Cameras Tilted with Respect to a Plane

We consider the particular case in which the cameras are aligned with the axes, have equal intrinsic parameters, zero relative rotation, and are tilted by the pitch angle with respect to the plane $z=0$.

Under this particular configuration, the projection matrix simplifies slightly, taking the form

$$\mathbf{K}\mathbf{R} = \begin{bmatrix} u_0 \cos \vartheta ... ...rtheta \\ \cos \vartheta & 0 & - \sin \vartheta \end{bmatrix}$$ (10.27)

Note that the RPY angle convention (Appendix A.1) was used for matrix $\mathbf{R}$.

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

$$u = u_{0} - \frac{ k_{u} (y - y_0) } { \cos \vartheta (x - x_0) - \sin \vartheta (z - z_0)}$$ (10.28)

Under the rectified-camera assumptions introduced above, namely equal orientation and equal intrinsic parameters, a condition that can always be achieved through rectification or by considering suitable image rows, the projective matrix (10.27) is the same in the two different reference frames and, from equation (10.28), the only difference between the cameras is in the numerator, due to their different positions of the pin-hole along the $y$ axis. It follows that the difference between the coordinates $u$ in the two images $d = u_1 - u_2$ (the disparity) is

$$d = u_1 - u_2 = \frac{ k_u b } { \cos \vartheta (x - x_0) - \sin \vartheta (z - z_0)}$$ (10.29)

where $b = y_1 - y_2$ has been redefined. Using relationship (10.28) in equation (10.29) gives the notable result

$$u_i = u_{0} - d \frac{y - y_i}{ b }$$ (10.30)

from which the coordinate $y$ of the point is finally obtained:

$$y = - b \frac{u_i - u_0}{d} + y_i$$ (10.31)

When the cameras are perfectly aligned, the only calibration parameter affecting coordinate $y$ is $b$.

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

$$v - v_0 = - \frac{k_v}{b k_u} ( \sin \vartheta (x - x_0) + \cos \vartheta (z - z_0) ) d$$ (10.32)

Thus, the system of equations is

$$\begin{matrix} \cos \vartheta (x - x_0) - \sin \vartheta (z... ..._0) = - \dfrac{v - v_0}{k_v} \dfrac{b k_u } { d } \end{matrix}$$ (10.33)

whose solution, yielding the remaining two three-dimensional coordinates of the given point, is

$$\begin{matrix} x - x_0 = \dfrac{b k_u}{d} \left( \cos \var... ...{k_v} \cos \vartheta + \sin \vartheta \right) \\ \end{matrix}$$ (10.34)

V-Disparity

A particular disparity case arises when observing a plane, namely the ground plane, which accounts for most of the points in the image. When the baseline lies along the $y$ axis, the disparity of the plane $z=0$ depends only on $v$, and this equation is that of a straight line.

The relationship between disparity and coordinate $v$ can be derived by solving for $x$ in the second equation and substituting it into the first equation of (10.33):

$$\begin{matrix} x - x_{0} = \tan \vartheta (z - z_{0}) + \d... ...{k_v}{k_u} \dfrac{z - z_{0} }{ b \cos \vartheta } \end{matrix}$$ (10.35)

From the first equation in (10.35), it can be seen that the disparity depends only on the distance $x$ if the height $z$ is fixed (for example, on the ground). The second equation shows that disparity $d$ grows linearly with coordinate $v$, with the known slope

$$d = \cos \vartheta \dfrac{ b }{z_{0}} (v - v_{d=0} )$$ (10.36)

in the classical case where $k_u \approx k_v$ (square pixels). The zero-disparity point $v_{d=0}$ mentioned above is located at

$$v_{d=0} = v_{0} - k_{v} \tan \vartheta$$ (10.37)

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