Red, to get half the Kv, the number of winds is increased by factor two -> wire resistance increases by factor 2. And the cross-sectional area of the wires is reduced by a factor 2 too, to make rome for the extra winds. -> wire resistance again reduced by factor 2. All in all, resistance now has increased by factor four. On ten cells, current is half the original current. And since P.loss = I²R, nothing has changed i.e. as far as the motor is concerned.
Losses the battery-leads are reduced by factor four, not 2. Asssuming we use batteries with half the capacity to keep the original amount of available energy, battery losses are halved because battery resistance doubles and battery current halves.
Vriendelijke groeten
Ron
On the losses in the battery leads and in the battery itself we seem to agree. I stand corrected on the losses in the motor!
The old textbook that I used as reference for the motor constants seems to have made a mistake on wheter k or K should be used to denote these constants. I should have looked looked it up in some more references.