资料来源：https://sites.cs.ucsb.edu/~lingqi/teaching/games101.html
笛卡尔坐标系中，空间坐标的位置变换是通过矩阵相乘实现的.
Linear Transforms = Matrices
\[\left [ \begin{matrix} x^{new} \\ y^{new} \end{matrix} \right ] = \left [ \begin{matrix} a & b\\ c & d \end{matrix} \right ] \left [ \begin{matrix} x\\ y \end{matrix} \right ]\] \[x^{new} = Mx\] \[x^{new} = ax+ by\] \[y^{new} = cx+ dy\]2D Scale Matrix
\[\left [ \begin{matrix} x^{new} \\ y^{new} \end{matrix} \right ] = \left [ \begin{matrix} s & 0\\ 0 & s \end{matrix} \right ] \left [ \begin{matrix} x\\ y \end{matrix} \right ]\]2D Reflection Matrix
\[\left [ \begin{matrix} x^{new} \\ y^{new} \end{matrix} \right ] = \left [ \begin{matrix} -1 & 0\\ 0 & 1 \end{matrix} \right ] \left [ \begin{matrix} x\\ y \end{matrix} \right ]\]Shear Matrix
\[\left [ \begin{matrix} x^{new} \\ y^{new} \end{matrix} \right ] = \left [ \begin{matrix} 1 & a\\ 0 & 1 \end{matrix} \right ] \left [ \begin{matrix} x\\ y \end{matrix} \right ]\]Rotation Matrix
\[\left [ \begin{matrix} x^{new} \\ y^{new} \end{matrix} \right ] = \left [ \begin{matrix} cos \Theta & -sin\Theta\\ sin \Theta & cos \Theta \end{matrix} \right ] \left [ \begin{matrix} x\\ y \end{matrix} \right ]\]Homogeneous coordinates
有些变换不能以矩阵形式表示
\[\left [ \begin{matrix} x^{new} \\ y^{new} \end{matrix} \right ] = \left [ \begin{matrix} a & b\\ c & d \end{matrix} \right ] \left [ \begin{matrix} x\\ y \end{matrix} \right ] + \left [ \begin{matrix} tx \\ ty \end{matrix} \right ]\]平移不是线性变换
但我们不希望平移成为一个特殊的案例
是否有一个统一的方法来表示所有的转换？ (以及额外消耗是多少？)
添加第三个坐标（w坐标）
2D point = (x, y, 1)
2D vector = (x, y, 0)
平移的矩阵表示
\[\left [ \begin{matrix} x^{new} \\ y^{new} \\ w^{new} \end{matrix} \right ] = \left [ \begin{matrix} 1 & 0 & tx \\ 0 & 1 & ty\\ 0 & 0 & 1 \end{matrix} \right ] \left [ \begin{matrix} x\\ y\\ 1 \end{matrix} \right ] + \left [ \begin{matrix} x+tx \\ y+ty \\ 1 \end{matrix} \right ]\]所以在齐次坐标下，有这些运算结果:
vector + vector = vector
\[\left [ \begin{matrix} xa \\ ya \\ 0 \end{matrix} \right ] + \left [ \begin{matrix} xb \\ yb \\ 0 \end{matrix} \right ] = \left [ \begin{matrix} xa+xb \\ ya+yb \\ 0 \end{matrix} \right ]\]point - point = vector
\[\left [ \begin{matrix} xa \\ ya \\ 1 \end{matrix} \right ] + \left [ \begin{matrix} xb \\ yb \\ 1 \end{matrix} \right ] = \left [ \begin{matrix} xa-xb \\ ya-yb \\ 0 \end{matrix} \right ]\]point + vector = point
\[\left [ \begin{matrix} xa \\ ya \\ 0 \end{matrix} \right ] + \left [ \begin{matrix} xb \\ yb \\ 1 \end{matrix} \right ] = \left [ \begin{matrix} xa-xb \\ ya-yb \\ 1 \end{matrix} \right ]\]point + point = point (两坐标中点) w = 2 w!=0
\[\left [ \begin{matrix} xa \\ ya \\ 1 \end{matrix} \right ] + \left [ \begin{matrix} xb \\ yb \\ 1 \end{matrix} \right ] = \left [ \begin{matrix} xa-xb \\ ya-yb \\ 2 \end{matrix} \right ] = \left [ \begin{matrix} xa-xb /w \\ ya-yb /w\\ 2 /w \end{matrix} \right ] =\left [ \begin{matrix} xa-xb /2\\ ya-yb /2\\ 1 \end{matrix} \right ]\]Affine map = linear map + translation
仿射变换 = 线性变换 + 平移
\[\left [ \begin{matrix} x^{new} \\ y^{new} \end{matrix} \right ] = \left [ \begin{matrix} a & b\\ c & d \end{matrix} \right ] \left [ \begin{matrix} x\\ y \end{matrix} \right ] + \left [ \begin{matrix} tx \\ ty \end{matrix} \right ]\]Using homogenous coordinates:
\[\left [ \begin{matrix} x^{new} \\ y^{new} \\ 1 \end{matrix} \right ] = \left [ \begin{matrix} a & b & tx \\ c & d & ty\\ 0 & 0 & 1 \end{matrix} \right ] \left [ \begin{matrix} x\\ y\\ 1 \end{matrix} \right ]\]Scale
\(S(sx,sy) = \left [ \begin{matrix} s & 0 & 0\\ 0 & s & 0\\ 0 & 0 & 1 \end{matrix} \right ]\) Rotation
\[R(\Theta) = \left [ \begin{matrix} cos\Theta & -sin\Theta & 0\\ sin\Theta & cos\Theta & 0\\ 0 & 0 & 1 \end{matrix} \right ]\]Translation
\[T(tx,ty) = \left [ \begin{matrix} 1 & 0 & tx\\ 0 & 1 & ty\\ 0 & 0 & 1 \end{matrix} \right ]\]Inverse Transform
逆变换，物体经过缩放矩阵缩放后，可以通过缩放矩阵的逆矩阵给还原，这就是逆矩阵的逆变换的作用 \(M^{-1}\)
Composite Transform
复合变换
矩阵乘法没有交换性 \(R45·T(1,0) != T(1,0)·R45\)
3D Transformations
\[3D point = (x, y, z, 1)^T\] \[3D vector = (x, y, z, 0)^T\]In general, (x, y, z, w) (w != 0) is the 3D point:
(x/wa,y/w,z/w)
\[\left [ \begin{matrix} x^{new} \\ y^{new} \\ z^{new} \\ 1 \end{matrix} \right ] = \left [ \begin{matrix} a & b & c & tx \\ c & d & d & ty \\ e & f & g & tz \\ 0 & 0 & 0 & 1 \end{matrix} \right ] \left [ \begin{matrix} x\\ y\\ z\\ 1 \end{matrix} \right ]\]Rotation around x-, y-, or z-axis
\[Rx(\alpha) = \left [ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & cos\alpha & -sin\alpha & 0 \\ 0 & sin\alpha & cos\alpha & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right ]\] \[Ry(\alpha) = \left [ \begin{matrix} cos\alpha & 0& sin\alpha & 0 \\ 0 & 1 & 0 & 0 \\ -sin\alpha & 0 & cos\alpha & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right ]\] \[Rz(\alpha) = \left [ \begin{matrix} cos\alpha & -sin\alpha & 0 & 0 \\ sin\alpha & cos\alpha & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right ]\]从Rx、Ry、Rz构成任何三维旋转？
所谓的欧拉角，就可以从任何三维旋转
在飞行模拟器中经常使用：滚动、俯仰、偏航 (roll, pitch, yaw)
所以有以下公式可以实现任意轴角度的旋转，替代上面三个不同轴的旋转矩阵
Rotation by angle α around axis n
\[N = \left [ \begin{matrix} 0 & -nz & ny\\ nz & 0 & -nx\\ -ny& nx & 0 \end{matrix} \right ]\] \[R(n,\alpha) = cos(\alpha)I + (1-cos(\alpha))nn^T+sin(\alpha)N\] \[R(n,\alpha) = cos(\alpha)I + (1-cos(\alpha))nn^T+sin(\alpha)\left [ \begin{matrix} 0 & -nz & ny\\ nz & 0 & -nx\\ -ny& nx & 0 \end{matrix} \right ]\]1
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Eigen::Matrix4f rotation = Eigen::Matrix4f::Identity();
float rad = (angle / 180.0f) * acos(-1);
Eigen::Matrix3f N = Eigen::Matrix3f::Identity();
N << 0, -axis.z(), axis.y(),
axis.z(), 0, -axis.x(),
-axis.y(),axis.x(), 0;
rotation.block<3,3>(0,0) = cos(rad) * Eigen::Matrix3f::Identity() +
(1-cos(rad))*axis*axis.transpose() +
sin(rad)* N;