Abstract
Review of the periodic table and existing research on isotopes of the various elements of the periodic table is conducted and the attempt is made here to visualize the process of element formation.
The role of Neutron in element formation is investigated and found to be vital for existence of elements . In the new light, the new model of nucleus is proposed which explains the stability of the nuclei and reason for multiple stable isotopes of elements.
Review of the Periodic Table
The inspection of the periodic table of elements reveal an interesting fact that for all elements other than Hydrogen, for element to be stable, number of Neutrons are always greater or equal, to the number of Protons in the nucleus. The periodic table is analyzed with respect to the abundance in nature for the elements.
The fusion of Hydrogen to produce Helium can be envisioned as follows:
1H + 1H + 1H + 1H → 2H + 2N + 2e+ (1)
where 2e+ later annihilating to produce additional gamma radiation.
Since tremendous force is needed to keep four protons together during the formation of helium and emit 2 positrons, it is unlikely. The mass of neutron is more than the mass of proton and this fact alone negates the likeliness of the above scenario.
- Neutrons are the building blocks for elements in nature.
- In the elements , other than hydrogen, neutron and proton form a pair (np) and keeps distinct identity.
- Excess neutrons stay at the center of the nuclei but keeps their distinct identity.
- Abundance of isotope is close to 100% when excess neutrons are odd count.
- If the excess neutron count is even, the isotope is radio active.
- Majority elements have only one Isotope.
- Only Potassium has three isotopes, the one with 2 excess neutron is radio active with 0.001 relative abundance.
- Elements He, C, O, Ne, Mg, Si, S are close to one in abundance with no excess neutron and maximum three isotopes.
- Element Be has one excess neutron and one isotope, 100% abundance.
- All other higher elements require four or more excess neutrons for the element to be abundant in nature and has up to 10 stable isotopes.
- Element Ca is exception with 0 excess neutron and 97% abundance. However it is unstable with >E+21 a Half life.
The nature prefers, for more complex nucleus, more neutrons than protons and creates a delicate balance to form a stable nuclei.
This balance of forces is so critical that in case of element F (the stable isotope is with 9 proton and 10 neutrons) The isotope 18F with 9 Protons and 9 Neutrons, with in 20 minutes decays and forms 18O which is stable with 8 protons and 10 Neutrons and gives up a e+ positron to convert proton to neutron and e+ and e- reaction produces Gama radiation.
Equation 18F → 18O + e+
Thus the existence of Neutron is vital to the existence of the universe itself, because without neutron, elements may not have been possible and hence the intelligence as we know today.
Structure of Neutron
The fact that Gravitational binding forces (Fg) of masses in the nucleus needs to be more than the destructive electromagnetic forces (Fem) created by the electrically positive environment of the nucleus, (Fem < Fg) excess neutrons are required in the nuclei. However, even at the center of the nucleus neutrons keep their distinct identity, rather than lumping together to form one heavy neutron. Therefore, neutron can not be just a fuzzy mass but a well defined structure able to hold induced charges with precise demarcation of boundary with an insulating layer, more like a free standing capacitor. Which suggests that at least three particles with two sets of characteristics are required to create a neutron. For our model here, that has to be two particles capable of holding charges and one particle which provides insulation between these particles, similar to that of dielectric layer in the capacitor. Thus in this postulated model the neutron looks and behaves neutral for all practical purposes. However, within nuclei keeps separate and distinct identity and does not combine with other neutrons to form a one central heavy body.
The proposal here defers from the model independent analysis predicting the structure having negative charge on the surface, Positive charge in the next lower level and than neutral mass. My model suggests that neutron is a free standing charged capacitor. This helps in keeping the distance between excess neutron in the nuclei. Otherwise there is nothing stopping these neutrons from lumping together to form a mini neutron star with in the nuclei. This model of Neutron is justified from the known fact that Neutron is heavier than Proton.
For elements other than Hydrogen, the nuclei becomes a chaotic environment. For Example, in case of
4He, two protons require their space and wants to be separate from each other as far as possible, and two neutrons are caught between the two protons as shown in figure 1. From the reference frame of protons it is envisioned that Neutrons spin on its axis at very high rotational speed. One clockwise and other anticlockwise.
Model of nuclei
A new model of nuclei is proposed as shown in the figure 3. A spherical shell of excess neutrons with one neutron at the center of the shell surrounded by paired proton neutron (pn) shell.
For the stable nuclei system the electromagnetic forces needs to be balanced and gravitational forces maximized. From the example of 4He above, it is envisioned that elements are built, in this model, by Heavy Hydrogen nuclei (pn) as a building block and just enough excess neutrons to provide needed gravitational force for stability.
Energy levels (orbitals) in pn outer shell follows the similar shell structure of electrons, with K,L,M,N,O being primary shell and s, p, d, f, g sub shells with similar total charge particle capacity. However for excess neutron shell it differs, where a single neutron stays at the center of the nuclei when in excess. The energy level for that central neutron in this model is called “Foundation neutron” (Fn).
The Table 4 shows the placement of neutrons in each shell.
Nuclei Stability Analysis using above model
For Odd atomic number nuclei:
In the ‘p-n shell’ the outermost pair has no symmetry, however the excess neutron shell is symmetrical for all odd atomic number elements giving the stability and abundance. Referring to table 2 and table 4 , isotopes, with one excess neutron, has relative abundance in nature of 1 or close to 1, The excess neutron takes the place of Fn. The isotopes with 2 excess neutrons are all radio active which can be attributed to the asymmetry of neutron in unfilled K shell, which can hold up to 2 neutrons. When the K shell is completely filled as in the case of 41K, the nuclei is stable. Table 2 outlines isotopes and abundance for majority of elements, all follow the same principle and model fits perfectly with the exception of 14N with atomic number 7. Iodine has 21 excess neutrons, as we see in table 4, the outer most excess neutron shell is symmetrical, making it stable. All other elements follows.
For Even atomic number nuclei:
The stability is obtained from the pn shell. However, excess neutron shell becomes asymmetric and depending on which sub shell, there is a room for additional excess neutrons and hence exhibits many stable isotopes with relative abundance. Theoretical probability calculation may prove this fact.
Instability and radio activity in heavy elements:
The neutron proton (np) pair in nuclei has capability to grow indefinitely. However as the heavier elements are built the inner excess neutron shell is large enough to interact with lower np shells, thus giving instability to the nuclei. e.g. for 238U there are 148 total neutrons and 54 excess neutrons. A nuclei with 111 excess neutron will completely fill up the O shell. All elements after 209Bi are radio active, this shows that interaction of neutrons form excess neutron shell(inner shell) with pn shell is destructive when the N shell is half filled. Thus there is a upper limit to the size of the inner excess neutron shell and this may be the reason for non existence of super heavy stable elements in nature.
Conclusion
This proposed model of nuclei, explains the stability and relative abundance of the nuclei. The nuclei(except for H and He) in this model is a shell with in a shell, where an innermost shell structure is formed with excess neutrons with one Fundamental foundation neutron (Fn) and the rest arranged in shell structure. Protons, form pair with neutron and form shell enclosing this excess neutron shell as shown in figure 3. The stability of the nuclei is a direct result of symmetrical placement of neutrons in the outer most shell of excess neutron shell. The same shell is responsible for many stable isotopes in even atomic number nuclei. The additional isotopes are possible if the outer shell of excess neutron shell holds odd number of neutrons. The nature of neutron is postulated here, the proof of which depends on future experiments.
References
Numerous Scientists’ and scholars’ exhausting work in development of periodic table, investigation of Isotopes is utilized and due credits are given. The list is too large to print here, but due credits are mentally given to all scientists.
Dear Mr. Rossi,
Please allow me to commend you on your achievements. Not only am I stunned by the potential of your recent invention but also the simplicity of your device, which I find to be scientifically elegant. In your efforts to bring the invention to market, even against powerful and unjust dogma, I can find no better example of entrepreneurial courage. Perhaps even Albert Einstein would commend you, he once said “Any intelligent fool can make things bigger, more complex, and more violent. It takes a touch of genius, and a lot of courage, to move in the opposite direction.”
I have been following both sides of your story over the internet. I can’t help but think that at no point in time would your invention have been communicated to more people (and the powerful criticize against it) than at this moment. Here in America a whispering of your invention has taken place in media institutions such as MSNBC, Forbes, Physics Today, Fox News, and many other.
However, I am concerned. Concerned, because people see hope in your invention. Many see something with life-alterning potential within their grasp. Many who see this are trying to follow in your footsteps. I understand your right to protect your invention, it’s your’s, no other’s. I also am greatful for your willingness to demonstrate in front of the public and to share in some of the technical details. But realize that without the knowledge you have people will go on searching in the dark until they find something because they see hope in you. Given the current circumstances, most of these people are common persons with little technical knowledge and limited resources. It is the safety of these people that worries me. You yourself have said in this blog that people should take every precaution (in fact you said a woman emailed you trying to replicate your results in her kitchen with her children). With the information you’ve provided (which is commendable) people believe that only a secret catalyst sits between them and a new tommorrow. Common people will try anything. How can you be sure that every precaution will be taken? Consider not only the obvious risks to common people (high temperatures and pressures) but also the unpredicable hazards. For example, consider the most toxic industrial substance is Nickel tetracarbonyl, deadly at just 30ppm. As you have indicated radiation maybe play in role in the operation of your invention, how can you be sure that if someone were duplicate what you have they wouldn’t fall victim the neutron and gamma ray bombardment?
I understand your right to protect your invention, but by operating behind closed doors the risks to common people grow over time. Therefore, I urge you to gain the support of the people and put to rest all doubt about your invention, let it be know what you have! Consider your Bologna brethen, Marconi, who was a master of public demonstration and communication, and actively sought the support of institutions. If it fame or fortune you seek, rest ashured that you will have both. The same media pathways that brought your invention to my attention guarantee your place in the history books. If it’s fortune you seek know that history always rewards those who act, America surely will. Eitherway, it’s not worth keeping the public in dark and risking the safety of common people, even just the safety of one individual. Please do the right thing, the honorable thing, and show the public what you really have. Tell us what not to do.
Sincerely,
A concerned citizen
Good morning again Ing. Rossi,
thanks for Your prompt reply.
Regarding the answer
“1- at 104.5 Celsius you can bet you have steam…
Probably You measured the steam pressure, please can you tell which was the measured value?
Thanks
Kind regards
Franco
Bhagirath Joshi wrote in December 6th, 2011 at 9:50 AM
“Dear WG:
This paper was published in 2008 by Oberaker et al Here they are revisiting Q(b) for 16S34 and did calculations using some model and compared with experimental data.
http://arxiv.org/pdf/nucl-th/0309081
Please read this and explain to me, since I did not understand completely. The Q(b) exp (page 4) they are comparing is not the one I provided,
Looks like with new tools available , work still continues in Q(b). “
Dear Joshi,
the experimental work is ready.
Unlike, the theoretical work is not ready. And I think they will never succeed to finish it if they keep their attempt with the current models of Nuclear Physics.
What the physicists are doing is to continue the theoretical attempt, trying to get values agree to the the experimental data. And they are using new mathematical tools.
I think they cannot be fully successful in their theoretical work by two reaons:
1- Their nuclear model is wrong
2- In my opinion there is need to consider the shimmy of nuclei due to unbalance of masses, not considered in their work.
I think the lack of shimmy in their work is responsible for some puzzles in their theoretical work. For instance, consider the following:
1- Stable 14Si29, with Z=14, N=Z+1 , has Q(b) = 0.
2- So, we would expect that stable 16S33, with Z=16 and N=Z+1, would have to have Q(b)=0 too, since both 14Si29 and 16S33 have Z=pair, and N=Z+1.
3- However 16S33 has Q(b) = -0,064
4- And obviously the question is: WHY ??:??
It’s seems to be impossible find a reasonable explanation by considering the current Nuclear Physics. After all, as both of them have Z=pair, and N=Z+1, then they will have to have the same Q(b) =0, if we consider the distribution of nucleons considered in Nuclear Physics.
Let’s see the explanation by considering the Hexagonal Floors model of Quantum Ring Theory.
1- 8O16:
Let’s start up by considering the oxygen nucleus 8O16.
See the figures in this link:
http://hexfloor.blogspot.com/
Fig. 1 shows the 8O16.
The central 2He4 produces fluxes of gravity, which captures protons, neutrons, and 1H2 nucleons. They form several hexagonal floors around the 2He4.
Actually the hexagonal floor is no flat, as shown in Fig. 1. Actually all the 1H2 oscillate, and the structure of 8O16 is like shown in the Fig. 2.
This is the reason why 8O16 has Q(b)=0. The nucleus has a spin. And since the 6 nucleons 1H2 oscillate, the nucleus has a chaotic rotation, and so it’s distribution of charge is symmetric because of such rotation.
2- 14Si28:
Look at the structure of 14Si28 shown in the Fig. 3. The nucleus has a spin about the Z-axis, and as happens in the case of the 8O16, 14Si28 has Q(b) = 0.
3- 14Si27:
Then let’s start our analysis from the nucleus 14Si27. It has Q(b)=0, which is very surprising if we consider the current Nuclear Model, because it has odd number of neutrons, and therefore it’s hard to expect a symmetric distribution of charge from the current nuclear model.
Now consider the 14Si27. Its structure is shown in the Fig. 4.
Note that there are 10 nucleons 1H2 symmetrically distributed about the central 2He4, and so they do not produce elec. quad. mom. And there is a proton and a nucleon 1H2, they both situated symmetrically regarding the central 2He4, and their position is coincident with the Z-axis.
So, 14Si27 has:
A) symmetric charge distribution
B) The proton and the 1H2 situated in the Z-axis do not produce unbalance of the nucleus. Therefore there is not asymmetry of charge due to shimmy.
Experiments show that nuclear magnetic moment of 14Si27 is (-)0,8554. Let’s see why.
Magnetic moment of neutron is -1.913
Magnetic moment of 1H2 is +0.857
Look at the Fig. 3: both the neutron and the 1H2 have no angular momentum, because they are situated in the Z-axis. Then the total magnetic moment is about -1,913+0,857 = -1,054 .
4- 14Si29:
With the introduction of one excess neutron in the structure of the 14Si28, the neutron takes the position coincident with the Z-axis, as shown in the Fig. 5.
Note that there is a symmetric distribution of charges due to the 12 nucleons 1H2 about the central 2He4, and there is balance of masses, because the excess neutron is situated coincident with the Z-axis, and therefore there is no shimmy.
Magnetic moment: -0.55529
Look at the Fig. 5 that such magnetic moment is due to the neutron in the Z-axis.
5- 14Si30:
The second excess neutron takes a position coincident with the Z-axis, as shown in Fig. 6, and so it has Q(b)=0.
6- 14Si33:
It has Q(b) = 0, because it has symmetric distribution of charge, and it has balance of masses (null shimmy), as shown in Fig. 7.
7- 16S32:
Its structure is shown in Fig. 8. A new hexagonal floor begins between the first and the second floors.
16S32 has symmetric distribution of charge, and its mass is balanced (no shimmy). That’s why it has Q(b) = 0.
8- 16S31:
It has Q(b) = 0, and its structure is shown in Fig. 9 : there is symmetrical distribution of masses regarding to the Z-axis, and there is also symmetrical distribution of charges.
9- 16S33:
This isotope has Q(b) = -0,064. Its structure is shown in Fig. 10.
Note that the excess neutron takes a position in the third floor. This fact brings the following consequences:
A) 16S32 has a spherical form. But with the addition of excess neutrons in the third floor, there 16S33 is submitted to a new tendency: it tends to get an ellipsoidal shape along the Z-axis, which is the feature of the positive elec. quad. mom: Q(b)>0.
The tendency to get that ellipsoidal shape is because the additional neutrons in the third floor diminishes the space available for the oscillations of the 1H2 nucleus (that oscillation shown in Fig. 2).
B) On another hand, the odd neutron in the third floor cause an unbalance of masses, and the shimmy tends to yield Q(b)<0.
C) As the tendency to form an ellipsoidal structure is weaker than the tendency due to the shimmy (because there is only one neutron), then 16S33 has Q(b)<0.
Magnetic moment of 16S33: +0.6438212
It has spin 3/2, and so the two nucleons 1H2 have not contrary magnetic moments. So, its total magnetic moment, according to Fig. 10, would be: -1.913 +2 x 0.857 = – 0,2.
However, the radius of rotation of the two 1H2 is longer that that of the neutron (the neutron is situated in the side of the hexagon, and each 1H2 is the vertex).
10- 16S34:
Its structure is shown in Fig. 11. We see that it has no excess neutrons in the third floor. That’s why it keeps the spherical shape. And there is balance of masses, there is no shimy, and it has Q(b)=0
11- 16S35:
This isotope shown in Fig. 12, and it has Q(b) = +0,047. Let’s see why.
A) There is one excess neutron in the third floor. So the nucleus exhibits a tendency to have Q(b)>0.
B) In the case of 16S33, the unbalance of masses was 1/33. In the case of 16S35 the unbalance of masses is 1/35. So the eccentricity due to the shimmy decreased.
C) The two opposite excess neutrons in the first and second floor may contribute for the ellipsoidal shape (because with the growth of quantity of neutrons the nucleus has other tendency: it suffers shrinkage because of the attraction due to the strong force between the neutrons and protons.
12- 16S36:
Fig. 13 shows that 16S36 do not have excess neutrons in the third floor. That’s why it keeps its spherical form, and has Q(b) =0.
13- 16S37:
Fig. 14 shows that 16S37 will have Q(b) > 0, and stronger than that of 15S35
14- 16S38:
Its structure is shown in Fig. 15, and it has two neutrons in the third floor. So it has a strong tendency to have Q(b) >0.
The sulfur isotopes with N>6 excess neutrons are submitted to two oppose tendency:
A) The growth of excess neutrons tends to yield an ellipsoidal shape, because of the growth of the distance between the first and second hexagonal floors.
B) But the growth of excess neutrons tends do yield a contrary effect: the shrinkage of the nucleus due to the attraction by protons and neutrons by the strong force.
C) The tendency to increase Q(b)>0 stops when the third floor is filled. But the shrinkage shrinkage do not stop. So, after 16S44 there is a tendency to have decrease of Q(b)>0 between 16S44 and 16S50. This is shown in the FIG. 2 of the link:
http://arxiv.org/PS_cache/nucl-th/pdf/0309/0309081v1.pdf
15- 16S50:
It has the tendency to have a spherical distribution of charge. And as it has balance of masses, it has the tendency to have Q(b) = 0. Its structure is shown in Fig. 16.
Regards
WG
Dear Paolo:
Thank you for the suggestion, but for now car applications is not a priority.
Warm Regards,
A.R.
Dear Franco Morici:
1- at 104.5 Celsius you can bet you have steam. The steam has been condensed and the water recycled.
2- The amount of hydrogen we charge is enormously redundant: only picograms of H react, so that when we charge the reactors most of the H goes away with the purge
3- how we get the pressure inside the reactor is a confidential information
4- Maybe it seems to you the 1 MW plant is not equipped with a real condensate drain… we are pretty sure it is equipped with a very real and efficient condensate drain. It has been constantly checked during the test that all the non condensed water was dropping in its reservoir (not by me, but by the Consultant of the Customer).
Warm Regards,
A.R.
Dear Eugeno Mieli:
Thank you very much, please stay in contact:
info@leonardocorp1996.com
Warm Regards,
A.R.
Dear Felize:
1- no. We are not yet ready to produce electric power, even if we are very close. I want also to say that an agreement we reached yesterday will accelerate this development.
2- the power output does not change with the pressure. 1 MW of thermal power remains 1 MW of thermal power, indipendently from the pressure of the steam.
Warm Regards,
A.R.
Dear Mr. Rossi,
Congratulations again for your improvements.
I would like to ask you the following questions:
1) Are you going to use the first 1MW plant for producing electrical power? If yes, have you redesigned the plant’s layout?
2) What is the (expected) thermal output power of each reactor during the production of a steam at 480°C?
Thank you in advance.
Best regards,
Felize
Gentile Andrea Rossi,
le scrissi (in inglese) molto tempo fa chiedendole come potevamo darle un sostegno concreto. Lei mi rispose che, fintanto che non fosse terminata la centrale da 1 MW, non potevamo fare nulla.
Allora, dopo il test del 28 Ottobre, ho pensato di portare le grandi aziende a conoscenza della sua invenzione e delle opportunità commerciali che oggi sono già sul mercato: come lei sa bene, ben pochi, anche oggi, sono a conoscenza della rivoluzione energetica determinata dall’e-Cat e degli antefatti che l’hanno preceduta.
A questo scopo ho costruito una brochure informativa, basandomi sul materiale presente in rete, cercando, se possibile, di essere fedele alle sue indicazioni: sono laureato in fisica e questo mi è stato utile nel mare magnum dei documenti trovati.
Fino ad oggi sono riuscito a far arrivare questo documento ai vertici di Finmeccanica (che in realtà avevo allertato, sul tema e-Cat, già da Aprile) e di Telecom Italia (informata a fine Novembre). So, per via informale, che Finmeccanica, alla fine non ha mostrato interesse per l’argomento, mentre, invece, Telecom forse lo sta valutando.
Sto cercando di fare lo stesso con Enel, ma devo trovare un canale giusto.
Spero che qualcuno di questi miei interventi, in un modo o nell’altro, vada a buon fine.
Se vuole copia della brochure che ho costruito mi faccia sapere a quale casella di posta inviargliela. La mia è eugenio.mieli@fastwebnet.it.
Grazie di quello che sta facendo per tutti noi,
Eugenio Mieli
Good morning Ing. Rossi,
please I would ask to You three main technical questions.
The first…
During test of 1MW plant of 28 October the produced steam did not was visible in any photo or any short film and there was not a valve to release steam in atmophere as witness of steam presence (measured temperature of 104.5°C is not garantee that has been produced only steam).
How can we be sure that the big heat excangers did not dissipate only or in part “overheated water” instead of steam?
The second…
How can You explain that the 1MW plant (52 E-cat modules I understood) had been loaded with only 1,7g of Hydrogen even if in the test ot 6 October 1.5g of Hydrogen loaded for a single module E-Cat.
Can You explain how did You get 2 bar minimum of Hydrogen pressure in each reactors declared as preferred pressure range on Your patent?
The third…
Why the 1MW plant seems be not equipped with a real condensate drain to get collection of condensate as good common practice?
Thank You for the reply. My questions are to clarify and to dissipate technical doubts, not to get from You sensible informations.
Kind regards
Franco
Caro Rossi, so che di lavoro ne hai già abbastanza e di progetti futuri uno sproposito,ma volevo suggerirti un prossimo passo dopo la conversione in elettricità tramite turbina. Cioè l’applicazione per veicoli elettrici. Attualmente la tecnologia per auto elettriche è molto avanzata ( celle a combustibile )il problema è il rifornimento di questa energia. Il tuo apparato,magari più ridimensionato si potrebbe integrare già su motori elettrici esistenti, tra l’altro l’idrogeno che ti serve per la reazione viene prodotto nelle celle a combustibile. Potrebbe essere un sistema chiuso. Credo che una collaborazione con qualche azienda automobilistica magari italiana sarebbe fantastico. Saluti
Dear Iggy Darlymple:
We have scheduled a manufacturing network with outsourcing manufacturers and our ones, which will be activated depending on the location of the Customers.
Warm Regards,
A.R.
Dear Dr Rossi,
Since I am a citizen of Florida I am excited about your plans for production here. I hope your efforts to produce in Massachusetts will not replace the Florida factory. Are you receiving support from the local and state government here? I’ve read nothing about the E-Cat in the Miami or the capitol newspaper. I’ve written the governor and Florida Trend Magazine pleading for their support of your efforts, but haven’t received a reply.
Sincerely,
Iggy Dalrymple
Dear Mr. Rossi,
I wonder if Star Scientific might have what is needed to make cold fusion?
http://www.starscientific.com.au/
Star Scientific make Muon Catalysed Fusion because they have found a easy way to make Pion in a cheap and easy way. Pions so mutates into muons that replaces the electron and make the atoms core get 200 times closer and opens up for fusion at low temperature. In their deterium –deterium or deterum-tritium fusion, they get strong radiation and nutrons, so strong shielding is needed.
I wonder if Ni –H fusion might be possible if muon could be used as catalyst. Then the hydrogen could get very close to the Ni core and fuse. This process might give very little radiation. If I am getting close to your secret I do not know, but if Star scientific have found a way to make Pions, then the might also make Ni-H fusion possible.
This is really interesting times with so many new revolutionary ways of making energy.
Regards
Svein
insight wrote in December 6th, 2011 at 4:58 AM
Dear Wladimir Guglinski,
“I ask you, If this happened, would be the spin of the nuclei slowed down too?”
I think yes.
But the best would be to ask it to the nuclear theorists, since they are the owners of the current prevailing nuclear model.
PS: I made a mistake when I wrote that sentence. The correct is:
“There is NO way to explain it by considering the current nuclear model of Nuclear Physics, because the alpha particle could be emitted in any direction. And as the nucleus has a spin, when the alpha particle leaves the nucleus it would be dragged by the spin, and it would leave out the nucleus by a tangential trajectory.”
Dear Viktor:
Thank you for your considerations. Of course I cannot give information about the phenomena inside the reactors.
Warm Regards,
A.R.
Dear Andrea Rossi,
I follow both sides of the story, want to believe in your product (already subscribed to two home eCATs), eagerly hunt for information on the web supporting your team’s theory, was even very happy to see special interest in LENR technology from 3 different NASA scientists recently.
On the other hand i also follow some ‘snake’-media, to balance my information input.
The ‘snakes’ normally stick to two points. one is related to your personality and history, which is not fair, especially not in the scientific world.
Their other argument comes back to the Coulomb-barrier.
With my half-scientific fairy-vision i imagine this Coulomb-barrier as a huge safe in a bank.
1. You can try to break this barrier with force, and this means hot fusion. Like blowing up the whole safe with dynamite, the whole street outside would hear it and see the smoke.
2. The other solution is to find the key or secret code and open the safe without any loud activity which could be realised by anyone outside of the bank… thus limited radiation and no gazillion celsius involved.
So, what is this secret code then?
a, The catalyst – I believe this is only used to get rid of oxides and clean the Nickel surface, perhaps to keep the Hydrogen on atomic level, but the catalyst is not necessarily involved in the LENR itself.
b, The frequency generator – i think the truth lies somewhere here. The frequencies of the H and Ni atoms or their electron clouds could be influenced so that they are happier near each other, and even want to have sex. It’s like red wine.
All in all, good luck in your commercial preparations, the sooner the customers will confirm their happiness, the quicker you will get rid of snakes.
Viktor
Dear A.Casali:
1- At the price we apply now the 1 MW plant pays back itself in 3 years. The expected life is 30 years. Anyway, we foresee to lower the price with the increase of the orders, to allow more economy scale. Our target is to arrive within the year to 1000 Euro/kW.
2- We are working on this issue, see point 1.
3- yes
4- As soon as possible, when the Customer will allow it. Consider that when the name of a Customer of us is published, he is buried under thousands of inquiries, emails, insults, together with good considerations. It is not an issue as easy as you can think. We all need to work in peace, and we have an army of enemies who want to forbid this. Among them the wannabe “third parties who have to make a real test” ( which turned out to be all competitors or consultants of them, what a strenge coincidence!)
5- I am holding nothing, we are already making our R&D with all our Consultants, and our Customers are accepting our technology already. They don’t care too much who is testing our plants, they care the plants work properly, that’s all they want.
Thank you for your direct and useful questons,
Warm Regards,
A.R.
Wladimir Guglinski
December 5th, 2011 at 5:59 PM
Your Comment:
Here we go again.
Dear Joshi,
My Explanation:
Dear WG:
This paper was published in 2008 by Oberaker et al Here they are revisiting Q(b) for 16S34 and did calculations using some model and compared with experimental data.
http://arxiv.org/pdf/nucl-th/0309081
Please read this and explain to me, since I did not understand completely. The Q(b) exp (page 4) they are comparing is not the one I provided,
Looks like with new tools available , work still continues in Q(b).
Sincerely
Bhagirath Joshi
Dear Dr. Rossi,
it has been a while since the last time i wrote you but in meantime things don’t seem to have been moving that much, at least from our point of view, i mean from the point of view of us internet followers.
With your permission i wolud like to ask you a few questions to try and understand where are things going.
1) At current price of 2000 Euro per Kwh and given a granted COP of 1:6, your 1MW plats don’t seem so appealing, explecially if we take into account the cost of electricity. What are you offering to your first customers to compensate the low convenience of your devices at current price? I mean are they going to be also partners or just siple clients? if just simple clients where is the real deal for them?
2) How is it going with the indusctrializzation of the production of the 1MW plants?
3) Is the industrialization process going to lower the price considerably in the short term?
4) You recently stated you have signed a contract with the first “normal” customer, when do you think you will be able to publish their name?
5) Considering the importance of university R&D for the acceptance of your technology in terms of certification and authorisations, not to speek about the performance improvements that they may bring to your great invention, why are you still holding on instead of letting the R&D start?
I know some of my questions go straight to the point and maybe you prefer not to disclose too many information but i felt the need to ask you anyway.
Thanks a lot for your patience as usual and keep going, a silent and hopeful army is supporting you.
Warm Regards,
ac.
Dear Wladimir Guglinski,
you do not need to go so far to find flaws in a model or in a scientific point of view: it is enough to sift the paper or even the posts. For example you said in an old post of this thread:
“There is now way to explain it by considering the current nuclear model of Nuclear Physics, because the alpha particle could be emitted in any direction. And as the nucleus has a spin, when the alpha particle leaves the nucleus it would be dragged by the spin, and it would leave out the nucleus by a tangential trajectory.”
and
” As there is not a central 2He4 in your model, the alpha particle could be emitted in any direction. And therefore many alpha particles would be emitted along a tangential direction.”
I ask you, If this happened, would be the spin of the nuclei slowed down too?
Bhagirath Joshi wrote in December 5th, 2011 at 10:39 AM
“Your Comment:
hose elec quad. moment are not regarding to STABLE nuclei.
They are regarding to EXCITED nuclei.
My explanation:
Precisely. The reason we have electric quad moment Null, is because neutrons are in the center and since they are not exactly balancing each other, when excited, exhibits the EQM. Other wise it does not.”
Here we go again.
Dear Joshi,
In your paper you say:
“Thus from the above analysis, it looks like that all nuclei, stable or otherwise, prefers to keep at least one neutron at the center of the nuclei from its excess count ”
I suppose such condition must be ALWAYS satisfied, otherwise you cannot explain the occurrence of isotopes, as it is your intension.
Stable 8018 has electric quadrupole moment Q(b)=0
It has two excess neutrons.
So, I suppose that one neutron occupies the center, and the other one occupies the n-shell.
So, its structure is asymetric.
Then I cannot see how stable 8O18 may have Q(b)=0, unless if the two excess neutrons occupy the same excess n-shell (but in such case the fundamental premise is not satisfied: one excess neutron must be in the center).
The same we can say about stables 4Be10, 6C14, 10Ne22, 12Mg26, 14Si30, 16S34, 18Ar38, 20Ca42, 22Ti46, 24Cr50, 26Fe54, 28Ni58.
All they have null electric quadrupole moment, Q(b)=0.
4Be10 = 0
6C14 = 0
10Ne22 = 0
12Mg26 = 0
14Si30 = 0
16S34 = 0
18Ar38= 0
20Ca42 = 0
22Ti46 = 0
24Cr50 = 0
26Fe54 = 0
28Ni58 = 0
But according to your theory they CANNOT have Q(b)=0
Regards
WG
Wladimir Guglinski
December 5th, 2011 at 6:21 AM
Your Comment:
If each author decides to reject some of those tracks, because his new nuclear model does not fit the experimental data, then we will never succeed to discover the true structure of the nucleus.
Because, as the tracks are the only way, then it is not possible to succeed in the enterprise if each author decides to reject some tracks.
My explanation:
I am not making any attempt to disregard any experimental data. In fact I am encouraging everyone to analyze my model and see what is causing the stability and prevents EQM in those isotopes.
I know there are doubts and I personally have many, but physically for me it is impossible to verify all available experimental data. But there are clues, like in excited state EQM is exhibited by these test nuclei you are putting forward as a test case, This may be the direct result of “Excess Neutron Shell ” as proposed by me.
If few scientists test my hypothesis, do some calculations and compare with their experimental data, Then two things can happen
1) The model is proven correct and we accept it
or 2) The model is drastically wrong and we throw it out and work on some another hypothesis.
The “Excess Neutron Shell” has one obvious advantage and that it prevents building unlimited number of isotopes and elements as preferred by nature.
I have sent my paper to Harvard, MIT, Oxford, Cambridge, Texas A & M, Kyoto university and many others to check. May be one day they will.
Sincerely,
Bhagirath Joshi
Wladimir Guglinski
December 5th, 2011 at 6:21 AM
Your Comment:
hose elec quad. moment are not regarding to STABLE nuclei.
They are regarding to EXCITED nuclei.
My explanation:
Precisely. The reason we have electric quad moment Null, is because neutrons are in the center and since they are not exactly balancing each other, when excited, exhibits the EQM. Other wise it does not.
Sincerely
Bhagirath Joshi
Bhagirath Joshi wrote in December 4th, 2011 at 10:44 AM
10Ne22 = -0.19(4) CER
12Mg26 = -0.21(2) CER
14Si30 = -0.05(6) or +0.01(6) CER
16S34 = +0.06(4) CER
20Ca42 = -0.19(8) CER
22Ti46 = -0.21(6) CER
24Cr50 = -0.36(7) CER
26Fe54 = -0.05(14) CER, +0.30(4) st TDPAD, TF
28Ni58 = -0.10(6) CER
Dear Joshi,
those elec quad. moment are not regarding to STABLE nuclei.
They are regarding to EXCITED nuclei.
Look:
10Ne22 = -0.19(4) CER
Ex= 1275
12Mg26 = -0.21(2) CER
Ex= 1809
14Si30 = -0.05(6) or +0.01(6) CER
Ex= 2235
16S34 = +0.06(4) CER
20Ca42 = -0.19(8) CER
Ex= 1525
22Ti46 = -0.21(6) CER
Ex= 889
24Cr50 = -0.36(7) CER
Ex= 783
26Fe54 = -0.05(14) CER,
Ex= 1408
+0.30(4) st TDPAD, TF
Ex= 6527
28Ni58 = -0.10(6) CER
Ex= 1454
Dear Joshi,
For the discovery of the true structure of the nucleus existing in the nature, we have to consider the tracks we have available: the known nuclear properties and the experimental data.
If the new nuclear model does not fit to some tracks, then something is wrong with that model.
If each author decides to reject some of those tracks, because his new nuclear model does not fit the experimental data, then we will never succeed to discover the true structure of the nucleus.
Because, as the tracks are the only way, then it is not possible to succeed in the enterprise if each author decides to reject some tracks.
regards
WG
Dear Martin,
Thank you!
Warm Regards,
A.R.
Dear Guglinski and Joshi,
As I have stated before,my scenario for the creation of excess energy in the Rossi device involves the emission of an electron type of particle from the nucleous of the nickel which then interacts with the lattice electrons.I have suggested that these particles are the result of decay processes instigated by the field effects of an additional charge carried into the lattice configuration by injected ionized hydrogen atoms.There are three types of electron like particles that have been observed in previous experimental work that are emitted by the nucleous.Heavy electrons(Muons)Beta particles(electrons)and internal conversion electrons.The first two are direct results of decay processes and are well known.The internal conversion electrons are the result of S orbital electrons which are forced into the nucleous by the negative electron fields of the upper orbital electrons when atomic reactions distort the normal configuration of the undisturbed atom.The S electron actually enters the nucleous and causes the ejection of an electron from the atomic configuration This electron then can interact with the lattice to create the excess energy.These electrons differ from Beta electrons in their energy spectrums since the Betas have a wide spectrum(caused by the variation of the amount of energy retained by the neutrinos in the decay)and the other electron energys are specific only for the energy levels involved with the reaction.I asked Rossi if he had installed a window of some sort into his core so he could could observe if visible photons could be seen during the operation of his unit.He said yes but could not reveal what was observed because it could reveal secrets.If Beta electrons or muons were emitted,generated light photons would look more like black body radiation and light caused by the interaction of conversion electrons would be much more spectrally pure.
Both of your theories may hold clues as to how the electron like particles are generated within the nucleous.
Dear Bhagirath Joshi. Your reply Dec 1st 2011 I find very informative and certainly something to think about and as you must be aware I am unable to comment on your technical neutron calculations in any way. I like many others know that Einstein was fascinated by ‘Time’ and to an extent, I believe, his understanding of time did nothing other than bring about confusion. He seemed unable to put it into an understandable concept and that is why wild speculating theories existed during his time. Your model is, I believe, one step forward in comparison to Einstein who although contributed greatly to physics was not always right. Regards Eric Ashworth.
Dear Andrea,
So much people asking things to do to get some (offcourse not enough) proof.
I think it is time for a commercial plan to check out THE motivation of THE askers.
Just give à price to do this extra service and we Will see if there are people who are motivated enough to Pay something. I will not pay anything because you don’t have to proof Nothing to me. I just wait for my 10 kw ecat! Hope you get THE patent and other certifications soon. Advantage of this aproach is that you earn money. This is normal because it delays THE building of 1 Mw. Plants.
Best regards,
Martin
Dear Neri Accornero:
We are not able to produce electric power, just make heat.
Warm Regards,
A.R.
Dear Tomasz Rojewski:
You are not wrong.
Warm Regards,
A.R.
Dear Claud:
This could be arranged, we have to think about.
Warm Regards,
A.R.
Dear H. Hansson:
There are problems of security and confidentiality. We cannot leave an E-Cat in others’ sites. We are selling only 1 thermal MW plants.
Warm Regards,
A.R.
Dear Mr. Rossi,
A web-camera in your labs seem to be a bad idea
I’m sure that if you ask around there will be many in your neighborhood ready to hosts a set-up with a single eCat show. After all the winter is coming.
Dear A.R. to show an e-cat working on a video livestream, a mere laboratory corner framing with a single device on a table would be enough!
It would be fabulous!
Dear mr Rossi
I think it’s not an argument
“I cannot open a camera on our test room, because we make also confidential operations. Only a plant in regular operation in our Customers’ concerns is possible.”
do you have onely one room? 😉
Dear Ing Rossi
http://www.ecat.com site should appear more attractive and frequently updated , as I already wrote to you a visible counter of the home system booking will help and, as others suggest, a live webcam on a single working module evenctually powering a small stirling motor generator that light a lamp surely will be impressive . I agree that you do not need other public test but you must differentiate from other appearing (false ?) competitors web sites.
Sincerely
Neri Accornero
Dear Wladimir Guglinski
December 4th, 2011 at 7:53 AM
Your Comment:
My understanding is that any nuclear model must be able to explain ALL the nuclear properties. Otherwise it cannot be correct, and so it cannot be accepted.
That’s why the prevailing nuclear model of current Nuclear Physics cannot be correct, since it is not able to explain several nuclear properties.
My Understanding:
You are absolutely correct. However, not a single nuclear model thus far is able to explain the upper limit to isotope building or able to calculate theoretically half life of unstable element. And yet it is accepted.
Now , In excited states the elements you are putting forward as a test case do exhibit the above property. One explanation from my model is that , when excited neutron goes through some changes and under these conditions these effects are observed. The elements multiple of Helium equivalent are very stable, (except 20Ca40 which is unstable and eventually disintegrates) and are able to compensate for the wobble of the neutron at the center in a limited way and this can only be understood by some theoretical calculations.
WG: please fill the values of elec. quad. moments of the STABLE nuclei, according to that nuclear table:
B. Joshi:
4Be10 =
6C14 =
10Ne22 = -0.19(4) CER
12Mg26 = -0.21(2) CER
14Si30 = -0.05(6) or +0.01(6) CER
16S34 = +0.06(4) CER
18Ar38=
20Ca42 = -0.19(8) CER
22Ti46 = -0.21(6) CER
24Cr50 = -0.36(7) CER
26Fe54 = -0.05(14) CER, +0.30(4) st TDPAD, TF
28Ni58 = -0.10(6) CER
CER = Coulomb Excitation Reorientation
TDPAD = Time Dependent Perturbed Angular Distribution
Thanking You,
Sincerely
Bhagirath Joshi
Dear Italo:
I cannot open a camera on our test room, because we make also confidential operations. Only a plant in regular operation in our Customers’ concerns is possible.
Warm Regards,
A.R.
Dear Ing. Rossi, it really is a great idea putting a webcam and a couple of temperature measurements on a running E-Cat!!
This wouldn’t be another unuseful test like those asked by skeptics or snakes, but a simple and clear demonstrations that there is a running reactor, ready for mankind, ready for each of us!
I think that it isn’t necessary putting the webcam on a sold plant, but only on one E-Cat module in your laboratory.
Kind regards, we trust you!
Italo R.
Bhagirath Joshi wrote in December 3rd, 2011 at 7:20 PM
“Also there may be anomalies in some situations with stable nuclei at present and the reason could be that the experimental data may be inconclusive or incomplete.”
Dear Bhagirath Joshi,
the experimental data are NOT inconclusive NOR incomplete.
The stable nuclei:
4Be10, 6C14, 10Ne22, 12Mg26, 14Si30, 16S34, 18Ar38, 20Ca42, 22Ti46, 24Cr50, 26Fe54, 28Ni58
all they have NULL electric quadrupole moment.
Stable nuclei with NULL elec. quad. moment are NOT listed in the nuclear table.
And stable nuclei with null elect. quad. mom. and also null magnetic dipole moment even does not exist in the nuclear table.
For instance: stables 2He4, 8016, and 20Ca40 are not listed in the nuclear table, because:
1 – they have null magnetic dipole moment
2 – they have null electric quadrupole moment
My understanding is that any nuclear model must be able to explain ALL the nuclear properties. Otherwise it cannot be correct, and so it cannot be accepted.
That’s why the prevailing nuclear model of current Nuclear Physics cannot be correct, since it is not able to explain several nuclear properties.
Before to be accepted by the scientific community, your model will be submitted to scrutiny by the physicits, as I did here.
And if your model cannot explain some phenomena, they will conclude that it cannot be correct.
And I dont think they will replace an unsatisfying model (the current model of Nuclear Physics) by another unsatisfying model.
Regards
WG
Dear Italo A. Albanese:
You are right, but to do this we must wait to have the right Customer, who allows us to put his plant online. I will do it, as soon as I will have this chance.
Warm Regards,
A.R.
Dear Andrea Rossi,
May I remember you this is not the first time you say “I’ll do a video live stream”?
Best regards,
Italo A.
Dear Kim P,
Great suggestion. Add a Clock on the wall and a digital temperature meter on the outlet/inlet.
But it should not be portrayed as a scientific test or something like that. Just as demonstration that comes as it is…boiling water day after day. The smart thing is in simplicity and time factor.
Wladimir Guglinski
December 2nd, 2011 at 11:43 PM
Your comments on Q(b).
My understanding:
It is important to understand certain phenomena for a stable isotope, because that is the only way to gain the insight into the structure of nuclei. Using unstable isotopes may give some insight, but the inherent flow with in which makes it unstable in the first place, may exhibit various moments or may not and it could send one on a goose chase. The proposed model may seem to fail there and it will, since the processes internal to that unstable isotope will exhibit different property. Does that mean the premises is falls?
Also there may be anomalies in some situations with stable nuclei at present and the reason could be that the experimental data may be inconclusive or incomplete.
The real test of my model is to use it to investigate nuclei behavior in solid.
The number of isotopes are limited and there is also limit to how heavy a stable isotope can be. The “Excess neutron shell model” as proposed by me shows that Gravity does play part in this.
The next thing to do is to see how one can test this model in solid to predict the half life of unstable isotope.
Thanking you
Sincerely
Bhagirath Joshi
Dear Mr. Rossi
I am agreed with Kim. It could be a very good idea also to keep high media’s attentions.
my compliments for your work
Regards
“Great idea, we will make it ( not the mental bricolage)”
HA HA HA HA HA
Bhagirath Joshi wrote in December 3rd, 2011 at 2:20 AM:
“My Explanation:
FYI
http://www.uni-due.de/physik/wende/englisch/nuclear-moments.pdf
In the above table Q(b) is not zero for these isotopes. “Excess Neutron shell” is asymmetric for these isotopes And it agrees with my model.”
Dear Bhagirath Joshi
please fill the values of elec. quad. moments of the STABLE nuclei, according to that nuclear table:
4Be10 =
6C14 =
10Ne22 =
12Mg26 =
14Si30 =
16S34 =
18Ar38=
20Ca42 =
22Ti46 =
24Cr50 =
26Fe54 =
28Ni58 =
Dear Kim Patterson:
Great idea, we will make it ( not the mental bricolage)
Warm Regards,
A.R.
With all due Resepect
You need to set up 1 unit and
run it 24 hours a day on video
live stream.
(This is a simple thing and this
is what the people want)
All else is mental masturbation
Respect
Kim