sailquik said..
There was some earlier references to angles sailed on speed courses. Almost by chance, I recently revisited a session I had in Albany in January 2017, and was somewhat surprised to see this. What surprised me a bit was how broad my speed runs (yellow lines) were, considering that the wind was only around 25 knots max. The pretty unique thing about this spot is that the water is glass flat because of the weed, so one can make speed runs on both tacks. The only other places I know of like this are Budgiewoi and Lake George, but often, one tack is more favoured due to various conditions.
Before viewing this, I would have guessed that my runs would have been more like 125-130 degrees off the wind, rather than closer to 135 degrees as this shows.




So if the wind was 25 knots (probably only peaks in the gusts), and speeds were 35 knots, can you calculate the Beta angle here Ian K?

10s:
time m [knots]
13:55:15 183.1 35.594+/-0.070
15:09:28 180.3 35.051+/-0.066
14:42:22 180.2 35.021+/-0.073
14:43:31 179.8 34.951+/-0.061
15:18:56 179.0 34.801+/-0.072

More trigonometry revision! Use the triangle for reference again. But this time ignore the right angle, it's not necessarily a right angle. And ignore the circle.


So Vt the true wind is 25. Vy the board speed is 35. Angle at C is 180 -135 = 45 degrees. Unknowns are angle A (the beta) Va the apparent wind and angles C and B. (Angle B is not a right angle as is now incorrectly shown in the diagram).

We need to solve the triangle. We can't use Pythagorus because there's no right angle. But we can use the cosine rule which is just Pythagorus's theorem modified for non right angles. (If angle C is 90 degrees Cos 90 = 0 and you're left with the familiar old Pythagorus)

Va ^2 = Vy ^2 + Vt ^2 - 2* Vy * Vt * Cos C

Va ^2 = 35 ^2 + 25 ^2 - 2*35*25 * Cos 45
Va = 24.75

Now we can solve for angle A using the Sine rule i.e.. sin A/Vt = Sin C/Va

A = 45.58 degrees = Beta.

Can now check how close you were in picking the theoretical angle for maximum speed which would be angle B = 90 degrees.
SinB/Vy = Sin 45/Va
Sin B = Vy * sin 45 / Va = 0.999948

B = 89.4 degrees.

Not far off 90 degrees. You're pretty spot on in picking the fastest point of sail Daffy!

sailquik said..

Thanks for the comments RacerX.

Can you list some of the specific references you allude to that back up your list of theories please?

That is a lot of homework, probably more than I have time for.

But in summary the lift line theory by Prandtl, gives the basics about span wise loading en.wikipedia.org/wiki/Lifting-line_theory

EXCEPT the calculations are not valid for low aspect ratio swept foils, like a windsurf sail.

A quick google pulls this up from 1954 naca.central.cranfield.ac.uk/reports/arc/rm/2935.pdf

A lot formula and some of it probably wrong but it makes the point that low aspect ratio and sweep make a difference! You can find other such references in lots of places, just no one has made those kind of studies for windsurf sail, much more fun to go for a blast!