Camera Transformation

To define a camera, consider:

• Position $\mathbf{e}$
• Gaze direction (look-at) $\mathbf{g}$
• Up direction, perpendicular to the look-at direction $\mathbf{t}$

We establish the camera as the origin, with up at $\mathbf{Y}$ and look-at towards $\mathbf{-Z}$. This approach enables us to transform everything in relation to the camera's position and orientation. Consequently, we construct a Camera Transform Matrix $\mathbf{M}_{view}$ to translate $\mathbf{e}$ to the origin, rotate $\mathbf{g}$ towards $\mathbf{-Z}$, and rotate $\mathbf{g} \times \mathbf{t}$ towards $\mathbf{X}$. $$\mathbf{M}_{view} = \mathbf{R}_{view} \mathbf{T}_{view}$$ Although direct rotation might seem complex, we can apply reverse rotation: $$\mathbf{R}_{view}^{-1}= \begin{bmatrix} x_{g\times t} & x_t & x_{-g} & 0 \cr y_{g\times t} & y_t & y_{-g} & 0 \cr z_{g\times t} & z_t & z_{-g} & 0 \cr 0 & 0& 0& 1 \end{bmatrix}$$ Given that the rotation matrix is orthogonal: $$\mathbf{R}_{view} = (\mathbf{R}_{view}^{-1})^{-1} = (\mathbf{R}_{view}^{-1})^T = \begin{bmatrix} x_{g\times t} & y_{g\times t}& z_{g\times t} & 0 \cr x_t & y_t & z_t & 0 \cr x_{-g} & y_{-g} & z_{-g} & 0 \cr 0 & 0& 0& 1 \end{bmatrix}$$ Thus, by transforming the object with $\mathbf{M}_{view}$, the objects' appearance remains consistent from the camera's viewpoint.

Projection Transformation

(a) Orthographic projection. (b) Perspective project. (c) Perspective with hidden lines.

Orthographic projection

In general, for a cuboid, we first translate it to the centre. Then, we use the parameters left, right, bottom, top, far, and near to define its dimensions as $[l,r] \times [b,t] \times [f,n]$. This allows us to transform it into a canonical cube of dimensions $[-1,1]^3$. $$\mathbf{M}_{ortho}= \begin{bmatrix} \frac{2}{r-l} & 0 & 0 & 0 \cr 0 & \frac{2}{t-b} & 0 & 0 \cr 0 & 0 & \frac{2}{n-f} & 0 \cr 0 & 0& 0& 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & -\frac{r+l}{2} \cr 0 & 1 & 0 & -\frac{t+b}{2} \cr 0 & 0 & 1 & -\frac{n+f}{2} \cr 0 & 0& 0& 1 \end{bmatrix}$$

Perspective Projection

Recall: $[x,y,z,1]^T = [kx,ky,kz,k]^T=[xz,yz,z^2,z]^T$ in homogeneous coordinates.

For perspective projection, we start by transforming the frustum into a cuboid, which is then followed by an orthographic projection. This involves:

• Keep near plate the same
• Keep z value of far plate the same
• Keep the centre of far plate the same.

Through similarity, the projection on the x and y axes can be represented as: $$x'=\frac{n}{z}x$$

$$y'=\frac{n}{z}y$$ So we can have part of our $\mathbf{M}_{per}$: $$\mathbf{M}_{per\to othro} = \begin{bmatrix} n & 0 & 0 & 0 \cr 0 & n & 0 & 0 \cr \# & \# & \# & \# \cr 0 & 0 & 1 & 0 \end{bmatrix}$$ Based on our rule, we need near plate and the centre of far plate remain the same. $$\mathbf{M}_{per\to othr}[x,y,n,1]^T=[x,y,n,1]^T == [nx,ny,n^2,n]^T$$

$$\mathbf{M}_{per\to othr}[0,0,f,1]^T=[0,0,f,1]^T == [0,0,f^2,f]^T$$

Solve, it: $$\mathbf{M}_{per\to othr} = \begin{bmatrix} n & 0 & 0 & 0 \cr 0 & n & 0 & 0 \cr 0 & 0 & n+f & -fn \cr 0 & 0& 1& 0 \end{bmatrix}$$ Then, we can have: $$\mathbf{M}_{per}=\mathbf{M}_{othr}\mathbf{M}_{per\to othr}$$ The refined explanation for the Perspective to Orthographic projection transformation is as follows: $$\mathbf{R}'=\mathbf{M}_{per\to othr}\mathbf{R}= \begin{bmatrix} x \cr y \cr z \cr 1 \end{bmatrix} \begin{bmatrix} n & 0 & 0 & 0 \cr 0 & n & 0 & 0 \cr 0 & 0 & n+f & -fn \cr 0 & 0& 1& 0 \end{bmatrix}= \begin{bmatrix} \frac{n}{z}x \cr \frac{n}{z}y \cr n+f -\frac{nf}{z} \cr 1 \end{bmatrix}$$ Specifically, when $z$ equals the near $n$ or far $f$ plane distances, the transformed $z'$ coordinate remains unchanged. However, for values of $z$ between $n$ and $f$, $z'$ becomes smaller, approaching $f$. This transformation, therefore, compresses the depth dimension, making objects appear further away as they move closer to the far plane.