by E.N. Tsyganov
(UA9 collaboration) University of Texas Southwestern
Medical Center at Dallas, Texas, USA
Abstract
Recent accelerator experiments on fusion of various elements have clearly demonstrated that the effective cross-sections of these reactions depend on what material the target particle is placed in. In these experiments, there was a significant increase in the probability of interaction when target nuclei are imbedded in a conducting crystal or are a part of it. These experiments open a new perspective on the problem of so-called cold nuclear fusion.
Introduction
Experiments of Fleischmann and Pons made about 20 years ago [1], raised the question about the possibility of nuclear DD fusion at room temperature. Conflicting results of numerous experiments that followed, dampened the initial euphoria, and the scientific community quickly came to common belief, that the results of [1] are erroneous. One of the convincing arguments of skeptics was the lack in these experiments of evidence of nuclear decay products. It was assumed that “if there are no neutrons, therefore is no fusion.” However, quite a large international group of physicists, currently a total of about 100-150 people, continues to work in this direction. To date, these enthusiasts have accumulated considerable experience in the field. The leading group of physicists working in this direction, in our opinion, is the group led by Dr. M. McKubre [2]. Interesting results were also obtained in the group of Dr. Y. Arata [3]. Despite some setbacks with the repeatability of results, these researchers still believe in the existence of the effect of cold fusion, even though they do not fully understand its nature. Some time ago we proposed a possible mechanism to explain the results of cold fusion of deuterium [4]. This work considered a possible mechanism of acceleration of deuterium contaminant atoms in the crystals through the interaction of atoms with long-wavelength lattice vibrations in deformed parts of the crystal. Estimates have shown that even if a very small portion of the impurity atoms (~105) get involved in this process and acquires a few keV energy, this will be sufficient to describe the energy released in experiments [2]. This work also hypothesized that the lifetime of the intermediate nucleus increases with decreasing energy of its excitation, so that so-called “radiation-less cooling” of the excited nucleus becomes possible. In [5], we set out a more detailed examination of the process. Quite recently, a sharp increase of the probability of fusion of various elements was found in accelerator experiments for the cases when the target particles are either imbedded in a metal crystal or are a part of the conducting crystal. These experiments compel us to look afresh on the problem of cold fusion.
Recent experiments on fusion of elements on accelerators
For atom-atom collisions the expression of the probability of penetration through a Coulomb barrier for bare nuclei should be modified, because atomic electrons screen the repulsion effect of nuclear charge. Such a modification for the isolated atom collisions has been performed in H.J. Assenbaum and others [6] using static Born-Oppenheimer approximation. The experimental results that shed further light on this problem were obtained in relatively recent works C. Rolfs [7] and K. Czerski [8]. Review of earlier studies on this subject is contained in the work of L. Bogdanova [9]. In these studies a somewhat unusual phenomenon was observed: the sub-barrier fusion cross sections of elements depend strongly on the physical state of the matter in which these processes are taking place. Figure 1 (left) shows the experimental data [8], demonstrating the dependence of the astrophysical factor S(E) for the fusion of elements of sub-threshold nuclear reaction on the aggregate state of the matter that contains the target nucleus 7Li. The same figure (right) presents similar data [7] for the DD reaction, when the target nucleus was embedded in a zirconium crystal. It must be noted that the physical nature of the phenomenon of increasing cross synthesis of elements in the case where this process occurs in the conductor crystal lattice is still not completely clear.
Figure 1. Up – experimental data [8], showing the energy dependence of the S-factor for sub-threshold nuclear reaction on the aggregate state of matter that contains the nucleus 7Li. Down – the similar data [7] for the reaction of DD, when the target nucleus is placed in a crystal of zirconium. The data are well described by the introduction of the screening potential of about 300 eV.
The phenomenon is apparently due to the strong anisotropy of the electrical fields of the crystal lattice in the presence of free conduction electrons. Data for zirconium crystals for the DD reactions can be well described by the introduction of the screening potential of about 300 eV. It is natural to assume that the corresponding distance between of two atoms of deuterium in these circumstances is less than the molecular size of deuterium. In the case of the screening potential of 300 eV, the distance of convergence of deuterium atoms is ~510ˆ12 m, which is about an order of magnitude smaller than the size of a molecule of deuterium, where the screening potential is 27 eV. As it turned out, the reaction rate for DD fusion in these conditions is quite sufficient to describe the experimental results of McKubre and others [2]. Below we present the calculation of the rate process similar to the mu-catalysis where, instead of the exchange interaction by the muon, the factor of bringing together two deuterons is the effect of conduction electrons and the lattice of the crystal.
Calculation of the DD fusion rate for “Metal-Crystal” catalysis
The expression for the cross section of synthesis in the collision of two nuclei can be written as
where for the DD fusion
Here the energy E is shown in keV in the center of mass. S(E) astrophysical factor (at low energies it can be considered constant), the factor 1/E reflects de Broglie dependence of cross section on energy. The main energy dependence of the fusion is contained in an expression
that determines the probability of penetration of the deuteron through the Coulomb barrier. From the above expressions, it is evident that in the case of DD collisions and in the case of DDμcatalysis, the physics of the processes is the same. We use this fact to determine the probability of DD fusion in the case of the “metal-crystalline” DD-catalysis. In the case of DDμ- catalysis the size of the muon deuterium molecules (ion+) is ~5×10ˆ13m. Deuterium nuclei approach such a distance at a kinetic energy ~3 keV. Using the expression (1), we found that the ratio of σ(3.0 keV)/σ(0.3 keV) = 1.05×10ˆ16. It should be noted that for the free deuterium molecule this ratio [ σ(3.0keV)/σ(0.03keV)] is about 10ˆ73. Experimental estimations of the fusion rate for the (DDμ)+ case presented in the paper by Hale [10]:
Thus, we obtain for the “metal-crystalline” catalysis DD fusion rate (for zirconium case):
Is this enough to explain the experiments on cold fusion? We suppose that a screening potential for palladium is about the same as for zirconium. 1 cmˆ3 (12.6 g) of palladium contains 6.0210ˆ23(12.6/106.4) = 0.710ˆ23 atoms. Fraction of crystalline cells with dual (or more) the number of deuterium atoms at a ratio of D: Pd ~1:1 is the case in the experiments [2] ~0.25 (e.g., for Poisson distribution). Crystal cell containing deuterium atoms 0 or 1, in the sense of a fusion reaction, we consider as “passive”. Thus, the number of “active” deuterium cells in 1 cmˆ3 of palladium is equal to 1.810ˆ22. In this case, in a 1 cmˆ3 of palladium the reaction rate will be
this corresponds to the energy release of about 3 kW. This is quite sufficient to explain the results of McKubre group [2]. Most promising version for practical applications would be Platinum (Pt) crystals, where the screening potential for d(d,p)t fusion at room temperature is about 675 eV [11]. In this case, DD fusion rate would be:
The problem of “nonradiative” release of nuclear fusion energy
As we have already noted, the virtual absence of conventional nuclear decay products of the compound nucleus was widely regarded as one of the paradoxes of DD fusion with the formation of 4He in the experiments [2]. We proposed the explanation of this paradox in [4]. We believe that after penetration through the Coulomb barrier at low energies and the materialization of the two deuterons in a potential well, these deuterons retain their identity for some time. This time defines the frequency of further nuclear reactions. Figure 2 schematically illustrates the mechanism of this process. After penetration into the compound nucleus at a very low energy, the deuterons happen to be in a quasi-stabile state seating in the opposite potential wells. In principle, this system is a dual “electromagnetic-nuclear” oscillator. In this oscillator the total kinetic energy of the deuteron turns into potential energy of the oscillator, and vice versa. In the case of very low-energy, the amplitude of oscillations is small, and the reactions with nucleon exchange are suppressed.
Fig. 2. Schematic illustration of the mechanism of the nuclear decay frequency dependence on the compound nucleus 4He* excitation energy for the merging deuterons is presented. The diagram illustrates the shape of the potential well of the compound nucleus. The edges of the potential well are defined by the strong interaction, the dependence at short distances Coulomb repulsion.
The lifetime of the excited 4He* nucleus can be considered in the formalism of the usual radioactive decay. In this case,
Here ν is the decay frequency, i.e., the reciprocal of the decay time τ. According to our hypothesis, the decay rate is a function of excitation energy of the compound nucleus E. Approximating with the first two terms of the polynomial expansion, we have:
Here ν° is the decay frequency at asymptotically low excitation energy. According to quantum-mechanical considerations, the wave functions of deuterons do not completely disappear with decreasing energy, as illustrated by the introduction of the term ν°. The second term of the expansion describes the linear dependence of the frequency decay on the excitation energy. The characteristic nuclear frequency is usually about 10ˆ22 sˆ-1. In fusion reaction D+D4He there is a broad resonance at an energy around 8 MeV. Simple estimates by the width of the resonance and the uncertainty relation gives a lifetime of the intermediate state of about 0.810ˆ22 s. The “nuclear” reaction rate falls approximately linearly with decreasing energy. Apparently, a group of McKubre [2] operates in an effective energy range below 2 keV in the c.m.s. Thus, in these experiments, the excitation energy is at least 4×10ˆ3 times less than in the resonance region. We assume that the rate of nuclear decay is that many times smaller. The corresponding lifetime is less than 0.3×10ˆ18 s. This fall in the nuclear reaction rate has little effect on the ratio of output decay channels of the compound nucleus, but down to a certain limit. This limit is about 6 keV. A compound nucleus at this energy is no longer an isolated system, since virtual photons from the 4He* can reach to the nearest electron and carry the excitation energy of the compound nucleus. The total angular momentum carried by the virtual photons can be zero, so this process is not prohibited. For the distance to the nearest electron, we chose the radius of the electrons in the helium atom (3.1×10ˆ11 m). From the uncertainty relations, duration of this process is about 10ˆ-19 seconds. In the case of “metal-crystalline” catalysis the distance to the nearest electrons can be significantly less and the process of dissipation of energy will go faster. It is assumed that after an exchange of multiple virtual photons with the electrons of the environment the relatively small excitation energy of compound nucleus 4He* vanishes, and the frequency of the compound nucleus decaying with the emission of nucleons will be determined only by the term ν°. For convenience, we assume that this value is no more than 10ˆ12-10ˆ14 per second. In this case, the serial exchange of virtual photons with the electrons of the environment in a time of about 10ˆ-16 will lead to the loss of ~4 MeV from the compound nucleus (after which decays with emission of nucleons are energetically forbidden), and then additional exchange will lead to the loss of all of the free energy of the compound nucleus (24 MeV) and finally the nucleus will be in the 4He ground state. The energy dissipation mechanism of the compound nucleus 4He* with virtual photons, discussed above, naturally raises the question of the electromagnetic-nuclear structure of the excited compound nucleus.
Fig. 3. Possible energy diagram of the excited 4He* nucleus is presented.
Figure 3 represents a possible energy structure of the excited 4He* nucleus and changes of its spatial configuration in the process of releasing of excitation energy. Investigation of this process might be useful to study the quark-gluon dynamics and the structure of the nucleus.
Discussion
Perhaps, in this long-standing history of cold fusion, finally the mystery of this curious and enigmatic phenomenon is gradually being opened. Besides possible benefits that the practical application of this discovery will bring, the scientific community should take into account the sociological lessons that we have gained during such a long ordeal of rejection of this brilliant, though largely accidental, scientific discovery. We would like to express the special appreciation to the scientists that actively resisted the negative verdict imposed about twenty years ago on this topic by the vast majority of nuclear physicists.
Acknowledgements
The author thanks Prof. S.B. Dabagov, Dr. M. McKubre, Dr. F. Tanzela, Dr. V.A. Kuzmin, Prof. L.N. Bogdanova and Prof. T.V. Tetereva for help and valuable discussions. The author is grateful to Prof. V.G. Kadyshevsky, Prof. V.A. Rubakov, Prof. S.S. Gershtein, Prof. V.V. Belyaev, Prof. N.E. Tyurin, Prof. V.L. Aksenov, Prof. V.M. Samsonov, Prof. I.M. Gramenitsky, Prof. A.G. Olshevsky, Prof. V.G. Baryshevsky for their help and useful advice. I am grateful to Dr. VM. Golovatyuk, Prof. M.D. Bavizhev, Dr. N.I. Zimin, Prof. A.M. Taratin for their continued support. I am also grateful to Prof. A. Tollestrup, Prof. U. Amaldi, Prof. W. Scandale, Prof. A. Seiden, Prof. R. Carrigan, Prof. A. Korol, Prof. J. Hauptmann, Prof. V. Guidi, Prof. F. Sauli, Prof. G. Mitselmakher, Prof. A. Takahashi, and Prof. X. Artru for stimulating feedback. Continued support in this process was provided with my colleagues and the leadership of the University of Texas Southwestern Medical Center at Dallas, and I am especially grateful to Prof. R. Parkey, Prof. N. Rofsky, Prof. J. Anderson and Prof. G. Arbique. I express special thanks to my wife, N.A. Tsyganova for her stimulating ideas and uncompromising support.
References
1. M. Fleischmann, S. Pons, M. W. Anderson, L. J. Li, M. Hawkins, J. Electro anal. Chem. 287, 293 (1990).
2. M. C. H. McKubre, F. Tanzella, P. Tripodi, and P. Haglestein, In Proceedings of the 8th International Conference on Cold Fusion. 2000, Lerici (La Spezia), Ed. F. Scaramuzzi, (Italian Physical Society, Bologna, Italy, 2001), p 3; M. C. H. McKubre, In Condensed Matter Nuclear Science: Proceedings Of The 10th International Conference On Cold Fusion; Cambridge, Massachusetts, USA 21-29 August, 2003, Ed by P. L. Hagelstein and S. R. Chubb, (World Sci., Singapore, 2006). M. C. H. McKubre, “Review of experimental measurements involving dd reactions”, Presented at the Short Course on LENR for ICCF-10, August 25, 2003.
3. Y. Arata, Y. Zhang, “The special report on research project for creation of new energy”, J. High Temp. Soc. (1) (2008).
4. E. Tsyganov, in Physics of Atomic Nuclei, 2010, Vol. 73, No. 12, pp. 1981–1989. Original Russian text published in Yadernaya Fizika, 2010, Vol. 73, No. 12, pp. 2036–2044.
5. E.N. Tsyganov, “The mechanism of DD fusion in crystals”, submitted to IL NUOVO CIMENTO 34 (4-5) (2011), in Proceedings of the International Conference Channeling 2010 in Ferrara, Italy, October 3-8 2010.
6. H.J. Assenbaum, K. Langanke and C. Rolfs, Z. Phys. A – Atomic Nuclei 327, p. 461-468 (1987).
7. C. Rolfs, “Enhanced Electron Screening in Metals: A Plasma of the Poor Man”, Nuclear Physics News, Vol. 16, No. 2, 2006.
8. A. Huke, K. Czerski, P. Heide, G. Ruprecht, N. Targosz, and W. Zebrowski, “Enhancement of deuteron-fusion reactions in metals and experimental implications”, PHYSICAL REVIEW C 78, 015803 (2008).
9. L.N. Bogdanova, Proceedings of International Conference on Muon Catalyzed Fusion and Related Topics, Dubna, June 18–21, 2007, published by JINR, E4, 15-2008-70, p. 285-293
10. G.M. Hale, “Nuclear physics of the muon catalyzed d+d reactions”, Muon Catalyzed Fusion 5/6 (1990/91) p. 227-232.
11. F. Raiola (for the LUNA Collaboration), B. Burchard, Z. Fulop, et al., J. Phys. G: Nucl. Part. Phys.31, 1141 (2005); Eur. Phys. J. A 27, s01, 79 (2006).
by E.N. Tsyganov
(UA9 collaboration) University of Texas Southwestern
Medical Center at Dallas, Texas, USA
If cop is 6 when the heater is on and is infinity when the e cat is in self sustain mode and it runs in this way for the 50% of the time ,with an efficency of 40%,could we imagine that with a reactor of 12 kWht(for example)we should have 91,2 kWhel in 24h? It would be very remunerative.If in Italy will not be possible use this tecnology before I have my hair completely white,I will come in USA.You have the first custom for the e cat able to generate electricity,Dott Rossi,if you want of course !
Dear Franco:
Again, the COP is, and will remain, 6.
Warm Regards,
A.R.
Dear Ing Rossi and Dr Fine,
it seems to me that declared E-Cat COP of 6 already takes into account the average ON-OFF duty cycle of 50%.
In other words making an example please consider a time period of 1 hour and suppose that during this time the output thermal energy produced by the E-Cat is 10kWh.
Within the overall “ON” period (for example half hour) the E-Cat consumes 3.33kW of electrical power whereas during the overall “OFF” period no electrical energy is consumed.
As consequence 3.33kW for first half hour and 0.0kW for the second one means 1.666kWh, therefore the COP to be considered should be 6.
I am reading Your comments about my opinion.
Kind Regards
Franco
Thanks.
For a COP of 6, 1 kWhe (electric) is used when the heaters are on, for each 6 kWht (thermal) produced. When the heaters are off, 0 kWhe (i.e. ZERO) are used for each 6 kWht produced. The combination of a COP of 6 and an COP of infinity gives you a total COP of 12, since you only use up 1 kWhe half of the time. I think we are talking about the same thing. As for production of electricity, at 600 (+/- a few) degrees, you should be able to get 40% efficiency with various technologies.
Best regards,
Joseph
Not as Confused
in NJ
Dear Dr Joseph Fine:
COP 6, as you correctly say, meand that you get 6 kWht consuming 1 kWhe.
About the production of electricityi any data are virtual, because we did not yet make t, so I prefer confirm the data only after we got them really, not virtually. The reason you are confused for is that we did not give precise data, yet.
Warm Regards,
A.R.
A.R.
OK. For a 12 kW thermal output, a COP of 6 means:
Either,
a) 12 kW/6 = 2 kW (electrical) input. Or,
b) 12 kW/3 = 4 kW (electrical) input for 50% of the time
Since 24 * 12 kW = 288 kW-Hrs/day (thermal), if 2 kW (electrical) heaters are used for 12 hr, only 24 kW-Hrs/day (electrical) energy are used. And 288/24 = 12 and not 6.
What part of COP do I not understand?
Joseph
Confused in
Dear Dr Joseph Fine:
1- the timing of the on-off of the resistances is estabilished by the control system and is not constant. The 50% is an average.
2- the max COP is anyway 6
3- the range is the one you said, moreless.
Warm Regards,
A.
Dear Andrea,
The electric heaters in the core operate about 50% of the time. The on and off intervals of the heaters are variable, but the duty cycle remains 50%.
With an E-Cat of 12 kW output power (as an example), 24 hrs * 12 kW = 288 kW-hrs of thermal energy are produced each day. For a 50% heater duty cycle and a COP of 6, do the heaters require only 12 hrs * 2 kW = 24 kW-hrs of electrical energy? Is this, in fact, a COP of 12 (288/24 = 12) and not 6 since the heaters do not operate continuously?
For a COP of 6, does that mean the heaters use electrical power that equals 1/3 of the output thermal power when they are on, but only use this for half the time.
Can the core heaters remain turned on 100% of the time? And, would the COP remain the same compared with a 50% duty cycle?
The control system turns the heaters on and off as required. This suggests there is a variable power output and/or variable output temperature. For a nominal 600 degree C operating temperature and a nominal 10 kW output, is there a range of temperatures or power for normal operation. That is, does the output temperature vary between 599 and 601 degrees C or 590 and 610 degrees C etc. Does the output power vary between 9.5 and 10.5 kW, 9.9 and 10.1 kW or some other range.
Best regards,
Joseph
Dear Francesco,
You are right, I wrote wrongly.
Warm Regards,
A.R.
Dear Michele Bruno:
Thank you, we are working at the highest level of our capacities.
Warm Regards,
A.R.
Ing. Rossi,
la sua invenzione è un sogno per tutta l’umanità. Per la prima volta ogni di noi potrà provvedere all’energia di cui necessità in maniera gratuita ed a bassissimo costo. L’energia che da sempre è stato uno strumento per cui l’oligarchia di potenti ha sottomessi i popoli, potrà finalmente diventare libera ed accessibile per tutti.
Questa è la svolta che tutti noi aspettavamo.
Il mondo cambierà in meglio.
Per cui grazie Andrea Rossi. Sono a sua disposizione nel caso dovesse aver bisogno di me.
Saluti
Caro ing. Rossi
Allow me a small and benevolent rebuke. She wrote to me:
———————————————————————————–
“”May 27th, 2012 at 6:54 AM
Dear Francesco:
Within a week we will have important news regarding the high temperature reactors.
Warm Regards,
A.R.””
———————————————————————————–
Then within a week.
But I’m not worried … maybe you’re very busy. But I wait patiently.
Warm Greetings
F.T.
Dear Gillana:
I said we are going to give important information regarding the high temperature reactors, but not this week. We are completing our tests: matter of a couple of weeks.
The earthquake did not stop our work, luckily. We continued to work.
Warm Regards,
A.R.
Dear Dr Joseph Fine:
I totally agree.
Warm Regards,
A.R.
Another potential application for the E-Cat would be producing liquid biofuels from cellulose. While electricity production is a major goal, producing transportation fuels is not unimportant. Producing butanol (vs. ethanol) would be a better choice, but chemists and engineers (and the market place) can work that out. The Purdue conversion process (See link) requires heat at 500 degrees C and Hydrogen. The E-Cat can provide heat at very low cost. So don’t throw away your lawnmower clippings yet, you may be able to fill up your car with… “New Fuel from New Fire”.
http://www.midwestenergynews.com/2012/06/06/cellulosic-biofuel-gets-better-but-theres-a-but/
Joseph
Egr. Dr. Andrea Rossi
Nel web ho letto alcuni giorni fa che avrebbe rilasciato delle notizie interessanti, in settimana, riguardo al suo reattore E-cat.
Oggi leggo che nel suo Journal of Phisics (non so esattamente dove), queste dichiarazioni saranno prorogate anche a causa del terremoto in emilia?
Ancora una volta la mia speranza (e forse di molti altri) di avere una conferma che qualcosa si sta muovendo nella sua invenzione è stata disattesa.
Caro Dottore, invece di annunciare che si annuncierà non sarebbe meglio annunciare e basta.
Cordiali saluti!
Gillana Giancarlo
Dear Luca Coppola
Thank you!
Warm Regards,
A.R.
Dear Marc Fuochi:
Thank you for your attention. We have registered your request.
Warm Regards,
A.R.
Dear Pietro F.:
Thank you, but I cannot enter in the particularsof what happens in the reactor. Interesting analisys, though.
Warm Regards,
A.R.
Buongiorno sig. Rossi,
penso che in quest’articolo possa trovare spunti interessanti per risolvere i problemi che sta incontrando:
http://ecatreport.com/e-cat/will-the-e-cat-have-hot-spots
Hello Mr. Rossi,
I contacted you on this site back in October to see if I could somehow help you in your extraordinary effort. You asked me to contact you in November, but other obligations prevented me from doing so.
Now I am back and excited to see the incredible progress your team has made!
I am still interested in being part of your team and helping in any way that I can. I’m in New Orleans, LA, and my background is in computer engineering / computer science / project management. I’m not sure how I could best use my talents to help with your efforts, but I’m sure you’ve got a place somewhere for an enthusiastic, hard-working techie 🙂
If it would help, I would be honored to travel and meet with you to discuss my qualifications.
Kindest regards,
Marc Fouchi
A.R.
Thanks for your answers. As for ‘c’ in furlongs per fortnight, you forgot the seconds per fortnight conversion. After a 16 hour day, I’m amazed you tried it and almost got it right the first time.
c = 3×10^8 m/sec;
1 furlong = 200 m (approx.)
1 fortnight = 2 weeks = 1.2096×10^6 sec.
c = 3×10^8 (m/sec) x (1 furl/200 m) x (1.2096×10^6 sec/fortnight) =
1.5*10^6 x 1.21×10^6 = 1.815 x 10^12 (more or less).
Joseph
Dear Dr. Rossi,
I won’t give you my opinion on all the subject above just because I’m not a scientist and I don’t even have the basic knowledge to deeply understand the process.
My first meeting with cold fusion was in 1989 when I took my Bachelor’s exam in a scientific lycee in Viareggio. I was so fond for physics that my first oral exam was on this subject, even if nuclear reactions were not in the program I studied everything I could about it. I was so excited on Pons&Fleishman announcements! When the examiners president asked me about fusion and cold fusion I could talk for a while and cut a very good figure!
Since then I followed any new announcement about cold fusion with a passionate eye.
After the exam I had the opportunity to realise my dream: become a Pilot. Now as a pilot I have the training and the habit to take rapid decisions , to evaluate options and to encourage teamwork and team problem solving. That’s what makes me so angry! I don’t know (nobody but you knows) if you ecat really works and how. The thing I know is that we are in dare need for energy and that governments are spending a fortune in research about fusion and other energy sources, like ITER (10 billion € for a prototype) or gas from bituminous schists. It would definitely be a “WISE OPTION” to consider to spend a minimal fraction of that amount to let you try! They have nothing to loose but a better future for mankind to gain. I’d like to say I’m rich enough to invest personally in your company but I’m not.. I can only hope that people in Italy will open their eyes (hopefully not too late) and wish you to fully succeed.
Good luck!
Luca
Dear Dr Joseph Fine:
The SSM (self sustained mode) is regulated by the control system based on a complex interaction between parameters. The longest period can be 2 hours, as an average the self sustained mode runs for the 50% of the total time. The ionizing electromagnetic emissions have no substantial delta between SSM and driven mode.
Now: you want I calculate, around midnight, after 16 hours of work, the speed of light in furlongs per fortnight. Oh, my guess!
Data:
1 Fortnight = 2 weeks = 1,209600 x 10^6 seconds
1 Furlong = 200 m
c = 3 x 10^8 m x s^-1
therefore
c = 1.803 x 10^12 Furlongs x Fortnight^-1
Now I go to sleep.
Good Night,
A.R.
Dear Andrea Rossi,
What is the longest time (in minutes, hours, days or fortnights*) that the new E-Cat has been run continuously in self-sustain mode? Or, looking at it another way, what is the longest time the new E-Cat has been run in driven (active heater) mode?
Is there any difference in X-ray levels – if above background – in the driven/(active htr.) or self-sustain modes?
* Fortnight = 2 weeks.
For example: Calculate the speed of light in furlongs per fortnight.
Joseph Fine
Dear Pekka Janhunen:
I will answer to your question when we will be free to explain exactly how the reactor works. For now just consider that 1 gram of matter equals 23 x 10^6 kWh.
We get less than 1/1000th of this energy, considering also that Ni is not the sole buddy of the box.
Warm Regards,
A.R.
Dear Andrea Rossi,
Can you help: I cannot get the numbers given in Hank Mills’ recent email interview at e-catnews.com to match. The interview describes 90 days (expected) operation at 10 kW with 1.5 grams of nickel. This translates to 34 MeV per nickel atom. This is somewhat more than what can be liberated by turning 62Ni or 64Ni to copper, even if all the nickel would turn to copper. Can you clarify, I suspect that some of these numbers or assumptions are wrong?
By the way your approach towards history sounds very sensible.
regards, pekka
Dear Pavel Vrbovsky:
Thank you!
Warm Regards,
A.R.
Dear Bernie Koppenhofer
>The Socially Responsible Scientific Entrepreneur Party.
Great! and the slogan will be “Less Politics, More Work” 😀
Dear Joseph Fine
>my birthday was May 29th so we are both Geminis.
Happy BirthWeek to you!
dear Andrea Rossi
Happy Birthday ! All best to you and yours family.
Pavel
Dear Charlie Sutherland:
Thank you!
Newton’s birthday is Dec 25th 1642 (Old Style), or Jan 1st 1643 (Gregorian Calendar).
Warm regards,
A.R.
Dear bernie Koppenhofer:
You are right! I will follow the suggestion,
Warm Regards,
A.R.
Dear Prof. Joseph Fine:
Thank you, thank you very much,
Warm Regards,
Andrea
Dear Giovanni Guerrini:
Right,
Warm Regards,
A.R.
Dear Giorgio Adorni Francia:
You are absolutely right, but I already said what I had to say. Obviously, they will produce nothing that will make any kind of work.
And the Tazpayer will pay…
Warm Regards,
A.R.
Dear dr. Rossi
someone in the press is approaching the “so-called cold atom”, a piezo-nonsense, with which snatch more money from italian taxpayers, with your research, in a rather dirty way. Perhaps it is desirable to have a word from you.
Best regards.
…vale a dire è energia sprecata,e non nucleare!
Caro dott Rossi,buon compleanno!
A proposito del suo sfogo,ho letto da qualche parte una frase che cerco di tenere sempre a mente:”mai discutere con un imbecille,prima ti porta al suo livello poi ti batte con l’esperienza!”
Cari saluti Giovanni G
Happy Birthday
Does anyone remember Newton’s birthday. Maybe we all will remember yours for working on Sunday.
Andrea,
Happy Birthday. Due to time zones, it may be Monday AM. So I apologize if I missed your birthday by a few hours. As a matter of fact, my birthday was May 29th so we are both Geminis.
Happy Birthday to you, Happy Birthday to me.
Best regards,
Joseph
PS I wonder when I’ll get wiser?
Ti ringrazio per la riconferma, Andrea,
con quel ” I was referring to the theoretical gurus” hai tranquillizzato un sacco di gente eh eh eh
Mr. Rossi, Antonella:
I see the seeds of a new political party that would attracted both Democrats and Republicans:
The Socially Responsible Scientific Entrepreneur Party. (:
PS Mr.Rossi, please do not let these nuts who say, “sellers, not scientists” detract you from your important work!!
Dear Antonella:
No, I was referring to the theoretical gurus unable to make anything working, searching taxpayers money, that insult persons who make real work by means of their money.
Warm Regards,
A.R.
Caro Andrea,
stavo pensando a questo:
“Some imbecile has written that we are sellers, not scientists:”
A me, come ad altri tuoi sostenitori, è capitato di rispondere con una frase simile a chi ti accusava di negligenza scientifica, quando rifiutavi di sottomettere l’E-cat alla sperimentazione ufficiale nelle Università. In questo caso il termine “scienziato” sta a definire il “mestiere” che uno fa. Quindi la risposta era “Rossi non fa lo scienziato, fa l’imprenditore”. E anche “Non è importante se la sua laurea non è in ingegneria, è ingegnere perchè fa l’ingegnere di mestiere”.
Insomma, spero di non essere inclusa nell’elenco degli imbecilli 🙂
Dear Francesco:
Thank you very much…by the way, I am celebrating by working even it is Sunday…
Warmest Regards,
A.R.
Dear Andrea Rossi
My Best Birthday wishes.
I hope you at least another hundred years of life to see a your central on Mars, to melt the ice and thus make the habitable planet.
You might impress even the extra-terrestrial, as well as you impressed us
Eh, eh …
Warm Regards
F. T.
Dear Antonella,
Thank you!
Warm Regards,
A.R.
Dear Andrea,
Happy Birthday!
I wish you a great new trip around the Sun.
Dear Andrea Manganelli:
Of course my reply was not related to you, your comment just triggered my considerations, like an acoustic wave can generate an avalanche: the acoustic wave has nothing to do with the snow.
About the Cyclone Engine, we are studying it.
Warm Regards,
A.R.
Caro ingegnere,
Comprendo il suo sfogo anche se non riguarda me.
Mi permetta di insistere, non voglio farle perdere tempo ma credo che sia un punto cruciale del processo.
Il motore Cyclone nella sua versione WHE mi sembra perfetto per il suo e-cat nella fase di produzione di energia elettrica.
Questa descrizione del motore dovrebbe far chiarezza:
Applications
The WHE is designed to run on heat as low as 500ºF from many different external sources .
Si tratta di un motore ” a vapore ” potrebbe essere la soluzione semplice e poco costosa.
La guardi da vicino, ecco il sito:
http://www.cyclonepower.com/whe.html
Non me ne voglia se insisto, la sua risposta, anche se condiVisibile nei concetti non mi ha convinto che lei abbia guardato davvero questa soluzione..
Cari saluti
andreamanganelli