How to extract the screw axis vector and the angle from the exponential coordinates?

Question

Given the 6-dimensional vector of the exponential coordinates of the homogeneous transformation: $S\theta$, where $S$ is the screw axis consisting of the pair $(\omega, v)$ and both of them are $3$ dimensional vectors, $\theta$ is the angle followed by the transformation around that screw axis. Also for a screw axis, either $v$ or $\omega$ is a unit vector.

How to separate the screw axis $S$ and the angle of rotation $\theta$ from the given exponential coordinates 6-vector $S\theta$? Note that the unit vector is not given. it may be $v$ or $\omega$.

Is the following the correct algorithm?

\begin{align*} \theta &= \|S\theta\| &\text{#which is the magnitude of $S\theta$ $6$-vector.} \\ S &= \frac{S\theta}{\|S\theta\|} &\text{#the screw axis is the $S\theta$ $6$-vector divided by its magnitude.} \end{align*}

Thanks a lot in advance.

Answer 1

to answer my question, the screw axis has 2 vectors each of 3 dimensions. these vectors are ω and v. As long as we consider the screw axis and not the corresponding twist, either one of those vectors should be a unit vector.

If the given screw axis S has non-zero ω vector then the ω vector is the unit vector, if ω is a zero vector (pure translation) then the unit axis is the other vector v.

Returning to the exponential coordinates Sθ, after the unit vector of the screw axis S is determined as aforementioned above, the angle theta θ and the axis S is extracted from the 6-vector of exponential coordinates Sθ as follows,

let Sθ = (ωθ, vθ)

Hence,

θ = ∥ωθ∥ if ω is the unit vector. In other words, ωθ vector is non-zero.

OR

θ = ∥vθ∥ if v is the unit vector. In other words, ωθ vector is zero.

Consequently the screw axis S can be determined simply as,

S= Sθ / θ