Idea on how to trick the conservative gravity - a very possible solution.

[broli]

After parametrization of the formulas I came to the conclusion. That the work W_rot due to the rotation of the structure and the work W_w due to the water displacement has a certain fixed relationship. I really think I did something wrong since this is just too simple to have missed for centuries  .

Anyways the relation in my case having the vacuum cylinder always volume being V= pi m^3. That is a diameter of 1m and height of 1m. I found out no matter how deep you go (thinking it'll increase the torque if I had the volume at some deeper level) there's always the same relation between the two work formulas.

Namely W_rot/W_w = r / (r - 0.25) ...r being the distance from the center to the center of the vacuum. As you can see W_r (the energy from the rotation) is bigger than the energy needed to displace the rod with pistons.

[TinselKoala]

From Simanek's site: http://www.lhup.edu/~dsimanek/museum/themes/buoyant.htm

"Consider a body of mass m and volume V that is less dense than water. If it is pushed underwater and released, it moves upward. As it moves upward from position A to position B, water must move downward. The volume V at position A which the body occupied must now be replaced with water, and the water that was in the space it now occupies at position B must be displaced. The net effect is that as the volume V of the body moves from position A to position B, an equal volume V of water moves from position B to position A.

(I couldn't get the table to format correclty so it's omitted--TK.)

Whenever a mass moves in a gravitational field, work may be done on the body by the field or on the source of the field by the body. That work is of size Fh where F is the size of the force exerted on the body by the field and h is the distance moved vertically. When a body moves up in the field (opposite to the gravitational force) it does work on the source of the field. When the body moves down in the field (in the direction of the gravitational force) the source of the field field does work on the body.

We tabulate the force and work on the body (mass m) and on the equal volume of water (mass M). Since the body is lighter than water, m < M. The net work done on the body moving up in the liquid is (M-m)gh = (Mg-mg)h. We identify Mg as the size of the Buoyant force, and mg as the force due to gravity acting on the body of mass m.

Mgh is the work done by gravity on the water during the displacement downward. -mgh is the work done on the lighter body being displaced upward. "

[TinselKoala]

In broli's figure 1, a 2 meter by 2 meter cylinder of water falls one meter. This is all the energy there is in the system, and this is where the buoyancy comes from. In figure 2, that same cylinder of water must be raised back up. If you think that the rotating path makes a difference in the energy, then yes, you have made a mistake in your calculations or assumptions!!

[Low-Q]

Ok here's a further addition. I ended up having to use a sealed container (black part on drawing). Which has water between the pistons/plates. the idea as Low-q mentioned is that buoyancy will pull the vacuum part of the container up by a very big force. When it has rotated 180° the pistons need to be pushed up. This is not free as Low-q mentions. But my argument is that the energy of the rotation is far above the energy needed to push the liquid 1 meter up. According to my primarily calculations that is...

1) finding out how much energy it takes to lift the mass of liquid between the pistons up 1m
2)finding out how much energy the buoyancy provides due it's torque

And comparing 1 and 2 I found out that using my dimensions I have 1.5x more energy from the rotation than lifting the pistons 1m up. I could have made a mistake and could be wrong. But please take not this is pure conceptual. Meaning every part is considered mass less. The liquid is considered non-viscous, that is when a object moves through it there's no resistive force in essence it can move freely. The last one is important in calculating the energy from the rotation. Pretty much assuming that the wheel has no resistance during rotation.

More will follow.

I think the problem is that you cannot use liquid. Liquid is not compressable so at the bottom the density of the liquid is not higher due to higher pressure. So the pressure acting on the piston at the bottom, will be greater than the top, and the piston cannot move upwards. Since the volume doesn't change from <^oto 360^o, the buoyancy will be the same at all levels of depth, but the pressure on the piston wont.
In a heavy gas the buoyancy will be greater the deeper the constant volume is present, or at least the buoyancy at greater depths will increase parallell to the increasing pressure acting on the piston at the bottom. So those forces cancels out. Left you have the a greater buoyncy at the right side of the wheel - which i also believe will be sufficient to defeat friction.

Br.

Vidar

[broli]

I have to point out that I did make a mistake in my calculation. The 1.5x came out of the radius which was really 1m so the end result should have been 1x in other words no gain in output. I realized the mistake when I was in my bed trying to sleep  .