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Imagine a vertical pipe (both ends opened) in the water. Drop several ping-pong balls into pipe and cover them with a cylinder. When you have enough balls, the cylinder will float. Now start adding weight into the top of a cylinder, and simultaneously start adding ping-pong balls via the bottom of pipe, to compensate the increasing weight.

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At the some weight (and balls columns height) the topmost row of balls start crunching due the buoyancy pressure. In this question @mms answered than one ping-pong ball can withstand 3 atmosphere pressure.

The question is: how to approximate the maximum height of ball column (or maximum weight is possible float) for the pipe radius R before the topmost row of balls start crunching?

Or, maybe will start crunching not the topmost row, but somewhere else because the sum of buoyancy pressure (from the balls below) and water pressure will be more the 3 atm. How to determine?

Community
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clt60
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  • Isn't crunching the better word than "popping"? The lowest row of balls feels of course the highest "buoyancy pressure" but the higest row will crunch first, because the force between weight and the ball column is tranferred via points, where the balls make contact. This kind of load force has nothing to do with tat 3 bars of hmogenous pressure. Is this homework? – Georg Nov 12 '11 at 16:20
  • Not a homework, just thinking about variable buoyancy pontoon what will have a constant height above water level depending on weight. And are your'e sure about the topmost row? Because at some depth the balls must withstand the pressure of the water PLUS the pressure (forces) what are transferred via the contact points... (edited poping -> crunching - thx for the suggestion) – clt60 Nov 12 '11 at 16:36
  • I'm quite shure. This point-forces are much harder for the ball than homogenous pressure. Its an old trick of wine-farmers to impress laymen: They put an raw egg between the grapes in the hydraulic press. This egg comes out unharmed, despite of several bars in the press. (Of course only successful with old-fashioned presses, in rotavators the chains will smasch the egg, even befor pressure develops) – Georg Nov 12 '11 at 17:03
  • Hm.. so.. probably the question is how much force can withstand the ball by contact-point forces... Is here a difference when the balls will be aligned in row (like vertical "oooooo" - 2 contact points only)? Any idea how to (approx) calculate the crushing force? – clt60 Nov 12 '11 at 17:45
  • just found this ftp://124.42.15.59/ck/2011-05/165/053/860/957/Dynamic%20crushing%20of%20thin-walled%20spheres%20An%20experimental%20study..pdf - i hope it will help me getting the answer ;) – clt60 Nov 12 '11 at 17:55
  • This case is easiest to calculate, but with a closest packing of PP-balls You have some wedge-effect, increasing the force in contact point, at the same time You have more contact points. All this is very complicated and unphysical, engineers love such things :=) BTW, why not using a closed caisson, with just air? Another possibility were balls/bricks of styrofoam. – Georg Nov 12 '11 at 18:13
  • Sure - the "most precise" method will be experimentally test it in real... :) :) – clt60 Nov 12 '11 at 18:29
  • @jm666 I am not able to access the link you provided, can you attach it here, i mean download and then attach it here, so all can see it – Jack Feb 01 '12 at 02:04
  • For the above link - the sever is closed now - i'm not able get the doc too. :( – clt60 Feb 02 '12 at 10:01
  • unfortunately, the doc is now available as paid - e.g. from here http://www.sciencedirect.com/science/article/pii/S0734743X07001832 :( – clt60 Jul 18 '12 at 16:47

1 Answers1

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In order to avoid ambiguity: I'll define the crushing pressure of a layer balls operationally:

  • take a flat*, rigid surface (optionally, you could corrugate it in the manner of a layer of dense-packed balls).
  • collect a single layer of balls (arrange them as they would be in the cylinder) on the rigid surface.
  • Place a flat slab on top of the balls.
  • Add weight until the balls collapse.

Take the net weight above the balls, divide by the area and obtain the crushing pressure.

The top layer of balls will collapse when the weight added to the top of the cylinder produces the crushing pressure defined above.

The logic is that: as long as the top layer of balls remains stationary at the surface of the water, the problem with the floating cylinder is exactly the same as for the fixed, rigid table.

Dave
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  • Partially true. But the it is a function of diameter too. Imagine only an "one-ball" diameter, so balls are as "ooooooo" (vertically) down to water. For enough weight the ball column can reach 30m depth, and this depth the balls crunched by water pressure itself, so, therefore one of the limitation factors are - not possible to have bigger depth as 30m. (upvoting anyway - as only answer - but not fully accept this) – clt60 Jul 18 '12 at 16:39
  • Assume that the overall pressure at 30m depth is 3atm. The "weight pressure" $W/A$ weight on top of the pontoon divided by the area of the pontoon will be 3atm in order to get the lowest layer of balls down to 30m; if it were less the pontoon would rise, if it were more, the pontoon would sink. This is true whether the area $A$ is small or large. Under these conditions we know that the lowest layer of ball(s) would collapse solely due to the water pressure. However, we also know that the top layer of balls is experiencing even more pressure, since all of the balls below press up. – Dave Jul 18 '12 at 22:01