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Consider the following rotation matrix:

$$R=\begin{pmatrix} -\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}} \\ \sqrt{\frac{3}{8}} &\frac{1}{2}& \sqrt{\frac{3}{8}} \\ -\frac{1}{\sqrt{8}}&\sqrt{\frac{3}{4}}&-\frac{1}{\sqrt{8}}\end{pmatrix}$$

Determine the angle and the axis of rotation by using the following equations: $$T=R+R^T-[\text{Tr}R-1]I\\\cos{\varphi}=\frac{1}{2}(\text{Tr}R-1)$$

Where $\text{Tr}R$ is the trace of $R$.

This is what I have done so far:

$$R+R^T-(\text{Tr}R-1)I=\begin{pmatrix} -\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}} \\ \sqrt{\frac{3}{8}} &\frac{1}{2}& \sqrt{\frac{3}{8}} \\ -\frac{1}{\sqrt{8}}&\sqrt{\frac{3}{4}}&-\frac{1}{\sqrt{8}}\end{pmatrix}+\begin{pmatrix}-\frac{1}{\sqrt{2}}&\sqrt{\frac{3}{8}}&-\frac{1}{\sqrt{8}} \\ 0&\frac{1}{2}&\sqrt{\frac{3}{4}} \\ \frac{1}{\sqrt{2}} & \sqrt{\frac{3}{8}}&-\frac{1}{\sqrt{8}}\end{pmatrix}-\left (-\frac{1}{\sqrt{2}}+\frac{1}{2}-\frac{1}{\sqrt{8}}-1\right )I \\ $$

$$=\begin{pmatrix} -\frac{2}{\sqrt{2}}&\sqrt{\frac{3}{8}}& \frac{1}{2\sqrt{2}} \\ \sqrt{\frac{3}{8}}&1& \frac{\sqrt{2}\sqrt{3}+\sqrt{3}}{2\sqrt{2}} \\\frac{1}{2\sqrt{2}}&\frac{\sqrt{2}\sqrt{3}+\sqrt{3}}{2\sqrt{2}}&-\frac{2}{\sqrt{8}}\end{pmatrix}=...?$$

How do I determine the axis of rotation from this matrix?

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  • $\begingroup$ Go back to your book, and find out what the "following equations", in particular the $T$, mean. $\endgroup$ – Christian Blatter May 17 '16 at 9:25
  • $\begingroup$ @ChristianBlatter In our lecture note ist says: "The rotation vector is one of the rows of matrix $T$". But all my rows are non-zero so how can it be just one of the rows? Do I have to decompose the matrix? $\endgroup$ – bluemoon May 17 '16 at 9:27
  • $\begingroup$ @ChristianBlatter By parallel do you mean they should be linearly dependent? $\endgroup$ – bluemoon May 17 '16 at 9:39
  • $\begingroup$ There is an error in your calculations. In reality all rows of T are multiples of the same vector, which is then the axis. $\endgroup$ – Christian Blatter May 17 '16 at 11:04
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Consider Rodrigues' formula for a rotation matrix:

$$ R=I+\sin\varphi K+(1-\cos\varphi)K^2 $$

with

$$ K=\pmatrix{ 0&-k_3&k_2\\ k_3&0&-k_1\\ -k_2&k_1&0 }\;, $$

where $k$ is the (unit) rotation axis.

To my mind it seems easier to just take the antisymmetric part of $R$, which is $\sin\varphi K$, and normalise to get rid of the $\sin\varphi$, but your lecture notes apparently take the other option and extract $K$ from the symmetric part:

$$ R+R^\top=2\left(I+(1-\cos\varphi)K^2\right) $$

and thus

\begin{align} R+R^\top-(\operatorname{Tr}R-1)I &=2\left(I+(1-\cos\varphi)K^2\right)-(\operatorname{Tr}R-1)I\\ &=2\left(I+(1-\cos\varphi)K^2\right)-2\cos\varphi I\\ &=2(1-\cos\varphi)(K^2+I)\;. \end{align}

$K^2$ has $-k_2^2-k_3^2$, $-k_3^2-k_1^2$ and $-k_1^2-k_2^2$ on the diagonal, so $K^2+I$ has $k_1^2$, $k_2^2$ and $k_3^2$ on the diagonal.

It follows that you must have made a mistake, as the diagonal elements should all have the same sign. Indeed it seems you forgot to subtract the multiple of the identity in the last line.

This doesn't give you the signs of the $k_i$; you're going to have to extract them from the other components of $K^2$, or use the antisymmetric part after all.

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    $\begingroup$ I took me 20 mins to really understand but it makes perfect sense now. Thanks for your help! $\endgroup$ – bluemoon May 17 '16 at 11:25
  • $\begingroup$ Using the symmetric part is more work, but has the advantage that it still works when the rotation angle is $\pi$. $\endgroup$ – bubba Dec 30 '16 at 1:11

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