Least strenuous rpm for motor and controller???
- Thread starter tengwa
- Start date
I have the 900w cyclone on one of my bikes and i am sure that one has an internal temp sensor, so i imagine yours will have too but not 100% about this. You will find best efficiency and lowest heat generation comes with using mid to high rpms rather than pulling high currents at low revs which will generate loads more heat, i have found with the cylcone though that you can ride it pretty much how you like, as ours, the 900w, has a reduction gearbox built into it so it tends to rev quite high all the time.
Simon.
Simon.
Idontwanttopedal
1 kW
I have the 1200 with gearbox it dose have over heat cut off and seems happier at high rpm
Farfle
100 kW
Any motor (barring certain motors like the turnigy outrunners that start heating due to eddy current losses at high rpms) will be happiest at full unloaded speed.
Farfle said:
all motors, especially the iron statored ones we use get heating and eddy current losses from rpm's, even the magic pie hubbies Farfle -ive graphed it
prolly just that it happens lots with high kv rc motors cos revs come so easily and for so little money.
would a motor not be happier with just a bit of load so that it produces good power at peak efficiency, say 90% of its unloaded speed.
surely an unloaded motor is somewhat unhappy with all that effort thrashing around for the production of zero power..
-but yea, 80-90% of the unloaded speed is generally a sweet spot.
I'm guessing that a motor would be happiest at peak efficiency. That's the point where the copper losses and parasitic losses are at parity. With a slow speed motor like a hub motor, hysteresis losses make up most of the parasitic losses (run at design speeds, of course). Hysteresis losses go up lineally with rpm but eddy current losses go up with the square of rpm and so, as rpm increases eddy current losses become more significant until they dominate the parasitic losses.
This is the way that I've rationalised the maximum efficiency point:
If you apply a constant input voltage to an unloaded motor the rotor accelerates until the generated torque falls to a level where it balances the parasitic torque. At that point, efficiency is zero.
If you then load the motor very slightly, the speed/phase volts fall until the torque again balances the load applied. Efficiency will have increased because you now have a torque output but parasitic losses are still dominant, so efficiency is still low.
Now increase the load to the point where the active/copper losses are at parity with the parasitic losses. This is the maximum efficiency point. Why? Because beyond it active/copper losses become dominant. Torque is proportional to phase current (ignoring saturation effects) but losses increase by the square of phase current.
Anyone take issue with this?
If you apply a constant input voltage to an unloaded motor the rotor accelerates until the generated torque falls to a level where it balances the parasitic torque. At that point, efficiency is zero.
If you then load the motor very slightly, the speed/phase volts fall until the torque again balances the load applied. Efficiency will have increased because you now have a torque output but parasitic losses are still dominant, so efficiency is still low.
Now increase the load to the point where the active/copper losses are at parity with the parasitic losses. This is the maximum efficiency point. Why? Because beyond it active/copper losses become dominant. Torque is proportional to phase current (ignoring saturation effects) but losses increase by the square of phase current.
Anyone take issue with this?
John in CR
100 TW
A properly sized and geared motor is always happy if you use it every day. Controllers are happiest as close to full duty cycle as possible, so the motor's BEMF is limiting the current not the controller limiting current by turning the flow on and off more often. My motors and controllers are happiest and least strained cruising WOT down a very slight grade. 
John in CR
100 TW
Miles said:
I don't understand why that has to be the point of peak efficiency. Peak efficiency occurs where the % copper losses+iron losses is lowest, but why would copper losses and iron losses need to be equal at that point? Aren't their individual curves different?
Yes, but that's where they cross.John in CR said:
That is the point (increasing the load) where direct losses start to dominate over (relatively) fixed losses. So, efficiency can only fall from there, because efficiency falls with generated torque. Above the parity point, 'fixed' losses start to dominate, so efficiency can only fall, too.
Dave_S
1 mW
tengwa said:
That sounds like your calculation was for the RPM of the output shaft? IIRC the Cyclone motor gearbox reduction ratio is 9.3:1, so motor RPM would be 6789 RPM if all my guesses were correct?
since i still don't know the rated rpm of the 1200w kit im thinking... wouldn't the motor be happiest with the gear that gives the rpm that requires the least turn on the throttle on a given speed.
Because if the rpm is too low or too high it will need more throttle to reach the given speed and then the ampere will be higher... or am i lost?
Because if the rpm is too low or too high it will need more throttle to reach the given speed and then the ampere will be higher... or am i lost?
liveforphysics
100 TW
tengwa said:
Lost.
Re-read what Miles wrote.
In fact, search all posts by Miles and read them. You will conclude your reading borh understanding your question and the answer.
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