MoneyBox/DensityAnalysis
(real real world 10c testing) |
m (first writeup, 10c coin analysis) |
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;$2: 20.5mm, 3mm |
;$2: 20.5mm, 3mm |
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| − | == How to calculate === |
+ | So, since my first litre of coins obtained from the MoneyBox was in 10c coins, I'll calculate that out first. |
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| − | There are three estimates easily possible. |
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| − | # 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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| − | 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. |
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| − | ;Volume of coin: (pi*r^2) * thickness |
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| − | ;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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| − | === 5c === |
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| − | ==== real coins ==== |
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| − | * 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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| − | ==== Squared coins ==== |
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| − | * 19.41^2 * 1.55 = 583.96 cubicmm |
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| − | * This gives us now only 1712 coins. |
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| − | ==== Hexed coins ==== |
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| − | * ((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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| − | ==== Conclusion ==== |
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| − | 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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=== 10c === |
=== 10c === |
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| − | ==== real coins ==== |
+ | The volume of a cylinder is determined as it's circular cross sectional area multiplied by it's height. Area is pi*(r^2). |
| − | * 1.98*pi*(11.8^2)= pi*275.7 = 866.12 cubic mm |
+ | So, for a 10c coin, the volume is 1.98*pi*(11.8^2)=pi*275.7 which is roughly 866.12 cubic mm |
| − | * This gives us 1154 coins to occupy 1litre of space. |
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| − | ==== Squared coins ==== |
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| − | * 23.6^2*1.98 = 1102.8 cubic mm. |
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| − | * This now gives us only 906 coins |
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| − | ==== Hexed coins ==== |
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| − | * ((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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| − | ==== Conclusion ==== |
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| − | In the real world, you can probably get about 1000 10c coins into a 1litre container - or $100. |
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| − | In an actual test however, 750 coins ($75) left little room available, and as coins become available, I doubt I'll get more than 800 in there. |
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| − | === 20c === |
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| − | ==== real coins ==== |
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| − | * 2.52*pi*(14.26^2)= pi*512.436 = 1609.86 cubic mm |
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| − | * This gives us 621 coins to occupy 1litre of space. |
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| − | ==== Squared coins ==== |
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| − | * 28.52^2*2.52 = 2049.744 cubic mm. |
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| − | * This now gives us only 487 coins |
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| − | ==== Hexed coins ==== |
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| − | * ((sqrt(3)*28.52^2)/2)*2.52 = 1775.13 cubic mm |
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| − | * This gives us 563 coins into our million cubic millimetre volume. |
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| − | ==== Conclusion ==== |
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| − | In the real world, you might get lucky to squeeze 600 20c coins into your 1litre container - or $120 |
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| − | === $1 === |
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| − | ==== real coins ==== |
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| − | * 2.8*pi*(12.5^2)= pi*437.5 = 1374.45 cubic mm |
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| − | * This gives us 727 coins to occupy 1litre of space. |
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| − | |||
| − | ==== Squared coins ==== |
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| − | * 25^2*2.8 = 1750 cubic mm. |
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| − | * This now gives us only 571 coins |
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| − | ==== Hexed coins ==== |
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| − | * ((sqrt(3)*25^2)/2)*2.8 = 1515.54 cubic mm |
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| − | * This gives us 659 coins into our million cubic millimetre volume. |
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| − | ==== Conclusion ==== |
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| − | In the real world, you're gonna be looking at around 600-650 coins. I'm not gonna even pretend I need to work out the dollar value here. |
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| − | === $2 === |
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| − | ==== real coins ==== |
+ | 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. |
| − | * 3*pi*(10.25^2)= pi*315.19 = 990.19 cubic mm |
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| − | * This gives us 1009 coins to occupy 1litre of space. |
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| − | ==== Squared 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. |
| − | * 20.5^2*3 = 1260.75 cubic mm. |
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| − | * This now gives us only 793 coins |
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| − | ==== Hexed coins ==== |
+ | With out 1million cubic mm litre, we can fit 906 coins in via this method. |
| − | * ((sqrt(3)*20.5^2)/2)*3 = 1091.84 cubic mm |
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| − | * This gives us 915 coins into our million cubic millimetre volume. |
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| − | ==== Conclusion ==== |
+ | 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. |
| − | * You're probably looking at the 950 coins mark if you're lucky, or $1900 work :) |
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| − | === Links === |
+ | Conclusion: You can probably get about 1000 10c coins into a 1litre container - or $100 |
| − | * http://www.drking.worldonline.co.uk/hexagons/misc/area.html |
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Revision as of 14:58, 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
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
