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Volume problem - Closest packing

  1. Jun 16, 2005 #1
    Ok I am trying to show the packing efficiency for 3 different types of packing.

    1) Hexagonal closest packing (efficiency = 74%)
    2) Cubic closest packing - (efficiency = 74%)
    3) Body-centered - (efficiency = 68%)

    I know the efficiency values because they are in my textbook and all over the internet. I need to come up with these numbers myself.

    efficiency = Volume of the spheres/Volume of a box around the spheres.

    the spheres in my model have a radius of 3cm, so volume of the spheres is easy... 4/3(pie)r^3 x (number of spheres)


    I'm trying to figure out a way to find the volume of a box around Hexagonal closest packing and cubic closest packing.

    Hexagonal closest packing: 13 balls in total in 3 layers. (see pictures below)

    bottom layer: 3 spheres
    middle layer: 7 spheres
    top layer 3 spheres

    Cubic closest packing : 13 balls in total in 3 layers

    bottom layer: 4 balls
    middle layer: 5 balls
    top layer: 4 balls

    I know that the volume of both must be the same since they each have 13 balls and each have the same packing efficiency.

    How do I find the volume of a box around these spheres?


    I've been trying to figure this out for hours so any help at all is appreciated........and sorry for the length of the question :redface:
  2. jcsd
  3. Jun 16, 2005 #2
    I can't think of a way to do this from scrath, but if you know the effeciency, you can solve for the volumes. That's kind of cheating though.

    [tex]\frac{\mbox{Volume of Spheres}}{\mbox{x}} = \mbox{effiency}[/tex]

    Solve for x.

    I'm curious to the full answer of this problem as well.
  4. Jun 16, 2005 #3


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    In most cases, a box around whole spheres is not the way to look at it. You will probably do better assuming the corners of the volume you select are alighned with the centers of some spheres and then count the number of whole spheres and fractional spheres contained within that volume. Simple cubic can be done either way because there is no interlacing of layers. That is not the case for close-packed configurations.
  5. Jun 16, 2005 #4
    ahh ok, thanks for the quick response ... I was trying to come up with something that would give me a precise answer.. but i guess thats not the most efficient way to solve it. :smile:

    So basically, just pick the lines of an imaginary cube, calculate the volume..

    Count the amount of spheres/partial spheres within that cube, multiply that by the volume of a sphere.

    then doctor up your count to make the efficiency close to 74% :smile:

  6. Jun 16, 2005 #5


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    You wont even have to doctor it up :smile: Your second configuration also goes by the name face-centered-cubic.
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