Curvature of the Earth - Math Central

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Question from Max:

Recently I read the answer to a question proposed by someone on this site.

The question : What is the rate of curvature per mile on Earth?
The answer given : Use Pythagoras' Theorem to solve for the answer, given a 1 mile side
and a side as the radius. The hypotenuse minus the radius is your answer of drop/mile or curve/mile.

My conjecture : Why go through all of that work if the distance is one? Something like
{1/diameter} would would fine for such a problem. Seems like a lot of work for no reason.

I understand the practical application of Pythagoras' Theorem in this certain situation, as you would need
to use a^2+b^2=c^2 for any distance greater than one [mile]..
It just seems excessive and unnecessary if you're solving for curve / one mile.

Drop/mile is a bad measure of curvature because it isn't a ratio. If we approximate the Earth as a sphere, half the distance means (for short distances) about 1/4 the drop.

You can measure curvature in degrees per mile - how fast does the direction of vertical change as you move? (The answer is about 0.00145 degree per mile.) So if you drive 100 miles in a straight line, the direction of "up" changes by about 1/7 degree. That's quite measurable with fairly simple instruments!

Good Hunting!
RD

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