Torque and RPM Graph for Castle Creation products ?

Ok I'm following it now.I thought what I said sounded a little funny.You'll have to forgive me,I can't tell the difference in those colors to good. :)

[What's_nitro?]

I see where I was wrong. The speed only drops about 3500 RPM. I'm sorry if I threw anyone off. After Patrick's explanation I can see how the graph represents % Eff. under varying load. You can take that RPM trace and figure the actual kV under that load, since I assume the 16V is regulated...

[Pdelcast]

Originally Posted by What's_nitro?

I see where I was wrong. The speed only drops about 3500 RPM. I'm sorry if I threw anyone off. After Patrick's explanation I can see how the graph represents % Eff. under varying load. You can take that RPM trace and figure the actual kV under that load, since I assume the 16V is regulated...

Yeah -- the 16.0V is regulated from a Sorensen DHP80-150 (0-80V, 0-150A) power supply. That's our main Dyno supply.

[What's_nitro?]

Originally Posted by Pdelcast

Yeah -- the 16.0V is regulated from a Sorensen DHP80-150 (0-80V, 0-150A) power supply.

[brushlessboy16]

what esc was this dyno run conducted on?

[Pdelcast]

Originally Posted by brushlessboy16

what esc was this dyno run conducted on?

A Phoenix-125 set to low timing.

[brushlessboy16]

thank you

[VintageMA]

I know this is an odd follow-up to this graph, and I understand the multiplication of the torque at the motor through the gearing to the wheels, but I have a follow-up as to how we can apply this to real setups?

Is there a way to calculate how much torque an application will need for a certain amount of acceleration or top speed? Obviously you would want to pick a motor where the torque your application needs doesn't fall off the top of the efficiency scale where it starts to drop back down from max eff%.

Also - I am confused with the oz/in ratings when comparing to servos. Picking a servo for an application we see ratings in the 125oz/in all the way up to 400oz/in. I know this already takes into account the reduction gearing in the servo and there is also much the throw of the servo compared with a spinning motor. But, when I look at that graph and see ratings for 25-30oz/in of torque for these huge motors is just seems very small in relation to a servo putting out 5-20 times more torque.

Does where I am going make sense? Can anyone help explain the relation a bit more?

[Pdelcast]

Originally Posted by VintageMA

I know this is an odd follow-up to this graph, and I understand the multiplication of the torque at the motor through the gearing to the wheels, but I have a follow-up as to how we can apply this to real setups?

Is there a way to calculate how much torque an application will need for a certain amount of acceleration or top speed? Obviously you would want to pick a motor where the torque your application needs doesn't fall off the top of the efficiency scale where it starts to drop back down from max eff%.

Also - I am confused with the oz/in ratings when comparing to servos. Picking a servo for an application we see ratings in the 125oz/in all the way up to 400oz/in. I know this already takes into account the reduction gearing in the servo and there is also much the throw of the servo compared with a spinning motor. But, when I look at that graph and see ratings for 25-30oz/in of torque for these huge motors is just seems very small in relation to a servo putting out 5-20 times more torque.

Does where I am going make sense? Can anyone help explain the relation a bit more?

Power is torque times RPM -- so to get horsepower, you actually take torque in oz/in, and multiply by the RPM. If a servo can make 120 degrees in .2 seconds (for example) and maintain 200oz in of torque (which servos CANNOT do -- they measure torque at stall (0 RPM)) -- Then you would get 100 RPM at 200 oz/in.

To find horsepower, multiply the RPM by the oz/in and divide by 1 million. So for a servo, that would be 100RPM * 200ozin / 1000000 -- or 1/50th horsepower. However, the best servos right now can't sustain that kind of output for very long.

Now, take a motor like the one in the graph -- it's turning 33,500RPM at 30oz-in of torque. 33,500 * 30 / 1000000 = 1 horsepower, or 50 times more power than the servo's motor is outputting. NOW, that said, that same motor can also output 240oz/in at about 15,000 RPM (240 * 15000 / 1000000 = 3.6 horsepower) on 16V.

Now to illustrate why RPM is important -- Take that motor (at 30ozin/33,500RPM) and put it in a servo. If we add gearing to bring the RPM down to about 100 RPM (similar to a good servo) we would multiply the torque by the same ratio as we divide the RPM. So, 33,500 / 100 = 335 (gear reduction.) Torque would then be 30ozin * 335, or about 10,000 ozin.

So, in the same application (RC Servo) a 1515/1Y motor would generate about 10,000 ozin of torque at 100RPM, and about 160,000 ozin of torque (about 800ft/lbs) at stall.


Make sense?

[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]

Keep in mind that where the efficiency @ low loads is poor, the motor is drawing very little current, and therefore isn't making much heat. For example even if it's 50% efficient at a 10 amp load, and 80% efficient at a 50 amp load, then you are still making twice as much heat with the 50 amp load. That said, a good motor shouldn't overheat at low loads.

[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 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?

[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...)