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Thread: Trickier still

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    Trickier still

    This is something I'm actually working to solve at the moment, similar to the <!post=close packing of the coins,407801>close packing of the coins<!/post> but in 3D.
    There's a big sphere of radius R to be covered with smaller touching spheres of radius r. <IMG SRC=http://members.aol.com/browrob549/emo/char023.gif>

    What value of r is appropriate to obtain a complete coverage of the large sphere, using such a close packing arrangement of small spheres? <IMG SRC=http://members.aol.com/browrob549/emo/bounce014.gif>

    The idea is to get maximum coverage or masking of the surface area of the big sphere, by the smaller ones.
    How much of the area is masked?
    How could additional "masking" be achieved? <IMG SRC=http://members.aol.com/browrob549/emo/talk002.gif>

    Alan

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    Re: Trickier still

    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">If the outer spheres are to be touching then you have only 3 arrangements.

    3 spheres touch with the centers in an equilateral triangular shape with sides of length 2r. These can be put together into 3 triangular faced "platonic solids":
    Tetrahedron (4 faces, 3 faces meet at each intersection)
    Octahedron (8 faces, 4 faces meet at each intersection)
    Icosahedron (20 faces, 5 faces meet at each intersection)
    [Note: 6 faces meeting at each intersection is a hexagon and is planar so it can not be a "solid" as it is 2-d not 3-d.]
    The tetrahedron can hold a sphere of radius = 0.155r [distance from center of outer sphere to center of inner sphere = r/cos30, so R = r/cos30-r]so r = 6.46R

    The octahedron from outer center to inner center = 0.414r [R = r/cos45-r]so r = 2.414R

    The icosahedron from outer center to inner center is R=0.90r [Icosahedron vertices form 3 "golden rectangles" = 1:1.618] so r = 1.11R

    It gets more complicated if you were to allow for the outer spheres to not touch completely or not touch the inner sphere</span hide>

    Steve

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    Re: Trickier still

    I was looking some more at lunch:

    You could cover it in the shape of a snub dodechahedron, an "Archimedean solid" and then you would have 60 smaller outer spheres at all the vertices. 12 more could be added at the center of each of the 12 pentagon faces , though these would not touch the "inner sphere" (you could have 12 "even smaller spheres" to fill in this area if desired, and have them touch the inner sphere: 1 sphere of radius R, 60 radius r, and 12 radius q (where q<r)).

    I have not looked at at the of the "radius calcs" for this "solution"

    Steve

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    Re: Trickier still

    Hi Steve

    These are very interesting ideas, particularly the snub dodecahedron, which I'd sort of "dreamed up" myself during my musings. The idea of using smaller spheres to fill interstices was what I was hinting at in the last part of the puzzle. I know from working for a place producing lead powders <img src=/S/sick.gif border=0 alt=sick width=15 height=15> that a major process challenge was to obtain a mix of three sizes, in the right proportions. When the powder mix was shaken appropriately, the spaces between the big particles were filled by the next biggest, and the even smaller spaces remaining filled by the tiny stuff. This meant more weight per drum.

    The other side of this problem is maximising the masking effect, which might be even better achieved by relaxing the "touching" requirement, and using a variety of littler fleas on the big fleas.

    Alan

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    Re: Trickier still

    You could probably start with any of the shapes, I chose this as an example since it had all the "triangle faces" which is how the spheres seem to fit the best.

    I guess if you got your "hands' around the calcs for this, starting with an inner sphere of R, you can determine the size of the snub dodecahedron and then the size of the "larger" outer spheres(r1), you could then calc the size of th next smaller spheres to cover the pentagons(r2). You could then look at each of the "spaces" among the triangles for radius r3, and then among the triange/pentagon spaces for radius r4, and then continually look at smaller and smaller ones. I am sure some pattern would emerge ...

    I am not sure how many different sizes are required. This shape would not expand out as much as a solid that had more different "face polygons" but those could also be done.

    Steve

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