Parking lot

CaptHenway

Just parking this here in case I need the math later.

Just as an aside, I did a little "Fun with Math!" last night to see how much 50 Pesos would have to be rolled down to produce a correct-weight $20 "planchet."

The diameter of a 50 pesos is 37.08 mm. Using the "pi radius squared" formula the area is 1,079.86 sq. mm.
The diameter of a $20 is 34.29. Using the formula the area is 923.47 sq. mm, approx. 85.5% the area of the 50 pesos.

If the 50 pesos coin were perfectly flat (but still unrolled) and you punched a 34.29 mm "planchet" out of it, it would weigh approx. 35.63 grams (41.6667 grams x 0.855). A $20 weighs 33.436 grams, so you would have to reduce the thickness of the 50 pesos by about 6.2% to get a $20 "planchet" of the right weight.

But of course a 50 pesos is not flat, and has raised designs concentrated near the centers. Punch a "planchet" out of it and it would weigh more than 35.64 grams. How much, I do not know, but let's say that you have to reduce the thickness of your 50 pesos by about 7% before you punch your "planchet" out of it. Then you could weigh it and adjust the weight with a file if necessary.

I do not know if rolling a 50 pesos down by 7% would obliterate the design of the 50 pesos enough to keep it from showing through on the overstrike. Obviously the raised designs would be compressed by more than 7%, but I think they would still be clearly visible, and would expect some of them to survive the overstriking.

Omega $20's have been well studied. I have never seen, or heard of, an understrike on an Omega $20.

TD