Originally Posted by
asgelle
Do I really have to type this again? Drag is non-linear. The resultant axial force is not (v cos(theta))^2. It's v^2 cos(theta).
It's hard to say it more concisely than this. I think there are a few reasons why some of you are not believing what Asgelle is saying:
1. It is counterintuitive that a pure crosswind can slow you down.
2. You are misapplying vector analysis in the problem. The vectors that are important in this problem are force vectors NOT wind velocity vectors. It's true that a pure crosswind has no vector component opposite to the direction of travel, but this does not mean that the force resulting from the crosswind has no vector component opposite the direction of travel.
3. You are treating the crosswind and headwind separately and then adding the results. This works for linear systems, but not for non-linear systems. Proper analysis requires you to sum the crosswind and headwind (due to movement of the bike) vectors, then calculate the resulting force and then, finally, determine the vector component of the force opposite the direction of travel.
Cheers,
Tomato