Closed ben-l-p closed 4 months ago
Found a bug in the conversion from a $\boldsymbol{\Psi}$ vector to a rotation matrix $\mathbf{R}$. This only applies for larger $\Psi$ angles, as the small angle correction (which most "normal" cases will use) is correct.
Intended expression: $$\mathbf{R} = \mathbf{I} + \sin||{\boldsymbol{\Psi}||} \tilde{\boldsymbol{\Psi}} + \frac{1-\cos{||\boldsymbol{\Psi}||}}{||\boldsymbol{\Psi}||^2}\tilde{\boldsymbol{\Psi}} \tilde{\boldsymbol{\Psi}}$$
Used expression: $$\mathbf{R} = \mathbf{I} + \frac{\sin||\boldsymbol{\Psi}||}{||\boldsymbol{\Psi}||} \tilde{\boldsymbol{\Psi}} + (1-\cos{||\boldsymbol{\Psi}||})\tilde{\boldsymbol{\Psi}} \tilde{\boldsymbol{\Psi}}$$
I'm bad at maths, back to the drawing board...
Found a bug in the conversion from a $\boldsymbol{\Psi}$ vector to a rotation matrix $\mathbf{R}$. This only applies for larger $\Psi$ angles, as the small angle correction (which most "normal" cases will use) is correct.
Intended expression: $$\mathbf{R} = \mathbf{I} + \sin||{\boldsymbol{\Psi}||} \tilde{\boldsymbol{\Psi}} + \frac{1-\cos{||\boldsymbol{\Psi}||}}{||\boldsymbol{\Psi}||^2}\tilde{\boldsymbol{\Psi}} \tilde{\boldsymbol{\Psi}}$$
Used expression: $$\mathbf{R} = \mathbf{I} + \frac{\sin||\boldsymbol{\Psi}||}{||\boldsymbol{\Psi}||} \tilde{\boldsymbol{\Psi}} + (1-\cos{||\boldsymbol{\Psi}||})\tilde{\boldsymbol{\Psi}} \tilde{\boldsymbol{\Psi}}$$