Torque and RPM Graph for Castle Creation products ?

[VintageMA]

Thanks Patrick - great informative explanation!!

That along with the graphs also helps a lot explain the downside of being either very over-geared or under-geared. You're not running in the peak efficiency of the motor and more power is being turned into heat than mechanical energy.

Also helps to explain how using too big a motor for the application can be inefficient, and also how using too small a motor can also be inefficient but in the other direction. If you ask a small motor to put out too much power you then surpass it's eff% range and you start to get more heat and less mechanical output - where with the larger motor you get the same inefficient result but just because there's not enough load on the motor.

Also explains why you shouldn't be running your motors at no-load - eff% is very low there and almost all energy going into the motor gets turned into heat - why they heat up in your hand very fast when testing.

[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)?

[SpEEdyBL]

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)?

1. Torque is directly proportional to current (independent of voltage), unless (as described earlier) the stator is oversaturated. However, a good motor is designed so this cant happen within it's specified voltage range. Think of voltage proportional to theoretical rpm i.e. volts x kv, current proportional to torque, and efficiency at high loads approx. = actual rpm/(volts x kv).

2. The equation for % efficiency of a motor is (VI - rI^2 - I0)/(VI) where V= volts, I = amps, r=resistance and I0 = amps at no load. So if you plot the graph as % efficiency vs. amps, you will see that if you use a higher voltage constant, you will get a higher max efficiency. But keep in mind this is only a theoretical equation and at some voltage, the efficiency starts to get worse. This voltage where peak efficiency is the highest is known as the sweet spot of the motor.

[Pdelcast]

Originally Posted by SpEEdyBL

2. The equation for % efficiency of a motor is (VI - rI^2 - I0)/(VI) where V= volts, I = amps, r=resistance and I0 = amps at no load. So if you plot the graph as % efficiency vs. amps, you will see that if you use a higher voltage constant, you will get a higher max efficiency. But keep in mind this is only a theoretical equation and at some voltage, the efficiency starts to get worse. This voltage where peak efficiency is the highest is known as the sweet spot of the motor.

That equation only models electrical losses -- not magnetic losses (magnetic losses are lumped in with the IO constant...) Keep in mind that magnetic losses are non-linear (especially at high power.) Also, the IO constant needs to be measured at the same RPM as the output -- because other losses (like windage) are also non-constant (windage losses are RPM cubed...)

[othello]

Great information. Thanks to both of you.

>Pdelcast: It would also be the same if you doubled the number of poles and kept the winding the same...

Meaning: The simple saying that a 2 pole motor "has" more torque than a 4 pole motor is simply wrong giving the same windings and rotor size. This is what i thought as you can not pack more magnetic material on the same rotor.

[SpEEdyBL]

Originally Posted by Pdelcast

That equation only models electrical losses -- not magnetic losses (magnetic losses are lumped in with the IO constant...) Keep in mind that magnetic losses are non-linear (especially at high power.) Also, the IO constant needs to be measured at the same RPM as the output -- because other losses (like windage) are also non-constant (windage losses are RPM cubed...)

Thanks Patrick. Whoops, forgot to multiply I0 by V, but oh well, its only a theoretical equation that will result in a graph similar to the one posted. I knew there were other variables of some sort. Otherwise it makes feigaos seem really efficient!

[Pdelcast]

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)?

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?