This really confuses me. I was buying a capsule for a 1 oz Gold Buffalo and I noticed it had the same 32.7mm diameter as a 1oz Gold Eagle. But when I looked at the thickness the Buffalo is thicker at 2.95mm compared to the Gold Eagle’s 2.87mm. If a 1 oz Gold Buffalo is .9999 fine and an AGE is only .9167 fine but has 1 oz of gold in it plus some alloy metal (copper & silver) making up the difference how can the Buffalo be a bigger coin? Shouldn’t the AGE be bigger since it’s got 1 oz of gold like the Buffalo and 2 grams of other metals in it? Maybe it’s the coin design? That’s all I can think of.
No definitely not counterfeit. It’s the official dimensions of the coins themselves. If APMEX says that a coin has X diameter and Y thickness that’s what they are. It’s not an individual coin I’m talking about. Every real Gold Buffalo will be larger than every real Gold Eagle and I’m just wondering how that’s possible. It would be like a 1 oz American Gold Eagle being bigger than a 1 oz American Silver Eagle even though logically that just can’t be since Gold is denser than silver and so gold coins are always smaller than silver coins of equal weight. A 1 oz American Silver Eagle just can’t be smaller than a 1 oz Gold Eagle.
Ok buddy.. I'm just giving you some April Fool's day aggrevation.. I hope you find the answer you seek! Peace
I'm certainly no expert, but I believe it has to do with the thickness/thinness of the field vs the thickness/thinness of the relief. If the field is much thinner on the Buffalo and the relief is much thicker, then there's your answer! Time to get out the micrometer.
Oh my mistake xD. It’s hard to tell someone’s demeanor online because it’s just text with no facial expressions or tone of voice.
The Gold Buffalo should be more dense since it’s .9999 fine gold and the Gold Eagle is only .9167 fine gold. I just don’t understand how the Buffalo can be denser and larger than an AGE when the AGE has exactly the same amount of gold plus additional copper & silver.