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作业1及笔记
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139 changes: 139 additions & 0 deletions 一些笔记.md
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# GAMES101 笔记

## 课程注解

### 旋转变换矩阵是正交矩阵


### 先平移再线性变换

$$
\begin{pmatrix}
a&b&c&t_x\\
d&e&f&t_y\\
g&h&i&t_z\\
0&0&0&1\\
\end{pmatrix}=
\begin{pmatrix}
a&b&c&0\\
d&e&f&0\\
g&h&i&0\\
0&0&0&1\\
\end{pmatrix}\begin{pmatrix}
1&0&0&t_x\\
0&1&0&t_y\\
0&0&1&t_z\\
0&0&0&1\\
\end{pmatrix}
$$


$$
变换=线性变换\times平移变换
$$


### 透视投影

![1689059400149](image/一些笔记/1689059400149.png)

对远端平面上的任意一点,x、y坐标的缩放如下图所示

n为原点与近端平面的距离

![1689059532544](image/一些笔记/1689059532544.png)

因此变换后的齐次坐标为

$$
\begin{pmatrix}
nx/z\\ny/z\\z'\\1
\end{pmatrix}==\begin{pmatrix}
nx\\ny\\zz'\\z
\end{pmatrix}
$$

设变换矩阵的为

$$
\begin{pmatrix}
n&0&0&0\\
0&n&0&0\\
A&B&C&D\\
0&0&1&0\\
\end{pmatrix}
$$

近端平面上任意一点的坐标不变(规定)为n

$$
\begin{pmatrix}
x\\y\\n\\1
\end{pmatrix}==\begin{pmatrix}
nx\\ny\\n^2\\n
\end{pmatrix}
$$

因此

$$
\begin{pmatrix}
A&B&C&D
\end{pmatrix}\begin{pmatrix}
x\\y\\n\\1
\end{pmatrix}=Ax+By+Cn+D=n^2\\
=>A=B=0
$$

远端平面上任意一点的z坐标不变(规定)为f

$$
\begin{pmatrix}
0&0&C&D
\end{pmatrix}\begin{pmatrix}
x\\y\\f\\1
\end{pmatrix}=Cf+D=f^2
$$

联立解得

$$
C=n+f\\
D=-nf
$$

因此透视投影的变换矩阵为

$$
\begin{pmatrix}
n&0&0&0\\
0&n&0&0\\
0&0&n+f&-nf\\
0&0&1&0\\
\end{pmatrix}
$$

对梯形中的任意一点,变换之后的坐标为

$$
\begin{pmatrix}
n&0&0&0\\
0&n&0&0\\
0&0&n+f&-nf\\
0&0&1&0\\
\end{pmatrix}\begin{pmatrix}
x\\y\\z\\1
\end{pmatrix}=\begin{pmatrix}
nx\\ny\\(n+f)z-nf\\z
\end{pmatrix}\\
其中\ f\le z\le n\lt0
$$

计算可知

$$
\frac{(n+f)z-nf}{z}\le z
$$

因此变换之后,梯形体内部的z坐标相比变换前变小了,也即变远了

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