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I found this site that calculates the drop off of the earth at any given distance from an observer. Is this formula correct?

http://earthcurvature.com/

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Yes, the formula is correct. For an angle $a$ in radians at the centre of the earth we have

$d = r a \\ r - h = r \cos (a) \\ \displaystyle \Rightarrow h = r(1 - \cos (a)) = r\left(1 - \cos \left(\frac {d}{r}\right) \right)$

For small values of $a$ we have $\displaystyle \cos (a) \approx 1 - \frac {a^2}{2}$, so $\displaystyle h \approx \frac {d^2}{2r}$.

If $d$ is measured in kilometres, this is approximately $7.85 \times 10^{-5} \space d^2$ km.

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  • $\begingroup$ But note that this has nothing to do with the horizon, as suggested in question title. $\endgroup$ – Ben51 Feb 16 at 16:03
  • $\begingroup$ @Ben51 Good point. I have edited the question title to remove the horizon reference. $\endgroup$ – gandalf61 Feb 16 at 16:19

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