Focusing on the Ground

Suppose we have a view camera and we wish to focus on a horizontal plane. More precisely, let the center of the lens \(O\) be \(a\) above the ground, and suppose the rear standard makes an angle \(\theta\) with respect to the horizontal. We wish to find the angle \(\alpha\) that the front standard must be tilted at to focus on the ground.

By the Scheimpflug rule. the rear standard, front standard, and horizontal intersect at a point \(S\). By the hinge rule, the front focal plane, the horizontal, and the plane through \(O\) parallel to the rear standard are concurrent; this is true if and only if the intersection point \(H\) of the horizontal and the plane parallel to the rear standard lies at a distance \(f\) from the front standard, where \(f\) is the focal length of the lens.

We have:
\begin{array}{lcl}
O'S = a / \tan{\alpha}\\
O'H = a / \tan{\theta}\\
SH = O'S - O'H = a(1/\tan{\alpha} - 1/\tan{\theta}) = a \left( \frac{\cos{\alpha}}{\sin{\alpha}} - \frac{\cos{\theta}}{\sin{\theta}} \right)
\end{array}
This means:
\begin{array}{lcl}
f = SH\sin{\alpha} = a\sin{\alpha} \left( \frac{\cos{\alpha}}{\sin{\alpha}} - \frac{\cos{\theta}}{\sin{\theta}} \right)\\
\frac{f}{a} = \cos{\alpha}-\frac{\cos{\theta}}{\sin{\theta}}\sin{\alpha}\\
\frac{f}{a}\sin{\theta} = \cos{\alpha}\sin{\theta} - \sin{\alpha}\cos{\theta} = \sin{(\theta-\alpha)}\\
\theta-\alpha = \arcsin{(\frac{f}{a}\sin{\theta})}\\
\alpha = \theta-\arcsin{(\frac{f}{a}\sin{\theta})}
\end{array}
In other words, the angle between the front and rear standards is \(\arcsin{(\frac{f}{a}\sin{\theta})}\). This is pretty neat; in particular, for small magifications we can say the ratio of  of the sines of the angles is approximately equal to the magnification of the camera.