i @ max eff = sqrt(io*V/R)

where

io = idle current

V = input voltage

R = motor resistance

kv doesn't come into play yet. I can choose a kv if I pick a prop, then decide on a required thrust. This works because once i have a prop and a required thrust, I know how much power the motor must output. And for best system, I want this power to be converted from electrical to mechanical at the highest efficiency. This will maximize g/W. So using the above equation, I can pick kv. However, the kv that comes out of this process is dependent mostly on those motor internal losses, io and R. That is, the math doesn't yield a certain kv for a certain prop. Generally, it shows that bigger prop requires lower kv. But it depends on how much thrust you want, and again those motor values. How is it that we all decided 10x4.7 requires ~700-900 kv, etc.? My math is telling me that for a 10x4.7, at 500 g thrust, I need 450 kv if io=0.5 and R=0.2.

If I choose R=0.1 and io=0.5 then the optimal kv = 575

If I choose R=0.1 and io=1.0 then the optimal kv = 700

If I choose R=1.0 and io=1.0 then the optimal kv = 300

So it's quite dependent on those constants; confusing me.