1
$\begingroup$

For rigid body rotation discrete time kinematics equation is as follows

$R_{k+1} = T_r \cdot R_k$

where:

$R$ is a rotation matrix

$T_r = e^{-[a\times]}$ is a transition matrix

$[a\times]$ means a skew symmetric matrix from vector $a$

$a=\omega \cdot \delta t$ is a rotation vector

$\omega$ is angular velocity vector in body frame

$\delta t$ is a time step.

Using Rodrigue's formula one can have a closed form solution for matrix exponent so no infinite series calculation required.

$T_r=I+sin(\theta)\left[e\times \right] + (1-cos(\theta))\left[e\times \right]^2$

where

$e = \frac{a}{||a||}$ is a unit vector of rotation

$\theta = ||a||$ is angle of rotation around that unit vector

When dealing with quaternions similar expression is

$q_{k+1} = T_q \cdot q_k$

where

$T_q = e^{\frac{1}{2}\begin{bmatrix}-\left[a\times\right] && a\\ -a^T&& 0 \end{bmatrix}}$

Hope I have put not many mistakes in the expressions above. But general idea should be clear... Is there a closed form solution for quaternion discrete kinematics transition matrix $T_q$?

$\endgroup$

1 Answer 1

1
$\begingroup$

Observe

$$ \begin{bmatrix}-[a\times] & a \\ -a^T & 0\end{bmatrix}^2=\|a\|^2\begin{bmatrix}-I_3 & 0 \\ 0 & -1\end{bmatrix} $$

implies

$$ \exp\Big(\frac{1}{2}\begin{bmatrix}-[a\times] & a \\ -a^T & 0\end{bmatrix}\Big)\,=\,\cos\!\Big(\frac{\theta}{2}\Big)I_4+\sin\!\Big(\frac{\theta}{2}\Big)\begin{bmatrix}-[e\times] & e \\ -e^T & 0\end{bmatrix}. $$

$\endgroup$
3
  • $\begingroup$ That's amazing! Thank you so much! Curious if that is a known formula or you have discovered it for the first time? If later is the case you deserve it to have your name like Rodriguez has his one :) Nice to notice it is even simpler for quaternion than for rotation matrix case. $\endgroup$
    – aliko
    Commented Aug 5, 2023 at 6:20
  • $\begingroup$ @aliko It is textbook how to exponentiate diagonalizable matrices, and this is block-diagonal with $2\times2$ blocks which we also know how to exponentiate. Indeed, the matrix with the $a$s in it is none other than the matrix representing right-multiplication-by-$a$ on the quaternions with respect to the ordered basis $\{i,j,k,1\}$, so we can expect the matrix exponential $\exp(\tfrac{1}{2}[\cdots])$ to be the matrix representing right-multiplication by the quaternion exponential $\exp(\frac{1}{2}a)=\cos(\theta/2)+\sin(\theta/2)e$. $\endgroup$
    – coiso
    Commented Aug 5, 2023 at 8:51
  • $\begingroup$ To block-diagonalize it, BTW, extend $e=\hat{a}$ to an orthonormal basis $\{\hat{a},b,c\}$ of $\Bbb R^3$, then extend further with $d=e_4$, so then the matrix with $a$s in it acts block-diagonally with two $2\times2$ blocks both $\theta=\|a\|$ times $90^\circ$ rotation matrices with respect to the basis $\{b,c,\hat{a},d\}$. $\endgroup$
    – coiso
    Commented Aug 5, 2023 at 9:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .