Originally Posted by
TallRider
prathmann, I'm fully with you on the calculation of F (drag force), but why would power depend on ground speed? When I think about power, it's driving the continual acceleration required to maintain speed against resistance (some rolling resistance and bearing drag, but mainly air resistance) which is trying to decelerate the rider. I'm not saying you're wrong (and your calculations are consistent with our experience), but theoretically, I still don't understand why ground speed is part of the multiplier for calculating power. Can you explain the rationale here a bit more?
The energy required to move something against a resisting force is equal to the size of the force times the distance that the object is moved. The simplest example is lifting a weight - say you lift a 10 lb weight up a 100' cliff. The energy required is then 1000 ft-lbs. Similarly, the energy required to overcome air resistance would be the drag force due to the air times the distance you are moving against that force. Power is energy per unit time, so the power needed is the drag force times the distance traveled divided by the time. But distance divided by time is just your ground speed. So the power is the drag force times your ground speed.
And yes, greg is correct the speed should have been 20.8 mph w/o wind as equivalent to 10 mph into a 20 mph headwind. (I was looking at 20 mph into a 10 mph wind.)