# Height of the center of projection

intro

## 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.

## Solution

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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