The breakthrough in Free Energy - a system that cannot be debunked

[Tarsier_79]

Hi Codwell

Lets pretend for a start you close a valve, then move your container, reopen for water movement.

Your input is the mass of the water, #1-#2, lifting 1 unit. 0.5 x 1 = 0.5

We have an output #2-#3 The water drops half a unit. 0.5 x 0.5 = 0.25

We have a theoretical nil transaction #3-#4, A weightless empty container drops half a unit.

We have a second output #4-#5 Water drops half a unit 0.5 x 0.5 = 0.25.

Seems like unity to me.

Lets consider the effect Wayne is trying to use in your diagram:

#2-#3  Here the buoyant object on the left of 1 unit rises 0.5 units! Remember for the effect to work, the buoyant object must be submerged at all times. So you are calculating a return of 0.5. How does the Buoyant object on the right lower, while the water drops and still stay submerged? We have to add energy to keep it displacing the water on the way down. This is the equivalent to 1 unit of buoyancy x 0.5 units of distance on the "pull down". -0.5

The same thing happens again #4-#5.

Your other suggestion interests me: Can we move the container, capture the energy of the flow, then still use the water once it settles on the other side?

Buoyancy isn't a magical thing where light things float. Buoyancy is displaced water. You move water from a depth to the surface, displacing it with a "buoyant object" The further you move the water, the more energy you have put in to do so.

Cheers

Kaine.

[broli]

Buoyancy isn't a magical thing where light things float. Buoyancy is displaced water. You move water from a depth to the surface, displacing it with a "buoyant object" The further you move the water, the more energy you have put in to do so.

I agree with that. It appears there is some sort of leverage effect going on in the old aquarium videos shown. However the Archimedes principle does not quantify this, as irregardless of the shape or depth of the air displacing the liquid the buoyant force should remain the same if the volume of air remains the same?

[broli]

Btw in the spirit of full disclosure the video made by Markus mentioned the lawsuit was dropped and he got an apology letter. However a quick google search shows that this is far from the truth: OSC Orders Display |Year=%|Search=|Page= (ok.gov)


The lawsuit by the Oklahoma Department of Securities is still going strong it seems and Wayne was ordered to pay up to 2million$ back to "members" that have put money into the invention.

[Tarsier_79]

I agree with that. It appears there is some sort of leverage effect going on in the old aquarium videos shown. However the Archimedes principle does not quantify this, as irregardless of the shape or depth of the air displacing the liquid the buoyant force should remain the same if the volume of air remains the same?

Agreed. As you push it lower in the water you add energy, and you can take energy back on the way up. The amount of energy you can get is determined by how far you push it down.... Hence why you can calculate energy of displacement on graph paper.

[broli]

Agreed. As you push it lower in the water you add energy, and you can take energy back on the way up. The amount of energy you can get is determined by how far you push it down.... Hence why you can calculate energy of displacement on graph paper.

So if you trade in force for distance then what is the big fuss about. I also went through the 10 year old thread here: [size=78%]Hydro Differential pressure exchange over unity system.[/size] Which does not offer much besides countless pages of text. No one has managed to replicate or show any energy production that exceeds the input. So now 10 years later he's back with an on going 2million$ lawsuit and still no working commercial device big nor small. Are we just rolling back time and starting all over?