 # Motor moment of inertia

I presented an experimental method to identify moment of inertia at the IndianaFIRST / Purdue FIRST forum this past fall.

Attached are moment of inertia results for the CIM and MiniCIM motors, the presentation, and a reference from the University of Colorado.

Has anyone tried a similar approach?
Has anyone tried a different approach?

Thanks!!

Inertia results 141224.xlsx (22.2 KB)
Purdue Inertia presentation.zip (2.41 MB)
M4Fall04 TorsionalPendulum.pdf (77.1 KB)

Inertia results 141224.xlsx (22.2 KB)
Purdue Inertia presentation.zip (2.41 MB)
M4Fall04 TorsionalPendulum.pdf (77.1 KB)

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The inertia measurement method described here is similar to the one used in many motor labs, including mine. The U. of Colorado instructional lab procedure cited is also similar to the one taught at most universities, although I recall it was a Physics lab when I was at Georgia Tech.

A simple calculation of rotor inertia using standard formulae would be useful to compare against the experimental results. Kollmorgen still supports a nice tool for that: link

I think the influence of motor rotor moment of inertia on limiting FRC robot drive train acceleration is very small compared to the mass of the robot. In round numbers, the FRC robot with 4 CIMs can reach ~10 ft/sec in about one second if there is no counter force. The CIM motor can reach full speed much faster than that, accelerating only its own rotor inertia – might take 5 msec or so. I hope Ether is following this. I think he posted on a closely related topic a while back.

*The equations on page 21 of the presentation don’t appear to agree with the equations on page 20.

Using the 7.754E-05 kg*m^2 value for Jrotor for CIM from the spreadsheet to compute the equivalent mass at the vehicle I get ~22.8 pounds for a vehicle geared for 5 ft/sec (vehicle speed at CIM free speed of 5310 rpm).

For a 4-CIM drivetrain, that would be ~91 lbs. Doesn’t seem right.

I appreciate the inquiry. Algebra always seems to be a challenge.
Today I started with the equation from page 20 and the assumption that I had improperly derived the equations on page 21.

Attached is the work.
Based upon this,I believe that I correctly provided the equations on page 21.

I have been working on a mobility simulation, in excel.
I do not have confidence in this work - perhaps this year we can do some testing to perform validation. I have also attached this work.

I have focused on reflecting the system mass as a motor shaft inertia - as opposed to reflecting the motor shaft inertia into a system mass. Assuming a 120 pound robot, 4 CIMs, 6 inch wheels, and a 12.75:1 toughbox; I believe that the motor inertia represents 13.5 percent of the overall system inertia reflected to the motor shaft.

I quickly ran the opposite computation, and I believe that the motor appears as ~2.2 kg mass, or 8.8 kg out of 54.4 kg.

Explanation of Page 21 equations.docx (31.9 KB)
Mobility sizing 141226.xlsx (2.61 MB)

Explanation of Page 21 equations.docx (31.9 KB)
Mobility sizing 141226.xlsx (2.61 MB)

I agree. I should have been clearer. My intended meaning was that the equations on page 21 do not obviously follow from direct substitution into the J[sub]2[/sub] = J[sub]1[/sub] * (T[sub]2[/sub]2 - T[sub]1[/sub]2) / T[sub]1[/sub]2 equation on page 20.

Here’s another way to derive the necessary equations.

I get the same numbers, using both m = J*(g/r)2 and m = J*(ω/V)2

The second equation clearly shows the inverse square relationship between equivalent mass and the vehicle speed at motor free speed (due to the combined effect of gear ratio and wheel diameter). Reducing the vehicle free speed from 10.9 to 5 fps changes the equivalent mass from 2.2 kg to 10.3 kg. For a 4 CIM drive that would be about 91 pounds!

These numbers are surprisingly large. But you appear to have taken great pains to conduct the test very carefully. Has anyone else done rotor inertia testing?

m=J(omega_div_V)^2.xls (14 KB)
m=J(g_div_r)^2.xls (14 KB)

m=J(omega_div_V)^2.xls (14 KB)
m=J(g_div_r)^2.xls (14 KB)