Some approximations for $\arccos(1/(1+x))$ 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:
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? 
 A: 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
$$d(h)
= 2\pi R \cdot \arccos\frac{R}{R+h}
= 2\pi R \cdot \arccos\frac{1}{1+\frac{h}{R}}.
$$
Since $h/R$  is small in this case, 
$$d(h)
\approx 2\pi R\sqrt{2\tfrac{h}{R}}
= \pi\sqrt{8R} \cdot \sqrt{h}
\approx 22441\sqrt{h},
$$
which only differs from Excel's calculation of the coefficient by $0.42\%$.
A: Nice approximation for $\arccos(R/(R+h))$, I like that! However the original formular for the visible ground distance is wrong - the correct formula is:
$$
d = R\arccos(R/(R+h)) 
$$
(in essence, lose the $2\pi$ - to get the length of a segment on a circle given the angle in rad, just multiply the radius by the angle in rad - no $2\pi$!)
With that simpler (and correct formula) the approximation gets: $d = 3,571\sqrt{h}$ with $h$ in m and $d$ in km. 
