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From: Bob Goethe
Date: 2016 Nov 12, 15:57 -0800
In section 46 of my user manual, I unpack an involved technique that Arvel Gentry developed to determine the optimal course to sail off the wind: to tack downwind.
Back in October, Paul Hirose said, "In the time / speed / distance problems, I think you are working way too hard." I am wondering if I am doing the same thing now in selecting optimal courses for tacking, either upwind or downwind.
Arvel's basic approach works equally well with sailing on or off the wind. The math only cares how far off the wind you are, i.e. angle theta, not whether the wind is from ahead or astern.
Please take a look at this and tell me what you think. This is first draft.
§48. Velocity Made Good
VMG = S x cos(q)
...where
VMG = velocity made good (toward your destination) in knots
S = boatspeed in knots
q = the angle off the wind that you are sailing, either upwind or downwind.
The equation implies that you can make the very best time by sailing directly into the wind or running directly before the wind. This is clearly not true when travelling under sail. For instance, 30° off the wind is as high as you can point. And we know that running dead downwind is slower than pointing even 10° higher than dead downwind.
The table below is solving the same problem as in section §45, Tacking Downwind, for the analysis of results for an 8 knot wind. The equation for VMG doesn't care whether you are sailing upwind or downwind. It only cares about your angle off the wind ahead or astern.
Course q° higher than dead downwind | Speed on course q
| S x cos(q) = VMG | Ideal Course |
10° | 4.3 kn | 4.2 kn |
|
20° | 5.3 kn | 5.0 kn |
|
30° | 6.0 kn | 5.2 kn | « |
40° | 6.7 kn | 5.1 kn |
|
50° | 7.0 kn | 4.5 kn |
|
This gives the same result as with the Arvel Gentry approach, with a bit less calculation to do.
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The table above is trying to solve a problem where the destination is 10° higher than dead downwind. In that situation, sailing 10° higher than dead downwind should give a VMG of the same as the speed on course theta. That is, it should be VMG = 4.3 knots. So maybe in the first line of the table, you don't use the S x cos(theta), but simply make the S = VMG in line 1 as a starting point.
Other than that bit of oddness, my testing of this equation seems to locate the "Ideal Course" whether tacking upwind or downwind as well as the Arvel method. You simply look for the angle theta which gives the highest VMG velocity.
It seems simple...simpler than what I have been doing up to now. Are there any pitfalls here I need to avoid? Or can I rewrite/simplify my section on tacking downwind?
Bob
