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Old 08-19-2003, 08:24 PM   #15
Azred
Drow Priestess
 

Join Date: March 13, 2001
Location: a hidden sanctorum high above the metroplex
Age: 55
Posts: 4,037
Question Mark

[img]graemlins/erm.gif[/img] Why not do it the easy way?

Obtain the volume of each US coin via water displacement, from which you can calculate each coin's value in units $/mL; measure the mass of the coin and you can also calcuate its density. Take any bottle of champagne and use the volume on the front, as well as the price, to calculate its value in $/mL. Now you can compare base value/volume directly.
Once you have these, take any glass and calculate the volume it holds by filling it with water and compare to mass to the empty glass. Multiply glass volume * coin density, divide by the mass of one coin and round down to the largest integer to deduce the maximum number of coins the glass may contain; from this you can figure the value of the coins in the glass. Finally, it is a simple calculation to figure out how much a glassful of champagne would cost (volume * value in $/mL).
In the end, you know how much a glassful of each coin in worth and you may compare these against the value of a glassful of champagne.

Don't forget to use half-dollar, "silver" dollar, Susan B. Anthony dollar, and Sacajawea dollar coins.

After all this, you'll most likely want to drink what is left of the champagne. [img]graemlins/awesomework.gif[/img] [img]graemlins/petard.gif[/img]
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