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<p>[QUOTE=&quot;Sunbird, post: 8279480, member: 116324&quot;]Hi all – I&#039;m surprised by the amount of &quot;missing&quot; volume / weight in circulation coins compared to a smooth cylinder/coin made of the same density alloy. For example, take the modern US quarter:</p><ol> <li>Diameter: <b>24.26 mm</b></li> <li>Thickness: <b>1.75 mm</b></li> <li>Volume if it was a perfectly smooth cylinder (no upsetting, stamped design, etc.): <b>0.809 cm³</b></li> <li>Mass of such a smooth cylinder made of 91.67% copper, 8.33% nickel: <b>7.22 grams</b> (Copper and nickel have almost identical densities, so the proportions don&#039;t matter much – I used 8.92 g/cm³ as the combined density.)<br /> </li> <li>Mass of actual quarter: <b>5.67 grams</b></li> <li>Missingness: <b>≈21.4%</b></li> </ol><p>That is, the mass is 21.4% shy of a featureless coin blank of the quarter&#039;s dimensions, which means that the volume is 21.4% short as well. I knew that it would be short, but I&#039;m having a hard time understanding how it can be <i>that</i> short. Looking at the coin, the field&#039;s recession from the rim doesn&#039;t look deep enough to account for such a drop in mass/volume. A 20% shave would require the field to be recessed a tenth of the coin&#039;s thickness/height on <i>each side</i>, and that the field was flat and smooth with no design on the coin.</p><p><br /></p><p>But the quarter has a lot of design on both sides that <i>rises up from the field</i>, which means there&#039;s more volume and mass there than there would be if the coin was featureless. This in turn means that the field&#039;s recession or sunkenness from the rim must be even deeper to make up for the volume and mass of the elevated design features. I don&#039;t see it, and my calipers can&#039;t measure the thickness of the coin at its thinnest points, recessed field to recessed field. I can only measure the rim thickness, which is the max. Anyone know what the thickness of a quarter is at its thinnest?</p><p><br /></p><p>The only other angle I can see is the reeded/milled edge. That would create some missing volume and mass compared to a smooth cylinder. If the highs and lows are equal in area around the rim, the average diameter of the coin would be reduced to the nominal diameter - (the depth of the lows/2). Maybe that ends up shaving half a millimeter from the diameter – no idea, but it&#039;s another aggregate hole in the coin cylinder.</p><p><br /></p><p>Anyway, I&#039;m surprised by how unintuitive this looks to me, as far as the amount of lost volume and mass, from just eyeing the coin. I&#039;ll note that the reality is similar with other circulation coins. Canadian steel coins have nominal densities of 5.62 - 6.36 g/cm³, even though mild steel is around 7.87 g/cm³ (not to mention the few percent of copper and nickel plating they have, with a density of 8.92 g/cm³). By &quot;nominal density&quot;, I mean the computed density of a perfect coin cylinder of the dimensions of the relevant coin, given the weight of the coin (I have a spreadsheet with nominal volumes and densities for dozens of coins, along with the usual measurements). It&#039;s always less than the density of the coin metals because coins are always lower in volume than a cylinder of the same dimensions. For some reason Mexican coins have the highest nominal density relative to their material, even though stainless steel is a bit lighter than mild steel (their coins are made of 430 stainless) – they must have less sunkenness from the rim.[/QUOTE]</p><p><br /></p>

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