Return to Revisions

Nickel value per volume:

$$\frac{5 ¢}{\pi(\frac{21.21}{2}\ \text{mm})^2 * 1.95\ \text{mm}} = \frac{7.257 ¢}{\text{mL}}$$

Dime:

$$\frac{10 ¢}{\pi(\frac{17.91}{2}\ \text{mm})^2 * 1.35\ \text{mm}} = \frac{29.40 ¢}{\text{mL}}$$

So, for large volumes with reasonable dimensions$\dagger$, if you had twice the volume of nickels as dimes, the nickels would worth about half as much as the dimes.

$\dagger$ I say reasonable dimensions because fringe effects come into play if you have to pay attention to the coins bumping into the wall.