How do hub motor specs translate into real world power?

On BMSBATTERY's website they do a few versions of the BPM hub motor.

 

Bafang BPM 48V500W Rear Driving EBike Hub Motor. RPM: 393, Code: 10.

 

Bafang BPM 36V500W Rear Driving EBike Hub Motor. RPM: 393, Code: 8.

 

Now both of these have the same RPM.

My question: Would the 48v hub produce more torque?

I'm guessing that because it would spin faster than the 36v, then it would need to be geared down more than the 36v to match it's RPM?

Does this make sense

It can be done through different windings rather than gears, I don't know if one would be better than the other though.

Hi If you want GRUNT for climbing hill or if you are a heavy person 48 volts

 

You can wind the motors different we sell 7 windings 9.4 rpm per volt

 

we use this as standard

 

 

or 5 windings 12.5 RPM per volt we use this mainly if we fit to a 20 Inch rim

 

motor power is battery volts + amps that the controller And BMS will Handel

 

so 48 volts + 20 amp controller say 1,000 watts motor 80 % good at converting the watts to motive power so approx 800 watts on the road

 

WE now only sell 48 volt Kits as they work we limit the speed to 15 MPH Electronically

 

Frank

As the code 10 has the same 500 watts and the same RPM on a 33.3% higher voltage, that voltage gain will almost certainly be expressed as more torque. It will be differently "geared" either electrically or mechanically than the code 8.

Yes its all in the winding of the motor. Theres a good write up at ebikes.ca on this see: Information about Hub Motors

 

 

One argument for higher voltage comes from better efficiency due to lower losses in the motor, wiring and controller. This is especially true for high power setups IE: Taking an example by Jeremy Harris over on ES, if total system resistance is 60mOhm and you are pulling say 2000w at 48v (42amps) then your losses are about 106w (0.06 * 42amp = 2.52v * 42amp)

 

If the voltage was now 36v then the motor could pull 55.5amps meaning losses are now 185w and at 24v a whopping 416w.

So, you're saying that for a given load, the higher voltage results in more of the electrical energy going into forward motion? And from this it can be taken as meaning an increase in torque?

So, you're saying that for a given load, the higher voltage results in more of the electrical energy going into forward motion? And from this it can be taken as meaning an increase in torque?

 

More of the energy appears as driving force which is an increase in efficiency, but not necessarily in torque. For a given set of windings as in Jeremy's example, the result is more speed, not more torque.

 

To turn the gain into torque instead of speed, either the motor's internal gearing must be changed to give the same output revs as at the lower voltage, or the windings must be changed to produce that result electrically.