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I'm computing the principal axes and moments of a triangle at its centroid. The triangle vertices are at the points $(12,0,0)$, $(0,24,0)$, $(0,0,36)$. According to my calculations, the relevant characteristic equation is $\lambda^3 - 112\lambda^2 + 2352\lambda = 0$. This has roots 84, 28, 0. It's the zero eigenvalue that's bothering me. The values 84 and 28 are moments of inertia about axes lying in the plane of the triangle. So, according to the perpendicular axis theorem, the moment of inertia around the axis perpendicular to this plane ought to be $84 + 28 = 112$. So, why is $112$ not a root of the characteristic polynomial ?? Puzzled.

If it helps -- the characteristic polynomial is derived from the matrix (the so-called "inertia tensor") whose rows are $[104,8,12]$, $[8,80,24]$, $[12,24,40]$. It's symmetric, obviously.

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How did you compute the inertia tensor? You would get zero along the direction normal to the triangle if you did $\int \mathbf r\mathbf r^T \mathrm dA$ instead of $\int (\mathbf r^T\mathbf r \mathbf 1_{3\times3} - \mathbf r\mathbf r^T) \mathrm dA$, where $\mathbf r$ is the position vector relative to the centroid. –  Rahul May 19 '12 at 13:25
    
Thanks. I believe I computed the inertia tensor correctly. I did it once using the equation you cited, and again using the parallel axis theorem. And I assumed it was correct, since it gave me two out of three principal moments correctly (the 84 and 28 values are correct according to my CAD software). But, anyway, I'll check again and maybe post my Mathematica code. –  bubba May 19 '12 at 22:47
    
Can you post the matrix that you derived the characteristic equation from? –  Rahul May 19 '12 at 23:00
    
> Can you post the matrix ... I did. See above. The matrix has rows [104 8 12], [8 80 24], [12 24 40]. –  bubba May 19 '12 at 23:38
    
Rahul -- please don't waste your time checking my inertia tensor calculations (unless you really want to). If I fouled up those calculations, I expect I can find and fix the problem. My real concern was more of a conceptual one, about the general process of finding principal moments by calculating eigenvalues. I was worried that maybe this process doesn't work for planar objects in 3D (or something like that). –  bubba May 20 '12 at 0:20
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The characteristic polynomial of $$\mathbf A = \begin{bmatrix}104 & 8 & 12 \\ 8 & 80 & 24 \\ 12 & 24 & 40\end{bmatrix}$$ is $\det(\lambda \mathbf 1_{n\times n} - \mathbf A)$, which according to Mathematica is equal to $$\lambda^3 - 224\lambda^2 + 14896\lambda - 263424.$$ This has roots $28$, $84$, and $112$ like you expect. Perhaps you mixed up two matrices when you were working it out.

Conceptually, the moment of inertia tensor is "morally" continuous in some physically intuitive sense, so there shouldn't be any reason to worry that it won't work for planar objects -- you can think of them as the limit of thinner and thinner solid bodies. On physical grounds, the only way you can get a zero moment about some axis is if the body is an infinitesimally thin rod lying along that axis. Any other object requires energy to spin it about the axis, and rotational kinetic energy is $\frac12 \mathbf\omega^T\mathbf I\mathbf\omega$ for angular velocity $\mathbf\omega$, so $\mathbf I$ cannot have a zero eigenvector there.

In general, though, I find it easier to think about the tensor $\mathbf \Sigma = \int \mathbf r\mathbf r^T \mathrm dV$, which a statistician might call the covariance matrix, because its eigenvalues correspond to how "stretched out" the body is in a given direction. The moment of inertia can be easily derived from this as $\mathbf I = \mathbf 1_{3\times3}\operatorname{tr} \mathbf \Sigma - \mathbf\Sigma$, but $\mathbf\Sigma$ corresponds more closely with my intuition: it scales naturally with the body under nonuniform scaling, while the moment of inertia tensor does not.

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Thanks very much, again. I'm embarrassed. I used Mathematica to obtain the characteristic polynomial, so, as you say, I must have mistyped something. I understand the idea of a triangle being the limit of thin plates. In my CAD package, though, the triangle and thin plates gave quite different answers. That's one thing that worried me. I need to figure out why. I have also found it easier to work with the "covariance" quantity. Does it have a standard name ?? I have seen it called the "mass tensor" and the "Euler tensor". –  bubba May 20 '12 at 5:27
    
I mixed up the inertia tensor and the "covariance" tensor. The characteristic polynomial I cited actually belongs to the latter. Bone-head mistake. Sorry. –  bubba May 20 '12 at 7:20
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