Notable Results

We can use the rotation matrix and the pin-hole equation (9.25) to demonstrate some notable results. From the system, we define the function $f_{pm}$ of $\mathbb{R}^{3}$ in $\mathbb{R}^{2}$, called perspective mapping, as:

$$f_{pm}(x,y,z) = \left( k_{u} \frac{r_{0} x + r_{1} y + r_... ..._{4} y + r_{5} z}{r_{6} x + r_{7} y + r_{8} z} + v_0 \right)$$ (9.30)

the explicit form of the pin-hole camera model. For simplicity, the pin-hole is assumed to coincide with the origin of the reference frame.

Vanishing Points and Calibration

Every image contains 3 vanishing points, closely related to the choice of reference axes.

Consider, for example, the first axis. In our reference frame, coordinate $x$ is the distance (the same reasoning applies to the other two coordinates). Let this coordinate tend to infinity while keeping the others constant. The result is the point

$$\lim_{x\to\infty}f_{pm}(x,y,z) = \left( k_{u} \frac{r_{0}}{r_{6}} + u_0, k_{v} \frac{r_{3}}{r_{6}} + v_0 \right)$$ (9.31)

The same result can be obtained using homogeneous matrices, with a more compact formalism.

Taking the perspective transformation (9.24) and successively letting $x\rightarrow\infty$, $y\rightarrow\infty$, and $z\rightarrow\infty$ tend to infinity, the image points (in homogeneous coordinates) obtained, representing the vanishing points in the three directions, are exactly the columns of matrix $[\mathbf{v}_x \mathbf{v}_y \mathbf{v}_z] = \mathbf{K} \cdot \mathbf{R}$, namely:

$$\begin{array}{rl} \mathbf{v}_{x} &= \mathbf{K} \mathbf{r}_... ...} \\ \mathbf{v}_{z} &= \mathbf{K} \mathbf{r}_{3} \end{array}$$ (9.32)

where the syntax $\mathbf{r}_{i}$ denotes the columns of matrix $\mathbf{R}$. This is a first example of camera calibration based on knowledge of the image, namely the positions of the vanishing points.

In particular, in the simplified case $u_0=0$, $v_0=0$, and $k_\gamma = 0$, the vanishing points are located at

$$\begin{array}{rl} \mathbf{v}_{x} &= \left( k_u \dfrac{r_0}... ...\dfrac{r_2}{r_8}, k_v \dfrac{r_5}{r_8} \right) \\ \end{array}$$ (9.33)

It should be noted that, since the 3 columns of $\mathbf{R}$ are orthonormal, knowing 2 vanishing points is sufficient to always obtain the third (see the previous section).

Horizon Line

If more than one variable is sent to infinity rather than just one, more than one point is obtained. For $x\to\infty$, but with $y = m x$, the vanishing point degenerates into a line whose equation is

$$k_{v} (r_{3} r_{7} - r_{4} r_{6}) u + k_{u} (r_{6} r_{1} - r_{7} r_{0}) v + k_{u} k_{v} (r_{4} r_{0} - r_{3} r_{1}) = 0$$ (9.34)

the horizon line.

Degenerate Points and Lines

Just as a point in the projected image degenerates into a line, a line with equation $au +bv +c =0$ becomes, in the projected image,

$$a k_u (r_{0} x + r_{1} y + r_{2} z) + b k_v (r_{3} x + r_{4} y + r_{5} z) + c(r_{6} x + r_{7} y + r_{8} z) = 0$$

that is,

$$(a k_u~r_{0} + b k_v r_{3} + c r_{6} ) x + (a k_u r_{1} + b... ...{4} + c r_{7}) y + (a k_u r_{2} + b k_v r_{5} + c r_{8}) z = 0$$ (9.35)

which represents the degenerate plane (with normal as given by the equation) in three dimensions that passes through the origin (the pin-hole).