack is to make up the necessary crosswind distance to get to the mark.

The reason for this is shown in the second attachment. If the boat alternates between any two points, whether going to windward or not, the locus of average speeds will fall on a straight line connecting the two points. The position along the line will depend on the relative time spent at each of the end points. For example, take A and A' on the symmetrical polar. If the boat spends half of its time at A and A', then it will be averaging in the direction of the blue arrow at the speed where the arrow intersects the dashed line. This holds for any two points, such as B and B'. It's easy to see that C and C' are the optimum points for going to windward, because the line connecting any other pair of points will fall short of C-C'. Likewise, D and D' are the optimum points for sailing downwind. When you apply the same logic to an asymmetrical polar, you get E-E'.

What really matters is advancing the line bridging the two lobes. One can go to any point on the line with no losses by choosing how much time to spend on each tack. So the red arrow perpendicular to E-E' is the direction to use for Vmg, because it's the rate at which the line advances that determines the progress to the mark. Note that for this case, Vmg would not be directly to windward, but 10 degrees off the wind.

So Vmg needs to be defined by the performance of the boat, not the direction of the wind. It is a coincidence that for a symmetrical polar Vmg is aligned with the wind direction. But for an asymmetrical polar, that won't be the case.

Now consider what happens with wind shear. One can define the wind by using a particular height, say, 10 m, as the reference wind direction and any deviation from that direction is the wind shear. Another choice would be masthead height, since that is where the apparent wind is typically measured. If one used CFD to calculate the forces on the boat, it really wouldn't matter what height was used as the reference height. When the shear was added to the reference direction to get the wind at a given height, you'd get the same answer. So the forces and moments on the boat would be the same regardless of the choice of reference height, and the calculated boatspeed polar would be the same. The polar might be asymmetrical as a result of the shear, but any choice of reference height would result in the same speed at a given direction through the water.

What will change is the grid used to plot the polar. If the grid is shown with true wind direction up, as is the usual convention, then the polar will be skewed on the grid when the reference height is changed. However if the polar is oriented with North up, then the speed curves will be the same but the grid will shift with the wind direction as the reference is changed.

Now suppose that the polar is oriented so Vmg, as defined by being perpendicular to the convex hull connecting the two windward lobes, is pointing up. Once again the grid will shift depending on the reference height. However, the speed curves will line up.

This leads me to the surprising conclusion that:
- Vmg should be defined by reference to the boat's performance, not the wind direction.
- The best choice for the true wind direction is parallel to Vmg.

In other words, instead of defining Vmg as being the velocity in the direction of the true wind, the true wind direction is defined by the Vmg.

This solves all of the paradoxes. The true wind direction does not depend on the height at which the wind is measured. Vmg is the rate of progress in the direction of the true wind. The same definitions work for both uniform wind and for sheared winds, for symmetric polars and asymmetric polars.

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### Mikko BrummerSenior Member

Wow... at least at a glance, one cannot but agree. And the situation with asymmetric polars is very real, not even uncommon - in addition to wind shear, even more often caused by waves coming from a different direction than the wind.

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### latestarterSenior Member

It is uncomfortable however the average wind direction for an Optimist will differ from an AC72.
You would not choose the height to measure the wind direction, it will be a function of the sail plan.
Similarly the wind speed varies with height so the relevant wind speed will vary with sail plan.

I am trying to get my head round the rest of the argument.

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### latestarterSenior Member

This an interesting idea you have floated, I have thought of little else since I got up this morning

I can understand the above but you have lost me on the next bit.

Surely this is a velocity diagram, the only way to advance the line bridging the two lobes would be to get a faster boat producing bigger lobes. Any point on the line represents an average speed rather than a position on the water. There should be only one point on the line that gets you to the windward mark.

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### tspeerSenior Member

Yes, that's what it would mean. A small boat would experience a different wind than a large boat. The wind over the height of the small boat's mast will have a relatively larger variation due to wind gradient, but a smaller variation due to wind shear, assuming a power law wind gradient. The larger boat will experience more shear, but relatively less variation in wind strength over its span.

But the strange thing is two different boats of the same mast height could be said to experience different winds because their performance affected differently by the same wind.

Mikko points out that asymmetric polars can be due to waves as well as the wind. That would certainly affect the Vmg, but is it fair to say that it affects the definition of the wind direction, too?

Maybe instead of trying to use the yacht's performance to define the wind, it would be better to simply decouple Vmg from the wind direction and accept that Vmg has to be defined based on the yacht's performance, not the wind.

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### tspeerSenior Member

You can view the velocity polar as being a snapshot of the distance traveled. If you plotted the distance traveled from the same point in a given amount of time at various courses, you'd get a plot of the same shape. So with time, the distance traveled polar is blowing up like a balloon and the line is moving along its normal. That was the sense in which I was talking about it advancing.

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### latestarterSenior Member

Isn't that the general situation regarding apparent winds, the true wind is constant but apparent wind varies with boat performance.

Some of my confusion is that Vmg means different thing at different times. There is the instant speed at any time, the average so far and the final Vmg for the beat when the windward mark is reached.

To get a better appreciation of what is going on I have redrawn one of your polars with the minimum of information. Having found E and E' from your polar everything else is not needed.

The figures quoted are the measurements in centimetres on my monitor so not the same as yours but it is all relative.

On starboard tack Vmg is 3.4 and laterally 9.2, on port tack 6.9 and 9.8 respectively.

I then drew on the vectors for equal times on port and starboard which gives a resultant to the right of the vertical. The extra time on starboard will bring it back to the vertical.

Your red arrow at right angles to the dotted line only works when the speeds are the same e.g. the blue arrow bisecting A-A'

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### tspeerSenior Member

Bingo. You've put your finger right on the piece I was missing.

Consider the case when the boat is not sailing at the optimum point. That's shown as point E'' in the attached diagram. The sailor wants to make a change in speed, dV. The circle indicates the change in dV applied in any direction. But the boat's performance only allows at best a change that is tangent to the polar, shown by the black arrow.

Now the component of dV that is parallel to E-E'' is of no consequence, because that direction will be compensated for when the boat is on the other tack. What matters is the change dVmg, which is perpendicular to E-E''.

So the sailor is trying to improve performance by maximizing dVmg. This moves the boat closer to E'. Once it arrives at E', no further improvement can be made on that tack. This is totally driven by the fact that the polar is not convex between E and E'. The wind direction plays only an indirect role in shaping the polar.

In fact, this definition of Vmg applies any time the polar is not convex. The second attachment illustrates a polar for a yacht with a spinnaker that has a hollow between two reaches. The best strategy is to alternate between F and F' instead of sailing directly to the mark. If the mark was at 135 deg, the boat would make a substantial gain using this strategy. The vector I've labeled as Vavg is the average velocity in the direction to a somewhat different mark, and Vmg is the velocity made good. When sailing at either F or F', one is trying to maximize Vmg. The overall course sailed will depend on the amount of time spent at F vs F'.

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9. ### Baltic BanditPrevious Member

So I've been mulling this over in my head, and I'm not convinced your first set of charts is accurate (and hence potentially the rest of the logic flowing from them).

Your initial chart shows essentially a symmetric polar but shifted because of assymetric ballast. but I'm not convinced that with assymetric ballast you get symmetric polar curves, particularly upwind.

The polar tells you essentially what angle you can sail at what speed in a particular wind. but because assymetric ballast will change your heel angle for a given wind power, your drag for that wind will change and change rather significantly.

For example in some boats, 30 degrees of heel are enough for the root of the keel to begin to experience surface effects. So if your ballast biases you by say 10 degrees, and there is enough wind to heel the boat 10 degrees on the side of positive ballast, then the side that is negatively ballasted will be seeing big drag jumps right out of the box and that will result in more than just a rotated asymmetry in the Polar

I think part of my problem lies with the assertion that
This will clearly be true if the wind forces on the rig are symmetrical, but you are exploring a case where specifically they are not. so I would contend that rather than using the same wind speed polar for both tacks, you have to instead use a wind speed polar that reflects the power being generated by the rig on that particular tack.

The "waves" example is a good one to look at. In an offset wave condition, well driven boats will set up their sail plans differently - with more power being generated on the tack heading into the waves and more point on the tack going laterally across the waves. Essentially because the waves are slowing down the hull, the AWS "seen" by the sailplan is different. Thus the sails are trimmed for "different wind speeds" is one way to look at this (although its also a case of corkscrewing rig motion and the need to reaccel vs maintain existing speed).

Well the same can be used for a sheered wind. Lets say the sheer is 15 degrees at midspan of the mast. the ideal would be a two piece sail - where the bottom sail is trimmed to the lower AWA and the upper section to the upper AWA. but because these two trim modes are BOTH relying on their relative angle to the keel for the vector direction of their max lift, the amount of power you can generate out of the rig differs on each tack. That is the same as the case on STB you are sailing in 15 knots of wind and on port you are sailing in 10 knots of wind (not unusual in a situation where you have a dying steady wind competing with a katabatic wind along a shoreline (Columbia river Gorge for example))

In that case your STB polar is the 15 knot curve and your port is the 10 knot curve. and those may well have very different optimal VMG points that will not be represented by the tangential line but instead by two different lines tangential to the wind direction.

In a sheered wind, your rig has the same issue. You can generate more power on one tack vs. the other. (because bringing the boom above centerline for the lower section is usually ineffective from a power/drag tradeoff). And it is maximixation of the power generating vector that is the driving force behind polars and why you need different polars for different wind speeds.

So I think by using the assym ballasted boat as your starting point, you are going down the wrong analaytic path.

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### Mikko BrummerSenior Member

This second case is known as VMC, velocity made on course, or best velocity towards a mark not lying straight upwind or downwind. Pretty common when on a course where you cannot quite lay the mark under spinnaker.

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### tspeerSenior Member

Interesting. The Occam site has a somewhat different definition of Vmc. They Take it as the component of the speed in the direction of the desired course, as in Figure 3 of this document.

They also talk about Vmg as being a special case of Vmc, but don't consider asymmetrical polars.

I realize it's been known for a long time that the optimum way to sail a course near the spinnaker cross-over is to sail at the points of tangency to the convex hull. But I've not seen anything about the component perpendicular to that line. What you're really doing in locating either of those two points of tangency is maximizing the perpendicular component.

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### tspeerSenior Member

I once asked San Honey, "What is the job of the inshore navigator?" His reply was, "The number one job of the inshore navigator is to estimate the wind shear. Because that determines the speed on the other tack."

In the light of this discussion, to optimize the course for the tack you're on, you need to need to know the angle for best Vmg on the opposite tack. Because that will determine the angle of the line comprising the convex hull and the direction of its perpendicular for you to determine the Vmg for the tack you're on.

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### tspeerSenior Member

Ignore the first set of charts. I actually constructed them originally to illustrate the effect of choosing different reference heights for the true wind measurement.

When I was composing the post, I grabbed the first graphic I'd done up that I thought I might to illustrate my point. But I hadn't fully grasped my own conclusions at that time, and improved my thinking as I did the second graphic.

None of these polars are real. I was simply drawing something that looked like a polar to help with the general discussion. The only thing that is important is they have two lobes and the lobes may be of different sizes and shapes.

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### latestarterSenior Member

TS you should put a health warning on the thread, I woke up the other night at 3.00am and could not get thoughts about assymetric polars out of my head.

Presumably no one would set out to create this situation. I am imagining movable water ballast in the port hull and the pumps fail, so you are trying to make the best of a bad job.

Looking at my revision of your diagram, taking the specific case of getting to a windward mark asap. This is achieved by maximising the intersection of E-E'' with the vertical, (G -G') which occurs when E-E' is reached.

I can not see the benefit of the large red arrow in the original diagram, it is only its component in the vertical that matters.

As you wrote, the dotted line will give an indication of the position when the percentage is longer on one tack than the other however the method breaks down at the ends.

For instance on starboard tack when you reach 100% the performance on the opposite tack becomes irrelevant. This method would advise you to use E whereas C is better.

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### philSweetSenior Member

Au contraire mon frere, you do do this on purpose. During the AC, Oracle is alleged to have used different foils port and stbd when there was a windshear relative to the course layout. They knew they would spend 60/40 going to windward mark and 30/70 going downwind and chose the foils accordingly. This would have generated some fairly wicked asymmetrical polars with respect to windward foiling ability.

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