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Odd Weight for a 1982 penny
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<p>[QUOTE="James Miller, post: 3653328, member: 106185"]To find the actual density of the penny, I measure it with a calibrated scale. Then we have this fixture that allows you to suspend the coin in liquid, but still weigh it. There will be a difference in weight. If you think about picking up a 10 pound rock. Then go into a lake and pick up a 10 pound rock. It's easier... So basically you first finding the specific gravity as it relates to the liquid. From there it's math to get to the actual density of the penny. I found it to be between 8.91g/ml and 9.02g/ml</p><p>To find the theoretical density of a coin, I just use the known composition percentages and known densities of materials used. So the theoretical density of the 1981 Canadian coin is 8.93g/ml while the American copper 1982 coin is 8.87g/ml </p><p>No matter what, my penny has more copper content percentage then it should for an American penny. That's the only way the numbers work.</p><p>I wasn't sure if you wanted me to be more in depth with my calcs... I'm an engineer and have access to lots of test equipment.[/QUOTE]</p><p><br /></p>
[QUOTE="James Miller, post: 3653328, member: 106185"]To find the actual density of the penny, I measure it with a calibrated scale. Then we have this fixture that allows you to suspend the coin in liquid, but still weigh it. There will be a difference in weight. If you think about picking up a 10 pound rock. Then go into a lake and pick up a 10 pound rock. It's easier... So basically you first finding the specific gravity as it relates to the liquid. From there it's math to get to the actual density of the penny. I found it to be between 8.91g/ml and 9.02g/ml To find the theoretical density of a coin, I just use the known composition percentages and known densities of materials used. So the theoretical density of the 1981 Canadian coin is 8.93g/ml while the American copper 1982 coin is 8.87g/ml No matter what, my penny has more copper content percentage then it should for an American penny. That's the only way the numbers work. I wasn't sure if you wanted me to be more in depth with my calcs... I'm an engineer and have access to lots of test equipment.[/QUOTE]
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