- #1

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Like now we have moment of inertia: (Ix,Iy,Iz)=(20,18,25), and hot to calculate the inertia tensor like

(Ixx,Ixy,Ixz

Iyx,Iyy,Iyz,

Izx,Izy,Izz)?

I have read about this page several times, but still have no idea.

- Thread starter kasoll
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- #1

- 3

- 1

Like now we have moment of inertia: (Ix,Iy,Iz)=(20,18,25), and hot to calculate the inertia tensor like

(Ixx,Ixy,Ixz

Iyx,Iyy,Iyz,

Izx,Izy,Izz)?

I have read about this page several times, but still have no idea.

- #2

- 9,975

- 3,130

$$I=\begin{pmatrix}

20 & 0 & 0 \\

0 & 18 & 0 \\

0 & 0 & 25

\end{pmatrix}$$Does this answer your question?

- #3

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The moment of inertia of a figure about a line is the sum of the products formed by multiplying the magnitude of each element (of area or of mass) by the square of its distance from the line. So the moment of inertia of a figure is the sum of moments of inertia of its parts.

Now we know that the moments of inertia of a figure about lines which intersect at a common point are generally unequal. The moment is greatest about one line and least about another line perpendicular to the first one. A set of three orthogonal lines consisting of these two and a line perpendicular to both are the principal axes of inertia of the figure relative to that point. If the point is the figure's centroid, the axes are the central principal axes of inertia. The moments of inertia about principal axes are principal moments of inertia.

- #4

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Thanks for the answer. But almost all the inertia tensor I have seen have non-zero valur as Ixy,Ixz,Iyz. I know to get principal axes moment of inertia from inertia tensor, which just looks like the one you mentioned._{xy}= I_{yz}= I_{zx}= 0. So in this case you would write

$$I=\begin{pmatrix}

20 & 0 & 0 \\

0 & 18 & 0 \\

0 & 0 & 25

\end{pmatrix}$$Does this answer your question?

But how to reverse the calculation? By multiply a matrix? And how to get the spercific matrix?

- #5

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- 3,130

If there are non-zero off diagonal matrix elements, then you solve the eigenvalue problem, find the eigenvectors and construct the transformation matrix ##T## from the direction cosines as explained on pages 8-9 in the reference that you quoted. Read it carefully. Equation (15) says that ##I'=(T)~(I)~(T)^T##. If you want to go back the other way, then ##(I)=(T)^T~(I')~(T)##.Thanks for the answer. But almost all the inertia tensor I have seen have non-zero valur as Ixy,Ixz,Iyz. I know to get principal axes moment of inertia from inertia tensor, which just looks like the one you mentioned.

But how to reverse the calculation? By multiply a matrix? And how to get the spercific matrix?

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