It's sounding like we could use a quick refresher on windings, torque, heating, etc.
First, the bottom line:
For two windings of
equal copper fill,
- both windings will have equal heating at the same torque and RPM.
- both windings will draw equal battery current at the same torque and RPM.
- the faster (fewer turns) winding will have a higher top speed for the same battery voltage.
- if the controller limits battery current but not phase current, both windings will have equal low-speed torque.
- if the controller limits phase current, then the slower (more turns) winding will have more low-speed torque (unless you can program the phase current to match the winding).
- the slower wind will be less demanding on the controller and the phase leads.
For a winding with
greater copper fill,
- the motor will generate more torque for the same amount of heating, or less heating for the same amount of torque.
- the motor will draw less battery current for the same torque and RPM.
- if copper fill differs only slightly (e.g. 65 vs 64 strands), then the points above regarding equal-fill windings still apply, if you replace "equal" with "nearly equal".
Now some details. First of all, when talking about current, be sure you are clear about whether it's
battery current or
phase current. We normally talk about battery current since that's easier to measure, but it's phase current that determines torque and resistive loss. The following is about phase current.
Equal copper fill. Suppose you have 64 strands of wire wrapped around a stator tooth, each with a resistance of 100 mOhm, and each is carrying 1 A. These 64 Amp-turns will generate a given flux in the stator, which will result in a given torque in the motor. Furthermore, there is (1 A)^2*(100 mOhm)*64 = 6.4 W of resistive loss.
The torque and loss in this scenario is
irrespective of the series-parallel combinations of these wire strands. It doesn't matter if you have 4 turns of 16 parallel strands, or 8 turns of 8 parallel strands, it only matters that each time a strand loops around the tooth, it is carrying 1 A of current.
It looks different from the outside, of course. The 4 x 16 winding will have a resistance of 100*4/16 = 25 mOhm, while the 8 x 8 winding will have a resistance of 100*8/8 = 100 mOhm. For the 4 x 16 winding, the controller will have to provide 16 A so that each of the 16 parallel strands gets 1 A, whereas the 8 X 8 winding only requires 8 A for its 8 parallel strands. But if you calculate I^2*R for these combinations, you'll see they're the same.
Greater copper fill. The 4 x 16 and 8 x 8 are the equal-fill case. Now let's look at the 5 x 13 winding, which wraps 65 strands around a tooth, or 1.5% more copper than the previous cases. Suppose you want the same torque as before, so you still need 64 Amp-turns of current. But now the current is distributed among 65 strands, so the current in each strand is 64/65 A. Now the resistive loss is (64/65 A)^2*(100 mOhm)*65. This works out to the 6.4 W from the previous case, reduced by the factor 64/65.
Remember, this is at the same torque, but the loss is reduced by the ratio of the copper fill.
