Aerodynamic Drag Vs HP Requirements
Hal - Your comments about aerodynamic drag, as it relates to HP requirements, are not correct, and under estimate the HP requirements by a factor of 2. Your comments are correct for relating wind drag to changes in speed, but not HP.
The drag, not the HP, increases as the square of the speed, i.e., if the speed doubles, the drag increases by the square of the speed, or a factor of 4 (2^2 = 4). The HP requirements are also affected by speed, but in this case even more so, as HP = drag force x velocity, or restated, as speed increases the HP required is now the cube of the change in speed. So, if the speed doubles the HP required increases by a factor of 8 (2^3 = 8).
Let's make an assumption and apply this to some of the numbers I provided earlier. Assuming it takes 5 hp to go 50 mph, how many HP does it take to go 60 mph? The speed changed by a factor of 1.2 (60 mph/50 mph = 1.2). Cubing the speed change gives a factor of ~ 1.73 (1.2^3 = 1.73), and applying this factor of 1.73 to 5 hp gives ~ 8.6 HP (1.73 x 5 = 8.6).
Think about that for a moment. You've done lots of work on your engine and say you've increased the power from 2.5 HP to 5 HP (100% increase is a huge change!) and moved you from 40 mph to 50 mph. Now you want to go 60 mph which will require 8.6 HP. Where in the world are you gong to find an additional HP increase of 72% on top of the 100% you've already gained? That's why I said previously that 60 mph is a very challenging target.