Mechanical resonant oscillation as basic overunity method

[perpetual]

There is no force required to hold the tube above the waterline. Tube is standing on supports above the water line by itself.

https://youtu.be/yqL-uKf0OhI?t=1452

Besides, like i said, it is nonsensical to talk of vacuum in a tube which is completely filled with water.

Only if air is intentionally let into the tube or tube is higher than 10 meters partial vacuum will begin to form, but never real vacuum cause, as i said, the moment significant partial vacuum starts to form water boils and space is filled with water vapor.

[Willy]

Want to measure the vacuum force inside this 'sealed, partially submerged' tube?
It is equal to the force required to hold that tube above the waterline.

It is equal to the force the tube exerts upon the supports.

[perpetual]

The force the tube exerts upon the supports is the weight of the tube itself, not water, so that is surely not the measure of the partial vacuum force.

"Besides, like i said, it is nonsensical to talk of vacuum in a tube which is completely filled with water."

Weight of the water in the tube is neutralized by the surrounding air pressure, water in the tube has no weight. That is why pressure at the bottom of the tube is 1 atmosphere in the first place.

We cannot talk of vacuum in a container which is completely filled with a fluid.

That is nonsense.

Partial vacuum is generated at the top only if tube is higher than 10 meters.

[perpetual]

So, are you still denying the energy gain.

Ok, let's say conditions are a 10 meter tall tube half meter wide, obviously fully filled with water above a water pool wide enough to support such large column of water.

We got a steel spindle, elongated, sharp pointed, water resistance minimized.

Spindle has 10kg but due to air chamber inside it has -10kg in water.

Spindle is inserted at the bottom of the tube and let go.

Over a pully -10kg weighing spindle is spinning - for the sake of argument - 100% efficient generator and all energy - the tiny losses due to water resistance is captured in the capacitors or batteries.

As spindle hits the top small valve is opened on it, it releases the air out, now it weighs 10kg and free falls back to the bottom again spinning the generator and all the energy is again captured.

Once back at the bottom pump is pumping the air released at the top through a small tube running vertically through the whole big tube back into the spindle.

Cycle repeats.

Question, what is the energy needed to pump the air back down into spindle.

First and logical assumption would be energy equal to energy needed to lift the displaced weight of water (10kg) to that height (10m).

And energy to lift 10kg to that height is equal to energy spindle generated IN ONE DIRECTION, whether falling up or down, two being exactly the same.

Thus, energy produced in one direction is completely free and efficiency of the system is around 200%.

Additional info. It is possible altho not very likely that energy needed to pump the air back down is even less due to the fact this is not exactly the same situation as when suction is created from above the tube. Here, pumping is done from the bottom through the tube going up. It is clearly seen in video below he fills this big tube with few breaths and it appears effortless.

https://youtu.be/6RnorkCkEqI?t=59

Of course this is far far from a 10m tube.

It is hard to be sure if energy needed to pump the air back down is less than spindle generates as it falls up or down, but it is definitely not more.

Nix

[Willy]

Yes, but I am not as yet satisfied that you have met the conditions of
our gentleman's  agreement as written below

1. To eliminate all confusion, you must first describe HERE IN, for me, that tube and the conditions it is under.

First, I will study your description of the device and conditions, to see if
I am understanding them correctly.