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| To you - I pledge my calculators, my codes, and my safety glasses. I'll be your Super Nerd forever! |
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#16
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Re: motor Ke and Kt Quiz
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If you designed it for FRC, I'd love to hear more about the process. |
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#17
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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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#18
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Re: motor Ke and Kt Quiz
There is.
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#19
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Re: motor Ke and Kt Quiz
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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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#20
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Re: motor Ke and Kt Quiz
Last edited by Ether : 21-05-2014 at 18:02. Reason: corrected power equation |
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#21
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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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#22
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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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#23
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Re: motor Ke and Kt Quiz
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"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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#24
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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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#25
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Re: motor Ke and Kt Quiz
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Last edited by Ether : 21-05-2014 at 22:39. Reason: added sketch |
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