Quote:
Originally posted by FotoPlasma
(1/4in)/2 = 1/8in
pi(1/8in)^2 = 0.049087385212341in^2
0.049087385212341in^2 * .125in = 0.0061359231515426in^3
1/0.0061359231515426in^3 = 162.9746617261 quarter inch holes per in^3
(I love my TI-89)
solve(((162.9746617261holes)/(0.098lb/in^3))=((x)/(1)),x)
According to my calculations, 1663.0067523071 holes per lb of aluminum...
If any of that looks wrong, please alert me to it immediately...
I hope that's right... it's kinda late...
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Two things, I think.
First, wouldn't it have been easier to find the weight of one 1/4" dia. x 1/8" thick piece of aluminum, and divide 1 lb. by that number? It seems like there's less math involved, and it omits having to make assumptions about the number of holes per in^3.
Therein, I think, lies the second problem. First, the number of holes should be measure in square inches, rather than cubic inches, because we're assuming that each hole goes entirely through the thickness of the aluminum. Second, while the combined area of 162... holes may approximate 1 in^2, the reality is that it may be a better approximation to treat each hole as a slightly larger square, thus insuring that you can properly punch the specified number of holes, and also providing you with a chance at having a piece of aluminum left when you're done.
If that doesn't make sense, I can try to clarify further
EDIT: Oh oh oh. I see what you did. It was just backward from what I did, kinda, but it's the same in the end.
Duh. Still, for what it's worth, assuming a 1/16" sliver of clearance on each side of the holes, you can fit 7 hole per square inch. 1600+ holes are required to lose one pound. That'd mean you'd have to continuously and uninterruptedly hole up 19.41 square feet of aluminum to lose 1 pound