Self Siphoning Water

[tbird]

hi guys,

i have a question.  i know you said before, the small tube on the right should enter above the water in the tank, but should the exit tube on the left have its entrance (top at the tank) above water?  is that just an art mistake and you really mean for it to be exposed to the water in the tank?  and those 2 thin black lines are just left over art that mean nothing?

point of interest...the left tube now looks like my old chicken waterer, if the top were all closed up.  water would stay inside until the outside level was lower than the lip of the pipe (we used a jar).  then air would go in and water would refill the outer area, covering the lip again.

tom

[tbird]

i don't think you understand my question.  must be my accent.  let me put it a different way.

the tube on the left has an elbow at the top that looks like it is connected to the tank just above tank water level.  also the elbow shows to have air in it, which would be logical since it enters the tank from the side above the water in the tank.  my question was, is that the way you meant for us to understand the picture or is that an art (drawing) error?

the same question for those 2 small lines, one on each pipe close to where they go to the tank.  are they just left over drawings that mean nothing?  or are they some kind of valve?

tom

[Low-Q]

  @aLL,
This is to explain basic hydraulic theory. In the design we're discussing,
vacuum is replacing pistons, hopefully. It might be an untested idea 

With the attached pic, if A is PiR^2 = 3.14 x 2^2 =
                                                      3.14 x 4 = 12.56^2 (square) inches

and if B is P x 1^2 =
                  3.14 x 1  = 3.14

What hapens is the pressure applied to A is transmitted to B increasing the potential.
An example is if 5psi of force is applied to A, then it is 5 x 12.56 = 62.8

Since B is 3.14^2 inches, we divide 62.8/3.14 (B) = 20 psi
The reason this happens is the piston focuses the force on the shaft connecting the two pistons. And the resulting pressure is the total pressure applied to piston A divided by piston B surface area.
It's simlar to gear ratios in transwmissions and drive axles/differentials on cars.
So what happens is because piston A has 4 times the surface area of piston B, the pressure is amplified 4 times.


                                                                                         Jim

Yes, but the volume are also reduced to 1/4... So 4 times the pressure does not help if the volume of the pressure are 1/4. Then you end up with the same potential energy in both volumes. No difference in potential energy will not be possible to harness.

Vidar

[FreeEnergy]

See the Open Source Vs. Patenting topic over here: http://www.overunity.com/index.php?topic=1821.0

[ResinRat2]

Hello Gentlemen,

I have not been commenting because I feel I am missing some detail, forgive me if this has already been addressed, but I can't see how you can have a vacuum and an opening to the atmosphere in the same system.

Nature hates a vacuum, so either air or water will fill it. Wouldn't the water just run out and the area at the top start to fill with air or water? Wouldn't the water just run out on the left (picture) side? What would you think would stop it from happening?

Again, sorry if I misunderstand, but I am certainly missing something vital here.