Quote:
Originally Posted by othello
Thanks Patrick for sharing this graph with us. Just saw it and a few questions came to my mind.
. When multiplying 16Vx2200kv you will get 35200rpm. Your graph starts above 36000rpm which would correspond to a 2250+kv motor at 16V. I guess this is due to slight input Voltage variations and/or that a 2200kv motor will never be exactly a 2200kv motor.
. I must admit that i badly missed an Amperage curve on this graph ;-) First it would have been interesting to correlate Amps to torque. And secondly it would have shown the Amp range at which efficiency is at its best.
Which leads me to a few theoretic questions.
If one changes input voltage to say 14,8V (4), 18,5V (5) or 22,2V (6s):
. Does torque stay at the same level if you feed this motor with a given Amperage (say 20A) for all 3 voltages?
. How will the efficiency curve shift with all 3 voltages? Can one say that when you apply higher voltage, the efficiency curve will rise earlier with lesser Amps and might fall off earlier with higher Amps when compared to lower voltage applied?
And there is one thing which always intrigued me when it comes to voltage / wind choice for a certain motor. Is there a slight advantage for lower wind motors when it comes to absolute high power levels. And if so ... why? Better copper filling? Can one really expect the same power output from a Castle 1515/1Y (2200kv) and maybe a future Castle 1515/2Y (1100kv)?
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AHHHA Speedy beat me to it. ^^^^
Hi Othello,
Yeah, the 2200Kv motors vary a bit from about 2200 to about 2260.
Let's see -- Amperage is not graphed, but can be calculated from torque, RPM and efficiency directly. Torque (oz in) * RPM / 1008400 = Horsepower, 1 Horsepower = 746 watts, divide by efficiency to get INPUT watts, then divide by 16.0 to get amps.
So, for example 33,500 RPM and 30 oz in torque = about 1 hp -- 746 output watts / .85 efficiency = 877 input watts, for about 55 Amps input.
Torque and Amps are fairly proportional through the flat area of the efficiency curve. (This is why motors are often rated at oz in / amp)
The voltage Vrs wind is exactly how you might think. A 2 turn (1100Kv motor) will have pretty much the same efficiency, input and output power as a 1 turn (2200 Kv) motor at the same output torque and RPM. The 2 turn motor will require twice the voltage, and half the amperage of the 1 turn, but will also have approximately four times the winding resistance. (It would also be the same if you doubled the number of poles and kept the winding the same...) It's all the same in the end -- what matters most is the motor size and efficiency -- that really governs output power, not the windings or number of poles. Make sense?