MoneyBox/DensityAnalysis
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| − | Density Analysis of australian coinage |
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| − | + | ...a Density Analysis of australian coinage |
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ie, how much value of a given coin can you fit inside, say, 1litre of space? |
ie, how much value of a given coin can you fit inside, say, 1litre of space? |
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;$2: 20.5mm, 3mm |
;$2: 20.5mm, 3mm |
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| − | == How to calculate === |
+ | == How to calculate == |
There are three estimates easily possible. |
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) |
# 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 |
# 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. (is this the most efficient possible?) |
+ | # Same as before, but assume coins are hexagonal and stacked in a hex lattice. (is this the most efficient possible in a 3d space?) |
In the real world, coins are unlikely to be stacked neatly, but with repeated shaking should form themselves into dense stacks. Thus I'd expect a real-world result somewhere between the 'squared' and 'hexed' results. |
In the real world, coins are unlikely to be stacked neatly, but with repeated shaking should form themselves into dense stacks. Thus I'd expect a real-world result somewhere between the 'squared' and 'hexed' results. |
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In the real world, you can probably get about 950 10c coins into a 1litre container - or $95. |
In the real world, you can probably get about 950 10c coins into a 1litre container - or $95. |
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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. |
+ | In actual testing, I achieved 940 coins - or, $94! :) |
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=== 20c === |
=== 20c === |
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==== Conclusion ==== |
==== Conclusion ==== |
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In the real world, you might get around 500 20c coins into your 1litre container - or $100 |
In the real world, you might get around 500 20c coins into your 1litre container - or $100 |
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| + | * In an actual test, I had 442 coins - or $88.40 |
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| + | (some $1 stacks had $1.20, so the actual number is closer to $89 |
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| + | * In a second jar, I had 469 coins - or $93.80 |
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| + | Clearly, internal stacking can have more than a 5% different in value! |
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| + | (my stacking method tends to have been 'shake and repeat' until it seems to work... |
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| + | === 50c === |
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| + | Aussie 50c coins are |
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| + | * huge |
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| + | * not round |
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| + | Due to the second of these, I've never bothered to do the math. Maybe one day I'll have a jar full and I can give real-world results however! :) |
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| + | |||
| + | ...Real world test got it to $185! |
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| + | |||
=== $1 === |
=== $1 === |
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Latest revision as of 21:54, 20 December 2012
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...a 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
[edit] 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. (is this the most efficient possible in a 3d space?)
In the real world, coins are unlikely to be stacked neatly, but with repeated shaking should form themselves into dense stacks. Thus I'd expect a real-world result somewhere between the 'squared' and 'hexed' results.
- Volume of coin
- (pi*r^2) * thickness
- Volume of squared coin
- d^2 * thickness
- Volume of hexed coin
- ((sqrt(3)*d^2)/2) * thickness
[edit] 5c
[edit] 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.
[edit] Squared coins
- 19.41^2 * 1.55 = 583.96 cubicmm
- This gives us now only 1712 coins.
[edit] Hexed coins
- ((sqrt(3)*19.42^2)/2)*1.55 = 506.245 cubic mm
- This now gives us 1975 coins.
[edit] Conclusion
In the real world, you might get around 1800 5c coins into your 1litre container - or approx $90
[edit] 10c
[edit] 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.
[edit] Squared coins
- 23.6^2*1.98 = 1102.8 cubic mm.
- This now gives us only 906 coins
[edit] 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.
[edit] Conclusion
In the real world, you can probably get about 950 10c coins into a 1litre container - or $95.
In actual testing, I achieved 940 coins - or, $94! :)
[edit] 20c
[edit] real coins
- 2.52*pi*(14.26^2)= pi*512.436 = 1609.86 cubic mm
- This gives us 621 coins to occupy 1litre of space.
[edit] Squared coins
- 28.52^2*2.52 = 2049.744 cubic mm.
- This now gives us only 487 coins
[edit] Hexed coins
- ((sqrt(3)*28.52^2)/2)*2.52 = 1775.13 cubic mm
- This gives us 563 coins into our million cubic millimetre volume.
[edit] Conclusion
In the real world, you might get around 500 20c coins into your 1litre container - or $100
- In an actual test, I had 442 coins - or $88.40
(some $1 stacks had $1.20, so the actual number is closer to $89
- In a second jar, I had 469 coins - or $93.80
Clearly, internal stacking can have more than a 5% different in value! (my stacking method tends to have been 'shake and repeat' until it seems to work...
[edit] 50c
Aussie 50c coins are
- huge
- not round
Due to the second of these, I've never bothered to do the math. Maybe one day I'll have a jar full and I can give real-world results however! :)
...Real world test got it to $185!
[edit] $1
[edit] real coins
- 2.8*pi*(12.5^2)= pi*437.5 = 1374.45 cubic mm
- This gives us 727 coins to occupy 1litre of space.
[edit] Squared coins
- 25^2*2.8 = 1750 cubic mm.
- This now gives us only 571 coins
[edit] Hexed coins
- ((sqrt(3)*25^2)/2)*2.8 = 1515.54 cubic mm
- This gives us 659 coins into our million cubic millimetre volume.
[edit] Conclusion
In the real world, you're gonna be looking at around 600 coins. I'm not gonna even pretend I need to work out the dollar value here.
[edit] $2
[edit] real coins
- 3*pi*(10.25^2)= pi*315.19 = 990.19 cubic mm
- This gives us 1009 coins to occupy 1litre of space.
[edit] Squared coins
- 20.5^2*3 = 1260.75 cubic mm.
- This now gives us only 793 coins
[edit] Hexed coins
- ((sqrt(3)*20.5^2)/2)*3 = 1091.84 cubic mm
- This gives us 915 coins into our million cubic millimetre volume.
[edit] Conclusion
- You're probably looking at the 800-850 coins mark if you're lucky, or $1600+ :)
