Perspective Projection

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Perspective projection, where the distance of the Center of Projection from the plane of projection is finite. It models the viewing system of our eyes, and is typically modeled using the pinhole camera model.

• CoP: The pinhole
• Plane of projection: The image plane (upside down). However, in computer graphics we can make the image upright by placing the plane of projection in front of the CoP.

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xy-plane

If we want to project on the plane $P^{\prime }=[x^{\prime },y^{\prime },d]$, this is done by using the perspective matrix

$${\mathbf{P}}_{PER}=\left[\begin{array}{cccc}d& 0& 0& 0\\ 0& d& 0& 0\\ 0& 0& d& 0\\ 0& 0& 1& 0\end{array}\right]$$

Which results in the projection

$$P^{\prime }=P_{PER}\cdot P=\left[\begin{array}{c}xd\\ yd\\ zd\\ z\end{array}\right]$$

And dividing by the homogeneous coordinate

$$P^{\prime }=\left[\begin{array}{c}\frac{xd}{z}\\ \frac{yd}{z}\\ d\\ 1\end{array}\right]$$

Arbitrary Plane

We have the following equations to find the projection matrix

$$\begin{array}{rl}1.\overrightarrow{O{P}^{\prime }}& =a\overrightarrow{OP}(\text{Colinear})\\ x^{\prime }& =ax,y^{\prime }=ay,z^{\prime }=az\\ 2.\vec{N}\overrightarrow{R_0{P}^{\prime }}& =0(\text{Orthogonality})\\ n_x(x^{\prime }-x_0)& +n_y(y^{\prime }-y_0)+n_z(z^{\prime }-z_0)=0\end{array}$$

Combining Eq. 1 and Eq. 2 yields

$$a=\frac{c}{n_{x}x+n_{y}y+n_{z}z}$$

Where $c=n_x{x}_0+n_y{y}_0+n_z{z}_0$

This finally yields the projection matrix

$$P_{PER,\Pi }=\left[\begin{array}{cccc}c& 0& 0& 0\\ 0& c& 0& 0\\ 0& 0& c& 0\\ n_x& n_y& n_z& 0\end{array}\right]$$

Since we have to divide with the homogeneous coordinate, this matrix will yield the same result.