MoneyBox/DensityAnalysis
m |
(hex version of 10c) |
||
| Line 33: | Line 33: | ||
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. |
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 |
+ | Hexagon stacking: |
| + | |||
| + | Same idea as above, but we work out the volume of a hexagon that is tangiential 6 times to the circle of the coin. We know the diameter of the coin, and from this the area of the hexagon in question is (sqrt(3)*d^2)/2 |
||
| + | |||
| + | So for our 10c coins, this gives us a volume of 955.036 cubic mm - fitting 1047 coins into our million cubic millimetre area. |
||
| + | |||
| + | Conclusion: In the real world, you can probably get about 1000 10c coins into a 1litre container - or $100 |
||
| + | |||
| + | |||
| + | === Links === |
||
| + | * http://www.drking.worldonline.co.uk/hexagons/misc/area.html |
||
Revision as of 23:51, 23 January 2004
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
20c
I've not caclulated this out, but from experience, $40 of 20c coins will fill a 1litre container nicely.
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.
Hexagon stacking:
Same idea as above, but we work out the volume of a hexagon that is tangiential 6 times to the circle of the coin. We know the diameter of the coin, and from this the area of the hexagon in question is (sqrt(3)*d^2)/2
So for our 10c coins, this gives us a volume of 955.036 cubic mm - fitting 1047 coins into our million cubic millimetre area.
Conclusion: In the real world, you can probably get about 1000 10c coins into a 1litre container - or $100
