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Thread: What happens when you have equal force on opposing pistons with a 2:1 gear train?

  1. #21
    Principle Engineer
    Join Date
    Mar 2011
    Posts
    175
    Okay, one last try considering the fluid entrained.

    If you took physics you would know that the net force acting on each piston is

    F=(Pa-Pf)*A and the net moment on the pinion is

    M=F*r-F*2r
    Pa atmospheric pressure
    Pf= fluid pressure

    The result of this imbalance is to move the cylinders in such a direction to make M=0.The only way for M= 0 is that F=0 and Pa=Pf
    which means that the motion would cause the fluid pressure, Pf to be equal to Pa. So initially Pa=Pf and now Pa starts to rise.
    As Pa rises the momentary imbalance in moment will cause the piston having the greater displacement to move a distance 2x , reducing the volume 2x*A while the the other piston would move x and increase the volume by x*A or a net DECREASE in volume of x*A and the system resulting reduction in fluid volume would cause Pf to increase to the new value of Pa.( which also supports the thermodynamic of increasing Pf as the volume decreases).

    By the OP mechanism, he would have the volume increase but the Thermodynamics would be violated since a fluid cannot increase in pressure with an increase in volume unless there is an outside source of heat.

    The case for the decrease in Pa is exactly the opposite, a reduction in Pa causes the fluid to expand increasing its volume and reducing Pf to Pa.

    No other scenario is possible for your configuration.
    Last edited by zeke; 11-15-2012 at 10:40 PM.

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