Re: How far to travel to equal 1 ft. drop in elevation?

Since you specify you want to assume the Earth is a perfect, smooth sphere, 
the problem becomes one of simple geometry:



If 'R' is the radius of the Earth (~4,000 miles), and 'h' is 1 foot, what 
we want to find is the angle 'a' in radians, which gives us the distance on 
the surface in units of the Earth's radius.

Let's convert the Earth's radius to feet:

4,000 miles * 5,280 ft/mi = 21,120,000 feet

Now the Secant of 'a' is equal to (R + h)/R, or 1 + 1/R, since 'h' is 1.

1 + 1/R = 1.00000004734848

The arcsecant of the above is 0.000307729 radians.

We multiply that by the Earth's radius in feet and get

6499.23 feet.

Or about one-and-a-quarter miles of travel for the beach-ball Earth to drop 
one foot.

Note that the multi-digit accuracy I used is unneccessary since we're 
already approximating the radius of the earth. Also, in real life the Earth 
bulges slightly at the equator, so the curvature is fractionally different 
walking north/south than east/west. The figure 1.25 miles per foot of 
curvature is as accurate an approximation that you need for this kind of 
question.

Cheers,
-Adams Douglas
 Senior Developer
 Dicon, Inc.