2.3.3. Screw Joints

Screw (Helical) Joint

A screw (helical) joint is a one-DOF joint that couples rotation about a fixed axis with translation along the same axis.

We choose a unit axis direction $\omega$ (right-hand rule) and pick any point $q$ on that axis. We also define the pitch $h$ as translation per radian of rotation (units: length/rad).

Because the point undergoes rotation about the axis exactly like a revolute joint, but also drifts along the axis as it rotates, the instantaneous point velocity is:
$\dot{p}(t)=\dot{\theta }(t)(\omega \times (p(t)-q)+h\omega ).$

Rotating about an axis through q contributes the same “moment arm” term as the revolute joint, while the additional translation along the axis contributes $h\omega$. Therefore the linear part of the twist must satisfy:
$v=-\omega \times q+h\omega ,{\omega }^{T}v=h(\Vert \omega \Vert =1).$

A twist stacks the linear and angular velocity components:
$\xi =\left[\begin{array}{c}v\\ \omega \end{array}\right]=\left[\begin{array}{c}-\omega \times q+h\omega \\ \omega \end{array}\right].$

Because $\hat{\omega }$ is the matrix that implements cross products (i.e., $\hat{\omega }x=\omega \times x$), the twist in $\mathfrak{se}(3)$ is:
$\hat{\xi }=\left[\begin{array}{cc}\hat{\omega }& v\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}\hat{\omega }& -\hat{\omega }q+h\omega \\ 0& 0\end{array}\right].$

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, plus a translation of magnitude $h\theta$ along $\omega$, plus the same induced translation term from rotating about an axis through $q$. Therefore, the transform is:
$e^{\hat{\xi }\theta }=\left[\begin{array}{cc}{e}^{\hat{\omega }\theta }& (I-e^{\hat{\omega }\theta })q+h\omega \theta \\ 0& 1\end{array}\right].$

If $h=0$, the screw joint reduces to a revolute joint; if $\omega =0$, it reduces to a prismatic joint (pure translation).