# 问题及解答

## [Def]欧拉角(Euler angles)

Posted by haifeng on 2012-07-17 13:34:57 last update 2012-07-17 15:58:57 | Edit | Answers (0)

$\mathbf{\Xi}=g(\psi)\mathbf{X},$

$g(\psi)= \begin{pmatrix} \cos\psi & -\sin\psi & 0\\ \sin\psi & \cos\psi & 0\\ 0 & 0 & 1 \end{pmatrix}.$

$\mathbf{\Xi\'}=g(\theta)\mathbf{\Xi},$

$g(\theta)= \begin{pmatrix} 1 & 0 & 0\\ 0 & \cos\theta & -\sin\theta\\ 0 & \sin\theta & \cos\theta \end{pmatrix}.$

$\mathbf{X\'}=g(\phi)\mathbf{\Xi\'},$

$g(\phi)= \begin{pmatrix} \cos\phi & -\sin\phi & 0\\ \sin\phi & \cos\phi & 0\\ 0 & 0 & 1 \end{pmatrix}.$

$\mathbf{X\'}=g\mathbf{X}=g(\phi)g(\theta)g(\psi)\mathbf{X},$

$\begin{split} & \begin{pmatrix} \cos\phi & -\sin\phi & 0\\ \sin\phi & \cos\phi & 0\\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0\\ 0 & \cos\theta & -\sin\theta\\ 0 & \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} \cos\psi & -\sin\psi & 0\\ \sin\psi & \cos\psi & 0\\ 0 & 0 & 1 \end{pmatrix}\\ =& \begin{pmatrix} \cos\phi\cos\psi-\cos\theta\sin\phi\sin\psi & -\cos\phi\sin\psi-\cos\theta\sin\phi\cos\psi & \sin\phi\sin\theta\\ \sin\phi\cos\psi+\cos\theta\cos\phi\sin\psi & -\sin\phi\sin\psi+\cos\theta\cos\phi\cos\psi & -\cos\phi\sin\theta\\ \sin\theta\sin\psi & \sin\theta\cos\psi & \cos\theta \end{pmatrix}. \end{split}$

$\psi\in[0,2\pi],\quad\theta\in[0,\pi],\quad\phi\in[0,2\pi].$

$\phi,\theta,\psi$ 分别称为 spin, nutation, precession.

Reference:

Moshe Carmeli, Shimon Malin, Theory of Spinors, An Introduction.

James Diebel, Representing Attitude: Euler Angles, Unit Quaternions, and Rotation Vectors, 2006.