Stability of light nuclei

by
Wladimir Guglinski
retired, author of the Quantum Ring Theory
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Abstract
Dr. Wilfried Nörtershäuser of the Helmhotz Center for Heavy Ion Research at the University in Mainz says on the 2009 experiment which had detected a neutron halo in 4Be11 with distance 7fm from the cluster:
“By studing neutron halos, scientists hope to gain further understanding of the forces within the atomic nucleus that bind atoms together, taking into account the fact that the degree of displacement of halo neutrons from the atomic nuclear core is incompatible with the concepts of classical nuclear physics”[ 2 ]
In the case of 4Be11, the halo neutron and the nuclear core are separated by the distance of 7fm, and so such isotope represents the experimental proof that the cohesion of nucleons within the light isotopes cannot be promoted by the strong nuclear force.
Such experimental discovery published in 2009 had been predicted years ago, because according to the new nuclear model proposed in Quantum Ring Theory, published in 2006, the cohesion of the nucleons within the light nuclei is not caused by the strong nuclear force.
Here in this paper the new nuclear model is submitted to a scrutinity so that to verify whether from its structure it’s possible to explain the stability of the light nuclei and to reproduce the nuclear properties as nuclear spins, electric quadrupole moments, and magnetic moments. Nuclear magnetic moments are calculated from two different and independent methods.  In the second, named “method of equilibrium between nucleons”, it’s presented the Lagrangian for nuclei with Z < 8.  The results obtained from them agree each other, and are corroborated by nuclear spins and electric quadrupole moments suplied by nuclear tables.
In this Part One are presented calculations on magnetic moments for the isotopes of lithium, beryllium, and boron. In the next paper Part Two will be exhibited  calculations for carbon, nitrogen, and oxygen.  In the paper Part Three the author will exhibit calculations for electric quadrupole moments.
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485 comments to Stability of light nuclei

  • http://www.e-catworld.com/2013/05/recording-of-rossi-smart-scarecrow-interview/

    Recording of Rossi Smart Scarecrow Interview
    Here is a recording of today’s interview that Sterling Allan and I conducted with Andrea Rossi on the Smart Scarecrow show.

  • Koen Vandewalle

    I wonder how many kWh of man made LENR would have been produced in the years coming and how many patents would have been granted in this matter without the endeavours of Andrea Rossi.
    Probably none !

  • Joe

    Wladimir,

    1. In QRT, what causes the sign of the spin of a nucleon to change abruptly and work against the Least Action Principle?

    2. What exactly is the effect of the gravitational flux n(o) on the sign of the spin of a nucleon, that an abrupt (working against the Least Action Principle) change in that sign should be IGNORED and treated as if there were NO change at all that had occurred, thereby affecting NEITHER the signs of the magnetic moments (intrinsic and induced) of that nucleon, NOR the F(SI) between that nucleon and any other nucleon?

    All the best,
    Joe

  • Giuliano Bettini

    Caro Andrea Rossi,
    siccome sono nato a Bologna (..tanti anni fa) mi permetto di chiedere:
    mi sembra che da Bologna e Ferrara sia stato portato via tutto.
    Tutto è stato portato negli Stati Uniti.
    Quindi per fare la ricerca e sviluppo si deve costruire nuovo hardware? e quale?
    Dear Andrea Rossi,
    just because I was born in Bologna (…. many years ago) I ask:
    it seems to me no hardware has been left there.
    For R & D which kind of new hardware is planned to be built there?
    Thanks
    Giuliano Bettini

  • Had a great interview with Andrea Rossi today, 1.5 hours.

    http://www.youtube.com/watch?v=0HwYXL-lmHY

    Interview with Andrea Rossi About 1 MW E-Cat Plant Delivery – We discussed the latest E-Cat plant status; the initial military customer plant status; the glowing-hot-cat, and safety considerations at this early stage, and their postponing of domestic product roll-out; the Hot-Cat research and work toward electricity generation; the Hot-Cat 3rd party test conditions and status, … (PESN; May 7, 2013)

  • Andrea Rossi

    Dear Steven N.Karels:
    1- yes
    2- yes
    3- yes
    Warm Regards,
    A.R.

  • Steven N. Karels

    Dear Andrea Rossi,

    I believe in re-stating the obvious to help clarify (or change) my understandings.

    For the Hot eCat:

    1. I assume from your posting that you are at a point in the Hot eCat development where you are routinely achieving excellent stability and control when the operating temperature is in the vicinity of 350 Celsius? Please verify.

    2. You have previously demonstrated in experimental testing temperatures in the vicinity of 1000 Celsius.

    3. You currently are working on future Hot eCats that may be able to continuously operate at temperatures higher than 350 Celsius, such as those around 600 Celsius. But this would be a future product whose current status is still in experimental development. Is this essentially correct?

  • Andrea Rossi

    Dear Marco:
    Yes, the system is perfect at 350 ( so far).
    Warm Regards,
    A.R.

  • Marco

    If the Hot cat is stable between 330 and 400 C, why do you limit the output to 350 C? Is it because the turbine that you are using for testing is limited to 350 C steam?

  • eernie1

    To DR Joseph Fine,
    Didn’t we discuss the use of Nickel isotopes and the method of centrifical separation on facebook some time ago?

  • Andrea Rossi

    Dear Manuel Cilia:
    interesting,
    Thank you.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear argon:
    You probably are not informed about the normal times of a scientific publication. Usually from tha date of a test and the date of the publication the average time is 6 months. I hope the publication will be made before this time is elapsed, though.
    I agree with you about the congratulations to Pekka Janhunen.
    Warm Regards,
    A.R.

  • argon

    Just wanted to put it up here also. My sincere congratulations to persistent JONP follower Mr Pekka Janhunen for getting his great invention “Solar Wind Sail” to orbit first time.
    You may follow the mission:
    http://www.estcube.eu/en/mission/what-is-estcube-1
    For Andrea, please make sure you use press independent testing group hard enough to publish report without additional delays. (I know you said you can’t, but there is no valid reason for report missing many weeks anymore)

  • Dear Dr Rossi,
    Please have a look at the follow company who is doing work on smaller scale CO2 gas turbines in the 10kw to 3Mw range. I think they are still in the prototype phase but it looks promising. they are based in the USA
    http://www.co2turbine.com/R744/Supercritical_CO2_Infinity_Turbine.html

  • Wladimir Guglinski

    Joe wrote in May 7th, 2013 at 3:54 AM
    Wladimir,

    In the last diagram of Figure 9 on page 12 of your article, why does the deuteron that is bound inside the nucleus have its electron orbiting only one proton, like in the case of a free deuteron? Should not the electron orbit each proton in a figure-8?

    Joe,
    it’s hard to make all the diagrams with all the accurated details.

    But you can consider that the electron in the last diagram of Fig. 9 is moving with an orbit with the shape of infinite symbol, by considering that the electron changes its spin when it is orbiting the proton at the left.
    With such consideration the electron keeps its µ=4,706, and the deuteron keeps its µ=+0,857.

    regards
    wlad

  • Wladimir Guglinski

    Dear Joe

    In 9th August 2012 John Arrington, from the Argonne National Laboratory, had sent an email to me, where he had written:

    I have not looked at your
    theory in any detail, in part because detailed nuclear structure is not my
    main focus, in part because I see no deficiencies with the present models

    In 3th October 2012 he had sent other email saying:

    Dear Wladimir,
    What I said is correct.
    1) I am not aware of any deficiencies in the current models, and in particular, not in the context of our recent measurement.

    In 16th April 2013 John had enphasized to me again that according to his opinion there is nothing wrong with the current nuclear models:

    “I said that I was not aware of any significant deficiencies in the current models describing the structure of light nuclei. […] So no, I have not changed my mind.”

    What do you think, Joe?

    For instance, consider that question in the Fig. 29-A, page 40, of the present paper.

    It’s very easy to understand that (by considering the classical current nuclear models) it’s IMPOSSIBLE to explain why the excited isotope 6C12 has spin i=2 and µ = 0.

    Please look at the Fig. 29-A again.

    Tell me:
    1- do you think is it possible to find an explanation for i=2 and µ = 0 of the excited 6C12, by considering the current nuclear models ?????

    2- do you think is it possible to explain i=2 and µ = 0 if 6C12 without to consider the flux n(o), as shown in Fig. 29 (page 40) ????

    3- Any honest scientist has the obligation to respond:

    A) it’s IMPOSSIBLE to solve the paradox of 6C12 by considering the current nuclear models

    B) it’s IMPOSSIBLE to solve the paradox of 6C12 without considering the flux n(o), as proposed in my theory.

    .

    If the nuclear theorists cannot understand such trivial question, then there is no honesty among the nuclear theorists.

    And then,as there is no honesty among the nuclear theorists, do you think is it possible to find a satisfactory solution so that to eliminate the inconsistencies of the current Classical Nuclear Physics ?????

    Regards
    WLAD

  • Joe

    Wladimir,

    In the last diagram of Figure 9 on page 12 of your article, why does the deuteron that is bound inside the nucleus have its electron orbiting only one proton, like in the case of a free deuteron? Should not the electron orbit each proton in a figure-8?

    All the best,
    Joe

  • Andrea Rossi

    Dear Steven N. Karels:
    1- yes
    2- from 330 through 400 Celsius
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Dr Joseph Fine:
    1- SSM is about 65% of the time
    2- yes
    3- we must trade COP for safety
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Gherardo:
    1- temperature of the wall of the heat exchanger: steam will have a tempersature depending on the flow rate
    2- see 1
    3- we prefer “activator” and E-Cat
    4- confidential
    5- confidential: that photo has been stolen and published without any authorization: so I have been told by the person we gave it (under NDA) 6 months ago
    Warm Regards,
    A.R.

  • Gherardo

    Dott.Rossi,
    forgive us but your answers generated questions…
    1) Besides “supporting” Steven and Joseph questions and looking for your answers, when you say 350, 600 and so on, you refer to the temperature of the external chamber of the cat or the usefull temperature of the vapor?
    2) If the answer to (1) is “it depends”, could you summarize a few of them for the cat, hot cat and cat & mouse?
    3) Why don’t we call cat and mouse, Tom & Jerry? Humor never killed science :-)
    4) Is there any guess/measure on what is the temperature of the Ni matrix for the various cats?
    5) How was the end of the destructive test? Melted on the floor? What’s the reactor structure made of? Ceramic?
    Thanks, Gherardo

    PS: I could not find the pictures, can you post the official link?

  • Joseph Fine

    Dear Andrea Rossi,

    The Oct 2012 Pordenone report stated that Hot Cat surface temperature was 1050 C.

    Assuming that heat transfer losses account for most of the temperature drop,

    1) Does a stable Hot-Cat temperature of 350 C refer to operation entirely in Self-Sustain mode or does it refer to the stable temperature using external drive and SSM together? ( I recall that SSM and external drive were each used about 50% of the time.)

    Or, 2) Is the 350 C temperature a requirement when used together with an external power generating system (steam, or S-CO2 turbine)?

    And 3) Can you trade operating temperature for COP? That is, for a COP of 6, is 350 C the maximum operating temperature or just a convenient operating point for demonstrating electrical power conversion (e.g. at approx. 30% efficiency)

    I apologize if you answered these questions before.

    Best regards,

    Joseph Fine

  • Steven N. Karels

    Dear Andrea Rossi,

    You posted “The stability temperature of the Hot Cat is around 350 Celsius” We know the Hot eCat is rated to output 600 Celsius. Can you please clarify what you mean by the stability temperature?

    1. Does this mean that the Hot eCat is stable under all conditions when the output temperature is 350 Celsius or less?

    2. Does this mean that the Hot eCat is conditionally stable under positive control from your control mechanism between output temperatures of 350 to 600 Celsius?

  • Wladimir Guglinski

    Joe,
    in my post of May 5th, 2013 at 10:05 AM I had written:

    “Consider the electron moving about one proton L (left) and one proton R(right).
    When the electron is in the side of the proton R the electron is orbiting in counter clockwise direction with intrinsic spin-up. When the electron is in the side of the proton L it is moving with clockwise direction with spin-up.

    However, we actually have to consider that the electron changes its spin from up to down when it leaves out the proton R and starts up to orbit the proton L.
    Let me explain why.

    In the paper Anomalous Mass of the Neutron it had been considered that the electron’s orbit about the protons in the structure of the free deuteron contributes for the magnetic moment of the deuteron.
    See Fig. 6 in that paper:
    http://www.journal-of-nuclear-physics.com/?p=516#more-516

    Now consider a deuteron within the nuclei. As the electron changes the direction of its orbit from counter clocwise in the proton R to the clockwise direction in the proton L, and the electron keeps the direction up of its spin in the whole orbit, then the total contribution of the electron’s orbit for the magnetic moment of the deuterons within the nuclei would have to be null.
    Thefore the magnetic moment of the deuterons within the nuclei would have to be µ=+2,793+2,793=+5,586, and not µ =+0,857 as detected in the experiments.

    Then we have to consider that, along its orbit with the shape of the infinite symbol about the protons L and R in the deuteron, the spin of the electron actually changes from to down.
    So, from such consideration, when the electron is orbiting the proton R with spin up, the electron’s orbit induces a NEGATIVE magnetic moment, and when it is orbiting the proton L with spin down it also induces a NEGATIVE magnetic moment.

    Then, concerning the mechanism which induces the flux n(o) we arrive to the following conclusions:

    1- The intensity of the flux n(o) induced by the rotation of the particles depends on:

    A) the CHARGE of the particle

    B) the direction (clockwise or counter clockwise) of its motion in the orbit.

    2- The intensity of the induced flux n(o) does NOT depend on the spin of the particle.

    regards
    wlad

  • Andrea Rossi

    Dear Gherardo:
    For the Hor Cat are valid the same considerations made for the E-Cat. The melting point of Ni is around 1500 C.
    Anyway that photo ( that has been published without our authorization, and we do not know how it has been leaked) is referred to a destructive test- experiment we made about 6 months ago. The stability temperature of the Hot Cat is around 350 Celsius
    Warm Regards,
    A.R.

  • Wladimir Guglinski

    Dear Joe,

    I suspect that the free deuteron changes its structure when it is captured by the flux n(o).

    In the paper Anomalous Mass of the Neutron it is calculated the magnetic moment of the free deuteron by considering the electron moving about one proton only.
    http://www.journal-of-nuclear-physics.com/?p=516#more-516

    When the free deuteron is captured by the flux n(o), the electron changes its trajectory moving in a orbit which shape is the infinite symbol, and the radii of the two protons have a big shrinking (from 0,89fm to 0,26fm).

    That’s why in the shell thickness 2b of the heavy nuclei (see Fig. 1) in the paper Anomalous Mass of the Neutron) detected by experiments the value of b is b= 0,55fm.

    Such shirinking in the proton’s radius is caused by the flux n(o).

    regards
    wlad

  • Joe

    Wladimir,

    1. Why do you not include the intrinsic magnetic moment of the electron when calculating the strength of the flux n(o) through the protons and neutrons?

    2. Does the intrinsic magnetic moment of the electron not contribute to its own n(o) in the same way that the intrinsic magnetic moment of the proton or neutron does for itself?

    3. You say that the flux of the deuteron is caused by the two protons only. Does this mean that it is two times stronger than the flux caused by only one proton, such as that of 3Li5?

    All the best,
    Joe

  • Gherardo

    Dott.Rossi,
    In the past for the first e-cat you said that the reaction in case of malfunction (forgive my inaccurate wording) would melt the core and the reaction would extinguish.
    I was looking at the picture with the hot-cat cylinder red hot for the 800+ degrees temperature reached on the outside and was thinking how is that runaway situation now for the hot-cat.
    Thanks, Gherardo

  • Andrea Rossi

    Dear Martyn Aubrey:
    I am in the USA, but thank you very much for your kind feeling.
    Warm Regards,
    A.R.

  • Wladimir Guglinski

    Joe wrote in May 4th, 2013 at 3:19 AM
    Wladimir,

    1. Can you post a detailed picture of 2He4? (Is it a disk with a proton and neutron moving clockwise, and another proton and neutron moving counter clockwise?)

    RESPONSE:
    Joe, I think there is no way to post pictures in Rossi’s blog.
    The detailed structure of 2He4 is shown in the page 120 of my book, in the paper No.9, New Nuclear Model, where it is calculated the binding energy of the 2He4.

    The position of the two deuterons is shown in the Figure 2 of this present paper, into the 8O16.

    The central 2He4 in the Fig. 1 is an “explosion” of the 2He4, so that to show how the two fluxes n(o) cross each deuteron of the 2He4.

    2. Can you explain this sentence from your previous post: “Due to repulsions, the two deuterons in the 2He4 have small oscillations along the flux n(o).”

    RESPONSE:
    The flux n(o) is like a string, and the ring of the nucleons (protons and neutrons) crossed by the flux oscillate regarding the ring.
    Imagine you tie the end of a string in the ceiling, and in the other end you introduce a ring. With the string strained, you move the ring with a small alternative up and down oscillation.

    3. Since QRT has the gravitational force as strong as the magnetic force, why can not a neutron be captured by the n(o) of 2He4 to form 2He5 in the same way that a deuteron is captured by both the F(M) and n(o) to form 3Li6? (There is one force per nucleon in each case.)

    RESPONSE:
    Within the nuclei there is attractive and repulsive gravity. The flux n(o) is formed by attractive gravitons, and it is surrounded by repulsive gravitons (otherwise two fluxes n(o) would attract one each other, and all the six fluxes n(o) of the 8O16 would met in one unique side).

    The flux n(o) is induced by the rotation of the body ring of protons and neutrons.

    See Fig. 9.
    In the neutron the electron is always moving in the same direction about the proton. There is tendency the flux n(o) induced by the electron to anihilate the flux n(o) of the proton, so the neutron induces a weaker flux (o).

    In the deuteron, the electron moves about the two protons with an orbit which shape is the symbol of infinite.
    Consider the electron moving about one proton L (left) and one proton R(right).
    When the electron is in the side of the proton R the electron is orbiting in counter clockwise direction with intrinsic spin-up. When the electron is in the side of the proton L it is moving with clockwise direction with spin-up.
    Then the total contribution of the electron in the deuteron is null, regarding the induction of the flux n(o). And therefore in the deuteron the flux n(o) is induced thanks to the contribution of two protons only, and so it is a flux n(o) very stronger thant that of the neutron in 2He5, and also stronger than that due to one unique proton in the 3Li5.

    regards
    wlad

  • Martyn Aubrey

    Bologna Tornado from the BBC – http://www.bbc.co.uk/news/world-europe-22417166

    “A tornado has battered northern Italy, causing millions of dollars worth of damage in two provinces near Bologna, and injuring at least 12 people.

    The regional governor of Emilia Romagna has asked the government to declare a state of emergency. Wendy Urquhart reports.”

    Dear Dr Rossi, I have just read on the BBC News page about the destructive tornado near Bologna.

    I hope yourself, your family and your team are all OK.

    All good wishes,
    Martyn Aubrey

  • Andrea Rossi

    Dear Pekka Janhunen:
    Thank you for your interesting suggestion.
    Warm Regards,
    A.R.

  • Dear Andrea Rossi,
    Returning to my earlier comment on vapourisation and condensation of nickel. Chemical vapour deposition (CVD) is a large family of industrial techniques which deals with related phenomena. Maybe a reactor can be long-lived if it periodically remakes its nickel nanostructure by a “CVD” process. Ionisation, plasma formation and free radical formation caused by X-rays certainly also modify the process. I wouldn’t be surprised if the E-cat would be already making use of such physics. But perhaps by consulting some experts on CVD techniques it might be possible to improve the process further.
    regards, /pekka

  • Andrea Rossi

    Dear eernie 1:
    The photos of the arrival need the permission of the Partner. We’ll see.
    Warmest Regards,
    A.R.

  • Andrea Rossi

    Dear Dr Joseph Fine:
    Thank you for your information. very interesting.
    Warm Regards,
    A.R.

  • Joseph Fine

    AR, The correct link to the S-CO2 turbine article should have been:

    http://www1.eere.energy.gov/solar/sunshot/csp_sunshotrnd_nrel_turbine.html

    The previous link had extra characters (like barnacles on the hull of a ship) and should be replaced by the above link. (I hope.)

    JF

  • Joseph Fine

    Dear Andrea Rossi and Readers,

    I recently saw this article on the development of a 10 MWe Supercritical-CO2 turbine. This application (“Sun-Shot”) is intended for solar power applications, but if you use multiple Hot-Cat systems to produce 25 to 30 MW-thermal at 350-600 degrees C, you might someday be able to use this/(these) 10 MW (Super-critical) Carbon-Dioxide turbine(s). (Of course, this would be a major investment. It does not have to happen tomorrow, but may be considered as an alternative for the future.)

    My estimate of the need for 25-30 MW-th (that is, 25 to 30 1-MW Hot Cats) is because conversion efficiency might be about 33-40%. For example, if you have a 40% conversion efficiency, you’d need 25 MW-th to produce 10 MW-electric.

    It is estimated that S-CO2 turbines will be significantly smaller than conventional steam turbomachinery. Hopefully, that means less expensive.

    turbine.http://www1.eere.energy.gov/solar/sunshot/csp_sunshotrnd_nrel_turbine.html

    You have to walk before you run. Also, you have to stand up before you walk. But you’re getting there.

    Mega-Regards,

    Joseph Fine

  • eernie1

    I would like to propose a number of what ifs. the Russians are considering the use of 63Ni(Beta- emitter with half life of 100 years)to produce a long life battery(30 years)through a betaelectric process using a semiconductor as the target.What if we can use a negative field(H-)to cause rapid electron capture in the nucleus and speed up the decay times to produce many more Betas which can react with the nickel lattice?What if another factor is the electron capture producing conversion electrons which assume an amount of the energy released in the de-exitation of the isomer(exited neutron)which produces the Beta electron plus a neutrino? The 63Ni transmutes to 63Cu.What if the the conversion electrons now with much more acceleration caused by the added energy exiting the nucleus are inhibited by the columbic field inside the nucleus? The electrons exiting the nucleus will maintain its energy(COE)but will undergo a decrease in acceleration.Therefore,its mass must increase(E=mv^2)and the electron acts like a muon(heavy electron).This results in an increase of interaction cross section with the lattice electrons and a conversion to thermal energy.Just saying what if.Andrea,I enjoyed the photos.Can we have some of the arrival?

  • Andrea Rossi

    Dear Steven N. Karels:
    You are right, we now are working on that.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Petter Ekfors:
    The next ones will be manufactured in the USA.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Gherardo:
    1- The photos are published on this blog, just look at the comment sent by Eng. Fulvio Fabiani
    2- We are not authorized to publish any photo regarding the shipping to the USA
    3- yes
    4- because inside there is not enough space
    5- nothing is final, but death, and we are pretty alive
    6- yes
    Warm Regards,
    A.R.

  • Gherardo

    Dott.Rossi,
    I have seen a picture from Daniele Passerini and some other about the transportation from Bologna to Ferrara (so was written. They say they are form The Journal but I wasn’t able to find a link from here.
    1) Which of them can you confirm?
    2) Those aren’t the US shipping. Can we expect more to come?
    BTW, the standard container originally shown in picture one year ago’ for the 1MW cat now looks smaller, about 4.5 to 5 meters, out of standard measures (6-12 meters) and thinner as can stay on a truck trailer with sideboards up.
    3) Is it actually smaller that the standard 20 foot container?
    4) Why you didn’t choose to keep all external “things” and the “hat” inside and protected?
    5) Is that the final version?
    6) Is it correct that you did optimize the packaging?
    Thanks, Gherardo

  • Petter Ekfors

    Dear Andrea Rossi,
    It was nice of you to publish pictures from the shipping of the Mega-Cat. There were also some pictures from your workshop where we could see another Mega-Cat. Earlier we have learnt at this site that another plant was to be delivered to its customer by the end of last month. So how many Mega-Cats have you built altogether by now?
    //Kind regards, Petter Ekfors

  • Joe

    Wladimir,

    1. Can you post a detailed picture of 2He4? (Is it a disk with a proton and neutron moving clockwise, and another proton and neutron moving counter clockwise?)

    2. Can you explain this sentence from your previous post: “Due to repulsions, the two deuterons in the 2He4 have small oscillations along the flux n(o).”

    3. Since QRT has the gravitational force as strong as the magnetic force, why can not a neutron be captured by the n(o) of 2He4 to form 2He5 in the same way that a deuteron is captured by both the F(M) and n(o) to form 3Li6? (There is one force per nucleon in each case.)

    All the best,
    Joe

  • Steven N. Karels

    Dear Andrea Rossi,

    After a short, well-deserved rest, I am sure you will be on to your next major achievement. Will be it a production version of a Hot eCat (600C)? Maybe an electric generating power unit? Can you give us any clues?

  • Davide C.

    Dear Dr. Rossi, good news (or rumors) about your hot-cat.
    http://oi44.tinypic.com/20ix4s0.jpg

  • Wladimir Guglinski

    Joe wrote in May 2nd, 2013 at 12:45 AM
    Wladimir,

    How does QRT explain the electric quadrupole moment of zero for 2He4? (Does it involve treating 2He4 like two protons and two neutrons, with the two protons performing a sphere? Or is it necessary to deconstruct 2He4 further into four protons and two electrons, with the four protons performing a sphere? In the latter case, how would the charge of the two electrons affect the electric quadrupole moment?)

    Joe,

    The reason why 2He4 has null quadrupole electric moment Q(b)=0 is the same why all the nuclei with Z=N=pair also have Q(b)=0.
    It is explained in the page 137 of my book Quantum Ring Theory, and my argument had been plagiarized in the article How Atomic Nuclei Cluster, published by the Journal Nature in 2012.
    http://www.zpenergy.com/modules.php?name=News&file=article&sid=3402

    Due to repulsions, the two deuterons in the 2He4 have small oscillations along the flux n(o).
    As 2He4 has null nuclear spin, and because the nucleus gyrates, the 2He4 has a chaotic spin. And because it has null magnetic moment, in the experiments there is no way to align it so that to measure its quadrupole electric moment.
    Due to the chaotic spin, the 2He4 nucleus behaves in average as having a spherical distribution of charges.

    Regards
    wlad

  • Andrea Rossi

    Dear Ing. Michelangelo De Meo:
    Very difficult to answer to generic questions. To answer is necessary a complex analisys, and this is not the place to make it. It is difficult, if not impossible, calculate the exponents at this stage.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Dr Joseph Fine:
    Thank you for your suggestions: always useful.
    Warm Regards,
    A.R.

  • Joseph Fine

    Andrea Rossi,

    I rechecked the dimensions of the Hot-Cat from last year (October 2012). At that time the module dimensions were as follows:

    (Each module is assumed to be rated at 10 KW-th.)
    //////////////////

    (from 2012 Pordenone Report)
    DESCRIPTION OF THE REACTOR

    The reactor is a cylinder which has the following dimensions:
    Length cm 33
    Diameter cm 8.6
    (See photos in the Penon Report attached)

    /////////////////

    In an earlier comment, I overestimated the dimensions of each module by claiming that a module had a radius of 9 cm and a length of 1 meter. These dimensions are overestimates. I apologize for not reading the report from 2012 before submitting my idea. I don’t know what dimensions you will select, but there seems to be enough room for plumbing, control and other equipment. Don’t forget to put in an access door so someone can go in and come out.

    Joseph Fine

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