Control Moment Gyro - MOI calculation & units of measure

[Cosmoneer]

Hi all,

I am not a physics major and have spent some time learning via the university of Google. I would really appreciate a second set of eyes on what I am calculating and tell me if I am correct, or if I have done something wrong along the way. I am trying to calculate MOI and need results in milliNewtonmeters. After that, since I am working on a small Control Moment Gyro, I need to calculate the velocity producible by the CMG gimbal rates and the RPM of the gyro wheel.

I have a gyro flywheel with a mass of 8.84 grams @ radius of 8.55mm. As I understand it, one of the formulas used to calculate MOI is simply

I=MR^2
(I = moment of inertia, M = Mass, R is radius to center of mass)

So...

I = 8.84 * (8.55)^2
I = 8.84 * 73.1025
I = 646.2261 gmm^2

At this point I am confident in my result, but now I need to convert this into mNM. I found I needed to first convert to KgM^2, so I used this converter:

http://www.endmemo.com/sconvert/gmm2kgm2.php

Which is basically:

I = 646.2261 * 0.000000001
I = 0.0000006462261 KgM^2

Then to convert to NM I found that I need to multiply by a simple conversion factor:

I = 0.0000006462261 * 0.101971621298
I = 0.0000000658967231420834778
I = 6.589672314208348e-8 Nm

And then convert to mNm:

I = 6.589672314208348e-8 * 1000
I = 6.589672314208348e-5 mNm
I = 0.0000658967231420834778 mNm

At this point I feel rather confident that I haven't made a mistake. It's this next part where I get fuzzy.I now need to know how much torque I will derive from the CMG when I apply torque on the Gyro via the gimbal. From what I can find, the formula for that is this:

L = I * w
(L = torque, I = Moment of Inertia, w = angular velocity)

So, I found that w can be represented as:
w= RPM * 0.10472

This gives radians per second of the gyro wheel. Which, if given an RPM of 1000, I would get this:

L = (0.0000658967231420834778mNm) * (1000 * 0.10472)
L = (0.0000658967231420834778mNm) * (104.72 rads/sec)
L = 0.0006900704847438981795216 mNm rads/sec

Does this looks ok? What I don't see is how much gimbal angle produces this? What else am I missing?

Thanks for any input and or insight you provide!-Andrew

 

[Dr.D]

This is not going to turn out well in the direction you are headed!

The expression that you have for I = M*R^2 where R is the radius to the CM is simply wrong. We should presume that your gyro is well balanced, so the CM should be extremely close to the axis of rotation, therefore, R is approximately zero for your expression.

In actual fact, the MMOI depends on the shape of the rotor. Without knowing what that shape is, there is little that can be said to help you other than the definitiion:
I = int(r^2 dm) taken over the entire mass of the rotor.

Also, look a the units. You say you want units of milliNewtonmeters, but this will never be appropriate for the MMOI. The units for I in the SI system are kg-m^2. You can divide by 1000 or not, but the general form will remain unchanged.

 

[Cosmoneer]

Hi Dr. D, thanks for the reply!

I am sorry I didn't clarify the shape of the mass and what specific MOI I am calculating (linear vs. angular.) For discussion purposes the shape of the rotor is a perfectly balanced ring (the actual rotor is made of multiple rings on different planes, but that is a different discussion.) I used I=M*R^2 because that is what is suggested for calculating MOI for a spinning ring mass.

http://www.pstcc.edu/departments/natural_behavioral_sciences/Web%20Physics/Chapter09.htm
(About two-thirds down the page)

Moment of Inertia of a Thin Ring:

The mass M at radius R (as shown above) can be distributed uniformly along a thin ring as shown below. Its mass moment of inertia ( I ) will still be given by

I = MR^2.

You are suggesting I leave my results as kg-m^2? Trust me, one less conversion is better for me. Heck, I was tempted to keep it at g-mm^2 because the units are so small. I only converted to mNm because Nm is what they use for space applications (see example link below.)

http://www2.l-3com.com/spacenav/pdf/datasheets/CMG.pdf
(2nd Page of PDF)

Angular Momentum: 4760 Newton-meter-seconds (Nms)

-Andrew

 

[SergioPL]

You are suggesting I leave my results as kg-m^2? Trust me, one less conversion is better for me. Heck, I was tempted to keep it at g-mm^2 because the units are so small. I only converted to mNm because Nm is what they use for space applications (see example link below.)

NM has units of kg-m^2/s^2, haven't you forgotten a squared frequency term (w²) in your formulae?

 

[Cosmoneer]

Thanks SergioPL & Dr. D. Clearly I missed that one.

https://en.wikipedia.org/wiki/Newton_(unit)

1 N = 1 kg ⋅ m/s²

I will post an update to my calculation to make sure the path to my answer is correct.
So please, help me put a bow tie on what I understand then...

• The MOI ("I") determines how much mass is involved during one full revolution (radian) of the rotor.
• The Angular velocity ("w") determines over what period the inertial mass is available to harness torque from its spin, usually represented by rads/sec, or in my case, RPM.

So if the rotor is precessed for 1 second, then I will yield an output torque (/s², due to the introduction of time through the application of precession), based on the speed of the rotor.
Therefor, if the spinning gyro rotor is SLOWLY precessed through an angle of 1 degree, that determines what direction and magnitude the torque is output.

 

[Dr.D]

While the formula given would correctly express the MMOI for a thin ring, it is not possible to make the rotor of a real gyro in the form of a thin ring. There must be some connection to the axle, and that means more mass must be involved.

 

[Cosmoneer]

Hi Dr. D,

You are right, a thin ring alone will not have a connection.

The rotor in my particular case is basically a cylinder with the bearing support end being closed in enough to hold the bearing on a shaft, but there is no solid cylinder to speak of. In this case, I had considered using the formula:
I = ¹/₂ * ( M * (ID² + OD²) )

As stated before, the mass I am dealing with is just under 9 grams. I placed both formulae into a spreadsheet and the difference is not very much, but the formula above yields the higher result, about 24gmm² more.

 

[Dr.D]

Is this rotor a hollow cylinder with both ends closed, one end closed, or something else?

Assuming that it is a hollow cylinder with at least one end closed, you need to compute the MMOI as the sum of terms for the cylindrical shell and separately for each end closure. It should make a difference.

To keep the units straight, I strongly suggest you go strict SI. Thus put the mass in kg, the radius in m, and the result will be in kg-m^2. It will greatly help in getting your eventual final calculations correct.

An aside: Many years ago while I was teaching, I spent a summer working for IBM in a printer development group. I needed to the MMOI for some small printer parts for my work, so I sent the parts to the lab where they had experimental measurement capability for such things. The results came back as some numbers with a unit I had never previously encountered: moiss, pronounced "moise." I asked about this and was told that this was the unit they always used for these small parts. It means milli-ounce-inch-second^2 which is in fact a dimensionally correct MMOI unit. Only at IBM!

 

[Cosmoneer]

Hi Dr. D,

The rotor itself is a single precision piece with numerous shapes on it, all of them open cylinders in some form or another. You just confirmed a suspicion I had about the solution I arrived at a couple of days ago.

Lets start with the main central cylinder. It is the widest and heaviest of the pieces. It is capped on one end with an open disk. This disk is then capped with a small cylinder with male threads. On the opposite end of the main cylinder is a a short/thin hoop with a slightly larger diameter than the main cylinder. Inside the main cylinder, feeding off the end-cap plate is a smaller diameter open cylinder that runs parallel with the main cylinder. The ID of this cylinder is the same ID as the plate and the threaded end piece. This inner cylinder is what the rotor bearing mates with.

What I did was weigh the entire rotor to obtain its mass. I then one by one, using a Dremel, I slowly & destructively removed each shape from the whole, weighing the remainder each step of the way. I removed them in this order:

1> Threaded End
2> Outer Hoop
3> Main Cylinder
4> Inner Cylinder
Remainder: Open End CapThe end cap and the hoop are the only two items that seem more "hoop-like" than cylindrical. But for calculation purposes, I eventually treated them as open cylinders, too.
Thanks for the suggestion to start with the Units of Measure I want to end with. Again, that makes perfect sense!-Andrew

 

[Dr.D]

I'm sure it is all quite clear to you, but I am unable to confidently follow your description to the point I think I know what this rotor looks like. Can you post a sketch, please? That would help a lot.

 

[Cosmoneer]

Dr. D,

The picture on the right has been doctored, as I don't have a disassembled motor handy to take a picture of. It is basically an outrunner brushless motor. the bearings are in the middle.-Andrew

 

[Dr.D]

I presume that the shiny silver object is an intact rotor? On that basis, it looks to me like you have at least five separate zones to consider:
1. Outer ring at the bottom
2. Upright cylinder
3. End cover disk, without the center
4. Ring inside the end cover disk
5. Inner threaded center solid cylinder.
I would suggest that you calculate the MMOI for each part separately, then add them up. If your dimensions are accurate and your mass values, you should be pretty close to the truth.

 

[Cosmoneer]

Dr. D,

Thanks for the guidance! I will go that route.-Andrew