Packing Pennies in a JarDate: 06/08/99 at 09:55:41 From: Greg Gibson Subject: Surface Area Question What is the area of a penny? If a jar has a height of 11 inches and a radius of 7 inches and it is full of pennies evenly to the top, how many pennies can fit in this jar? I don't have anything with which to measure the height of a penny, and I'm really not sure which formula to use once I know the actual dimensions. Can you help? Date: 06/08/99 at 12:14:56 From: Doctor Peterson Subject: Re: Surface Area Question Hi, Greg. What you're really interested in is not the surface area of a penny, but its volume; and not really its volume, but the volume of the space it will take up when it fits in a jar with other pennies, including the wasted air space around it that no penny will fit into. If the pennies are all lying flat so no space is wasted between layers, you could approximate this by picturing each penny as a hexagon, to include the "corner" space between pennies when they form an approximately hexagonal array. The area of this hexagon would be six times that of a triangle with height r (the radius of the penny) and base 2r/sqrt(3), and the volume is that times the height h of the penny: V = 6 * 1/2 * r * 2r/sqrt(3) * h = 2 r^2 h sqrt(3) = 3.464 r^2 h +--ooooooooo--+ /ooo ooo\ /o \ / o\ oo \ / oo o \ / o /o \ / o\ /o \ / o\ +-o-----------X-----------o-+ \o /|\ o/ \o / | \ o/ o / | \ o oo / |r \ oo \o / | \ o/ \ooo | ooo/ +--ooooooooo--+ 2r/sqrt(3) (By the way, this gives a crude approximation of pi.) You'll have to find a way to measure the penny; I would suggest stacking up enough of them to make several inches, and dividing that distance by the number of pennies. If you really want accuracy, you would have to find how many pennies really fit into a given volume experimentally, to take account of how they fit; you could do this by filling a known container and, if you don't want to actually count, weighing them and dividing by the weight of one penny. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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