# combined free speed of 3 motors

*The combined stall torque and stall current of 3 motors (or combination of motors and gearmotors) is simply the sum of the stall torques and stall currents, respectively.

But what about free speed? If anyone has published the math for this, I didn’t find it with a quick search.

So I took a crack at it. Comments?*
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I’ve been wrestling with this problem this week as well. I’ll be following along closely and will give it some more thought soon.

These topics and posts are very ‘math’ inspirational. Too bad FIRST

did not exist in the '70s. I could have used the inspiration. Thank you Ether.

I’m not sure I understand the question. Are the motors tied together in a gearbox or other connected load? If so, there isn’t really a free speed. Free speed means there’s no load on the motor. Ganging them together in a gearbox means that any motor that isn’t trying to spin as fast as the others (based on voltage and its free speed) will present some load on them. The maximum speed of the system, barring additional loads, will be somewhat less than the lowest free speed of the three. How much less will be determined by friction in the system. Whenever you’re ganging motors together, you want their free speeds to be as closely matched as possible, taking into account any difference in gearing.

If the motors are not tied together, each will have a free speed specified by its motor performance data.

Yes.

If so, there isn’t really a free speed. Free speed means there’s no load on the motor.

Free speed in this context means “no load on the output shaft of the gearbox” (and ignoring friction in the gearbox with no load on it).

The maximum speed of the system, barring additional loads, will be somewhat less than the lowest free speed of the three.

False. It will be somewhere between the slowest and fastest.

Whenever you’re ganging motors together, you want their free speeds to be as closely matched as possible, taking into account any difference in gearing.

The equation shown works whether the individual free speeds are matched (with a speed reducing gearhead on the faster motors) or not. If speed reduction is used on any of the motors, then use the Free Speed and Stall Torque of the gear+motor for that motor.

Sorry, I should have said somewhat less than the highest free speed of the three.

Correct. And the equation provided allows you to calculate that value.

I can see how the concepts of free speed, stall torque, stall current, and free speed current can all apply if you only consider specific voltages for each. However, I can’t seem to figure out what the motor velocity or motor torque constants mean in the context of having multiple motors, as you will have multiple voltages (ie, you can change the voltage being applied to each motor independent of the voltage being applied to the other motor). But I went ahead and threw in the numbers for getting a reasonable model of a gearbox with two motors (adding a third would just make equations messier).

two_motor_model.pdf (80.1 KB)

two_motor_model.pdf (80.1 KB)

The derivation I provided in the original post applies if the same voltage is applied to all the motors, and the individual motor stall torque and free speed values are all at the same spec voltage. The resulting “combined” Free Speed (together with the simple-to-calculate Stall Current and Stall Torque) allow you to view the combination as equivalent to a single motor with those characteristics. Treating the combination as equivalent to a single motor allows you to use all the single-motor math you are already familiar with.

However, using all that single motor math will cause you to result in suboptimal results (if you attempt to optimize in any way, beyond just modelling the system). Also, has anyone gone through the proof that you can validly treat the two motors as one, given that you input the same voltage?

I’m having trouble extracting an interpretation of the above. Could you explain what you mean? Perhaps give an example?

Also, has anyone gone through the proof that you can validly treat the two motors as one, given that you input the same voltage?

I thought that’s what I did in the attachment to the original post. At what point in the derivation do you disagree?

To be clear, the derivation assumes the motors will be used to drive a load (like a drivetrain or heavy arm for example) with a reflected inertia much larger than the rotor inertia, so the dynamic contribution of the rotor inertia can be ignored.

If you had two different motors, both attached to something (perhaps a shooter wheel), and you wanted to create an optimal controller for the wheel (given some cost function), then I doubt (although I may be wrong) that that optimal controller would have you always apply the same voltage to the motors. Hence, it is suboptimal to make this simplification, although there certainly may be situations where it is convenient, and therefore valuable, to make the simplification (such as if you were limited for space on your Digital Sidecar and wanted to be able to use a y-cable to control two different motors).

I see how you would be driving a large load (in my equations, the moment of inertia was primarily a place holder for such a load), but I do not see how a derivation of a formula for the free speed logically leads to the idea that all the single motor formulas must be valid.
I agree that you can calculate a stall torque, stall current, free speed, resistance, and free current for the motor combination, but just because those numbers may be meaningful does not necessitate that the single motor equations are true.
In other words, just because you have formulas for combined stall torque, stall current, free speed, resistance, and free current, why is it that these equations are still valid for values of V, I, omega, and Torque other than those at stall torque and free speed:
V = I*R + omega / Kv
Torque = Kt * I
I may be missing something, but I have not seen any proof that these equations continue to hold, and I like to see proofs :).

Ether,

Joe Johnson and I worked through this for two motors some time ago. Joe had a post on it back in either 2002 or 2003. His terminology and mine were different, but recently I made sure our two analysis methods came to the same conclusion. They did and yours matches our conclusions, but adds one more motor.

We did this analysis to try to kill the myth that you had to match free speeds. Joe went into detail about loading conditions that will cause one motor to actually be pushing the other, therefore the slower motor would contribute negative work to the system. Again, your analysis corroborates our conclusions back then.

I attached my hand calculations to prove Joe and my analysis methods were the same. I now use Joe’s terminology because it is easier to put in Excel.

*Many teams that put 3 motors on a drivetrain gearbox want to know the acceleration characteristics when full voltage is applied to all 3 motors. Finding the single-motor “equivalent” motor curve permits the use of existing spreadsheets and models that don’t support multiple mixed-motors on the gearbox.

I may be missing something, but I have not seen any proof that these equations continue to hold, and I like to see proofs

Go through the attachment to post#1 step-by-step and tell me at what step you disagree or are unconvinced, and I will explain the justification and make explicit any tacit assumptions.

Additionally, we have been using this model for several years to determine our robots’ acceleration profile and time to distance at different spots of the field. The analysis has routinely been within 5% of actual robot performance.

Thanks Paul. I thought I remembered seeing something like that, but couldn’t locate it.

yours matches our conclusions, but adds one more motor.

Notice the nice symmetry in the solution. It’s easy to see how to extend this to N motors, instead of 3.

Perhaps it will help to present the math in a slightly different way. Attached is a derivation for the torque vs speed curve for the 3-motor combination directly from the torque vs speed curves for the individual motors.

It’s straightforward algebra. It follows naturally from the assumptions that 1) all three motor speeds are the same (since they are mechanically linked) and 2) the individual motor torques are additive.

As you can see, the combination behaves exactly like a single motor with torque = Tstall*(1-Speed/Sfree), where:

Tstall = T1s + T2s + T3s … and … Sfree = (T1s + T2s + T3s)/(T1s/S1f + T2s/S2f + T3s/S3f)
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*

I bookmarked this post the last time this came up, just so I could reference it quickly anytime I needed to:

http://www.chiefdelphi.com/forums/showpost.php?p=24239&postcount=3

This is definitely the right step getting to where I want to be, however, although I do see the linear relationship between net speed and torque, I do not see how that necessarily allows us to use the single motor equations (V=IR + v/Kv and I = KtI) for the combined motors. I have worked out some of the equations and attached the derivations. Hopefully you will be able to see where I am going and can help get it that last few steps.

multiple_motors.pdf (83.8 KB)

multiple_motors.pdf (83.8 KB)