244 lines
7.6 KiB
Markdown
244 lines
7.6 KiB
Markdown
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---
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title: "3D基本变换和观测变换(viewing transform)"
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date: 2022-05-10T09:10:25+08:00
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---
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## 3D基本变换
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### 平移矩阵
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通常处理三维中对模型进行平移的行为
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<br>$T(t_x,t_y,t_z) = \begin{pmatrix} 1 &0 &0 &t_x \\\\ 0 &1 &0 &t_y \\\\ 0 &0 &0 &t_z \\\\ 0 &0 &0 &1 \end{pmatrix} $<br>
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$t_x,t_y,t_z$通常表示对应轴上平移的距离
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--------------------------
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### 缩放矩阵
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通常处理三维中对模型进行缩放的行为
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$S(s_x,s_y,s_z) = \begin{pmatrix} s_x &0 &0 &0 \\\\0 &s_y &0 &0 \\\\ 0 &0 &s_z &0 \\\\ 0 &0 &0 &1 \end{pmatrix} $
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$s_x,s_y,s_z$通常表示对应xyz轴的缩放比例
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-------------------------------------------
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### 旋转矩阵
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这个矩阵通常处理三维中对模型绕坐标轴进行旋转的行为
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-------------
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绕x轴旋转的矩阵为
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$\begin{pmatrix} 1 &0 &0 &0 \\\\ 0 &cos(r) &-sin(r) &0 \\\\ 0 &sin(r) &cos(r) &0 \\\\ 0 &0 &0 &1 \end{pmatrix} $
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---------------------
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绕y轴旋转的矩阵为
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$\begin{pmatrix} cos(r) &0 &sin(r) &0 \\\\ 0 &1 &0 &0 \\\\ -sin(r) &0 &cos(r) &0 \\\\ 0 &0 &0 &1 \end{pmatrix} $
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-----------------------
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绕z轴旋转的矩阵为
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$\begin{pmatrix} cos(r) &-sin(r) &0 &0 \\\\ sin(r) &cos(r) &0 &0 \\\\ 0 &0 &1 &0 \\\\ 0 &0 &0 &1 \end{pmatrix} $
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----------------------
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对于给定三个旋转角度的旋转，通常使用欧拉角
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$R_{xyz}(\alpha,\beta,\gamma)=R_x(\alpha)R_y(\beta)R_z(\gamma)$
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此时的三个旋转方向将被称为roll,pich,yaw
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-----------------------
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对于围绕某一特定点进行旋转的行为，则将该点平移至原点处后视为绕特定轴旋转
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---------------------------
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### 视角变换矩阵
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这个矩阵通常用来定义相机对应的视角朝向，利用这个矩阵来将相机位置移动到原点，便于后续的模型进行平移旋转等变换
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```c++
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Eigen::Matrix4f view = Eigen::Matrix4f::Identity();
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Eigen::Matrix4f translate;
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translate << 1, 0, 0, -eye_pos[0],
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0, 1, 0, -eye_pos[1],
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0, 0, 1, -eye_pos[2],
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0, 0, 0, 1;
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view = translate * view;//移动相机位置到顶点
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```
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在上述代码中，eye_pos(x,y,z,1)往往为相机的位置
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-----------------------------------------------------------
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### 正交/投影矩阵
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#### 正交矩阵
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正交矩阵将使摄像头置于坐标系原点，看向-Z轴方向，可以在Y轴上平行移动。
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最终结果将表现为在XY轴平面上的2D图像，让模型坐标归一化到[-1,1]之间
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----------------------------------
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总体流程
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为了将一个$[l,r]\times[b,t]\times[f,n]$的长方体转换为符合canonical(正则、规范、标准)的正方体，我们需要进行两步操作
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第一步
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平移这个长方体到坐标系的原点
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对应矩阵$M_{translate}=\begin{bmatrix} \frac{2}{r-l} &0 &0 &0 \\\\ 0 &\frac{2}{t-b} &0 &0 \\\\ 0 &0 &0 &\frac{2}{n-f} \\\\ 0 &0 &0 &1 \end{bmatrix}$
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第二步
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缩放这个长方体到符合正则、规范、标准的正方体
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对应矩阵$M_{scale}=\begin{bmatrix} 1 &0 &0 &-\frac{r+l}{2} \\\\ 0 &1 &0 &-\frac{t+b}{2} \\\\ 0 &0 &1 &-\frac{n+f}{2} \\\\ 0 &0 &0 &1 \end{bmatrix}$
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由第一二步可得出
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正交矩阵为$M_{ortho}=\begin{bmatrix} \frac{2}{r-l} &0 &0 &0 \\\\ 0 &\frac{2}{t-b} &0 &0 \\\\ 0 &0 &0 &\frac{2}{n-f} \\\\ 0 &0 &0 &1 \end{bmatrix}\times\begin{bmatrix} 1 &0 &0 &-\frac{r+l}{2} \\\\ 0 &1 &0 &-\frac{t+b}{2} \\\\ 0 &0 &1 &-\frac{n+f}{2} \\\\ 0 &0 &0 &1 \end{bmatrix}$
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图示
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--------------------
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#### 投影矩阵
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投影矩阵将使模型满足自然界透视效果，如物体近大远小、所有的平行线变得不再平行，总会交于一点
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------
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推导过程
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将远平面与近平面连线形成的梯形“挤压”到成为一个正方体
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24R8DF[AB98E9]@1.png)
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挤压的过程对梯形做横切面可知，计算挤压后的坐标点值实质上为计算相似三角形
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如下图，可知挤压后的坐标与之前的坐标存在的数学关系为
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$$y^{'}=\frac{n}{z}y$$ $$x^{'}=\frac{n}{z}x$$
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由此可知经过“挤压”后的坐标为
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$$M_{persp->ortho}\times\begin{pmatrix}
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x \\\\ y \\\\ z \\\\ 1
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\end{pmatrix}=\begin{pmatrix}
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nx \\\\ ny \\\\ {未知} \\\\ z
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\end{pmatrix}$$
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所以
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$$M_{persp->ortho}=\begin{pmatrix}
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n &0 &0 &0 \\\\ 0 &n &0 &0 \\\\ ? &? &? &? \\\\ 0 &0 &1 &0
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\end{pmatrix}$$
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又由远平面在被“挤压”后相当于近平面做正交投影得到的远平面，可知
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1. 任何在近平面上的点的坐标在“挤压”的过程中不发生改变
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2. 任何在远平面的点的坐标中的Z值不发生改变,即$$\begin{pmatrix}
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0 \\\\ 0 \\\\ f \\\\ 1
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\end{pmatrix}\rArr\begin{pmatrix}
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0 \\\\ 0 \\\\ f \\\\ 1
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\end{pmatrix}==\begin{pmatrix}
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0 \\\\ 0 \\\\ f^2 \\\\ f
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\end{pmatrix}$$
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在将上述坐标公式中Z的值以n替换之后可得
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$$\begin{pmatrix}
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x \\\\ y \\\\ n \\\\ 1
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\end{pmatrix}\rArr\begin{pmatrix}
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x \\\\ y \\\\ n \\\\ 1
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\end{pmatrix}==\begin{pmatrix}
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nx \\\\ ny \\\\ n^2 \\\\ n
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\end{pmatrix}$$
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据此，可推测第三行未知坐标值符合以下关系
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$$\begin{pmatrix}
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0 &0 &A &B
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\end{pmatrix}\begin{pmatrix}
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x \\\\ y \\\\ n \\\\ 1
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\end{pmatrix}=n^2$$
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因此
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$$An+B=n^2$$
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联立性质2推导的
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$$Af+B=f^2$$
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可得出
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$$A=n+f$$
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$$B=-nf$$
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所以
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$$M_{persp}=M_{ortho}M_{persp->ortho}=$$
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$$\begin{pmatrix} \frac{2}{r-l} &0 &0 &0 \\\\ 0 &\frac{2}{t-b} &0 &0 \\\\ 0 &0 &0 &\frac{2}{n-f} \\\\ 0 &0 &0 &1 \end{pmatrix}$$
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$$\times$$
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$$\begin{pmatrix} 1 &0 &0 &-\frac{r+l}{2} \\\\ 0 &1 &0 &-\frac{t+b}{2} \\\\ 0 &0 &1 &-\frac{n+f}{2} \\\\ 0 &0 &0 &1 \end{pmatrix}$$
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$$\times$$
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$$\begin{pmatrix} n &0 &0 &0 \\\\ 0 &n &0 &0 \\\\ 0 &0 &n+f &-nf \\\\ 0 &0 &1 &0 \end{pmatrix}$$
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----------------------------------------------
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#### 涉及FovY的投影矩阵
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FovY表示视域，即摄像机在固定时能看到的最大角度或最低角度的范围
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Aspect ratio 表示纵横比，投影平面的长宽比
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对应的相似三角形关系不变，参数改变
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可得如下关系
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$$\tan{\frac{fovY}{2}}=\frac{t}{|n|}$$
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$$aspect=\frac{r}{t}$$
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因此
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$$t=near\times tan(\frac{fovY}{2})$$
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$$r=aspect\times near\times tan(\frac{fovY}{2})$$
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$$l=-aspect\times near \times tan(fovY/2)$$
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带入上述由l,b,n,f构成的矩阵可得
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$$M_{persp->ortho}$$
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$$=$$
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$$\begin{pmatrix} \frac{2}{r-l} &0 &0 &0 \\\\ 0 &\frac{2}{t-b} &0 &0 \\\\ 0 &0 &\frac{2}{n-f} &0 \\\\ 0 &0 &0 &1 \end{pmatrix}$$
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$$\times$$
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$$\begin{pmatrix} 1 &0 &0 &-\frac{r+l}{2} \\\\ 0 &1 &0 &-\frac{t+b}{2} \\\\ 0 &0 &1 &-\frac{n+f}{2} \\\\ 0 &0 &0 &1 \end{pmatrix}$$
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$$=$$
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$$\begin{pmatrix}
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\frac{\frac{\cot{FovY}}{2}}{apsect*near} &0 &0 &0 \\\\ 0 & \frac{\frac{\cot{FovY}}{2}}{near} & 0 & 0 \\\\ 0 & 0 & \frac{2}{near-far} & 0 \\\\ 0 & 0 & 0 & 1
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\end{pmatrix}$$
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$$\times$$
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$$\begin{pmatrix}
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1 &0 &0 &0 \\\\ 0 &1 &0 &0 \\\\ 0 &0 &1 &-\frac{near+far}{2} \\\\ 0 &0 &0 &1
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\end{pmatrix}$$
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$$=$$
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$$\begin{pmatrix}
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\frac{\frac{\cot{FovY}}{2}}{apsect * near} &0 &0 &0 \\\\ 0 &\frac{\frac{\cot{FovY}}{2}}{near} &0 &0 \\\\ 0 &0 &\frac{2}{near-far} &-\frac{near+far}{near-far} \\\\ 0 &0 &0 &1
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\end{pmatrix}$$
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----------------
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#### 视口变换
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经过MVP矩阵计算后得到的一个正则的正方体需要将X轴和Y轴上的坐标映射到屏幕坐标[0,width]$\times$[0,height]
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变换时需要先将[-1,1]缩放到屏幕大小[width,height]，再进行平移使得原点坐标与屏幕原点对齐
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变换矩阵
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$$M_{viewport}=\begin{bmatrix}
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\frac{width}{2} &0 &0 &\frac{width}{2} \\\\ 0 &\frac{height}{2} &0 &\frac{height}{2} \\\\ 0 &0 &1 &0 \\\\ 0 &0 &0 &1
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\end{bmatrix}$$
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