## Sunday, September 3, 2017

### 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.