(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.

  • $\begingroup$ Welcome to Robotics Greg Schmit and thanks for the disclosure. On Robotics, questions asking for homework help must include a summary of the work you've done so far to solve the problem, and a description of the difficulty you are having solving it. Please edit your question to add this information. Take a look at How to Ask and tour for more information on how stack exchange works. For advice on how to write a good question, see the Robotics question checklist. $\endgroup$
    – Mark Booth
    Apr 11, 2017 at 13:49

1 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.

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.

  • $\begingroup$ I wonder if maybe in the Adjoint map, 0 and [p]R are in reversed locations. IOW, that Adjoint appears to be the transpose. $\endgroup$ Apr 13, 2017 at 2:35
  • $\begingroup$ Maybe because you write a twist by listing the linear component first? $\endgroup$ Apr 13, 2017 at 2:54
  • $\begingroup$ Oh, I think I see what you mean. Yes, I learned to write twist as the column vector (transpose) of [ v w ], so v on top and omega on bottom. $\endgroup$ Apr 13, 2017 at 3:16
  • 1
    $\begingroup$ Link to edX course is broken $\endgroup$
    – chutsu
    Apr 26, 2022 at 10:30

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