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 is the length of the short side and is the length of the long side:
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,
Therefore, for an A paper we have
For the diagonal we use Pythagoras’ theorem, so
We focus now on the short segment with quadrance , but the resulting formula is equally applicable to the long segment and the diagonal . Simply multiply with the appropriate factor.
For the exact dimensions you can use the fact that A0 is 1.0 , A1 is A0 folded in half, etcetera. I usually use A4 paper.
The segment with quadrance lies on the ground plane. The focal point of the camera is a point above the plane. The lines connecting the endpoints of and the focal point have quadrances and . This forms a triangle and the spread between and is .
We can now apply the cross law
If we drop a line from the focal point to the ground plane and define our height quadrance . This line forms two new triangles with the endpoints of the segment with quadrance . The spreads between the line with quadrance and the two other lines with quadrances and are and respectively. We can apply the spread law to these right triangles
We can solve these for and and insert them into the equation we derived from the cross law
We can rearrange this equation by taking a square root and multiplying by
with all terms of on one side.
If we solve for we get
Note that there are 2 solutions. It turns out that only relevant solution in our context is for the spread being acute, and therefore the one with a minus sign.
So, the height quadrance for a given segment with quadrance on the ground plane, visual spread , and spreads with the gravity direction and is
I can not imagine how to derive this in classical trigonometry.