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"Fineness" is not the actual amount of silver in a coin
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<p>[QUOTE="rush2112, post: 1818438, member: 25570"]I know most of you know better, but there is a coin dealer on that popular online auction site, who was using the "fineness" or "purity" of an old Mexican coin as the actual amount of silver in the coin he had listed. Fineness is only part of the formula to calculate the actual amount of silver or gold in a coin.</p><p> </p><p>If you have the weight and fineness you can figure out how the asw or agw.</p><p> </p><p>Actual gold or silver weight in a coin measured in in troy oz. calculation:</p><p> </p><p>Actual gold or silver weight in grams divided by 31.10 = troy oz.</p><p> </p><p>Actual gold or silver weight in a coin measured in grams calculation:</p><p> </p><p>Actual gold or silver gram weight of a coin divided by fineness = actual silver or gold weight in grams.</p><p> </p><p>If there is a problem with this calculation please correct this post. It worked for me when I tried it several times to make sure it was right.</p><p> </p><p>READ BELOW, the second part of this calculation was wrong, Tom B corrected it.[/QUOTE]</p><p><br /></p>
[QUOTE="rush2112, post: 1818438, member: 25570"]I know most of you know better, but there is a coin dealer on that popular online auction site, who was using the "fineness" or "purity" of an old Mexican coin as the actual amount of silver in the coin he had listed. Fineness is only part of the formula to calculate the actual amount of silver or gold in a coin. If you have the weight and fineness you can figure out how the asw or agw. Actual gold or silver weight in a coin measured in in troy oz. calculation: Actual gold or silver weight in grams divided by 31.10 = troy oz. Actual gold or silver weight in a coin measured in grams calculation: Actual gold or silver gram weight of a coin divided by fineness = actual silver or gold weight in grams. If there is a problem with this calculation please correct this post. It worked for me when I tried it several times to make sure it was right. READ BELOW, the second part of this calculation was wrong, Tom B corrected it.[/QUOTE]
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