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I read that for DC electric motors you should generally use the motor with the highest "motor size constant" you can (motor size constant Km = Kt^2 / R, where Kt is motor torque constant i.e. N*m/A, and R is the resistance of the motor windings). Furthermore, Km is independent of motor windings since if you take a motor with 6 turns of parallel double wire and rewire it with 12 turns of single wire, this will double the Kt but quadruple the motor winding resistance by effectively doubling the length of the wire and halving the cross-sectional area. Thus if Km is the right evaluation criterion, then you should be indifferent between winding schemes for a given motor.

But this seems confusing to me in the context of a project I'm working on.

Lets say you have an electric vehicle and are trying to decide whether you should wind your motor with 6 turns of parallel double wire (6T-2W), or 12 turns single wire (12T-1W). You will choose your gearing so that your vehicle will have the same top speed regardless of winding. Since the Kv of the 6T-2W motor is twice that of the 12T-1W motor (Kv = motor velocity constant rpm/V, Kv = 1/Kt), if your gearing ratio for the 6T-2W setup is 1:1, your gearing ratio for the 12T-1W setup will be 1:2. At the wheels, the gearing perfectly offsets the effect of the winding scheme. Thus your vehicle speed and torque for a given applied voltage and current will be the same regardless of which winding scheme you use. However, the 12T-1W winding scheme will have 4x the winding resistance of the 6T-2W scheme, and will therefore generate more heat through copper losses and have a lower maximum current draw/torque output at the wheel. Doesnt this mean that even though the two winding schemes result in the same motor size constant Km, you should use the 6T-2W winding?

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  • \$\begingroup\$ You have probably glossed over a few factors and also not taken into account the relative cost or importance of some factors. You need to be sure that power in for all schemes is the same when comparing them. Obviously V & I & R interrelate. And with due design one scheme may allow direct drive and another need a gearbox or gearing/. Also gear trains of equal power throughput will have N times more torque in sections that operate at N times less speed affecting mechanical design aspects. There will often be a "best" solution BUT the measure of "best" will depend on circumstances. \$\endgroup\$ – Russell McMahon Jul 16 '15 at 7:34
  • \$\begingroup\$ No, you keep the gearing the same, and double the voltage with the 12T motor. This preserves similarity. P=V**2/R. The 12T motor has 4*R ... and 2*V. Now you can see that the turns are the primary winding of a voltage-motion transformer! Which winding you choose depends on the best (most efficient, safest, cheapest) source of voltage. \$\endgroup\$ – Brian Drummond Jul 16 '15 at 9:47
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Changing the windings does not affect anything fundamental about the motor. The motor will always produce its highest efficiency and power output at its highest speed, which is generally limited by mechanical constraints (flying apart, burning up). So regardless of the windings the motor uses it will have the same top speed and max torque characteristics. All that has changed is the effective operating voltage.

The way you would choose the windings is to ensure that at maximum required speed and maximum required torque the terminal voltage on the motor is less than (and normally roughly equal too) whatever your power supply is. If you are using a 12V battery, then you'll have less turns of thicker wire. If you are using a 230V supply, then you'll have lots of thin fine wires.

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