Lecture 3 Transformation1. Why study transformation2. 2D transformations3. Homogeneous coordinates4. 3D Transforms5. Viewing transformation

1. Why study transformation

1.1 Modeling

1.2 Viewing

2. 2D transformations

Reflection Matrix

$\left[\begin{array}{l}{x}^{\prime }\\ y^{\prime }\end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{l}x\\ y\end{array}\right]$

Shear Matrix

$\left[\begin{array}{l}{x}^{\prime }\\ y^{\prime }\end{array}\right]=\left[\begin{array}{ll}1& a\\ 0& 1\end{array}\right]\left[\begin{array}{l}x\\ y\end{array}\right]$

Rotation Matrix

${\mathbf{R}}_{\theta }=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$

Linear Transforms = Matrices

$x^{\prime }=ax+by$
$y^{\prime }=cx+dy$

$\left[\begin{array}{l}{x}^{\prime }\\ y^{\prime }\end{array}\right]=\left[\begin{array}{ll}a& b\\ c& d\end{array}\right]\left[\begin{array}{l}x\\ y\end{array}\right]$

${\mathbf{x}}^{\prime }=\mathbf{M}\mathbf{x}$

3. Homogeneous coordinates

Translation cannot be represented in matrix form
Solution: Homogenous Coordinates
Add a third coordinate (w-coordinate)

• 2D point = (x, y, 1)T
• 2D vector = (x, y, 0)T

$\left(\begin{array}{l}{x}^{\prime }\\ y^{\prime }\\ w^{\prime }\end{array}\right)=\left(\begin{array}{ccc}1& 0& t_x\\ 0& 1& t_y\\ 0& 0& 1\end{array}\right)\cdot \left(\begin{array}{l}x\\ y\\ 1\end{array}\right)=\left(\begin{array}{c}x+t_x\\ y+t_y\\ 1\end{array}\right)$

why point and vector are different?
because the translation should not change the direction of vector
Valid operation if w-coordinate of result is 1 or 0
•vector + vector = vector
•point – point = vector
•point + vector = point
•point + point = ?? (meaningless) (mid point)

3.1 Affine Transformations

Affine map = linear map + translation
$\left(\begin{array}{l}{x}^{\prime }\\ y^{\prime }\end{array}\right)=\left(\begin{array}{ll}a& b\\ c& d\end{array}\right)\cdot \left(\begin{array}{l}x\\ y\end{array}\right)+\left(\begin{array}{l}{t}_x\\ t_y\end{array}\right)$

Using homogenous coordinates:
$\left(\begin{array}{l}{x}^{\prime }\\ y^{\prime }\\ 1\end{array}\right)=\left(\begin{array}{ccc}a& b& t_x\\ c& d& t_y\\ 0& 0& 1\end{array}\right)\cdot \left(\begin{array}{l}x\\ y\\ 1\end{array}\right)$

Inverse Transform

inverse transform = invers matrix

3.2 Composite Transform

Transform Ordering Matters
$R_{45}\cdot T_{(1,0)}\ne T_{(1,0)}\cdot R_{45}$
Note that matrices are applied right to left:
$T_{(1,0)}\cdot R_{45}\left[\begin{array}{l}x\\ y\\ 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}\cos 45^{\circ }& -\sin 45^{\circ }& 0\\ \sin 45^{\circ }& \cos 45^{\circ }& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{l}x\\ y\\ 1\end{array}\right]$

Decomposing Complex Transforms

4. 3D Transforms

Use homogeneous coordinates again:
•3D point = (x, y, z, 1)T
•3D vector = (x, y, z, 0)T

Use 4×4 matrices for affine transformations
$\left(\begin{array}{l}{x}^{\prime }\\ y^{\prime }\\ z^{\prime }\\ 1\end{array}\right)=\left(\begin{array}{cccc}a& b& c& t_x\\ d& e& f& t_y\\ g& h& i& t_z\\ 0& 0& 0& 1\end{array}\right)\cdot \left(\begin{array}{l}x\\ y\\ z\\ 1\end{array}\right)$

Scale
$\mathbf{S}\left(s_x,s_y,s_z\right)=\left(\begin{array}{cccc}{s}_x& 0& 0& 0\\ 0& s_y& 0& 0\\ 0& 0& s_z& 0\\ 0& 0& 0& 1\end{array}\right)$

Translation
$\mathbf{T}\left(t_x,t_y,t_z\right)=\left(\begin{array}{cccc}1& 0& 0& t_x\\ 0& 1& 0& t_y\\ 0& 0& 1& t_z\\ 0& 0& 0& 1\end{array}\right)$

Rotation around x-, y-, or z-axis
${\mathbf{R}}_x(\alpha )=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& \cos \alpha & -\sin \alpha & 0\\ 0& \sin \alpha & \cos \alpha & 0\\ 0& 0& 0& 1\end{array}\right)$
${\mathbf{R}}_y(\alpha )=\left(\begin{array}{cccc}\cos \alpha & 0& \sin \alpha & 0\\ 0& 1& 0& 0\\ -\sin \alpha & 0& \cos \alpha & 0\\ 0& 0& 0& 1\end{array}\right)$
${\mathbf{R}}_z(\alpha )=\left(\begin{array}{cccc}\cos \alpha & -\sin \alpha & 0& 0\\ \sin \alpha & \cos \alpha & 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)$
Compose any 3D rotation
${\mathbf{R}}_{xyz}(\alpha ,\beta ,\gamma )={\mathbf{R}}_x(\alpha ){\mathbf{R}}_y(\beta ){\mathbf{R}}_z(\gamma )$
So-called Euler angles; Often used in flight simulators: roll, pitch, yaw

Rotation by angle αaround axis n
$\mathbf{R}(\mathbf{n},\alpha )=\cos (\alpha )\mathbf{I}+(1-\cos (\alpha ))\mathbf{n}{\mathbf{n}}^T+\sin (\alpha )\left(\begin{array}{ccc}0& -n_z& n_y\\ n_z& 0& -n_x\\ -n_y& n_x& 0\end{array}\right)$
一般情况下，说明沿某个轴的方向旋转，都默认该轴是经过原点的。
若是沿任意轴（不经过原点）旋转，则先平移再旋转再平移。

5. Viewing transformation

5.1 View / Camera transformation

Think about how to take a photo
-Find a good place and arrange people (model transformation)
-Find a good “angle” to put the camera (view transformation)
-Cheese! (projection transformation)
MVP变换

Define the camera first

• Position $\vec{e}$
• Look-at / gaze direction $\hat{g}$
• Up direction $\hat{t}$ (assuming perp. to look-at)

How about that we always transform the camera to

• The origin, up at Y, look at -Z
• And transform the objects along with the camera

Transform the camera by $M_{view}$

So it’s located at the origin, up at Y, look at -Z

$M_{\text{view}}$ in math

• Let’s write $M_{\text{view}}=R_{\text{view}}{T}_{\text{view}}$
• Translate e to origin
  $$T_{view}=\left[\begin{array}{cccc}1& 0& 0& -x_e\\ 0& 1& 0& -y_e\\ 0& 0& 1& -z_e\\ 0& 0& 0& 1\end{array}\right]$$
• Rotate g to $-z,t$ to $Y,(g\times t)$ To $X$
• Consider its inverse rotation: $X$ to $(g\times t),Y$ to $t,Z$ to $-g$
  $R_{view}^{-1}=\left[\begin{array}{cccc}{x}_{\hat{g}\times \hat{t}}& x_t& x_{-g}& 0\\ y_{\hat{g}\times \hat{t}}& y_t& y_{-g}& 0\\ z_{\hat{g}\times \hat{t}}& z_t& z_{-g}& 0\\ 0& 0& 0& 1\end{array}\right]$
  因为旋转矩阵是正交矩阵，所以求逆=求转置。
  $R_{view}=\left[\begin{array}{cccc}{x}_{\hat{g}\times \hat{t}}& y_{\hat{g}\times \hat{t}}& z_{\hat{g}\times \hat{t}}& 0\\ x_t& y_t& z_t& 0\\ x_{-g}& y_{-g}& z_{-g}& 0\\ 0& 0& 0& 1\end{array}\right]$

Summary

-Transform objects together with the camera
-Until camera’s at the origin, up at Y, look at -Z
•Also known as ModelView Transformation
•But why do we need this? For projection transformation!

5.2 Projection transformation

正交投影和透视投影

5.2.1 Orthographic projection

A simple way of understanding

• Camera located at origin, looking at - $Z$, up at $Y$ (looks familiar?)
• Drop Z coordinate
• Translate and scale the resulting rectangle to $[-1,1{]}^2$ #这样做可以方便以后的计算
  
  存在问题：如何区分物体的前后

In general

• We want to map a cuboid $[l,r]\times [b,t]\times [f,n]$ to the "canonical (正则、规范、标准)”cube [-1, 1]3

Slightly different orders (to the “simple way”)

• Center cuboid by translating
• Scale into “canonical” cube
  
  定义的立方体中，l是left，r是right，在x轴上；b是bottom，t是top，在y轴上；
  f是far，n是near，在z轴上，但是这里注意z轴上是远的地方数值更小，因为是看-z方向。（然而，在一些API中，例如OpenGL中是定义的左手系，就不存在这个问题；但是这样的话x和y叉乘就不是z）

Transformation matrix

• Translate (center to origin) first, then scale (length/width/height to 2)

$M_{\text{ortho}}=\left[\begin{array}{cccc}\frac{2}{r-l}& 0& 0& 0\\ 0& \frac{2}{t-b}& 0& 0\\ 0& 0& \frac{2}{n-f}& 0\\ 0& 0& 0& 1\end{array}\right]\left[\begin{array}{cccc}1& 0& 0& -\frac{r+l}{2}\\ 0& 1& 0& -\frac{t+b}{2}\\ 0& 0& 1& -\frac{n+f}{2}\\ 0& 0& 0& 1\end{array}\right]$

5.2.2 Perspective projection

•Most common in Computer Graphics, art, visual system
•Further objects are smaller
•Parallel lines not parallel; converge to single point

How to do perspective projection

• First “squish” the frustum into a cuboid $(\mathrm{n}->\mathrm{n},\mathrm{f}->\mathrm{f})\left({\mathrm{M}}_{\text{persp->ortho }}\right)$
• Do orthographic projection (Mortho, already known!)
  

In order to find a transformation

• Find the relationship between transformed points (x’, y’, z’) and the original points $(x,y,z)$
  $$y^{\prime }=\frac{n}{z}y{x}^{\prime }=\frac{n}{z}x\text{ (similar to y’) }$$

In homogeneous coordinates,
$$\left(\begin{array}{l}x\\ y\\ z\\ 1\end{array}\right)\Rightarrow \left(\begin{array}{c}nx/z\\ ny/z\\ \text{ unknown }\\ 1\end{array}\right)\begin{array}{l}\text{ mult. }\left(\begin{array}{c}nx\\ ny\\ \text{ by }z\\ \text{ still unknown }\\ z\end{array}\right)\end{array}$$

So the “squish” (persp to ortho) projection does this
$$M_{persp\to ortho}^{(4\times 4)}\left(\begin{array}{l}x\\ y\\ z\\ 1\end{array}\right)=\left(\begin{array}{c}nx\\ ny\\ \text{ unknown }\\ z\end{array}\right)$$

Already good enough to figure out part of ${\mathrm{M}}_{\text{persp->ortho }}$
$$M_{persp\to ortho}=\left(\begin{array}{llll}n& 0& 0& 0\\ 0& n& 0& 0\\ ?& ?& ?& ?\\ 0& 0& 1& 0\end{array}\right)$$

Observation: the third row is responsible for $z^{\prime }$

• Any point on the near plane will not change
• Any point’s z on the far plane will not change

下面是推导和原理

最后

以上就是和谐手链为你收集整理的Lecture 3 Transformation1. Why study transformation2. 2D transformations3. Homogeneous coordinates4. 3D Transforms5. Viewing transformation的全部内容，希望文章能够帮你解决Lecture 3 Transformation1. Why study transformation2. 2D transformations3. Homogeneous coordinates4. 3D Transforms5. Viewing transformation所遇到的程序开发问题。

如果觉得靠谱客网站的内容还不错，欢迎将靠谱客网站推荐给程序员好友。