Experimental Evidence of a Breach of Unity

by
Rosemary Ainslie, Donovan Martin,
Evan Robinson, Mario Human,
Alan Macey, Riaan Theron

Abstract
This First part of a two-part paper deals with results from a circuit that was designed to determine whether the amount of energy dissipated in a circuit could exceed the amount of energy delivered from a battery supply. If so, then this result would prove the basis of a magnetic field model that predicts an exploitable non-conservative field condition. This model is the subject of the second paper.
The distinction is drawn that the energy that is dissipated in a circuit is sourced from the induced potential difference in the circuit material itself.
Correspondingly then, the proposal is that the circuit material itself may be used as a supplementary and secondary energy supply source that has not, typically, been factored into the analysis of power conversion.
This raises the question as to whether Kirchhoff’s Laws exceed Faraday’s Laws of Induction. And if not, then there is also a potential for the conservation of potential difference at a supply that may be exploited to enhance energy efficiencies.
Test results show that this is, indeed, possible and that the inductive and conductive components of circuit material may be factored in as a potential energy supply source depending on the circuit design and intention.

190 comments to Experimental Evidence of a Breach of Unity

  • Andrea Rossi

    Dear Ivan:
    Our indistrial plants are already a reality and we are manufacturing them. For the domestic ones we are waiting the certifications.
    Warm Regards,
    A.R.

  • Ivan

    Dear Dr Rossi,
    When is your product being a reality?, In Sydney Australia, the KWH of electricity at peak time is now 48Cents of Australian dollar. ($1 AUD = 1.02 USD)
    How many people are suffering cold because they can not pay exorbitant rates.
    We need your technology….!NOW!
    Thanks
    Ivan

  • Wladimir Guglinski

    Joe wrote in July 3rd, 2012 at 4:48 PM

    “Bhagirath and Wladimir,

    How do you define the quark and the gluon?”

    Dear Joe
    quarks are made by agglutination of massless particles of the aether. The quark has the form of a ring (that’s why my theory is named Quantum Ring Theory”). A flux n(o) crosses the ring of the quarks (of course not strong as the flux n(o) that crosses the proton).
    The quark is similar to the particle and antiparticle that compose the photons.

    Perhaps gluons are other name for the particles p(+) and p(-) proposed in my theory.

    Or maybe gluons are other particles of the aether. Probably they work in partnership with the gravitational particles.

    Regards
    wlad

  • Kim

    We need some fire works Andrea!

    Give us something we can get out
    teeth into.

    Respect
    KIm

  • Joe

    Bhagirath and Wladimir,

    How do you define the quark and the gluon?

    All the best,
    Joe

  • Andrea Rossi

    Dear Robert Tanhaus:
    Russia and Japan have been licensed.
    Warm Regards,
    A.R.

  • Robert Tanhaus

    Dear Mr. Rossi,

    some time ago you said that you have no licensee for Russia and Japan.
    Has that changed?

    BR
    Robert Tanhaus

  • Andrea Rossi

    Dear Joe:
    1- no, it is impossible, for intrinsic reasons, as I explained: once the Ni melts, the operation stops and Ni melts around 1500 C
    2- no
    Warm Regards,
    A.R.

  • Joe

    Dr Rossi,

    1. In my last question, what I really wanted to know was whether you believe that the process occurring in the E-Cat would affect the melting temperatures of the reactants. Is there a remote possibility that, for example, the E-Cat could ultimately sustain a temperature of 2000C – well above the melting temperature of nickel (1455C)?

    2. In your new 600C process, is there a change in the rate of transmutation for nickel into copper? If so, is it higher or lower? (Lower would be better for business, of course.)

    All the best,
    Joe

  • Wladimir Guglinski

    Bhagirath Joshi wrote in July 2nd, 2012 at 5:06 PM
    You already mentioned that it does not levitate in vacum. This is due to lack of air particles.

    Dear Joshi,
    theoretical calculations pointed out that the ionization of the air, itself, is unable to explain the lifter levitation.

    In spite of the lifter does not levitate in the vaccuum, however the experiments in the vaccuum showed that there is a thrust in the lifter.
    Without such thrust, the lifter could not levitate even in the air.

    The levitation is due to two causes:
    – the thrust due to the ionization of the air
    – the thrust which occurs in the vaccuum

    I suggest you to read about the theories in
    http://jnaudin.free.fr/lifters/theories.htm

    You will see written there:
    The calculations indicate that ionic wind is at least three orders of magnitude too small to explain the magnitude of the observed force of the capacitor”

    regards
    WLAD

  • Bhagirath Joshi

    Wladimir Guglinski and Joe ant other readers.

    While this is interesting discussion, until the reality is introduced in the discussion, one does not get a full picture of the process. From the conservation of energy and hence the mass one can argue that for something to exist an opposite must exist. This is where everything gets complicated.

    Here are the equations.

    0 = matter + antimatter ——- equation 0
    where matter -> Known universe
    anti matter unknown universe
    matter = sum of all ( neutron ) ———- equation 1

    neutron = proton + electron ———- equation 2
    proton = non charged particle + (positive Charge) ——- equation 3
    electron = non charged particle + ( -ve charge) —- equation 4

    and you cam keep breaking down the system to smallest particle.

    But we already have systems built with these neutron, proton and electrons.

    You may like to look at this from outside in. But best way to look at the system is from inside out,

    Looking at equation 1, the known universe is represented by sum of all neutrons. Let us build the system with these neutrons. I prefer to build elements with neutrons for one reason that being electrically neutral at nuclear distances, neutrons can come very close to each other. and you may see this readily.

    let me represent now n -> neutron, p -> proton, e -> electron

    n -> 1H1 -> p + e
    2n -> 1H2 -> p + n + e
    3n -> 1H3 -> p + 2n + e -> 2He3 -> 2p + 2e + n
    4n -> 2He4 -> 2p + 2e + 2n
    5n -> 2He5 -> 2p + 2e + 3n -> 3Li5 -> 3p + 3e + 2n
    6n -> 3Li6 -> 3p + 3e + 3n

    and so on.. off course some elements are not possible due to the forces with in, and you see that at every step the mass and energy is conserved in the known universe. at nuclei level when the symmetry seems to be compromised it is recovered at the molecular level.

    Now the problem is, since we are dealing with only the known universe. If you look at the equation 0, it is evident that the same processes may be going on in the unknown universe.. so the issue here is that we need to find that unknown universe.

    It is amazing that Cern ( European agency) wants to announce the Higgs Bozon finding on Wednesday (USA’s independence Day) !!!!!!!! Is it really god particle … or it is 0?

    Bhagirath Joshi

  • Bhagirath Joshi

    Question: Nobody knows why the lifter levitates.
    Answer: The lifter levitates due to the force from ionized air. on the aluminum surface of the device 21 Kv +ve charge is applied. The air particles collide with the device and are immediately gets +vely charged. The ground is negative and these charged air particles rush to the ground, which gives thrust to the lightweight device and it levitates. If you hold burning match under the levitator you will see that the flame is bent due to the air current. You already mentioned that it does not levitate in vacum. This is due to lack of air particles.

    Bhagirath Joshi

  • Andrea Rossi

    Dear Nixter:
    We are already producing plants. Your choice is right and easy.
    Warm Regards,
    A.R.

  • Nixter

    Engineer Andrea Rossi,

    There is an interesting new Poll at E-Cat World, http://www.e-catworld.com/

    Who Will Produce the First Mass Market LENR Product?

    Leonardo Corp. (A. Rossi)

    Other

    I voted that Leonardo Corp. (A. Rossi) will be first.

    Do you think I am right? You are perhaps the only person in the world who knows the answer to this.

  • Italo R.

    Dear dr. Rossi, as you probably know, in these days and weeks there are conferences in many part of the world about LENR.
    Scientists and physics will talk about theories, experiments, researches and R&D.
    It seems that they don’t know that there already is a man who is already selling 1 MW LENR plants…

  • Hank Mills

    Hello Joe,

    I read an interesting theory about why the E-Cat may tend to become unstable when very high COPs are produced. I will try to share the theory here. This is total speculation and is probably very wrong. Please do not take this as fact.

    THE FOLLOWING IS ONLY SPECULATION THAT IS PROBABLY VERY WRONG.

    The theory is that at lower temperatures a Casimir effect is taking place in the micro-cavities in the micro-nickle powder. The size and dimension of these shapes allow for molecular hydrogen to enter, and then the Casimir effect adds energy which produces atomic hydrogen. The atomic hydrogen is then cycled back into molecular hydrogen and then back into atomic hydrogen. Each time it takes place energy is added. The result is free energy being extracted from the Zero Point Energy field or the virtual photon flux. The free energy is in the form of heat.

    According to this theory, at high COP (6) this free energy effect is producing most of the excess heat. Also, it is doing so without the production of many gamma rays because very few transmutations are taking place. However, at very high COPs fusion actually takes place. Protons actually enter the nucleus of nickle atoms and produce a series of nuclear reactions which results in the production of copper.

    So at high COP we have free energy producing most of the heat, and at very high COP we have fusion, fission, and positron-electron annihilation producing most of the heat plus gamma rays.

    Why do I think that there is a .001% chance this theory may be correct?

    Andrea Rossi has stated that there are four ways that energy is being produced in an E-Cat. One of them is fusion, another is fission, another is positron electron annihilation, and the last one is the effect that produces the most energy.

    The above theory seems to fit in my opinion. However, please do not consider it to be accurate or correct. It is probably wrong.

    If the theory is correct, it would be pretty amazing, because it would mean the E-Cat is not only a fusion, fission, and matter/anti-matter reactor, but also extracts energy from the zero point energy field.

  • Andrea Rossi

    Dear Joe:
    1- yes
    2- If it happens, the E-Cat stops, because if the reactants melt, there are no more powders, but liquis metals and grains, with which the process does not work. Therefore the process is intrinsecally safe.
    Warm Regards,
    A.R.

  • Joe

    Dr Rossi,

    1. Have you been able to identify the specific phenomenon that is hindering your progress in achieving stability at COP>6?

    2. Is there a possibility that the E-Cat reaction could raise the melting temperatures of the various reactants?

    All the best,
    Joe

  • Joe

    Wladimir,

    In QRT, how are the quark and gluon defined?

    All the best,
    Joe

  • Hank Mills

    Hello Everyone,

    I want to make one thing clear.

    Andrea Rossi has never stated that he has produced stable heat production at 1000 C.

    As far as I am aware, he has only stated that the new E-Cat designs are operating with stability at temperatures above 600 C. How far above 600C are they operating at? We do not know!

    However, an individual who has shared very accurate information in the past (Francesco CH) has hinted on the E-Cat World website that 1000 C. temperatures may be Rossi’s goal. Of course this is just speculation, and we do not know for sure if this is the case.

    ONCE AGAIN ROSSI HAS NOT STATED 1000 C. TEMPERATURES ARE BEING PRODUCED.

    As many people know, I’m very excited about the E-Cat technology, and I like to share the latest news about the technology. But if I went beyond my bounds by mentioning the possibility of 1,000 C. heat production I apologize. I did not intend to mislead anyone, or make anyone think that Rossi had made such a claim when he had not done so. The mention of the figure — from someone who has given accurate info in the past — perhaps made me get a little too anxious to tell others about the possibility.

    Again, I apologize.

    Please forgive me for my excessive enthusiasm about this technology. I should be more careful to specify what is KNOWN, what is NOT KNOWN, and what is only speculation.

    Thank you.

    Hank Mills

  • Andrea Rossi

    Dear Guru Gurovich
    I think only to work to make real products. I am mot interested to other issues.
    Warm regards
    AR

  • Guru Gurovitsch

    Dear Mr. Rossi,

    One consideration:

    How much folks know name of first astronaut Gagarin ?
    Answer is: all folks

    How much folks know name of 4th astronaut ?
    Answer is: maybe 2 from 1000.

    This is what I am aiming at You. A lot of inventors make a note: Invention is 2% of idea and 98% of sweat. You are working 16 hours per day for some years.
    Some of your not so angelic competitors are planning to present a big humbug at IFFC-17 in Korea in August.

    Mr. Rossi, please don’t allow them to steal all spotlights and all fame.
    After 20 years of numbers of laboratories apparatuses capable of 20 milliWatt output, You are the real inventor of first truly useful LENR device.
    You deserve not only commercial success (and some profit), however you deserve a “immortality” (glory) too.
    You are the truly Gagarin, not Ting-Gong-Kung a 338th astronaut in chartlist.

    Every day of delay of this knowledge to unleash to whole public cost world around 10 billion US dollars. Yes, this silence cost world 10 billions of dollars per every day mostly in wrongly allocated capital into obsolete technologies and consecutive effects.

    Please, allow to whole world to know, that very powerful, clean and safe LENR is real and proven after 20.000 hour test, i.e.: as first in world, not 338th.

    With Best Regards Guru

  • Wladimir Guglinski

    Dear Joe,
    let me explain the similarity between the creation of the photon, the neutrino, and the bosons W and Z.

    PHOTON:
    A photon is generated by an excited atom by the following steps:

    1- A particle is created, from the agglutination of some massless particles of the aether. The quantity of particles agglutinated depends on the energy involved in the process. Light photons have small quantity of massless particles of the aether.

    2- As the particle is extracted from the “sea” of massless particles of the aether, its creation induces the creation of its antiparticle.

    3- There is a delay between the birth of the two corpuscles. As after the creation of the particle it starts to move with the speed of light, the delay provokes a distance “d” between the particle and the antiparticle.
    Such distance d is responsible for some properties of the light. For instance, when a photon enters into a polarizator, its polarization occus when there is resonance, which depends on the distance “d” and the distance “D” between the planes of the crystal. If the resonance does not occur, the photon does not crosse the polarizator.

    In my book Quantum Ring Theory it is shown that by considering such distance “d” in the photon it’s possible to explain the EPR paradox, without to take in consideration the entanglement.

    NEUTRINO:
    Its creation follows the following steps:

    1- A postitron is created, from the aggutination of massless particles of the aether.

    2- The creation of the positron induces the creation of an electron.

    3- The electron moves about the positron

    4- They positron-electron move with helical trajectory

    When an antineutrino is created, the electron is created firstly, and the positron gyrates about the electron.

    In the case of the photons and the neutrino, the repulsive gravity avoids the direct interaction of the particle and antiparticle, and so they do not anihilate one each other.
    BOSON W:
    Its creation follows the following steps, in the neutron decay:

    1- A boson W(+) is created.
    2- A boson W(-) is generated from the creation of the boson W(+)
    3- The repulsive gravity is not able to avoid their direct interaction. So they anihilate one each other at once they are created (even before their lifetime expires). There is no way to detect them in the neutron’s decay.

    In the case of the formation of the neutron, when an electron is captured by the proton and interacts with its quark up, the boson W(-) is created firtly.

    BOSON Z:
    The boson Z is created in the collisions between neutrinos and electrons.
    As the boson Z has no charge, and its time life is very short, there is no way to detect it from the collisions of neutrinos and electrons.

  • Wladimir Guglinski

    ERRATA:

    In my last post, where it is written:

    However the boson Z does play any role in the interaction electron-neutrino.

    the correct is:

    However the boson Z does NOT play any role in the interaction electron-neutrino.

  • Andrea Rossi

    Dear Andy H.:
    I cannot post updates of the work in progress regarding the certifications, it is under NDA. We will give information when the certification will have been granted.
    Warm Regards,
    A.R.

  • Andy H.

    Dear Mr. Rossi,
    We understand that the certification is the last thing before production. Would you be willing to post frequent updates on the certification progress?

  • Wladimir Guglinski

    Joe wrote in June 27th, 2012 at 7:08 PM

    1. Do you believe that the weak force is a fundamental force or a composite/partial force?

    2. Since you believe that the weak particle plays no role in the decay of a neutron, what do you believe the role of the weak particle to be?

    3. Have scientists not witnessed the expulsion of a weak particle from a nucleus?

    4. Have scientists not witnessed the decay of the weak particle into an electron (antielectron) and an antineutrino (neutrino)?

    Dear Joe
    Yukawa proposed a model of neutron where a meson jumps between two protons.
    The physicists started to look for finding the meson, and detected it in experiments.
    But we know today that Yukawa’s model of neutron is wrong. The meson does not exist within the structure of the neutron, and it does not play any role in the interactions within the neutron.

    The weak force has a range of interaction extremelly short, a conclusion based on Fermi theory.
    In the page 157 of my book Quantum Ring Theory it’s written:
    As we see, the conclusion based on the Fermi’s theory implies that the weak force can be a consequence of the interaction (ue) shown in Figure 1, where d is very short”

    “d” is the distance between the electron and the quark up within the structure of the neutron proposed in my theory.

    The structure of neutron in my theory is (u,d,ue), where ue is the quark up and the electron tied through the spin-fusion mechanism.

    The beta-decay of neutron occurs as follows:

    1- Before the decay, the electron is moving without the helical trajectory. When the decay occurs, the electron recovers its helical trajectory, and an antineutrino is emitted.
    The antineutrino does not exist within the structure of neutron. And so it does not play any role in the interactions within the neutron. It is only a particle emitted when the electron recovers its helical trjectory.

    2- When the neutron’s decay occurs, the interaction ue is broken. As consequence of this break, a boson W is created, with very short time life.
    The boson W did not exist within the neutron. It does not play any role within the neutron’s structure. It does not play any role in the decay of neutron, nor before neither after the decay. It is only a particle created when the interaction ue is broken.

    However, note that, as the boson W is created as consequence of the weak interaction between the quark up and the electron, its characteristics can be predicted successfully from the rules of the Standard Model.

    In the collisions between electrons and neutrinos, as they interact via weak force, the boson Z is created.
    However the boson Z does play any role in the interaction electron-neutrino. The boson Z is only a partice created during the interaction, with a time life very short. That’s why its characteristics are predicted successfully by the Standard Model.

    As happened in the case of Yukawa theory, where the meson was wrongly considered as playing a role within the neutron, also the existence of bosons W and Z was wrongly intepreted by the physicists. These two bosons do not mediate the weak interaction, as was wrongly interpreted by the theorists. They are only particles produced during the weak interaction.

    Regards
    WLAD

  • Joe

    Wladimir,

    1. Do you believe that the weak force is a fundamental force or a composite/partial force?

    2. Since you believe that the weak particle plays no role in the decay of a neutron, what do you believe the role of the weak particle to be?

    3. Have scientists not witnessed the expulsion of a weak particle from a nucleus?

    4. Have scientists not witnessed the decay of the weak particle into an electron (antielectron) and an antineutrino (neutrino)?

    All the best,
    Joe

  • Andrea Rossi

    Dear Nixter:
    What I am doing is for me just my duty, my job. I do not consider it anything exceptional, if I think to the complexity of the Universe. We are just surviving. About the specialized magazines: surely we will give them information for the public.
    Warm Regards,
    A.R.

  • Nixter

    Engineer Andrea Rossi,

    Can you disclose your plans for final product roll out of the home E-Cats when production finally begins? Will pre production units be given to a select few for review, publications like Consumers Reports, Popular Mechanics, etc, etc? Or will you only first ship to paying customers in the general public. Have you considered extending any privileges or premiums to those who signed up for the E-Cat list early? What will determine the order in which potential purchasers are contacted? Can you tell us what the time-line will be from production start, to first marketable units coming off the line?

    Although I may sound cool, calm and relaxed, I am actually extremely excited by your E-Cat Technology, its impact will extend beyond our Earthly boundaries and effect humanity more profoundly than we can imagine. I wonder,… is it possible that you are so involved with developing your inventions, that you have become mentally “conditioned” to think of it as normal and mundane? Therefore underestimating its true potential to alter the course of Human history by orders of magnitude previously though impossible?

    Simply put, do you realize how huge this is?

  • Dear Wladimir Guglinski,

    A few weeks back now you stated, “if you have any material why not put it on the web”. My response was, I will but it could take a few weeks and it has. As you are aware I have consistently stated that your helical trajectory theory is correct and I have also stated that I am more mechanical than an academic theorist. I believe that your Quantum Ring Theory is rejected for the same reasons that cold nuclear fusion is and of which has had such a hard time and my technology also and to name another Gaston Naesens 714X nitrogen therapy of which I am a personal witness to its validity but only from the higher levels of involvelment do these rejections filter down. Your theory Wladimir is in the same category as my technology. It is basically before its time. This is what I have been told with regards my technology. We are now, I believe, a hiatus in physics because we cannot move forward in certain departments without destroying mysteries and this the crux of the problem for people who are involved in innovative technologies or new theories that challenge the old accepted ones. Perhaps my web site will help some people to understand why certain areas in physics is considered taboo. The understanding of the fission/fusion interaction is the key to understanding structure. The understanding of the e-cat process is to do with with fission and fusion or transmutation which is to do with structure, This is a valid part of the problem, not just the financial aspects. What I say is that you can structure physical material (the gaseous/liquid diminsions) by understanding ‘the static and mobile mechanics of energy interaction’. You can do this also with the aether dimension to form four dimensions of structure in rapid progression of aether to form fizzing and fusing spheres of light. Both of these activities will provide propulsion, both of these activities require a strict timing sequence and this is where ‘Time’ enters the energy equation in the manipulation of aether but as you must realize, as stated previously, this subject is taboo. See http://triteckindustries.ca Regards Eric Ashworth

  • Wladimir Guglinski

    Dear Joe,
    there is another serious problem with the current nuclear model of Nuclear Physics.

    In the page 643 of their book Quantum Physics[ 2 ] (in Portuguese), Eisberg and Resnick write:

    Analyzis of alpha particles scattering by the nuclei corresponding to small values of A have showed that the radii of such nuclei are a little small than 10F. […] Analyzis of the tax emission of alpha paticles by radioactive nuclei having higher A have showed that the radii of those nuclei, defined in the same way, are aproximatelly equal to 9F.

    But with a radius equal to 9F would be impossible to explain the aggregation of light nuclei, because the distance between nucleons would be larger than 2fm.

    How did the nuclear physicists solve such paradox ?

    Simple. They rejected the experiments with alpha particles scattering. They replaced the alpha particles scattering by electron scattering.

    By this way they obtained the following equation of the radius of nuclei:

    R = Ro.A^1/3 , where Ro= 1,2fm

    Let’s analyse the beryllium nucleus, as follows.

    The 4Be8 would have a radius 1,2×2 = 2,4fm

    As the nuclei have a nuclear spin, the nucleons 1H2 are submitted to centripetal acceleration.

    The centripetal force on a nucleon 1H2 with mass m would be:

    Fc = m.W².R = m.W².2,4

    But the radius 7fm obtained in John Arrington experiment the centripetal force is:

    Fc = m.W².7 , which is 3 times stronger than the force considered in current Nuclear Physics.

    So, there two serious points to be noted:

    – As the strong force actuates in the range of 2fm, it cannot agglutinate the 4Be nucleus.

    – And what is the worst: there is a strong centripetal force trying to expel the 1H2.

    As the strong force is unable to respond for the 4Be nucleus aggregation, the situation becomes worst because besides the repulsion between the 1H2 and the central 2He4 there is a centripetal force trying to expel the 1H2.

    Only by considering the flux n(o) it is possible to explain the aggregation of beryllium nucleus.

    Regards
    WLAD

  • Wladimir Guglinski

    Joe wrote in June 26th, 2012 at 1:58 PM


    1. Does a central body capture, by way of its gravitational flux n(o), only those bodies that are smaller than itself? Is this the reason for which 2He4 will capture 1H2 but not another 2He4? If so, there must be a different mechanism that allows 1H2 to

    i) capture another 1H2 (same size)

    AND

    ii) form a central body (2He4) (there are no other types of central body).

    Dear Joe,
    I dont have all the answers.

    Some years ago I was thinking about such questions, and I imagined that perhaps in the future the technology will be developed so that to get a nucleus with a central 2He4-2He4 (two 2He4 coupled in the center). That would be a new element not existing in nature.

    One 1H2 cannot capture other 1H2 because when they met together they form a 2He4.
    Besides, as the flux n(o) of the 1H2 which captures the other 1H2 is weaker than the flux n(o) of 2He4, the 1H2 cannot keep the other 1H2.

    And when one 2He4 captures other 2He4, the flux n(o) cannot keep it, because there is repulsion between 4 protons.

    An interesting phenomenon occurs in the alpha decay of heavy nuclei. The central 2He4 is emitted, and the nucleus is broken. But immediatelly another central 2He4 is formed, by the fusion of two 1H2.


    2. In response #2 of June 20, 2012, you state, “The gravitational flux n(o) of the 2He4 is stronger than that of the 1H2. And when the 1H2 is captured by a 2He4, the flux n(o) of the 1H2 reinforces the flux produced by the 2He4. The more nucleons 1H2 are captured, more stronger the flux n(o) into the 2He4 becomes.”

    The term ‘stronger’ connotes that a bigger nucleus would necessarily be a more stable one. But in reality, are not bigger nuclei generally less stable?

    Stability is consequence of several conditions:

    – if the number Z of protons is pair or odd

    – if the number of hexagonal floors is pair or odd (for instance, 20Ca40 has 3 hex. floors, and this is the reason why it is very stable; when the number of floors is pair, the nucleons 1H2 of one floor takes a place in front of other 1H2 of the another floor)

    – the saturation of the flux n(o) within the 2He4 (it cannot grow anymore) and the increase of repulsion due to the growth of the quantity of protons

    All the best,
    WLAD

  • Andrea Rossi

    Dear Damiano:
    We have to wait for the certification, after which we have already prepared the line of production for mass scale.
    The industrial plants are already in production.
    Warm Regards,
    A.R.

  • Damiano

    Hello Andrea

    Do you have an estimate for the production of the smaller heating units ?
    (the ones that just produce heat for a household)

    Thanks

    Damiano

  • Joe

    Wladimir,

    1. Does a central body capture, by way of its gravitational flux n(o), only those bodies that are smaller than itself? Is this the reason for which 2He4 will capture 1H2 but not another 2He4? If so, there must be a different mechanism that allows 1H2 to

    i) capture another 1H2 (same size)

    AND

    ii) form a central body (2He4) (there are no other types of central body).

    2. In response #2 of June 20, 2012, you state, “The gravitational flux n(o) of the 2He4 is stronger than that of the 1H2. And when the 1H2 is captured by a 2He4, the flux n(o) of the 1H2 reinforces the flux produced by the 2He4. The more nucleons 1H2 are captured, more stronger the flux n(o) into the 2He4 becomes.”

    The term ‘stronger’ connotes that a bigger nucleus would necessarily be a more stable one. But in reality, are not bigger nuclei generally less stable?

    All the best,
    Joe

  • Andrea Rossi

    Dear Wladimir Guglinski:
    A valid product produces richness and profits: richness and profits produce taxes.
    Warm Regards,
    A.R.

  • Wladimir Guglinski

    Dear Andrea Rossi,

    obviously the most great resistance by governments of all countries against the implementation of eCats will be the lack of taxes , because the users of eCats no longer will pay taxes for energy consupmtion.

    Perhaps the way to avoid such resistance is starting up a negociation with the governments, so that to establish taxes to be payed by the users of the eCats.

    regards
    WLAD

  • Joseph Fine

    Martin, A.R.

    This article, (see below) is in response to Martin’s comment on Fukushima. Almost nothing good can be said about the accident, or multiple accidents. However, Crystalline Silico-Titanates (CSTs) have been used to remove nearly all of the radioactive Cesium from contaminated water, and that’s radioactive Cesium that didn’t have to be dumped in the ocean. So it felt good to hear that some good has come out of this, even if it only kept some radioactive waste out of the ocean. The best way to deal with serious accidents is never to have one.

    https://share.sandia.gov/news/resources/news_releases/fukushima_cleanup/

    To this end, there is a long technology ramp-up to replace the world’s nuclear reactors.

    Let’s hope it is not so long.

    Joseph Fine

  • Andrea Rossi

    Dear Martin:
    We are already manufacturing the industrial 1 MW plants and we are close to the manufactiruing of high temperature plants. We have a Licensee for Japan.
    Warm Regards,
    A.R.

  • Martin

    Dear Andrea,

    Today i watched a documentry about the nuclear disaster in Japan.
    Very heavy stuff. The people of Japan are suffering a lot. They can’t go
    Back to There homes for at least 30 year and it may be longer.
    Can you give them à fast perspective to get rid of (hot) nuclear power?
    It is very important that your invention is coming fast to prevent more damage!
    Can you give us an update about THE progress of THE ecat?

    Best regards,

    Martin

  • Wladimir Guglinski

    Joe wrote in June 24th, 2012 at 3:05 PM


    1. Is the electric particle from QRT massless because it is composed of a particle-antiparticle pair? If so, then

    No, Joe.
    The particles of the aether are not massless when they are alone. But you cannot find a particle of the aether without its antiparticle.

    The composition composed by paticle/antiparticle behaves as massless.
    I think a particle/antiparticle of the aether do not anihilate one each other because there is repulsive gravity between them, which avoid them to interact directly.

    I avoid to say that the particles of the aether are not massless because the physicists would interpret it wrongly.


    ii) why does the electric particle have an electric charge at all since a combination of particle and antiparticle would always yield a particle that is neutral?

    I suggest you to ask it to God.


    2. How can there not be available to the nucleus enough energy to form larger central units since, as you stated 4 days ago in response #2, “The gravitational flux n(o) of the 2He4 is stronger than that of the 1H2. And when the 1H2 is captured by a 2He4, the flux n(o) of the 1H2 reinforces the flux produced by the 2He4. The more nucleons 1H2 are captured, more stronger the flux n(o) into the 2He4 becomes.” Does not a stronger flux into 2He4 provide 2He4 with more and sufficient energy to combine it with other bodies to form a larger central body?

    Look at the figure 9.1 in my article published last month in JNP:
    http://www.journal-of-nuclear-physics.com/files/How%20repulsive%20gravity%20contributes%20for%20cold%20fusion%20occurrence.pdf

    That figure 9.1 is concerning the proton. But as the 2He4 is composed by 4 protons and 2 electrons, you can understand what happens when 1H2 is captured by the nucleus 2He4, by looking at that figure.

    Suppose that, in that figure 9.1, instead of a proton, the blue particle in the center is the 2He4.

    As you can see in the figure, there is agglutination of repulsive gravitons G about the flux n(o) of gravitons g.
    And you may realize that such agglomeration of repulsive gravitons G take place between the central 2He4 and the nucleons 1H2 captured by it.
    So, there is not a direct attraction between the central 2He4 and the nucleon 1H2.

    The nucleon 1H2 is captured by the flux n(o), and not by its attraction with 2He4. The structure of beryllium 4Be8 detected by John Arrington experiment corroborates such hypothesis, since there is a distance 7fm between the 2He4 and the 1H2, and so the strong force cannot respond for the agglutination of 4Be8.


    3. In the quote of you just above, you seem to suggest that the amount of flux of “dynamic” gravity depends on the mass of the particles involved, namely 1H2 and 2He4. But if flux is not dependent on mass as you have stated in response #3 on June 20, then why do 1H2 and 2He4 have different amounts of flux?

    Dear Joe, English is not my first language, and I’m afraid you are not understanding well my words.

    Of course two protons have a stronger flux n(o) than one unique proton.

    There are two sort of dynamic gravity:

    1- strong force- it’s the force of interaction between two particles with relavive motion. For instance, when a proton with speed V interacts with other proton with speed V in contrary direction. The velocity of interaction is 2V. In this sort of interaction the velocity is the stronger agent of interaction. This is the interaction known in current Nuclear Physics.

    2- gravitational flux – it’s the interaction between particles when one is captured by the flux n(o) of another particle. It’s not known in current Nuclear Physics.


    Also, in QRT, what is the difference between a neutron and a hydrogen atom?

    Joe, I suggest you to read two articles of mine in Peswiki:

    Zitterbewegung Hydrogen Atom of Quantum Ring Theory
    http://peswiki.com/index.php/PowerPedia:Zitterbewegung_Hydrogen_Atom_of_Quantum_Ring_Theory

    Cold fusion, Don Borghi’s Experiment, and hydrogen atom
    http://peswiki.com/index.php/PowerPedia:Cold_fusion%2C_Don_Borghi%27s_Experiment%2C_and_hydrogen_atom

    All the best,
    WLAD

  • Joe

    Wladimir,

    1. Is the electric particle from QRT massless because it is composed of a particle-antiparticle pair? If so, then

    i) what is the identity of that subparticle?

    ii) why does the electric particle have an electric charge at all since a combination of particle and antiparticle would always yield a particle that is neutral?

    2. How can there not be available to the nucleus enough energy to form larger central units since, as you stated 4 days ago in response #2, “The gravitational flux n(o) of the 2He4 is stronger than that of the 1H2. And when the 1H2 is captured by a 2He4, the flux n(o) of the 1H2 reinforces the flux produced by the 2He4. The more nucleons 1H2 are captured, more stronger the flux n(o) into the 2He4 becomes.” Does not a stronger flux into 2He4 provide 2He4 with more and sufficient energy to combine it with other bodies to form a larger central body?

    3. In the quote of you just above, you seem to suggest that the amount of flux of “dynamic” gravity depends on the mass of the particles involved, namely 1H2 and 2He4. But if flux is not dependent on mass as you have stated in response #3 on June 20, then why do 1H2 and 2He4 have different amounts of flux?

    Also, in QRT, what is the difference between a neutron and a hydrogen atom?

    All the best,
    Joe

  • Wladimir Guglinski

    Joe wrote in June 22nd, 2012 at 9:19 PM


    1. In the Standard Model, all charged fundamental particles – electron, quark, W – have mass. Yet, you propose a charged particle that is massless. Do you believe that charge is independent of mass? If so, then what is the definition for charge in QRT?

    In general people think about mass in the sense of Newton, which is a concept applied to matter (a body formed only by particles as protons, neutrons, electrons).

    The Newtonian concept of mass cannot be applied for particles composed by matter-antimatter.

    For instance, the photon is composed by particle/antiparticle. The photon has no mass in the sense of Newton, because the photon is not matter, since the combination particle/antiparticle is not matter.

    But the photon has mass in another sense, since the particle and the antiparticle that compose the photon have mass.
    However, when a particle with mass “m” and an antiparticle with mass “m” form a photon, their mass is not 2m.
    The photon is an exotic particle, because when it moves it behaves as it should have no mass. In the sense of Newton, it is massless.
    But when a photon is stopped in its trajectory, the two particle try to exhibit their masses. But such mass is vanished immediatelly (the particle and the antiparticle anihilate one each other).
    In this sense, the particles that compose the aether also have mass.

    The concept of mass of a particle must be understood by considering the sort of its interaction with the aether. A particle with mass at rest can have mass zero if it moves together with its antiparticle.

    In QRT, the neutrino is formed by electron+positron, and its mass is almost zero, in spite of the electron and the positron have a big mass.


    2. Why can two 1H2 overcome their repulsive gravitational and electric fields to form one 2He4, but two 2He4 not do likewise to form an even larger central body?

    In the page 120 of the book Quantum Ring Theory it is shown that there is need three sort of energy, so that to agglutinate two 1H2, in order to get a 2He4:

    1- Energy Er – so that win the Coulomb repulsion
    2- Energy Ed- so that to perforate the secondary field
    3- Energy Em – so that to match the principal field of the two 1H2

    There is not available into the nuclei sufficient energy so that to match the two principal fields of 1H2

    3. Two days ago, in response #3, you stated, “The difference of masses between proton and neutron has not influence in the case of dynamic gravity.” Does this imply that the mass of 1H2 and 2He4 also plays no role in determining the strength of the gravitational attraction between these two? If that is the case, then one 1H2 is attracted to another 1H2 as easily as it is to one 2He4, which of course would disrupt QRT.

    No, I only explained to you that you cannot appply your understanding of stationary concept of gravity within the nuclei, because inside there the gravity that takes place is the dynamic gravity.
    You are reasoning by trying to apply the concepts of current Nuclear Physics to the nuclear models of QRT.
    Such way of yours makes no sense, because QRT works by laws different of that considered in current Nuclear Physics.


    Also, if an electron loses its spin in “spin-fusion”, where does that spin go? Spin angular momentum can not simply vanish.

    According to QRT, quantum-spin is the combination of two motions:
    1- intrinsic-spin, which is the rotation of the particle about its own axis
    2- zitterbewegung, which is the rotation of the particle about the axis of its helical trajectory

    When one electron loses its spin (its helical trajectory is vanished), a neutrino is emitted. So, the total angular momentum is kept. This happens when a proton capturates one electron, and they form a neutron.

    When one electron without spin recovers its spin again, an antineutrino is emitted (the electron also recovers its helical trajectory). This happens in the beta decay of neutron.

    All the best,
    WLAD

  • Joe

    Wladimir,

    1. In the Standard Model, all charged fundamental particles – electron, quark, W – have mass. Yet, you propose a charged particle that is massless. Do you believe that charge is independent of mass? If so, then what is the definition for charge in QRT?

    2. Why can two 1H2 overcome their repulsive gravitational and electric fields to form one 2He4, but two 2He4 not do likewise to form an even larger central body?

    3. Two days ago, in response #3, you stated, “The difference of masses between proton and neutron has not influence in the case of dynamic gravity.” Does this imply that the mass of 1H2 and 2He4 also plays no role in determining the strength of the gravitational attraction between these two? If that is the case, then one 1H2 is attracted to another 1H2 as easily as it is to one 2He4, which of course would disrupt QRT.

    Also, if an electron loses its spin in “spin-fusion”, where does that spin go? Spin angular momentum can not simply vanish.

    All the best,
    Joe

  • Andrea Rossi

    Dear mark Saker:
    We need at least one more month to get sure results.
    Warm Regards,
    A.R.

  • Mark Saker

    Dear Andrea,

    Phew! :)

    So have you decided upon a date to end the extended period high temperature reactor test?

    Thanks

    Mark

  • Wladimir Guglinski

    Joe wrote in June 20th, 2012 at 9:52 PM


    1. Since the secondary field is electric, it can not be induced by the action of a principal field that is either gravitational or electric. A rotating gravitational field induces a gravitomagnetic field. And a rotating electric field induces an electromagnetic field. Neither one is coulombic.

    Dear Joe,
    you are wrong.
    You are thinking about macroscopic phenomena, produced by interaction of flux of electrons, or by gravitational interactions of bodies formed by protons and neutrons.

    I am not speaking about them. I am speaking about fluxes of particles of the aether.
    Quantum Ring Theory proposes the laws of induction between the flux of electric massless particles of the aether and the flux of gravitons (named n(o) in my QRT).
    It’s the unification of gravity and electromagnetism.
    It is a similar work of that made by Faraday for the laws of electromagnetism.
    The difference is that he used experiments.
    I used my nuclear model so that to discover the laws that unify gravity and electromagnetism.

    2. How do each of two 1H2 overcome the repulsive gravitational field of the other in order to form one 2He4?

    There is need to know the laws that rule the interactions between the particles of the aether.
    One attractive graviton g do not intereact directly with other graviton g.
    The repulsive gravitons G are captured by attractive gravitons g only in certain special condictions (for instance, gravitons G agglutinate around the flux n(o) formed by gravitons g).

    3. You say that gravity is independent of mass at the smallest scale.

    I did not say that

    Also, why do you not believe that a neutron is composed of 3 quarks?

    Because from the model of neutron n=p+e it’s possible to explain several phenomena that it’s impossible to explain from the current model of quarks.

    Besides, QRT proposes the spin-fusion concept, according to which the electron (or any lepton) loses its spin in the interaction with the quarks of proton when they form the neutron.
    There are several high energy reactions which cannot explained by current Standard Model.

    For instance:
    The mesons pi and Rho have the same structure, but:

    • The meson pi(+) has structure ud’ , its rest mass is 140MeV , and its time decay is 2,6×10^-8s
    • The meson Rho(+) has structure ud’, its rest mass is 770MeV , and its time decay is 0,4×10^-23s
    • The meson pi(0) has structure (uu’+dd’)/2^1/2 , its rest mass is 135MeV, and its time decay is 0,8×10^-16s
    • The meson Rho(0) has structure (uu’+dd’)/2^1/2 , its rest mass is 770MeV, and its time decay is 0,4×10^-23s.

    Note the following:
    1- The masses of pions pi(0) and pi(+) have a difference of 5MeV.
    But Rho(0) and Rho(+) have the same mass 770MeV
    Why ????
    2- The pions pi(0) and pi(+) have different time decays: 2,6×10-8s and 0,8×10^-16s.
    But the mesons Rh(0) have the same time decay: 0,4×10^-23s s
    Why ????

    There is no way to explain it from the current Standard Model of Modern Physics, because:

    A) If we use an argument so that to explain the difference of mass 5MeV between pi(0) and pi(+), however the same argument would have to be applied to the masses of Rho(0) and Rho(+) and they would have to exhibit a difference of mass too. But Rho(0) and Rho(+) have the same mass !!!

    B) If we use an argument so that to explain the difference of time decay between 0 and +, however the same argument would have to be applied to the time decay of Rho0 and Rho+ , and they would have to exhibit a difference of time decay. But Rho0 and Rho+ have the same time decay !!!

    In my book to be published in the upcoming months in London, I show that such paradox is eliminated when we consider the spin-fusion, as follows:
    – there is one electron hidden into the structure of the meson pi(-)
    – there is a positron in the meson pi(+)
    – those mesons have spin zero because the electron and the postitron lose their spin in their interaction with the quarks of the meson, thanks to the spin-fusion phenomenon proposed in my theory.


    Does the strong interaction not have a place in QRT?

    The structure of beryllium nucleus detected by John Arrington experiment (where a nucleon 1H2 has distance 7fm from the central 2He4, proved that strong force cannot agglutinate the nuclei..
    In QRT the nuclei are agglutinated by the flux n(o)which captures the several 1H2.

    In QRT the strong force is considered as a dynamic sort of gravitational interaction, but it does not respond for the agglutination of nuclei.

    All the best,
    WLAD

  • Andrea Rossi

    Dear Mark Saker:
    As you see, you can post without problems. Probably some message from you contained links that our robot interpreted as spam.
    We for sure have not censored your comments.
    Warm Regards,
    A.R.

  • Mark Saker

    Dear Andrea,

    I do not seem to be able to post from my usual e-mail address. Have I been blocked? For what reason? :(

    Have you decided upon a date to end the extended period high temperature reactor test?

    Thanks

    Mark

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