# Valid Inertia Tensor

1. Jul 22, 2008

### KLoux

Hello,

Recently I was using multibody dynamics software at work and when I ran the simulation it warned me that the values that I specified for the inertia tensor of one of the bodies was physically impossible. It turned out that I had mistyped one of the values and it was an easy fix, but it got me thinking - how did it know?

After some searching I found a source (which I cannot remember now) that said the sum of any two principle moments of inertia for an object must be greater than the third. After some testing of different shapes I think that this is true, but why? Is there a derivation anywhere? None of my textbooks mention this at all and the one source that I did manage to find didn't provide an explanation.

-Kerry

2. Jul 22, 2008

### arildno

Hmm..this is one of Euler's original results, I think.

3. Jul 22, 2008

### robphy

4. Jul 22, 2008

### Crosson

IIRC, this is a consequence of the fact that the trace of a matrix is equal to the sum of its eigenvalues.

5. Jul 22, 2008

### D H

Staff Emeritus
Two exceptions to the rule: A point mass and an infinitely thin rod. A point mass has a null inertia tensor. The inertia tensor for an infinitely thin rod about its center of mass is a degenerate inertia tensor in which one of the principal moments is zero and the other two are equal to each other. The test fails for both of these cases, but neither case is real.

The eigenvalues of the inertia tensor of a set of three non-colinear point masses about the common center of mass will obey the relation 2*max(lambdai) < sum(lambdai). This is the basis for the relationship. If you somehow conjoin two objects that obey the relationship, the inertia tensor for the composite object will also obey the relationship.

6. Jul 23, 2008

### KLoux

Thanks for the link, but I must be missing what I'm supposed to be seeing?

What does this mean physically? If you're using principle axis as your reference frame, then the sum of the principle moments of inertia is equal to the sum of the principle moments of inertia? Can you elaborate?

Does this mean that the test is actually 2*max(lambdai) < sum(lambdai)? Reading your post made me remember that I forget to mention - I think it is important that you are using the CG of the object as your reference. You can sometimes pass the test even if your inertia tensor is not physically valid if you're using some other frame.

Thanks for all the replies!

-Kerry

7. Jul 23, 2008

### D H

Staff Emeritus
Your test is lambdai < lambdaj+lambdak. Adding lambdai to both sides of the expression yields 2*lambdai < sum(lambdaj), where j=1,2,3. This is trivially true for all but the largest eigenvalue. Hence your test is the same as mine.

8. Jul 23, 2008

### KLoux

Yes, yes... I should have actually looked at your equation - thanks for pointing this out

Ah, the first two hits are links to a paper that contains a proof! I missed them because the paper isn't free (but luckily I still have friends at a university with a subscription to the American Journal of Physics). The generalized perpendicular axis theorem was exactly what I was looking for! Thank you robphy.

-Kerry