Height of the focal point


A-series paper size

First we look at the dimensions of an A series paper size. The crucial property is that if you fold it in half, the ratio of the long and short sides stays the same. If a is the length of the short side and b is the length of the long side:

\frac{a}{b} = \frac{\frac{1}{2}b}{a}


b^2 = 2 a^2

We do not have to take square roots here to get the lengths since we are using rational trigonometry. Separations between points are represented as quadrances. So,

A = a^2, B = b^2

Therefore, for an A paper we have

B = 2 A

For the diagonal C we use Pythagoras’ theorem, so

C = A + B = A+ 2 A = 3 A

We focus now on the short segment with quadrance A, but the resulting formula is equally applicable to the long segment B and the diagonal C. Simply multiply with the appropriate factor.

For the exact dimensions you can use the fact that A0 is 1.0 m^2, A1 is A0 folded in half, etcetera. I usually use A4 paper.

Central projection

The segment with quadrance A lies on the ground plane. The focal point of the camera is a point above the plane. The lines connecting the endpoints of A and the focal point have quadrances D_1 and D_2. This forms a triangle and the spread between D_1 and D_2 is p_{12}.

We can now apply the cross law

(D_1 + D_2 - A)^2 = 4 D_1 D_2 (1-p_{12})

If we drop a line from the focal point to the ground plane and define our height quadrance H. This line forms two new triangles with the endpoints of the segment with quadrance A. The spreads between the line with quadrance H and the two other lines with quadrances D_1 and D_2 are q_1 and q_2 respectively. We can apply the spread law to these right triangles

\frac{1-q_1}{H} = \frac{1}{D_1}

\frac{1-q_2}{H} = \frac{1}{D_2}

We can solve these for D_1 and D_2 and insert them into the equation we derived from the cross law

(\frac{H}{1-q_1}  + \frac{H}{1-q_2} - A)^2 = 4 \frac{H^2}{(1-q_1)(1-q_2)} (1-p_{12})

We can rearrange this equation by taking a square root and multiplying by (1-q_1)(1-q_2)

[(1-q_1)+(1-q_2) \pm 2 \sqrt{(1-p_{12})(1-q_1)(1-q_2)}]H = (1-q_1)(1-q_2) A

with all terms of H on one side.


If we solve for H we get

H = \frac{(1-q_1)(1-q_2)}{(1-q_1)+(1-q_2) \pm 2 \sqrt{(1-p_{12})(1-q_1)(1-q_2)}} A

Note that there are 2 solutions. It turns out that only relevant solution in our context is for the spread p_{12} being acute, and therefore the one with a minus sign.

So, the height quadrance H for a given segment with quadrance A on the ground plane, visual spread p_{12}, and spreads with the gravity direction q_1 and q_2 is

H = \frac{(1-q_1)(1-q_2)}{(1-q_1)+(1-q_2) - 2 \sqrt{(1-p_{12})(1-q_1)(1-q_2)}} A

I can not imagine how to derive this in classical trigonometry.