What would be the calculations behind the skyscraper view of a sunset as a demonstration of a rotating earth? This is my first post here.  I have been challenged to proved proof that the earth is not flat and that we live on a spinning ball.  This was a fairly easy task for me and I proved several observable proofs.  But the flat earth believer wanted more.  He wanted the mathematical equations and THEN he wanted me to demonstrate that the math worked.  Well, my math skills are a bit rusty on this sort of level since I have been out of college.  Although I could probably do this, I figure it would be a greater ease for me to just ask one of the math majors or grad students that frequent online mathematics forums for the equations.  So I think what I will do is list my ideas one at a time in this mathematical forum.  And now, here is the first one.
The rapper B.O.B. has come out saying that he thinks the earth is flat.  Neil Tyson gave him a reply and PBS offered an Op Ed piece explaining some simple tests anyone can do to prove the earth is round.  One of them was to lay on the beach on your back and with your head pointed towards the sun set (do this on the pacific coast of course.  The moment you see the sun set, immediately stand up and you can see the sun set again -- or so says the op ed piece from PBS.  I have herd that something similar can be done with large buildings.  If you watch the sun set on the ground level, (let's say from the point of view of a 6ft 4 man.  Or some measured eye level from the ground) and then you take an elevator to the top floor, you will be able to see the sun set a second time.
The problem is this, if you know all of the variables, how long would it take to see the sun set a second time.
I hope I have posed this question well enough.  Let me know if you have any questions.  After this one is answered, I have at least one more question I will start in a second thread.
 A: 
The radius of the Earth is about $6371$ kilometers.  Let the radius of the Earth be $R$ and the height of the observer be $h$, with the Earth as shown in the figure rotating counter-clockwise.  Initially the sun is just visible from the surface of the Earth. After the Earth has rotated through an angle $\theta$, the sun is just visible to the observer.  From right-angle geometry, 
$$\cos \theta = \frac{R}{R+h}$$
so $$\theta = \cos^{-1} \left( \frac{R}{R+h} \right)$$
This formula gives us an answer in radians, which we can convert to time by observing that the Earth rotates $2 \pi$ radians in 24 hours.
For a numerical example, suppose the height of the observer is $2 \text{ meters} = 2 \times 10^{-3} \text{ kilometers}$.  Then $\cos \theta = R/(R+h) = 0.999999686$, so $\theta = 0.000792$ radians.  Multiplying by $24 / (2 \pi)$ hours / radian, this is $0.00303$ hours, which converts to $10.9$ seconds.
There is a danger of a catastrophic loss of precision in the calculation of $\cos \theta$ because the value is just a little less than $1$, but an ordinary pocket calculator seems able to handle the calculation in this example, provided three significant digits in the answer are sufficient.
EDIT: The above analysis assumes the observer is located on the Earth's equator.  For an observer at some other latitude, say a latitude of $\lambda$, where $\lambda = 0$ at the equator, we need to use a different value for $R$.  Instead of the Earth's radius, we should use the radius of the circle of latitude passing through the observer's location.  This radius is $R' = R \cos \lambda$, where $R$ is the radius of the Earth, as before.  But we should also correct the observer's height so we get the projection of his height in the plane of the circle of latitude.  This height is $h' = h \cos \lambda$.  So the end result is that the factors of $\cos \lambda$ cancel out in the calculation of $\cos \theta$, and the original formula works without taking the latitude into account.
