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BALL FOUNTAIN-need help!

  1. Nov 9, 2006 #1
    BALL FOUNTAIN--need help!

    Hi, my partner and I are doing a senior project and we are creating a ball fountain. A ball fountain is simply a perfect sphere sitting on a ring thats perfectly matched to the sphere with water shooting up from the bottom and the sphere can begin to spin. Now my partner and I have the sphere and the perfectly matched ring, but we are not sure how much water pressure we need from a water pump. water pumps are very expensive so we want to make sure that we get it right the first time. The ball weighs 11.5 lbs. and we need to know how much pressure and flow we need to support it to "hover" over the ring. If you need me to be more specific with this at all please let me know. your replys are greatly appreciated.

    here is a link to get an idea of what a ball fountain looks like.

    http://www.dmg-raj.com/images/Ball Fountain - 2.jpg

    Thanks again
  2. jcsd
  3. Nov 9, 2006 #2
    You can try to find a windshield washer fluid pump for a car somewhere....
  4. Nov 9, 2006 #3
    thanks for the idea but I am really seeking more information, but thanks though
  5. Nov 9, 2006 #4
    krickon, can you give any more information about obtaining the matched sphere and ring? I quite like those "ball fountain" sculpture things, and I'd like to put together the biggest one I possibly can (for example my national science centre, Questacon, has one in a courtyard with probably at least a 60cm or so diameter marble ball.. Ideally I'd like to go larger).

    As for your question, I'd roughly estimate the answer this way: divide the mass of the ball by the cross-sectional area of the ring (to get the pressure exerted by the ball's weight) and perhaps add the pressure required to maintain the height of water (maybe gravity multiplied by water density and the height difference between the tops of the ring and reservoir). Flow shouldn't be an issue, especially if the shapes are well matched.

    You might want to just try the windshield washer pump, since you can likely pick some up from a car wreckers for a few dollars.
    Last edited: Nov 10, 2006
  6. Nov 9, 2006 #5

    Chi Meson

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    Give us the diameter of the ring. The minimum water pressure inside the ring must be equal to the weight of the ball divided by the area of the ring.

    And, you'll have to go with a higher pressure pump and adjust the pressure with a valve, since many other factors will contribute to the actual pressure at the ring. I'm guess that a 1 psi pump will be adequate.
  7. Nov 10, 2006 #6


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    Since this is a physics rather that mathematics question- and is school work, I am moving this thread to the "Physics-Homework" forum
  8. Nov 10, 2006 #7


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    You could use a raised bucket of water with a pipe to experiment and find the required water pressure (although there is a difference between static and dynamic pressure). By recording the overflow from the ball fountain as a function of time you should get a good idea of the required flow rate.
  9. Nov 10, 2006 #8
    I think the concept for this kind of fountain is that it will work in the limit of zero flow, unlike levitating a bernoulli ball. Analogously, while a rocket exhaust must maintain a certain flow rate, a hovercraft can in principle levitate above the ground with zero air flow (provided it could also seal to the ground well enough to maintain pressure). Hence, unless the ball fits the ring poorly, the required flow should be produced very trivially by any pump that can provide enough pressure.

    Unless I'm wrong about reducing the problem to hydrostatics, I don't see many other factors contributing to the pressure.. though I certainly agree about err-ing on the side of the better pump.
  10. Nov 10, 2006 #9

    Chi Meson

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    I was considering the length and size of the pipe used to carry the water, height difference between pump and output, as well as the connecting fixtures (any right angles or whatnot), plus any other "unknown unknowns" (Rummy was great wasn't he?).

    Still I kinda think now that all these considerations are not even necessary. It doesn't seem that the pressure will be the issue, so I think that even the smallest pump do (even the car windshield fluid pump as already suggested). A simple adjustable valve will be necessary to get the desired flow rate.
  11. Nov 15, 2006 #10
    ok so the width of the bottom of the ring is 5.5 inches. and the top of the ring is 7.5 inches. I need to know the psi and flow. the ball weighs roughly 11.5 lbs. does anyone know the answer to this problem? Thank you.
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