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Thread: Basic Maths Question

  1. #1
    Cricket Web: All-Time Legend Matteh's Avatar
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    Basic Maths Question

    Sure someone like Manan/Pickup will be able to have a decent stab.

    If i have a hundred small balls and i want to join them together to make a large ball, how many balls would you put on each layer of the ball to make it actually ball shaped?

    All i'm getting with my efforts are pyramids, which aren't strictly balls.
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    ret
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    hmm, interesting Q

    a ball is a sphere, therefore it could hv a ball in the middle and that ball is surrounded by other balls .... the middle layer would hv maximum balls, say n number of balls and then the next layers on its two sides would be n-1, something like the equation below:

    ...., n-2, n-1, n, n-1, n-2, ....

    so when you take, n = 10 ... the equation gets solved
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    International Regular Josh's Avatar
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    My 13 years of maths taught me one thing

    The word "Pythagoras"

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    State Vice-Captain nikhil1772's Avatar
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    Dude,that's anything but a basic maths question... And I am not quite sure I understood the problem correctly,and if I think what I think it means...then I am as useless to you as the problem itself.
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    If you need to just make a sphere you if you can attach each ball with three others to form triangles you eventually make a hollow sphere-like shape.

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    Quote Originally Posted by Josh View Post
    My 13 years of maths taught me one thing

    The word "Pythagoras"
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    Quote Originally Posted by Josh View Post
    My 13 years of maths taught me one thing

    The word "Pythagoras"
    Ancient Babylonians ftw.
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    Cricket Web Staff Member Burgey's Avatar
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    Look, I haven't done maths since 1987 when I left school, though now I'm doing times tables with my 6 yo twins.
    But seriously, if you're not into your maths as part of your job or you have a pretty big interest in it, how is the question posed in this thread a "basic" one? Seems damn hard to me, frankly.
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    Cricket Web Staff Member / Global Moderator Neil Pickup's Avatar
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    Quote Originally Posted by Burgey View Post
    Look, I haven't done maths since 1987 when I left school, though now I'm doing times tables with my 6 yo twins.
    But seriously, if you're not into your maths as part of your job or you have a pretty big interest in it, how is the question posed in this thread a "basic" one? Seems damn hard to me, frankly.
    It's not basic if you bloody teach it... why else do you think I haven't answered?

    Going to sit down and have a (google-aided) crack now, and see what I get.
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    ret
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    the diameter of a circle is the longest chord in a circle, so this would be like making two opposite pyramids from a single base, so let's say the diameter has 10 balls there the next chords on either side of the diameter would have 9 balls and so on like shown below

    1+2+3+4+5+6+7+8+9+10+9+8+7+6+5+4+3+2+1 = 100 balls


    from the eq that i gave in my earlier post

    ...., n-2, n-1, n, n-1, n-2, ....; taking n =10

  14. #14
    Cricket Web Staff Member / Global Moderator Neil Pickup's Avatar
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    Having researched it, no it isn't basic. It's an application of the Kepler Conjecture.

    Basically, good layers to use are:
    (hexagonally): 1 [blue], 7 [green], 19 [beige], 37 [pink]
    (broadly triangular): 3 [olive], 12 [red], 27 [turquoise]



    This is two sequences: one with the formula 3n(n+1)+1, the other more simply 3n.

    With these formulae you can extend the sequences thus:
    1, 7, 19, 37, 61, 91, 127
    3, 12, 27, 48, 75, 108, 147

    This provides a fairly exhaustive list of numbers of balls in the layers to construct your sphere, maintaining of course a symmetrical layout.

    Therefore, your smallest sphere would have one layer of one sphere.
    The second would be a 1 - 3 - 1 = making five spheres.
    The next would contain 1 - 3 - 7 - 3 - 1 = making fifteen.

    This pattern continues as follows: 1, 5, 15, 34, 65, 111, 175, 260, 369, 505, 671, 870, 1105, 1379... and loads of people have found this sequence before, in loads of other things, and it turns out be n(n+1), also known as Sequence A006003.

    I actually enjoyed finding all that out. Is that wrong?

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    Cricket Web Staff Member / Global Moderator Neil Pickup's Avatar
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    Quote Originally Posted by ret View Post
    the diameter of a circle is the longest chord in a circle, so this would be like making two opposite pyramids from a single base, so let's say the diameter has 10 balls there the next chords on either side of the diameter would have 9 balls and so on like shown below

    1+2+3+4+5+6+7+8+9+10+9+8+7+6+5+4+3+2+1 = 100 balls


    from the eq that i gave in my earlier post

    ...., n-2, n-1, n, n-1, n-2, ....; taking n =10
    No, it wouldn't. That would give you a slice through a sphere.

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