In robot kinematics, we have $e^{(\theta*twist)}$, where $twist$ is 6*1 vector. How do I get the 4 by 4 transformation matrix by using product of exponentials?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Actually, the 6x1 vector is sometimes better referred to as the coordinates of the twist. The twist itself is a 4x4 matrix, element of $SE(3)$, found by $$ \begin{align} A &= \begin{bmatrix} \widehat{\omega} & v \\ 0 & 0 \end{bmatrix} \\ v &\triangleq -\omega \times q \end{align} $$ where $\omega$ is the unit vector pointing along the axis of rotation (for a rotating joint), $\widehat{\omega}$ is the 3x3 skew symmetric matrix derived from $\omega$, and $q$ is any point along the axis of rotation. The matrix exponential is then found by $$ \begin{align} e^{A \theta} &= \begin{bmatrix} R & (I-R)(\omega \times v) + \omega \omega^T v \theta \\ 0 & 1 \end{bmatrix} \\ R &= e^{\hat{\omega} \theta} = I + \hat{\omega} \sin\theta + \hat{\omega}^2 (1-\cos\theta) \end{align} $$
For a more detailed explanation, see section 3.2 of Murray, Li, and Sastry.
