Perpetual Plumbing!!

[Newton II]

Here is a buoyancy device which eats the ball at the top and shits it out from the bottom causing perpetual plumbing!   This device makes use of momentum and collission unlike other buoyancy devices.  If you use a lengthy tube,  the momentum gained by the ball should be sufficient enough to cause perpetual plumbing!   Will it work?

http://jestr.org/downloads/volume1/fulltext12.pdf

[tinman]

And how exactly is the water level achieved as shown in the pic,with out flowing around to the side of the U shaped pipe,into the ball holding cavity?.

[Newton II]

And how exactly is the water level achieved as shown in the pic,with out flowing around to the side of the U shaped pipe,into the ball holding cavity?.

Water is standing on mercury below it.   There is no empty space below the water.   Since mercury is a heavy liquid, the level of mercury is far less than level of water in the tube.    The weights of several balls acting  on the bottom most ball pushes it to some extent into the mercury,  eventhough the ball is lighter than mercury and floats in both liquids. 

There is nothing wrong in figure or idea.    The question here  is,   the momentum of ball falling from the top is sufficient enough to push the bottom most ball through balance portion of mercury to the bottom of water,  so that from there ball floats to the top and cycle repeats.

[TinselKoala]

Please study this section of Simanek's Museum:

http://www.lhup.edu/~dsimanek/museum/buoy4.htm

[Newton II]

Please study this section of Simanek's Museum:

http://www.lhup.edu/~dsimanek/museum/buoy4.htm

The important difference between Simanek's drawing and the one above is,   Simanek's drawing shows only one ball which gains and loses energy as he has explained in that section.   But the drawing above  has several balls which have their own weight and total weight altogether act on the bottommost ball  pushing it to some extent inside mercury and storing initial potential energy in it.   Now you need energy only to cover the balance portion of mercury and not the entire portion.   If the falling ball (this ball falls on another solid ball making a collission and not directly on mercury) from the top provides this balance energy then bottommost ball should cover the balance portion of mercury  and move towards top of water.

Since mercury is a very heavy liquid,  the energy required to push the ball through it will be high.  So, whether this system works -  I don't know.