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Maximize volume of ball in wine glass

  1. Oct 13, 2007 #1

    rcgldr

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    Given a wine glass with angle theta, height h, what diameter ball, when placed in the wine glass will displace the most volume?
     
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  3. Oct 13, 2007 #2

    tony873004

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    hmmm. I guess you'd need to know the density of the ball and the density of wine. A ball that sinks would displace more volume than say a ping pong ball which would virtually sit on the top of the liquid, or a wood ball which would also float, but would have a significant amount of its mass below the surface.

    A wine glass usually has a curved side. This sounds more like a Martini glass. Is the shape of this glass like an upside-down pyramid with a circular base?

    Does the ball need to fit completely in the glass, or can the ball sit on the rim of the glass, with its bottom protruding down into the liquid, which would make me want to say "on the glass" rather than "in the glass"?

    If the ball is not allowed to sit on the rim of the glass, and the glass is an upside down circular pyramid, and the ball is assumed to sink, then you just need to compute the largest ball that will fit in the glass.
     
  4. Oct 13, 2007 #3

    Hurkyl

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    Your choice.

    Amongst balls that fit in the glass (depending on just what you mean by that)... why do you think the largest one displaces the most volume?
     
  5. Oct 13, 2007 #4

    tony873004

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    Is my intuition failing me?? :)
     
  6. Oct 13, 2007 #5

    Hurkyl

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    Well, a large ball can't fit in very deeply, so it's not clear that it can displace more wine!
     

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    Last edited: Oct 13, 2007
  7. Oct 13, 2007 #6

    tony873004

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    ok, I was thinking that it would have to completely fit in the glass, slightly smaller than your 2nd image. I guess I wasn't thinking outside the box, er... wine glass.
     
  8. Oct 13, 2007 #7

    Hurkyl

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    That's why I said it depends on what you meant by "in the glass". :smile:
     
  9. Oct 14, 2007 #8

    uart

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    In terms of setting up equations there are really three separate cases (as in three regions for which the volume displaced is governed by three different formulas).

    The first case is where the ball is small enough to be completely submerged (the least interesting case).

    The second case is where the ball is large enough that its top protrudes from the top of the glass but not so large that it can't fit in the glass with its contact points tangential to the glass. This is the most interesting case and probably the region where the maximum lays.

    The third case is where the ball is too large to contact the sides of the glass tangentially and so it more or less just sits on top of the glass.



    BTW. It's an interesting problem Jeff, where did this one come from?
     
    Last edited: Oct 14, 2007
  10. Oct 14, 2007 #9

    Gokul43201

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    Before tackling this puzzle, might I recommend Jeff try a few martinis?
     
  11. Oct 14, 2007 #10

    Hurkyl

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    Has a pretty answer too, at least if I haven't flubbed it up.
     
  12. Oct 14, 2007 #11

    uart

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    A pretty answer! Dam it looks like my solution can't be correct. I found the displaced volume to be a cubic function of "r" in the second region I outlined above. My solution is a pretty messy looking root of a quadratic with lots of cosecants and cotan's. Maybe there are some trig identities in there that I haven't yet seen to simplify it.

    I'm guessing that you haven't posted your solution yet because you don't want to spoil everyone else fun whereas I haven't posted my solution yet because it's so ugly that I'm embarrassed to. :)


    BTW Hurkyl, just a quick numerical check to see if my ugly solution could be correct. What do you get if you take theta = pi/3 (that's the full angle at the base of the glass ok) and height h = 10cm ? I get the radius of 5cm gives maximum displacement for this numerical case. How does that compare with your solution?
     
  13. Oct 14, 2007 #12

    Hurkyl

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    I concur. Try drawing the solution.
     
  14. Oct 14, 2007 #13

    Hurkyl

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    Reviewing my work, I think I made an error and my solution only works for that particular angle. :frown: (but it still looks like you shouldn't have to solve a cubic!)
     
  15. Oct 14, 2007 #14

    uart

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    Oh yeah that was like a special case where exactly half the ball was below the top of the glass and exactly half above. But that is not always the case with my solution. If for example I use theta=2*pi/5 (72 degrees) and again h=10cm then I get r=6.55cm (to 2 dp) to displace maximum volume. I don't notice gemoetrically special about this case when I graph it.

    Edit. I first posted this before I read your last post. No I didn't need to solve a cubic, the function was a cubic and I needed to find the turning points so it was only solving a quadratic. The solution was ugly because the coefficients of the quadratic where messy trig functions of the angle.
     
    Last edited: Oct 14, 2007
  16. Oct 14, 2007 #15

    arildno

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    Well, it's getting late, so I'll just doodle down a few trivialities first, that all have jotted down, I presume:

    Assuming theta is the angle between the axis and the cone side (rather than twice this, i.e, the opening angle), the angle between the tadius from the ball centre (i.e, radius) to the tangential point of contact and the axis must be pi/2-theta, since the radius is orthogonal to the tangent line made by the cone's side.
    Thus, the contact circle has radius Rsin(pi/2-theta)=Rcos(theta), where R is the radius of the ball. I'm sure I need this somewhere, perhaps I'll finish this tomorrow. :smile:
     
  17. Oct 14, 2007 #16

    Hurkyl

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    It factors. And I didn't use the angle until the very end of the problem; the arithmetic is probably a lot simpler that way.
     
  18. Oct 14, 2007 #17

    mathwonk

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    thats a cocktail glass, a (red) wine glass is shaped more like a tulip to maximize surface area wrt volume.
     
  19. Oct 14, 2007 #18

    rcgldr

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    Yes, I meant martini glass.

    My dad, who was a civil engineer (he passed away back in 1990) mentioned this problem to me and my half brothers. I'm not sure of the original source, but my guess is that it was typical of the type of math puzzles that math and engineering students got involved with back in the 1940's.

    One of my half brothers mentioned that a 2 dimensional solution should probably work, but I never got a follow up from him. Assuming he did solve it, I just wanted a "second opinion" to see if the solution is correct. I'm assuming that what is needed is the derivative of the integrals to calculate the displaced volume.
     
    Last edited: Oct 14, 2007
  20. Oct 15, 2007 #19

    uart

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    Ok I'll post my slightly ugly solution. It was even worse but it looks a bit better now that I at least factored out h. I think it's correct, though I'm sure that someone who's a whiz with trig identities could pound it a bit more into shape.


    [tex] \frac{r}{h} = \frac{\cot^{2}(q) - \sqrt{ \cot^{4}(q) - \csc^{4}(q) + 3 \, \csc^{2}(q) - 2 \, \csc(q)}} {2 + \csc^{3}(q) - 3 \, \csc(q)} [/tex]


    PS. [tex]q = \theta /2[/tex] is the angle measured from the axis of symmetry at the bottom of the glass.
     
    Last edited: Oct 15, 2007
  21. Oct 15, 2007 #20

    uart

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    Ok I decided not to be so lazy and have a go at simplifying it. I got it down to,

    [tex] \frac{r}{h} = \frac{\sin(q) - \sin^{2}(q)} {2 \, \sin^{3}(q) - 3 \, \sin^{2}(q) + 1} [/tex]
     
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