Computer Graphics- Cheatsheet

Homogeneous coordinates

Adding a fourth term: \(w\) makes it easier to work in 3D euclidian space. $$({x \over w},{y \over w},{z \over w}) \Rightarrow (x,y,z,w)$$

Why? Here's a long list of reasons, but a couple key points:

• 3D scaling, rotation, and translation operations can be represented as a single linear transform (4x4 matrix)
• A direction (Infinity) can be represented in 3D space: (x,y,z,0)

Transformations

Translation

for a point \(p\): $$ T(t)p = \begin{bmatrix} 1 & 0 & 0 & t_x\\ 0 & 1 & 0 & t_y\\ 0 & 0 & 1 & t_z\\ 0 & 0 & 0 & 1\\ \end{bmatrix} \begin{bmatrix} p_x\\ p_y\\ p_z\\ 1\\ \end{bmatrix} = \begin{bmatrix} p_x + t_x\\ p_y + t_y\\ p_z + t_z\\ 1\\ \end{bmatrix} $$ for a vector \(v\): $$ T(t)v = \begin{bmatrix} 1 & 0 & 0 & t_x\\ 0 & 1 & 0 & t_y\\ 0 & 0 & 1 & t_z\\ 0 & 0 & 0 & 1\\ \end{bmatrix} \begin{bmatrix} v_x\\ v_y\\ v_z\\ 0\\ \end{bmatrix} = \begin{bmatrix} v_x\\ v_y\\ v_z\\ 0\\ \end{bmatrix} $$ inverse transform: \(T^{-1}(t) = T(-t)\)

Scaling

$$ S(t)p = \begin{bmatrix} s_x & 0 & 0 & 0\\ 0 & s_y & 0 & 0\\ 0 & 0 & s_z & 0\\ 0 & 0 & 0 & 1\\ \end{bmatrix} \begin{bmatrix} p_x\\ p_y\\ p_z\\ 1\\ \end{bmatrix} = \begin{bmatrix} s_xp_x\\ s_yp_y\\ s_zp_z\\ 1\\ \end{bmatrix} $$

Rotation

When applying individual rotations (yaw, pitch, roll), the order of rotation operations is important. To get around this problem use axis-angle representation.

If you only need to rotate around a single axis, then applying the rotation transform can be straightforward. $$R_z(\alpha )=\stackrel{\text{yaw}}{\left[\begin{array}{cccc}\cos \alpha & -\sin \alpha & 0& 0\\ \sin \alpha & \cos \alpha & 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]}{R}_y(\beta )=\stackrel{\text{pitch}}{\left[\begin{array}{cccc}\cos \beta & 0& \sin \beta & 0\\ 0& 1& 0& 0\\ -\sin \beta & 0& \cos \beta & 0\\ 0& 0& 0& 1\end{array}\right]}$$

The inverse of a rotation matrix corresponds to its transpose

$$R_x(\gamma )=\stackrel{\text{roll}}{\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& \cos \gamma & -\sin \gamma & 0\\ 0& \sin \gamma & \cos \gamma & 0\\ 0& 0& 0& 1\end{array}\right]}$$

Coordinate Systems

Local space, World space, View space, Clip space, Screen space

More on Coordinate Systems

Perspective vs Orthographic Projection

perspective / orthographic projection. near plane / far plane. clipping.

Screen Space

You can use the Viewport Tranform to get to Screen Space (actual pixels)

Combining Transformations

To transform a vertex coordinate to clip coordinates: $$V_\text{clip} = M_\text{projection}\cdot{M_\text{view}}\cdot{M_\text{model}}\cdot{V_\text{local}}$$

Other Useful References

Coordinate systems we commonly reference for development.

Pinhole camera model

$$ \begin{align} P & = \overbrace{K}^{\text{Intrinsic Matrix}} \times \overbrace{\left[R|\mathbf{t}\right]}^{\text{Extrinsic Matrix}} \\ & = \overbrace{ \underbrace{ \begin{bmatrix} 1 & 0 & x_0\\ 0 & 1 & y_0\\ 0 & 0 & 1\\ \end{bmatrix} }_{\text{2D Translation}} \times \underbrace{ \begin{bmatrix} f_x & 0 & 1\\ 0 & f_y & 1\\ 0 & 0 & 1\\ \end{bmatrix} }_{\text{2D Scaling}} \times \underbrace{ \begin{bmatrix} 1 & s/f_x & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{bmatrix} }_{\text{2D Shear}} }^{\text{Intrinsic Matrix}} \times \overbrace{ \underbrace{ \begin{bmatrix} 1 & 0 & 0 & t_x\\ 0 & 1 & 0 & t_y\\ 0 & 0 & 1 & t_z\\ \end{bmatrix} }_{\text{3D Translation}} \times \underbrace{ \begin{bmatrix} ... \end{bmatrix} }_{\text{3D Rotation}} }^{\text{Extrinsic Matrix}} \end{align} $$

Source: Kyle Simek's excellent computer vision blog

FAQs

How do I get the camera direction from a given view matrix?

Column 2 of the view matrix is the camera's -Z direction. \begin{bmatrix} u_x & v_x & -n_x & -\text{eye}_x\\ u_y & v_y & -n_y & -\text{eye}_y\\ u_z & v_z & -n_z & -\text{eye}_z\\ 0 & 0 & 0 & 1\\ \end{bmatrix} Where \(u\), \(v\), and \(n\) are the normalized vectors for the camera referential. \(u\) is the up vector, \(n\) is the direction the camera is looking at, and \(v\) is perpendicular to both \(n\) and \(u\).