变换 Transformations

通过矩阵变换 即相机外参数 包括旋转矩阵和平移矩阵 使得物体进行移动。各类变换矩阵定义如下

缩放矩阵

$$\begin{bmatrix} \color{red}{S_1} & \color{red}0 & \color{red}0 & \color{red}0 \\ \color{green}0 & \color{green}{S_2} & \color{green}0 & \color{green}0 \\ \color{blue}0 & \color{blue}0 & \color{blue}{S_3} & \color{blue}0 \\ \color{purple}0 & \color{purple}0 & \color{purple}0 & \color{purple}1 \end{bmatrix} \cdot \begin{pmatrix} x \\ y \\ z \\ 1 \end{pmatrix} = \begin{pmatrix} \color{red}{S_1} \cdot x \\ \color{green}{S_2} \cdot y \\ \color{blue}{S_3} \cdot z \\ 1 \end{pmatrix}$$

位移矩阵

$$\begin{bmatrix} \color{red}1 & \color{red}0 & \color{red}0 & \color{red}{T_x} \\ \color{green}0 & \color{green}1 & \color{green}0 & \color{green}{T_y} \\ \color{blue}0 & \color{blue}0 & \color{blue}1 & \color{blue}{T_z} \\ \color{purple}0 & \color{purple}0 & \color{purple}0 & \color{purple}1 \end{bmatrix} \cdot \begin{pmatrix} x \\ y \\ z \\ 1 \end{pmatrix} = \begin{pmatrix} x + \color{red}{T_x} \\ y + \color{green}{T_y} \\ z + \color{blue}{T_z} \\ 1 \end{pmatrix}$$

旋转矩阵

• 沿x轴旋转 $$\begin{bmatrix} \color{red}1 & \color{red}0 & \color{red}0 & \color{red}0 \\ \color{green}0 & \color{green}{\cos \theta} & - \color{green}{\sin \theta} & \color{green}0 \\ \color{blue}0 & \color{blue}{\sin \theta} & \color{blue}{\cos \theta} & \color{blue}0 \\ \color{purple}0 & \color{purple}0 & \color{purple}0 & \color{purple}1 \end{bmatrix} \cdot \begin{pmatrix} x \\ y \\ z \\ 1 \end{pmatrix} = \begin{pmatrix} x \\ \color{green}{\cos \theta} \cdot y - \color{green}{\sin \theta} \cdot z \\ \color{blue}{\sin \theta} \cdot y + \color{blue}{\cos \theta} \cdot z \\ 1 \end{pmatrix}$$
• 沿y轴旋转 $$\begin{bmatrix} \color{red}{\cos \theta} & \color{red}0 & \color{red}{\sin \theta} & \color{red}0 \\ \color{green}0 & \color{green}1 & \color{green}0 & \color{green}0 \\ - \color{blue}{\sin \theta} & \color{blue}0 & \color{blue}{\cos \theta} & \color{blue}0 \\ \color{purple}0 & \color{purple}0 & \color{purple}0 & \color{purple}1 \end{bmatrix} \cdot \begin{pmatrix} x \\ y \\ z \\ 1 \end{pmatrix} = \begin{pmatrix} \color{red}{\cos \theta} \cdot x + \color{red}{\sin \theta} \cdot z \\ y \\ - \color{blue}{\sin \theta} \cdot x + \color{blue}{\cos \theta} \cdot z \\ 1 \end{pmatrix}$$
• 沿z轴旋转 $$\begin{bmatrix} \color{red}{\cos \theta} & - \color{red}{\sin \theta} & \color{red}0 & \color{red}0 \\ \color{green}{\sin \theta} & \color{green}{\cos \theta} & \color{green}0 & \color{green}0 \\ \color{blue}0 & \color{blue}0 & \color{blue}1 & \color{blue}0 \\ \color{purple}0 & \color{purple}0 & \color{purple}0 & \color{purple}1 \end{bmatrix} \cdot \begin{pmatrix} x \\ y \\ z \\ 1 \end{pmatrix} = \begin{pmatrix} \color{red}{\cos \theta} \cdot x - \color{red}{\sin \theta} \cdot y \\ \color{green}{\sin \theta} \cdot x + \color{green}{\cos \theta} \cdot y \\ z \\ 1 \end{pmatrix}$$ 聚合矩阵 $$\begin{bmatrix} \cos \theta + \color{red}{R_x}^2(1 - \cos \theta) & \color{red}{R_x}\color{green}{R_y}(1 - \cos \theta) - \color{blue}{R_z} \sin \theta & \color{red}{R_x}\color{blue}{R_z}(1 - \cos \theta) + \color{green}{R_y} \sin \theta & 0 \\ \color{green}{R_y}\color{red}{R_x} (1 - \cos \theta) + \color{blue}{R_z} \sin \theta & \cos \theta + \color{green}{R_y}^2(1 - \cos \theta) & \color{green}{R_y}\color{blue}{R_z}(1 - \cos \theta) - \color{red}{R_x} \sin \theta & 0 \\ \color{blue}{R_z}\color{red}{R_x}(1 - \cos \theta) - \color{green}{R_y} \sin \theta & \color{blue}{R_z}\color{green}{R_y}(1 - \cos \theta) + \color{red}{R_x} \sin \theta & \cos \theta + \color{blue}{R_z}^2(1 - \cos \theta) & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$ 明显的 这样的旋转方式会造成万向节死锁问题，解决方案为使用四元数

译注对四元数的理解会用到非常多的数学知识。如果你想了解四元数与3D旋转之间的关系，可以来阅读我的教程。如果你对万向节死锁的概念仍不是那么清楚，可以来阅读我教程的Bonus章节。
现在3Blue1Brown也已经开始了一个四元数的视频系列，他采用球极平面投影(Stereographic Projection)的方式将四元数投影到3D空间，同样有助于理解四元数的概念（仍在更新中）：https://www.youtube.com/watch?v=d4EgbgTm0Bg

矩阵相乘即可简单的获取顺序化的变换矩阵。

$$Trans . Scale = \begin{bmatrix} \color{red}1 & \color{red}0 & \color{red}0 & \color{red}1 \\ \color{green}0 & \color{green}1 & \color{green}0 & \color{green}2 \\ \color{blue}0 & \color{blue}0 & \color{blue}1 & \color{blue}3 \\ \color{purple}0 & \color{purple}0 & \color{purple}0 & \color{purple}1 \end{bmatrix} . \begin{bmatrix} \color{red}2 & \color{red}0 & \color{red}0 & \color{red}0 \\ \color{green}0 & \color{green}2 & \color{green}0 & \color{green}0 \\ \color{blue}0 & \color{blue}0 & \color{blue}2 & \color{blue}0 \\ \color{purple}0 & \color{purple}0 & \color{purple}0 & \color{purple}1 \end{bmatrix} = \begin{bmatrix} \color{red}2 & \color{red}0 & \color{red}0 & \color{red}1 \\ \color{green}0 & \color{green}2 & \color{green}0 & \color{green}2 \\ \color{blue}0 & \color{blue}0 & \color{blue}2 & \color{blue}3 \\ \color{purple}0 & \color{purple}0 & \color{purple}0 & \color{purple}1 \end{bmatrix}$$

GLM

相关变换矩阵使用库glm其中有mat4类型的矩阵用于旋转变换

glm::mat4 trans = glm::mat4(1.0f);
//平移
trans = glm::translate(trans, glm::vec3(-0.1f, 0.0, 0.0));
//旋转
trans = glm::rotate(trans, glm::radians(90.0f), glm::vec3(0.0, 0.0, 1.0));
//缩放
trans = glm::scale(trans, glm::vec3(0.5, 0.5, 0.5));

glsl中也有对应mat4的变量 实际上是一个 float[16]

uniform mat4 transform;
[...]
//左乘变换矩阵完成变换操作
gl_Position = transform * vec4(aPos, 1.0);

同样通过uniform变量从C代码中传入

glm::mat4 trans = glm::mat4(1.0f);
int transLocation = glGetUniformLocation(shader.ID, "transform");
glUniformMatrix4fv(transLocation, 1, GL_FALSE, glm::value_ptr(trans));