Easy Gravity wheel

[Low-Q]

@Dr. You also know that 1-1=0. You do the math too. Take 10 Newton times (1m - 1m) you get 0 joule. No matte what number you multiply with zero you'll always get zero. There is your proof. If you have another equation please let me look at it.

[quantumtangles]

Interesting to see the discussion unfold.

I confess I think gravity wheels cannot work because net forces (emphasis added) are always zero.

However, consider fluids.

If you have a dense fluid (for example seawater of density 1020kg/m3) and you wish to recirculate a less dense working fluid (say castor oil or dry-cleaning fluid of density circa 860 kg/m3), in this case less dense working fluid should rise to the top of the seawater tank.

Working fluid can then fall onto the cups of an impulse turbine via gravity before rising to the top of the seawater tank due to lower density.

The tricky part is getting working fluid to re-enter the base of the seawater tank.

I concluded air pressure (via an air compressor) and a one way valve would be fairly efficient.

Pressure relief valves also prevent P1V1=P2V2 equalisation.

So when one says gravity is a conservative force (which of course it is) what precisely does this mean in practical terms (when one should be able to cause a fluid to strike a turbine via gravity, and then 'float it' up through a dense substrate so it can fall onto the turbine all over again?

I did some maths in a post concerning this (recirculating fluid turbine).

It is on topic because the turbine is a 'wheel' and the machine works using gravity.

It is off topic because the maths is complicated for the uninitiated.

Please let me know your views 

[Low-Q]

Interesting to see the discussion unfold.

I confess I think gravity wheels cannot work because net forces (emphasis added) are always zero.

However, consider fluids.

If you have a dense fluid (for example seawater of density 1020kg/m3) and you wish to recirculate a less dense working fluid (say castor oil or dry-cleaning fluid of density circa 860 kg/m3), in this case less dense working fluid should rise to the top of the seawater tank.

Working fluid can then fall onto the cups of an impulse turbine via gravity before rising to the top of the seawater tank due to lower density.

The tricky part is getting working fluid to re-enter the base of the seawater tank.

I concluded air pressure (via an air compressor) and a one way valve would be fairly efficient.

Pressure relief valves also prevent P1V1=P2V2 equalisation.

So when one says gravity is a conservative force (which of course it is) what precisely does this mean in practical terms (when one should be able to cause a fluid to strike a turbine via gravity, and then 'float it' up through a dense substrate so it can fall onto the turbine all over again?

I did some maths in a post concerning this (recirculating fluid turbine).

It is on topic because the turbine is a 'wheel' and the machine works using gravity.

It is off topic because the maths is complicated for the uninitiated.

Please let me know your views 

You are talking about a buoyancy effect. For oil to enter the bottom of a tank filled with water, you must spend the energy required to displace the same volume in water, minus the displacement of same volume of oil. If you displace 1 litre of water (1kg) at 10m depth, with 1 litre of oil (0.8kg), the required energy to do this operation is 1kg - 0.8kg * 9.81 * 10m = 19.62 Joule. The very same energy you can take out of the oil on its way up to the surface.

It wont work. You need to supply energy/heat to alter the fluids density on demand. That would be the only way to make a working gravity wheel. Heating one side, and cool the other side. Then the temperature will displace the weight you need in order to achieve rotation.

Vidar

[Dr]

@Vidar: You are not taking into account weight transfer, which is completely different than movement! Take  a 10 ft. dia. wheel, at the 12 oclock position fasten a pivot point and hang a 5 ft. lever. at the end of the lever fasten a 5lb. weight. Where does the wheel see the weight? You see the weight in the center of the wheel! The wheel see,s the weight at the 12 oclock position! I am absolutely convinced that Bessler had a working gravity driven wheel, and I think he found a way to transfer the weight rather than move it.

[Low-Q]

@Vidar: You are not taking into account weight transfer, which is completely different than movement! Take  a 10 ft. dia. wheel, at the 12 oclock position fasten a pivot point and hang a 5 ft. lever. at the end of the lever fasten a 5lb. weight. Where does the wheel see the weight? You see the weight in the center of the wheel! The wheel see,s the weight at the 12 oclock position! I am absolutely convinced that Bessler had a working gravity driven wheel, and I think he found a way to transfer the weight rather than move it.

What equation prooves that this wheel will selfrun just by placing a lever on it?

Vidar