Cameras Tilted with Respect to a Plane

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

Under these conditions, the projection matrix is simplified slightly and takes the form

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

It should be noted 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 identical orientations and 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. From equation (10.28), the only difference between the cameras is therefore in the numerator, due to the 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 defined again. Using relation (10.28) in equation (10.29) yields 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 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 $z=0$ is a function only of $v$, and this equation is that of a straight line.

The relationship between disparity and coordinate $v$ can be derived from the value of $x$ in the second equation and by substituting it into the first of equations (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 of equations (10.35), it can be seen that the disparity depends only on the distance $x$ when the height $z$ is fixed (for example, on the ground), while the second shows that the 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 field of view and the pitch (it is clearly the same coordinate as the vanishing point).