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Unread 21-05-2014, 13:27
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Re: motor Ke and Kt Quiz

Quote:
Originally Posted by Richard Wallace View Post
Paul -- this motor is your design, and I want to go on record saying you did a fine job. It is a worthy successor to the venerable Globe, and about three orders of magnitude more flexible for a range of applications powering FRC mechanisms. You may have noticed I ordered a few of them yesterday -- will share test data (and THEN my own answer to Russ's question) when I can.
Paul, did you design this motor for FRC? or was it a motor already in production (or not in production but available as a configurable item) that you picked out?

If you designed it for FRC, I'd love to hear more about the process.
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Unread 21-05-2014, 14:04
Rauhul Varma Rauhul Varma is offline
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Re: motor Ke and Kt Quiz

I’ll give it a try:

So from the given data we can write an equation relating current I to speed W:
I = (Istall-Ifree)(1-W/Wfree)+Ifree

We can also write an equation that relates T to W:
T = (Ts)(1-W/Wfree)

Kt = T/I, however if you divide these two equations you get Kt as a function of W, and this is because the equations have different roots. Theoretically they have the same roots as at T=0->I=0, but they do not because the equation for T only takes into account external loads on the motor and ignores the torques due to friction forces in the motor.

So the equation for T really is:
T = (Ts)(1-W/Wfree)-Tloss

Solving I=0 for W gets:
W(I=0) = Wfree(1+Ifree/(Istall-Ifree))

Then plugging that W into Tloss = (Ts)(1-W/Wfree) to solve Tloss gives:
Tloss = -(Tstall*Ifree/(Istall-Tfree))

Plugging Tloss back into T = (Ts)(1-W/Wfree)-Tloss gives us:
T = Ts(1-W/Wfree+Ifree/(Istall-Ifree))

Now since these two equations have the same root, we can divide them and get a constant:
Kt = T(W)/I(W) = -Tstall(Istall/(Ifree-Istall))/Istall

Plugging in for a bag motor yields:
Kt = 0.0102

——————

Ke is much simpler to solve for:
Ke = V/W

However you need to take into account the voltage drop through the motor’s windings at W, so we first need to solve for internal motor resistance which is:
R = Vspec/Istall

We now can rewrite Ke to include this drop:
Ke = (Vspec-I(W)*R)/W

Since we are given Wfree and Ifree we can plug those in along with R and get:
Ke = (Vspec-Ifree*Vspec/Istall)/Wfree

Plugging in for a bag motor yields:
Ke = 0.0078

As to why they are different… I would guess bad measurement, but its more likely I did the calculation wrong

Last edited by Rauhul Varma : 21-05-2014 at 14:41. Reason: Accidentally wrote Tstall instead of Istall
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Unread 21-05-2014, 14:22
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Re: motor Ke and Kt Quiz

Quote:
Originally Posted by Ether View Post
By all means Paul, please do jump in. I think there may be at least a small audience of interested students following the thread.

There is.
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Unread 21-05-2014, 15:59
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Re: motor Ke and Kt Quiz

Quote:
Originally Posted by AdamHeard View Post
Paul, did you design this motor for FRC? or was it a motor already in production (or not in production but available as a configurable item) that you picked out?

If you designed it for FRC, I'd love to hear more about the process.
A little bit of both. CCL, the manufacturer of the CIM motor, did all the detail design work. They did not have any motor like the BAG. In our VEXpro product discussions, before we even made one product, we all agreed that losing the Globe motor from the KoP was a big loss to FRC teams. We then finalized what we wanted from a power, torque at max power, and packaging perspective. I ran some quick numbers based on DC motor design formulas (I will get the exact book details that I prefer at another time b/c I loaned it out to someone recently) to get a rough idea of what was possible.

I then took the rough design to CCL and said, "construct this like the CIM motor, use all ball bearings, and hit these power and torque numbers". I was fairly certain they could do it, but haven't had to design a DC motor in a long time so left it to them.

But this definitely is a custom motor specific to the FRC application.

The MiniCIM was a similar process except we literally took a picture of a dismantled CIM and said make it only "this" long. I gave them torque and power numbers so it would match up 1:1 with a CIM and contribute the most power possible in the meat of the CIM speed profile.
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Unread 21-05-2014, 16:13
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Re: motor Ke and Kt Quiz

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Originally Posted by Gregor View Post
There is.
OK, I'll toss this out for discussion


Attached Thumbnails
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Last edited by Ether : 21-05-2014 at 18:02. Reason: corrected power equation
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Unread 21-05-2014, 16:15
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Re: motor Ke and Kt Quiz

So, for most small, high Flux DC motors there is this phenomenon called magnetic field saturation. It happens when the permanent magnet doesn't have enough field strength to drive the high amount of Flux through the moving parts of the motor. I am simplifying here to try to keep it short.

The bottom line is that at high current (aka, near stall), the magnetic field saturates producing less actual torque than ideal and this is one of the reasons for the difference between Kt and Ke.

However, it is not the only difference. This can really only be found with motor testing.
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Unread 21-05-2014, 16:35
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Re: motor Ke and Kt Quiz

From Ether's 3rd attachment, Ke = Kt + (other losses /Iw).
If there are "other losses" these will account for the difference.

Of course, there's a lifetime of study in "other losses"...
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Unread 21-05-2014, 16:36
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Re: motor Ke and Kt Quiz

Quote:
Originally Posted by Paul Copioli View Post
at high current (aka, near stall), the magnetic field saturates producing less actual torque than ideal...However, it is not the only difference.
Thanks Paul. I think that completes the explanation.

"Kt equals Ke" is a theoretical result which is only approximately true for real-world motors.

Firstly, the "linearity" (straight lines) assumption (notice the asterisks) used to compute Kt would not be valid when the magnetic field saturates.

Second, "other_losses" such as bearing friction, windage & viscous damping, eddy currents, and hysteresis in the power equation account for the difference between Ke and Kt.


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Unread 21-05-2014, 20:34
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Re: motor Ke and Kt Quiz

If Ke = Kt (which would be a perfect world scenario)
Then the max torque the motor could output would be:
Tstall = Ke*Istall

Which for a bag motor comes out to:
Tstall =~ .32 Nm

This is less than the .4 Nm in the motor specs, and given we have come to the conclusion that the real world Kt is always less than Ke, wouldn't the true stall torque of the bag motor be even less than .32 Nm?

(I also don't understand why our calculated Kt is larger than Ke or is that a result of the curve actually being nonlinear?)
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Unread 21-05-2014, 22:17
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Re: motor Ke and Kt Quiz

Quote:
Originally Posted by Rauhul Varma View Post
we have come to the conclusion that the real world Kt is always less than Ke
In the real world, Kt and Ke are not constant. So without enough data to compute them both at the same operating point, comparing them is like comparing apples and oranges.



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Last edited by Ether : 21-05-2014 at 22:39. Reason: added sketch
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