MoneyBox/DensityAnalysis
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;$2: 20.5mm, 3mm |
;$2: 20.5mm, 3mm |
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| + | == How to calculate === |
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| − | === 20c === |
+ | There are three estimates easily possible. |
| + | # Count of coins for maximum displacement within our 1litre limit (1litre divided by volume per coin) |
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| + | # Same as before, but assume coins are square with edge length equal to diameter. Simulates stacks arranged in a grid |
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| + | # Same as before, but assume coins are hexagonal and stacked in a hex lattice. |
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| − | I've not caclulated this out, but from experience, $40 of 20c coins will fill a 1litre container nicely. |
+ | In the real world, coins are unlikely to be stacked, and in my estimate, are likely to be slightly more efficiently packed in than even the simulated hex lattice. |
| − | === 10c === |
+ | ;Volume of coin: (pi*r^2) * thickness |
| + | ;Volume of squared coin: d^2 * thickness |
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| + | ;Volume of hexed coin: ((sqrt(3)*d^2)/2) * thickness |
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| − | The volume of a cylinder is determined as it's circular cross sectional area multiplied by it's height. Area is pi*(r^2). |
+ | === 5c === |
| − | So, for a 10c coin, the volume is 1.98*pi*(11.8^2)=pi*275.7 which is roughly 866.12 cubic mm |
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| − | 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. |
+ | ==== real coins ==== |
| + | * 1.55*pi*9.705^2 = 145.99*pi = 458.638 cubic mm |
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| + | * This gives us 2180 coins to occupy 1litre of space. |
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| − | 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. |
+ | ==== Squared coins ==== |
| + | * 19.41^2 * 1.55 = 583.96 cubicmm |
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| + | * This gives us now only 1712 coins. |
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| − | With out 1million cubic mm litre, we can fit 906 coins in via this method. |
+ | ==== Hexed coins ==== |
| + | * ((sqrt(3)*19.42^2)/2)*1.55 = 506.245 cubic mm |
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| + | * This now gives us 1975 coins. |
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| − | 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 ==== |
| + | In the real world, you might get lucky to squeeze 2000 5c coins into your 1litre container - or just on $100 |
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| − | Hexagon stacking: |
+ | === 10c === |
| − | 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 |
+ | ==== real coins ==== |
| + | * 1.98*pi*(11.8^2)= pi*275.7 = 866.12 cubic mm |
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| + | * This gives us 1154 coins to occupy 1litre of space. |
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| + | So, it will take 1154 10c coins will occupy the volume of a litre. |
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| − | So for our 10c coins, this gives us a volume of 955.036 cubic mm - fitting 1047 coins into our million cubic millimetre area. |
+ | ==== Squared coins ==== |
| + | * 23.6^2*1.98 = 1102.8 cubic mm. |
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| + | * This now gives us only 906 coins |
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| − | Conclusion: In the real world, you can probably get about 1000 10c coins into a 1litre container - or $100 |
+ | ==== Hexed coins ==== |
| + | * ((sqrt(3)*23.6^2)/2)*1.98 = 955.036 cubic mm |
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| + | * This gives us 1047 coins into our million cubic millimetre volume. |
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| − | === 5c === |
+ | ==== Conclusion ==== |
| + | In the real world, you can probably get about 1000 10c coins easily into a 1litre container - or $100 with a little room left over |
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| − | Volume per coin = 1.55*pi*9.705^2 = 145.99*pi = 458.638 cubic mm |
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| − | This gives us 2180 coins to occupy 1litre of space. |
+ | === 20c === |
| − | |||
| − | Squared circle estimate gives us: 19.41^2 * 1.55 = 583.96 cubicmm |
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| − | |||
| − | This gives us now only 1712 coins. |
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| − | |||
| − | The hexed circle estimate: volume = ((sqrt(3)*19.42^2)/2)*1.55 = 506.245 cubic mm |
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| − | This now gives us 1975 coins. |
+ | I've not caclulated this out, but from experience, $40 of 20c coins will fill a 1litre container nicely. |
| − | Conclusion: In the real world, you might get lucky to squeeze 2000 5c coins into your 1litre container - or just on $100 |
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=== Links === |
=== Links === |
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Revision as of 12:14, 24 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
Contents |
How to calculate =
There are three estimates easily possible.
- Count of coins for maximum displacement within our 1litre limit (1litre divided by volume per coin)
- Same as before, but assume coins are square with edge length equal to diameter. Simulates stacks arranged in a grid
- Same as before, but assume coins are hexagonal and stacked in a hex lattice.
In the real world, coins are unlikely to be stacked, and in my estimate, are likely to be slightly more efficiently packed in than even the simulated hex lattice.
- Volume of coin
- (pi*r^2) * thickness
- Volume of squared coin
- d^2 * thickness
- Volume of hexed coin
- ((sqrt(3)*d^2)/2) * thickness
5c
real coins
- 1.55*pi*9.705^2 = 145.99*pi = 458.638 cubic mm
- This gives us 2180 coins to occupy 1litre of space.
Squared coins
- 19.41^2 * 1.55 = 583.96 cubicmm
- This gives us now only 1712 coins.
Hexed coins
- ((sqrt(3)*19.42^2)/2)*1.55 = 506.245 cubic mm
- This now gives us 1975 coins.
Conclusion
In the real world, you might get lucky to squeeze 2000 5c coins into your 1litre container - or just on $100
10c
real coins
- 1.98*pi*(11.8^2)= pi*275.7 = 866.12 cubic mm
- This gives us 1154 coins to occupy 1litre of space.
So, it will take 1154 10c coins will occupy the volume of a litre.
Squared coins
- 23.6^2*1.98 = 1102.8 cubic mm.
- This now gives us only 906 coins
Hexed coins
- ((sqrt(3)*23.6^2)/2)*1.98 = 955.036 cubic mm
- This gives us 1047 coins into our million cubic millimetre volume.
Conclusion
In the real world, you can probably get about 1000 10c coins easily into a 1litre container - or $100 with a little room left over
20c
I've not caclulated this out, but from experience, $40 of 20c coins will fill a 1litre container nicely.
