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Perspective

You should have known the homogeneous coordinates

Here we just provide a simple affine projection matrix, first:

3D affine transformations

$\begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & -1/c & 1 \end{bmatrix}$

note: the $c$ is a constant number.

Perspective Projection Derivation

Orthographic projection

First of all, move the object’s center onto origin $(0,0,0)$ .

$M_{translate} = \begin{pmatrix} 1 & 0 & 0 & \frac{-(l+r)}{2}\\ 0 & 1 & 0 & \frac{-(b+t)}{2}\\ 0 & 0 & 1 & \frac{-(f+n)}{2}\\ 0 & 0 & 0 & 1 \end{pmatrix}$

Then, normalize the object.

$M_{normalize} = \begin{pmatrix} \frac{2}{r-l} & 0 & 0 & 0\\ 0 & \frac{2}{t-b} & 0 & 0\\ 0 & 0 & \frac{2}{n-f} & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}$

And multiply them:

$M_{ortho} = M_{normalize} \cdot M_{translate}$

Step into the deep.

perspective-projection

Through observing the figure, we see that there is a similar triangle
and then:

$y'=\frac{n}{z}y$

Similarly:

$x'=\frac{n}{z}x$

Then we have a mapping matrix:

$\begin{pmatrix} x\\y\\z\\1 \end{pmatrix}\rightarrow \begin{pmatrix} \frac{n}{z}x\\\frac{n}{z}y\\\text{unknown}\\1 \end{pmatrix}\rightarrow \begin{pmatrix} n & 0 & 0 & 0\\ 0 & n & 0 & 0\\ ? & ? & ? & ?\\ 0 & 0 & 1 & 0 \end{pmatrix}\cdot \begin{pmatrix} x\\y\\z\\1 \end{pmatrix}$

Here, we need know some rules:

Points of the near plane are constant.
The coordinates $(0, 0, f, 1)$ of the centric point of the far plane are unchanged.
The values of points’ $z$ coordinate on the far plane are always $f$.

Because of rule 3, we can get a equaltion:

$\begin{pmatrix} n & 0 & 0 & 0\\ 0 & n & 0 & 0\\ ? & ? & ? & ?\\ 0 & 0 & 1 & 0\\ \end{pmatrix}\cdot \begin{pmatrix} x\\y\\n\\1 \end{pmatrix}= \begin{pmatrix} nx\\ny\\n^2\\n \end{pmatrix}\underset{normalize}{=} \begin{pmatrix} x\\y\\n\\1 \end{pmatrix}$

Suppose the thrid row of the above matrix are $(A, B, C, D)$ .

$\begin{align*} \therefore Ax + By + Cn + D = n^2 \end{align*}$

And, it’s clear that

$A = 0\\ B=0$

Then we could get:

$Cn + D = n^2 \tag{1}$

According the rule 2, the coordinates of the centric point of the
far plane are $(0, 0, f, 1)$ .

Take it in the above matrix and get the following matrix;

$\begin{pmatrix} n & 0 & 0 & 0\\ 0 & n & 0 & 0\\ ? & ? & ? & ?\\ 0 & 0 & 1 & 0\\ \end{pmatrix}\cdot \begin{pmatrix} 0\\0\\f\\1 \end{pmatrix}= \begin{pmatrix} 0\\0\\f^2\\f \end{pmatrix}\underset{normalize}{=} \begin{pmatrix} 0\\0\\f\\1 \end{pmatrix}$

Like the above equaltion, get the following equaltion:

$Cf+D = f^2 \tag{2}$

There are two equaltions with two varibles.

$\left\{\begin{matrix} Cn+D = n^2\\ Cf+D =f^2 \end{matrix}\right. \Rightarrow \left\{\begin{matrix} C = n+f\\ D = -nf \end{matrix}\right.$

Finally, we get the perspective projection matrix:

$M_{persp} = M_{normalize} \cdot M_{translate} \cdot \begin{pmatrix} n & 0 & 0 & 0\\ 0 & n & 0 & 0\\ 0 & 0 & n+f & -nf\\ 0 & 0 & 1 & 0 \end{pmatrix}$

Rodrigoues’ Rotation Formula

Rotation by angle $\alpha$ around axis $n$ .

$R(n,\alpha)=\cos(\alpha)I+(1-\cos(\alpha))nn^T +\sin(\alpha) \underset{N}{\underbrace{\begin{pmatrix} 0 & -n_z & n_y \\ n_z & 0 & -n_x \\ -n_y & n_x & 0 \end{pmatrix}}}$