Is this scaling of aircraft size and power (at least roughly...) correct?

Question

I use to fly a very light plane, with a 100-hp engine. In order to calculate the power needed by a similar plane, but 50% larger, I reason as follows:

In general, the power required by an airplane is proportional to its weight times its speed. The weight scales with the cube of the linear dimension, so the larger plane would weigh 1.5 cubed = 3.375 times more. Since the airspeed is squared in the formula for lift, the exponent for the proportionality of lift should be 1/2.

I have now two exponents for the scaling, 3 for the weight (or mass, in this case) and 1/2 for the speed. As the weight and speed are multiplied in order to get the proportionality of the power required, I add the exponents, and the result is that the larger plane would need an engine of 1.5 ^ 3.5 = 4.13 x 100 = 413 hp.

Is this OK?

Answer 1

For a first order of magnitude check your result looks good.

Experience tells that the mass scales not quite with the length cubed, so an exponent between 2.3 and 2.6 gives better results. Wing area goes up with the square of the length increase, so your wing loading will become $$\frac{m_{large}}{A_{large}} = \frac{m_{small}^{2.4}}{A_{small}^2} = 1,176\cdot\frac{m_{small}}{A_{small}}$$ This requires 8.44% higher speeds. To achieve them, thrust must go up with the square of the speed increase and power with its cube, since power is thrust times speed. Therefore, the power increases by a factor of 3.375 if the mass scales with the exponent 2.4. Power loading will now be $$\frac{P_{large}}{A_{large}} = \frac{3.375\cdot P_{small}}{A_{small}^2} = 1,5\cdot\frac{P_{small}}{A_{small}}$$

Answer 2

Assuming that length, width, and height of all dimensions are scaled up uniformly, and assuming that weight scaled with L^3, and assuming that the lift and drag coefficients of your wing remain the same, then I don't see anything wrong with your number.