- #1
bugatti79
- 794
- 1
Hi Folks,
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##...?
Thanks
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##...?
Thanks