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<p>[QUOTE="Earle42, post: 2492480, member: 78261"]OK kids (sorry - former math teacher) let's do the math:</p><p>From the link here:</p><p><a href="https://www.cointalk.com/threads/national-bicentennial-medal.216522/" class="internalLink ProxyLink" data-proxy-href="https://www.cointalk.com/threads/national-bicentennial-medal.216522/">https://www.cointalk.com/threads/national-bicentennial-medal.216522/</a></p><p><br /></p><p>and the above pics we see the closest match from this list to the above pics is the </p><p>- 1-5/16" / Gold, Proof finish / 1.167 troy ounces of gold / $400.00</p><p><br /></p><p>Let's see what a medal of pure gold of same dimensions should weigh.</p><p>1. We don;t know the exact thickness (which is needed) but can estimate.</p><p>2. The pics seem to show a 1 5/16 " match for diameter = 1.3125" </p><p>3. Therefore radius = .65625</p><p>4. Volume of a circular, cylindrical surface in simplisitc form is its height time pi times its radius squared.</p><p>5. The pics seem to indicate a thickness of just under 1/8th inch. So lets use a full 1/8th (.125) and also half of that (.0625) and see if the pic of the coin's weight on the scales is somewhere in between.</p><p>For the .125 (1/8) thickness we get:</p><p>Height (thickness in this case) X 3.14 X Radius^2 (which means radius times radius)</p><p><br /></p><p>.125 X 3.14 X .6525^2 = .1671 cubic inches of gold</p><p><br /></p><p>The thinner calculation:</p><p>0.625 X 3.14 X .6525^2 = .0836 cubic inches of gold</p><p><br /></p><p>according to:</p><p><a href="http://www.aqua-calc.com/calculate/volume-to-weight" target="_blank" class="externalLink ProxyLink" data-proxy-href="http://www.aqua-calc.com/calculate/volume-to-weight" rel="nofollow">http://www.aqua-calc.com/calculate/volume-to-weight</a></p><p><br /></p><p>The first calculation gives a weight of 1.87 oz</p><p><br /></p><p>The thinner calculation gives: .93 oz</p><p><br /></p><p>The pic of the coin on the scales fits between these two - and is actually closer to the thicker one where we already know the thickness is not quite 1/8.</p><p><br /></p><p>If I had to make an educated guess, I would say the coin is gold.</p><p><br /></p><p>Please correct any errors for me in the above.[/QUOTE]</p><p><br /></p>
[QUOTE="Earle42, post: 2492480, member: 78261"]OK kids (sorry - former math teacher) let's do the math: From the link here: [url]https://www.cointalk.com/threads/national-bicentennial-medal.216522/[/url] and the above pics we see the closest match from this list to the above pics is the - 1-5/16" / Gold, Proof finish / 1.167 troy ounces of gold / $400.00 Let's see what a medal of pure gold of same dimensions should weigh. 1. We don;t know the exact thickness (which is needed) but can estimate. 2. The pics seem to show a 1 5/16 " match for diameter = 1.3125" 3. Therefore radius = .65625 4. Volume of a circular, cylindrical surface in simplisitc form is its height time pi times its radius squared. 5. The pics seem to indicate a thickness of just under 1/8th inch. So lets use a full 1/8th (.125) and also half of that (.0625) and see if the pic of the coin's weight on the scales is somewhere in between. For the .125 (1/8) thickness we get: Height (thickness in this case) X 3.14 X Radius^2 (which means radius times radius) .125 X 3.14 X .6525^2 = .1671 cubic inches of gold The thinner calculation: 0.625 X 3.14 X .6525^2 = .0836 cubic inches of gold according to: [url]http://www.aqua-calc.com/calculate/volume-to-weight[/url] The first calculation gives a weight of 1.87 oz The thinner calculation gives: .93 oz The pic of the coin on the scales fits between these two - and is actually closer to the thicker one where we already know the thickness is not quite 1/8. If I had to make an educated guess, I would say the coin is gold. Please correct any errors for me in the above.[/QUOTE]
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