0
$\begingroup$

Given a quaternion obtained by compositions of rotations and translations, is it safe to take the $3\times 3$ upper left matrix as the "full rotation" and the last column of the matrix as the "full translation" ?

Using block-matrices, I have : $$ \left[ \begin{matrix} R & t\\ 0 & 1 \end{matrix} \right] = \left[ \begin{matrix} 1 & t\\ 0 & 1 \end{matrix} \right] \times \left[ \begin{matrix} R & 0\\ 0 & 1 \end{matrix} \right] $$ where $R$ would be the rotation and $t$ the translation.

Am I right?

$\endgroup$
2
$\begingroup$

Note that what you are describing are not quaternions, they are homogeneous transformations. A quaternion would be used to perform the same function as your $R$, however, you would not store it (or use it) as a matrix. Additionally, quaternions do not encode translations, only rotations.

To answer your question, yes you can decompose a homogeneous transformation into its constituent rotation and translation parts and the decomposition that you show above is correct.

$\endgroup$
  • $\begingroup$ Thanks for your answer. I was quite convinced that the $4\times 4$ matrix I show is a "quaternion", welI I was mistaken. $\endgroup$ – Sylvain B. Nov 4 '16 at 15:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.