# Calculating different distances between two buildings.

I am trying to work out what the different distances would be , measured at ground level and at the top, between two perfectly straight, plumb buildings each a kilometer high (like the trade towers were, only taller) . For the sake of argument assume the buildings are one kilometer apart (although I think the difference between the two measurements will be the same no matter what the actual distance apart, within reason)

Obviously the way to calculate it is to use the radius of the earth (6371 kilometers) as each side of an equilateral triangle with the base measuring one kilometer, and then calculate what the base of the triangle would be with each side 6372 kilometers, subtended by the same angle.

I am not expert enough to make use of the Windows 7 calculator to figure this out (although I do have an elementary grasp of trigonometry) , so any assistance on how to use the calculator to do the math would be welcome , as would the actual results of the calculation.

I do appreciate that a perfectly straight, plumb building a kilometer high is an engineering impossibility … it would be built with stepped sides, getting narrower the higher you went , but let’s just pretend .

You’ve got two similar triangles. The sides of similar triangles are in the same proportion. The smaller triangle has a long side of 6371 km, and the large triangle has a long side of 6372 km.

Left as an exercise for the reader:
a) What is the ratio of the long side of the big triangle to the long side of the small triangle?
b) What is the ratio of the short side of the big triangle to the short side of the small triangle?
c) What is the length of the short side of the big triangle?
d) Does the answer you get make sense?

Nope. The farther apart they are, the bigger the difference between the top and bottom distances.

The “small angle approximation” says that if you have a thin sliver of a triangle with two unit sides, the third side is equal to the (small) angle in radians.

Earth is about 40,000 km around, so our angle in radians is 2pi1 km/40,000 km, or 0.00016. So our distance is that times 1 km is ~16 cm.

Thanks both of you. I make it to be 15.6 cm using Chronos’ method. Sure beats trying to work out the sine of the subtended angle …