Buoyancy/gravity wheel - another approach?

[sm0ky2]

you guys are getting close to the answer.
allow me to add some clarity.

the object displaces the water equal to its spacial volume. regardless of the number of locks, or chambers it travels to, when it reaches the top, an equal volume of its' displacement will have moved to the bottom.
giving you a net = 0 potential energy in gravity.
The weight of the displaced water at the top of the column (E = mgh) is exactly what can be lifted by the buoyant force. Now, there are losses due to friction, water resistance, etc. that make the actual value of PE slightly less than the displaced mass over the height.
Now - you CAN lower the mass of the buoyant object to much less than the displaced water mass that is traveling down.
  However -
the EXACT amount of energy between the potential energy of the water mass at the height, and the potential energy of the buoyant mass at the same height,
will represent itself in Buoyant Force over Distance !!!!!!!   <------

all in all, Energy in = Energy out minus losses.

the math is long and convoluted, but at the end of the day
i must conclude that this type of buoyant system cannot be overunity.

That is not to say we cannot extract buoyant force in another way, but this type of cyclical system, all energy can be accounted for.

[norman6538]

possible buoyancy motor idea for discussion

  This is based on the following video.
https://www.youtube.com/watch?v=rxFXsoqbfrk
and was discussed somewhere in overunity.com but I lost track
of where that was.

Given a 3 lb weight that barely floats. The volume of the object is enough
that it can be pushed down and submerged by 1 ft lb. which means the object
displaces less than 3 lb of water and floats.

Given that the 3 lb weight is submerged into the bottom of a sealed 10 ft tube
of water it will rise to the top.

Then when removed from the water with maybe 3 ft lbs of work
it will have 30 ft lbs of potential work.
We then subtract the work in of 1 ft lb plus 3 ft lbs from 30 ft lbs we have
26 usable ft. lbs of work....

This will work because in water the 3lb weight does not have its full weight
because of buoyancy whereas in air out of water it has the full 3lbs of weight.

This would not work if the float had little weight like foam because
there would be few ft. lbs of potential energy when out of the water
and it would take more ft lbs of work to submerge it than it weighs due to the
large volume and low weight.


Of coarse I have not considered the valve changing work required.

I tested the idea with a plastic slide film container and several nuts from
a bolt. With 2 nuts it floats and with 3 it sinks. So in a sealed tube it would rise
many feet giving many "foot nuts" of work.

Do you see any flaws in this idea? Do you have any suggestions?

My former assessment of the pyramid idea was discussed here
http://overunity.com/8539/buoyancygravity-wheel-another-approach/msg455188/#msg455188


Norman 
   

[TheCell]

The example is very straight forward.
As long as the water level does not significantly drop during the raise of the floated weights (which the valves are supposed to prevent) there is a surplus on the energy side. The tube can be as long as you like therefore the energy needed to push the weight under water to get it into the tube is negligible.
The energy to operate the valves is only friction related. The valve which is opened or closed does not be under stress from the upper or lower side which would cause friction caused by water pressure. The valve could be constructed as such that it only slices a cookie of water sidewards, so the overall water level does not depend on whether a valve is closed or not.
The valves are essential I think for functioning.

As long as the water level does not significantly drop during the raise of the floated weights...
And this is the point: each 'floated weight' displaces water . No energy surplus

[norman6538]

Thank you TheCell. I am working on testing the water levels. That is critical.
I will post the results.

Norman

The example is very straight forward.
As long as the water level does not significantly drop during the raise of the floated weights (which the valves are supposed to prevent) there is a surplus on the energy side. The tube can be as long as you like therefore the energy needed to push the weight under water to get it into the tube is negligible.
The energy to operate the valves is only friction related. The valve which is opened or closed does not be under stress from the upper or lower side which would cause friction caused by water pressure. The valve could be constructed as such that it only slices a cookie of water sidewards, so the overall water level does not depend on whether a valve is closed or not.
The valves are essential I think for functioning.

As long as the water level does not significantly drop during the raise of the floated weights...
And this is the point: each 'floated weight' displaces water . No energy surplus

[norman6538]

By using some simple tube and a small eye drop bottle I have seen how

There will be water creep down from the top tube.
This is how it occurs.

When the submerged object is moved
into the first tube with its top closed water will be displaced by the
submerged object below that part of the tube which is the tub. Then when
the bottom of the middle tube is closed and connected to the top tube and the submerged object is moved up water from the top tube will drop down and replace the submerged object. And so we have water creep down.

There was someone purporting to use some displacers that might compensate
for this. ie. If that displaced water was moved off to the side or up 1 foot instead
of down then we could prevent the water creep down.

I used several props to show me what happens. 1. a M&M candy tube. 2. A small
eye drop bottle as a displacer. 3. A 1 qt container.  I filled item 1, the tube with water
and then turned item 3 upside down to cover it tight . Then flipped the container
right side up and filled it with water and lifted item 1 up and pushed item 2 into the tube displacing the water and adding it to item 3.  Then pushed item 1 down to
seal against item 3 and flipped it upside down draining the water. Then pulled
item 2 out and you can see how much water was displaced into item 3 by item 2. 

So this leads me to this question

I have a 1 qt displacer object.
If I displace one quart of water into the upper tube a foot above
the valve so that it can later be dropped down below the valve.
How much work does it take to do that?

If that work is less then the potential ft lbs of the rising submerged object
then we can do this thing....

The limit will be the slowly rising submerged object and the lifted distance.

Norman