2.3.1. Revolute Joints

Revolute Joint

A revolute joint is a one-DOF joint that produces pure rotation about a fixed axis.

We choose a unit axis direction $\omega$ (following the right-hand rule) and pick any point $q$ on that axis so we can measure the radius from the axis to a moving point $p$.

Because the instantaneous motion of a point under pure rotation is tangent to its circular path (perpendicular to both $\omega$ and the radius vector $r=p-q$), its velocity is:
$\dot{p}(t)=\omega \times (p(t)-q).$

Rotating about an axis that passes through $q$ induces a translational component at the origin equal to “angular velocity $\times$ moment arm,” so the linear part of the twist must satisfy:
$v=-\omega \times q=q\times \omega ,{\omega }^{T}v=0.$
A twist stacks the linear and angular velocity components:
$\xi =\left[\begin{array}{c}v\\ \omega \end{array}\right]=\left[\begin{array}{c}-\omega \times q\\ \omega \end{array}\right].$
Because $\hat{\omega }$ is the matrix that implements cross products ($\hat{\omega }x=\omega \times x$), the twist in $\mathfrak{se}(3)$ is:
$\hat{\xi }=\left[\begin{array}{cc}\hat{\omega }& v\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}\hat{\omega }& -\hat{\omega }q\\ 0& 0\end{array}\right].$
Then, $\hat{\xi }$ is the infinitesimal generator of the motion in homogeneous coordinates. The instantaneous and finite kinematics satisfy:
$\left[\begin{array}{c}\dot{p}(\theta )\\ 0\end{array}\right]=\hat{\xi }\left[\begin{array}{c}p(\theta )\\ 1\end{array}\right],\left[\begin{array}{c}p(\theta )\\ 1\end{array}\right]=e^{\hat{\xi }\theta }\left[\begin{array}{c}p(0)\\ 1\end{array}\right].$
Exponentiating the generator yields a rotation $e^{\hat{\omega }\theta }$ about the axis, and the induced translation for an axis passing through $q$. Therefore, the transform is:
$e^{\hat{\xi }\theta }=\left[\begin{array}{cc}{e}^{\hat{\omega }\theta }& (I-e^{\hat{\omega }\theta })q\\ 0& 1\end{array}\right].$

If the axis passes through the origin, then $q=0$, so no translation is induced and $v=0$.

Example

Finding $se(3)$

Find $\hat{\xi }$, such that $\dot{\bar{p}}=\hat{\xi }\bar{p}.$

Solution. $\hat{p}={\left[\begin{array}{cc}p& 1\end{array}\right]}^T,$ and use the given relation $\dot{\hat{p}}=\left[\begin{array}{cc}\hat{\omega }& v\\ 0& 0\end{array}\right]\hat{p}.$

Start from first principles by defining: