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I was trying to calculate the maximum ground distance you can see on mountains, with your elvation given.

After some simple geometry, I was able to come up with the following formula:

Let $h$ be your elevation, $d(h)$ be the maximal distance you can see, then

$$d(h)=2\pi R\arccos\frac{R}{R+h}$$

where R is the radius of earth. We take $R=6378100m$ as its value.

But when I plot it in excel, here's what I got:enter image description here The unit for the vertical axis is km while the unit for the horizontal axis is m.

Amazingly, for $d\in(0,20000m)$ (essentially the maximum elevation you can achieve without paying millions to board a spaceship) , $d(h)$ can be approximated by

$$d(h)\approx22345\sqrt{h}$$

here

with a $r^{2}$ value of 1!.

Only when $h>5\cdot10^{5}m$ dose the the approximation begin to deviate away significantly.

Does any one have a explanation of this from a numerical prespective?

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1 Answer 1

up vote 10 down vote accepted

The Taylor series for the cosine is $$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dotsb.$$

Truncating this series after the $x^2$ term gives the rather good approximation $$\cos x \approx 1 - \frac12 x^2,$$ from which, by substituting $\sqrt{2y}$ for $x$, we get $$\cos \sqrt{2y} \approx 1 - y$$ and thus $$\arccos (1-y) \approx \sqrt{2y}.$$

Since, for small values of $y$, $$\dfrac{1}{1+y} = 1 - \dfrac{y}{1+y} \approx 1 - y,$$ it follows that $\sqrt{2y}$ is also a good approximation for $\arccos \dfrac{1}{1+y}$ when $y$ is small.


To verify this approximation, consider the original equation $$\begin{aligned} d(h) =& 2\pi R\arccos\frac{R}{R+h} \\ =& 2\pi R\arccos\frac{1}{1+\frac{h}{R}}. \end{aligned}$$

Since $h/R$ is small in this case, $$\begin{aligned} d(h) \approx& 2\pi R\sqrt{2h/R} \\ =& \left(\pi\sqrt{8R}\right)\sqrt{h} \\ =& 22441\sqrt{h}, \end{aligned}$$ which only differs from Excel's calculation of the coefficient by $0.42\%$.

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