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Gold Prospecting
Metal Detecting for Gold
Specific Gravity test help - Gold and quartz
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<blockquote data-quote="Hawkear" data-source="post: 647374" data-attributes="member: 4728"><p>I have never been good at remembering things like formulae which many use to calculate gold content from wet and dry spec weights and use the following estimation process.</p><p></p><p>The example at the start of this topic assumed a dry weight of 25.68 gr and I assume that the author meant a weight loss of 2.99 gr when weighed suspended in water. To estimate the gold content, I use the following method.</p><p></p><p>A 2.99 gram loss in weight would equate (Archimedes principle) to a volume of 2.99 cc for the specimen which if all quartz would weigh just 2.99 x SG (2.62) = 7.83 gr. That would mean that the rest of the weight must be the gold content ie. 25.68 -7.83 = 17.85 gr of gold.</p><p>This is of course a minimum first estimation and can be refined further as we know that the gold itself would occupy some of the specimen volume we attributed to quartz in this first estimation.</p><p></p><p>For a second estimation, we can then calculate that 17.85g of gold would occupy a volume of 17.85 divided by its SG, 17.85 divided by 19.3 = 0.92 cc. This reduces the amount of quartz volume to 2.99 – 0.92 = 2.07 cc. Then again this volume of quartz would only weigh 2.07 x 2.62 = 5.42gr. Therefore our estimation of gold should be increased to 25.68 – 5.42 = 20.26gr of gold.</p><p></p><p>A third iteration of this process can be done to improve the gold estimation to 20.60gr and yet a fourth improves the estimation to 20.65gr.</p><p></p><p>No point in going any further due to the inherent weighing, specimen mineral and SG variations.</p></blockquote><p></p>
[QUOTE="Hawkear, post: 647374, member: 4728"] I have never been good at remembering things like formulae which many use to calculate gold content from wet and dry spec weights and use the following estimation process. The example at the start of this topic assumed a dry weight of 25.68 gr and I assume that the author meant a weight loss of 2.99 gr when weighed suspended in water. To estimate the gold content, I use the following method. A 2.99 gram loss in weight would equate (Archimedes principle) to a volume of 2.99 cc for the specimen which if all quartz would weigh just 2.99 x SG (2.62) = 7.83 gr. That would mean that the rest of the weight must be the gold content ie. 25.68 -7.83 = 17.85 gr of gold. This is of course a minimum first estimation and can be refined further as we know that the gold itself would occupy some of the specimen volume we attributed to quartz in this first estimation. For a second estimation, we can then calculate that 17.85g of gold would occupy a volume of 17.85 divided by its SG, 17.85 divided by 19.3 = 0.92 cc. This reduces the amount of quartz volume to 2.99 – 0.92 = 2.07 cc. Then again this volume of quartz would only weigh 2.07 x 2.62 = 5.42gr. Therefore our estimation of gold should be increased to 25.68 – 5.42 = 20.26gr of gold. A third iteration of this process can be done to improve the gold estimation to 20.60gr and yet a fourth improves the estimation to 20.65gr. No point in going any further due to the inherent weighing, specimen mineral and SG variations. [/QUOTE]
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Gold Prospecting
Metal Detecting for Gold
Specific Gravity test help - Gold and quartz
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