Quote:
Originally Posted by TomcatViP
The 1/3 coeff i s good (came from dEc/dt=SUM(P)). Seems at least this went trough the mind of the bloggers. 
However at high speed (and we will talk abt what is high speed), drag does nit increase linearly, but rather as a square function at the rate of the maximum local speed on the extrados of the wing (I am taking into account wing drag only) which is already significantly higher than the plane frwd speed.
To be rigorous also, at speed higher than Mach0.3, you'll need to make the conversion btw local press, ro and speed. The relation btw the Power and the speed is not true anymore if you don't add a term in ^2 to reflect the wet surface and the viscous drag effect.
So there is no linear relation btw speed and power, hence no guess work on the estimate gain in speed. A close look at a pressure plot of any airfoil will give you a hint. Usually an honest guy will use this equation in reverse, to have an idea of what is the ABSOLUTE MINIMUM of POWER you'll need for any increase of speed.  |
Sorry if I jumped over some steps: I'm not saying there is a 1/3 linear relationship between power and speed. Actually it's to the power of 3. In addition I agree that drag does not go up linearly but by the square. I'm only assuming the coefficients as constant, not the drag.
Thrust :T=(P x n)/v
Drag: D=0.5 * ra*v^2*(Cdo+Cdi)*S
This gives: P/v^3=0.5 * ra*(Cdo+Cdi)*S*1/n
Where n is prop efficiency, ra is density, Cdo and Cdi drag coefficients and S wing area. Since these are assumed to remain constant I substitute this with constant K
Therefore P/v^3=K in both cases
So entering the numbers we get:
910/460^3=990/v^3
Solving this for v we get:
v=460 x (990/910)^(1/3)= 473 Km/h