MoneyBox/DensityAnalysis

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Density Analysis of australian coinage
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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?
  
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.
 
# 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)
 
# 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.
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# Same as before, but assume coins are hexagonal and stacked in a hex lattice. (is this the most efficient possible in a 3d space?)
   
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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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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 coin: (pi*r^2) * thickness
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==== Conclusion ====
  
==== 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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In the real world, you might get around 1800 5c coins into your 1litre container - or approx $90
   
 
=== 10c ===
  
=== 10c ===
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==== Conclusion ====
  
==== Conclusion ====
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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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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! :)
   
 
=== 20c ===
  
=== 20c ===
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==== Conclusion ====
  
==== Conclusion ====
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In the real world, you might get lucky to squeeze 600 20c coins into your 1litre container - or $120
 +
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...
  +
  +
  +
=== 50c ===
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Aussie 50c coins are
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* huge
  +
* 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! :)
  +
  +
...Real world test got it to $185!
  +
   
 
=== $1 ===
  
=== $1 ===
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==== Conclusion ====
  
==== 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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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.
   
   
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==== Conclusion ====
  
==== Conclusion ====
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* You're probably looking at the 950 coins mark if you're lucky, or $1900 work :)
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* You're probably looking at the 800-850 coins mark if you're lucky, or $1600+ :)
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=== Links ===
  
=== Links ===

Latest revision as of 21:54, 20 December 2012

Contents

...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.

  1. Count of coins for maximum displacement within our 1litre limit (1litre divided by volume per coin)
  2. Same as before, but assume coins are square with edge length equal to diameter. Simulates stacks arranged in a grid
  3. 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+ :)


[edit] Links

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