Easy question about an aircraft's top speed

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The discussion revolves around using the drag equation to calculate the top speed of a theoretical aircraft. The user initially struggles to rearrange the equation for velocity and expresses confusion about the relationship between power and speed. They note that while drag increases with the square of velocity, power requirements increase with the cube of velocity, leading to discrepancies in their calculations. After posting for help, the user quickly resolves their issue independently. The conversation highlights the complexities of drag and power dynamics in aerodynamics.
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I found an equation on Wikipedia that I believe will give me the top speed of a theoretical aircraft. Mission accomplished? Well... maybe not.

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
 
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Just answered my own question

Geez after 2 hours of tearing my hair out I post a message and 5 minutes later I figure it out. DOH!

thanks anyway,

Al
 
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