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by
Valeriy Y.Tarasov
E-mail: vytarasov@yahoo.com
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Read the whole article
Download the ZIP file
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Abstract
The h-space theory is a variant of unified physical theory – a theory of everything.
This theory was built de novo, as the existing physical theories are incompatible and so unsuitable for unification.
A new approach is needed, and has been developed by re-evaluating the definitions of primary physical concepts.
The starting point for the re-evaluation was the following equation – Et = mvL, where energy – E, time – t, length – L, mass – m, velocity – v.
Analysis of these physical concepts resulted in the construction of a unique equation of the primary concepts such as space, length, energy and velocity.
From this, models could be developed that explain all well-known physical phenomena.
In addition, h-space theory predicts phenomena rejected by the current mainstream theories, such as limits to gravitational and electrostatic interactions, and the possibility of cold fusion (as a consequence of the electric charge definition, a modification of Coulomb’s law and the definitions of elementary particles in h-space theory).
The final section of this article describes a number of experimental tests that could be used to verify the h-space theory.
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Read the whole article
Download the ZIP file
.
I too believe that the fall in oil prices has nothing to do with the recent developments of LENR on the media. Rather it has to do with the global recession, with new climate change policies, with the price war (the Arabs could plan to make non-competitive the american shale) and with speculation. Oil traders are among the most informed people in the world about everything that happens in the world, in any sector can influence prices. Look at these sites, for example:
http://www.321energy.com/reports/flynn/current.html
http://www.321energy.com/archives.php?c=oil
http://www.insidefutures.com/articles/articles.php?author=12
If you do a search on 321energy.com + cold fusion, you will only find old articles, a clear sign they know the issue but so far the idea has not made them even remotely tickled. Everything can change tomorrow morning of course, but they are too afraid Wlad receive a tip-off about magnetic prospection oil shortage problems and he could elaborate a new theory, so they won’t even barely touch the topic
In the price of hydrocarbons should be also calculated the cost of global warming,but it is very difficoult because it is a factor in future which we don’t know the amount.
I hope that some economists,with scientists,make projections to have a real price of oil,gas and coal.
This is a collective cost and it would be a “real carbon tax” on hydrocarbons.
Regards G G
Joe wrote in December 10th, 2014 at 3:06 AM
Wladimir,
You imagine that the proton loses its electric field by becoming a neutron in order to avoid this “disturbance”.
—————————————————————————
Joe,
let us speak about what the proton enjoys.
We know from the Newton’s laws that Nature does not appreciate changes:
1- If a body is at rest, it continues at rest, unless a force is applied on it
2- If a body movies with constant linear velocity, it continues moving with constant velocity unless a force is applied on it
A particle moves with helical trajectory (zitterbewegung).
When the particle is near to be at rest regarding the aether, the radius of its helical trajectory tends to infinitum.
When the velocity of the particle is near to the speed of light, the radius of its helical trajectory tends to zero
When a particle is being accelerated (as happens with the proton in the LHC), the radius of its helical trajectory is changing (decreasing).
Consider the proton being accelerated in the LHC.
The Nature does not appreciate changes. Therefore, the Nature tries to avoid the change in the radius of the helical trajectory , and (in order to avoid that “disturbance” due to the change in the radius of the helical trajectory) the zitterbewegung of the proton extracts particles of the aether, and emits them in the form of photons (Maxwell’s law).
In the Rossi-Effect, the proton is accelerated by the orbit of the electron 2s1, and therefore the proton is extracting particles from the aether, and it emits them as photons, in order to avoid the “disturbance” in the radius of its helical trajectory.
To each absolute velocity of the proton in the aether correspond a specific radius of the helical trajectory.
Then a second disturbance appears, when the proton is near to cross the plane of the 2s1 orbit: the electron tries to change the radius of the helical trajectory of the proton.
We have:
1- in a determined instant “t” the proton has a determined instantaneous velocity V and a radius R of the helical trajectory.
2- But the action of the electron 2s1 changes the radius R, and now the orbit has a new radius Rc. However, with that instantaneous velocity V the proton cannot have another different radius. With the velocity V the radius must be kept as R.
3- That discrepancy between the instantaneous velocity V and the new radius Rc requires an extraction of an electron from the aether, so that to transform the proton in a neutron
regards
wlad
Wlad,
In the Mosier-Boss reaction, have you forgotten about the energy used to cause the fusion of the deuterons or the mass change which can be translated as energy that must be accounted for? More than enough to create the energetic particles. the contribution of the strong force is just a part of the overall energy balance.
I think the 2s1 electron must be more energized so it can sling shot the neutron past the influence of the 2px electron much like the influence of gravity on the space vehicles as they pass a massive body.
Regards.
Dr Peter Forsberg:
You made a good point.
Warm Regards,
A.R.
Joe wrote in December 10th, 2014 at 3:06 AM
Wladimir,
1. ) —————————————-
free proton has never been observed to undergo beta+ decay. And by free, we mean not bound to other masses. But obviously free protons are handled by scientists with the use of electric and magnetic fields. And energies associated with these fields can be greater than E = 2*m*c^2, yet with no beta+ decay of the proton ever witnessed.
The scientists at the LHC are not complaining about protons transmuting into neutrons in their device as the protons are being accelerated by fields. If free protons could undergo beta+ decay, especially as quickly as you imagine that they do, the LHC would never have been built.
——————————————————–
And in the LHC the protons are crossing the orbit of an electron, whose radius is in the order of the Bohr radius 10^-11m ?
And is the proton moving in a positive elecrosphere of a nucleus ????
Tell them to improve the following in the LHC, and they will observe the decay+ of the proton no LHC:
1- They have to create a positive electric field in the LHC accelerator (equivalent to that of the electrosphere of the Ni nucleus), within which the proton must be moving.
2- They must put in the LHC accelerator an electron moving in an orbit with radius R= 10^-11m, perpendicular to the motion of the proton.
3- The proton must cross the plane of the electron’s orbit in the center of the orbit.
.
1.A) ———————————————–
Here is an energy comparison:
1 MeV (1.602×10^−13 J): about twice the rest energy of an electron
versus
14 TeV: the designed proton collision energy at the Large Hadron Collider (which has operated at half of this energy since 30 March 2010)
————————————————
In the LHC, what is replacing the electric charge of the electron 2s1, so that to induce a helical trajectory in the proton’s motion?
The charge of the electron in the Rossi-Effect causes a disturbance in the proton’s zitterbewegung.
In the LHC, what is causing the disturbance in the proton’s zitterbewegung?
.
2. ) —————————————
You state that the proton has a helical trajectory along the z-axis due to interaction with the aether. You also state that a second helical trajectory caused by the Coulomb interaction between proton and rotating 2s1 electron is superimposed upon the proton, and that this causes a “disturbance” in the proton’s original helical trajectory. You imagine that the proton loses its electric field by becoming a neutron in order to avoid this “disturbance”.
———————————————
And you imagine that within a nucleus with excess of protons, one the protons captures an electron from the aether, and loses its electric field by becoming a neutron in order to avoid the “disturbance’ caused by the excess positive charge of the nucleus?
.
i) —————————————–
If the proton were never to become a neutron, and the “disturbance” therefore never avoided, what would happen to the trajectory of the proton?
———————————————
If a nucleus with excess protons do not capture electrons from the aether, and therefore the disturbance of the nucleis is never avoided by the positron emission, what would happen to the nucleus?
.
ii) ———————————————–
If the two overlapping helical trajectories were to have no negative effect upon the successful crossing of the plane of rotation by the proton, could the strength of the Coulomb interaction between proton and 2s1 electron still prevent the proton from entering the target nucleus by pulling it back to the plane of rotation?
—————————————————-
it seems yes
.
Joe,
do you think the acceleration of the beam of protons in the LHC did not have occurred to me earlier?
I already had reflected about the difference between the conditions in the LHC and in the cold fusion experiments.
You have to keep in mind that in cold fusion there are many special conditions. You cannot compare them with the ordinary conditions with which the physicists had used to deal with along a century.
regards
wlad
I agree with Andrea Rossi and Robert Cuerto. The fall in the price of oil has nothing to do with ECat. I wish it did. It has more to do with world politics and international economic trends.
But I’m concerned about the fall in the price of oil, since it makes the ECat less competitive in different ways. If this is a problem depends on how far the price will fall, and for how long it will stay low.
Regards
Peter
Greg Leonard:
Thank you for your opinion. I do not agree.
Warm Regards,
A.R.
Dear AR and Robert Curto
I believe the evidence is against your opinion.
Fracking has been making an impact for years now. It may well have been part of the reason for the Rockefellers to divest themselves of their oil portfolio – but that did not cause a sharp slide in the oil prices.
The current slide in oil prices is largely down to the giant hedge funds coming out of oil (I had not realised what influence they wield).
Again, that slide started exactly when the Lugano report was leaked. This is not coincidence, it is cause and effect.
regards
Greg Leonard
Wladimir,
1. A free proton has never been observed to undergo beta+ decay. And by free, we mean not bound to other masses. But obviously free protons are handled by scientists with the use of electric and magnetic fields. And energies associated with these fields can be greater than E = 2*m*c^2, yet with no beta+ decay of the proton ever witnessed.
Here is an energy comparison:
1 MeV (1.602×10^−13 J): about twice the rest energy of an electron
versus
14 TeV: the designed proton collision energy at the Large Hadron Collider (which has operated at half of this energy since 30 March 2010)
The scientists at the LHC are not complaining about protons transmuting into neutrons in their device as the protons are being accelerated by fields. If free protons could undergo beta+ decay, especially as quickly as you imagine that they do, the LHC would never have been built.
2. You state that the proton has a helical trajectory along the z-axis due to interaction with the aether. You also state that a second helical trajectory caused by the Coulomb interaction between proton and rotating 2s1 electron is superimposed upon the proton, and that this causes a “disturbance” in the proton’s original helical trajectory. You imagine that the proton loses its electric field by becoming a neutron in order to avoid this “disturbance”.
i) If the proton were never to become a neutron, and the “disturbance” therefore never avoided, what would happen to the trajectory of the proton?
ii) If the two overlapping helical trajectories were to have no negative effect upon the successful crossing of the plane of rotation by the proton, could the strength of the Coulomb interaction between proton and 2s1 electron still prevent the proton from entering the target nucleus by pulling it back to the plane of rotation?
All the best,
Joe
eernie1 wrote in December 9th, 2014 at 1:38 PM
Wlad,
In the Mosier-Boss experiment where 2 deuterium atoms fuse to create a tritium atom plus a neutron with approx. 10Mev energy to balance the energy equation, the high energy balance is required because the nucleons are under the influence of the strong force which requires the high energy to balance the reaction.
—————————————————–
Eernie,
this also does not work.
The binding energy of the deuteron is 2,2MeV (the proton and the neutron are bound with 2,2MeV). A neutron cannot be emitted with energy higher than 2,2MeV when the tritium is formed.
regards
wlad
eernie1 wrote in December 9th, 2014 at 1:38 PM
Wlad,
In the case of the Halo electron of the 3Li7, the strong force is almost completely neutral because of the distance between the neutron and the nucleon cluster. Therefor the neutron can exit the nucleus without the high energy required by the cluster nucleons and can exit the nucleus as a thermal neutron which can then easily react with the 28Ni nuclei.
—————————————
Eernie,
as I said, before to cross the plane of the 2s1 orbit, the neutron has attraction with the orbit, and it is accelerated.
After crossing the plane of the orbit, the neutron continues to have attraction, and so it is disaccelerated, and therefore it will not hit the Ni nucleus.
Your proposal could work if the neutron would be emitted with high energy by the 3Li7 decay, and so the attraction force of the 2s1 orbit would not be able to stop the neutron.
However the neutron exits the 3Li7 with low energy, since it is weakly bound.
regards
wlad
Joe wrote in December 9th, 2014 at 2:12 PM
Wladimir,
1. ) ————————————–
You write,
” […] positive charge was favouring the electron capture by the proton.”
It is not enough to just simply have a positive charge. Energy states, both initial and final, must be taken into account.
From Wikipedia, Beta Decay:
“β+ decay can occur when the mass of the initial atom exceeds that of the final atom by at least twice the mass of the electron.”
This, of course, is referring to beta+ decay occurring INSIDE the nucleus. But outside the nucleus, beta+ decay has never been observed. The reason given, from the same article, is the following:
“β+ decay cannot occur in an isolated proton because it requires energy due
to the mass of the neutron being greater than the mass of the proton.”
——————————————————
yes,
however such condition do not consider that the proton is crossing the orbit of one electron.
The energy required is E= 2.mc² , necessary to extract a pair positron-electron from the aether. Such energy can be supplied by the electromagnetic energy of the electron 2s1 moving in its orbit.
2. ) ——————————————–
You write,
“And when the proton crossed the plane of the 2s1 orbit, the proton experienced a disturbance, helping the capture of the electron by extracting it from the aether.”
What specifically is that “disturbance”?
————————————————–
As you know, the proton moves by helical trajectory (zitterbewegung), due to the interaction with the aether.
But suppose the proton is NOT moving with zitterbewegung along the z-axis of the nucleus Ni, going toward the direction of the 2s1 orbit. In this case, when the proton is near to the plane of the electron orbit 2s1, the circular motion of the electron applies a Coulomb attraction force on the proton, and by this way the motion of the electron produces a circular motion of the proton around the z-axis. Combined with the rectilinear motion of the proton along the z-axis, such attraction with the electron produces a helical trajectory in the motion of the proton.
Remember that here we are considering that the proton does not move with zitterbewegung.
But the proton has its zitterbewegung. And therefore the helical trajectory induced by the electron 2s1 on the proton causes a disturbance in the proton’s zitterbewegung.
The best way for the proton to avoid the disturbance in its zitterbewegung is by losing its electric field, so that do not have Coulomb interaction with the electron 2s1. And the way for losing its electric field is by a capture of one electron from the aether, becoming a neutron.
regards
wlad
Ing. Michelangelo De Meo:
Another giant enters in the LENR R&D. Thank you for the information.
Warm Regards,
A.R.
Dear Dr. Rossi,
Here is another great news, the birth of a Research Institute Energie ” Energiforsk ” announced by Elforsk .
Magnus Olofson after the publication of Lugano E-cat test report have said that there was now a good reason to launch research efforts around LENR…
http://www.lenr-forum.com/forum/news/index.php/News/39-Elforsk-announce-the-birth-of-Energiforsk-Energie-Research-institute/
Dr. Rossi, I believe you are correct about the oil price drop.
It has nothing to do with the E-Cat.
It will be many years before the E-Cat has a good effect on oil prices.
Robert Curto
Ft. Lauderdale Florida
USA
Wladimir,
1. You write,
” […] positive charge was favouring the electron capture by the proton.”
It is not enough to just simply have a positive charge. Energy states, both initial and final, must be taken into account.
From Wikipedia, Beta Decay:
“β+ decay can occur when the mass of the initial atom exceeds that of the final atom by at least twice the mass of the electron.”
This, of course, is referring to beta+ decay occurring INSIDE the nucleus. But outside the nucleus, beta+ decay has never been observed. The reason given, from the same article, is the following:
“β+ decay cannot occur in an isolated proton because it requires energy due
to the mass of the neutron being greater than the mass of the proton.”
2. You write,
“And when the proton crossed the plane of the 2s1 orbit, the proton experienced a disturbance, helping the capture of the electron by extracting it from the aether.”
What specifically is that “disturbance”?
All the best,
Joe
Wlad,
I am curious about your complication of the Rossi effect by insisting that a neutron must decay to a proton in the 3Li7 nucleus before it could be ejected. Isn’t it more simple to assume that the lightly bound Halo neutron of the 3Li7 is made to exit the nucleus by applying an external force(electron electrosphere of 2p1 electrons created by exciting the 2s1 valence electron). In the Mosier-Boss experiment where 2 deuterium atoms fuse to create a tritium atom plus a neutron with approx. 10Mev energy to balance the energy equation, the high energy balance is required because the nucleons are under the influence of the strong force which requires the high energy to balance the reaction. In the case of the Halo electron of the 3Li7, the strong force is almost completely neutral because of the distance between the neutron and the nucleon cluster. Therefor the neutron can exit the nucleus without the high energy required by the cluster nucleons and can exit the nucleus as a thermal neutron which can then easily react with the 28Ni nuclei.
Regards.
ERRATA:
Joe,
in my last reply to you,
where I wrote:
But here we are facing a new phenomenon: when the neutron decays in the 3Li7, the proton has no time available for capturing an electron from the aether.
the correct is:
But here we are facing a new phenomenon: when the neutron decays in the 3Li7, the nucleus has no time available for capturing an electron from the electrosphere.
Joe wrote in December 9th, 2014 at 1:18 AM
Wladimir,
Your quote states that beta+ decay occurs inside nuclei. The newborn proton that crosses the plane of rotation of 2s1 is obviously not inside a nucleus, or even an entire atom. It is in transit between two atoms, Li and Ni.
————————————————-
Joe,
because the physicists had noted that beta+ decay occurs inside nuclei does not mean that it cannot occur in the transit between Li and Ni.
The physicists had never studied the cold fusion phenomenon, we are only in the beginning of the discoveries.
Besides, of course the beta+ decay must occur within the nuclei IN NORMAL CONDITIONS, because the protons are within the nuclei.
But here we are facing a new phenomenon: when the neutron decays in the 3Li7, the proton has no time available for capturing an electron from the aether.
We cannot apply all the rules known in the Standard Model, since cold fusion is impossible to occur from the known rules of that model.
Note that in NORMAL CONDITIONS the 4Be7 decays in 3Li7 by electron capture (from the electrosphere), It has 53,2 days half-life.
But there is not time available to capture the electron from the electrosphere, because the proton exits the 4Be7.
When the proton exits the 4Be7, the 4Be7 will not capture an electron from the electrosphere (as occurs in normal decay of 4Be7), since the proton exited the nucleus, and the excess positive charge disappeared for the 4Be7.
But the excess positive charge continues, because there is an excess of one proton, and it is moving toward the Ni nucleus.
Therefore there is favourable conditions for the capture of the electron by the proton (from the aether), because:
1- When the neutron transmuted to proton in the 3Li7, the electrosphere of the newborn 4Be7 became positive.
2- Exiting the electrosphere of the 4Be7, the proton would transfer the positive charge excess for the electrosphere of the Ni.
3- Therefore excess positive charge was favouring the electron capture by the proton. And when the proton crossed the plane of the 2s1 orbit, the proton experienced a disturbance, helping the capture of the electron by extracting it from the aether.
regards
wlad
Dear Readers of Journal of Nuclear Physics,
Meanwhile the long reviewing process, the presented here article “h-Space Theory” has been updated. The major corrections and more detailed explanations were made in the sections presenting the mechanism of gravitational attraction and the determination of electric charge of elementary particles and ions. To download the updated version of article the following link can be used:
http://h-theory.narod.ru/h-space_theory_v2.pdf
Best wishes,
Valeriy Tarasov
Christen: good idea.
Warm Regards,
A.R.
Greg Leonard:
I am pretty sure the shift in the oil price is due to the yield of the fracking shale gas production.
Warm Regards,
A.R.
Subject: on the fission of the Pd nucleus in the Mosier-Boss experiment
Dear Joe,
what do you think about the mechanism prposed for the acceleration of the proton (and neutron), when applied for the explanation of fission prodution of the nucleus Pd in the Mosier-Boss experiment?
As I said earlier:
==============================================================
Andrea Rossi already told in the JoNP that he had tested all the sort of combinations, and he concluded that Ni is the best fuel for the eCat. For instance, he said that 46Pd is not a good element. Then probably Andrea Rossi already had tested 50Sn, and concluded that it is also not a good element to react with 3Li7.
However it is easy to understand why Andrea Rossi did not succeed to get a god COP by using 46Pd and 50Sn combined with 3Li7. We can see it in the Fig. 1 ahead,
FIG. 1:
http://peswiki.com/index.php/Image:Fig._1-50Sn_as_RECEPTOR_in_eCat.png
After the decay 3Li7 -> 4Be7, the proton of the 4Be7 is pulled by the orbit of the electron 2s1, and when the proton crosses the plane of the orbit the proton by electron capture transmutes to neutron, which is now pushed by the orbit 2s1 with a repulsion force.
So, in the reactor using Ni-3Li7 as fuel, the final acceleration on the neutron is not so high, because the radius orbit of the electron 2s1 is not large.
But in the reactor using Pd-3Li7 the final acceleration on the neutron is very faster, because the radius orbit of the electron 2s1 is very larger. And so the neutron trespasses the 46Pd nucleus, and is not captured.
==============================================================
The neutron is not captured by the Pd nucleus, and most of them cross the nucleus, and exit it.
But some of the neutrons hit the Pd causing its fission.
.
Did you read about fision in the Mosier-Boss experiment?
====================================================================
III. CONCLUSIONS
In this communication, the use of microscopic examination, sequential etching, and LET spectrum
analysis to identify the energetic charged particles and neutrons responsible for the tracks in CR-39 detectors used in Pd/D co-deposition experiments were discussed. These analytical techniques gave complementary results .
The energetic particles were identified as 2.45 MeV neutrons, 3-10 MeV protons, 2-15 MeV alphas, and 14.1 MeV neutrons.
[…]
The only known source of these long range alphas is fission reactions. This suggests that the new elements observed on the cathodes are caused by fissioning of the Pd nucleus
http://newenergytimes.com/v2/conferences/2012/ICCF17/papers/Mosier-Boss-Its-Not-Low-Energy-Paper-ICCF17-pp.pdf
===============================================================
Do you think the authors of cold fusion theories can find an acceptable theory based on the tunneling mechanism, able to explain the fission?
regards
wald
Dear AR,
You say that all different power sources will be integrated, but there will be a substantial shift away from the common fossil fuels.
We have already seen the slide in crude oil prices which started when the Lugano test report was leaked. This adjustment happened in just a few months.
The lower oil price is causing most of the producer countries to lack the finances to run their economy as they have in the past – which will lead to considerable social upheaval.
The lower oil price has also affected the economic argument for several wind power systems.
We live in interesting times.
Dear Andrea,
I suggest you, yo avoid misunderstandings, the use of SSM acronym for self sustain mode and SSP acronym for start stop mode of operation in the e-cat system.
Wladimir,
Your quote states that beta+ decay occurs inside nuclei. The newborn proton that crosses the plane of rotation of 2s1 is obviously not inside a nucleus, or even an entire atom. It is in transit between two atoms, Li and Ni.
All the best,
Joe
Achi:
Thank you for your kind consideration.
About your questions:
1 + 2: I will be able to answer after at least 1 year of operation of the 1 MW plant
3: no, if the control system works properly in the specific phases
4- I do not think so, but I could be wrong
5- no
Thank you for your attention,
Warm Regards,
A.R.
Italo R.:
Beyond the “Spectre” syndrome, we must understand that all the energy sources will integrate, with a slow shift, gradual in time, toward the most convenient in specific sectors of market. This is in the interest of all, therefore History will tend to this, as always happens: History goes where the common interest is.
Warm Regards,
A.R.
Dear Dr. Rossi, what do you think will happen in the sector of world energy when your 1 MW plant will demonstrate unequivocally that it is capable to generate more energy than it receives in input?
You said many times that all forms of energy must be integrated among themselves, but I am afraid that the arrival of the E-Cat will tremendously upset the economic balances consolidated till now. Do you think that there will be this derangement, or the effect of the presence of E-Cat in the world will be soft because it will be necessarily introduced gradually?
Kind Regards,
Italo R.
Dear Mr Rossi,
I would like to thank you for all the hard work you and your team are putting in to bring LENR to the world. It feels good knowing that such a revolutionary tech is just on the horizon.
I would like to ask you about the different modes your ecats have.
I know of the start stop mode where power is cycled, and self sustained mode where there is no power supplied to drive the device, and I would also assume that there is a mode where power is constantly being supplied.
Of these modes, which do you believe the LENR effect to be the most powerful? Most able to sustain a load?
Do you believe that the reaction changes effect with the different modes?
Do you think that could be more than one type of reaction occuring?
Are there other “modes” in which the ecat runs?
I understand if you cannot answer my questions but I appreciate your reading them.
Thank you,
Achi
Hank Mills:
Yes. But I must repeat that only after at least one year of regular operation we will be able to know the data of a reliable operation.
Warm Regards,
A.R.
Silvio Caggia:
Yes, you are.
Warm Regards,
A.R.
Charles Jameson wrote in December 7th, 2014 at 7:06 PM
Dear Wladimir
Although I am a retired engineer with no experience in nuclear physics I commend you on the simplicity and clarity of your theory. Your diagrams are a big help and I think I do have a basic understanding of what you are proposing.
—————————————————
Dear Charles,
we, engineers,
we know that we cannot design a machine through mathematical equations.
first we have to do the design of the the machine by doing a drawing its components, and by doing a drawing showing how the components work all they together.
If the components of the machine do not work harmoniously, it makes no sense to apply the mathematics on it. First of all, we have to succed in puting all the pieces woring well toghether.
And later we will use mathematical equations, so that to calculate the diameter of axes, gears, pistons, electric generators, relays, coils, etc, in order to be sure that the machine will not break.
The atomic nucleus is an mechanical-electromagnetic machine. We cannot describe its working by mathematical equations without to know how the components of the machine are working within it.
However this is precisely what the physicists are trying to do.
They know that within the machine there are protons and neutrons. But they do not know how protons and neutrons are bound within the machine, they do not know how protons and neutrons are distributed, and how they move within there. They also do not know exactly how is produced the electric field of each proton, and the shape of this electric field.
Nevertheless, they are sure that they can describe accurately the working of the machine by mathematical equations.
That’s why it is impossible to explain cold fusion from the current nuclear models
regards
wlad
Joe wrote in December 8th, 2014 at 5:57 PM
1) —————————————
I am talking about both the proton and electron entering through the hole of the Ni nucleus together along the z-axis.
——————————————
The electron resulted of the neutron decay cannot be pulled by the orbit 2z1, because there is Coulomb repulsion between two electrons.
.
2) —————————————-
Why do you say that the Coulomb barrier is keeping the proton from entering the nucleus when there is supposed to be no Coulomb barrier at the hole?
——————————————-
The proton can enter through the hole of the barrier only when the z-axis of the nucleus is aligned toward the direction of a vector magnetic field , because when this occurs the nucleus stops to gyrate chaotically, and so the shape of its Coulomb force stops to be spherical.
In normal conditions (when the conditions for cold fusion occurrence are not satisfied), the z-axis of the nucleus is not aligned toward any direction (the z-axis gyrates chaotically), and so the Coulomb barrier takes the spherical shape. In this case, the proton can enter within the nucleus only via HOT fusion.
.
2. ) —————————————-
You write,
“the same sort of electron created by electron capture within the nuclei. It is extracted from the particles of the aether, as also happens in the positron capture.”
This is illogical. The usual electron capture is from the electrosphere.
———————————————
Ok,
then instead of to call it electron capture, let us call it positron emission:
“Positron emission or beta plus decay (β+ decay) is a particular type of radioactive decay and a subtype of beta decay, in which a proton inside a radionuclide nucleus is converted into a neutron while releasing a positron and an electron neutrino”.
http://en.wikipedia.org/wiki/Positron_emission
The mechamism is the following:
1- A proton captures an electron from the aether, and is converted into a neutron.
2- When the electron is captured from the aether, a positron is created together, and is emitted.
In other words, an electron is captured (created) from the aether, and a positron is emitted.
.
3) ———————————————
What you are describing is the creation of an electron from aether particles. This goes against conservation of charge. When an electron is created, it is always done with an accompanying positron. This is known as “pair creation”.
————————————————
Yes, and while the electron is captured by the proton, the positron exits the nucleus
regards
wlad
Dear Andrea Rossi,
You suggested, more than a time, to read a book written by Norman D. Cook to understand the phisics behind e-cat.
I red it.
Are you sure to agree with the author’s ideas about LENR?
It seems to me that Cook interprets C.F. not as Cold Fusion but as Clean Fission…
Am I wrong?
Dear Andrea,
So, to avoid confusion and any misunderstanding, that is 2 hours of self sustained operation (constant temperature of the individual reactor) with no electrical power to the drive?
Thanks!
Hank Mills:
The longest period of ssm we got so far with the E-Cats is 2 hours, but only after the end of the test period of the 1 MW plant in the factory of the Customer we will have reliable numbers.
Warm Regards,
A.R.
Herb Gillis:
This is one of the tasks of the control system.
Warm Regards,
A.R.
Andrea Rossi:
Are you presently able to select (entirely at your choice) which of the two SSM modes the reactor will be in when external drive is stopped?
Kind Regards; HRG.
Wladimir,
1. In one of my comments, I wrote,
“I guess this means that the newborn electron would also exit through the “hole”. Maybe both newborns [proton and electron] would cross the plane of rotation together with opposite helicities and re-constitute a neutron in the target nucleus.”
You responded,
“Joe,
what you say makes no sense.
The proton cannot cross the positive Coulomb barrier because the proton has positive charge, and so there is repulsion on the proton.
The electron has negative charge, and it has not repulsion with the Coulomb barrier, and so it can cross it in any point.”
I am talking about both the proton and electron entering through the hole of the Ni nucleus together along the z-axis. Why do you say that the Coulomb barrier is keeping the proton from entering the nucleus when there is supposed to be no Coulomb barrier at the hole?
2. You write,
“the same sort of electron created by electron capture within the nuclei. It is extracted from the particles of the aether, as also happens in the positron capture.”
This is illogical. The usual electron capture is from the electrosphere. What you are describing is the creation of an electron from aether particles. This goes against conservation of charge. When an electron is created, it is always done with an accompanying positron. This is known as “pair creation”.
All the best,
Joe
Dear Andrea,
That is a huge accomplishment! Start Stop Mode is impressive, but Self Sustain Mode is phenomenal beyond measure. I hope you can answer this one question: it would be like a Chrismas present for many of us.
In tests of the individual reactors that compose the plant, how long is the period of self sustain (in which the reactor maintains a constant or increasing temperature without input) that has been deemed to be safe for use with minimal risk of thermal runaway?
Thank you for sharing this amazing news!
Frank Acland:
SSM is an acronym that can be used for either definition.
Warm Regards,
A.R.
Hank Mills:
The 1 MW plant can use either of these methods.
Warm Regards
Dear Andrea,
The acronym SMM can represent two different modes of operation: start stop mode and self sustain mode.
Start Stop mode is when the drive is activated with electrical power, the reactor chamber is heated, the nuclear reactions are simulated, and then the input to the drive (resistances) are turned off. When the drive is turned off, the anonamalous nuclear reactions are still talking place and are contributing to the output of the device. The reactor cools, but it does not drop as quickly as it should if the fuel charge was not in place producing a diminishing quantity of excess power. However, the device as a whole is slowly dropping in temperature and the reactions are dying off until the drive is turned back on. When the drive is off, the reactions are not truly sustaining. They are present, but diminishing.
Self Sustain Mode is when – after the reactor has been heated to operating temperature – the input to the drive is cut off. However, the reactions have been simulated to such a high level they are self sustaining with no external input. This means the temperature of the reactor is staying the same or increasing. Synonyms for the word sustaining could be: unending, constant, continual, enduring, perpetual, etc. This mode is different than start stop in which the reactions produce excess energy after the input is cut off, but drop in frequency so the reactor (if no more input was applied) would drop to room temperature. The small current drawback is that self sustain mode has a slightly increased chance of thermal run away that is not virtually zero like when the start stop mode is used.
Please let me know if these two explanations are accurate and which mode is used in the new one megawatt plant.
Of course, a plant using either of these methods would be far superior to any other energy source on the planet. The lower COP of a plant using Start Stop mode would still be far higher than what hot fusion scientists hope to achieve decades from now with teens of billions of dollars in funding.
Dear Andrea,
Would it be more accurate to define ‘ssm’ as self-sustain mode, or start-stop mode?
Thank you,
Frank Acland
Dave Roberson:
With ssm we mean that part of operation during which the electric power that drives the E-Cats is turned off. That’s all I can say. The cycles are regulated by the complex control system.
Warm Regards,
A.R.
Dear Mr. Rossi,
Congratulations to you for the progress that you and your team has made during the recent months. I look forward to the day when your designs become widely available and change the world energy equation in a positive manner.
I have been following your efforts for several years and believe that I understand how your devices behave, at least to a limited extent. There are plenty of questions concerning the SSM mode of operation and I would like to explore one characteristic of that mode if I may. Is it true that you generally operate in that mode by using pulse width modulation of the drive power source, whether this source is electrical or gas energized? This would be consistent with the first third party test report. The radiated power waveform that they published indicated that the internal heating was not like a resistor load but instead demonstrated an increasing rate of temperature rise during the external heating pulse.
After a measured application of drive power, your device is then placed into the SSM state as it begins to cool down toward the temperature at which the cycle repeats. I suspect that you refer to the period of time during which the device coasts lower in temperature as the SSM condition. Is this correct? It certainly makes a great deal of sense to call it by that name since the internal core is generating a great deal of additional heat energy during that time frame. Also, I have reason to believe that the newly generated heat power causes the rate of temperature decay to be reduced when compared to the response of a resistive load that has no extra internal power generation. The first third party report contained a graph that clearly suggests this type of activity.
It appears that people are confused by the definition of the SSM and I hope that you will clarify that operation by answering my questions above. Perhaps after that you will be subject to fewer additional questions concerning the matter.
Thank you for your attention and I wish you much success in your important endeavor.
ing. Michelangelo De Meo:
Very interesting, this is clearly the development of the work of Brian Ahern, that I always said is the one that is making a good replication. This is positive, because adds evidence to the replicability of our work from third parties, even if with different methods. As for the competition, they are at the point we were 10 years ago, but it is obvious that sooner or later we will have competition. Now, thank to the enormous fight made by us, giants are entering in the battlefield: MIT, NASA, Bill Gates, Lockeed Martin etc etc. When the play gets hard, hard guys go into play, and we are not soft, believe me.
Warm Regards,
A.R.
Hello Dr. Rossi,
increase scientists around the world who are studying the LERN . Reading your last report you notice that you are light years ahead of other competitors .
http://e-catalyzer.it/e-catalyzer/fusione-fredda-aggiornamenti-sul-nanor