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Author Topic: quentron.com  (Read 989909 times)

Offline profitis

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Re: quentron.com
« Reply #1965 on: February 01, 2014, 12:41:30 PM »
this reversable cyle.totaly compatible with and predictable by college textbook rules @sarkeizen

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Re: quentron.com
« Reply #1965 on: February 01, 2014, 12:41:30 PM »

Offline Philip Hardcastle

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Re: quentron.com
« Reply #1966 on: February 01, 2014, 03:43:38 PM »
Alternative Sebby Diagram to help people understand how asymmetry makes it work.

Offline MarkE

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Re: quentron.com
« Reply #1967 on: February 01, 2014, 08:51:55 PM »
Alternative Sebby Diagram to help people understand how asymmetry makes it work.
May we begin with a completely symmetric version without the 6eV mesh?  Would you agree that such a version will not perform any work on its own?  Is it not obvious to you that you must then add any work to that system that you take out?

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Re: quentron.com
« Reply #1967 on: February 01, 2014, 08:51:55 PM »
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Offline Philip Hardcastle

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Re: quentron.com
« Reply #1968 on: February 02, 2014, 09:59:29 AM »
May we begin with a completely symmetric version without the 6eV mesh?  Would you agree that such a version will not perform any work on its own?


That is of course what I have always said. The point was before in trying to explain complex surface and interfacial physics the audience did not seem to appreciate the other issues, so when a few days ago someone asked me to explain it again I came up with the dual vacuum diagram, it has the simple advantage that absent the 6eV mesh it is by mere logic determined to be in balance, this takes away the complications.


The significant thing therefore is what happens when you self bias a 6eV platinum mesh off the 5eV collector, think about it, it is obvious.



Offline MarkE

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Re: quentron.com
« Reply #1969 on: February 02, 2014, 10:24:14 AM »

That is of course what I have always said. The point was before in trying to explain complex surface and interfacial physics the audience did not seem to appreciate the other issues, so when a few days ago someone asked me to explain it again I came up with the dual vacuum diagram, it has the simple advantage that absent the 6eV mesh it is by mere logic determined to be in balance, this takes away the complications.


The significant thing therefore is what happens when you self bias a 6eV platinum mesh off the 5eV collector, think about it, it is obvious.
Since eV are a measure of energy, inserting a 6eV source inserts a tiny bit of energy.  It seems obvious to me that virtually nothing happens.

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Re: quentron.com
« Reply #1969 on: February 02, 2014, 10:24:14 AM »
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Offline Philip Hardcastle

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Re: quentron.com
« Reply #1970 on: February 02, 2014, 10:42:42 AM »
Since eV are a measure of energy, inserting a 6eV source inserts a tiny bit of energy.  It seems obvious to me that virtually nothing happens.


Then you have not brought your intellect to the task.


What is the state of charge of the 6eV mesh when it comes into contact with a metal of 5eV work function, what is the chemical potential of the 5eV metal when it reaches equilibrium without the mesh?


I think you need to see the mesh as a region that is more negative than the collector, this being so it has a few effects, the one that is well understood, and the reason for S3 grids in pentodes, is that it suppresses secondary electron emissions from the collector (plate).


The other effect is that the presence of an electric charge on the mesh in proximity to the collector will cause the chemical potential at the 5eV metal surface to be lowered, but at the same time the charge of the whole collector rises, that is to say the chemical potential at the left hand surface of the 5eV metal must therefore rise, and if so it means that the equilibrium of the left hand side is disturbed, electron must flow from the left hand side 5eV, this then means that the left hand side 1eV chemical potential must rise from the equilibrium point it would have absent the grid on the right hand side becoming charged, and of course the increase of chemical potential in the 1eV on the left hand side translates to a rise in chemical potential on the right hand side 1eV surface, so it (the RHS emitter) more readily emits electrons into the RHS gap.


Now we know that the grid has a charge higher than the RHS 5eV metal surface, so we know that hot tail electrons from the RHS 1eV surface has to do work to get past the mesh, so hot tail electrons present in the population are selectively (by dint of their own energy) allowed to reach the 5eV RHS surface only after they have cooled (done work) getting past the charged grid of the 6eV mesh.


I suppose this is what you mean by it is obvious that virtually nothing happens? :-X

Offline MarkE

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Re: quentron.com
« Reply #1971 on: February 02, 2014, 12:15:41 PM »

Then you have not brought your intellect to the task.


What is the state of charge of the 6eV mesh when it comes into contact with a metal of 5eV work function, what is the chemical potential of the 5eV metal when it reaches equilibrium without the mesh?
If you introduce 1E-18J, which is what ~6eV is then unless you are at an atomic scale, which as soon as you say the word "mesh" you aren't, you haven't done much of anything.  Are you certain you are saying what you intend to say?  Are you saying that you want to introduce 6eV into the system or a material with a 6eV work function?  The latter would make more sense but still doesn't get you anywhere.
Quote


I think you need to see the mesh as a region that is more negative than the collector, this being so it has a few effects, the one that is well understood, and the reason for S3 grids in pentodes, is that it suppresses secondary electron emissions from the collector (plate). 


The other effect is that the presence of an electric charge on the mesh in proximity to the collector will cause the chemical potential at the 5eV metal surface to be lowered, but at the same time the charge of the whole collector rises, that is to say the chemical potential at the left hand surface of the 5eV metal must therefore rise, and if so it means that the equilibrium of the left hand side is disturbed, electron must flow from the left hand side 5eV, this then means that the left hand side 1eV chemical potential must rise from the equilibrium point it would have absent the grid on the right hand side becoming charged, and of course the increase of chemical potential in the 1eV on the left hand side translates to a rise in chemical potential on the right hand side 1eV surface, so it (the RHS emitter) more readily emits electrons into the RHS gap.


Now we know that the grid has a charge higher than the RHS 5eV metal surface, so we know that hot tail electrons from the RHS 1eV surface has to do work to get past the mesh, so hot tail electrons present in the population are selectively (by dint of their own energy) allowed to reach the 5eV RHS surface only after they have cooled (done work) getting past the charged grid of the 6eV mesh.


I suppose this is what you mean by it is obvious that virtually nothing happens? :-X
Really, nothing happens.  We can and do put metals with different work functions near each other all the time and they sit there happily in thermodynamic equilibrium all day long even though there is an internal electric field that results from the difference in work functions.

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Re: quentron.com
« Reply #1971 on: February 02, 2014, 12:15:41 PM »
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Offline Philip Hardcastle

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Re: quentron.com
« Reply #1972 on: February 02, 2014, 12:28:49 PM »
What I thought was obvious, and I accept your statement as being confused on the issue, was that it was the work functions of the materials. The chemical potentials will of course come to equilibrium in a circuit with no persistent current (the diagram shown but without the mesh). Clearly a circuit producing a persistent current must, ipso facto, have a non uniform chemical potential of end surfaces.


As to your saying nothing happens that simply goes against the proved use of a suppressor grid in a pentode tube.


Sorry MarkE but if you cannot see it from that diagram then I cannot make you see it, no amount of argument from me is going to sway your view.

Offline MarkE

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Re: quentron.com
« Reply #1973 on: February 02, 2014, 12:35:34 PM »
What I thought was obvious, and I accept your statement as being confused on the issue, was that it was the work functions of the materials. The chemical potentials will of course come to equilibrium in a circuit with no persistent current (the diagram shown but without the mesh). Clearly a circuit producing a persistent current must, ipso facto, have a non uniform chemical potential of end surfaces.


As to your saying nothing happens that simply goes against the proved use of a suppressor grid in a pentode tube.


Sorry MarkE but if you cannot see it from that diagram then I cannot make you see it, no amount of argument from me is going to sway your view.
Mr. Hardcastle pentodes do not spontaneously drive energy into external circuits.  External power supplies supply operating power to pentodes.  If you think differently, then you can devise and publish an experiment that you think will show otherwise.

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Re: quentron.com
« Reply #1973 on: February 02, 2014, 12:35:34 PM »
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Offline Philip Hardcastle

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Re: quentron.com
« Reply #1974 on: February 02, 2014, 01:18:34 PM »
Mr. Hardcastle pentodes do not spontaneously drive energy into external circuits.  External power supplies supply operating power to pentodes.  If you think differently, then you can devise and publish an experiment that you think will show otherwise.


I have all the proof I need backed up by others doing the same experiment, all you are showing in your statement is that you are prejudiced by the view that the 2nd is absolute.


MarkE, with due respect because I can see you are a professional in science, get a pentode, wire it as shown in my $10 experiment, heat it in a fancy oven and measure it with all the care you can, then tell me what you know.

Offline MarkE

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Re: quentron.com
« Reply #1975 on: February 02, 2014, 02:33:06 PM »

I have all the proof I need backed up by others doing the same experiment, all you are showing in your statement is that you are prejudiced by the view that the 2nd is absolute.


MarkE, with due respect because I can see you are a professional in science, get a pentode, wire it as shown in my $10 experiment, heat it in a fancy oven and measure it with all the care you can, then tell me what you know.
Mr. Hardcastle I have not made any statements about the 2nd ( I presume you mean the Second Law of Energy ).  I have simply stated the fact that two materials separated by a vacuum can have different work functions with the result that there is a static electrical field, and yet the system is in thermodynamic equilibrium.  IE, if one were to connect an external passive circuit, the existence of that field would not cause any work to be done on the external circuit.  Matters are quite the opposite:  because of the existence of non-zero work functions, external work must be expended in order to pass current through those materials and the intervening vacuum.

I am not familiar with your $10. experiment.  I would have to know more about the intent and construction in order to comment on it.

What are you trying to determine with the experiment?
How is the experiment designed to test what it is that you are trying to determine?
What are the error tolerances for the measurements?
What null experiments are included to determine that the experiment as designed can determine what it is that you want to determine?
What experiment results have been measured to date?


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Re: quentron.com
« Reply #1975 on: February 02, 2014, 02:33:06 PM »
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Offline Philip Hardcastle

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Re: quentron.com
« Reply #1976 on: February 03, 2014, 12:17:14 AM »
@MarkE, I thought you already knew of the experiment.


In any case you will not do the experiment, and you will never agree with me that the addition of the mesh is significant, and no matter what evidence I provide to you verbally you will never concede I am right.


However I do provide a modified diagram for significant power generation at room temp rather than the sebby 550C.


My last comments on the subject are that the experiment is already posted here under some heading (sebithenco I believe), that you are welcome to contact me on my email (pjhardcastle@gmail.com) if you actually want to do the experiment, and that in any case your support in any way is academic for others of significant scientific credentials are interested.


The problem with the experiment is that though the results are conclusive, the ones from uni physics that have done it are reticent to go public because of the prejudice that pervades science, one physicist did tell his colleagues of the production of power from an isothermal oven using the $10 experiment Vacuum tube (Philips E180F Pemtode equivalent) and was essentially attacked, so he has been silenced. This is something that almost always happens when a discovery contradicts the status quo.

Offline MarkE

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Re: quentron.com
« Reply #1977 on: February 03, 2014, 12:27:23 AM »
@MarkE, I thought you already knew of the experiment.


In any case you will not do the experiment, and you will never agree with me that the addition of the mesh is significant, and no matter what evidence I provide to you verbally you will never concede I am right.


However I do provide a modified diagram for significant power generation at room temp rather than the sebby 550C.


My last comments on the subject are that the experiment is already posted here under some heading (sebithenco I believe), that you are welcome to contact me on my email (pjhardcastle@gmail.com) if you actually want to do the experiment, and that in any case your support in any way is academic for others of significant scientific credentials are interested.


The problem with the experiment is that though the results are conclusive, the ones from uni physics that have done it are reticent to go public because of the prejudice that pervades science, one physicist did tell his colleagues of the production of power from an isothermal oven using the $10 experiment Vacuum tube (Philips E180F Pemtode equivalent) and was essentially attacked, so he has been silenced. This is something that almost always happens when a discovery contradicts the status quo.
Mr. Hardcastle, if you have an experiment that shows something remarkable I find it rather incredible that you say that it has been successfully replicated by academics and they are anything but anxious to be the first to report such a discovery.  Something rings very wrong.

In terms of work function creating an energy source, that is a misunderstanding on your part.  Work function describes field required to move an electron from within a solid to surrounding vacuum.  One can visualize that as an energy hill to overcome.  Different materials have different work functions and so it is true that when placed in local proximity to one another a static electric field gradient exists.  But that gradient exists in complete thermodynamic equilibrium.  One might imagine this as a sort of roller coaster, any traverse around a closed path will not yield a net gain or loss in total gradient.

If you are confident that you have a discovery then I encourage you to properly document:  Your hypothesis, your experiment design, your null result checks, and your measurements into a paper for peer review.

Offline Philip Hardcastle

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Re: quentron.com
« Reply #1978 on: February 03, 2014, 12:45:58 AM »
Mr. Hardcastle, if you have an experiment that shows something remarkable I find it rather incredible that you say that it has been successfully replicated by academics and they are anything but anxious to be the first to report such a discovery.  Something rings very wrong.

In terms of work function creating an energy source, that is a misunderstanding on your part.  Work function describes field required to move an electron from within a solid to surrounding vacuum.  One can visualize that as an energy hill to overcome.  Different materials have different work functions and so it is true that when placed in local proximity to one another a static electric field gradient exists.  But that gradient exists in complete thermodynamic equilibrium.  One might imagine this as a sort of roller coaster, any traverse around a closed path will not yield a net gain or loss in total gradient.

If you are confident that you have a discovery then I encourage you to properly document:  Your hypothesis, your experiment design, your null result checks, and your measurements into a paper for peer review.


Sorry MarkE but your comments that you believe academics would run to ring the bell of discovery are wrong, I know for I have had the conversations many times, you are merely speculating. The history of science is that people are attacked for holding a contrary view, in fact almost all discoveries are ignored for some time until some critical mass is achieved. That is the way it is and your comment about it ringing wrong is part of the problem, you assume I am wrong because others are not on the front page of the news. You should try for yourself to get a science journal to return your call if you send them a note saying you have just found a way of converting ambient heat to power with 100% efficiency, or that you have breached the Kelvin interpretation, if they reply they will tell you you must have made a mistake.


You and others on these sites also tell me I must have made a mistake.


The next bit where you give me a lecture is downright rude, the fact is I know this science clearly better than you do, I never said anywhere that work function is an energy source, I fully understand all the science. I will not take you to task with what you said or gesture to educate you for I have better things than to engage in a Profitis / Sarkeizen style exchange, and I am sure you do to.


 

Offline Philip Hardcastle

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Re: quentron.com
« Reply #1979 on: February 03, 2014, 01:08:35 AM »
Here is the one pager I posted somewhere else.

 

OneLink