Quote:
Originally Posted by Captain Awesome
Drag is proportional to the square of the speed of the moving object.
FD=1/2ρv*2CDA
FD is the force of drag, which is by definition the force component in the direction of the flow velocity
ρ is the mass density of the fluid
v is the velocity of the object relative to the fluid
A is the reference area
CD is the drag coefficient
Since everything else is the same except for velocity, the drag is directly proportional to the square of the speed.
So the drag on a car going 203MPH versus 220MPH could be expressed as
41209/48400 = 85.1%
In theory, it should take just under 15% less horsepower to get the identical car moving 203MPH instead of 220MPH.
So, there must be other losses at work here...
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Nah, you just forgot that power equals the force vector dotted with the velocity vector. Neglecting the other forces such as friction (actually quite small in this instance) losses through the drive train, etc., it takes around 27% more horsepower to go from 203 to 220mph. Since the average car loses around 20% through the drive train, you should need around 34% more horsepower.