Convert Twist from frame B to frame A [closed]

Question

(full disclosure: this is homework)

I have a twist expressed in frame B:

$\zeta_b = \begin{bmatrix}1\\3\\-2\\0\\-2\\4\end{bmatrix}$

And a general transformation matrix:

$g_{ab} = \begin{bmatrix}-0.4749 & 0.8160 & 0.3294 & -1.5\\-0.2261 & -0.4749 & -0.8505 & -1\\-0.8505 & -0.3294 & 0.4100 & 2\\0 & 0 & 0 & 1\end{bmatrix}$

How would I go about converting my twist into frame A?

I suspect I would break $\zeta_b$ into its component $\omega$ and $v$ vectors using the knowledge that:

$\zeta = \begin{bmatrix}v \\ \omega\end{bmatrix} = \begin{bmatrix}-\omega \times q \\ \omega\end{bmatrix}$ (where $q$ is a point on $\omega$)

But I am unsure.

Answer 1

Hint: First, write the transformation matrix as $$ T = \begin{bmatrix} R &p\\0_{1\times3} &1 \end{bmatrix}. $$

Now we use the relations $\omega_a = R\omega_b$ and $q_a = Rq_b + p$. Then since $v_a = -\omega_a \times q_a$, we can derive $v_a$ in terms of all the known quantities.

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In fact, two twists representing the same screw motion described in different reference frames are related to each other by an adjoint map. That is given $S_a = (\omega_a, v_a)$ and $S_b = (\omega_b, v_b)$, we have $$ S_a = [Ad_{T_{ab}}](S_b)\\ S_b = [Ad_{T_{ba}}](S_a), $$ where $[Ad_T] = \begin{bmatrix} R & 0\\ [p]R & R \end{bmatrix}$ and $[p]$ is the skew-symmetric matrix constructed from $p$.

I found the material of this edX course really helpful for me to understand this topic. Maybe you want to check it out.