\[\newcommand{\u}[1]{\boldsymbol{\mathsf{#1}}} \renewcommand{\b}[1]{\boldsymbol{#1}} \newcommand{\t}[1]{\textsf{#1}} \newcommand{\m}[1]{\mathbb{#1}} \def\RR{\bf R} \def\bold#1{\bf #1} \def\mbf#1{\mathbf #1} \def\uv#1{\hat{\usf {#1}}} \def\dl#1{\underline{\underline{#1}}} \newcommand{\usf}[1]{\boldsymbol{\mathsf{#1}}} \def\bs#1{\usf #1}\]

Finding Angle and Axis of Rotation given a Rotation Tensor

Determining \((\theta, \u{e})\) given rotation tensor \(\u{R}\)

We proceed first by computing the value of the trace of \(\u{R}\), \(\t{tr}(\u{R})\).

\[\begin{align} (\u{R})_{,ij} = R_{ij} &= \left[a_{i}a_{j}-(\delta_{ij}-a_{i}a_{j})\cos\theta + \epsilon_{ikj}a_{k}\sin\theta\right] \\ \t{tr}(\u{R}) = R_{ii} &= a_{i}a_{i}-(\delta_{ii}-a_{i}a_{i})\cos\theta + \underbrace{\epsilon_{iji}a_{k}\sin\theta}_{=0} \\ &= a_{i}a_{i} + (3-a_{i}a_{i})\cos\theta \\ &= \left\|\u{a}\right\|^2 + (3-\left\|\u{a}\right\|^2)\cos\theta \\ \t{tr}(\u{R}) &= 1+ (3-1)\cos\theta \\ &= 1+2\cos\theta \end{align}\]

Hence, \(\cos\theta = (\frac{\t{tr}(\u{R}) -1}{2})\).

We next compute \(\u{a}\):

\[\begin{align} & \frac{1}{2} (\u{R} - \u{R}^{\t{T}}) = \t{skew}(\u{R}) \\ & \frac{1}{2} (R_{ij} - R_{ji}) = \epsilon_{ikj}a_{k}\sin\theta \\ & \frac{1}{2} \epsilon_{irj} (R_{ij} - R_{ji}) = \epsilon_{irj}\epsilon_{ikj}a_{k}\sin\theta \\ & \frac{1}{2} \epsilon_{irj} (R_{ij} - R_{ji}) = (\delta_{rk}\delta_{rk}jj - \delta_{rj}\delta_{jk})a_{k}\sin\theta \\ & \frac{1}{2} \epsilon_{irj} (R_{ij} - R_{ji}) = (3\delta_{rk} - \delta_{rk})a_{k}\sin\theta \\ & \frac{1}{2} \epsilon_{irj} (R_{ij} - R_{ji}) = 2a_{r}\sin\theta \\ & \boxed{a_{r} = \frac{1}{4\sin\theta}(\epsilon_{irj})(R_{ij}-R_{ji})} \end{align}\]

And its components are

\[\begin{align} a_{1} &= \frac{1}{4\sin\theta}\left[\epsilon_{i1j}(R_{ij}-R_{ji})\right] \\ &= \frac{1}{4\sin\theta}\left[\epsilon_{213}(R_{23}-R_{32})+\epsilon_{312}(R_{32}-R_{23})\right] \\ &= \frac{1}{4\sin\theta}\left[ R_{32}-R_{23}-R_{23}+R_{32}\right] \\ &= \frac{2}{4\sin\theta}\left[ R_{32}-R_{23}\right] = \frac{1}{2\sin\theta}\left[ R_{32}-R_{23}\right] \end{align}\] \[\begin{align} a_{2} &= \frac{1}{4\sin\theta}\left[\epsilon_{123}(R_{13}-R_{31})+\epsilon_{321}(R_{31}-R_{13})\right] \\ &= \frac{1}{2\sin\theta}\left[ R_{13}-R_{31}\right] \end{align}\] \[\begin{align} a_{3} &= \frac{1}{4\sin\theta}\left[\epsilon_{132}(R_{12}-R_{21})+\epsilon_{231}(R_{21}-R_{12})\right] \\ &= \frac{1}{2\sin\theta}\left[ R_{21}-R_{12}\right] \end{align}\]

Remark : Notice that, as expected, the axis of rotation is not defined when \(\theta =0\), i.e, any direction is an axis of rotation in this case.

Example:

Let’s recover \(\theta\) and \(\u{a}\) from the last example. We have

\[\begin{align} \left[\u{R}\right] = \begin{bmatrix}\frac{2}{3} & -\frac{1}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{2}{3} & -\frac{1}{3} \\ -\frac{1}{3} & \frac{2}{3} & \frac{2}{3}\end{bmatrix} \end{align}\] \[\begin{align} & \t{tr}(\u{R}) = 3\times\frac{2}{3} = 2 \\ & \cos\theta = (\frac{2-1}{2}) = \frac{1}{2} \quad \Rightarrow \quad \theta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3} \end{align}\]

The axis \(\u{a}\) is

\[\begin{align} & a_{1} = \frac{R_{32}-R_{23}}{2\sin\theta} = \frac{(\frac{2}{3} + \frac{1}{3})}{2\sin(\frac{\pi}{3})} = \frac{1}{\sqrt{3}} \\ & a_{2} = \frac{R_{13}-R_{31}}{2\sin\theta} = \frac{1}{\sqrt{3}} \\ & a_{3} = \frac{R_{21}-R_{12}}{2\sin\theta} = \frac{1}{\sqrt{3}} \end{align}\]

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