初始基向量为i ^ ( 1 , 0 ) , j ^ ( 0 , 1 ) \hat{i}(1,0),\hat{j}(0,1) i ^ ( 1 , 0 ) , j ^ ( 0 , 1 ) i ^ ( 3 , − 2 ) , j ^ ( 2 , 1 ) \hat{i}(3,-2), \hat{j}(2,1) i ^ ( 3 , − 2 ) , j ^ ( 2 , 1 )
仅需取出向量的坐标，将它们分别于矩阵的特定列相乘，然后将结果相加即可
逆时针旋转90°的例子
当两个向量变为线性相关时，两个向量共线，二维空间会被挤压到一维
物体本身的长度、角度、大小不会变化包括：
Identity（不变）
Translation（平移）
Rotation（旋转）
以及他们的组合
保持角度
Identity（不变）
Translation（平移）
Rotation（旋转）
Isotropic Scaling（均衡缩放）
以及他们的组合
线性变换满足如下方程：
L(p + q) = L(p) + L(q)
aL(p) = L(ap)
Identity（不变）
Rotation（旋转）
Scaling（缩放）
Reflection(对称)
Shear(切变、剪切)
保持直线以及直线与直线的平行
Grid lines remain parallel and evenly spaced and such that the origin remains fixed.
它保持网格线平行且等距分布，并且保持原点不动
The objective of HC is to use a 4D column matrices to represent both points and vectors in 3D
齐次坐标的本质是使用四维数组来表示三维空间中的点和向量
If we multiply a homogeneous coordinate by an affine matrix, w is unchanged; by an projective matrix, the value of w will change
w=0时齐次坐标表示一个方向而不是一个点
[ x ′ y ′ z ′ 1 ] = [ 1 0 0 t x 0 1 0 t y 0 0 1 t z 0 0 0 1 ] ⋅ [ x y z 1 ]
\begin{bmatrix}
x'\\
y'\\
z' \\
1
\end{bmatrix} = \begin{bmatrix}
1 & 0 & 0 & t_x\\
0 & 1 & 0 & t_y\\
0 & 0 & 1 & t_z\\
0 & 0 & 0 & 1
\end{bmatrix}\cdot
\begin{bmatrix}
x\\
y\\
z\\
1
\end{bmatrix}
⎣ ⎢ ⎢ ⎡ x ′ y ′ z ′ 1 ⎦ ⎥ ⎥ ⎤ = ⎣ ⎢ ⎢ ⎡ 1 0 0 0 0 1 0 0 0 0 1 0 t x t y t z 1 ⎦ ⎥ ⎥ ⎤ ⋅ ⎣ ⎢ ⎢ ⎡ x y z 1 ⎦ ⎥ ⎥ ⎤
P ′ = [ a 0 0 0 b 0 0 0 a ] ⋅ [ P x P y P z ]
P' = \begin{bmatrix}
a & 0 & 0\\
0 & b & 0\\
0 & 0 & a
\end{bmatrix}\cdot
\begin{bmatrix}
P_x\\
P_y\\
P_z
\end{bmatrix}
P ′ = ⎣ ⎡ a 0 0 0 b 0 0 0 a ⎦ ⎤ ⋅ ⎣ ⎡ P x P y P z ⎦ ⎤
取[-y, x]为Q作为垂直于P的向量，则：
P ′ = P c o s θ + Q s i n θ
P' = Pcos\theta + Q sin\theta
P ′ = P c o s θ + Q s i n θ { P x ′ = P x c o s θ − P y s i n θ P y ′ = P y c o s θ + P x s i n θ
\left\{\begin{matrix}
P_x' = P_xcos\theta - P_ysin\theta\\
P_y' = P_ycos\theta + P_xsin\theta
\end{matrix}\right.
{ P x ′ = P x c o s θ − P y s i n θ P y ′ = P y c o s θ + P x s i n θ P ′ = [ c o s θ − s i n θ s i n θ c o s θ ] P
P' = \begin{bmatrix}
cos\theta & -sin\theta\\
sin\theta & cos\theta\\
\end{bmatrix}P
P ′ = [ c o s θ s i n θ − s i n θ c o s θ ] P
R z ( θ ) = [ c o s θ − s i n θ 0 s i n θ c o s θ 0 0 0 1 ]
R_z(\theta) = \begin{bmatrix}
cos\theta & -sin\theta & 0\\
sin\theta & cos\theta & 0\\
0 & 0 & 1\\
\end{bmatrix}
R z ( θ ) = ⎣ ⎡ c o s θ s i n θ 0 − s i n θ c o s θ 0 0 0 1 ⎦ ⎤ R x ( θ ) = [ 1 0 0 0 c o s θ − s i n θ 0 s i n θ c o s θ ]
R_x(\theta) = \begin{bmatrix}
1 & 0 & 0\\
0 & cos\theta & -sin\theta\\
0 & sin\theta & cos\theta\\
\end{bmatrix}
R x ( θ ) = ⎣ ⎡ 1 0 0 0 c o s θ s i n θ 0 − s i n θ c o s θ ⎦ ⎤ R y ( θ ) = [ c o s θ 0 s i n θ 0 1 0 − s i n θ 0 c o s θ ]
R_y(\theta) = \begin{bmatrix}
cos\theta & 0 & sin\theta\\
0 & 1 & 0\\
-sin\theta & 0 & cos\theta\\
\end{bmatrix}
R y ( θ ) = ⎣ ⎡ c o s θ 0 − s i n θ 0 1 0 s i n θ 0 c o s θ ⎦ ⎤
p r o j A P = ( A ⋅ P ) A p e r p A P = P − ( A ⋅ P ) A
proj_AP = (A\cdot P)A\\
perp_AP = P - (A\cdot P)A
p r o j A P = ( A ⋅ P ) A p e r p A P = P − ( A ⋅ P ) A
因为与A重合的旋转后不会改变，所以仅需计算垂直的向量即可。为了计算p e r p A P perp_AP p e r p A P A × P A\times P A × P
rotation of p e r p A P perp_AP p e r p A P P ′ = [ P − ( A ⋅ P ) A ] c o s θ + ( A × P ) s i n θ + ( A ⋅ P ) A
P' = [P - (A\cdot P)A]cos\theta + (A \times P)sin\theta + (A\cdot P)A
P ′ = [ P − ( A ⋅ P ) A ] c o s θ + ( A × P ) s i n θ + ( A ⋅ P ) A
We could find that translation, rotation and isotropic scale preserve the correct normal, but shear or isotropic scale will make the normal incorrect.
If M is a 3×3 matrix with which we transform a vertex position, then the same matrix M
N , T N,T N , T N ′ , T ′ N',T' N ′ , T ′
N ⋅ T = N T T = 0
N\cdot T = N^T T = 0
N ⋅ T = N T T = 0 N ′ ⋅ T ′ = N ′ T T ′ = 0
N'\cdot T' = N'^T T' = 0
N ′ ⋅ T ′ = N ′ T T ′ = 0 N T ( M − 1 M ) T = N T M − 1 T ′ = 0
N^T (M^{-1} M)T = N^TM^{-1} T'= 0
N T ( M − 1 M ) T = N T M − 1 T ′ = 0 N ′ T = N T M − 1
N'^T = N^TM^{-1}
N ′ T = N T M − 1 N ′ = ( M − 1 ) T N
N' = (M^{-1})^T N
N ′ = ( M − 1 ) T N
So the matrix is the transpose of inverse of M
大拇指指向为x轴正方向，食指指向为y轴正方向，中指指向为z轴正方向，OpenGL采用右手坐标系，DirectX为左手坐标系
When the focal point is at infinity, the rays are parallel and orthogonal to the image plane, When xy-plane is the image plane, ( x , y , z ) − > ( x , y , 0 ) \bm{(x,y,z) -> (x,y,0)} ( x , y , z ) − > ( x , y , 0 )
为了后续裁剪的方便，我们需要将物体的坐标转换到canonical view volume (CVV) 中，转换后的坐标被称为normalized device coordinates (NDC) ，这里l，r，b，t，n，f采用对应的x轴y轴z轴坐标值。投影就是将长方形的视域体压缩到指定范围即可，从x轴开始，视域体中的点的x坐标范围在[l, r]，想把它变换到范围在[-1, 1]：
l ⩽ x ⩽ r 0 ⩽ x − l ⩽ r − l 0 ⩽ x − l r − l ⩽ 1 − 1 ⩽ 2 ( x − l ) r − l − 1 ⩽ 1 − 1 ⩽ 2 x r − l − r + l r − l ⩽ 1 x ′ = 2 r − l x − r + l r − l
l\leqslant x \leqslant r \\
0\leqslant x - l \leqslant r - l \\
0\leqslant \frac{x - l}{r - l} \leqslant 1 \\
-1\leqslant \frac{2(x - l)}{r - l} - 1 \leqslant 1 \\
-1\leqslant \frac{2x}{r - l} - \frac{r + l}{r - l} \leqslant 1 \\
x' = \frac{2}{r - l}x - \frac{r + l}{r - l}
l ⩽ x ⩽ r 0 ⩽ x − l ⩽ r − l 0 ⩽ r − l x − l ⩽ 1 − 1 ⩽ r − l 2 ( x − l ) − 1 ⩽ 1 − 1 ⩽ r − l 2 x − r − l r + l ⩽ 1 x ′ = r − l 2 x − r − l r + l
y同理y ′ = 2 t − b y − t + b t − b
y' = \frac{2}{t - b}y - \frac{t + b}{t - b}
y ′ = t − b 2 y − t − b t + b
z需要在z=n时映射为-1，z=f时映射为1− 1 = n A + B 1 = f A + B A = 2 f − n B = − f + n f − n
-1 = nA + B \\
1 = fA + B \\
A = \frac{2}{f - n} \\
B = -\frac{f + n}{f - n} \\
− 1 = n A + B 1 = f A + B A = f − n 2 B = − f − n f + n
所以z’为
z ′ = 2 f − n z − f + n f − n
z' = \frac{2}{f - n}z - \frac{f + n}{f - n}
z ′ = f − n 2 z − f − n f + n
所以正交投影矩阵为
[ 2 r − l 0 0 − r + l r − l 0 2 t − b 0 − t + b t − b 0 0 2 f − n − f + n f − n 0 0 0 1 ]
\begin{bmatrix}
\frac{2}{r - l} & 0 & 0 & -\frac{r + l}{r - l}\\
0 & \frac{2}{t - b} & 0 & -\frac{t + b}{t - b}\\
0 & 0 & \frac{2}{f - n} & -\frac{f + n}{f - n}\\
0 & 0 & 0 & 1
\end{bmatrix}
⎣ ⎢ ⎢ ⎡ r − l 2 0 0 0 0 t − b 2 0 0 0 0 f − n 2 0 − r − l r + l − t − b t + b − f − n f + n 1 ⎦ ⎥ ⎥ ⎤
通常情况下摄像机定位在原点并且沿着z轴方向观看，所以l与r，b与t是轴对称的，我们可以定义宽度为w = r - l，高度为h = t - b，矩阵可以简化为
[ 2 w 0 0 0 0 2 h 0 0 0 0 2 f − n − f + n f − n 0 0 0 1 ]
\begin{bmatrix}
\frac{2}{w} & 0 & 0 & 0\\
0 & \frac{2}{h} & 0 & 0\\
0 & 0 & \frac{2}{f - n} & -\frac{f + n}{f - n}\\
0 & 0 & 0 & 1
\end{bmatrix}
⎣ ⎢ ⎢ ⎡ w 2 0 0 0 0 h 2 0 0 0 0 f − n 2 0 0 0 − f − n f + n 1 ⎦ ⎥ ⎥ ⎤
在OpenGL中n，f为正值，表示距离相机的远近，所以要对投影矩阵中的n,f取负
[ 2 r − l 0 0 − r + l r − l 0 2 t − b 0 − t + b t − b 0 0 − 2 f − n − f + n f − n 0 0 0 1 ]
\begin{bmatrix}
\frac{2}{r - l} & 0 & 0 & -\frac{r + l}{r - l}\\
0 & \frac{2}{t - b} & 0 & -\frac{t + b}{t - b}\\
0 & 0 & -\frac{2}{f - n} & -\frac{f + n}{f - n}\\
0 & 0 & 0 & 1
\end{bmatrix}
⎣ ⎢ ⎢ ⎡ r − l 2 0 0 0 0 t − b 2 0 0 0 0 − f − n 2 0 − r − l r + l − t − b t + b − f − n f + n 1 ⎦ ⎥ ⎥ ⎤
DirectX坐标轴z轴朝里，且将z值映射为[0,1]所以，DirectX的正交投影矩阵为
[ 2 r − l 0 0 − r + l r − l 0 2 t − b 0 − t + b t − b 0 0 1 f − n − n f − n 0 0 0 1 ]
\begin{bmatrix}
\frac{2}{r - l} & 0 & 0 & -\frac{r + l}{r - l}\\
0 & \frac{2}{t - b} & 0 & -\frac{t + b}{t - b}\\
0 & 0 & \frac{1}{f - n} & -\frac{n}{f - n}\\
0 & 0 & 0 & 1
\end{bmatrix}
⎣ ⎢ ⎢ ⎡ r − l 2 0 0 0 0 t − b 2 0 0 0 0 f − n 1 0 − r − l r + l − t − b t + b − f − n n 1 ⎦ ⎥ ⎥ ⎤
The camera is at the origin point and looks along the z-axis
The image plane is paraellel to the xy-plane at distance d
( x , y , z ) − > ( ( − d / z ) x , ( − d / z ) y , d ) \bm{(x,y,z) -> ((-d/z)x, (-d/z)y, d)} ( x , y , z ) − > ( ( − d / z ) x , ( − d / z ) y , d )
在透视投影中将点投影的近平面，则x与y的变化如下
x ′ = n z x y ′ = n z y
x' = \frac{n}{z}x \\
y' = \frac{n}{z}y \\
x ′ = z n x y ′ = z n y
然后用正交投影的公式，就可以把x和y的坐标转换到[-1, 1]的范围内
x ′ = 2 r − l n z x − r + l r − l y ′ = 2 t − b n z y − t + b t − b
x' = \frac{2}{r - l}\frac{n}{z}x - \frac{r + l}{r - l} \\
y' = \frac{2}{t - b}\frac{n}{z}y - \frac{t + b}{t - b}\\
x ′ = r − l 2 z n x − r − l r + l y ′ = t − b 2 z n y − t − b t + b
这里有个z，无法直接用矩阵表示，我们可以两边都乘以z
x ′ z = 2 n r − l x − r + l r − l z y ′ z = 2 n t − b y − t + b t − b z
x'z = \frac{2n}{r - l}x - \frac{r + l}{r - l}z \\
y'z = \frac{2n}{t - b}y - \frac{t + b}{t - b}z\\
x ′ z = r − l 2 n x − r − l r + l z y ′ z = t − b 2 n y − t − b t + b z
对于变换后的z’来说，z’并不依赖于x和y，为了统一我们也将z’乘以z，因此z’z = az + b，z=-n时，z’=-1；z=-f时，z’=1，于是z’z如下
z ′ z = f + n f − n z − 2 f n f − n
z'z = \frac{f + n}{f - n}z - \frac{2fn}{f - n}
z ′ z = f − n f + n z − f − n 2 f n
最终透视投影矩阵如下
[ 2 n r − l 0 − r + l r − l 0 0 2 n t − b − t + b t − b 0 0 0 f + n f − n − 2 f n f − n 0 0 1 0 ]
\begin{bmatrix}
\frac{2n}{r - l} & 0 & -\frac{r + l}{r - l} & 0\\
0 & \frac{2n}{t - b} & -\frac{t + b}{t - b} & 0\\
0 & 0 & \frac{f + n}{f - n} & - \frac{2fn}{f - n}\\
0 & 0 & 1 & 0
\end{bmatrix}
⎣ ⎢ ⎢ ⎡ r − l 2 n 0 0 0 0 t − b 2 n 0 0 − r − l r + l − t − b t + b f − n f + n 1 0 0 − f − n 2 f n 0 ⎦ ⎥ ⎥ ⎤
通过矩阵计算得到的结果是(x’z, y’z, z’z, z)。可以看出，w分量此时等于变换之前的z，然后，你应用通常的步骤去除以齐次坐标，得到(x’, y’, z’, 1)
在OpenGL中n，f为正值，表示距离相机的远近，所以要对投影矩阵中的n,f取负
x ′ = − n z x y ′ = − n z y
x' = \frac{-n}{z}x \\
y' = \frac{-n}{z}y\\
x ′ = z − n x y ′ = z − n y
− x ′ z = 2 n r − l x + r + l r − l z − y ′ z = 2 n t − b y + t + b t − b z − z ′ z = − f + n f − n z − 2 f n f − n
-x'z = \frac{2n}{r - l}x + \frac{r + l}{r - l}z \\
-y'z = \frac{2n}{t - b}y + \frac{t + b}{t - b}z \\
-z'z = -\frac{f + n}{f - n}z - \frac{2fn}{f - n} \\
− x ′ z = r − l 2 n x + r − l r + l z − y ′ z = t − b 2 n y + t − b t + b z − z ′ z = − f − n f + n z − f − n 2 f n
最终OpenGL透视投影的矩阵如下
[ 2 n r − l 0 r + l r − l 0 0 2 n t − b t + b t − b 0 0 0 − f + n f − n − 2 f n f − n 0 0 − 1 0 ]
\begin{bmatrix}
\frac{2n}{r - l} & 0 & \frac{r + l}{r - l} & 0\\
0 & \frac{2n}{t - b} & \frac{t + b}{t - b} & 0\\
0 & 0 & -\frac{f + n}{f - n} & - \frac{2fn}{f - n}\\
0 & 0 & -1 & 0
\end{bmatrix}
⎣ ⎢ ⎢ ⎡ r − l 2 n 0 0 0 0 t − b 2 n 0 0 r − l r + l t − b t + b − f − n f + n − 1 0 0 − f − n 2 f n 0 ⎦ ⎥ ⎥ ⎤
DirectX坐标轴z轴朝里，且将z值映射为[0,1]，因此z’z = az + b，z=n时，z’=0；z=f时，z’=1，于是z’z如下
z ′ z = f f − n z − f n f − n
z'z = \frac{f}{f - n}z - \frac{fn}{f - n}
z ′ z = f − n f z − f − n f n
最终DirectX透视投影的矩阵如下[ 2 n r − l 0 − r + l r − l 0 0 2 n t − b − t + b t − b 0 0 0 f f − n − f n f − n 0 0 1 0 ]
\begin{bmatrix}
\frac{2n}{r - l} & 0 & -\frac{r + l}{r - l} & 0\\
0 & \frac{2n}{t - b} & -\frac{t + b}{t - b} & 0\\
0 & 0 & \frac{f}{f - n} & - \frac{fn}{f - n}\\
0 & 0 & 1 & 0
\end{bmatrix}
⎣ ⎢ ⎢ ⎡ r − l 2 n 0 0 0 0 t − b 2 n 0 0 − r − l r + l − t − b t + b f − n f 1 0 0 − f − n f n 0 ⎦ ⎥ ⎥ ⎤
2 n t − b = n h 2 = c o t f o v 2 2 n r − l = 2 n w = n h 2 1 a s p e c t = c o t f o v 2 1 a s p e c t a s p e c t = w h
\frac{2n}{t - b} = \frac{n}{\frac{h}{2}} = cot\frac{fov}{2}\\
\frac{2n}{r - l} = \frac{2n}{w} = \frac{n}{\frac{h}{2}} \frac{1}{aspect} = cot\frac{fov}{2}\frac{1}{aspect}\\
aspect = \frac{w}{h}\\
t − b 2 n = 2 h n = c o t 2 f o v r − l 2 n = w 2 n = 2 h n a s p e c t 1 = c o t 2 f o v a s p e c t 1 a s p e c t = h w
所以矩阵可以表示为
[ c o t f o v 2 a s p e c t 0 0 0 0 c o t f o v 2 0 0 0 0 f + n f − n − 2 f n f − n 0 0 1 0 ]
\begin{bmatrix}
\frac{cot\frac{fov}{2}}{aspect} & 0 & 0 & 0\\
0 & cot\frac{fov}{2} & 0 & 0\\
0 & 0 & \frac{f + n}{f - n} & - \frac{2fn}{f - n}\\
0 & 0 & 1 & 0
\end{bmatrix}
⎣ ⎢ ⎢ ⎢ ⎡ a s p e c t c o t 2 f o v 0 0 0 0 c o t 2 f o v 0 0 0 0 f − n f + n 1 0 0 − f − n 2 f n 0 ⎦ ⎥ ⎥ ⎥ ⎤
屏幕映射主要是将裁剪空间内的坐标映射成屏幕坐标，以显示到屏幕上。
x将从[ − 1 , 1 ] [-1, 1] [ − 1 , 1 ] [ x 1 , x 2 ] [x_1, x_2] [ x 1 , x 2 ]
{ − A + B = x 1 A + B = x 2 x ′ = x 2 − x 1 2 x + x 1 + x 2 2
\left\{\begin{matrix}
-A + B = x_1\\
A + B = x_2
\end{matrix}\right. \\
x' = \frac{x_2 - x_1}{2}x + \frac{x_1 + x_2}{2}
{ − A + B = x 1 A + B = x 2 x ′ = 2 x 2 − x 1 x + 2 x 1 + x 2
同理y’y ′ = y 2 − y 1 2 y + y 1 + y 2 2
y' = \frac{y_2 - y_1}{2}y + \frac{y_1 + y_2}{2}
y ′ = 2 y 2 − y 1 y + 2 y 1 + y 2
在OpenGL中x 1 = 0 , x 2 = w , y 1 = 0 , y 2 = h x_1 = 0, x_2 = w, y_1 = 0, y_2 = h x 1 = 0 , x 2 = w , y 1 = 0 , y 2 = h
[ w 2 0 0 w 2 0 h 2 0 h 2 0 0 1 0 0 0 0 1 ]
\begin{bmatrix}
\frac{w}{2} & 0 & 0 & \frac{w}{2}\\
0 & \frac{h}{2} & 0 & \frac{h}{2}\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{bmatrix}
⎣ ⎢ ⎢ ⎡ 2 w 0 0 0 0 2 h 0 0 0 0 1 0 2 w 2 h 0 1 ⎦ ⎥ ⎥ ⎤
在DirectX中x 1 = 0 , x 2 = w , y 1 = h , y 2 = 0 x_1 = 0, x_2 = w, y_1 = h, y_2 = 0 x 1 = 0 , x 2 = w , y 1 = h , y 2 = 0
[ w 2 0 0 w 2 0 − h 2 0 h 2 0 0 1 0 0 0 0 1 ]
\begin{bmatrix}
\frac{w}{2} & 0 & 0 & \frac{w}{2}\\
0 & -\frac{h}{2} & 0 & \frac{h}{2}\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{bmatrix}
⎣ ⎢ ⎢ ⎡ 2 w 0 0 0 0 − 2 h 0 0 0 0 1 0 2 w 2 h 0 1 ⎦ ⎥ ⎥ ⎤
