Inclined Chambers with Respect to a Plane

Let us examine the particular case in which the cameras are aligned with respect to the axes, have identical intrinsic parameters, exhibit no relative rotation, and are inclined at a pitch angle with respect to the plane $z=0$.

In this particular condition, the projection matrix simplifies slightly, taking the form

$$\mathbf{K}\mathbf{R} = \begin{bmatrix} u_0 \cos \vartheta ... ...rtheta \\ \cos \vartheta & 0 & - \sin \vartheta \end{bmatrix}$$ (9.25)

It is worth noting that the RPY angle system (appendix A.1) has been used for the matrix $\mathbf{R}$.

The horizontal coordinate $u$ of a generic point $(x,y,z)$ in world coordinates is therefore given by:

$$u = u_{0} - \frac{ k_{u} (y - y_0) } { \cos \vartheta (x - x_0) - \sin \vartheta (z - z_0)}$$ (9.26)

With the assumptions of rectified cameras discussed previously, namely the same orientation and identical intrinsic parameters, a condition that can always be achieved through rectification or by considering appropriate rows of the image, The projection matrix (9.25) remains the same in the two different reference frames, and by examining equation (9.26), the only difference between different cameras is found in the numerator due to the varying position of the pin-hole along the $y$ axis.

It follows that the difference in coordinates $u$ in the two images $d = u_1 - u_2$ (disparity) is given by

$$d = u_1 - u_2 = \frac{ k_u b } { \cos \vartheta (x - x_0) - \sin \vartheta (z - z_0)}$$ (9.27)

, having redefined $b = y_1 - y_2$.

Using the relationship (9.26) in equation (9.27) yields the remarkable result

$$u_i = u_{0} - d \frac{y - y_i}{ b }$$ (9.28)

, from which we can finally derive the coordinate $y$ of the point. It seems you've entered "

$$y = - b \frac{u_i - u_0}{d} + y_i$$ (9.29)

". Please provide the specific text or content you would like me to translate from Italian to English, and I'll be happy to assist you! In the case where the cameras are perfectly aligned, the only calibration parameter that affects the coordinate $y$ is the sole $b$.

The coordinate $v$ of the point can instead be expressed as

$$v - v_0 = - \frac{k_v}{b k_u} ( \sin \vartheta (x - x_0) + \cos \vartheta (z - z_0) ) d$$ (9.30)

From which the system of equations follows:

$$\begin{matrix} \cos \vartheta (x - x_0) - \sin \vartheta (z... ..._0) = - \dfrac{v - v_0}{k_v} \dfrac{b k_u } { d } \end{matrix}$$ (9.31)

the solution that allows us to obtain the remaining two three-dimensional coordinates of the given point is

$$\begin{matrix} x - x_0 = \dfrac{b k_u}{d} \left( \cos \var... ...{k_v} \cos \vartheta + \sin \vartheta \right) \\ \end{matrix}$$ (9.32)

V-Disparity

A particular case of disparity occurs when observing a plane, that of the ground, which, due to the number of points, predominates in the image. In the case where the baseline is along the axis $y$, the disparity of the plane $z=0$ is solely a function of $v$, and this equation turns out to be that of a straight line.

The disparity relation from the coordinate $v$ can be derived from the value of $x$ from the second equation and substituting it into the first of the equations (9.31):

$$\begin{matrix} x - x_{0} = \tan \vartheta (z - z_{0}) + \d... ...{k_v}{k_u} \dfrac{z - z_{0} }{ b \cos \vartheta } \end{matrix}$$ (9.33)

From the first of the equations (9.33), it can be seen that the expression for disparity depends solely on the distance $x$ when the height $z$ is fixed (for example, at ground level). From the second equation, it is evident that the disparity $d$ increases linearly with the coordinate $v$ following a known slope

$$d = \cos \vartheta \dfrac{ b }{z_{0}} (v - v_{d=0} )$$ (9.34)

in the classical case where $k_u \approx k_v$ (square pixel). The point of zero disparity $v_{d=0}$, as mentioned earlier, is located at

$$v_{d=0} = v_{0} - k_{v} \tan \vartheta$$ (9.35)

and depends only on the vertical aperture and the pitch (which is obviously the same coordinate as the vanishing point).