Height of the center of projection

Given a line segment on the ground floor with known length, what is the height of the centre of projection of the camera?

Central projection

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

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