Notable Results

We can use the rotation matrix and the pin-hole equation (8.15) to demonstrate some notable results. We define, from the system, the function $f_{pm}$ of $\mathbb{R}^{3}$ in $\mathbb{R}^{2}$ called perspective mapping, defined as:

$$f_{pm}(x,y,z) = \left( k_{u} \frac{r_{0} x + r_{1} y + r_... ..._{4} y + r_{5} z}{r_{6} x + r_{7} y + r_{8} z} + v_0 \right)$$ (8.21)

the function of the pin-hole camera model written explicitly. For simplicity, it is assumed that the pin-hole coincides with the origin of the reference system.

Vanishing Points and Calibration

For each image, there are 3 vanishing points, closely related to the choice of reference axes.

Let us consider the first axis as an example. In our reference system, the coordinate $x$ represents the distance (the discussion is similar for the other two coordinates). We take this coordinate to infinity while keeping the other two constant. What we obtain is the point

$$\lim_{x\to\infty}f_{pm}(x,y,z) = \left( k_{u} \frac{r_{0}}{r_{6}} + u_0, k_{v} \frac{r_{3}}{r_{6}} + v_0 \right)$$ (8.22)

.

Using homogeneous matrices, it is possible to achieve the same result with a more compact formalism.

Taking the perspective transformation (8.14) and sequentially sending $x\rightarrow\infty$, $y\rightarrow\infty$, and $z\rightarrow\infty$, the image points (in homogeneous coordinates) obtained, representing the vanishing points in the three directions, are exactly the columns of the matrix $[\mathbf{v}_x \mathbf{v}_y \mathbf{v}_z] = \mathbf{K} \cdot \mathbf{R}$, namely:

$$\begin{array}{rl} \mathbf{v}_{x} &= \mathbf{K} \mathbf{r}_... ...} \\ \mathbf{v}_{z} &= \mathbf{K} \mathbf{r}_{3} \end{array}$$ (8.23)

where we have indicated with the syntax $\mathbf{r}_{i}$ the columns of the matrix $\mathbf{R}$. This is a first example of camera calibration that leverages knowledge of the image, specifically the position of the vanishing points.

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

It seems that you've provided a placeholder or a command for a math block (

$$\begin{array}{rl} \mathbf{v}_{x} &= \left( k_u \dfrac{r_0}... ...\dfrac{r_2}{r_8}, k_v \dfrac{r_5}{r_8} \right) \\ \end{array}$$ (8.24)

). If you have specific content or equations that you would like to translate from Italian to English, please provide that text, and I'll be happy to assist you!

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

Horizon Line

If we send to infinity not one variable but multiple ones, we obtain more than one point. For $x\to\infty$ but with $y = m x$, the vanishing point degenerates into a line described by the equation

$$k_{v} (r_{3} r_{7} - r_{4} r_{6}) u + k_{u} (r_{6} r_{1} - r_{7} r_{0}) v + k_{u} k_{v} (r_{4} r_{0} - r_{3} r_{1}) = 0$$ (8.25)

, known as the horizon line.

Degenerate Points and Lines

As a point in the projected image degenerates into a line, a line described by the 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

$$(a k_u~r_{0} + b k_v r_{3} + c r_{6} ) x + (a k_u r_{1} + b... ...{4} + c r_{7}) y + (a k_u r_{2} + b k_v r_{5} + c r_{8}) z = 0$$ (8.26)

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