Expert: Dmitry Livshits - 8/24/2011

Question
Hi Dmitry,

Silver American Eagles are supposed to be 1oz of pure silver.  According to the dimensions of the coin and the density of pure silver, I calculated that the coin should weigh 1.3oz, not 1oz.  My math follows:

The coin has a radius of 20.3mm and a thickness of 2.98mm.  Since V=(3.14)(r^2)(height), its V is 3856mm^3, which is equal to 3.856cm^3.

If mass = density*volume, and the density of silver is 10.49g/cm^3, then .the mass should equal (3.856cm^3)*(10.49g/cm^3) which = 40.45g.  Since 31.1g = 1oz, dividing 40.45g by 31.1g gives a coin that, if pure silver, should weigh 1.3oz, not 1oz.

Why the discrepancy?

Karl

Answer
Hi Karl,
My apologies for the delay in answering your question.
The discrepancy is due to the physical dimensions of the coin. If it had been a blank disk with the specified dimensions, your calculation would certainly apply.
However, when measuring the thickness of the coin from the edge, it does not account for the fact that the rim is slightly higher than most of the design of the coin. Although the features in relief will be about the same height, most of the coin surface are flat fields that would have a slightly smaller thickness than the edge.
For many coins this would not account for a significant difference, but it does for a coin this large and with features struck in a particularly high relief compared to the lower relief fields.
Similar story for diameter - the reeded edge means that the weight will be slightly lower than that of a coin with equal diameter but a plain edge. The reeds cut in about 1/2mm on the 2010 and 2011 coins I just examined, meaning half the edge by surface area is that much less in diameter than the actual measurement.
On a separate note - there are also .925 silver unofficial/privately minted 1 ounce a.s.w. versions with the same design out there. Such examples will always be a few grams heavier than US minted official .999 silver rounds.
Thanks for the question! =)