Games 101: 旋转矩阵

旋转矩阵

本文主要介绍了旋转矩阵的推导，分为两种方式：

• 旋转坐标
• 旋转坐标轴
  以下坐标系都是右手坐标系

旋转坐标

已知坐标点$A(x_a,y_a)$, 旋转$\theta$角后变为坐标点$B(x_b,y_b)$,求解旋转矩阵.

$$\begin{array}{r}\begin{array}{rl}{x}_a& =r_a\cdot cos(\alpha )=r_b\cdot cos(\alpha )\\ y_a& =r_a\cdot sin(\alpha )=r_b\cdot sin(\alpha )\\ x_b& =r_b\cdot cos(\alpha +\theta )\\ & =r_b\cdot cos(\alpha )\cdot cos(\theta )-r_b\cdot sin(\alpha )\cdot sin(\theta )\\ & =x_a\cdot cos(\theta )-y_a\cdot sin(\theta )\\ y_b& =r_b\cdot sin(\alpha +\theta )\\ & =r_b\cdot cos(\alpha )\cdot sin(\theta )+r_b\cdot sin(\alpha )\cdot cos(\theta )\\ & =x_a\cdot sin(\theta )+y_a\cdot cos(\theta )\end{array}\end{array}$$

即：

$$\begin{array}{r}\begin{array}{c}\left[\begin{array}{c}{x}_b\\ y_b\end{array}\right]=\left[\begin{array}{cc}cos(\theta )& -sin(\theta )\\ sin(\theta )& cos(\theta )\end{array}\right]\cdot \left[\begin{array}{c}{x}_a\\ y_a\end{array}\right]\end{array}\end{array}$$

推广到三维以及其他轴，可以得到三维坐标点的旋转矩阵为：

$$\begin{array}{r}\begin{array}{c}{R}_x(\alpha )=\left[\begin{array}{ccc}1& 0& 0\\ 0& cos(\alpha )& -sin(\alpha )\\ 0& sin(\alpha )& cos(\alpha )\end{array}\right]R_y(\beta )=\left[\begin{array}{ccc}cos(\beta )& 0& sin(\beta )\\ 0& 1& 0\\ -sin(\beta )& 0& cos(\beta )\end{array}\right]R_z(\theta )=\left[\begin{array}{ccc}cos(\theta )& -sin(\theta )& 0\\ sin(\theta )& cos(\theta )& 0\\ 0& 0& 1\end{array}\right]\end{array}\end{array}$$

旋转坐标

坐标轴$O{X}_1{Y}_1$绕z轴旋转$\theta$角后变为坐标轴$O{X}_2{Y}_2$，点A在坐标轴$O{X}_1{Y}_1$中坐标为$(x_a,y_a)$， 在坐标轴$O{X}_2{Y}_2$中$(x_b,y_b)$,求解旋转矩阵.
坐标轴$O{X}_1{Y}_1$,两个方向的单位向量为$\overrightarrow{x_1},\overrightarrow{y_1}$，坐标轴$O{X}_2{Y}_2$,两个方向的单位向量为$\overrightarrow{x_2},\overrightarrow{y_2}$

分解单位向量：

$$\begin{array}{r}\begin{array}{rl}\overrightarrow{x_1}& =\left|\overrightarrow{x_1}\right|\cdot cos(\theta )\cdot \overrightarrow{x_2}-\left|\overrightarrow{x_1}\right|\cdot sin(\theta )\cdot \overrightarrow{y_2}=cos(\theta )\cdot \overrightarrow{x_2}-sin(\theta )\cdot \overrightarrow{y_2}\\ \overrightarrow{y_1}& =\left|\overrightarrow{y_1}\right|\cdot sin(\theta )\cdot \overrightarrow{x_2}+\left|\overrightarrow{x_1}\right|\cdot cos(\theta )\cdot \overrightarrow{y_2}=sin(\theta )\cdot \overrightarrow{x_2}+cos(\theta )\cdot \overrightarrow{y_2}\end{array}\end{array}$$

向量变换：

$$\begin{array}{r}\begin{array}{rl}\overrightarrow{r_a}& =x_a\cdot \overrightarrow{x_1}+y_a\cdot \overrightarrow{y_1}\\ & =x_a\cdot (cos(\theta )\cdot \overrightarrow{x_2}-sin(\theta )\cdot \overrightarrow{y_2})+y_a\cdot (sin(\theta )\cdot \overrightarrow{x_2}+cos(\theta )\cdot \overrightarrow{y_2})\\ & =(x_a\cdot cos(\theta )+y_a\cdot sin(\theta ))\cdot \overrightarrow{x_2}+(-sin(\theta )\cdot x_a+cos(\theta )\cdot y_a)\cdot \overrightarrow{y_2}\\ \overrightarrow{r_a}& =x_b\cdot \overrightarrow{x_2}+y_b\cdot \overrightarrow{y_2}\end{array}\end{array}$$

即

$$\begin{array}{r}\begin{array}{c}\left[\begin{array}{c}{x}_b\\ y_b\end{array}\right]=\left[\begin{array}{cc}cos(\theta )& sin(\theta )\\ -sin(\theta )& cos(\theta )\end{array}\right]\cdot \left[\begin{array}{c}{x}_a\\ y_a\end{array}\right]\end{array}\end{array}$$

推广到三维以及其他轴，可以得到三维坐标轴的旋转矩阵为：

$$\begin{array}{r}\begin{array}{c}{R}_x(\alpha )=\left[\begin{array}{ccc}1& 0& 0\\ 0& cos(\alpha )& sin(\alpha )\\ 0& -sin(\alpha )& cos(\alpha )\end{array}\right]R_y(\beta )=\left[\begin{array}{ccc}cos(\beta )& 0& -sin(\beta )\\ 0& 1& 0\\ sin(\beta )& 0& cos(\beta )\end{array}\right]R_z(\theta )=\left[\begin{array}{ccc}cos(\theta )& sin(\theta )& 0\\ -sin(\theta )& cos(\theta )& 0\\ 0& 0& 1\end{array}\right]\end{array}\end{array}$$