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*.

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