A rigid body is simulated to move in six degrees of freedom. At every past timestep, I know its displacement vectors $x^{GLOB}$ in the global coordinate frame, as well as 3x3 matrices $R$ that describe a rotation from world global orientation to the body local orientation. $$ x^{GLOB}_{i-2}=R_{i-2} \times x^{BODY}_{i-2} $$ $$ x^{GLOB}_{i-1}=R_{i-1} \times x^{BODY}_{i-1} $$ $$ x^{GLOB}_i=R_i \times x^{BODY}_{i} $$ I can use the finite difference method to approximate the linear accelerations at timestep $i-1$. $$ \ddot{x}^{GLOB}_{i-1} \approx \frac{x_{i-2}^{GLOB}-2x_{i-1}^{GLOB}+x_{i}^{GLOB}}{(dt)^2} $$ I can't figure out what is the analogue for the rotation matrices. I know that the result should be a angular acceleration vector $\alpha^{GLOB}_{i-1}$. The angular velocity vector $\omega^{GLOB}_{i-1}$ I found from the handbook "Intermediate dynamics for engineers" by O'Reilly, as well as this question. $$ \omega_{0R}=ax(R^T\dot{R}) $$ Would simply taking a finite difference of the angular velocity give me an acceleration?
Asked
Active
Viewed 51 times
