Oddball SBA Dollar- Wrong Planchet?

And because I was push on, I have the DATA:

Strip Width Thickness
50 Cent . . 12 9/16-inch (319.08mm) .0685-inch (1.740mm) also 1 $
25 Cent . . 12 13/16-inch (325.43mm) .0545-inch (1.384mm)
10 Cent . . 12 11/16-inch (322.26mm) .041 -inch (1.041mm)
The strip must be manufactured within a permitted tolerance of .0015-inch (.038mm) thickness for clad strip of any denomination.

What you say about SAM?

PS: I hope the Mint will not cut my access to the data for this to publish because I was push at extreme. by this guy SAM.

Jon is one of the best of bests.

Interesting thread and I'm enjoying the lesson.

This coin is 27% underweight. (5.92g/8.1g)
It sure seems it has a weak strike near the rims from not having enough metal to fill the devices.
I don't see any of the classic signs of damage from acid, grinding, etc.

How can this not be an error?

Edit:
I like Brandmeisters evaluation and the reverse does have a copper look to it.

met, looks like you have a very cool find, one few of us make first hand

Silviosi, you can almost always get an engineer to pontificate about math, so let me begin with the traditional:



I just used simple algebra to reach my first estimate. The Mint published specs are a total composition of 8.33% nickel, balance copper. The clad layer spec is 25% nickel, 75% copper. The core is 100% copper. The only way to make that math work is for the clad layers to be one-third of the coin. Could the number published by their marketing and website folks be imprecise? Sure, but I wouldn't bet on it, because some random coin enthusiast would be emailing the U.S. Mint like 10 minutes after they published that webpage, asking about the numbers.

However, that estimate is based on simple ratios. It ignores a lot of factors, for example, the difference in density between nickel and copper. If we want to back into a mass estimate by volume and density:

Thickness t, diameter 2.65cm, therefore volume of t*pi*(d/2)^2 = t*5.51 = V. Density of .25(8.9 g.cm^-3 Ni)+.75(8.96 g.cm^-3 Cu) = 8.945 gram/cm^3. So the mass of a clad layer by V(t)*D = t*49.29, when Thickness t is expressed in centimeters (not mm).

The published thickness of a SBA $1 coin is 2mm (.2cm). So by my rough cut math, that makes a clad layer (.2)/6 thick, or 0.0333 cm. By the equation above, that gives a mass of 1.643g. Clearly a different answer than 8.1/6=1.35g. If you would like to give a different value for the thickness of a clad layer, that's fine with me. I'm sure the actual equations used by mechanical engineers at the U.S. Mint are way, way more detailed than my back-of-the-napkin math.

Neither of those estimates gets to the 2.1g mass lost by this coin. But as someone who has worked in manufacturing, I will say that once you are so far out of materials specs, you can get wildly unpredictable results. That metal went through a half dozen different machines that performed blanking, upsetting, annealing, cleaning, surface prep, striking. We don't really even know if the sheet metal roll was constructed correctly to start with. Measuring it against the strict product tolerances of a properly manufactured coin might be a doomed endeavor.

Spence, the idea of stapling that coin into a 2x2 sends a shiver down my spine. Even in AU grade, a missing-clad SBA $1 coin retails for about $200-300. Raw and ungraded, it is probably still a $100+ coin. Risking a staple scratch seems, well, risky.

https://sullivannumismatics.com/sea...ing+clad+SBA

https://www.hyattcoins.com/1979-d-m...tails-1.html

So, we have a verdict here by some very well educated people that study errors like this one.

So, I'll have to say that this is a great find here. Congrats.