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Math Help - kissing spheres in a cylinder

  1. #1
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    Problem on geometry. Solution needed!

    Look at attachment.
    It images a closed cylinder, containing four balls according to the circumstances that the four balls are equal and each ball contact with the above side, below side, side of wall and two of the neighbour balls. How much is the cugabe of the four balls.
    Attached Thumbnails Attached Thumbnails kissing spheres in a cylinder-cylinder.jpg  
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  2. #2
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    Cugabe?
    What is that in English?

    Looks like your problem here is similar to a previous problem that was asked in this forum a few days ago---about a golf ball inside a box.
    Yours is about 4 equal spheres inside a cylinder.
    Is "cugabe" then about the percentage of the volumes of the 4 spheres to the volume of the cylinder? Is "cugabe" about occupancy?
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  3. #3
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    Sorry, I am not very good at mathematical vocabulary.
    Therefore I thought space taken up by something - volume.
    So I would like to calculate the volume of a spehere (actually there're 4 spheres).
    Last edited by totalnewbie; July 27th 2005 at 01:58 PM.
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  4. #4
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    Sphere, of radius r.
    Volume = (4/3)*pi*r^3
    That formula is shown everywhere in books, internet, etc.

    I am sure the volume of the 4 spheres independently or without relation to the cylinder is not what you want to find with your posted problem here. Because at any positions or locations, touching or not touching, the volume of 4 equal spheres of radius r is always 4*(4/3)pi*r^3 = (16pi/3)*r^3.

    Did you copy the problem from somewhere? What is the whole or complete problem? The answer should have relation to the enclosing circular cylinder.
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  5. #5
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    Yes, you are right, the answer should have relation to the enclosing circular cylinder. In a word, How much of the volume of the cylinder do the spheres fill ?
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  6. #6
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    There you go. It is about occupancy then. It is about the percentage on the volume of the circulat cylinder that the 4 equal spheres occupy or fill up.

    4 equal spheres of radius r.
    >>>Their centers are in a horizontal plane because their tops and bottoms are controlled by the horizontal top and bottom circular sides or bases of the enclosing cylinder.
    >>>Volume = (16pi/3)r^3


    Circular cylinder of radius R.
    >>>Height = diameter of spheres = 2r
    >>>Volume = [pi*R^2]*2r = 2pi(r)(R^2)

    We will find the relation between r and R. We will express one into the other, say R in terms of r, so that we can cancel them out from the ratio or comparison of the volumes.

    The figure shown as posted is the horizontal plane through the centers of the 4 spheres. If we connect the 4 centers with straight lines, we will have a square with its two diagonals.
    The side of this square is 2r.
    Each of the diagonals is 2r plus the distance between the two corresponding smaller circles.
    We divide this square into 4 equal smaller squares. Each of these 4 smaller squares encloses a quarter of the smaller circles.

    In any of these smaller squares:
    >>>side = r
    >>>diagonal = r+x
    where x is the distance from smaller circle to center of the whole figure. (Actually, x is the radius of the imaginary circle that is tangent to all of the 4 smaller circles.)
    By Pythagorean Theorem,
    r^2 +r^2 = (r+x)^2
    2(r^2) = (r+x)^2
    Take the square roots of both sides,
    [sqrt(2)]r = r+x
    (1.4142)r = r +x
    x = (1.4142)r -r
    x = (0.4142)r ------***

    Back to the whole figure.
    If we extend a diagonal of the whole or bigger square, on both ends so that it will reach from one side of the biggest circle---the cirle of the cylinder--- to the opposite side, then this extended line is the diameter of the circular cylinder. It is the 2R. From this, we see that
    2R = r +r +x +x + r +r
    2R = 4r +2x
    R = 2r +x
    Plugging in the value of x,
    R = 2r +(0.4142)r
    R = (2.4142)r ----***

    Hence, the volume of the cylinder in terms of r is
    = 2pi(r)(R^2)
    = 2pi(r)(2.4142 r)^2
    = 2pi(r)(5.8284 r^2)
    = 11.6568 pi r^3

    The volume of the 4 spheres
    = (16pi/3)r^3
    = 5.3333 pi r^3

    (Volume of 4 spheres) / (volume of cylinder)
    = (5.3333 pi r^3) / (11.6568 pi r^3)
    = 5.3333 / 11.6568
    = 0.4575
    In percentage,
    = 45.75 %

    Therefore, the volume of the 4 equal spheres occupies 45.75 percent---close to half--- of the volume of the cylinder. ------answer.
    Last edited by ticbol; July 22nd 2005 at 02:17 PM.
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  7. #7
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    Than you very much for the help!
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  8. #8
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    ticbol, I found shorter successful way.
    Maybe someone want to see this way. So I write it down.

    R - sphere radius
    r - cylinder radius
    h - cylinder height

    First of all you should do the illustration (drawing).
    Next, mark the centres of the 4 spheres and link them with line. So you get a square. d= The diagonal of the square.

    Volume(cylinder)=pi*r^2*h
    Volume(4 spheres)=16*pi*R^3/3
    h=2*R
    d=2*R*square_root(2)
    Looking at the drawing you see that 2r-2R=d -> r=R*[1+square_root(2)]

    And the final thing to do is divide Volume(4 spheres) by Volume(Cylinder) and you get the answer 16/6(1+square_root(2))^2. It is approximately 0.4575

    The same answer but shorter way.
    Last edited by totalnewbie; July 23rd 2005 at 02:18 PM.
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  9. #9
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    Yes, that is right. Good work!
    Although, "r=[1+square_root(2)]^2" is not correct.
    Your 2r -2R = d is correct.

    (I found that one too after I posted my answer but I did not show it anymore for fear of overkill. No need to go through the "x" route. Well, that is Math, many ways to only one answer.)
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  10. #10
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    Quote Originally Posted by ticbol
    Although, "r=[1+square_root(2)]^2" is not correct.
    Yes, I did not notice that before (mistake of writing).
    The right is the following: r=R*[1+square_root(2)]
    Thus I made the correction on my previous post.
    Last edited by totalnewbie; July 23rd 2005 at 02:21 PM.
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