GAMES101_ 字节宝

使用计算机synthesize(合成) manipulate（操作) 可视化信息

Rasterization 光栅化
- project geometry primitives (3D triangles / polygons) onto the screen 将几何图形（3D三角形 / 多边形）投射到屏幕上
- break projected primitives into fragments (pixels) 将投影图元分解到片段(像素)
- gold standard in video games (real-time applications)
curves and meshes 曲线和栅格
- 怎样represent geometry in CG
ray tracing 光线追踪
- shoot rays from camera though each pixel
  - calculate intersection and shading 交叉点和阴影
  - continue to bounce the rays till they hit light sources
- gold standard in animations / movies (offline离线 application)
animation simulation
- key frame animation 关键帧动画
- mass-spring system 弹簧振子系统

No clear boundaries

noting: 只记了part

unit vector
- 单位向量，
- 用来代表方向
dot product in graphics vec{a}cdotvec{b} = |vec{a}|cdot|vec{b}|costheta Find angle between two vectors (e.g. cosine of angle between light source 光源 and surface表面) Finding projection of one vector on another measure how close two directions are decompose分解 a vector determine forward or backward
cross product in graphics

Orthonormal bases and coordinate frames 正交基底和坐标系
- Critical issue is transforming between these systems/ bases

In Graphics, pervasively used to represent transformations

representing transformations using matrices
rotation R_{theta} = begin{bmatrix} costheta & -sintheta \ sintheta & costheta end{bmatrix} R_{-theta} = begin{bmatrix} costheta & sintheta \ -sintheta & costheta end{bmatrix} = R_{theta}^{T} = R_{theta}^{-1}(by quad definition)
scale matrix begin{bmatrix} x^{‘} \ y^{‘} end{bmatrix} = begin{bmatrix} s_{x} & 0 \ 0 & s_{y} end{bmatrix}begin{bmatrix} x \ y end{bmatrix}
reflection matrix 反射（镜像）矩阵 begin{bmatrix} x^{‘} \ y^{‘} end{bmatrix} = begin{bmatrix} -1 & 0 \ 0 & 1end{bmatrix}begin{bmatrix} x \ y end{bmatrix}
shear matrix begin{bmatrix} x^{‘} \ y^{‘} end{bmatrix} = begin{bmatrix} -1 & a \ 0 & 1end{bmatrix}begin{bmatrix} x \ y end{bmatrix}
Linear transforms 线性变换：可以用一个矩阵表示的变换 x’ = ax by y’ = cx dy begin{bmatrix} x^{‘} \ y^{‘} end{bmatrix} = begin{bmatrix} a & b \ c & d end{bmatrix}begin{bmatrix} x \ y end{bmatrix}

Why homogeneous coordinates for example: translation begin{bmatrix} x^{‘} \ y^{‘} end{bmatrix} = begin{bmatrix} a & b \ c & d end{bmatrix}begin{bmatrix} x \ y end{bmatrix} begin{bmatrix} t_{x} \ t_{y} end{bmatrix}
Affine transformation 仿射变换仿射变换：先线性变换再加上一次平移 begin{bmatrix} x^{‘} \ y^{‘} \ 1 end{bmatrix} = begin{bmatrix} a & b & t_{x}\ c & d & t_{y} \ 0 & 0 & 1 end{bmatrix} begin{bmatrix} x \ y \ 1 end{bmatrix} S(s_{x}, s_{y}) = begin{bmatrix} s_{x} & 0 & 0 \ 0 & s_{y} & 0 \ 0 & 0 & 1 end{bmatrix} R(alpha) = begin{bmatrix} cosalpha & -sinalpha & 0 \ sinalpha & cosalpha & 0 \ 0 & 0 & 1 end{bmatrix} T(t_{x}, t_{y}) = begin{bmatrix} 1 & 0 & t_{x} \ 0 & 1 & t_{y} \ 0 & 0 & 1 end{bmatrix}
transform ordering matters
- matrix multiplication is not commutative 可交换的

decomposing complex transforms translate center to origin rotate translate back which means T(c) · R(alpha) · T(-c) 分解：变换可以分解，注意先后顺序是从右到左 2D变换矩阵（缩放，旋转，平移变换）

3D point = (x,y,z,1)^T^
3D vector = (x,y,z,0)^T^
begin{bmatrix} x^{‘} \ y^{‘} \ z_{‘} \ 1 end{bmatrix} = begin{bmatrix} a & b & c & t_{x}\ d & e & f & t_{y}\g & h & i & t_{z} \ 0 & 0 & 0 & 1 end{bmatrix} begin{bmatrix} x \ y \z \ 1 end{bmatrix}
三维空间中的齐次变换，最后一行和二维变换类似，是0 0 0 1，平移还是在矩阵最后一列
对于仿射变换，是先应用线性变换，再加上平移
what is order? linear transform first or translation first? scale S(s_{x}, s_{y},s_{z}) = begin{bmatrix} s_{x} & 0 & 0 & 0\ 0 & s_{y} & 0 & 0\0 & 0 & s_{z} & 0 \ 0 & 0 & 0 & 1 end{bmatrix} T(t_{x}, t_{y}, t_{z}) = begin{bmatrix} 1 & 0 & 0 & t_{x}\ 0 & 1 & 0 & t_{y} \0 & 0 & 1 & t_{z} \ 0 & 0 & 0 & 1 end{bmatrix} R_{xyz}(alpha, beta, gamma) = R_{x}(alpha)R_{y}(beta)R_{z}(gamma)

这个从“歪”的坐标轴旋转回正的坐标轴，不太好写。但是这个变换的逆过程，即：从正的坐标轴旋转到“歪”的坐标轴，是好写的，于是我们先写从“正”坐标轴变换到“歪”坐标轴的变换矩阵，再求其逆矩阵，就可以得到待求的变换矩阵。又因为旋转矩阵是正交矩阵，所以他的逆矩阵就只需要转置一下就可以得到了！注意，不但相机要做这个变换，其他物体也要做这个变换，因为我们想让相机看到的景物相对不变。（以上部分个人认为非常巧妙和关键！）
Projection (投影) transformation 3D to 2D Orthographic (正交) projection 没有近大远小平行投影首先定义空间中一个立方体，将其translate，使其中心在原点，再scale成标准立方体（边长为2 再次提醒，注意
css 图像处理对象存储

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