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Well... I don't have the dimensions of a bottle of Dom handy so I'll make one assumption that Dom comes out with a new bottle that is hexagonal in shape (and the hexagon just happens to be some multiple of the coin diameter) but contains the same 750ml. of good stuff. Given that assumption the container would hold:
Quarters (3.5cm x .155cm) -
Volume of the bounding hexagonal column - .94 cc
Number of coins in a 750ml (750cc) hexagonal container - 798
Value of coins - $199.00 - too much
Penny (1.905cm x .155cm) -
Volume of the bounding hexagonal column - 0.51 cc
Number of coins in a 750ml (750cc) hexagonal container - 1466
Value of coins - $14.66 - not enough
Those were the only dimensions for us coins I could find in a quick search... but frome these two you can extrapolate a conclusion:
A nickel is a bit larger than a penny but much smaller than a quarter... so lets guestimate that 1000 could fit in our container (high side estimate)... this would be worth $50.00... not enough
A dime is smaller than a penny... but lets say it's the same size (high side estimate)... so 1466 dimes would be worth $146.00... so the dime is our winner since more than 146 dollars worth would fit into our ideal container.
With a profile of a Dom bottle these calculations could be run taking into account bottle diameter... but it's unlikely the result would change.
EDIT... just looked at previous posts... the only problem you guys have is that coins stacked efficiently form hexagonal shapes... not square shapes (try it out yourself). What you get is a bunch of columns of coins, each of which occupies a hexagonal volume that is:
(((Coin Radius / Sin(60degrees)) ^ 2) * 2.598) * Column Height
[ 08-19-2003, 10:13 AM: Message edited by: Thoran ]
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