2
$\begingroup$

I have a coordinate system $0$ which I'd first like to rotate about its $z$-axis which gives me system $1$, and afterwards rotate system $1$ about its $y$-axis which gives me system $2$. See picture: enter image description here

Both rotation angles are $90^{\circ}$ and shall be realized with quaternions $q_1$ (first rotation) and $q_2$ (second rotation). So I have \begin{align} q_1=\left(\cos\left(\frac{1}{2}\frac{\pi}{2}\right),0,0,\sin\left(\frac{1}{2}\frac{\pi}{2}\right)\right) \newline q_2=\left(\cos\left(\frac{1}{2}\frac{\pi}{2}\right),0,\sin\left(\frac{1}{2}\frac{\pi}{2}\right),0\right) \end{align} Let's take only the $x_0$-axis for now whicht is $x_0=(1,0,0)$. I'd like to express every new coordinate axis of system 1 and 2 in the 0-system. \begin{align} x_1=q_1\cdot x_0 \cdot q_1^{-1} = (0,1,0) \end{align} is fine. But with the next transformation I'm in trouble, as it returns \begin{align} x_2=q_2\cdot x_1 \cdot q_2^{-1} = (0,1,0) \end{align} which is logical because a rotation of the y-axis around the y-axis has no effect. But also \begin{align} x_2=q_2\cdot q_1 \cdot x_1 \cdot q_1^{-1} \cdot q_2^{-1} = (0,0,1) \end{align} or \begin{align} x_2=q_2\cdot q_1 \cdot x_0 \cdot q_1^{-1} \cdot q_2^{-1} = (0,1,0) \end{align} is wrong, as it should be $x_2=(0,0,-1)$, shouldn't it?

Is there anything essential I've missed studying quaternions, and could anybody get this right for me please? Thank you very much!

$\endgroup$
4
$\begingroup$

Ok I got it by my own... it's just the same way like transforming with matrices. $x_2$ has to be multiplied first with $q_2$ and then with $q_1$ to express it in the 0-system! That's it!

$\endgroup$

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.