How can I calculate the size of a pile that contains different types of elements? Let's say I have a pile of coins. Let's say a coin has a radius of 1cm and a thickness of 0.20cm. I've heard the angle of repose is pretty low for coins, so that would be less than 10deg.
How would I figure out the size (height and diameter) of a pile of those coins? 
Now, what if the pile was a mixture of different sized coins? How could I figure out the size of a pile of mixed coins, say the coin above and another coin that is thicker and larger?
Thank you!
 A: I think it is illusory to try to calculate such figures by a direct approach for a small number of coins since it won't have a fixed shape anyway.
Also, I fear there is no simple answer for a moderately large number of coins, as it won't act as granular material. There is simply too many possible configurations for stacking these coins.

Instead for a (very very) big number of coins, the statistical effect will be dominant, coins would be assimilable to granular material and assemble in a perfect cone, which I guess, has a specific density $\rho$ when stacked this way. The density and angle of repose would change depending of the proportion of mixed coins in your stack.
So in fact for different mix of coins you'll get different densities and angles.

Since basically such a stack is a cone determined by the base angle, if you know how many coins you poured, you can calculate the total mass $M$.
From this you can estimate the volume of the cone $V=\frac M\rho$ and now $V=\frac 13 \pi r^2H$ but since $\tan(\theta)=\frac Hr$ you'll have $V=\frac 13 \pi r^3 \tan(\theta)$ and can calculate $r$ then $H$ from there.
Note, that this is ideal case, for such heaps, the summit won't be completely pointy, so you have to calculate the volume of a truncated cone, but it's almost the same calculation.
I think that's how industry proceed for heaps of other materials (like gravel, sand, grain, chalk, coconuts, ...), they pour a known quantity $M$ of the material and then measure the angle of repose and $r,H$ from which they calculate $V,\rho$. 
Then theoretically $(\rho,\theta)$ are constant data for a specific material,  independently of the height of the heap. So now you can calculate $M\leftrightarrow(r,H$) in both ways.
