intro

## Central projection

The segment with quadrance lies on the ground plane. The center of projection of the camera is a point above the plane. The lines connecting the endpoints of and the camera center 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 central projection 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.

## Solution

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.

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