MoneyBox/DensityAnalysis

Density Analysis of australian coinage

ie, how much value of a given coin can you fit inside, say, 1litre of space?

So what are we dealing with:

• 1litre = 1000cm^3 of volume = 1,000,000 cubic mm

Australian coin sizes: (see: http://www.worldmints.com/ccoin_ram.asp) (note that diameter is the specified, but thickness is the maximum legal allowed)

5c

diameter: 19.41mm, thickness: 1.55mm

10c

23.6mm, 1.98mm

20c

28.52mm, 2.52mm

50c

31.51mm (across flats), 2.8mm

$1

25mm, 2.8mm

$2

20.5mm, 3mm

So, since my first litre of coins obtained from the MoneyBox was in 10c coins, I'll calculate that out first.

10c

The volume of a cylinder is determined as it's circular cross sectional area multiplied by it's height. Area is pi*(r^2). So, for a 10c coin, the volume is 1.98*pi*(11.8^2)=pi*275.7 which is roughly 866.12 cubic mm

So, it will take 1154 10c coins will occupy the volume of a litre. That's a theoretical maximum of course, and for a real-world litre container, there will be spaces between the coins.

So, let's assume coins stacked in piles, and those stacks arranged in a grid. ie, as if the coins were square rather than round. In this case, each coin takes up the volume of 23.6^2*1.98 = 1102.8 cubic mm.

With out 1million cubic mm litre, we can fit 906 coins in via this method.

In the real world, coins even dumped in randomly, will arrange themselves much more efficiently than this method, but the math to work it out would be quite horrible, and would need to know the container shape. If we worked out the volume according to piles arranged in a honeycomb, then that would probably get us much closer.

Conclusion: You can probably get about 1000 10c coins into a 1litre container - or $100