- #1
bugatti79
- 786
- 4
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...?
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...?