矩阵

快速计算向量位置，下文矩阵全是列主序四阶矩阵

• 定义向量$\overrightarrow{vec} = (x,y,z,w)$
• 定义矩阵 $mat4 = \begin{bmatrix} m_{11} & m_{12} & m_{13} & m_{14} \\ m_{21} & m_{22} & m_{23} & m_{24} \\ m_{31} & m_{32} & m_{33} & m_{34} \\ m_{41} & m_{42} & m_{43} & m_{44} \\ \end{bmatrix}$

四阶矩阵

向量与矩阵相乘

$$\begin{pmatrix} x = x * m_{11} + y * m_{21} + z * m_{31} + w * m_{41} ,\\ y = x * m_{12} + y * m_{22} + z * m_{32} + w * m_{42} ,\\ z = x * m_{13} + y * m_{23} + z * m_{33} + w * m_{43} ,\\ w = x * m_{14} + y * m_{24} + z * m_{34} + w * m_{44} ,\\ \end{pmatrix} = mat4 \times \overrightarrow{vec}$$ $$\begin{pmatrix} x = x * m_{11} + y * m_{12} + z * m_{13} + w * m_{14} ,\\ y = x * m_{22} + y * m_{22} + z * m_{23} + w * m_{24} ,\\ z = x * m_{33} + y * m_{32} + z * m_{33} + w * m_{34} ,\\ w = x * m_{44} + y * m_{42} + z * m_{43} + w * m_{44} ,\\ \end{pmatrix} = \overrightarrow{vec} \times mat4$$

矩阵与矩阵相乘

$$\begin{bmatrix} m_{a11} & m_{a12} & m_{a13} & m_{a14} \\ m_{a21} & m_{a22} & m_{a23} & m_{a24} \\ m_{a31} & m_{a32} & m_{a33} & m_{a34} \\ m_{a41} & m_{a42} & m_{a43} & m_{a44} \\ \end{bmatrix} \times \begin{bmatrix} m_{b11} & m_{b12} & m_{b13} & m_{b14} \\ m_{b21} & m_{b22} & m_{b23} & m_{b24} \\ m_{b31} & m_{b32} & m_{b33} & m_{b34} \\ m_{b41} & m_{b42} & m_{b43} & m_{b44} \\ \end{bmatrix}$$

• $$m_{c11} = m_{a11} \times m_{b11} + m_{a21} \times m_{b12} + m_{a31} \times m_{b13} + m_{a41} \times m_{b14}$$
• $$m_{c12} = m_{a11} \times m_{b21} + m_{a21} \times m_{b22} + m_{a31} \times m_{b23} + m_{a41} \times m_{b24}$$
• $$m_{c13} = m_{a11} \times m_{b31} + m_{a21} \times m_{b32} + m_{a31} \times m_{b33} + m_{a41} \times m_{b34}$$
• $$m_{c14} = m_{a12} \times m_{b41} + m_{a22} \times m_{b42} + m_{a32} \times m_{b43} + m_{a42} \times m_{b44}$$
• $$m_{c21} = m_{a12} \times m_{b11} + m_{a22} \times m_{b12} + m_{a32} \times m_{b13} + m_{a42} \times m_{b14}$$
• $$m_{c22} = m_{a12} \times m_{b21} + m_{a22} \times m_{b22} + m_{a32} \times m_{b23} + m_{a42} \times m_{b24}$$
• $$m_{c23} = m_{a12} \times m_{b31} + m_{a22} \times m_{b32} + m_{a32} \times m_{b33} + m_{a42} \times m_{b34}$$
• $$m_{c24} = m_{a12} \times m_{b41} + m_{a22} \times m_{b42} + m_{a32} \times m_{b43} + m_{a42} \times m_{b44}$$
• $$m_{c31} = m_{a13} \times m_{b11} + m_{a23} \times m_{b12} + m_{a33} \times m_{b13} + m_{a43} \times m_{b14}$$
• $$m_{c32} = m_{a13} \times m_{b21} + m_{a23} \times m_{b22} + m_{a33} \times m_{b23} + m_{a43} \times m_{b24}$$
• $$m_{c33} = m_{a13} \times m_{b31} + m_{a23} \times m_{b32} + m_{a33} \times m_{b33} + m_{a43} \times m_{b34}$$
• $$m_{c34} = m_{a13} \times m_{b41} + m_{a23} \times m_{b42} + m_{a33} \times m_{b43} + m_{a43} \times m_{b44}$$
• $$m_{c41} = m_{a14} \times m_{b11} + m_{a24} \times m_{b12} + m_{a34} \times m_{b13} + m_{a44} \times m_{b14}$$
• $$m_{c42} = m_{a14} \times m_{b21} + m_{a24} \times m_{b22} + m_{a34} \times m_{b23} + m_{a44} \times m_{b24}$$
• $$m_{c43} = m_{a14} \times m_{b31} + m_{a24} \times m_{b32} + m_{a34} \times m_{b33} + m_{a44} \times m_{b34}$$
• $$m_{c44} = m_{a14} \times m_{b41} + m_{a24} \times m_{b42} + m_{a34} \times m_{b43} + m_{a44} \times m_{b44}$$

模型矩阵

位移

$$\begin{pmatrix} x + x_1 , \\ y + y_1 , \\ z + z_1 , \\ w , \\ \end{pmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ x_1 & y_1 & z_1 & 1 \\ \end{bmatrix} \times \overrightarrow{vec}$$

缩放

$$\begin{pmatrix} x \times x_1 , \\ y \times y_1 , \\ z \times z_1 , \\ w , \\ \end{pmatrix} = \begin{bmatrix} x_1 & 0 & 0 & 0 \\ 0 & y_1 & 0 & 0 \\ 0 & 0 & z_1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \times \overrightarrow{vec}$$

旋转

旋转

视图矩阵

通过目标点，视点，上方向，三个点计算视图矩阵。x,y 方向不变，z 方向相反。

• 目标点:$\overrightarrow{target} = (tx,ty,tz,tw)$
• 视点:$\overrightarrow{eye} = (ex,ey,ez,ew)$
• 上方向:$\overrightarrow{up} = (ux,uy,uz,uw)$

1. 以目标点为起点，视点为终点计算$\overrightarrow{d} = \overrightarrow{eye} - \overrightarrow{target} = (dx,dy,dz,dw)$
2. 将$\overrightarrow{d}$归一化$\overrightarrow{d} *= 1 / |\overrightarrow{d}|$
3. 利用$\overrightarrow{d}$与上方向叉乘求出$\overrightarrow{a} = \overrightarrow{d} \times \overrightarrow{up} = (ax,ay,az,aw)$
4. 将$\overrightarrow{a}$归一化$\overrightarrow{a} *= 1 / |\overrightarrow{a}|$
5. 利用$\overrightarrow{d}$与上方向叉乘求出$\overrightarrow{b} = \overrightarrow{d} \times \overrightarrow{a} = (bx,by,bz,bw)$
6. 利用$\overrightarrow{d}$取反求出$\overrightarrow{c} = -\overrightarrow{d}$
7. $(\overrightarrow{a},\overrightarrow{b},\overrightarrow{c})$即为视图矩阵中的$(\overrightarrow{x},\overrightarrow{y},\overrightarrow{z})$

正交投影矩阵

在裁剪空间内，物体大小不会因为距离远近而改变，通过上下边界，左右边界，近平面，远平面计算正交投影矩阵。

• 上边界:top
• 下边界:bottom
• 左边界:left
• 右边界:right
• 近平面:near
• 远平面:far

1. 计算裁剪空间的三轴方向长度与相机世界的三轴方向上的 1 个单位比例关系。
   • 裁剪空间(用 1 表示) / 相机世界(用 2 表示，默认为 1)
   • $w_1 = \frac{w_1}{w_2} = \frac{2}{right - left}$
   • $h_1 = \frac{h_1}{h_2} = \frac{2}{top - bottom}$
   • $p_1 = \frac{p_1}{p_2} = \frac{2}{far - near}$
2. 通过比例关系求出矩阵与裁剪空间的位移量
   • $x = \frac{right + left}{2} \times \frac{2}{right - left} = \frac{right + left}{right - left}$
   • $y = \frac{top + bottom}{2} \times \frac{2}{top - bottom} = \frac{top + bottom}{top - bottom}$
   • $z = \frac{far + near}{2} \times \frac{2}{far - near} = \frac{far + near}{far - near}$

• $$orthographicProjection = \begin{bmatrix} \frac{2}{right - left} & 0 & 0 & 0 \\ 0 & \frac{2}{top - bottom} & 0 & 0 \\ 0 & 0 & \frac{2}{far - near} & 0 \\ \frac{right + left}{right - left} & \frac{top + bottom}{top - bottom} & \frac{far + near}{far - near} & 1 \\ \end{bmatrix} =$$

透视投影矩阵

在裁剪空间内，物体大小会因为距离远近而改变，通过上下边界，左右边界，近平面，远平面计算正交投影矩阵。

• fov:摄像机视锥体垂直视野角度
• aspect:摄像机视锥体宽高比
• near:摄像机近裁剪面到视点的距离
• far:摄像机远裁剪面到视点的距离

1. 相机可视区域为四棱台形状，转化为长方体形状的投影盒子
   1. $\frac{top}{n} = \tan(\frac{fov}{2})$
   2. $top = \tan(\frac{fov}{2}) \times n$
   3. $bottom = -top$
   4. $\frac{right}{top} = aspect$
   5. $right = top \times aspece$
   6. $left = -right$
   7. 设$p_1 = ( x_1 , y_1 , z_1 , w_1)$,则$p_1$在投影盒子里的投影(x 与 y 发生偏移，z 位置不变)为：
   8. $\frac{x_2}{x_1} = \frac{-n}{z_1}$
   9. $x_2 = \frac{-n}{z_1} \times x_1$
   10. 同理可得 $y_2 =\frac{-n}{z_1} \times y_1$
   11. $p_2 = ( \frac{-n}{z_1} \times x_1 , \frac{-n}{z_1} \times y_1 , z_1 , w_1)$
2. 将投影盒子与裁剪空间进行映射
   1. 设裁剪空间中$p_3 = ( x_3 , y_3 , z_3 , w_3 )$与投影盒子中的$p_2$对应
   2. 设$p_4 = p_3 \times -z_1$
   3. $w_3 = w_1$
   4. $\frac{ x_3 - (-1) }{ 1 - (-1) } = \frac{ \frac{-n}{z_1} \times x_1 - left }{ right - left }$
   5. $x_4 = -z_1 \times x_3 = \frac{ 2 \times n}{ right - left } \times x_1 + \frac{ right + left }{ right - left } \times z_1$
   6. 同理可得$x_4 = -z_1 \times x_3 = \frac{ 2 \times n}{ top - bottom } \times y_1 + \frac{ top + bottom }{ top - bottom } \times z_1$
   7. 为了方便矩阵运算，逆向求$z_4$的结果 $$projectionMatrix = \begin{bmatrix} \frac{ 2 \times n}{ right - left } & 0 & 0 & 0 \\ 0 & \frac{ 2 \times n}{ top - bottom } & 0 & 0 \\ 0 & 0 & k & -1 \\ \frac{ right + left }{ right - left } & \frac{ top + bottom }{ top - bottom } & b & 1 \\ \end{bmatrix}$$
   8. $z_4 = -z_1 \times z_3 = k \times z_1 + b$
      • 当$z_1 = -n$时,$z_3 = -1$
      • 当$z_1 = -f$时,$z_3 = 1$
      • 求出$z_4 = -\frac{far + near}{far-near} \times z_1 - \frac{2 \times near \times far}{far - near}$
3. 最后求出 $$projectionMatrix = \begin{bmatrix} \frac{ 2 \times n}{ right - left } & 0 & 0 & 0 \\ 0 & \frac{ 2 \times n}{ top - bottom } & 0 & 0 \\ 0 & 0 & - \frac{ far + near }{ far-near } & -1 \\ \frac{ right + left }{ right - left } & \frac{ top + bottom }{ top - bottom } & - \frac{ 2 \times near \times far }{ far - near } & 1 \\ \end{bmatrix}$$