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Hi,
if anybody understands the basic operation of this device,
please let me know and post some more explanations.

http://de.youtube.com/watch?v=pXatO92If-I

How can he move the water the distance 2X with less work with
multiple containers ?

Many thanks.

Regards, Stefan.

It doesn't take any work to empty and refill a container, as long as you use the energy that you take out when emptying it to refill it again.

The easiest way I can think of is a spare container connected by a siphon tube, and held up by a counterweight via a cam-shaped pully that keeps it balanced at every height.  Then you can move the spare container slowly up or down with essentially no work, and the main container is filled or emptied via the siphon tube.

The trouble with this free energy idea lies elsewhere -- the "output unit" must be heavy enough to stay underwater as the tank is emptied, but buoyant enough to flip around an axle and lift the heavy internal weight.  You can't have both.

Quote from: Mr.Entropy on January 21, 2009, 11:14:56 PM
The trouble with this free energy idea lies elsewhere -- the "output unit" must be heavy enough to stay underwater as the tank is emptied, but buoyant enough to flip around an axle and lift the heavy internal weight.  You can't have both.

The "output unit" must simply float at the surface of the water.  When the water level lowers, the output unit will fall below the uni-directional latch.  When the water refills, the output unit will seek to float at the surface and can be flipped with minimal energy.

For the mass of the output unit = m; Volume of the entire output unit = v; And density of the water = p;

m < vp

Problems arise when the temperature changes... density changes.  Thus the mass and volume should be designed in a way that will allow a "window" of temperature changes.  The window is very small if the device runs underground.  And when the device runs underground, the output wind can be used to regulate the temperature of a building (geothermal) & provide electricity via a wind turbine.

If the output unit is slightly above the surface of the liquid, the dimensions of the outer tank and the amount of water emptied will be adjusted so that the top axle of the output unit will fall below the uni-directional latch.

Another YOUTUBE Free Energy movie

(Compressor water drive turbine -- that drive an electric generator.
last power the pump again, additional LAMPS !!
MUST SEE ! (and reconstruct it ?)

http://de.youtube.com/watch?v=0p3kkaDlWx4&feature=channel
Gustav Pese

Mr Palmer

Thankyou for sharing this concept [which I am presently studying [Your video]

Mr Pese

That fellows video has/is being discussed in this forum [some feel cavitation is playing a part]

I will find the link and post it

Chet

Where does the air with these 4 check valves go and
where does the water go ?

Please explain in more detail all the stages of air in and air out and
the water movements.

Many thanks.

Quote from: hartiberlin on January 22, 2009, 10:33:31 PM
Where does the air with these 4 check valves go and
where does the water go ?

Please explain in more detail all the stages of air in and air out and
the water movements.

Many thanks.

I would have a lock on the inner mass of the output unit.  Liquid from the tank would empty into subsequent containers via gravity... from top to bottom.  At this point, the output unit is below the uni-directional latches and the lock on the inner mass will be disengaged via potential from gravity (this lock might be magnetic, but other designs could be used).  The mass will fall and air flows out of the left side of the output unit (a rotating union will be necessary).  The energy must then be converted so that it can directly lift the containers ( I know that in a hydraulic system, forces can be multiplied... output unit may need to be adjusted if pneumatics can not be used in a similar way).  The distance that the containers must be lifted is dependent on the number of containers used.  (aside from the specifics of converting the Output Work into lifting the containers via an automated process, just realize that the Input Work can be reduced to a reasonable level by adding containers and adjusting dimensions).  Any excess energy may be harvested via a wind turbine and the inner mass will again lock into place as it comes to rest.  Once the containers are at the desired height, the bottom container's door will open automatically due to hydrostatic pressure.  As each previous container empties, the next container up will automatically empty in a similar fashion (I have done some design work on the containers and this will offer a challenge in building the unit).  As the tank is filled, a buoyant force will engage a lever which flips the output unit, resetting it to its initial state.

The overall process can be adjusted as needed and invention will be necessary, but the main point is to realize this:
If the entire process can be mechanized to automatically occur in sequence as a result of potential within the system, there exists enough output work to reset the output unit.

Check Valve Operation:
The air would come in through a vent on the right side.  And the air would exit through an exhaust on the left side.  The check valves are arranged for this.  As the inner mass falls, air can only be pulled in from the upper right check valve and air will exit out of the lower left check valve.  When the unit flips, the same will hold true.  The output unit can be designed in other ways, but I have chosen this type of output unit for ease in overall manipulation of output force.

My goal is to help build a device that will prove to the world that a system can be intelligently designed to remain in a constant state of non-equilibrium by harnessing forces which are in opposition.  If a device of this kind were to be of a purpose besides academic, it would be placed underground, where many output units would produce a net flow of air which could regulate temperatures of a building.  The units would produce a flow of air (wind) which could then be harnessed via a wind turbine.

These devices might also be used for creating wind for transportation.

The concepts are simple:
1. Output work is constant.
2. Work needed to reset the output unit is variable.

The specifics get a little bit complex.
I have another free energy device that would be easier to build if we have the right materials, but it is also somewhat complicated.  I will try to have a video available within a month or two. 

Are the checkvalves for the air input and output or wherefore ?

Please post drawings of each single step showing where the
water sits and where the air sits in the unit.

Also who is paying for the energy to compress air and push it
into the unit ?

I can get some drawings together.  The mass which falls is sealed and will pull air from the right and push air through the left.

Without check valves, the mass would fall and the air would circulate internally.  The check valves ensure that the air will flow in along the desired path.

First a "mass" is flipped (AKA) lifted to create a Potential Energy to move air through a moving and flipping passage submerged  in liquid.
Second a liquid is allowed to freely drain downhill, and then is lifted again by the energy obtained from draining downhill.

If many things won't work, will they work better better together?  Confuse the issue?

Quote from: truth on January 24, 2009, 05:15:35 PM
First a "mass" is flipped (AKA) lifted to create a Potential Energy to move air through a moving and flipping passage submerged  in liquid.
Second a liquid is allowed to freely drain downhill, and then is lifted again by the energy obtained from draining downhill.

If many things won't work, will they work better better together?  Confuse the issue?

You can start with the output unit flipping if you would like.  Then the liquid drains into the partitions, which allows the output unit to fall below the uni-directional latches.  The partitions are lifted via the output unit.  And the liquid from the partitions will fall back into the original container, resetting the output unit.

Simply compare the Potential from the output unit to the work needed to exchange the liquid out of and back into the original container via my partitioning scheme.

PE from output unit=
m * d - f * d
m = mass inside of the output unit.
d = distance which m will fall
f = friction encountered per distance

Work needed to exchange water:
M * X * 2/n + F * X * 2/n
M = Mass of liquid exchanged
X = height of liquid in original container
n = number of partitions
F = friction encountered while lifting the partitions
X * 2/n = distance partitions must be lifted.

Notice that exchanging the liquid in the original container is the mechanism which resets the output unit.  And the reset work is decreased by increasing the number of partitions, so long as the exchange between original container and partitions are automated to occur in series due to potential from the previous state.

Please explain where that "lever kicks the output unit" is in your math model?

Gravity is a wonderful energy storage force. A rock can be buried in a mountain for millions of years, but when the matter supporting it is eroded it rolls down the mountain. Funny thing is it never rolls back up.

You may be the first guy ever to try and create wind as his preferred output energy.    ;D

I read the way you responded to folks on utube and don't want to insult you, but you have several fatal errors in you concept.

Quote from: truth on January 24, 2009, 11:50:10 PM
Please explain where that "lever kicks the output unit" is in your math model?

Gravity is a wonderful energy storage force. A rock can be buried in a mountain for millions of years, but when the matter supporting it is eroded it rolls down the mountain. Funny thing is it never rolls back up.

You may be the first guy ever to try and create wind as his preferred output energy.    ;D

I read the way you responded to folks on utube and don't want to insult you, but you have several fatal errors in you concept.

As the water rises, it can lift a buoyant object which triggers the lever (much like a toilet).  It does not have to be in the math model because it is enacted by the reset work.

Please explain my fatal errors, after all, that is why I posted.  Most people on youtube are only saying I am wrong based upon their preconceptions.

MR Palmer

Thank you for bringing your idea here

I personaly feel water can change things

@Truth please comment on the flaws

@TinselKoala where are you /[this is what you do,comment on gravity devices][well one of many]

@ Hanns

@All help this man out ,please

         Chet

At 2:59 in the video it is shown that an unpartitioned "giving" container is lifted its full height ABOVE the top of the receiving container, which would be necessary if the drain hole is in the bottom of the unpartitioned upper container. So it takes 2x the work to lift it this high, to replace the water in the receiving container. The scheme of adding partitions finally reduces to the "giving" container, unpartitioned, at the SAME level as the receiving container, and the water just "pushed over" horizontally. SO the receiving container is re-filled with not 2x the work but with exactly 1x the work received.
The 2x figure from 2:59 in the video is a red herring, as are the multiple partitions. You could use a siphon nearly as effectively. The partition scheme, as the number of partitions increases, converges on the UNITY figure, and will not ever go below that. So, at the end of the day, you can reduce the amount of work it takes to reset the apparatus all the way down to just the amount of work you get out of it (neglecting losses.)
You get out, what you put in, less losses. There is no mechanism for energy gain in this device. Yes, it is possible to make the amount of work to reset variable, especially if you start out with an unnecessarily high figure (like the 2x at 2:59 in the video) but you can never make it smaller than the amount of work you get out of the cycle.

@ramset: I am near Toronto, Canada.

Tinsel Koala

I meant where are you  for your critique [I'm sure you knew that] which is greatly appreciated

I thought you were in Australia??

Any way none of my business

Thanks again for your response here

        Chet

Quote from: TinselKoala on January 26, 2009, 02:22:52 PM
At 2:59 in the video it is shown that an unpartitioned "giving" container is lifted its full height ABOVE the top of the receiving container, which would be necessary if the drain hole is in the bottom of the unpartitioned upper container. So it takes 2x the work to lift it this high, to replace the water in the receiving container. The scheme of adding partitions finally reduces to the "giving" container, unpartitioned, at the SAME level as the receiving container, and the water just "pushed over" horizontally. SO the receiving container is re-filled with not 2x the work but with exactly 1x the work received.
The 2x figure from 2:59 in the video is a red herring, as are the multiple partitions. You could use a siphon nearly as effectively. The partition scheme, as the number of partitions increases, converges on the UNITY figure, and will not ever go below that. So, at the end of the day, you can reduce the amount of work it takes to reset the apparatus all the way down to just the amount of work you get out of it (neglecting losses.)
You get out, what you put in, less losses. There is no mechanism for energy gain in this device. Yes, it is possible to make the amount of work to reset variable, especially if you start out with an unnecessarily high figure (like the 2x at 2:59 in the video) but you can never make it smaller than the amount of work you get out of the cycle.

@ramset: I am near Toronto, Canada.

Partitioning reduces the "y" coordinate distance which liquid must be lifted to empty and refill a reservoir.

When there are the theoretical infinite number of containers, the liquid is merely pushed out of the original container and back into the original container.  Hence, the "y" coordinate distance which the liquid must be lifted is zero.

Each individual partition must be lifted 2 * height of a partition.

distance = y-coordinate distance
With one giving container, the distance is 2*X
With two giving containers, the distance is X
With four giving containers, the distance is X/2
With six giving containers, the distance is X/3

Hence, the total distance is 2 * X / n; where n is the number of partitions.  I would also add a distance, c, which will allow the liquid from a partitioned container to "fall" instead of be "pushed".

Total distance = 2 * X / n + c

The UNITY figure will be determined by a minimum size for a partition (based upon the dimensions of the system) and a minimum size for c.  For a very large height of the system, the number of partitions will be quite large, which will reduce the distance to a very small amount.

This device is for learning and will lead to much more efficient perpetual motion devices.

Please note that a constructed device will teach the world about "cooperative systems".  Where each sub-system performs as we currently understand the laws of physics, but sub-systems can work in cooperation to defy our current laws of physics.  Hence, a system which is in constant non-equilibrium whose entropy merely oscillates within a window of max and min values.

http://www.youtube.com/watch?v=bIA2rZQgO_c (http://www.youtube.com/watch?v=bIA2rZQgO_c)

Is an easy overview of my partitioning discovery.  This is the foundation for my cooperative system.

Quote from: GavinPalmer1984 on January 26, 2009, 05:11:49 PM
Partitioning reduces the "y" coordinate distance which liquid must be lifted to empty and refill a reservoir.

True, but it also reduces the mass to be lifted against gravity. In ideal world (without any unwanted losses, your concept potential energy would remain the same (m*h*g), no matter how many partitions there are...)
In a real world, losses are adding to the overall efficiency in any conversion, so a device with many partitions would introduce cumulative losses (wasted energy made through many many pipes, drains, ...).

So, less containers, the better... I'm not kidding, seriously...

Quote
When there are the theoretical infinite number of containers, the liquid is merely pushed out of the original container and back into the original container.  Hence, the "y" coordinate distance which the liquid must be lifted is zero.

When there are "infinite" numbers of containers, the "Y" (h) is approaching zero, and the "m" is approaching zero, too. And the "g" factor approaches a finite number. Overall, the whole "potential" is conserved.
In reality, it gets much worse (due to losses which are CUMULATIVE!) than with a "single container".

Quote
Each individual partition must be lifted 2 * height of a partition.

distance = y-coordinate distance
With one giving container, the distance is 2*X
With two giving containers, the distance is X
With four giving containers, the distance is X/2
With six giving containers, the distance is X/3

Hence, the total distance is 2 * X / n; where n is the number of partitions.  I would also add a distance, c, which will allow the liquid from a partitioned container to "fall" instead of be "pushed".

Total distance = 2 * X / n + c

The UNITY figure will be determined by a minimum size for a partition (based upon the dimensions of the system) and a minimum size for c.  For a very large height of the system, the number of partitions will be quite large, which will reduce the distance to a very small amount.

This device is for learning and will lead to much more efficient perpetual motion devices.

Simply, NO. Sorry.

What the fuck are the "more efficient perpetual motion devices"?
I'd really like to see just one of those...
   

Quote
Please note that a constructed device will teach the world about "cooperative systems".  Where each sub-system performs as we currently understand the laws of physics, but sub-systems can work in cooperation to defy our current laws of physics.  Hence, a system which is in constant non-equilibrium whose entropy merely oscillates within a window of max and min values.

Ah, well... Why there are so many FE enthusiasts questioning the laws of physics?
The "laws of physics" are the best we (the Earthlings) can do at the moment to describe the Nature...

No problem if somebody wants to break them. Nobody would cry after them. Lol!

If you have to offer anything better, good.  Simply, just do it.

Quote from: spinner on January 27, 2009, 06:43:40 AM
True, but it also reduces the mass to be lifted against gravity. In ideal world (without any unwanted losses, your concept potential energy would remain the same (m*h*g), no matter how many partitions there are...)
In a real world, losses are adding to the overall efficiency in any conversion, so a device with many partitions would introduce cumulative losses (wasted energy made through many many pipes, drains, ...).

The mass is not reduced when adding containers.  I am not addressing potential energy.  I am simply emptying and refilling a container efficiently.  Ass you add containers, the distance decreases and the mass of the liquid remains the same.  The mass of the containers increases when adding containers, but an opposing force can negate the mass of the containers.  Pipes and drains are merely conduits for the liquid to pass through.  The system can be designed so that pipes open in an automatic, sequential manner so that the net energy needed to empty and refill a container is:

Work = g * M * d + f * d
g = gravity
M = mass of liquid exchanged + a constant for the containers
f = friction encountered during lift
d = 2 * X / n + c
X = height of original container
c = constant amount needed so that liquid can fall from one side to the other.

Quote from: spinner on January 27, 2009, 06:43:40 AM
So, less containers, the better... I'm not kidding, seriously...

So, no... please try to understand that the mass can be constant and the distance is reduced.  You might assume that energy is lost due to opening valves, etc.  But then you have specified  an inefficient design.  There exists potential in each stage which can trigger an event to occur.  Designing the system intelligently will allow you to empty and refill a container with my above equations for work.

Quote from: spinner on January 27, 2009, 06:43:40 AM
When there are "infinite" numbers of containers, the "Y" (h) is approaching zero, and the "m" is approaching zero, too. And the "g" factor approaches a finite number. Overall, the whole "potential" is conserved.
In reality, it gets much worse (due to losses which are CUMULATIVE!) than with a "single container".

Simply, NO. Sorry.

Yes, "Y"(h) approaches zero and "m" for each container approaches zero.  But "m" * n = mass of the liquid being exchanged.  Gravity is constant.  You keep talking about potential in odd areas... I am addressing work needed to empty and refill a container.

Quote from: spinner on January 27, 2009, 06:43:40 AM
What the fuck are the "more efficient perpetual motion devices"?
I'd really like to see just one of those...
   
Ah, well... Why there are so many FE enthusiasts questioning the laws of physics?
The "laws of physics" are the best we (the Earthlings) can do at the moment to describe the Nature...

No problem if somebody wants to break them. Nobody would cry after them. Lol!

If you have to offer anything better, good.  Simply, just do it.

More efficient perpetual motion devices will output more KW-Hours for less cost (in materials).  And they will not need replacement parts as frequently.
I do not break a law.  I break our interpretation.  The laws exist at each moment in time, but multiple systems can work cooperatively to achieve what our laws say is impossible.
I encourage you to simply understand this:

1. Through the partitioning system, work needed to empty and refill a container is reduced by decreasing the distance which the mass must be lifted.
2. An output unit which resets to its initial state due to an emptying and refilling of its container produces a constant amount of output energy.
3. If you transfer the liquid, initiate the flip of the output unit, and lock/engage events with potential within the processes, statements 1 and 2 allow for perpetual motion.
    more sources of energy which may be harnessed for intermediate steps:
    - There is potential in each individual container being filled and lowered.
    - There is potential in the rising reservoir.
    - There is potential in the actual rotation of the output unit.
    - There is potential in the falling output unit.

The way you would eliminate costs for transferring liquid is to design the system with triggered events which occur under certain conditions.  And before the next cycle, each trigger is reset.  I have solved further problems dealing with these more specific designs (utilizing rising water or falling objects)... but I want the readers to agree with statements 1, 2, and 3.  Once the readers understand and agree with statements 1, 2, and 3,  I hope the online community will assist in designing/building a unit.

GVP

Gavin,

I believe I understand all the parts, but not the whole.

You show the output unit in a container that allows the output unit to freely fall and flip.  But the container must also contain a (large) number of partitions to reduce the amount of work needed to lower and raise the water level.

How do the partitions not interfere with the raising and flipping of the output unit?

Thanks,

M.

Gavin, as spinner and I and others have tried to explain, your partition system is just wrong. Look at the top of the water in the receiving container. It is this height that determines the energy required to lift the water. You are still having to lift the entire container full of water, so that the top of the water is back to the starting level. The work required to lift this water is mgh, plus losses. Your partition system doesn't work to reduce the work to nothing, it can only reduce it to mgh. So there's no possible gain from this system.
It is equivalent to lifting the water to h (not 2h) and just sliding it over. You cannot do better than this. If you don't believe me, get into your bathtub with some containers and start doing experiments.
And please, stop talking about "intelligent designs" and "better working perpetual motion machines." You are talking to some very experienced builders and researchers, who have been dealing with devices like yours, and more complex ones, for a long time. It is just possible, that these people do in fact know whereof they speak. And, since your design is very similar to designs of gravity wheels and buoyancy drives that have been around for many years, without any of them successfully working, it is just possible that you don't know whereof you speak, especially since you are making that fundamental and rather obvious error with the partitions.

Quote from: mondrasek on January 27, 2009, 11:55:28 AM
Gavin,

I believe I understand all the parts, but not the whole.

You show the output unit in a container that allows the output unit to freely fall and flip.  But the container must also contain a (large) number of partitions to reduce the amount of work needed to lower and raise the water level.

How do the partitions not interfere with the raising and flipping of the output unit?

Thanks,

M.

The whole is complicated.  I have not addressed any specifics.  I just want a consensus.  Help me win the others with a consensus

Quote from: TinselKoala on January 27, 2009, 05:04:27 PM
Gavin, as spinner and I and others have tried to explain, your partition system is just wrong. Look at the top of the water in the receiving container. It is this height that determines the energy required to lift the water. You are still having to lift the entire container full of water, so that the top of the water is back to the starting level. The work required to lift this water is mgh, plus losses. Your partition system doesn't work to reduce the work to nothing, it can only reduce it to mgh. So there's no possible gain from this system.
It is equivalent to lifting the water to h (not 2h) and just sliding it over. You cannot do better than this. If you don't believe me, get into your bathtub with some containers and start doing experiments.
And please, stop talking about "intelligent designs" and "better working perpetual motion machines." You are talking to some very experienced builders and researchers, who have been dealing with devices like yours, and more complex ones, for a long time. It is just possible, that these people do in fact know whereof they speak. And, since your design is very similar to designs of gravity wheels and buoyancy drives that have been around for many years, without any of them successfully working, it is just possible that you don't know whereof you speak, especially since you are making that fundamental and rather obvious error with the partitions.

This is not a battle of egos.  I respect everyone here.  You and spinner do not fully understand my partitioning scheme.

http://www.youtube.com/watch?v=bIA2rZQgO_c (http://www.youtube.com/watch?v=bIA2rZQgO_c)

The link will show the partitioning scheme.  Each partition from the original container will drop a distance which is the height of a partition (X / n).  Then each partition must be lifted 2 * X / n.

For example.  With 10 partitions.  1/10th of the receiving container will empty into the top partition (giving container).  The next highest 1/10th of the receiving container will empty into the next lowest partition.  Repeat until all partitions are full.  All of the partitions must then be lifted 2 * X / 10.  The bottom partition will be emptied back into the receiving container.  The next highest partition will be emptied afterward and so on until the receiving container is full.  I have had many intelligent people vouch for this partitioning scheme.  Please try to understand it.

Distance = 2 * x / n

Quote from: GavinPalmer1984 on January 27, 2009, 05:48:21 PM
The whole is complicated.  I have not addressed any specifics.  I just want a consensus.  Help me win the others with a consensus

Gavin,

Unfortunately I cannot help you win the others with a consensus.  I believe this idea is flawed for the reasons that others have specified.  I only brought up my additional concern in the case that you had not noticed it and could not grasp the others' points.

I have heard several ideas "proved" in their respective pieces that could not work as explained once those working pieces were assembled into a system.  In those cases usually, like this one, one of the pieces also did not work.  But the inventor did not understand the explanation of that part's flaw.  So they ignored it and went on.  I am trying to show how the whole system does not work, even if the flawed part *did* work as envisioned.

But please continue until you are satisfied either way.  I would gladly accept being shown wrong, since it would mean free energy!  And we all could benefit from that.

I have not given it a lot of thought but might have to agree that your idea would work with waves and/or tides.  If so, the next question would be if this is a more efficient way of harvesting that energy than already employed or being developed.  Could be a part of the solution that way as well.

M.

People are claiming that I can not reduce the distance below X, where I claim that the distance can be reduced to 2 * X / n.

Just view my clip and tell me what does not make sense:
http://www.youtube.com/watch?v=bIA2rZQgO_c (http://www.youtube.com/watch?v=bIA2rZQgO_c)

I do not think that people are grasping my point.  Do not approach this with a preconception.  I have had experts at the National Science Foundation tell me that the partitioning scheme is very intelligent.  Once you guys understand the partitioning scheme, it follows that a perpetual system may be created.

The guys at the NSF could not fathom a device which produces more energy than what is needed to reset it to the initial state.

Quote from: GavinPalmer1984 on January 27, 2009, 05:48:21 PM
The whole is complicated.  I have not addressed any specifics.  I just want a consensus.  Help me win the others with a consensus

G'day Gavin and all,

It is not a question of consensus, it is a question of physics. Your physics are wrong. When all is said and done to drop a gallon of water 1 m you have to lift that gallon of water 1 m to start the cycle afresh regardless of the number of partitions. There is no free energy in this. In fact, as others have correctly pointed out partitioning costs additional energy.

Hans von Lieven

BTW what is the National Science Foundation that has sanctioned your idea, surely not the US agency of the same name.

Quote from: hansvonlieven on January 27, 2009, 07:11:28 PM
G'day Gavin and all,

It is not a question of consensus, it is a question of physics. Your physics are wrong. When all is said and done to drop a gallon of water 1 m you have to lift that gallon of water 1 m to start the cycle afresh regardless of the number of partitions. There is no free energy in this. In fact, as others have correctly pointed out partitioning costs additional energy.

Hans von Lieven

BTW what is the National Science Foundation that has sanctioned your idea, surely not the US agency of the same name.

The liquid only drops a distance of:

X / n

then is lifted a distance:

2 * X / n

And yes, the US NSF.  You guys should not assume that I am wrong.  I realize that I must convey my understanding to you, but please come into this with an open mind.  Go to my youtube post of partitioning and ask me questions instead of insisting I am wrong.

Gavin,

I'm with you on the partitions.  You are only dropping the entire X high column of water by a distance of one partition, or X/n.  You use this small drop to move the entire X high column of water *sideways*.  You can then return the water to it's original location if you raise it by 2 * X/n and allow it to drop X/n and shift sideways again. 

Really, you are not lowering the water level in the top half of the main container so much as shifting it sideways to a secondary container, leaving only half of the original container filled.  This has the same effect as lowering the water with regards to how the entire system is proposed to work.

But I'm still having trouble with how the rest of the system will work.  By adding the partitions you have effectively made many separate containers, not one big one through which the output unit can lower and flip.  If you are holding back the exact details of how this is designed to happen, I don't understand why.  But that is your prerogative.

M.

Mondrasek,

You got it!  The amount of work needed to empty and refill the container can be reduced to a very small amount.  The output unit produces a constant amount of energy.

I will be adding another video to show how we might decrease the energy needed to empty and refill the reservoir even more.  And finally, I will begin presenting specifics.  I will appreciate any help in discovering better designs.

Do you now see that perpetual motion is viable?  I want a group of people to help design, then build, a working system (possibly at a university).

Building and designing this thing as an individual is overwhelming.  I am merely the individual who brought up the feasible idea.  Everyone can help to create some final products.

Quote from: GavinPalmer1984 on January 28, 2009, 12:29:42 PM
Mondrasek,

You got it!  The amount of work needed to empty and refill the container can be reduced to a very small amount.  The output unit produces a constant amount of energy.

I will be adding another video to show how we might decrease the energy needed to empty and refill the reservoir even more.  And finally, I will begin presenting specifics.  I will appreciate any help in discovering better designs.

Do you now see that perpetual motion is viable?  I want a group of people to help design, then build, a working system (possibly at a university).

Building and designing this thing as an individual is overwhelming.  I am merely the individual who brought up the feasible idea.  Everyone can help to create some final products.

Gavin,

As I said from the start, I believe I understood all the pieces.  Including the way you planned to shift the water.  But I have yet to see you present a system that uses this effect that could work in it's entirety.  So no, I do not see that perpetual motion as you are describing it is viable.  Those specifics would appear to also be necessary from the start, for me at least.  Without them this idea appears to have huge gaps in logic.

M.

@Gavin
You're very determined with your idea... Nothing wrong with that!
But, maybe there's a possibility that you are the one who is missing something...?

Quote
People are claiming that I can not reduce the distance below X, where I claim that the distance can be reduced to 2 * X / n.
Just view my clip and tell me what does not make sense:
http://www.youtube.com/watch?v=bIA2rZQgO_c

That was never a real issue... You can reduce the distance to lift the same amount of liquid (mass) with the help of partitioning all way to the practical limits...
But it seems you're totally missing the "closing the loop" principle (so you can evaluate energy exchanges and a possibility for a surplus (OU)... ).
Your shifting mass or "drain, lift, drain" cycle uses reduced height difference with more partitions, but it also provides reduced energy potential (what you can get out of the system) in the first place...
I've seen the YT video showing "partitioning principle". Try to draw "partition containers" on the same side of the original container. Do you see a flaw now?  How about a combined "Center of Gravity" for all the partitioning containers wrt. the original one?

You are actually reducing a height needed with the partitioning scheme... But you are loosing the same amount of a "useful" energy which you can get out of it at the end, too.

A single reservoir (in a cycle) has a m(2h)g potential, with 2 partitions you have a m(2/2)hg, with 3 part. m(2/3)hg,...  8 partitions = m(2/8)hg , etc...

Quote
I do not think that people are grasping my point.  Do not approach this with a preconception.

Yes, you're right.  But I think that you should avoid having your own (personal) preconceptions, too! It seems you have a "fixed idea" about your concept (which may not be a good thing afterall...)

Quote
I have had experts at the National Science Foundation tell me that the partitioning scheme is very intelligent.  Once you guys understand the partitioning scheme, it follows that a perpetual system may be created.

I certainly don't know what or who or what the "NSF" is. And, regarding the "partitioning scheme", it is is interesting, but what - exactly - is it good for? Sorry to say, but I only see an overcomplicated mass handling... For me, absolutely nothing which can be linked to any kind of "Perpetual motion" possibility.. Sorry.

Quote
The guys at the NSF could not fathom a device which produces more energy than what is needed to reset it to the initial state.

This partitioning principle is far from being an overunity device. Please, see previous posts. Especially the ones made by the people like Hans and TinselKoala... They certainly know what they're talking about!

Quote
The liquid only drops a distance of:      X / n    , then is lifted a distance:     2 * X / n

And yes, the US NSF.  You guys should not assume that I am wrong.  I realize that I must convey my understanding to you, but please come into this with an open mind.  Go to my youtube post of partitioning and ask me questions instead of insisting I am wrong.

Yes, liquid drops a distance of X/n, then is lifted a distance of 2X/n... And then it is drained back to an original position with X/n . So, (X/n)-(2X/n)+(X/n) = complete cycle (which should be zero under ideal conditions...) Or, UNITY wrt. to all Energy interchanges (what you get OUT is what you put IN - minus losses)..

Quote
Mondrasek, you got it!  The amount of work needed to empty and refill the container can be reduced to a very small amount.  The output unit produces a constant amount of energy.

I will be adding another video to show how we might decrease the energy needed to empty and refill the reservoir even more.  And finally, I will begin presenting specifics.  I will appreciate any help in discovering better designs.

Do you now see that perpetual motion is viable?  I want a group of people to help design, then build, a working system (possibly at a university).

Building and designing this thing as an individual is overwhelming.  I am merely the individual who brought up the feasible idea.  Everyone can help to create some final products.

Gavin, If a "Perpetuum Mobile", or "OverUnity" would be THAT simple, It would already be invented centuries ago....   ;)

Please, don't take my post as a debunking attack. Like many (all) people here (or anywhere), I'd like to see the success!

Cheers!

Ignore potential energy inside of the partitioning scheme.  I have not yet addressed exactly how to use that potential.  Just realize the cost in emptying and refilling a container is very small when using my partitioning scheme.

The key point which I want spinner to understand:
The output from the overall system will come from the output unit.  The potential for the output unit is a constant amount.  And the output unit resets with a change in liquid depth.  So essentially, the reset mechanism can be reduced with partitioning.

I do not think that anyone has thought to do the partitioning scheme (even if they have) and combine it with a device which resets due to a change in liquid depth.

So I repeat:
The output comes from the output unit.  And the input to the system comes from lifting the partitions.
Output is constant:
m * d
and Input is variable(I am sure this can be reduced even more):
M * X / n

Where d < X and m < M

I will begin sharing more specific work soon.  Continue to ask questions please.

Quote from: GavinPalmer1984 on January 28, 2009, 05:06:04 PM
Ignore potential energy inside of the partitioning scheme.  I have not yet addressed exactly how to use that potential.  Just realize the cost in emptying and refilling a container is very small when using my partitioning scheme.

Ok... But I don't see the point.

Quote
The key point which I want spinner to understand:
The output from the overall system will come from the output unit.  The potential for the output unit is a constant amount.  And the output unit resets with a change in liquid depth.  So essentially, the reset mechanism can be reduced with partitioning.

I do not think that anyone has thought to do the partitioning scheme (even if they have) and combine it with a device which resets due to a change in liquid depth.

So I repeat:
The output comes from the output unit.  And the input to the system comes from lifting the partitions.
Output is constant:
m * d
and Input is variable(I am sure this can be reduced even more):
M * X / n

Where d < X and m < M

I will begin sharing more specific work soon.  Continue to ask questions please.

How the Hell is your "output constant" while "input is variable"  ??? Are you sure?

Ok, let's wait for the specifics....

Q: So, the catch is in the "output unit", not the "partitioning concept"?

The output unit has potential energy and exhausts that energy.  The output unit is in a container, and when that container is emptied half-way and refilled, the output unit resets, and the potential energy of the output unit goes back to what it was initially.

The original video demonstrates an output unit which can reset in a fashion described above.

Gavin, you are still missing the point that your partitioning scheme CANNOT reduce the work needed to reset the unit, past the amount of work gotten out of the output unit. It must do this, in order to repeat cycle or produce energy. But it cannot. Your partitioning scheme reduces the work from mg2h (which is unneccessarily high anyway) to mgh. But it cannot reduce it further.
It just occurred to me: Have you studied calculus, particularly differential calculus? Because you are thinking that an infinite series converges on Zero, when in fact your series converges on One.
For your device to function as intended, you must be able to reduce the work needed to reset the system to a value less than the work extracted. What you have described does not fulfill that criterion, because you are wrong about your partitioning system. Look at the top of the water in the receiving container. The height which this level rises, is the height that goes into the work equation mgh, no matter the number of partitions or how thin they are.

Quote from: TinselKoala on February 01, 2009, 02:49:57 PM
Gavin, you are still missing the point that your partitioning scheme CANNOT reduce the work needed to reset the unit, past the amount of work gotten out of the output unit. It must do this, in order to repeat cycle or produce energy. But it cannot. Your partitioning scheme reduces the work from mg2h (which is unneccessarily high anyway) to mgh. But it cannot reduce it further.
It just occurred to me: Have you studied calculus, particularly differential calculus? Because you are thinking that an infinite series converges on Zero, when in fact your series converges on One.
For your device to function as intended, you must be able to reduce the work needed to reset the system to a value less than the work extracted. What you have described does not fulfill that criterion, because you are wrong about your partitioning system. Look at the top of the water in the receiving container. The height which this level rises, is the height that goes into the work equation mgh, no matter the number of partitions or how thin they are.

Hi, TinselKoala!
Of course!
Gavin's problem is that he sees only the benefits of his partitioning principle....
Energy transactions should always be considered in it's "complete cycle", or in a closed loop, if we want to understand what is really happening....

Cheers!

Spinner and Tinsel,

  The top h * 1/n of the receiving container will fall a distance of h * 1/n into the first partition.  And the next h * 1/n will fall a distance of h * 1/n into the second partition.  Once all of the partitions are full, THE TOP OF THE TOP PARTITION IS ONLY h * 1/n BELOW THE RECEIVING CONTAINER'S TOP.  The partitions only need to be lifted 2 * h * 1/n.  Please look at the partitioning video and ask questions.

http://www.youtube.com/watch?v=bIA2rZQgO_c (http://www.youtube.com/watch?v=bIA2rZQgO_c)

I am very good at math.

Understand the partitioning scheme before you try to make a claim to the minimum work convergence.  You assume min. distance is h, when in fact it is 2 * h * 1/n.

All,

I'm with Gavin on this one.  His partition idea appears to allow the water in the container to move sideways, and not just down.  While the water in the main container does become only half as high, it does not do this by dropping a distance h.  It drops only one partition, or h/n.  It shifts *sideways* out of the main container, into a secondary partitioned container(s).  So to get the water back into the main container the secondary partitioned container must be raise only 2 * h/n.

The main container would have to have as many outlets as the number of partitions in the secondary container.  When lowering the main container, all of these outlets could be opened simultaneously.  In order to refill the main container (after raising the secondary partitioned container 2 *h/n) the partitions must be opened only one at a time, starting from the bottom.

Gavin, I also have realized that the partitions only need to exist in the secondary container and therefore would not be in the main container where I thought they would interfere with the output device falling and flipping.  All in all, I can't see why the whole idea is not possible in theory so far.  Very intriguing.

Thanks,

M.

Quote from: mondrasek on February 02, 2009, 12:45:29 PM
The main container would have to have as many outlets as the number of partitions in the secondary container.  When lowering the main container, all of these outlets could be opened simultaneously.  In order to refill the main container (after raising the secondary partitioned container 2 *h/n) the partitions must be opened only one at a time, starting from the bottom.

Gavin, I also have realized that the partitions only need to exist in the secondary container and therefore would not be in the main container where I thought they would interfere with the output device falling and flipping.  All in all, I can't see why the whole idea is not possible in theory so far.  Very intriguing.

Thanks,

M.

The main container would have as many outlets as there are partitions, and inlets as their are partitions. (unless an outlet for one stage could be an inlet for another stage or vise versa).  And just to clarify:

The top outlet from the main container opens first.  Once the top partition fills, The next lowest outlet from the main container opens.  This repeats downward until all of the partitions are full.  The partitions are lifted 2 * h/n and the bottom-most inlet on the main container allows the bottom partition to empty back into the main.  Once the bottom partition is empty, the next highest inlet is opened which will allow the next highest partition to enter the main.  Repeat upward.  Note that Opening and Closing of inlet/outlet should be autonomous and harness the Potential within the process itself (ie. not from output unit).

I am glad you are with me M.  Go ahead and ponder this one (I will do another youtube for it soon):

The outer partitions can be surrounded by a fluid so that each partition has enough mass to negate the air within the partition.  The air which exists in the partitions can be directed so that the output unit will only need to provide a small amount of air to lift 2 to 3 partitions only.  The air from the lower partitions will provide a buoyant force for an upper partition.  This concept is relatively new for me.  I will show a diagram ASAP.

Reset work = m * d * g
d = 2 * h/n
m = 2 * M/n  (this is with the added system which I will address soon)
M = mass of liquid exchanged in and out of main container
g = gravitational constant

Remember that this device will only serve to prove that perpetual motion is possible.  There will be many more-efficient devices to be discovered because of our work here.  We will lay the grounds for a new study of meta-physics (or interconnections of sub-systems).

After all, I am of the belief that the world is cyclical and on-going.  So that entropy oscillates within a window (whose min and max may oscillate as well?).  My theory and belief is that the world has existed for all time and will exist for all time.  And this device is a demonstration of my understanding of meta-physics ( a.k.a. nature ).  But let's stay focused on the task at hand :)-`

Quote from: GavinPalmer1984 on February 02, 2009, 10:06:06 AM
Spinner and Tinsel,

  The top h * 1/n of the receiving container will fall a distance of h * 1/n into the first partition.  And the next h * 1/n will fall a distance of h * 1/n into the second partition.  Once all of the partitions are full, THE TOP OF THE TOP PARTITION IS ONLY h * 1/n BELOW THE RECEIVING CONTAINER'S TOP.  The partitions only need to be lifted 2 * h * 1/n.  Please look at the partitioning video and ask questions.

http://www.youtube.com/watch?v=bIA2rZQgO_c (http://www.youtube.com/watch?v=bIA2rZQgO_c)

I am very good at math.

Understand the partitioning scheme before you try to make a claim to the minimum work convergence.  You assume min. distance is h, when in fact it is 2 * h * 1/n.

Gavin, your "partitioning principle" or the accompanying equations were never a real issue.... Your YT video explains it well, and most people actually understands it...

How F....d up you think we really are?  ;)

But you're obviously missing the point....

"We" never claimed that a min. distance is "h"....

And "we" agreed that a minimal distance is "2*h*1/n" with your part. principle.....

Now, try to evaluate the energy exchanges of your "partitioning scheme" in practice....
Do it with a real values, e.g. m(water)=1000kg, h=1m, g=10, n=10... (easy calcs)....
You can neglect (dismiss) all the possible losses (focusing for an ideal mechanical conversion)

So, a complete cycle energy exchange should be ZERO(0). No doubt about that. A conservation of energy never failed when observing the mass transfers in a gravity field (so far...)

Btw, the only equation you need (evaluating the mass transfers in a gravity field, or an ideal mechanical energy transactions) is a "m*g*h".... A gravitational potential energy says that you can get out of this (closed loop) cycle only what your input was..... Easy. Primary school stuff. Which never failed (to date...)

The complete (idealized) energy eq. for a partitioning scheme principle is: "Ep = n((m/n)*(h/n)*g)" = (mgh)/n
which equals to a known formula: "Ep = mgh" (without partitions ( n=1))....

Now, Wp is the maximum (theoretical) limit of any possible mechanical energy transaction (neglecting any and all the "real life" losses (we're - of course - not considering stuff like the atomic/molecular, zpe, "magic", etc... potentials )).

I do hope you'll see the flaw in your reasoning ("h" wrt. "h/n").
Or (much better), you'll show us something real which will shows us a mistake in "OUR" understanding....

Anyway, good luck!

Each container has sides and a base. If you split the water volume in 8 containers, you are lifting 8 times more container bases sections then with one. That is additional weight to factor into your idea.  You will also increase the transfer times since each of the 8 containers will have less starting head pressure. You will be transferring at the minimal speed 8 times. Taking the added bases plus the time factor all calculated as an energy expense, this idea is "dead in the water".

@ wattsup,

The weight of secondary partitioned containers could be balanced with a counterweight, just like the weight of an elevator is.  The added counterweight would add inertia to the whole system and therefore slow the acceleration of the raising water/container when whatever force is applied to do so.  But it would not raise the amount of force or energy required (minus additional friction... blah, blah, blah).

I don't understand your issue with how much time it takes to transfer the water.  Sure that fact would mean less "output energy' in a given amount of time (longer cycle time), but it would not decrease the "output energy" expected from the "output unit" in each cycle, right?

M.

Quote from: spinner on February 04, 2009, 06:03:07 AM
And "we" agreed that a minimal distance is "2*h*1/n" with your part. principle.....

Now, try to evaluate the energy exchanges of your "partitioning scheme" in practice....
Do it with a real values, e.g. m(water)=1000kg, h=1m, g=10, n=10... (easy calcs)....
You can neglect (dismiss) all the possible losses (focusing for an ideal mechanical conversion)

if the volume of the container is 1 m^3 then,
mass(water) = 1000 kg
work = W = m * d * g
d = 2 * h * 1/n = 2 * 1 * 1/10 = 0.2m
W = 1000 * 0.2 * 10 = 2000 J

If you have 100 partitions
W = 200 J

if you have 1000 partitions
W = 20 J

if you have 10000 partitions
W = 2 J

note that the size of the partitions are unrealistic with h = 1m.  But you should get the idea that the work needed to empty and refill a container is reduced by adding partitions.

Quote from: spinner on February 04, 2009, 06:03:07 AM
Btw, the only equation you need (evaluating the mass transfers in a gravity field, or an ideal mechanical energy transactions) is a "m*g*h".... A gravitational potential energy says that you can get out of this (closed loop) cycle only what your input was..... Easy. Primary school stuff. Which never failed (to date...)

Yes for the sub-systems.
The output unit outputs work = m1 * d1 * g.  And if you lift m1 up, and there are no losses, you can never get more output than input.
The partitioning work = m(water) * d * g.  And if you harness the falling of that water from a higher position to a lower position, you are once again correct.

BUT for the cooperation of the two sub-systems, our current understanding is WRONG.
The partitioning scheme resets the output unit by doing the work = m(water) * d * g via emptying and refilling a container;  creating potential, m1 * d1 * g in the output unit.  I have NOT even addressed using the falling m(water) as a source of energy.

Quote from: spinner on February 04, 2009, 06:03:07 AM
The complete (idealized) energy eq. for a partitioning scheme principle is: "Ep = n((m/n)*(h/n)*g)" = (mgh)/n
which equals to a known formula: "Ep = mgh" (without partitions ( n=1))....

Now, Wp is the maximum (theoretical) limit of any possible mechanical energy transaction (neglecting any and all the "real life" losses (we're - of course - not considering stuff like the atomic/molecular, zpe, "magic", etc... potentials )).

I do hope you'll see the flaw in your reasoning ("h" wrt. "h/n").
Or (much better), you'll show us something real which will shows us a mistake in "OUR" understanding....

Anyway, good luck!

Now listen up...  I will repeat:
Your analysis is right if you look at one of the trees (sub-system = partitioning scheme = output unit).  But if you look at the forest (partitioning scheme AND output unit), you can see what I have been stating all along... work needed to reset the output unit is dependent upon the number of partitions (aka variable) and the work produced by the output unit is constant.

Output unit is reset by emptying and refilling the reservoir.

So... you (and I hope I have changed your mind) and the modern ignorant physicists claim that the maximum amount of work that I can get out of this system (the forest = partitions and output unit) is equal to:
mgh/n  (idealized)

BECAUSE you think that I am going to harness the falling m(water)...  but NO!!  I am going to use the water to create a buoyant force on the output unit!

With hope that I have brought about an understanding, Gavin.

@mondrasek

I look at this as an energy in / energy out. Your first full container is emptied into one container that is then lifted to re-fill the first container. Compare that with the same full container emptied into 10 containers that is then requiring more energy to lift because each base adds to the weight, so you need more energy to lift it. You will also have to energize 10 latches to hold it there instead of one. Does not make sense.

@GavinPalmer1984

"I am going to use the water to create a buoyant force on the output unit!"

Any force of that type would render the highest and lowest container out of the fill level, meaning the top output container would not fill and there will be water left in the big container because the bottom output container will be too high. Either that or your explanation is really not clear of what you mean by buoyant force.

Usually guys do not make thing to do nothing unless you are working this as a perpetual  motion scheme. But here is nothing perpetual about it because you will never get more energy empty in the big container then you will get back by re-filling it due to the losses of lifting the output containers. So what's the big deal.

@wattsup
I think we have a language barrier.

Quote from: wattsup on February 04, 2009, 01:18:32 PM
@mondrasek

I look at this as an energy in / energy out. Your first full container is emptied into one container that is then lifted to re-fill the first container. Compare that with the same full container emptied into 10 containers that is then requiring more energy to lift because each base adds to the weight, so you need more energy to lift it. You will also have to energize 10 latches to hold it there instead of one. Does not make sense.

As mondrasek and I have said, the mass of the containers themselves can be negated by counter-weight / counter-forces.
You should not assume that we will need to energize latches.  And as I have mentioned, there is energy in my process which I have not even accounted for (aka. falling water, rising water in the partitions).

Quote from: wattsup on February 04, 2009, 01:18:32 PM
@GavinPalmer1984

"I am going to use the water to create a buoyant force on the output unit!"

Any force of that type would render the highest and lowest container out of the fill level, meaning the top output container would not fill and there will be water left in the big container because the bottom output container will be too high. Either that or your explanation is really not clear of what you mean by buoyant force.

Usually guys do not make thing to do nothing unless you are working this as a perpetual  motion scheme. But here is nothing perpetual about it because you will never get more energy empty in the big container then you will get back by re-filling it due to the losses of lifting the output containers. So what's the big deal.

Go look at the youtube video:
http://www.youtube.com/watch?v=bIA2rZQgO_c (http://www.youtube.com/watch?v=bIA2rZQgO_c)

You must have just come in here and started rambling without reading through the previous posts.  If you had read through the previous posts, you would address problems specifically and point to times in the video clips as reference.  You quote my conclusion...  what is your deal?  Please do not waste other people's time.  I have thought this thing out for over 2 years, and I am intelligent.  Please respect my work by putting forth more effort before claiming:

"But here is nothing perpetual about it because you will never get more energy empty in the big container then you will get back by re-filling it due to the losses of lifting the output containers. So what's the big deal. "

A respectful person will come in and ask questions instead of LEAPING to conclusions... you ignored mondrasek's previous reply to your misunderstanding.  Let me know if your native language is not English so that I might understand your lack of communication skills.

@GavinPalmer1984

On your video at the 0:50 mark, which is the container divided by two, you fill both containers and then you say that both containers only rise by x (d=x). That is not so. They both still rise at 2x and no matter how many containers you use, they will "always" have to rise at 2X otherwise you will not be able to fully empty the containers.

This thread is only 5 pages long so I did read it through but quickly cause I like spending my time on testing stuff.

No matter how many containers you use, they will always have to rise by 2x, not relative to your big container but to themselves. That's where the first 2x came from. 2 times the distance of the filled contained. So you can lift it once, or with 100 containers but you will always have to lift them at 2 x. So again, what's the big deal.

Quote from: wattsup on February 04, 2009, 03:53:42 PM
@GavinPalmer1984

On your video at the 0:50 mark, which is the container divided by two, you fill both containers and then you say that both containers only rise by x (d=x). That is not so. They both still rise at 2x and no matter how many containers you use, they will "always" have to rise at 2X otherwise you will not be able to fully empty the containers.

Each container will need to rise 2 * H/n where H/n = the height of one of the small containers and H = the height of the large container.  In the video, X = H.

Quote from: wattsup on February 04, 2009, 03:53:42 PM
This thread is only 5 pages long so I did read it through but quickly cause I like spending my time on testing stuff.

No matter how many containers you use, they will always have to rise by 2x, not relative to your big container but to themselves. That's where the first 2x came from. 2 times the distance of the filled contained. So you can lift it once, or with 100 containers but you will always have to lift them at 2 x. So again, what's the big deal.

You are right.  They must all rise two times the height of themselves.  So that every container will rise 2 * H/n.

So the work needed to reset the output unit = M * D * g
M = mass of water
D = 2 * H/n
g = gravitational constant
n = number of partitions

You may also say work = n * (M/n * D * g )
This means that each container with mass = M/n must be lifted a distance D = 2 * H/n.  The work required to lift one partition is = M/n * D * g.

Using more partitions decreases the distance which the water must travel in order to empty and refill the original container.

"Using more partitions decreases the distance which the water must travel in order to empty and refill the original container."

Exactly....... but since each container still has to rise at double its height, using one container or 100 containers will still require the same amount of work.

But the drawback in using so many containers is the time require to empty the large container will increase since each consecutive level will have the minimal head pressure to work with and that same will apply to the many containers emptying into the big container. More time.

Quote from: wattsup on February 04, 2009, 04:32:23 PM
"Using more partitions decreases the distance which the water must travel in order to empty and refill the original container."

Exactly....... but since each container still has to rise at double its height, using one container or 100 containers will still require the same amount of work.

You are wrong.  Just slow down and think through the problem.

If you use one container which is 1 meter high, and 1000kg of water, and assume gravity = 10m/s^2:
Work = 1000kg * 2 meters * gravity = 20000 J

If you have 2 partitions:
Work = 2 * (500kg * 1 meter * gravity) = 10000 J

If you have 4 partitions:
Work = 4 * (250kg * 0.5 meter * gravity) = 5000 J

Notice how the mass remains 1000kg.  And the distance decreases.

@GavinPalmer1984

So we're OK on the 2x thing.

Now, I get your math. Took me a little TIME to figure it out. So here goes.,,,,

The first calculation is based on gravity (9.8 m/s² or 32 ft/s²) so we say 10. 10 meters per second squared. This second is not calculated in your formula. Lifting the first 1000kg in one second means the speed of rise is the fastest.

The 2x500kg speed of rise will be halved because it has half the distance to rise IN THE SAME SECOND. So in the first you are lifting 1000kg in one second one meter high and you cannot compare it to lifting the same 2x500kg that only travels 1/2 a meter in the same second. To make the same comparison the 2 x 500kg would have to rise the 1/2 meter in 1/2 a second. And that makes them equal. As so on............

Time is on times side.

@wattsup

The time it takes to empty and refill the container does not matter because I am only proving that perpetual motion is possible.  I have previously stated that there will be more efficient perpetual motion devices to come... especially when people stop believing in our current interpretation of the laws of physics.

So yes, adding more partitions will increase the time it takes to empty and refill the container.

And it will decrease the total work needed to empty and refill the container.

@GavinPalmer1984

This is not about the time it takes to empty the containers. It is only about the time it takes to make the lift. That is the only work here.

Try this little quiz I made.

Put the following in order of least to most energy required.

1) Lifting 1000kg - 1 meter high - in one second
2) Lifting 1000kg - 1 meter high - in three seconds
3) Lifting 1000kg - 1/2 meter high - in one second
4) Lifting 1000kg - 1/2 meter high - in one half second

Hope this will get the point across.

@Gavin
It seems to me we're just having a big misunderstanding....

If it is of any help for you (so you can go forward with explaining your device/concept), I'd like to confirm that I understand:

>>The partitioning scheme (which enables a proportionally the lowest energy input to actually completely  drain and then fully refill a certain amount of water(mass close loop displacement)).

>> In this thought experiment, we're not considering any losses

>>and (as Gavin said) the energy (mass in a gravity field) mechanical equivalents are not the prime concern of this project...

Is this a consensus? ;)


So, I'd like you to continue with your explanations....
Several posts ago, I asked you about the whole picture (not just the part. scheme).

How is yours output unit supposed to operate?

You are mentioning that the real thing is - indeed -  happening in this "output unit".

Buoyancy... Mind you, the buoyancy is just a gravity "in disguise"....



Looking forward to hear more,
Cheers!

Quote from: wattsup on February 04, 2009, 10:28:29 PM
@GavinPalmer1984

This is not about the time it takes to empty the containers. It is only about the time it takes to make the lift. That is the only work here.

Try this little quiz I made.

Put the following in order of least to most energy required.

1) Lifting 1000kg - 1 meter high - in one second
2) Lifting 1000kg - 1 meter high - in three seconds
3) Lifting 1000kg - 1/2 meter high - in one second
4) Lifting 1000kg - 1/2 meter high - in one half second

Hope this will get the point across.

This does bring up a valid point in that more energy is required to lift the masses at faster rates.  But If we assume that the partitions have much time to rise, then the system can work.  The trick is transferring the work of the output unit to the lifting of the partitions in an intelligent way... I know it can be done.  (I will put up a new link explaining how to decrease the work of the partitioning even more... previously mentioned concerning partitions within a tank)

Think about it this way... The distance is decreased by increasing the partitions.  Lets assume that we want to lift the partitions in 3600 seconds.  As that distance decreases, the amount of energy needed to lift those partitions will also decrease.

Thank you for bringing this point up, as I have overlooked it previously.  I will have a link to another video up within 4 days, outlining the benefits of using intelligently designed partitions within a separate container of liquid... possibly decreasing the mass which is lifted to that of 2 * M/n.  Each partition would be lifted by an opposing buoyant force.

And so that I address wattsup's point directly, the design will have to allow the energy from the output unit to be expended over a large amount of time... and I know this is possible.

So yes, time is a bigger consideration than I have previously accounted for, but I have already thought of an idea to efficiently convert the output unit energy into a usable source of resetting the partitions (no matter how many partitions you use)

Thank you wattsup... please continue your efforts with myself.  And as a note to others who have something for me, it seems that the quiz style remark sparked my ability to understand where the author is coming from.

Quote from: spinner on February 05, 2009, 06:31:32 AM

>>The partitioning scheme (which enables a proportionally the lowest energy input to actually completely  drain and then fully refill a certain amount of water(mass close loop displacement)).

Exactly... and the partitioning scheme is not fully developed.

Quote from: spinner on February 05, 2009, 06:31:32 AM
>> In this thought experiment, we're not considering any losses

Yes and No.  Consider them so that the design avoids as much loss as possible. ( I am posting to get help from the online community... there are many brilliant minds out there)

Quote from: spinner on February 05, 2009, 06:31:32 AM
>>and (as Gavin said) the energy (mass in a gravity field) mechanical equivalents are not the prime concern of this project...

The mass of the output unit is the source to lift the partitions.  The emptying and refilling (half-way) of the output unit's container is the source to reset the output unit.

Quote from: spinner on February 05, 2009, 06:31:32 AM
So, I'd like you to continue with your explanations....
Several posts ago, I asked you about the whole picture (not just the part. scheme).

How is yours output unit supposed to operate?

The whole design is a work in progress.  I bring to the table many ideas which are possible.  I like to start broad so that others may work out other designs based upon my original concepts.

I think that the best output unit will be one in which the mass pushes air out of its container, which is used in conjunction with the partitioning scheme (with the partitions inside of another container), and air goes into the bottom of the partitioning container.  The air will then be used to provide buoyant force upon the bottom partitions.

I will not go into the "newer" partitioning model just yet... let me get a new video up with the 2.0 idea which will not only decrease the distance which M(water) will travel, but will also decrease the amount of work which the mass(output unit) must contribute to lifting all of the partitions(because only 2-3 of the bottom partitions will require a buoyant force... all other partitions will have received buoyant force from a lower partition).

I am taking baby steps... it is the best way in my opinion.

@GavinPalmer1984

This forum has always been a very good place to test your ideas and I am the last one to knock anyone down if the idea holds up or not. As long as we can all learn more. Better having it analyzed here then shooting it out in the real world because the remarks and objections will not be as quaint. Actually they can be pretty rough here too. lol

Now that you appreciate the "real" task at hand if there is something else that we are not aware of, then I am sure guys here will be as helpful as they can. Please think it over very carefully and don't assume anything because nature will not assume anything and will only react in the way it can, not in the way one wishes. (Don't I know it.)

All the best in your further endeavors.

Quote from: wattsup on February 04, 2009, 10:28:29 PM
@GavinPalmer1984
Try this little quiz I made.

Put the following in order of least to most energy required.

1) Lifting 1000kg - 1 meter high - in one second
2) Lifting 1000kg - 1 meter high - in three seconds
3) Lifting 1000kg - 1/2 meter high - in one second
4) Lifting 1000kg - 1/2 meter high - in one half second

Hope this will get the point across.

Energy required for the partition to get up to max velocity:
W1 = (m * d1 * g) + (1/2 * m * v^2)

Energy required for the partition to move with acceleration = 0, aka constant velocity:
W2 = (m * d2 * g)

Energy required for the partition to slow down to v = 0 and reach destination:
W3 = 0

So the overall work will be:
W = (m * d * g) + (1/2 * m * v^2)
d1 + d2 = d ~= 2 * H/n

So:
Work(1) = 1000kg * 1m * 10 + 1/2 * 1000kg * 1
Work(1) = 10000J + 500J = 10500J
Work(2) = 10000J + 56J   = 10056J
Work(3) = 5000J + 125J   =   5125J
Work(4) = 5000J + 500J   =   5500J

Realistic dimensions would mean that the height (y-dimension) of the output unit's holding container would be roughly equal to the depth(z-dimension).  The width(x-dimension) would be larger than the output unit's width.  The height and depth would be roughly twice the height of the output unit.  So each partition would have a width and depth which is equal to the output unit's container's width and depth.

Let's use these dimensions:
Output Unit:
(y-coordinate)height = 22 meters
(x-coordinate)width = 3.7 meters
(z-coordinate)depth = 3.7 meters

Let us assume that the mass of the output unit itself is zero and only the falling mass (piston) is accounted for.  We will also ignore resistive forces for now.  There are efficient geometric shapes which should be used... I may not be using the most efficient shapes.

mass of piston < Volume of output unit  * density of water
m(p) < V(ou) * p(h2o)
this is where you would account for a temperature window so that the output unit is always buoyant no matter the temperature... temperature affects density.
m(p) < 22 * 3.7^2 * 980kg/m^3
m(p) < 295,156.4 kg

The mass will be able to fall roughly 1/2 of the output unit's height.
Output Work = m * d * g
OW = 295,156.4 * 10 * 9.81
OW = 28954842.84 J - (f1 * 10)
f1 = friction on inner walls over 10 meters

Output Unit's Container:
height = 45 meters
width = 4 meters
depth = 45 meters

mass of water exchanged = m(h2o)
remember, water exchanged must be greater than the height of the output unit.
m(h2o) = Volume of h2o * density of h2o
m(h2o) = 4 * 45 * 22 * 980
m(h2o) = 3,880,800

Partition:
width = 4 meters
depth = 45 meters

ignore inner-operations for now (there is potential energy which can be harvested from other sources within the process)
Input Work = work needed to exchange the m(h2o) = IW
IW = (m * d * g) + (1/2 * m * v^2) + (f2 * d)
d = 2 * 22/n
n = number of partitions
m = m(h2o) = 3,880,800kg
f2 = friction encountered while lifting

We need:
OW > IW
OW = 28,954,842.84 J - (f1 * 10)
IW = (m * d * g) + (1/2 * m * v^2) + (f2 * d)
28,954,842.84 - (f1 * 10) > (3,880,800 * 2 * 22/n * 9.81) + (1/2 * 3,880,800 * v^2) + (f2 * 2 * 22/n)

Ignoring friction:
28,954,842.84 > (3,880,800 * 2 * 22/n * 9.81) + (1/2 * 3,880,800 * v^2)
n = 100;
v = 0.22 m/s;
height of a partition = 0.22 meters;

28,954,842.84 > (3,880,800 * 2 * 22/n * 9.81) + (1/2 * 3,880,800 * v^2)
28,954,842.84 > (16751085.12) + (93915.36)
28,954,842.84 > 16,845,000.48
Excess work =  12,109,842.36 J

with friction:
28,954,842.84 - (f1 * 10) - staticfriction1 > (3,880,800 * 2 * 22/n * 9.81) + (1/2 * 3,880,800 * v^2) + (f2 * 2 * 22/n) + staticfriction2
28,954,842.84 - (f1 * 10) - staticfriction1 > (1675108512 / n) + (1,940,400 * v^2) + (f2 * 44/n) + staticfriction2
n = 100;
v = 0.22 m/s;
height of a partition = 0.22 meters;

28,954,842.84 - (f1 * 10 meters) - staticfriction1 > (16751085.12) + (93,915.36) + (f2 * 0.44 meters) + staticfriction2
28,954,842.84 > 16,845,000.48 + (f2 * 0.44) + (f1 * 10) + staticfriction1 + staticfriction2

Relatively bad coefficients of kinetic and static friction are 1.0 and higher.
Force of friction = Normal force * coefficient of static friction + normal force * coefficient of kinetic friction.
Unrealistic worst-case friction:
normal force is related to the area of contact between materials and the pressure exerted.
width of seal = 0.05m
the piston is cube shaped = 4 sides that are 1m long
use 2 seals
normal pressure might be 75,000 N/m^2
for staticfriction = kinetic friction (I am using bad coefficients so this is safe)
f1 = Area * Pressure * coefficient of friction= (0.05m * 1m * 4 * 2) * (75,000) * 1.0 = 30,000 N = staticfriction1

28,954,842.84 > 16,845,000.48 + (f2 * 0.44) + (f1 * 10) + f1 + staticfriction2
28,954,842.84 > 16,845,000.48 + (30,000 * 10) + 30,000 + (f2 * 0.44) + staticfriction2
28,954,842.84 > 17,175,000.48 + (f2 * 0.44) + staticfriction2

28,954,842.84 - 17,175,000.48 > f2 * 0.44 + staticfriction2
11,779,842.36 > f2 * 0.44 + staticfriction2


if f2 + staticfriction2 / 0.44 < 26,772,369, then this device can definitely be perpetual.

check this out:
OW = m1 * d1 * g - f1 * d1
IW = m2 * d2 * g + f2 * d2

Doubling the dimension:
OW = 2*(m1 * d1 * g - f1 * d1)
IW = 2*(m2 * d2 * g + f2 * d2)

All of the masses double.
d1 will double.
d2 can stay the same... aka double the partitions.

If you ignore friction, doubling dimensions and partitions will quadruple OW but only double the IW.
The work done by friction for OW will quadruple because of the increased distance for f1 and the area of contact for f1.

I will not make the claim that the work done by friction for the OW will increase with doubling the dimensions... that just limits the innovation of the design.

Also, I know that there are better geometric shapes which should be used... I happened to use bad ones, and still brought forth great evidence that perpetual motion is possible.

I should add that any other losses within the system are a direct result of the design and implementation to achieve the desired (basic) operation.  For example, creating electricity from kinetic energy has inherent losses.  And storing / transferring that electricity has large inefficiencies.  The designers should conclude, that the use of electricity to maintain a desired state of the device should be avoided... Unless the electrical conversion/storage/transfer processes are increased to an efficiency which is at least equivalent to the inherent resistant forces (friction) within a system which avoids the use of electricity (compare direct mechanical work through pulleys and levers to that of an electric motor) for the same purpose.

Gavin,

I believe there is a problem with how you think the output device will work.  Once it has been lowered below the catches and the water level returned to full height, the output unit will not flip.  Even though it is buoyant and will want to rise, the majority of the weight will be at the bottom.  Much like the ballast in a boat, having the CG of the output unit at the bottom will keep the unit from being able to flip back to the surface.  It will want to rise, but only straight up, not flip.  You will not be able to reset the weight in the output unit using this system as shown so far.

Thanks,

M.

Quote from: mondrasek on February 07, 2009, 05:25:37 PM
Gavin,

I believe there is a problem with how you think the output device will work.  Once it has been lowered below the catches and the water level returned to full height, the output unit will not flip.  Even though it is buoyant and will want to rise, the majority of the weight will be at the bottom.  Much like the ballast in a boat, having the CG of the output unit at the bottom will keep the unit from being able to flip back to the surface.  It will want to rise, but only straight up, not flip.  You will not be able to reset the weight in the output unit using this system as shown so far.

Thanks,

M.

I think you are right.  The mass will need to be smaller and will be determined by the amount of displaced water which is lower than and/or level with the mass.

But the point still stands (although a hole has been shot in the master-plan):
output unit is constant.
reset work is variable.

I appreciate the help... maybe there is a better output unit design... the previous detailed example will not work because the mass in the output unit is too large.

For output unit:
From some previous calculations which are not shown on this forum yet, the most output is produced when the mass's height is half of the height of the container which it will slide within.  Having said that, the new and accurate measurement for the mass of the output unit will be figured as follows:

Assume that the output unit itself has no mass.
v = volume of output unit
m = sliding mass within the output unit.
p = density of liquid which output unit floats within

m < 1/2 * v * p

therefore, previous calculation's ammendments:

previously Ignoring friction:
28,954,842.84 > (3,880,800 * 2 * 22/n * 9.81) + (1/2 * 3,880,800 * v^2)
n = 100;
v = 0.22 m/s;
height of a partition = 0.22 meters;

28,954,842.84 > (3,880,800 * 2 * 22/n * 9.81) + (1/2 * 3,880,800 * v^2)
28,954,842.84 > (16751085.12) + (93915.36)
28,954,842.84 > 16,845,000.48
Excess work =  12,109,842.36 J

now ignoring friction with the more realistic mass estimation:
1/2 * 28,954,842.84 > (3,880,800 * 2 * 22/n * 9.81) + (1/2 * 3,880,800 * v^2)
n = 200;
v = 0.11 m/s;
height of a partition = 0.11 meters;

14,477,421.42 > 8,375,542.56 + 23,478.84
14,477,421.42 > 8,399,021.4
Excess work = 6,078,400.02 J

Notice how the mass of the output unit's work was halved, I then compensated by increasing the number of partitions and decreasing the velocity.  Please notice that increasing the partitions will decrease the reset work.  And also notice that I have better  calculations for the mass within the output unit (although I ignored the mass of the output unit itself because I am just proving a point)

please reply.

And I will add that with less mass, the top of the output unit will float above the liquid's water level, thus increasing the amount of water which most be emptied in order for the top of the output unit to fall below the uni-directional latches.  But do remember that the partitioning can minimize the work needed to exchange the water no matter how much water needs to be emptied and refilled.

Hi Gavin,

I had a closer look at your last video
http://www.youtube.com/watch?v=bIA2rZQgO_c

and you are totally right with your theory !

Great results...


If you look at this video for only just these 2 frames at:
1:57 min and
1:59 min
You can clearly see, that only 2 containers are lifted !
All the other containers of water just stay at their place !

So just let us calculate, how much energy we need to lift these
2 containers.

The lower left lowest water container  must be lifted excatly 1.10 Meter, if the height of the
main container is 1 Meter.
and the right lowest water container must be lifted 0.9 Meter.

So the energy required to lift the left container is:
Eleft = mass/11 x 10m/s^2 x 1.1 Meters
and the energy rquired to lift the right container is:
Eright = mass/11 x 10m/s^2 x 0.9 Meters

So now assuming we have a mass of 1000 KG in the main container we get:

The main output container can extract a lifting work for e.g.
a ship of about:
Eoutput= 1000 Kg x 10 m/s^2 x 1 meter= 10.000 Joules

Energy input into the 2 left and right containers is :

Eleft = 1000Kg /11 x 10m/s^2 x 1.1 Meters= 1000 Joules
plus
Eright = 1000Kg/11 x 10m/s^2 x 0.9 Meters= 818
so the total input energy is just only 1818 Joules
and the output energy is 10.000 Joules.

So with 11 Containers we have a COP of 5.5 !

Not too bad !

The only thing which must be seen is, that the boat which will
swimm on the output main tank must not sink too much into the water, so a
wooden or otherwise light substance must be used which does not drag the water down
but can lift the full M x G x H weightenergy.by pulling these external weights up with ropes
from the buoyancy of the swimming boat.

Well done Gavin ! Great system.

Regards, Stefan.

 :)

I have a question:
How much energy will be used opening, closing, and controlling flow from all of the valves required to move this water back and forth?

The buoyant part in the main tank must either weigh more than two partitions full of fluid plus the buoyant force, or have a buoyant force of more than the weight of two partitions full of fluid in order to provide the energy for the lift.

Is this lift to be done with the main tank full or empty?

This is a great idea, but I am not sure all of the energy required to operate and reset is currently being calculated correctly.

Quote from: truth on February 10, 2009, 10:30:46 PM
:)

I have a question:
How much energy will be used opening, closing, and controlling flow from all of the valves required to move this water back and forth?

The buoyant part in the main tank must either weigh more than two partitions full of fluid plus the buoyant force, or have a buoyant force of more than the weight of two partitions full of fluid in order to provide the energy for the lift.

Is this lift to be done with the main tank full or empty?

This is a great idea, but I am not sure all of the energy required to operate and reset is currently being calculated correctly.

The energy used opening and closing one gate will depend on the design.  If the energy needed to empty and close an inlet/outlet is less than the available potential within a partition and other available non-used potential energy, then the current equations will suffice.  I have one design for the inlets and outlets which would work, but would depend upon the partitions moving at separate times.  There are still obstacles, but I see that this type of device can be made perpetual.

dimensions from previous example:
Output Unit:
(y-coordinate)height = 22 meters
(x-coordinate)width = 3.7 meters
(z-coordinate)depth = 3.7 meters

Output Unit's Container:
height = 45 meters
width = 4 meters
depth = 45 meters

I MADE A MISTAKE ON THE OUTPUT UNIT'S CONTAINER.  THE DEPTH ONLY NEEDS TO BE SLIGHTLY LARGER THAN THE HEIGHT OF THE OUTPUT UNIT FOR THE ROTATION TO TAKE PLACE!!!
new depth = 23 meters

Output Unit's Container = 45 * 4 * 23 = 4140 m^3
Output Unit                  = 23 * 3.7 * 3.7 = 314.8 m^3

the Output Unit is roughly 1/13 * 1/2 = 1/26 of the mass of the Output Unit's Container.  This is using still inefficient geometries.  The Output Unit's Container would best be shaped somewhat like a half-cylinder which would decrease the mass of the liquid which must be emptied and refilled.

The Output Unit will usually be a proportion of the Output Unit's container.  It is in no relation to a constant amount of partitions.  Its relationship to partitions would be variable, depending on the number of partitions used.

The lifting of the partitions should be done with the Output Unit Container's liquid at minimum level... which means that all of the partitions are full.

How, exactly, is your "OU" (output unit) supposed to operate???

The overall energy potential limits are known. So far, no surplus....
:)

Quote from: spinner on February 12, 2009, 06:41:36 AM
How, exactly, is your "OU" (output unit) supposed to operate???

The overall energy potential limits are known. So far, no surplus....
:)

If you assume that the inlets/outlets are operated by potential within the partitions and that you can transfer the potential from the output unit in a manner that can allow you to dissipate the energy over any amount of time, then there is a surplus in the output unit when using the partitioning scheme.  I want you to see this before we go any further.

The potential of the Output Unit will be something like:
m * x * g
m > 1/26 * M
M = mass of the liquid within the output unit's container. (note that you do not have to empty the whole container and the geometries can be made more efficient which would better the ratio of m : M)
x > 1/5 * H
H = height of the output unit's container
g = gravity.
It simply follows that the Output Unit has a constant amount of output and the partitioning has a variable amount of input which can be reduced to a number close to zero if the dimensions are chosen properly.  But we only need the input to be less than the output.

Assuming we empty the whole container:
IE = mass * distance * gravity + 1/2 * mass * velocity^2
IE = M * 2 * H/n * g + 1/2 * M * V^2
Because we can dissipate the OE over any amount of time 1/2 * M * V^2 can go to zero
IE = M * 2 * H/n * g
OE = 1/5 * 1/26 * M * H * g
with above assumptions:
M * 2 * H/n * g < 1/5 * 1/26 * M * H * g
2 * H/n < 1/130 * H
2 * H < 1/130 * H * n
260 < n

So there you have it.  But realistically, the number of containers will be much less because I assumed we empty the whole output container and the output container's geometry is inefficient.  N would increase depending on the friction encountered and velocity chosen.

@hartiberlin

lift 2 containers and n is even:
(2 * M/n) * (2 * (H + H/n)) * g = 2 * M * 1/n * 2 * g * ( H + H/n)
mass = 2 * M/n
distance = 2 * H + 2 * H/n
g = gravity

lifting all the containers:
2 * H/n * M * g

2 * H * 1/n * M * g ~~ 2 * M * 1/n * 2 * g * (H + H/n)
H ~~ 2 * (H + H/n)

Therefore, lifting all of the containers would be more efficient AND less friction would be encountered.  But thank you for the recognition.  And also notice that even using your less efficient method of partitioning will still decrease the reset work when you increase the number of partitions by decreasing the mass.  The distance which the containers must be lifted will always be greater than H.

I should really post my addition which decreases the mass along with decreasing the distance. (the idea of submerging the partitions within another container of liquid and using the rising air from one partition intelligently so that the air in the bottom partition can be used to provide a buoyant force for at least half of the partitions. And the air in the second lowest partition can provide a buoyant force to the other half of partitions.)

The addition would decrease the work to roughly:
2 * M/n * 2 * H/n * g
mass = 2 * M/n
distance = 2 * h/n
g = gravity

All, have a look at this principle,
it is still in German language, but have a look at the embeded FLASH animation.

Maybe you can use this for your output principle ?

http://www.overunity.de/index.php?topic=74.0

Have a look at the second posting there with the
embeded FLASH animation, wait a few seconds until the
arrow comes up and click the arrow.
Then you will see the great animation there.


Regards, Stefan.

Quote from: hartiberlin on February 13, 2009, 02:09:35 AM
All, have a look at this principle,
it is still in German language, but have a look at the embeded FLASH animation.

Maybe you can use this for your output principle ?

http://www.overunity.de/index.php?topic=74.0

Have a look at the second posting there with the
embeded FLASH animation, wait a few seconds until the
arrow comes up and click the arrow.
Then you will see the great animation there.


Regards, Stefan.

I like it.  I have attempted to do something similar with buoyancy only, but I encountered problems with water pressure.  I will give you my interpretation of the device.  I will not assume that you are using hydraulics because it is not necessary, although it is capable of being a hydraulic system.  I will use your middle point as a reference and pretend that it does not actually exist.

It looks like your large interior mass (LM) is lifting the counterweight (cw).  The narrow cylinder is buoyant, therefore its mass (nm)is less than the mass of the liquid it displaces.

LM > 4 * cw
nm < liquid displaced by narrow cylinder... This provides a mechanism to ensure that the flip will not begin until a sufficient amount of offset has been reached... it is vital.
I will assume nm = 0 for now.
liquid displaced by narrow cylinder's mass is  = qm

So I guess you are trying to ensure that part of LM will remain on the upper portion of the output unit, so that the upper mass of the output unit (UM) is larger than the bottom mass of the output unit (BM)

UM > BM will enable a flip.

I will go ahead and tell you that at 90 degrees in the flip, the masses will start moving again... this means you need a sufficient offset to ensure a complete 180 degree flip.  This is what the buoyant cylinder achieves.  The buoyant cylinder also provides kinetic energy which can be harnessed in multiple ways.

When the buoyant cylinder rises to the top, the BM will increase by the amount equal qm.  As LM falls, the masses of BM and UM are affected.

I also suggest that the counterweight only needs to move from the lower half to the upper half.  It does not need to go beyond the mid-point.

I will use druckmasse as a time reference and assume that the qm is at the BM and I ignore equivalent liquid masses:
At druckmasse = 31: I would call this max BM:
volume of (1/8 * LM) * density of liquid = vM
BM = qm + cm + 3/8 * LM + vM
UM = -qm + nm + 5/8 * LM
and nm = 0;
UM = -qm + 5/8 * LM
BM = UM + cm + qm + vM - 1/4 * LM
BM > UM iff (cm + qm + vM > 1/4 * LM)

At druckmasse = 32: I would call this equilibrium of cm and LM:
BM = qm + 1/2 * cm + 1/2 * LM
UM = -qm + nm + 1/2 * cm + 1/2 * LM
and nm = 0;
UM = -qm + 1/2 * cm + 1/2 * LM
BM = UM + qm

At druckmasse = 33: This should be max UM
BM = qm + 5/8 * LM
UM = -qm + nm + cm + 3/8 * LM + vM
and nm = 0;
UM = -qm + cm + 3/8 * LM + vM
UM = BM + cm - qm + vM - 1/4 * LM
UM > BM iff (cm + vM > 1/4 * LM + qm)

And to be realistic:
(cm + vM > 1/4 * LM + qm) AND
(cm + qm + vM > 1/4 * LM) AND
(qm > 0) AND
cm < 1/4 * LM AND
2 * cm < cm + 1/4 * LM < 1/2 * LM

if qm = 0, then the equations are satisfied.  But qm must be greater than zero to ensure that we do a successful 180 degree turn.  Substitute (qm - nm) = qm for realistic buoyant cylinder. qm will be constant.

vM - (1/4 * LM - cm) > qm
qm = displaced volume of nm * density of liquid - nm  (liquid is at BM and nm is at UM)
vM = displaced volume of 1/8 * LM * density of liquid
p = density of liquid
Volume (1/8 * LM) * p > Volume of nm * p - nm
This will allow us to begin choosing a valid qm.

A low density matter for LM will be ideal for increasing the mass of vM which will allow for a larger qm.  This helps to ensure that the device achieves 180 degree flip.

Good job.  Let me know if I completely altered your idea.  I know how this would work because I have thoroughly investigated such a device.  It is amazing that we are nearly on the same page.  Also note that the buoyant cylinder would be the main source of output energy via a sliding mass within it and/or it being connected to another pulley system.  I was waiting to disclose my idea for a more efficient device like you have outlined, but your idea uses the same concepts in a better way.  I congratulate your work because I know how unique it is.

You might consider investigating hydraulics instead of pulleys.  Pulleys will have more losses than hydraulics.

I started with my initial design because I wanted to get my partitioning scheme out on the internet.  This is definitely a more efficient perpetual motion endeavor.  There are complications in implementing the device.  I will think about your device and we should definitely collaborate to improve your design with some of my ideas.

I would add that there is no reason why you should have the cm and pulleys in air.  I would have the whole thing operating within the reservoir... pulleys, cm, cables, and all.  The buoyant cylinder would be your output unit.  I am no expert on pulleys, so this might be the crux.

I wrote a physics journal and they write back:

Dear Mr Palmer,

I cannot accept your manuscript "Cooperative Systems" [PoF 090-0170-l] for publication as a Letter in the Physics of Fluids. You propose a perpetual motion machine, which is known to be impossible.

You need a free lift of the water. That is available in tides. In fact there is an interesting "pumping trick" of pumping into the tidal lagoon at high tide, an energy cost which is repaid with interest at low tide. You might wish to read Chapter 14 (pages 81-87) with Appendix G (pages 311-321) of David JC MacKay's book "Sustainable Energy - without the hot air", a free book available on the internet at www.withouthotiar.com

Yours sincerely,
John Hinch
Associate Editor for Letters
Physics of Fluids

_______________________________

I responded with:

Mr. Hinch,
I do not say that I need a free lift.

The lift of partitions will require work =
1/2 * m * v^2 + m * g * d + f * d
m = mass of liquid exchanged (assume partition's mass is countered)
v = max velocity of the lift
g = gravity
f = friction encountered (decreases with distance)
d = distance which the partitions must be lifted.
d = 2 * X/n
X = height of the container amount of liquid exchanged.

Please notice that as the number of partitions increase, the distance which the mass is lifted decreases.  You and the majority of people may assume that perpetual motion is impossible, but we once assumed the earth was flat as well.  You can't see that the earth is NOT flat unless you first acknowledge that our current understanding may be wrong.  I know for a fact that I can defend this proposal... I have not had one person disprove my process.  I beg you to investigate this further.  Please point to your areas of confusion.  I suggest that you stay away from the laws of thermodynamics and conservation of energy when addressing this manuscript.  There is no way to disprove my process with a law which I am attempting to prove inaccurate.

please,
  Gavin Palmer

____________________________

@hartiberlin

Did you ever read through my suggestions on your device?  I hope I helped.

So, Gavin are you surprised with the answer?

The "Physics Journal editor" answered you... (that is a success, most of similar, controversial letters never get any replies....)

What did you expect, claiming a "Perpetual motion", on paper, without any proof...???

The partitioning principle....  Aha.

And it seems that a "Flat Earth" stuff wasn't very helpfull, either....



If you want to break "The Laws"... Do that with a decent "proof of a concept" device.
Cheers!

HartiBerlin has a model that I may try to construct with my additions which I mentioned previously.  Tesla mentioned that everything can be done in the opposite way.  I had a device that was opposite to Harti's, but water pressure doubles every 33 feet.  The pressure of air is not as substantial.

I was using a device that rotated due to buoyancy with a gravity engaged imbalancer, while Harti's rotates due to gravity and has a buoyancy engaged imbalancer.  My design didn't account for the water pressure.  I do not see any complications with Harti's model, but I will find out if there are any issues.  I still think my design would have worked if I had furthered the design, but I was trying for something simple... and Harti's design retains simplicity.

I hope to let you know how it goes.

Bump...

Remember my concept of partitioning:

http://www.youtube.com/watch?v=bIA2rZQgO_c

And this is an example of how partitioning works in general:

http://www.physorg.com/news184268881.html

==================================

I have had other ideas as well that are similar.  Have you ever seen somebody wake surf?

http://www.youtube.com/watch?v=Q9Wy98Btfzk&feature=related

You have many devices that can ride a wake and all of them are in parallel acting as batteries, each generating energy.

At a certain number of wake-riding-energy-gathering-devices along a certain length wake, there would be enough energy to fuel the wake-creating device.

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