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Given a rotation matrix $Q \in \mathbb R^3$, how to find a magnitude of rotation. I guess it makes sense to say that identity matrix $I$ is considered 0 magnitude and the rotation of $-I$ is considered maximum magnitude. It would also makes sense that $mag(Q) = mag(Q^T)$.

Is there a way of computing this without going through the trouble of finding Euler angles and then deriving metric using identity $\cos^2(\alpha) + \cos^2(\beta) + \cos^2(\gamma) = 1$?

Gary
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Joonatan Samuel
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  • hmm, I have doubts whether $-I$ we can treat as a rotation matrix in 3d ... its determinant $=-1$ – Widawensen Nov 30 '16 at 20:18
  • I am pretty sure it is. Any orthogonal matrix can be considered a rotation matrix since the rotation has to follow property $Q Q^T = Q Q^{-1} = I$, it follows that determinant be +- 1. Also, you can verify that vector under this transformation remains constant length, since no scaling is done. Therefore it is rotation of some sort. But to make the matter concrete, consider vector $v = [1, 1, 1]$ – Joonatan Samuel Dec 01 '16 at 10:20
  • I was taught that determinant must be +1. But maybe it is just more general approach.., – Widawensen Dec 01 '16 at 13:08
  • What's the connection between the Euler angles, the direction cosines identity and the rotation magnitude ??? –  Dec 01 '16 at 13:10
  • On negative determinant. https://en.wikipedia.org/wiki/Rotation_matrix#Properties_of_a_rotation_matrix – Joonatan Samuel Dec 01 '16 at 13:28
  • Maybe it becomes clearer for what I am searching if I explain my necessity for it. I have data coming in from couple of devices accelerometers which are moving in unison, however the coordinate frames are not the same so I find a rotation Q such that the coordinate frames match minimizing square error. Next I am looking to make an automatic check of confidence on the coordinate frame we are in. One part of that is that, if rotation doesn't change too much we are more confident that it is correct. – Joonatan Samuel Dec 01 '16 at 13:36
  • @JoonatanSamuel: indeed, if your rotation determinant turns negative, you can suspect a mechanical problem, such as the Universe changing chirality. –  Dec 01 '16 at 13:59

2 Answers2

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I found it useful to think about the transformation rotation matrix $Q \in R^{3 \times 3}$ affects given vector $v$. That means what does $$\vec u = \vec v \times Q^T $$ look like. After that I can compare the initial vector $\vec v$ and compare it to $\vec u$ by dot product between them since $$cos\ \phi = \frac{\vec u \cdotp \vec v}{||\vec u|| \ ||\vec v||}$$.

$ \vec v \in R^3$; $\vec v$ should be unit vector, let us choose $[1, 0, 0]$. giving the metric from $[-1; 1]$ of $$metric(Q) = (\vec v Q^T) \cdot \vec v;\ v = [1, 0, 0]$$

This simplifies to $metric(Q) = Q_{11}$, which agrees with maximum of 1, when $Q = I$, minimum of -1 when $Q = -I$ and $metric(Q) = metric(Q^T)$

Joonatan Samuel
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You don't really define what is meant by magnitude and I don't think that the Euler angles representation is appropriate for that purpose.

I suggest to use the axis/angle representation, and this method: https://en.wikipedia.org/wiki/Rotation_matrix#Determining_the_angle.