Exponential Coordinate Representation of Rigid-Body Motions

Every displacement of a rigid body can be expressed as a displacement along a fixed screw axis $S$ (Chasles-Mozzi theorem).

Exponential coordinates of a homogeneous transformation $T$ :

S\theta \in \mathbb{R}^6

With:

With $S = (\omega, v) = (S_{\omega}, S_v)$ as screw axis:

if $\left\| S_{\omega} \right\| = 1$ ( $\left\| \omega \right\| = 1$ ) then for any distance $\theta \in \mathbb{R}$ moved along $S$ :

e^{[S]\theta} = \begin{bmatrix} e^{[\omega]\theta} & (I\theta + (1 - \cos \theta) [\omega] + (\theta - \sin \theta) [\omega]^2) v \\ 0 & 1 \end{bmatrix}

if $S_\omega = 0$ ( $\omega = 0$ ) and $\left\| S_v \right\| = 1$ ( $\left\| v \right\| = 1$ ) then:

e^{[S]\theta} = \begin{bmatrix} I & v\theta \\ 0 & 1 \end{bmatrix}

Every $SE(3)$ transformation can be expressed as an exponential function of a twist ( $se(3)$ ).

Exp: $[S]\theta \in se(3) \rightarrow T \in SE(3)$

Matrix exponential is like integration

Given: $T = (R, p) \in SE(3)$

Find:

$\theta \in [0, \pi]$
$S = (\omega, v) \in \mathbb{R}^6$
- at least one of $\left\| \omega \right\|$ or $\left\| v \right\|$ is unity

Such that:

e^{[S]\theta} = T

Algorithm:

if $R=I$ then $\omega=0$ , $v = p / \left\| p \right\|$ and $\theta = \left\| p \right\|$
otherwise:
- calculate $\omega = \hat{\omega}$ and $\theta$ for $R$ (see matrix log on SO(3))
- then: $v$ is calculated as $v = G^{-1}(\theta)p$
- with: $G^{-1}(\theta) = \frac{1}{\theta}I - \frac{1}{2}[\omega] + \left( \frac{1}{\theta} - \frac{1}{2} \cot \frac{\theta}{2} \right) [\omega]^2$

Log: $T \in SE(3) \rightarrow [S]\theta \in se(3)$

Matrix log is like differentiation

$T_{sb'} = e^{[S_b]\theta}T_{sb}$ is the configuration achieved if $S$ is expressed in the fixed frame $\{s\}$ .

$T_{sb''} = T_{sb}e^{[S_b]\theta}$ is the configuration achieved if $S$ is expressed in the body frame $\{b\}$ .

Notes taken from:

Modern Robotics: Mechanics, Planning, and Control by Kevin M. Lynch and Frank C. Park, Cambridge University Press, 2017