1982 Clipped Cent - Big Clip And Blakesley Effect

So looking at Coins and Canada, the weight should be 2.5 grams.

(2.5-2.3)/2.5 x 100 = 8%

Not bad. Thanks for that Errorcoins222.
Let me know if there are any math errors.

Nice clip! Tough year year to find too. I only have one 1982 1 cent clip and yours is way nicer. Good find.

Another way to estimate would be to calculate the relative area. 19.1 mm is the diameter, so (pi)(r)^2 = (3.14)(19.1/2)^2 = 287 square mm total area of coin.

Area of football-shaped clip is trickier. I just used formula for area of an ellipse. (pi)(A)(B) where A and B are half the height and width of the ellipse. See below.

I drew lines in the image, put a ruler on my computer screen, and measured the relative distances. Overall diameter 9 inches, total height of clip = 1.75 inches, total width of clip = 5.5 inches.

1.75/9*19.1mm = 3.71mm, divided by 2 = 1.85mm = A.
5.5/9*19.1mm = 11.2mm, divided by 2 = 5.6mm = B.

(pi)(1.85)(5.6) = 32.5 square mm, area of clip.
32.5/287 = 11%.

So by my rough calculation, the clip is 11%. The "football" area would be slightly smaller than a true ellipse, so it's probably more like 10%.

Yes, too much time on my hands, but taking a lunch break from mowing the lawn.

Thanks kbbpll, your analysis is excellent,
we used to use trapazoid areas for profile estimates in
oil traps. Volume of an upside down dome. I can see that
this would be good for a perfectly round coin as your arc
suggests, but the 1982 cent was 12 sided and this may
decrease the area clipped under your football shaped area
assumptions.

I agree Okie, the size of the rim is thick
and thus would contribute more to the weight of the
clipped portion, skewing the percentage. The coin has
had 8% of its mass/weight removed prior to striking, what
percentage of the volume of the coin is missing may be a
different number, pushing 11% as kbbpll
suggests. It's a big clip.

Nice large clip SilverDon , I have quite a few clips but I don't have a 1982.