Originally Posted by

**rpenmanparker**
You don't use the forces exerted by the separate wind vectors, you have to use the vector sum of the forward motion (rearward wind) and crosswind. Then apply the force of that wind vector. Since force is proportional to the square of the wind speed, the crosswind factor gets "locked into" the value so to speak. The compnent in the opposite to the dorection to travel is greater than just from headwind. A simple example: 2 mph forward motion, 2 mph perfect crosswind so the resulting wind is 2 X root 2 at 45 deg to forward direction. Since force is proportional to the square of the wind speed, the total force of the wind on the bike and rider is proportional to 8. The component of that force pushing opposite to the direction of travel is 8 divided by root 2 or 4 X root 2. But if there were no crosswind, the force would be proportional only to the headwind component = 2 squared or 4. So the actual impeding force would be a factor of root 2 times the headwind force. That is the effect of the crosswind. Simple, huh? Sorry to be so obtuse but I am typing on my phone. Equations aren't so easy.