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Author Topic: Overunity electrolysis - 31 times more effective gas production than with DC  (Read 119458 times)

Offline sparks

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One could apply AC at any frequency they like, but then have the problem that H2 and O2 would both evolve at each electrode.  The energy required to completely strip eight electrons is huge.  The likelihood that any such process would improve efficiency is next to nil.


  The potential between plates would have to be intense and brief.   The accelerating force is the electric field between the capacitor plates.  Electric fields tend to accelerate electrically charged particles like electrons.  The energy used to charge the cell is used to create the electric field.  This energy is in potential form.  The discharge of the capacitor into the inductance converts it from potential energy stored in the capacitor to potential energy stored in the magnetic field surrounding the inductor.  The cell voltage could reach very high potential creating a very intense electric field permeating the fluid.   This voltage threshold depends on the smoothness of the plates and the k of the insulating film as well as the time of the imposed emf.  We need only to deform the electron orbitals enough to lay the hydrogen protons bare.  We need not ionize them but move them within their orbitals.  The hydrogen nuclei without the electronic insulation effect are accelerated away from each other and the oxygen neucleus  due to like charges repel.  The hydrogen atoms don't need ionization energy to remove themselves from the oxygen bond.  The coulumb force becomes the force driving the molecular fracturing not the applied emf.  Monatomic hydrogen would form first as the hydrogen proton meets electron.  Then hydrogen gas.  The oxygen would now be left with too many electrons in it's deformed valence shell and upon field relaxation become stable as monatomic oxygen until it reacted to form o2 or o3. 
   

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Offline ltseung888

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In normal electrolysis case 1, the cathode and anode plates are well separated so to facilitate the collection of gases.
 
In Browns Gas production, the cathode and anode plates are often placed close together.  The gas coming out is "mixed".
 
Can the Browns Gas have a different composition (or electron cloud distribution) as case 1?  This will have very important implications.  Different Browns Gas production methods may have very different gas compositions.
 
More on this later.

Offline MarkE

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  The potential between plates would have to be intense and brief.   The accelerating force is the electric field between the capacitor plates.  Electric fields tend to accelerate electrically charged particles like electrons.  The energy used to charge the cell is used to create the electric field.  This energy is in potential form.  The discharge of the capacitor into the inductance converts it from potential energy stored in the capacitor to potential energy stored in the magnetic field surrounding the inductor.  The cell voltage could reach very high potential creating a very intense electric field permeating the fluid.   This voltage threshold depends on the smoothness of the plates and the k of the insulating film as well as the time of the imposed emf.  We need only to deform the electron orbitals enough to lay the hydrogen protons bare.  We need not ionize them but move them within their orbitals.  The hydrogen nuclei without the electronic insulation effect are accelerated away from each other and the oxygen neucleus  due to like charges repel.  The hydrogen atoms don't need ionization energy to remove themselves from the oxygen bond.  The coulumb force becomes the force driving the molecular fracturing not the applied emf.  Monatomic hydrogen would form first as the hydrogen proton meets electron.  Then hydrogen gas.  The oxygen would now be left with too many electrons in it's deformed valence shell and upon field relaxation become stable as monatomic oxygen until it reacted to form o2 or o3. 
 
"We need only to deform the electron orbitals enough to lay the hydrogen protons bare.  We need not ionize them but move them within their orbitals. "  Do you mean you want to ionize the hydrogen?  How else are you going to increase the polarization?

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Offline MarkE

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In normal electrolysis case 1, the cathode and anode plates are well separated so to facilitate the collection of gases.
 
In Browns Gas production, the cathode and anode plates are often placed close together.  The gas coming out is "mixed".
 
Can the Browns Gas have a different composition (or electron cloud distribution) as case 1?  This will have very important implications.  Different Browns Gas production methods may have very different gas compositions.
 
More on this later.
If there were a unique gas composition that could have been shown by spectral analysis a long time ago.

Offline sparks

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      Actually want to increase the water molecule polarization.  The two "shared" electrons move within the oxygen valence orbital.  So you could say the hydrogen and oxygen atoms become ionized.  The hydrogen atoms take on a plus one charge while the oxygen atom takes on -2 charges.  [size=78%]The oxygen atom itself becomes polarized with excess electrons moved towards the positive plate. This leaves the two hydrogen protons and the positive pole of the oxygen atom all working to fracture the molecule.[/size](http://www.brooklyn.cuny.edu/bc/ahp/SDPS/graphics/PolarWater.GIF)

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Offline ltseung888

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 Let me quickly review our discussion here.

1.    The Lead-out or Bring-in Energy Machine diagram is in reply 109.  It is the most important slide in our discussion.  The Lead-out Energy Machine is NOT the impossible Perpetual Motion Machine.  It requires energy to run.  It brings-in energy from the environment to run and contributes to Global Cooling.
The simplest example is the water bottle rocket where the temperature of the bottle is cooler after firing.  James Kwok tried to build a machine Hidro (reply 112) to take advantage of this principle.

2.    Browns Gas has the property of Implosion.  On ignition, the gas mixture implodes instead of explodes.  However, adding water vapor or other gases will turn the gas mixture to exhibit explosion.
A simple Implosion working diagram was presented on reply 130.
Some engineering ideas on how to modify the standard 4 stroke engine to use implosion is in reply 106:

 https://www.youtube.com/watch?v=ndpk3qAJS-8&list=PL7AF455C0A317C691&index=3

3.    Three scenarios related to electrolysis and Browns Gas are presented in reply 117.  Scenario A is the classical electrolysis where the cathode and anode are well separated to collect the resulting gases.  The reaction can be reversed by feeding the resulting hydrogen and oxygen gases into a fuel cell.  No Lead-out or Bring-in Energy from the environment is involved.  Energy for combustion must be supplied to do the electrolysis.  Scenario B is to use DC Pulsing to break the water molecule up into some form of HHO gas with much less energy.  Again, no lead-out or Bring-in Energy from the environment is involved.  Scenario C is to add Implosion to Scenario B.  Implosion brings-in the kinetic energy of the air molecules, cooling the environment.  Scenario C can become the basis of a Bring-in Energy machine running forever after starting as in reply 109.  However, there is much engineering difficulty in ensuring that Implosion does not change into Explosion.

In the next post, I shall describe the more interesting Scenario D of “resonance breaking” of the dipole water molecules.

Offline MarkE

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@MarkE,
Thank you very much for spending much time in doing the diagram.  I obviously do not understand some of your terms or accounting.  Thus I simplified it to my understanding.  Please explain your diagram in detail so that all can benefit.
Lawrence, ATM are quantities of the surrounding atmosphere environment, CYL are quantities of the cylinder, ST1 are quantities at State 1, ST2 are quantities at State 2, and ST3 are quantities at State 3.  P is pressure, V is volume, E is energy.  You incorrectly misaccount the amount of work performed displacing the air in the cylinder in State 1 with HHO in State 2.  The actual work that must be performed is not just the change in energy state within the cylinder, but the change in energy state of the SYSTEM, which includes the environment. 

It is total nonsense that you count on the environment for energy as part of your system to provide what you call LEAD IN energy, but ignore the energy that you have to drive into the environment when filling the cylinder with HHO.  When you drive the HHO into the cylinder, the energy in the cylinder increases by only the pressure difference times the cylinder volume, but the energy in the surrounding atmosphere into which you expel an identical volume of air increases by the starting pressure of the air times the cylinder volume.  By filling the cylinder with HHO you compress a gas spring in the surrounding atmosphere.  When the HHO burns and the gas volume changes to a many times smaller liquid volume, the atmospheric gas spring pushes the piston back to the top of the cylinder and in doing so relaxes.  Ignoring the water volume, the system: volume, pressure, and energy at State 3 is the same as at State 1.  I have explained your error a number of times now.  The diagram explains the actual energy relationships in each state in sufficient detail that any high school physics student can easily understand.

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Offline MarkE

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      Actually want to increase the water molecule polarization.  The two "shared" electrons move within the oxygen valence orbital.  So you could say the hydrogen and oxygen atoms become ionized.  The hydrogen atoms take on a plus one charge while the oxygen atom takes on -2 charges.  [size=78%]The oxygen atom itself becomes polarized with excess electrons moved towards the positive plate. This leaves the two hydrogen protons and the positive pole of the oxygen atom all working to fracture the molecule.[/size](http://www.brooklyn.cuny.edu/bc/ahp/SDPS/graphics/PolarWater.GIF)
Well I think you have a big problem.  You can apply so much field that you partially or fully ionize the molecule.  What I don't see any means to do is to apply an external field that increases the polarization within the molecules.  How do you think that you can increase the polarization within individual molecules by applying an external field?

Offline MarkE

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Let me quickly review our discussion here.

1.    The Lead-out or Bring-in Energy Machine diagram is in reply 109.  It is the most important slide in our discussion.  The Lead-out Energy Machine is NOT the impossible Perpetual Motion Machine.  It requires energy to run.  It brings-in energy from the environment to run and contributes to Global Cooling.
The simplest example is the water bottle rocket where the temperature of the bottle is cooler after firing.  James Kwok tried to build a machine Hidro (reply 112) to take advantage of this principle.
Post 132 clearly illustrates the fallacy of post 109. James Kwok was convicted of securiteis crimes.  His HidroCo is moribund and no longer promotes his claims.
Quote

2.    Browns Gas has the property of Implosion.  On ignition, the gas mixture implodes instead of explodes.  However, adding water vapor or other gases will turn the gas mixture to exhibit explosion.
A simple Implosion working diagram was presented on reply 130.
Some engineering ideas on how to modify the standard 4 stroke engine to use implosion is in reply 106:

 https://www.youtube.com/watch?v=ndpk3qAJS-8&list=PL7AF455C0A317C691&index=3

3.    Three scenarios related to electrolysis and Browns Gas are presented in reply 117.  Scenario A is the classical electrolysis where the cathode and anode are well separated to collect the resulting gases.  The reaction can be reversed by feeding the resulting hydrogen and oxygen gases into a fuel cell.  No Lead-out or Bring-in Energy from the environment is involved.  Energy for combustion must be supplied to do the electrolysis.  Scenario B is to use DC Pulsing to break the water molecule up into some form of HHO gas with much less energy.  Again, no lead-out or Bring-in Energy from the environment is involved.  Scenario C is to add Implosion to Scenario B.  Implosion brings-in the kinetic energy of the air molecules, cooling the environment.  Scenario C can become the basis of a Bring-in Energy machine running forever after starting as in reply 109.  However, there is much engineering difficulty in ensuring that Implosion does not change into Explosion.

In the next post, I shall describe the more interesting Scenario D of “resonance breaking” of the dipole water molecules.
The energy associated with the difference in energy between the gas volume at any pressure and disassociated water is a source of energy loss in any electrolysis system.  In the most ideal case of a machine that were to completely recover that energy difference, an electrolysis / combustion machine would be limited to the Carnot efficiency of the heat engineutilysing the hydrogen combustion as its input heat source.

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Offline ltseung888

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Lawrence, ATM are quantities of the surrounding atmosphere environment, CYL are quantities of the cylinder. P is pressure, V is volume, E is energy. You incorrectly misaccount the amount of work performed displacing the air in the cylinder in State 1 with HHO in State 2. The actual work that must be performed is not just the change in energy state within the cylinder, but the change in energy state of the SYSTEM, which includes the environment.
 
 It is total nonsense that you count on the environment for energy as part of your system to provide what you call LEAD IN energy, but ignore the energy that you have to drive into the environment when filling the cylinder with HHO. When you drive the HHO into the cylinder, the energy in the cylinder increases by only the pressure difference times the cylinder volume, but the energy in the surrounding atmosphere into which you expel an identical volume of air increases by the starting pressure of the air times the cylinder volume. By filling the cylinder with HHO you compress a gas spring in the surrounding atmosphere. When the HHO burns and the gas volume changes to a many times smaller liquid volume, the atmospheric gas spring pushes the piston back to the top of the cylinder and in doing so relaxes. Ignoring the water volume, the system: volume, pressure, and energy at State 3 is the same as at State 1. I have explained your error a number of times now. The diagram explains the actual energy relationships in each state in sufficient detail that any high school physics student can easily understand.
 

@MarkE,
 
Thank you for your explanation. 
 
 
 
From a pure Physics point of view of work performed by the piston, it is clear:
 
1.   Work done by the piston to fill the cylinder with HHO is the (HHO-Atm)pressure x Volume of the Cylinder.  HHO pressure can be 101% of Atm pressure.  The work done can then be 1% of Atm x Volume of Cylinder.
 
2.   Work done by the piston to fill the vacuum in the implosion process is Force x Displacement.  The force is Atm, Pressure x area of the Cylinder.  The Displacement is the height of the cylinder.  The work done is then equal to Atm pressure x area x height which is equal to Atm Pressure x Volume of Cylinder.  This is 100% times of stage 1.
 
From the pure Mechanical Consideration, the work done by the HHO gas to fill the cylinder is 1% of the work done by the Atm Pressure.  We assume the HHO gas has been generated and stored in a tank with HHO pressure 1% more than Atm.  The Energy used to generate the gas is NOT considered in this example.
 
 
 
May be we should let others comment on such calculations.  I consider the subject closed.  The work done or energy supplied by the Environment in one cycle = Atm Pressure x Volume of Cylinder.  This is the Bring-in Energy.
 

Offline MarkE

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@MarkE,
 
Thank you for your explanation. 
 
 
 
From a pure Physics point of view of work performed by the piston, it is clear:
 
1.   Work done by the piston to fill the cylinder with HHO is the (HHO-Atm)pressure x Volume of the Cylinder.  HHO pressure can be 101% of Atm pressure.  The work done can then be 1% of Atm x Volume of Cylinder.
Once again, you stubbornly focus on a subset of the system and ignore the system as a whole leading to your accounting error.  The ideal gas law still holds.  One does not get to displace the volume of air that one displaces with HHO without performing work on the external atmosphere equal to Patmosphere_initial*V of the cylinder.  Your fundamental error on which your false conclusions are based is easily seen by accounting for the identical loss of PHHO*VCYLINDER from whatever source you use for the HHO that you drive into the cylinder.  Why do you present yourself as unable to understand these obvious facts?
Quote

2.   Work done by the piston to fill the vacuum in the implosion process is Force x Displacement.  The force is Atm, Pressure x area of the Cylinder.  The Displacement is the height of the cylinder.  The work done is then equal to Atm pressure x area x height which is equal to Atm Pressure x Volume of Cylinder.  This is 100% times of stage 1.
The work released is (ignoring the water volume) identical to the work performed above.  The surrounding atmosphere acts as a gas spring Lawrence.  No more and no less.  The atmosphere is passive.  It does not generate energy.  It stores energy that you stubbornly refuse to account.
Quote

From the pure Mechanical Consideration, the work done by the HHO gas to fill the cylinder is 1% of the work done by the Atm Pressure.  We assume the HHO gas has been generated and stored in a tank with HHO pressure 1% more than Atm.  The Energy used to generate the gas is NOT considered in this example.
You insist on this absolutely incorrect accounting.  You are dead wrong Lawrence for the reasons that I have explained repeatedly.
Quote

 
 
May be we should let others comment on such calculations.  I consider the subject closed.  The work done or energy supplied by the Environment in one cycle = Atm Pressure x Volume of Cylinder.  This is the Bring-in Energy.
Then you insist on rejecting obvious fact in favor of a fantasy that you wish to entertain.

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Offline ltseung888

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 Scenario D – “resonance break up” of Water molecules into some gas mixtures.
 
This is the most controversial part of the discussion.  Let me first quote assumptions that will not violate the Laws of Physics.  These assumptions may or may not have been verified experimentally.
 
1.    Water Molecules are electrical dipoles.  The oxygen end is more negative than the hydrogen end.  These dipoles can be compared with magnetism in an iron rod.  Tapping an iron rod in the North and South direction can turn the iron rod into a magnet.
2.    DC Pulsing water can be compared with tapping the iron rod.  The electric field is externally supplied by the DC voltage.  The pulsing is similar to tapping.  It will change the Electron Cloud distribution of the “water particles”.  I consider “water particles” as any cluster of ions, atoms, molecules, crystal structures etc.
3.    The frequency of the Pulse is important.  In the field of Physics, there is the occurrence of resonance.  A continuous small external stimulation or vibration may build up to a huge internal vibration (resulting in destruction of structure in some cases.
4.    I assume that the DC Pulsing at the right frequency may alter the Electron Cloud Distribution and even break the hydrogen and oxygen bonds of the water particles.  This DC Pulsing may be done by an external Pulsing Source or may be via LCR resonance circuits. L is inductance or coils, C is capacitance and R is resistance.[/font]
5.    Radio stations rely on our tuning to the right frequency.  Missing that frequency means not receiving that station.  If we miss the “resonance frequency”, we may not generate much gas.  The parameters governing such “resonance frequency” have not been scientifically laid out yet.  It is a “hit or miss” operation.  One technique by experimenters is to observe the sudden increase in rate of “gas mixture” production.
6.    The Indian paper claimed such a behavior.  There are at least 10 other similar claims on the Internet.  Most readers are familiar with claims related to the Stan Meyer or the Browns Gas generators.  The Indian paper is done in a more traditional scientific fashion with peer reviews and details of the establishment.  The results can be checked with access to the actual source.
7.    My assumption is that the “gas mixture” generated in the Stan Meyer type device where the cathode and anode are placed very close to each other is different from the hydrogen and oxygen gases generated from classical electrolysis.  The Electron Clouds are different.  The “gas structures” are different.  Thus the energy contents associated are different.
8.    I further assume that the microscopic level or mechanism of breaking up of water molecules in classical electrolysis is different from DC Pulsing.  The classic Faraday energy model does not apply.
9.    I took the hint that – Browns Gas will lose its unique properties such as implosion when left for hours in a jar unused.  One explanation is the receiving of electromagnetic waves from the environment.  I assume the “gas mixture” has less energy than the hydrogen and oxygen gases from classical electrolysis.
 
Let me pause at this point and wait for comments.

Offline MarkE

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Scenario D – “resonance break up” of Water molecules into some gas mixtures.
 
This is the most controversial part of the discussion.  Let me first quote assumption that will not violate the Laws of Physics.  These assumptions may or may not have been verified experimentally.[/font]
 
1.    Water Molecules are electrical dipoles.  The oxygen end is more negative than the hydrogen end.  These dipoles can be compared with magnetism in an iron rod.  Tapping an iron rod in the North and South direction can turn the iron rod into a magnet.[/font]
2.    DC Pulsing water can be compared with tapping the iron rod.  The electric field is externally supplied by the DC voltage.  The pulsing is similar to tapping.  It will change the Electron Cloud distribution of the “water particles”.  I consider “water particles” as any cluster of ions, atoms, molecules, crystal structures etc.[/font]
3.    The frequency of the Pulse is important.  In the field of Physics, there is the occurrence of resonance.  A continuous small external stimulation or vibration may build up to a huge internal vibration (resulting in destruction of structure in some cases).[/font]
4.    I assume that the DC Pulsing at the right frequency may alter the Electron Cloud Distribution and even break the hydrogen and oxygen bonds of the water particles.  This DC Pulsing may be done by an external Pulsing Source or may be via LCR resonance circuits. L is inductance or coils, C is capacitance and R is resistance.[/font]
5.    Radio stations rely on our tuning to the right frequency.  Missing that frequency means not receiving that station.  If we miss the “resonance frequency”, we may not generate much gas.  The parameters governing such “resonance frequency” have not been scientifically laid out yet.  It is a “hit or miss” operation.  One technique by experimenters is to observe the sudden increase in rate of “gas mixture” production.[/font]
6.    The Indian paper claimed such a behavior.  There are at least 10 other similar claims on the Internet.  Most readers are familiar with claims related to the Stan Meyer or the Browns Gas generators.  The Indian paper is done in a more traditional scientific fashion with peer reviews and details of the establishment.  The results can be checked with access to the actual source.[/font]
7.    My assumption is that the “gas mixture” generated in the Stan Meyer type device where the cathode and anode are placed very close to each other is different from the hydrogen and oxygen gases generated from classical electrolysis.  The Electron Clouds are different.  The “gas structures” are different.  Thus the energy contents associated are different.[/font]
8.    I further assume that the microscopic level or mechanism of breaking up of water molecules in classical electrolysis is different from DC Pulsing.  The classic Faraday energy model does not apply.[/font]
9.    I took the hint that – Browns Gas will lose its unique properties such as implosion when left for hours in a jar unused.  One explanation is the receiving of electromagnetic waves from the environment.  I assume the “gas mixture” has less energy than the hydrogen and oxygen gases from classical electrolysis.[/font]
 
Let me pause at this point and wait for comments.
You rely on a series of assumptions and claims made over the years none of which have ever been successfully proven.  come up with a reliable replication of the claim of your choice and then there will be evidence worth talking about.  The Indian paper is poorly written, lacking a host of data that is necessary for the claims made to hold any credibility.  Stan Meyer was found by an Ohio court to have committed an "egregious fraud".  Nearly 20 years later, no one has successfully reproduced what Stan Meyer claimed to have done.  You do your own credibility no favors when you insist on invalid accounting and keep adding to a history of promoting unworkable ideas.

Offline ltseung888

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 Scenario D continued – “resonance break up” of Water molecules into some gas mixtures.

10. Zero point energy is often referred to the energy that will still be present when the temperature reaches absolute zero degrees.  The assumption is that thermal energy is zero.  However, in any space, there will always be electromagnetic waves and gravitational forces.  I assume that electromagnetic wave is the “Lead-out” or “Bring-in” Energy Source in this case.


11. The “gas mixture” may contain different levels of electromagnetic waves.  At the “resonance frequency”, small stimulus from the outside may excite the “internal set up of the gas mixture” to break up the hydrogen and oxygen bonds.  This break up will release energy much larger than the sum of the outside stimulus.  This is the “Lead-out or “Bring-in” Energy Source.  In other words, the Energy is already available in the surrounding environment and the “water or gas mixture”.  It is a matter of releasing them.  Such energy can easily be replenlished by the electromagnetic waves from the surrounding (or from Zero Point energy as some would like to call it).


12.  If a “resonance setup” can release such energy, the Stan Meyer car is theoretically possible.  The Stan Meyer device does not use implosion as it deliberately fed exhaust gas back into the combustion chamber – making use of explosion instead of implosion.


13. If the “resonance setup” is to be detected via the “hit and miss” method, the chance of getting a reproducible “hit” and accepted by the scientific community will be like winning the lottery.  Skeptics will keep saying – fraud and scam.


14. The successful team is likely to be one with Government Support.  They do not need to worry about mortgages and living expenses.  They can buy the most expensive and accurate scientific instruments.  (I learned the lesson of using the US$120 Atten Oscilloscope to try to get accurate results from the tiny Joule Thief Energy experiments).  They can have trained scientists and engineers to tune the equipment and correctly analyze the results.  They can build special scientific apparatus to do mid-stage verifications.  They can focus on “resonance tuning” for months or years without the wife or husband nagging.  They do not need to publish results and get jeered at.  They do not need to worry about venture capitalists or investors wanting to see results for their money or suing them in court.


15. I would expect the successful team will have to overcome the objections from other scientists who claim that since Perpetual Motion Machine is theoretically impossible, they are waiting their time and the Government Resources.  They may have to use the Lead-out or Bring-in Energy theory to overcome this first hurdle.  They have to study and re-think electrolysis; implosion; chemical reactions; electron cloud distributions; kinetic theory of gases; confirm the adiabatic energy exchange of the water bottle rocket; structure of “water gas”; resonance and resonance circuits, etc.  A whole research team or a new department at a University may need to be set up.  Many PhD students may be used to check out the hundreds of overunity claims on the Internet.


16. I expect my contribution will just be the Lead-out or Bring-in Energy theory.  Other posts will be treated as “speculation” from an old man who spends much time fishing, talking about politics and religion.  These posts will not be taken seriously.  This is perfectly acceptable.  Divine revealation may surprise us all.  A "speculation" may turn out to be the scientific truth.
 
More to come later.

Offline ltseung888

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You rely on a series of assumptions and claims made over the years none of which have ever been successfully proven.  come up with a reliable replication of the claim of your choice and then there will be evidence worth talking about.  The Indian paper is poorly written, lacking a host of data that is necessary for the claims made to hold any credibility.  Stan Meyer was found by an Ohio court to have committed an "egregious fraud".  Nearly 20 years later, no one has successfully reproduced what Stan Meyer claimed to have done.  You do your own credibility no favors when you insist on invalid accounting and keep adding to a history of promoting unworkable ideas.
Buy a toy water rocket.  Measure its temperature before and after the firing. 

 

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