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http://en.wikipedia.org/wiki/Drag_equation

D=.5P(V2)ACd

(V2 is supposed to mean V squared, and d is a subscript of C)

D is the force of drag,

ρ is the density of the fluid,

v is the velocity of the object relative to the fluid,

A is the reference area, and

Cd is the drag coefficient

I am trying to solve the equation on this page for V, but I've been out of college for 10 years so maybe that's my problem:

I came up with

V=SQRT((2D)/(PACd))

It looks correct to me but somehow I don't think I'm getting it correct because of the paragraph at the bottom of the page.

"Another interesting relation, though it is not part of the equation, is that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula. Since power is the rate of doing work, exerting four times the force at twice the speed requires eight times the power."

When I run my equation it's never an 8 to 1 ratio for power to speed. I've hosed something but I can't figure out what.

Any help would be greatly apreciated!!!!!

Thanks,

Al