Originally Posted by
echappist
Can someone do a quick calculation converting grams of drag to power needed to overcome that drag?
the equation reads
P = F v (of what?) = 1/2 x fluid density x (fluid velocity)^2 x CdA x velocity (of what?).
1st question. The drag in grams should actually be multiplied by gravitational acceleration, so it gives a reading of force. Does this mean the drag already accounts for fluid velocity?
2nd question. So say you are rolling along at 25mph into a 2mph wind, which velocity do you put in for the F v part?
Power is work / time. (P = W/t)
Work is force * distance. (W = fd)
Drag is a force, so P = fd/t ==> P= f * (d/t) ==>
Power = drag * speed (in m/s)
"Speed" would be the scalar of the sum of bike velocity and wind velocity, projected in the opposite direction of bike velocity. If you are going 25mph into a 2mph headwind, velocity for the above equation would be 27mph = 12.07m/s