A possibility (which I think I alluded to in my post) would be to take the speed of the true wind and the speed of the wind that the carrier generates, then treat them as vectors. The number of degrees the carrier has to turn right from the wind's direction would be calculated so that the apparent direction of the wind is the parallel with the landing area's centerline.
High-school trig would give the solution. Is this close to how they do it?
In ye olden days, the bridge (and Combat on small boys) would generate the solution (or were supposed to) by doing what you describe on what's called a MOBOARD. It's a polar plot to make the trig less mathy and more accessible to the watch team. Often when two people do the same calculation, they come up with a different answer (see above posts and see below). More modern ships now have a computer that does all of that for them, but not all ships have that capability.
Yes. OODs know how to do this.
No, many OODs do not know how to do this. Probably made worse by the system on the DDGs. As has been mentioned in previous posts, especially by the helo guys, the pilots can usually resolve a 90% solution faster by looking at the relative winds and the ship's gyro (if they know the ship's speed) than the bridge can incorrectly come up with a solution.
I think the biggest hurdle is getting confused with recipricals. As aviators, we're naturally self-centered, so we think of vectors as center out, where Shoes tend to mess up the idea that the true wind is a vector from and they end up getting the solution 180 out.
