Proper Dimensions For Bullion Coins (Specifically Palladium)?

[TripleB]

I bought a 1 ounce Palladium bullion coin (Canadian Maple Leaf) because I thought it was cool and may appreciate in value as part of my VP.

I ordered it from one of the few reputable bullion internet dealers and it arrived today. I busted out my period table which lists the density of Palladium at 12 grams per Cubic Centimeter, my kitchen scale, which confirmed the weight at the correct 1 Troy Ounce (= 31 grams), and my caliper.

I took the measurements in my caliper, converted to metric, used the cylinder volume formula, and the calculated density of the coin is around 9.5 grams/cm^3.

That's significantly different from 12. Nickel has a density of 8.9 and Silver 10.5... so in theory if this is a counterfeit coin, it could be an alloy of nickel and silver.

I'm sure this company would make good if it is counterfeit, and they'd probably be amazed anyone would counterfeit a palladium coin (although if you think about it, it's a good one to counterfeit because the dimensions aren't something I could easily find on Google in a few minutes).

I actually found one website that lists the alleged dimensions of this coin, and it's actually bigger in diameter by 4mm and thinner... interestingly enough, with those listed dimensions using my formula, it also gives a density of 9.5 grams/cm^3.

Is there any way I can see a listing of what this coin is supposed to measure? Or what the density is supposed to be? Wikipedia lists Palladium at 12 grams/cm^3 and I can imagine the volume estimation I performed is off slightly due to the face of the coin having markings, but that shouldn't lead to a nearly 30% difference in density.

This is all just academic because as I was measuring my palladium coin, while on my boat, it fell off the side into the water and is lost. However, in case I buy another one, this information might be useful 

[Greg]

I don't fully know the situation but using the cylinder volume formula I think might be bad for this because it is not uniform thickness throughout the material. The edges I believe are higher and which would mean the inner volume is lower to yield closer to the 12 g/cc.

You could always do the archimedes approach, get a cup of water with a known amount of fluid, drop your coin in, and see how high the water level changes (kinda difficult if you're just eyeballing it, unless you do this in a graduated cylinder, etc. something long and thin). That should give you potentially a better indication of volume to hopefully get close to the 12 g/cc you want.


EDIT: Disregard the numbers below and check the further down post for the Canadian Maple coin which is a pure gold coin. The methodology below though still holds true.


Also food for thought, here is the same example for an American Gold Eagle (1 oz.)

From wikipedia:

http://en.wikipedia.org/wiki/American_Gold_Eagle

3.270 diameter (cm)
0.287 thickness (cm)
33.930 mass (g)

Calculated using cylindrical volume --> (diameter/2)^2 * pi * thickness yield:
2.410 volume (cm^3)

Density of Elemental Gold:
http://en.wikipedia.org/wiki/Gold

19.300 density of elemental gold (g/cm^3)

Calculated Density (taking the mass given from wikipedia and dividing by calculated volume) yields:
14.077 density of 1 oz. American Gold Eagle (g/cm^3)

This shows there is a:
5.223 different in density
27.1% % difference in density

If everything else was equal, 12 g/cm^3 for your 1 oz. palladium at this difference in density would yield:
12*(1-.271) = 8.748 g/cm^3 <-- which is semi-close to your 9.5g/cm^3 you calculated. This is a simplification however since the coins don't have the same designs so the % difference in density would be different from the 27.1% for the American Gold Eagle.

[TripleB]

I don't fully know the situation but using the cylinder volume formula I think might be bad for this because it is not uniform thickness throughout the material. The edges I believe are higher and which would mean the inner volume is lower to yield closer to the 12 g/cc.

You could always do the archimedes approach, get a cup of water with a known amount of fluid, drop your coin in, and see how high the water level changes (kinda difficult if you're just eyeballing it, unless you do this in a graduated cylinder, etc. something long and thin). That should give you potentially a better indication of volume to hopefully get close to the 12 g/cc you want.

That's a Mensa-level suggestion, bro 

I'll do this (which is your suggestion, just using household stuff) - I'll find a clear container that I can fill with water and I'll weigh it. I'll put the coin in and I'll mark the second level.

I'll take the coin out and fill it with water until it hits the second level. Thus, I added an equivalent volume of water to the container as the coin.

I'll weigh it and find how much water = the volume of the coin. I know the density of water so I can find the volume of this marginal water based on the marginal weight. That's the volume of the coin.

And 99 out of 100 high school science students wouldn't be able to improvise this, and instead they are forced to memorize vocabulary words and in some case, even memorize the periodic table. Whereas this is actually a worthwhile critical thinking exercise and experiment.

[Greg]

I modified my original post which might help to add more confidence that your 9.5 g/cm^3 is actually a false assumption and that it is probably a real coin.

And yes, "Mensa" level. What a racket. Who is the actual smart one then? The person you paid to be told he was smart, or the organization being able to profit from telling a person they are smart? I'm not betting on me for this one.

I also agree about the 99 in 100 students. What you wrote is actually useful information. Critical thinking is one of the best things you can learn (hopefully) from education. Remembering the number of protons in calcium does me diddly.

[TripleB]

Greg:

An American Gold Eagle is only 90% gold with the other 10% being lighter elements. The Palladium Coin is 99.995% Palladium.

I'll run the numbers on the Canadian Maple Leaf Gold and see what the delta is and report back.

[Greg]

Greg:

An American Gold Eagle is only 90% gold with the other 10% being lighter elements. The Palladium Coin is 99.9% Palladium.

Whoops, I'm an idiot. I'll try it again with a Canadian Maple. Hopefully my point still holds though.

[TripleB]

Greg:

An American Gold Eagle is only 90% gold with the other 10% being lighter elements. The Palladium Coin is 99.9% Palladium.

Whoops, I'm an idiot. I'll try it again with a Canadian Maple. Hopefully my point still holds though.

That's what happens when you stop paying your Mensa dues 

[Greg]

Greg:

An American Gold Eagle is only 90% gold with the other 10% being lighter elements. The Palladium Coin is 99.9% Palladium.

Whoops, I'm an idiot. I'll try it again with a Canadian Maple. Hopefully my point still holds though.

That's what happens when you stop paying your Mensa dues 

For a 1 oz. Gold Canadian Maple

3.000 diameter (cm)
0.279 thickness (cm)
1.972 volume (cm^3)
31.103 mass (g)
15.771 density of 1 oz. Canadian Gold Maple (g/cm^3)
19.300 density of elemental gold (g/cm^3)
3.529 difference in density (grams)
18.3% % difference in density

There's still then a 18% difference in density from elemental vs. the calculated volume method.

And Mensa is more powerful than I imagined or cared. They can take my intelligence if I don't register with them and they can track me. Sounds familiar to the gun debate hah.

[TripleB]

Thanks, Greg! Looks like I have little to fear after all. I wasn't really worried, since it's a reputable dealer, just more curious.

[Greg]

Thanks, Greg! Looks like I have little to fear after all. I wasn't really worried, since it's a reputable dealer, just more curious.

You're certainly welcome. Always glad to stretch my engineering muscles again, I don't get that at work as much as I'd like hah.