Another wiew on gravity wheel - vacuum controlled under-water wheel

[myrmex]

Say you werre to create hydrogen at the very bottom can using spark gap inside trapped water , would it change the pressure inside the can ?

Since you change the pressure it would try to go up to release it , once at the top it then release and loose its hydrogen where you can capture it to recycle it . Would these action yield enough energy to rotate the waterwheel ?

[TinselKoala]

Well, if you have enough energy to make a spark why not just use that energy to turn the device? Or better yet, just use the spark energy for something else, and forget the buoyancy device...

No, that can't be what you meant.

[myrmex]

Well people already create hydrogen , when it creates it the bubble need to "fight" trough pressurized water , the buyoancy system is a system that use the difference in pressure , so if you were to add a small buyoancy system to an already working hydrogen  system it's energy that is freely tapped ( the action of the hydrogen fighting trough water atom to reach the top of the can ). Yea i don't know if it would be worth it ...

[Low-Q]

I think it's getting pretty complicated.

You make nice drawings, though. I wonder if it could be animated?
Do you know about Phun?
http://www.phunland.com/wiki/Home
You might be able to get this working in Phun, if you play around a little.

I still don't think it will work in real life, though, because I don't think the overbalanced (or overbuoyant) condition will continue all the way around the wheel. There will be a point (several actually) where all forces  balance and the wheel will stop, because inertia won't be able to carry it past that point.

I see that CLaNZeR is building a similar idea on another thread. If anybody can get one working, it will be he.

(EDIT to add there isn't going to be much power available from buoyancy changes, as you no doubt have already calculated. All that complex mechanism, and the drag of pulling stuff through the water, subtracts a lot of power from the system, and the bigger (more buoyant) it gets, the more drag there is...)

Hi,

I some times over do it with my drawings . The idea is to have membranes/pistons that doesn't get influenced by waterdepth, as you pointed out in my first drawing. How the practical solution is, can ofcourse be different. Using large gears and tooth rods instead. Less loss.
Drag in water is mentioned. This drag is relativly low loss, even if it is drag. Drag isn't pure friction, and does not create loss as heat very much. It is just a little time consuming.

Anyway, I have tried to understand how this machine does NOT work. If it worked, it would be to good to be true - honestly. So I have twisted my mind to find the flaw in the design.

For those who doesn't understand the principle, try to look at the two pistons in the four vacuum chambres, both vertical one and horizontal one. As the red vacuum chambres rotat around its axis, another axis to the left of the vacuum chambre axis is controlling the movement of the pistons by the brown mechanism. Both vacuum chambres axis and piston control axis is linked so they both run in the same speed in the same directions.

Br.

Vidar

[Low-Q]

Now I have made a simple animation.

here is the construction:

The red tube contains only vacuum.
The black "thing" is a piston which is controlled by a mechanism that moves it inside the vacuum tube. That mechanism should be a magnetic hinge or lock as the vacuum tube is sealed.
It is a vacuum tube so the piston inside can move more freely. With air, it would cost a lot of friction due to unlinearity in the compression and decompression process.

Be focused:
Look close to the black piston. Its centre and axis is aways following the black circle.
Look close to the red vacuum tube. Its centre and axis is fixed at the rightmost side of the circle.
Look close to the sum of all movements, and how the free space inside the tube is distributed at any time.
The black circle is just the path of the magnetic positioning lock of the piston, where the piston centre is following this path, and are not fixed to the main axis where the vacuum tube are fixed.
The piston will therefor slide from one side to the other if the vacuum tube as it turns around.
When the vacuum tube is in vertical position there is the exact same volume of emty space at the top and the bottom. When it is horizontal, there is no emty space at the left side, and only emty space at the right side.

The idea of how it works:
The wheel is not working by the weight of the black piston alone - at least I assume it doesn't. It requires zero energy to complete one round, but it also don't give any energy.
The clue is now how this machine keeps the buoyancy on the rightmost side of the red vacuum tube, which means this machine must run in something thicker than air - water or other liquids.
As assumed, the piston does not provide any work at all. But try to see how a tube with lots of empty space on the right side will act in water - it will be forced to flip over counter clockwise due to buoyancy at the right side.
There is allways more empty space at the right side of the wheel, and it will therefor allways try to stabilize itself by flipping over to a vertical position.
The idea is then how to move mass to one side to another without using energy, so the final energy consumtion is therefor negative - which in turn means more energy to take out than what it consumes -> OU.

OK! This is not a claim, but If you can't find a reason why this thing shouldnt work, you can try building one and see what's happening.