Matrix Rotation Query: Inertia Tensor D (+90y, -90z)

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SUMMARY

The discussion centers on the rotation of an inertia tensor D using standard right-hand rule notation in Cartesian coordinates. The user is applying a +90-degree rotation about the y-axis and a -90-degree rotation about the z-axis, represented by matrices N_y and N_z respectively. The user is seeking clarification on the correct transformation of the inertia tensor, questioning whether the new system tensor is calculated as N_R D N'_R or N'_R D N_R. The consensus leans towards the former, N_R D N'_R, as the correct formulation for the transformed tensor.

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bugatti79
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I have an inertia tensor D in the old Cartesian system which i need to rotate through +90 in y and -90 in z to translate to the new system. I am using standard right hand rule notation for this Cartesian rotation.


D= \mathbf{\left(\begin{array}{lll}I_{xx}&I_{xy}&I_{xz}\\I_{yx}&I_{yy}&I_{yz}\\I_{zx}&I_{zy}&I_{zz}\\\end{array}\right)}, N_y(+90)=\mathbf{\left(\begin{array}{lll}0&0&1\\0&1&0\\-1&0&0\\\end{array}\right)}, N_z(-90)=\mathbf{\left(\begin{array}{lll}0&1&0\\-1&0&0\\0&0&1\\\end{array}\right)}


If we let


N_R=N_z N_y (I am pre-multiplying N_y by N_z because that is the order) and the transpose N'_R=N_R^T.


Is the the new system tensor N_RDN'_R or N'_RDN_R...?
 
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bugatti79 said:
Is the the new system tensor N_RDN'_R or N'_RDN_R...?

This page would seem to indicate the former.
 

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