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Old 06-27-09 | 06:56 PM
  #48  
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Shimagnolo
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From: Zang's Spur, CO
Originally Posted by trekker pete
no, drag increases at, uhhh, I think it's the square of speed.
Correct. The acceleration is the derivative of the formula I gave earlier:

acceleration = g - ((Cd * rho * V^2 * A) / (2 * m))

where

V = *current* velocity,
m = mass of the falling object,
g = acceleration due to gravity,
Cd = drag coefficient,
rho = density of the fluid through which the object is falling, and
A = projected area of the object.

And for a vehicle coasting down a descent (ignoring friction of bearings and tires):

acceleration = (sin(theta) * g) - ((Cd * rho * v^2 * A) / (2 * m))

where theta is the angle, (measured as zero for a level surface)
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