Ping pong ball tube

[Comassion]

First off, you can number me among the skeptics - I don't think it's possible to produce a free energy device.

Second, I had an idea for one anyway, and I'm curious as to why it wouldn't work.

The idea just uses gravity and buoyancy to attempt to keep a ping pong ball in motion, and a valve that probably doesn't exist, so treat that as a hypothetical for now.

Fill the tube with water and leave it open at the top.  At the bottom of the tube, place the hypothetical valve such that the ping-pong ball can be pushed into the bottom of the tube without letting the water out.

You probably get the idea now - let the ball float up to the top of the tube, and use a device at the top to kick it out of the tube (depending on how fast it goes up, a clever design might be able to bounce it out just by using the upward motion of the ball if it's moving up fast enough when it breaks the surface of the water.)

So now the ball falls down next to the tube (and here is where someone could have it interact with devices to produce the energy needed to run a mechanism at the bottom to push the ball back into the tube).

Given that the tube can be quite tall, there would be room for several such devices between the top of the tube and the bottom - hopefully enough to power the device at the bottom to push the ball back into the tube.


Thoughts?

[Creativity]

to push the ball in throughout the valve, u will need to displace the volume of water equal to the ping pong ball volume. By pushing the ball in, the water level will rise, so will the pressure. Not to mention the pressure that the valve will have to hold and if to push anything inside the tube, this pressure will have to be overcome.
no go.

[Comassion]

I can understand that, thanks for the explanation.

Also, I found this (thanks to another thread on the forum) that goes into far more detail.

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

[vrstud]

Has anyone seen this?

http://www.frank.germano.com/oceanpower.htm

[allcanadian]

@Creativity
Quote:
"to push the ball in throughout the valve, u will need to displace the volume of water equal to the ping pong ball volume. By pushing the ball in, the water level will rise, so will the pressure. Not to mention the pressure that the valve will have to hold and if to push anything inside the tube, this pressure will have to be overcome.
no go."
I have found that there are people who throw their hands in the air and declare "that is impossible" and there are people who use deductive reasoning to simplify and solve what would seem to be impossible problems. The only problem is that usually the solution is so simple it is embarrassing for all involved.

Consider the following---
You have a thin walled hollow tube extending downward to the bottom of a tank of water, the tube has neutral boyancy and there is a valve on the bottom of the tube so no water can enter. Next a cylindrical hollow weight with an O-ring seal having a slight positive boyancy in water is dropped down the tube and work is extracted from this drop until it reaches the bottom of the tube. Now everyone tells me I cannot remove the hollow weight from the tube without displacing an equal volume of water, equal to the hollow weight volume---it is impossible. In fact it is easy, I open the valve, hold the hollow weight in place at the bottom of the tank and lift the tube, a portion of the hollow tube is now above the tank water level a distance equal to the length of the hollow weight. The weight is now in the tank and basically no water has been displaced, the valve is closed and the weight is given a little kick to the side and it floats to the surface. To reinsert the weight into the tube a set of arms attached to the tube grabs the weight, as the volume of the weight is lifted "out" of the water the tube gets heavier and falls "in" displacing an amount of water equal to that lost by the weight rising----net change in displacement of water = zero. I imagine someone will state the obvious, the weight cannot fall in the tube due to air pressure below it, we will place an air channel through the weight with a valve in it.
I am not claiming anything here gains anything, all I am saying is that all the experts stating that water must be displaced when a ball or hollow weight enters the tank equal to the ball or weight volume are wrong. In this example there is basically no displacement of the water because every action is countered with an opposite reaction, weight enters as tube leaves(rises)--weight leaves as tube enters(falls). I imagine some will say I am cheating or bending the rules, to this I would say ---people who succeed do not follow the rules they make their own.
PS-- I found this simple solution within five minutes of reading your post:)
Regards
AC