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Gold/Mining/Energy : JAB International (JABI)
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To: Martin Wormser who wrote (637)10/29/1997 7:09:00 PM
From: DDS-OMS  Read Replies (1) of 4571
 
Martin,

To expand on your explanation of the so called "nuggets"--the specific gravity, and therefore its weight per given volume of the given pure "rock" is known--so by displacement of a given amount of water, the volume of the rock can be accurately measured. If, for example, a cubic foot of the pure rock (with no gold in it) weighs 50 lbs, and a cubic foot of the "nugget" weighs 51 lbs, then the rock contains 1 pound, or 12 oz t of gold--or possibly a mixture of gold, paladium or other noble metal. Instead of milling this cubic foot of rock to recover 12 oz of pure gold valued @~$315/0z, the rock or nugget can be sold for possibly twice as much as its actual gold content.

Regards,
Gary
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