Math CentralQuandaries & Queries


Question from Jimmy, a parent:

Both batteries died in my scientific calculator and I have lost my formula for the height of the curvature of the earth between two points on the surface. I used degrees and miles. I only had to enter the distance between the two points on the surface and the formula gave me the height the earth raised between the two points.

Great circle distance over radius, $\left(\frac{d}{3963} \right)$ = angular distance in radians.

"Highest point" is halfway, $\left(\frac{d}{7916} \right)$ radians.

Height is $\left(1 - \cos\left(\frac{d}{7916}\right)\right)\times3963$ miles (using radian mode on your calculator!)

For short distances (say under 2000 miles) this is rather well approximated by taking

$R\approx4000,$ and $\cos(A) \approx 1 -\frac{A^2}{2}$ (again, this formula only holds using radians)

using which it simplifies to

\[h \approx \frac{d^2}{8R} \approx \frac{d^2}{32000}\]

So over 10 miles $h \approx \frac{1}{320}$ mile or about sixteen feet.

Good Hunting!

Jimmy wrote back

I appreciate your reply but the responses on the web site have no relationship to my question.

If the base of one antenna tower is 100 miles from the base of another antenna tower, how much does the earth rise between them.

I needed a formula using miles and degrees.

RD's expression

\[h \approx \frac{d^2}{8R} \approx \frac{d^2}{32000}\]

is the easiest to use. For a distance of 100 miles between the bases of the towers the "highest point" between them is

\[h \approx \frac{100^2}{32000} = 0.3125 \mbox{ miles, or approximately } 1650 \mbox{ feet.}\]

If you want to use the expression

\[ \left(1 - \cos\left(\frac{d}{7916}\right)\right)\times3963 \]

I get $\left(1 - \cos\left(\frac{100}{7916}\right)\right)\times3963$ to be 1670 feet using the radian mode on my calculator. If you want to use the angle in degrees then the expression is

\[ \left(1 - \cos\left(\frac{d}{7916}\times \frac{180}{\pi}\right)\right)\times3963 .\]


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