Estimating Height of SpaceX Crane; Height of line-of-sight at $5.87$ miles We have a camera on a tower with a height of $350/5280$ mi. For convenience we'll assume the tower is $90$ degrees to the line-of-sight going to the horizon. We're really interested in estimating the height of a crane at a distance of $5.87$ miles from the tower. But to simplify here, we simply want to calculate the height of the crane where it intersects the line-of-sight from the camera. In other words what is the height of the line-of-sight at a distance of $5.87$ miles from a starting point $350$ ft above the earth?
We start with a circle representing earth:  x^{2}+(y+7917.5)^{2}=7917.5^{2}.  $x^{2}+(y+3958.75)^{2}=3958.75^{2}$. Our camera is at height of 350/5280 at a distance from (0,0) of approx 32.4. The crane is at a distance from (0,0) of approx 26.53. So I guess the question is: How do I calculate $y$ at a given $x$?
Labpadre Sapphire camera: https://www.youtube.com/watch?v=eVDUmtGyBd4

 A: If I look straight toward the horizon from the top of your tower, say from point $Q$, then my line of sight is perpendicular to the radius of the Earth passing through the point $P$ at which my line of sight hits the horizon. So there is a right triangle with vertices $P, Q,$ and the center of the Earth, with long leg of length 3958.8 miles, the radius of Earth, and hypotenuse of length 3958.8+35/528 miles. Note that you’ve incorrectly used the diameter of Earth in your proposed circle equation.
This means that the angle in our triangle at the center of the Earth has a cosine of $\frac{3958.8}{3958.8+35/528},$ so the angle is about $.332^\circ.$ So the distance along the surface of the Earth from the tower to the horizon is $.332/360*2\pi*3958.8\approx 22.9$ miles. (I can’t replicate your claim about 32.4 miles, and my number is in agreement with the standard heuristic of around $1.22\sqrt h$ for this distance in miles from a height of $h$ feet.)
Thus, the distance from the cranes to my view of the horizon is about $22.9-5.87\approx 17.0$ miles. It looks like the gray tower in the scene sees almost the same horizon as I do from the camera, so if its height is $h,$ then we have another right triangle with vertices at the center of the Earth, the horizon, and the top of the tower, with hypotenuse of length $3958.8+h$ and a leg of length $3958$. The angle $\theta$ at the center of the Earth is so small that the (curved) distance along the Earth is almost exactly the same as the (straight) distance giving the line of sight from the gray tower, i.e. $17$ miles, which means $17.0^2+3958.8^2\approx (3968.8+h)^2,$ so $h$ solves the quadratic equation $h^2+7917.5h-289=0.$ This makes $h\approx 0.0365013*5280= 193$ feet.
So if the gray tower has the same line of sight to the horizon as our camera, it’s about $193$ feet tall. I can’t tell for sure whether the dark gray narrow stripe in the background is land or cloud; if it’s cloud then the gray tower is taller than that, looks like by about a ninth, so maybe $215$ feet. The other structures in the scene can be estimated from this.
Thanks for the problem! Think I’ll use this on my precalculus students sometime.
