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Will the pressures even out because the elasticity

  1. Jan 16, 2009 #1
    If I have a soft base (mud or sand) and place 4 rocket balloons (long narrow ones used for twisting into shapes, animals, etc.) all inflated to the same pressure, but each balloon of a different diameter (1", 2", 3" and 4") and place platform (a thin, but rigid sheet of plastic) on top of the balloons the platform will be at an angle; but if I then place a weight (10 pounds) on top of the platform it will level out. What I want to know: is there a different load value on the soft base beneath each of the balloons? Let's say: 20 psi, 40 psi, etc. or will the pressures even out because the elasticity of the balloon or its conformity to the shape of the soft base spread the load in some other way?
     
  2. jcsd
  3. Jan 17, 2009 #2

    nvn

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    For the same initial internal air pressure and wall thickness, larger cylinders are under higher tension but have a higher thinness ratio. The higher thinness ratio is probably the dominant effect here. Cylinders having a higher thinness ratio (with the same initial internal air pressure) ovalize more easily under an external applied load. The initial external applied force is approximately the same on each of your cylinders [1]. The larger cylinders ovalize more, until their stiffness approximately equals the stiffness of the smaller balloons. This greater ovalization creates a larger surface contact area. The final external load resisted by each balloon is roughly the same for each balloon [1], unless the ovalized balloons are no longer widely spaced. Hence, the final force on the soft base is roughly the same for each balloon. The same force applied to a larger contact area (underneath the larger balloons) causes a lower bearing stress (pressure) on the soft base. A lower bearing pressure on the soft base implies that the final air pressure inside the larger balloons is lower than in the smaller balloons. A lower bearing pressure on the soft base underneath the larger balloons might also cause less imprintment into the soft base.

    Therefore, I currently think the final force on the soft base is roughly the same for each balloon [1], in which case the final air pressure inside each balloon would be inversely linearly proportional to the final contact area of each balloon on the soft base.

    [1] Note that my above answer assumes that either the weight on the rigid plastic sheet is uniformly distributed (not a concentrated weight), or the rigid plastic sheet is a theoretical rigid-body plate. If this is not the case, the answer might depend on the dimensions and all material properties, which you might want to post, if you wish.
     
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