ORIGINAL: Red B.
ORIGINAL: ron_van_sommeren
Mind you, a motor with Kv=1000 on 5cells is will give the same results the same motor wound for Kv=500 on 10cells.
In principle yes, but in practice no! Doubling the number of cells from 5s to 10s reduces the current to half and thus all losses in the motor, speed controller and cables are reduced by a factor 4! The losses in the battery due to its internal resistance is reduced by a factor 2. Also, because of the lower current demand less strain is put on the battery, thus increasing its lifespan.
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 increased 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