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Solve for c/a ratio

  1. Sep 24, 2006 #1
    I'm trying to solve for the ideal c/a ratio of hcp packing. why doesn't c/2 =a/2, looks like they pass through the same amoutn of radius's so I'm not sure the geometry to use. Used this pic http://www.engr.ku.edu/~rhale/ae510/lecture2/sld013.htm [Broken] but it is more confusing to me how 1.67 comes out. Help please?
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Sep 24, 2006 #2


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    Look at the middle picture in the link you provided.

    In the c-direction, there is an extra layer of atoms (triangle) between the top and bottom layers (hexagonal). The a-length is the side of the hexagon, but also the distance among adjacent atoms in the 'basal' plane. The dimension 'a' is in the closest-packing direction. The packing is less efficient in the c-direction.
  4. Sep 24, 2006 #3
    I understand what you're saying about the picutre now, but I don't understand what you're using to compare C and A. I know where a is and i know where c is but i don't know the other length of triangle built between them, so i'ts unclear to me as to what the link is between thier length
  5. Sep 24, 2006 #4


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    It's not obvious.

    The trick is in realizing that the middle plane of atoms occupy positions directly above the centroids of the triangles in the base plane. This follows directly from a symmetry argument.

    1. Each basal plane has nearest neighbor atoms making equilateral triangles. So, a=2R (where R is the sphere radius).

    2. Each atom at height c/2 above the basal plane is positioned directly above the centroid of the triangles in the base plane. For an equilatreral triangle, the distance from a vertex to the centroid is two-thirds the length of the median, and is hence [itex](2/3)*(a\sqrt{3}/2) = a/\sqrt{3} [/itex].

    3. Each atom in the base plane has a nearest neighbor in this middle plane. So, the distance from the corner atom in the base plane to the nearby atom in the mid-plane is 2R.

    4. This distance can also be calculated from Pythagoras, giving:
    [tex]4R^2 = a^2 = (a/\sqrt{3})^2 + (c/2)^2[/tex]

    That should get you home.
  6. Sep 24, 2006 #5
    thank you I was feeling like an idiot that I couldn't girue this out, I'm glad it doesn't appear to really be a trivial process. I appreciate the help,
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