矩阵Matrix到欧拉角Euler转换

本文主要是介绍矩阵Matrix到欧拉角Euler转换，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

参考文献：

http://www.geometrictools.com/Documentation/EulerAngles.pdf

但是这里的公式不能直接用，原因是左右手系空间不同，我这边采用Direct3D默认的右手系，参考：

https://docs.microsoft.com/en-us/windows/win32/direct3d9/d3dxmatrixrotationyawpitchroll

所以需要自行推导右手系公式，已知各个轴旋转矩阵公式：

$R(\theta_{x})=\begin{bmatrix} 1 & 0 & 0\\ 0 & cos(\theta_{x}) & sin(\theta_{x})\\ 0 & -sin(\theta_{x}) & cos(\theta_{x}) \end{bmatrix}$ ， $R(\theta_{y})=\begin{bmatrix} cos(\theta_{y}) & 0 & -sin(\theta_{y})\\ 0 & 1 & 0\\ sin(\theta_{y}) & 0 & cos(\theta_{y}) \end{bmatrix}$ ， $R(\theta_{z})=\begin{bmatrix} cos(\theta_{z}) & sin(\theta_{z}) & 0\\ -sin(\theta_{z}) & cos(\theta_{z}) & 0\\ 0 & 0 & 1 \end{bmatrix}$

欧拉角变换顺序为YXZ，则先计算YX矩阵

$R(\theta_{y})\cdot R(\theta_{x})=\begin{bmatrix} cos(\theta_{y}) & sin(\theta_{y})\cdot sin(\theta_{x}) & -sin(\theta_{y})\cdot cos(\theta_{x})\\ 0 & cos(\theta_{x}) & sin(\theta_{x})\\ sin(\theta_{y}) & -cos(\theta_{y})\cdot sin(\theta_{x}) & cos(\theta_{y})\cdot cos(\theta_{x}) \end{bmatrix}$

最终YXZ矩阵

$R(\theta_{y})\cdot R(\theta_{x})\cdot R(\theta_{z})=\begin{bmatrix} cos(\theta_{y})\cdot cos(\theta_{z})-sin(\theta_{y})\cdot sin(\theta_{x})\cdot sin(\theta_{z}) & cos(\theta_{y})\cdot sin(\theta_{z})+sin(\theta_{y})\cdot sin(\theta_{x})\cdot cos(\theta_{z}) & -sin(\theta_{y})\cdot cos(\theta_{x}))\\ -cos(\theta_{x})\cdot sin(\theta_{z}) & cos(\theta_{x})\cdot cos(\theta_{z}) & sin(\theta_{x})\\ sin(\theta_{y})\cdot cos(\theta_{z})+cos(\theta_{y})\cdot sin(\theta_{x})\cdot sin(\theta_{z}) & sin(\theta_{y})\cdot sin(\theta_{z})-cos(\theta_{y})\cdot sin(\theta_{x})\cdot cos(\theta_{z}) & cos(\theta_{y})\cdot cos(\theta_{x})) \end{bmatrix}$

可以直接得知 $sin(\theta_{x})=r12$ ，即 $\theta_{x}=arcsin(r12)$ ，然后需要分三种情况

$\theta_{x}\in \left (-\frac{\pi }{2}, \frac{\pi }{2}\right )$ ，可知 $tan(\theta_{y})=\frac {sin(\theta_{y})\cdot cos(\theta_{x})}{cos(\theta_{y})\cdot cos(\theta_{x})}$ ，即 $\theta_{y}=arctan(\frac {-r02} {r22})$ ，同理 $\theta_{z}=arctan(\frac {-r10} {r11})$
当 $\theta_{x}=\frac{\pi}{2}$ ，则 $sin(\theta_{x})=1$ ，YXZ矩阵可简化为
$R(\theta_{yxz})=\begin{bmatrix} cos(\theta_{y})\cdot cos(\theta_{z})-sin(\theta_{y})\cdot sin(\theta_{z}) & cos(\theta_{y})\cdot sin(\theta_{z})+sin(\theta_{y})\cdot cos(\theta_{z}) & 0\\ 0 & 0 & 1\\ sin(\theta_{y})\cdot cos(\theta_{z})+cos(\theta_{y})\cdot sin(\theta_{z}) & sin(\theta_{y})\cdot sin(\theta_{z})-cos(\theta_{y})\cdot cos(\theta_{z}) & 0 \end{bmatrix}$
根据两角和公式，可得
$R(\theta_{yxz})=\begin{bmatrix} cos(\theta_{y}+\theta_{z}) & sin(\theta_{y}+\theta_{z}) & 0\\ 0 & 0 & 1\\ sin(\theta_{y}+\theta_{z}) & -cos(\theta_{y}+\theta_{z}) & 0 \end{bmatrix}$ ，即 $\theta_{y}+\theta_{z}=arctan(\frac {r01}{r00})$ ，且结果不唯一
当 $\theta_{x}=-\frac {\pi}{2}$ ，则 $sin(\theta_{x})=-1$ ，YXZ矩阵简化为
$R(\theta_{yxz})=\begin{bmatrix} cos(\theta_{y})\cdot cos(\theta_{z})+sin(\theta_{y})\cdot sin(\theta_{z}) & cos(\theta_{y})\cdot sin(\theta_{z})-sin(\theta_{y})\cdot cos(\theta_{z}) & 0\\ 0 & 0 & -1\\ sin(\theta_{y})\cdot cos(\theta_{z})-cos(\theta_{y})\cdot sin(\theta_{z}) & sin(\theta_{y})\cdot sin(\theta_{z})+cos(\theta_{y})\cdot cos(\theta_{z}) & 0 \end{bmatrix}$
可得
$R(\theta_{yxz})=\begin{bmatrix} cos(\theta_{y}-\theta_{z}) & -sin(\theta_{y}-\theta_{z}) & 0\\ 0 & 0 & -1\\ sin(\theta_{y}-\theta_{z}) & cos(\theta_{y}-\theta_{z}) & 0 \end{bmatrix}$ ，即 $\theta_{y}-\theta_{z}=arctan(\frac {-r01}{r00})$

基于以上思路，就能实现D3DXMATRIX到欧拉角的转换代码

D3DXVECTOR3* D3DXMatrixToEulerAngles(D3DXVECTOR3* pOut, const D3DXMATRIX* pM)
{if (pM->_23 < 0.999f) // some fudge for imprecision{if (pM->_23 > -0.999f) // some fudge for imprecision{pOut->x = asin(pM->_23);pOut->y = atan2(-pM->_13, pM->_33);pOut->z = atan2(-pM->_21, pM->_22);}else{// WARNING.  Not unique.  YA - ZA = atan2(-r01,r00)pOut->x = -D3DX_PI * 0.5f;pOut->y = atan2(-pM->_12, pM->_11);pOut->z = 0.0f;}}else{// WARNING.  Not unique.  YA + ZA = atan2(r01,r00)pOut->x = D3DX_PI * 0.5f;pOut->y = atan2(pM->_12, pM->_11);pOut->z = 0.0f;}return pOut;
}

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