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Author Topic: Buoyancy/gravity wheel - another approach?  (Read 21040 times)

Offline norman6538

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Re: Buoyancy/gravity wheel - another approach?
« Reply #30 on: July 08, 2015, 05:03:13 PM »
I really want to test this buoyancy principle in a simple way. The key will be to have
2 low friction valves. For each cycle they have to open and close so that means
4 valve movements. If those 4 valve movements take less work than the floated
object then the device will be OU. And as said before each additional vertical foot
is that much more work available so it seems to me it can be done if the tube is
tall enough.

A friend suggested for the water valve a plate between 2 orings would slide in and
out and would not displace much water. or just a ball valve which probably is
a ball with a hole through it in between 2 orings/seals.

In the video you can see the boy push and pull the a knob that opens and closes the
valve so it is not a ball valve but like the plate described above.

But the water displacement puzzles me.  if you push a buoyant object down into
the tub of water it will make the water level rise. Then when you slide that
object under the tube which is filled with water also will the tub water level
drop?  And likewise when a submerged object is removed the water level
will drop - so at the top of the tube will the water lever drop when removed?


I probably will just have to make it and ignore the valve work requirement losses
and then adjust the height to match those losses.


Norman



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Offline sm0ky2

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Re: Buoyancy/gravity wheel - another approach?
« Reply #31 on: July 08, 2015, 07:44:48 PM »
in an open system, the water will be displaced, and transferred to the surface level. it will reflect as a rise in water level at the top of the tank and tube. However, in this set-up, an open system will result in all the water from the tube flowing down into the lower basin.

Because the tube remains sealed at either end, water does not leave the tube. What occurs is a change in pressure, equal to the buoyant force x volume displaced. when the float and object leave the top of the tube, the pressure drops again as the "hole" is filled with water. the lower end of the tube is sealed when this takes place so the water does not fall down, a small amount will be transferred into and out of the tube to the upper basin, but this is negligible and recycled each time an object floats upwards.
If properly designed and executed, there should be no loss in water height. ( very small amounts of water may travel on the surface of the objects as they leave the system, but compared to the volume displaced, this is very small)

theoretically, the energy value for opening and closing the valves can be negated, or treated as 0.
even further, the work-function of the buoyant force as it travels up the water column, can be converted to perform this task, without affecting the potential gravitational energy of the mass once it reaches the top.

Offline sm0ky2

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Re: Buoyancy/gravity wheel - another approach?
« Reply #32 on: July 08, 2015, 07:54:32 PM »
buoyancy, by default, is an overunity process. Once the mass has been removed from the fluid medium in which it is buoyant, the object has gained a gravitational potential proportional to its' change in height.

This is the power source that operates every hydroelectric power station on earth.
The same power source that brings divers and their gear to the top of the ocean from great depths using only the tiny energy of a CO2 cartridge.
     - which is but a fraction of the energy that would be consumed if you lifted the diver with a mechanical force over distance.

The only time buoyancy systems are in thermodynamic equilibrium, is when you cycle the system by forcing the buoyant object back down through the fluid medium, without changing its' buoyant state. - this is a specific case when Energy In = Energy Out. And both sides of the cycle can be treated as a potential gradient. ALL other buoyant systems, are OU by nature.




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Re: Buoyancy/gravity wheel - another approach?
« Reply #32 on: July 08, 2015, 07:54:32 PM »
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Offline RomanEmpire

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Re: Buoyancy/gravity wheel - another approach?
« Reply #33 on: July 08, 2015, 09:04:20 PM »
I think that the water level drops because the spheres steal the space of the water when them rises. In the system shown in the video there is a water displacer in the upper side that reintegrates the water ( i think ). The water level will drop and must be reintegrate, this is the only problem to solve or the cost to be payed. we need to understand if energy expenditure is greater than that obtainable (not hard to believe that it is more)

Offline norman6538

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Re: Buoyancy/gravity wheel - another approach?
« Reply #34 on: July 08, 2015, 10:15:48 PM »
Thanks guys for the excellent water level discussion. I'm pondering them all.
I was thinking that if the water level drops with each cycle then where does it go?

It seems like when the floating object rises into a given section the level will rise
then fall - that would make the most sense to me....

Norman

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Re: Buoyancy/gravity wheel - another approach?
« Reply #34 on: July 08, 2015, 10:15:48 PM »
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Offline sm0ky2

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Re: Buoyancy/gravity wheel - another approach?
« Reply #35 on: July 08, 2015, 11:47:57 PM »
Thank you guys. This sheds some light in the issue.
has anyone done the experimentation on this?
take a look at this illustration, and I will walk through each step of the cycle, as I gently insert my foot into my mouth.....

1) As the object is placed into the lower basin, water is displaced, rising its' water level by that amount. lets call it X

2) When loch1 is opened, there is a negative pressure head, keeping the water pushed up inside the water column. As the object floats up into the "load chamber", the water that is displaced by the object is also pushed into the lower basin, but at the same time the water level in the lower basin drops by X, so the lower basin is still +X

3) loch1 is closed, loch2 opened. now the water displaced by the object is transferred from the "float chamber" to the "load chamber". an amount of X

4) loch 2 is closed, and loch 3 is opened. the object displaces water in the upper basin, and at the same time, that amount of water (X) is transferred into the "float chamber", so there is no noticeable change in water height of the upper basin.
until the object is removed. Then we have mass (Y), which is less than displacement (X) moved to the top. and displaced mass (X) moved to the bottom.

5) the difference between the mass, over height, is equal to the buoyant force over distance for the height of the column.

So,. I retract my earlier assumptions, and now state, verifiably, that yes, water is being moved down the column by the displaced buoyant mass.

the energy would be "free" if the source of water came from an already elevated point, such as the top of a waterfall, or the pressure from your local water tower to fill the column.  But the obvious assumption would be that the mass of the elevated water would possess more potential energy than the buoyant mass you float to the same height.

For instance, micro-hydro turbines inline with your sink faucet to extract energy from your city's water pressure?? there are places in your house where you don't need the full pressure, just the water that comes out. so extracting the pressure as electrical energy won't affect the way you live. just lower your energy costs.

we might have to take this project back to the lab and figure out a way to keep the water level up, or use a different source of fluid medium that is replenished as we displace it.


Offline norman6538

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Re: Buoyancy/gravity wheel - another approach?
« Reply #36 on: July 09, 2015, 03:20:29 PM »
My version of the water levels and movements.

When the floating object moves from the tub into the bottom section of the
tube it will displace that water in the tube down into the tub and
raise the tub water level a second time....

Then the lower valve is closed sealing it off from the tube to the tub.
and the upper valve is opened and the floating object will rise
and push the water above to below the floating object changing nothing.
there is no pressure change and no water level changes until the floating object pops above the top water level and is removed...then that level will drop....
and in each cycle it will drop more.... and that should be a problem.

So the net is the tub will rise and eventually overflow and the tube will
drop its level and need refilling.


Does this make sense now?

Norman

Free Energy | searching for free energy and discussing free energy

Re: Buoyancy/gravity wheel - another approach?
« Reply #36 on: July 09, 2015, 03:20:29 PM »
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Offline Low-Q

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Re: Buoyancy/gravity wheel - another approach?
« Reply #37 on: July 09, 2015, 07:59:42 PM »
Thanks guys for the excellent water level discussion. I'm pondering them all.
I was thinking that if the water level drops with each cycle then where does it go?

It seems like when the floating object rises into a given section the level will rise
then fall - that would make the most sense to me....

Norman
Norman,
Water level at the bottom reservoir will rise when the object submerge. When the first valve opens, and the object goes into next reservoir, and the valve behind it close, the water level at the bottom reservoir will remain at the new hight. Now the object will bring with it its displaced water, but when the object leaves the top reservoir, that object will not displace water anymore, and water level of the top reservoir will drop to a lower level, and remain this new level. Then the cycle repeats. Result is increasing level in the bottom reservoir, and reduced level in the top reservoir. Finally the water level is the same, and the process stops.

The energy we put in by filling up the tube with water, is the energy we get out from the buoyant objects.

Vidar.

Offline norman6538

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Re: Buoyancy/gravity wheel - another approach?
« Reply #38 on: July 09, 2015, 10:14:02 PM »
Norman,
Water level at the bottom reservoir will rise when the object submerge. When the first valve opens, and the object goes into next reservoir, and the valve behind it close, the water level at the bottom reservoir will remain at the new hight. Now the object will bring with it its displaced water, but when the object leaves the top reservoir, that object will not displace water anymore, and water level of the top reservoir will drop to a lower level, and remain this new level. Then the cycle repeats. Result is increasing level in the bottom reservoir, and reduced level in the top reservoir. Finally the water level is the same, and the process stops.

The energy we put in by filling up the tube with water, is the energy we get out from the buoyant objects.






Thanks Vidar for your excellent reply. I made one mistake in my earlier post -
...the tub water level will rise a second time. (wrong cause the object
leaving reduces the water level but is simultaneously replaced from the tube.)

But Vidar consider this. Everything you say is correct with respect to buoyancy but
net the potential energy of gravity from the lifted object. Every foot it rises beyond the work in is free except for replacing the water in the tube. So, imagine a 100 foot tall tube would give quite a bit of extra gravity power unless the power to replace
the water is too great.

The original pyramid idea did not consider the water replacment in the tube.

Norman

Free Energy | searching for free energy and discussing free energy

Re: Buoyancy/gravity wheel - another approach?
« Reply #38 on: July 09, 2015, 10:14:02 PM »
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Offline Low-Q

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Re: Buoyancy/gravity wheel - another approach?
« Reply #39 on: July 09, 2015, 11:52:55 PM »
Norman,
If the tube is 100ft tall you have to supply a respective amount of water the same hight in advance. So the net energy from lifting mass and water will combined end up in zero.
It is only the weight of displaced water that is the 'active medium' here. The weight of the objects are not interesting because it is only the difference in buoyancy in water vs. buoyancy in air that is interesting. The difference here is therfor the weight of displaced water, and only this.

The objects can be made of osmium or hydrogen. Doesnt matter. And for each object you rise 100ft you must supply the displaced water.

Vidar

Offline sm0ky2

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Re: Buoyancy/gravity wheel - another approach?
« Reply #40 on: July 10, 2015, 03:22:33 AM »
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.

Free Energy | searching for free energy and discussing free energy

Re: Buoyancy/gravity wheel - another approach?
« Reply #40 on: July 10, 2015, 03:22:33 AM »
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Offline norman6538

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Re: Buoyancy/gravity wheel - another approach?
« Reply #41 on: February 01, 2016, 02:27:22 PM »

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 
   

Offline TheCell

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Re: Buoyancy/gravity wheel - another approach?
« Reply #42 on: February 01, 2016, 07:28:33 PM »
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

Offline norman6538

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Re: Buoyancy/gravity wheel - another approach?
« Reply #43 on: February 01, 2016, 08:31:48 PM »
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

Offline norman6538

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Re: Buoyancy/gravity wheel - another approach?
« Reply #44 on: February 02, 2016, 01:45:30 AM »
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


 

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