The inverse of inertia tensor in 3D is an inertia tensor itself. The example of the sphere is missleading since it is some scalar factor times the identity matrix.
oh... then, i don't need to do: b_inertia = m_inertia_inverse = inertia.inverse(); right??
The inverse of inertia tensor in 3D is an inertia tensor itself. The example of the sphere is missleading since it is some scalar factor times the identity matrix.
inverse_inertia = inertia.inverse();
Anyways, if it aids:
By choosing the correct coordinate system for the object (For basic objects, the obvious one), you can always diagonalise the inertia tensor, and so it, and its inverse can be stored in a vector instead of a matrix, and then the inverse is just the component wise inverse:
(Ix, Iy, Iz)^-1 = (1/Ix, 1/Iy, 1/Iz)
oh
inverse_inertia = inertia.inverse();
Anyways, if it aids:
By choosing the correct coordinate system for the object (For basic objects, the obvious one), you can always diagonalise the inertia tensor, and so it, and its inverse can be stored in a vector instead of a matrix, and then the inverse is just the component wise inverse:
(Ix, Iy, Iz)^-1 = (1/Ix, 1/Iy, 1/Iz)
hm... so i can use vectors .... or i can use float inverseInertia = 1.0f / Ix; instead of vectors and do the multiplication between them...
