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.