### Table of Contents

# Value of US Coins by Volume

The task is to fill a 32 fluid ounce bottle with coins that are “*silver*”, i.e., no one cent pieces.

For these calculations, all values have been converted to metric - grams and mm - and the final weight is presented in both kilograms and pounds.

Converting 32 fl. oz. to mm^{3} results in a desired volume of 946,353 mm^{3}.

## Summary

The lowest value option is the result of filling the vessel exclusively with nickels. This results in a theoretical maximum value of $68.68, but a more realistic value of approximately $27 due to the space between the coins.

The highest value option results from filling the vessel with dollar coins for a theoretical maximum of $858.58, but a more realistic value of approximately $343.

**The more likely scenario is a mix of nickels, dimes, and quarters. Assuming an even distribution of all three denominations results in a theoretical value of $205.95 with a more realistic value of approximately $82.**

## Specifications of US Coins

The size and mass of the coins are taken directly from the US Mint.

Denomination | Value | Weight (g) | Diameter (mm) | Thickness (mm) |
---|---|---|---|---|

Cent | 1 | 2.500 | 19.05 | 1.52 |

Nickel | 5 | 5.000 | 21.21 | 1.95 |

Dime | 10 | 2.268 | 17.91 | 1.35 |

Quarter | 25 | 5.670 | 24.26 | 1.75 |

Half | 50 | 11.340 | 30.61 | 2.15 |

Dollar | 100 | 8.100 | 26.49 | 2.00 |

## Volume by Coin

The volume of each coin is calculated by using the formula for the volume of a cylinder:

π * r^{2} * thickness

## Value by Volume

Following is a table similar to the previous table that shows the volume occupied by each coin along with value in cents per mm^{3}, i.e., the denomination of each coin divided by the volume.

Denomination | Value | Volume (mm^{3}) | cents / mm^{3} |
---|---|---|---|

Cent | 1 | 433.2349000 | 0.002 308 |

Nickel | 5 | 688.9787539 | 0.007 257 |

Dime | 10 | 340.1064000 | 0.029 403 |

Quarter | 25 | 808.9274000 | 0.030 905 |

Half | 50 | 1,582.1770000 | 0.031 602 |

Dollar | 100 | 1,102.2590000 | 0.090 723 |

From the data in this table, one can see that the currency value by volume for dimes, quarter dollars, and half dollars is nearly identical with approximately ^{3}/_{10} of one cent per cubic millimeter, whereas a nickel has less than ^{1}/_{4} the value per volume of the other three coins and a dollar coin has approximately three (3) times the value per volume. The lowly one cent piece comes in at ^{1}/_{15} the value by volume of dimes, quarter dollars, and half dollars.

## Number of Coins in Target Vessel

Using the conversion value of 946,353 mm^{3} in 32 fl. oz., one may calculate the theoretical number of coins that will fit within the target vessel; the fractional portion of the coins has been truncated.

Denomination | Value | Volume (mm^{3}) | Coins that will fit in the vessel |
---|---|---|---|

Cent | 1 | 433.2349000 | 2,184 |

Nickel | 5 | 688.9787539 | 1,373 |

Dime | 10 | 340.1064000 | 2,782 |

Quarter | 25 | 808.9274000 | 1,169 |

Half | 50 | 1,582.1770000 | 598 |

Dollar | 100 | 1,102.2590000 | 858 |

## Value of Coins in Vessel

Multiplying the number of coins that will fit in the vessel times the value of each coin, the dollar value of the vessel may be calculated.

Denomination | Value | Volume (mm^{3}) | Coins | Value |
---|---|---|---|---|

Cent | 1 | 433.2349000 | 2,184 | $21.84 |

Nickel | 5 | 688.9787539 | 1,373 | $68.68 |

Dime | 10 | 340.1064000 | 2,782 | $278.25 |

Quarter | 25 | 808.9274000 | 1,169 | $292.47 |

Half | 50 | 1,582.1770000 | 598 | $299.07 |

Dollar | 100 | 1,102.2590000 | 858 | $858.56 |

## Derated Value

The total number of coins that will fit in the vessel is in actuality significantly less the number calculated above. The above numbers apply only if the coins were molten and could be poured into the irregular vessel without leaving any space. Because the coins will not be molten, the actual number of coins must be reduced to reflect the air gaps within the vessel.

It is well known that *cubic close packing*, also known as *face centered cubic*, or *hexagonal close packing* (HCP) when using uniform spheres within a volume will result in a density of ^{π}/<sub>3√2 ≅ 0.74048, whereas a strictly jammed sphere packing with the lowest density is a diluted fcc crystal with a density of 0.49365. Moreover, recent research predicts analytically that irregular packing of spheres cannot exceed a density of 64.3% and irregular sphere packing where the smaller spheres have a radius less than 0.41421 of the larger spheres allow for the smaller spheres to be placed into the interstitial spaces between the larger spheres results in a maximum density higher than that of uniform spheres.

In this example we are not working with spheres within a volume but with cylinders within a volume. The orientation of the cylinders within the vessel will greatly influence the ratio of open space to filled space. Based upon the information above, it is reasonably safe to assume that the minimum packing density of the vessel will be at least 40%.

Derating the number of coins to 40% of their maximum value results in these figures:

Denomination | Value | Volume (mm^{3}) | Coins | Derated | Value |
---|---|---|---|---|---|

Cent | 1 | 433.2349000 | 2,184 | 873 | $8.74 |

Nickel | 5 | 688.9787539 | 1,373 | 549 | $27.47 |

Dime | 10 | 340.1064000 | 2,782 | 1,113 | $111.30 |

Quarter | 25 | 808.9274000 | 1,169 | 467 | $116.99 |

Half | 50 | 1,582.1770000 | 598 | 239 | $119.63 |

Dollar | 100 | 1,102.2590000 | 858 | 343 | $343.42 |

## Evenly Distributed Nickel, Dime, Quarter

As one cent pieces are excluded from consideration and both half dollar and dollar coins being less common, consider the condition where the an equal number of nickels, dimes, and quarters are deposited into the vessel.

One may arrive at an aggregate volume value for these three coins by adding the individual volumes together:

688.9788 + 340.1064 + 808.9274 = 1,838.0126 mm^{3}

and the aggregate value of this virtual coin is the sum of the values of the individual coins:

5 + 10 + 25 = $0.40

Dividing the volume of the virtual coin into the volume of the vessel results in 514 of the $0.40 coins for a value of $205.95.

Derating the quantity of coins to only 40% of the theoretical maximum number of virtual coins results in 205 total virtual coins at a value of $0.40 for a tital value of $82.38.

The resultant table for individual coins and the virtual coin - the more realistic distribution of coins - is shown here, without any derating, so that the reader may use whatever density derate is desired.

Denomination | Value | Coins | Maximum Value |
---|---|---|---|

Cent | 1 | 2,184 | $21.84 |

Nickel | 5 | 1,373 | $68.68 |

Dime | 10 | 2,782 | $278.25 |

Quarter | 25 | 1,169 | $292.47 |

Half | 50 | 598 | $299.07 |

Dollar | 100 | 858 | $858.56 |

Virtual 5/10/15 | 40 | 514 | $205.25 |

## Weight

Purely as a thought experiment, the total weight of the vessel for each coin may be calculated. These values assume the theoretical maximum number of coins and should be derated as necessary to account for the spaces between the coins:

Denomination | Value | Weight (g) | Coins | Mass (Kg) | Weight (lbs) |
---|---|---|---|---|---|

Cent | 1 | 2.500 | 2,184 | 5.46 | 12.0 |

Nickel | 5 | 5.000 | 1,373 | 6.87 | 15.1 |

Dime | 10 | 2.268 | 2,782 | 6.31 | 13.9 |

Quarter | 25 | 5.670 | 1,169 | 6.63 | 14.6 |

Half | 50 | 11.340 | 598 | 6.78 | 15.0 |

Dollar | 100 | 8.100 | 858 | 6.95 | 15.3 |

Virtual 5/10/15 | 40 | 12.938 | 514 | 6.65 | 14.6 |