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

[Low-Q]

Here is a new and better picture of my idea.

The idea is that the torque from the buoyancy in cylinder B, C, and D should defeat the differential force that exist between 2-6, 4-8 and 3-7 and still have torque to spear.
I will point out that this wheel is not going to be tested in water, but in a heavy gas. If the gas have the same atmospheric pressure and density at the bottom as on the top, will it take energy to displace gas at the bottom and consume the same amount at the top?
And the question is: Will it make any difference in using gas instead of water?

Here is an example of a heavy gas that can be used:
http://www.youtube.com/watch?v=lpWjR8r_Uvc

Here is the model. The green numbered arrows, represent the force that acts on the pistons.

[TinselKoala]

That linkage would be mechanically feasible for one pair of pistons, as you showed in your earlier animation, but with several like here, the linkage is impossible. It requires parts to interpenetrate, at the axis.

Now, ignoring that, the substitution of a heavy gas to eliminate the pressure gradient doesn't change anything, because the pressure gradient in water isn't what prevented the submerged wheel from operating in the first place. It's the fact that, in all these configurations, as broli has discovered, the Work Out equals the Work In, once you have found all the masses that are being moved and account for them.

Keep thinking, though.

[Low-Q]

That linkage would be mechanically feasible for one pair of pistons, as you showed in your earlier animation, but with several like here, the linkage is impossible. It requires parts to interpenetrate, at the axis.

Now, ignoring that, the substitution of a heavy gas to eliminate the pressure gradient doesn't change anything, because the pressure gradient in water isn't what prevented the submerged wheel from operating in the first place. It's the fact that, in all these configurations, as broli has discovered, the Work Out equals the Work In, once you have found all the masses that are being moved and account for them.

Keep thinking, though.

I have been blind on the thaught that a gas (maybe water) in the same gas (or water) is relatively weightless, thus it isn't more work done by moving it upwards than sideways. The second thing that made me blind, was the thaught of a volatile gas (maybe water) which didn't "care" where the gas (maybe water) came from to fill up the cylinders rear side as the vacuum volume are decreasing gradually from the rightmost part to the leftmost part of the wheel. If we make a bubble of vacuum in this tank of heavy gas, I have been thinking of a net force working on the linkages due to vacuum in both cylinders and the possible difference in atmospheric pressure on the upper and lower pistons.

Do you have any comments on these thaughts?

Br.
Vidar

[Low-Q]

What happens if we attach a bubble or a springload right in the center on the rods?
This bubble contains a volume equivalent to the mass of displaced water + pressure differance + weight of the rod and pistons.
If spring loaded the force upwards will be adjusted accordingly.

This bubble or spring load is following the path of the linkages. The buoyancy of this volume, or force of the spring load, will equalize the force needed to displace water, the pressure difference, and the weight of the rod and pistons, regardless of angle of the system.

This bubble or spring load have only one task: To equalize the factors that stop this wheel from spinning.

I have attached a picture to show you how it can be solved with a weight. Now it will never require force or energy to lift the water upwards. So left there will be only the torque from the buoyancy from the empty space on the right side.

Did I miss anything?



Br.

Vidar

[Low-Q]

Further we can put this model in a wheel:

There is some small text in the picture.