To clearify the title, I am interested in the problem where we are trying to cover the surface of a sphere with cylinders. Let the sphere have radius R, the cylinders radius $r\ll R$ and cylinder height $h\ll r$. Here $r$, $R$ and $h$ are fixed, we are not allowed to change the radius of the coins to obtain a better filling.
Just as a starting point I thought maybe under perfect circumstances every $d^2=(2r)^2=4r^2$ unit of the surface could contain a coin. This would give a filling degree of $\pi r^2 /4r^2=\pi/4$. Thus, for the first layer we have
$\displaystyle\hspace{1cm}\frac{4\pi R^2}{4r^2} = \pi\Bigl(\frac{R}{r}\Bigr)^2$
coins. This would add an height $h$ to the surface, thus the next layer would have an height $R+h$. Giving a total number of coins of
$\displaystyle\hspace{1cm} \sum_{i=0}^{n-1}\pi\Bigl(\frac{R+ih}{r}\Bigr)^2 = \frac{\pi}{6r^2}(n+1)(h^2n(2n+1)+6hnR + 6R^2) $
Questions
- Is this a too simplistic approach, is there a better more correct way of solving this problem?
- Now I have calculated the number of coins used in $n$ layers (supposedly). What, if I want to flip the question. Given an huge stack of $M$ amount of money, where each coin is worth $1$, how many layers can we get?
The densest packing of circles on a plane is hexagonal packing; see e.g. Wikipedia. There’s no simple solution for the densest packing on a sphere, but as you did for the square lattice, we can approximately put a hexagonal lattice on the sphere – if we’re talking about the Earth, the error we make in doing so is negligible compared to the effects of the topography.
The density of the hexagonal packing is $\frac\pi{\sqrt{12}}\approx0.9069$, considerably denser than the density $\frac\pi4\approx0.7854$ of the square packing. The corresponding number of coins in the first layer would be
$$ \frac{4\pi R^2}{\sqrt{12}r^2}=\frac{2\pi}{\sqrt3}\left(\frac Rr\right)^2\;. $$
As for finding the number of layers you can get with a certain number of coins: The count you obtained for $n$ layers is cubic in $n$, so equating it to a given number of coins yields a cubic equation that can be solved in closed form.