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by
Valeriy Y.Tarasov
E-mail: vytarasov@yahoo.com
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Download the ZIP file
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Abstract
The h-space theory is a variant of unified physical theory – a theory of everything.
This theory was built de novo, as the existing physical theories are incompatible and so unsuitable for unification.
A new approach is needed, and has been developed by re-evaluating the definitions of primary physical concepts.
The starting point for the re-evaluation was the following equation – Et = mvL, where energy – E, time – t, length – L, mass – m, velocity – v.
Analysis of these physical concepts resulted in the construction of a unique equation of the primary concepts such as space, length, energy and velocity.
From this, models could be developed that explain all well-known physical phenomena.
In addition, h-space theory predicts phenomena rejected by the current mainstream theories, such as limits to gravitational and electrostatic interactions, and the possibility of cold fusion (as a consequence of the electric charge definition, a modification of Coulomb’s law and the definitions of elementary particles in h-space theory).
The final section of this article describes a number of experimental tests that could be used to verify the h-space theory.
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Read the whole article
Download the ZIP file
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Gherardo, another funny joke is if you read ECAT backward in italian: TACE, in english: he shuts up. So, the italian phrase “ECAT? Rossi tace”, in english is “The ECAT? Rossi shuts up”. In fact, that’s what really happens.
Steven N. Karels wrote in December 21st, 2014 at 7:47 AM
Dear Wlad,
Note the negative energy got 62Ni and 64Ni.
If these are correct, then the following may be observed
1. The overall energy release is limited by the amount of hydrogen available in the eCat.
2. Aluminum plays a role in the energy generation as it is converted to silicon.
3. This would explain why the reaction stops at 62Ni.
4. It is not clear why The Report indicates 64Ni was consumed.
5. Why did iron not show a shift in its isotopes (or was it just not reported)?
6. Is the iron a magnetic catalyst for the other parts of the reactions?
Thoughts?
——————————————————
Dear Steven
In my opinion the best fuel for the eCat is composed by isotopes of Ni and Li.
But of course Andrea Rossi could not put in he reactor only Ni and Li isotopes, because it would be very easy to everybody to conclude what are the reactions in the reactor, after the publication of the Lugano Report.
In order to confuse everybody who is reading the Lugano Report, Andrea Rossi put some other elements in the fuel.
For instance, he put Lithium Aluminum Hydride, because LiAlH4 produce free hydrogen by heating, and by this way maybe Andrea had intended to suggest that hydrogen contributes for the reactions.
But perhaps hydrogen plays a role as a catalyst, in spite of I cannot see how.
In the last page of the Lugano Report is written:
“Sample 2 was the fuel used to charge the E-Cat. It’s in the form of a very fine powder. Besides the analyzed elements it has been found that the fuel also contains rather high concentrations of C, Ca, Cl, Fe, Mg, Mn and these are not found in the ash.”
So, we conclude that Fe, Ca, Mg, Cl, indeed had fusion with 3Li7.
However, Andrea Rossi already had tested those elements (Fe, Ca, Mg, etc.), and he concluded that, in spite of they are able to produce heat by cold fusion, nevertheless the COP is very lower than that obtained by using Ni.
So, the best fuel is Ni+Li. That’s why Andrea Rossi calls it nickel technology.
The other elements found in the fuel of the Lugano Report had been put in the reactor with the aim of confusing us.
.
However, the controversy about the reactions must be solved by a the following procedure:
1) Put the fuel Ni-Li in 12 eCats.
2) The first eCat will be turned off after 15 days, and the ash will be analysed.
3) The second eCat will be turned off after 30 days, and the ash will be analysed.
4) The last eCat will be turned off after 6 months, and the ash will be analysed.
By this way the controversy about the reactions can be eliminated, and we will be able to unveil some puzzles, as for instance whether the cascade reactions Ni-Li stops in the 62Ni, or whether the 62Ni isotope continues the cascade giving the reaction 62Ni + n -> 63Ni -> 63Cu.
regards
wlad
Dear Andrea,
Do you use the same prepared powder for both the low temperature Ecat and the high temperature hot cat devices.
Kind Regards,
Keith Thomson.
Gherardo:
Thank you and same wishes to you, extended, as usual, to all our Readers, from the Team I work in and me,
A.R.
Dott.Rossi
Merry Christmas and Happy New Year to you, your co-workers and forum members.
2014 wishes were fruitfull so this year I hope “may 2015 be the year of the E-Cat”.
Today I found funny that ECAT acronim is also used by Escambia County Area Transit near Pensacola (Florida).
All the best, Gherardo
Dr Joseph Fine:
Whereas I cannot comment in positive or negative issues related to the operation of the E-Cat, I totally share the quote from Sir Arthur C. Clarke.
I too wish you, and again to all our Readers, a healthy 2015, after a peaceful and Merry Christmas.
A.R.
Silvio Caggia:
Ideas are never dismissed, they always work.
Warm Regards,
A.R.
Dear Wlad,
These are the reactions I would expect in an eCat based on lithium acting on the various other isotopes present in the eCat.
Start Result Energy Released
58Ni + 7Li 59Ni+6Li 1.7492963
59Ni + 7Li 60Ni + 6Li 4.1377298
60Ni+7Li 61Ni+6Li 0.5702163
61Ni + 7Li 62Ni + 6Li 3.3465222
62Ni + 7Li 63Ni + 6Li -0.41204
64Ni + 7Li 65Ni + 6Li -1.151829
27Al + 7Li 28Al + 6Li 0.821346
28Al 28Si + e 4.2961227
6Li +1H 7Li 0.0069433
7Li + 1H 4He + 4He 0.0186231
54Fe + 7Li 55Fe + 6Li 2.0482114
55Fe + 7Li 56Fe+ 6Li 3.9473333
56Fe + 7Li 57Fe + 6Li 0.3961208
57Fe + 7Li 58Fe + 6Li 2.7946144
Note the negative energy got 62Ni and 64Ni.
If these are correct, then the following may be observed
1. The overall energy release is limited by the amount of hydrogen available in the eCat.
2. Aluminum plays a role in the energy generation as it is converted to silicon.
3. This would explain why the reaction stops at 62Ni.
4. It is not clear why The Report indicates 64Ni was consumed.
5. Why did iron not show a shift in its isotopes (or was it just not reported)?
6. Is the iron a magnetic catalyst for the other parts of the reactions?
Thoughts?
Joe,
concerning your opinion:
“2. I still think that the target nucleus would need a magnetic dipole moment that is greater than that of the source nucleus.”
you have also to consider that the two nuclei Ni and 3Li7 are very far away one each other (a distance at least two times longer than the distance between 3Li7 nucleus and the orbit 2s1).
As the magnetic force decreases with the square of the distance, even if the intensity of the magnetic moment of the 2s1 and the magnetic moment of the nucleus Ni was the same, however due to the larger disance the magnetic attraction force between proton and the Ni nucleus would be 4 times weaker than the magnetic attraction force between the proton and the orbit 2s1.
Besides,
there is yet the Coulomb attraction between the proton in the newborn 4Be7 and the electron 2s1
So, concerning the extraction of the proton from the 4Be7, the magnetic moment of the Ni nucleus has not any influence in the mechanism of the extraction.
regards
wlad
ERRATA:
In my comment of December 21st, 2014 at 4:46 AM
where it is written;
If this is the case, then the radius of the 2s1 orbit is the maximum radius so that to be possible for a nucleus to have reaction with 3Li7.
the correct is:
If this is the case, then the radius of the 2s1 orbit in the Ni nucleus is the maximum radius so that to be possible for a nucleus to have reaction with 3Li7.
Dear Andrea Rossi,
Did you dismiss the idea of onion-cat or this technology is embedded within the “dog-bone” hot-cat?
It really looks like a leek-cat (it.m.wikipedia.org/wiki/Allium_ampeloprasum)… 🙂
Joe,
one cold fusion researcher of the Martin Fleischmaan Memorial Project wrote in their page in the Facebook:
“- We now understand WHY we need the alumina tubes on the feed wires”
However,
I suspect that Andrea Rossi uses permanent magnets within those alumina tubes. This can be his secret.
Indeed, let us think about those alumina tubes, as follows:
1- There is no need them so much long
2- Why alminium ? Well, because aluminium is not magnetic. By using aluminium the it is easier to hide the secret (for instance, aliminium does not attract small iron pieces).
3- However aluminium has interesting properties when it interacts with permanent magnets:
http://terpconnect.umd.edu/~wbreslyn/magnets/is-aluminium-magnetic.html
By this way, when the electric power is turn off, and the magnetic pulses stop, the excited Ni isotopes continue being aligned toward the axis of the reactor, and this is the reason why the eCat works in self sustained model along some hours. When the 58Ni, 60Ni, 62Ni, 64Ni lose their excitation, the self sustained mode is ended.
regards
wlad
Joe,
actually there is one more step along the transmutation 59Co -> 62Ni:
59Co + n -> 60Co -> 60Ni ====> 60Ni + n -> 61Ni ===> 61Ni + n -> 62Ni
I dont know yet why the stable 59Co has reaction with 3Li7, while the stables 63Cu and 65Cu do not have reaction with 3Li7.
However there is a difference of two protons between 27Co and 29Cu. Then the radius of the 2s1 orbit in 29Cu is larger, and therefore the proton (when it exists the 3Li7) has too much acceleration, and so the neutron is not captured by 29Cu , in spite of the energy of the neutron is not enough to cause the fission of the 29Cu.
If this is the case, then the radius of the 2s1 orbit is the maximum radius so that to be possible for a nucleus to have reaction with 3Li7.
So, Ni is in the limit for a receptor to have fusion with the emitter 3Li7.
regards
wlad
Wladimir,
1. You write,
“Only 61Ni produces cold fusion, however there is only 1,14% of 61Ni in the fuel, and there is no way to have self sustained mode.”
So then what is the minimal proportion of 28Ni61 that would allow an SSM? And how do you calculate this proportion?
2. You write,
“The magnetic moment of the 2s1 electron is very stronger than the magnetic moment of the nucleus Ni.”
And that is the problem. The induced magnetic moment of the electron will not allow the weaker magnetic moment of the target nucleus to steal the approaching nucleon from it.
3. What do you think is the reason for the lack of gamma rays observed in the E-Cat?
All the best,
Joe
Andrea Rossi,
Is it somewhat valid to think of the energy releasing interactions as being glancing blows between two (or more) nuclei? That is, two (or more) nuclei interact at shallow angles such that the heavier nucleus strips off neutrons from the lighter nucleus.
This is more like atomic etching or removing of material.
Conceptually (?), this is some type of atomic-scale milling machine where a collection of heavier nuclei (with sufficient kinetic energy) become a machine tool, milling machine or battering ram. In this case, the material removed from the lighter atoms* – which are being battered – are neutrons. Most, if not all, of the removed neutrons are then absorbed (or fused) into the milling machine tool or the heavier nuclei.
(* Lighter atoms are “damaged” more by collisions with heavier atoms just as lighter cars are damaged more by any collisions with heavier trucks.)
This is not only miraculous, to quote Sir Arthur C. Clarke:
“Any sufficiently advanced technology is indistinguishable from magic.”
Best wishes for a Christmas season full of blessings.
And also, a happy, healthy New Year to you, your family, your team and your readers.
Joseph Fine
Joe wrote in December 20th, 2014 at 3:48 PM
Wladimir,
1. —————————————–
Why would the E-Cat not work full time in the SSM just because the fuel might not be 100% 28Ni61? It would produce less power but it should last just as long.
——————————————–
No, Joe,
when the electric power is turn off, after a time the isotopes 58Ni, 60Ni, 62Ni, 64Ni lose the excitement, and without excitement they have nuclear magnetic moment zero. Therefore they cannot be aligned toward the axis of the reactor, and they do not produce cold fusion.
Only 61Ni produces cold fusion, however there is only 1,14% of 61Ni in the fuel, and there is no way to have self sustained mode.
2. ——————————————
I still think that the target nucleus would need a magnetic dipole moment that is greater than that of the source nucleus.
———————————————-
What you think makes no sense.
The magnetic moment of the 2s1 electron is very stronger than the magnetic moment of the nucleus Ni.
The proton of the 3Li7 is extracted by the electron’s orbit, and not by the nuclear magnetic moment of the Ni nucleus.
The only task of the nuclear magnetic moment of the Ni is to put the z-axis of the Ni aligned toward the axis of the reactor
.
3. ———————————————
The most important question is, at what point in this whole process is excess energy being created?
————————————————
7Li -> 6Li
58Ni + n -> 59Ni -> 59Co
59Co + n -> 60Co -> 61Ni ==> 61Ni + n -> 62Ni
60Ni + n -> 61Ni
61Ni + n -> 62Ni
62Ni + n -> 63Cu
64Ni + n -> 65Ni -> 65Cu
Note that the isotopes 58Ni(68,1%) + 60Ni( 26,2%) +61 Ni(1,1%) have as a final result the isotope 62Ni.
So, while 62Ni is transmutting to 63Cu, however 95,4% of the total Ni isotopes are transmutting to 62Ni, and this is the reason why after the 32 days of the eCat working the ash had 98,7% of 62Ni.
If the eCat continues working , after some months the percentage of 62Ni will be near to zero.
Also note that the eCat worked only 32 days.
Andrea Rossi already had reported that in earlier experiments he had found Cu in the ash. This is because the eCat had worked along 6 months, and there was time available for the 62Ni to transmute to 63Cu.
The transmutation 64Ni -> 65Cu is not representative, since 64Ni has only 0,9% of the fuel.
In the Lugano Report is said in the page 28:
“Even if that particular reaction is excluded, since no gammas are observed, we can tentatively use this number for each step towards 62Ni, and the information from ICP-AES that there is about 0.55 gram Ni in the fuel. We find then that there is about 2.2MWh available from the Nickel transformations. Accordingly, from Nickel and Lithium together there is about 3 MWh available, which is twice the amount given away in the test run.”
However,
these numbers used are referred to hot fusion.
In the case of cold fusion, the reactions do not give the quantity of energy given in hot fusion.
So, instead of 2.2MWh available from Ni transformations, for cold fusion actuallthe value must be inferior than 2.2MWh.
regards
wlad
orsobubu wrote in December 20th, 2014 at 1:14 PM
Wlad, I would love you succeeded in explain the Rossi effect without having access to the complete experimental datasets, on the basis of theoretical deduction. I already know mrs. Pamela will reply this way:
“Hi, Wlad, thanks for the suggestions. We have several experiments in line waiting to be realized. When we have time to test your suggestion I’ll warn you. Hug.”
————————————————
dear orsobubu,
actually it not is the Pamela’s style.
In 2009 she sent a reply, concerning her experiment in the US Navy:
“Subject: RE: absence of gamma-rays in your experiment, and neutron’s background
Date: Mon, 13 Apr 2009 10:29:47 -0700
From: pam.boss@navy.mil
To: wladimirguglinski@hotmail.com
CC: m_bernstein@acs.org; hestenes@asu.edu; canmarrai@gmail.com
Like many, we have very few funds and resources. But we will consider your suggestions and see what we can do as time and money permits.
Regards,
Pam”
Wladimir,
1. Why would the E-Cat not work full time in the SSM just because the fuel might not be 100% 28Ni61? It would produce less power but it should last just as long.
2. I still think that the target nucleus would need a magnetic dipole moment that is greater than that of the source nucleus.
3. The most important question is, at what point in this whole process is excess energy being created?
All the best,
Joe
Thomas Florek:
Thank you for this delighting gift that, of course, goes with your and my best wishes for a wonderful Christmas to all our Readers
Wlad, I would love you succeeded in explain the Rossi effect without having access to the complete experimental datasets, on the basis of theoretical deduction. I already know mrs. Pamela will reply this way:
“Hi, Wlad, thanks for the suggestions. We have several experiments in line waiting to be realized. When we have time to test your suggestion I’ll warn you. Hug.”
Good, since you frequent this JONP place you’ve made lots more friends
Hi Andrea,
Here is our ragtime-flavored Holiday greeting.
The very best Christmas wishes for you:
http://www.youtube.com/watch?v=nKo7Rt027MQ
-thomas
ERRATA:
In my comment of December 20th, 2014 at 5:11 AM
where it is written:
As after the reaction 60Ni-3Li7 the 60Ni transmutes to 62Ni, the isotope 62Ni requires excitation, so that the eCat continue producing heat.
the correct is:
As after the reaction 61Ni-3Li7 the 61Ni transmutes to 62Ni, the isotope 62Ni requires excitation, so that the eCat continue producing heat.
Email sent to Pamela Mosier-Boss
From: wladimirguglinski@hotmail.com
To: pam.boss@navy.mil
Subject: Your version of Fleischmann-Pons experiment improved by using 105Pd
Date: Sat, 20 Dec 2014 10:51:55 -0200
Hi, Pamela
I have strong reasons in believing that the alignment of the nuclear magnetic moment of the Pd nuclei toward an external magnetic field is responsible for the cold fusion occurrence in the Fleischman-Pons experiment (and of couse also in your version of their experiment. performed by you in the US Navy in 2009).
Therefore only the stable isotope 105Pd contributes for the cold fusion occurrence in your experiment, since only 105Pd has non-null nuclear magnetic moment.
All the other isotopes 102Pd, 104Pd, 106Pd, 108Pd, 110Pd, have null nuclear magnetic moment, and so they do not contribute for the cold fusion occurrence.
I had proposed for the cold fusion researchers of the Martin Fleischmann Memorial Project to perform the following experiment, so that to test such hypothesis:
========================================================
THE ALIGNMENT OF NUCLEI AS INDISPENSABLE PRE REQUISITE FOR THE COLD FUSION OCCURRENCE CAN BE TESTED BY EXPERIMENT, as follows:
1- Fleischmann-Pons experiment will be performed in two different vessels A and B
2- In both vessels will be used an external source of magnetic field, so that to eliminate the influence of the magnetic fields of the Earth and the Sun
3 – Magnetic pulses cannot be used in any of the two vessels A and B, in order DO ONT EXCITE the nuclei in both experiments
4- The vessel A will be filled with the isotope 105Pd and deuterium. As the nucleus 105Pd has nuclear magnetic moment, the 105 Pd nuclei will be aligned toward the vector magnetic field, and the cold fusion must occur .
5- The vessel B will be filled with the isotopes 102Pd, 104Pd, 106Pd, 108Pd, 110Pd, and deuterium. As all those nuclei have null magnetic moment, and as they will not be excited (because there is not magnetic pulses applied), they cannot be aligned toward the vector magnetic field, and therefore COUD FUSION CANNOT OCCUR.
========================================================
However, the researchers of the MFMP responded the following to me:
“Given that Pure Pd is over $5833 100g, where do you propose one gets affordable 105Pd?
If you are able to provide the raw materials, we may be able to conduct an experiment.”
So, they have no money for performing the experiment.
Then I would like to know:
May you be interested to make it in the laboratory of the US Navy?
Can you make it?
If I am right, the heat produced by using 100% of 105Pd must be 5 times greater than in the experiment made by you in 2009.
regards
W Guglinski
Wladimir Guglinski:
As you know, I cannot give information, in positive or in negative, related to the fuel isue, more than I already did.
Warm Regards,
A.R.
John Atkinson:
Than you for your kind attention.
We have specialists in our Team who deal with this problem.
Warm Regards,
A.R.
eCat full time in self sustained model
Joe,
I suspect that by using a fuel composed by 100% of 61Ni the eCat can work full time in the self-sustained-mode.
The reason why the reactor filled with fuel composed by 58Ni, 60Ni, 61Ni, 62Ni, 64Ni, after some hours stops to work in the self-sustained-mode is because after some time the excited isotopes 58Ni, 60Ni, 62Ni, 64Ni begin to lose their excitation, and so the reactions Ni-3Li7 stop.
Then there is need to apply magnetic pulses again in the coils of the reactor, in order to excite the isotopes 58Ni, 60Ni, 62Ni, 64Ni, and that’s why there is need to turn on again the electric power, supplying electric current to the coils of the reactor.
However, the eCat filled with 100% of the isotope 61Ni will work in the self sustained model only while there is yet 61Ni isotopes available within the reactor.
As after the reaction 60Ni-3Li7 the 60Ni transmutes to 62Ni, the isotope 62Ni requires excitation, so that the eCat continue producing heat.
So, when the fuel composed by 60Ni is totally converted to 62Ni, there is need to apply again magnetic pulses, in order to excite the 62Ni.
regards
wlad
As time passes the e cat gets closer and closer to a world marketed alternative energy source. There will come a time if it has not already accrued,when big oil will threaten or may even sabotage your efforts.The middle East may also try , as illustrated by North Korea’s treats and computer hacks towards Sony . Since a great deal is dependent on computer operation of the e cats when working in unison, have you and Industrial Heat made plans or possibly consulted with the US for such a contingency? It seems at this point it may be the only great treat to your tremendous success and saving this world from itself..
Andrea Rossi wrote in December 18th, 2014 at 11:41 AM
Frank Acland:
It is an interesting idea: under a theoretical and technological point of view, I do not see why not. The issue is in the price: if the electric energy supplied by batteries will be competitive with other sources, the coupling between high efficiency batteries and the E-Cats will be surely possible. Have you an idea of the cost per kWh supplied by this new generation of batteries? I am curious.
———————————————-
I think the eCat spends so much electric power because there is need to excite 98% of the Ni isotopes used as fuel in the eCat.
The natural abundance of the stable Ni isotopes is the following:
58Ni = 68,08%
60Ni = 26,22%
61Ni = 1,14%
62Ni = 3,63%
64Ni = 0,93%
Only 61Ni has magnetic moment, and so it does not require excitation so that to be aligned by magnetic field along the azis of the reactor, in order to have fusion with 3Li7.
All the other Ni isotopes have null magnetic moment, and all they require excitation so that to be aligned, in order to have fusion with 3Li7.
By using a fuel composed by 100% of 61Ni the electric power could be reduced drastically, because there is no need to excite the 61Ni (the electric power for the excitation of the 98% of the other Ni isotopes would be saved).
However, I do not know how much extra cost would be necessary to obtain a fuel composed of 100% of 61Ni.
regards
wlad
Frank Acland:
Very interesting, thank you for the information. I am learning.
Warm Regards,
A.R.
Dear Andrea,
There is a great deal of research and development and investment going into battery technology worldwide. I think the trend will be for lower cost per kWh as time goes by.
I am sure electric vehicles are going to become more and more popular as time goes by. Lack of charging stations for EVs are one obstacle to their widespread use. You should talk with Elon Musk of Tesla. I think together you might be able to work towards a solution!
Kind regards,
Frank Aclan
Frank Acland:
Thank you for this interesting information. As you can se, this is a price good for usual batteries utilization, but very high for us. Nevertheless, the use of batteries as storage of energy is very interesting. By the way: I tested a Tesla, is very funny: same acceleration of a Ferrari, but too short autonomy if you want to get driving fun.
Warm Regards,
Andrea
Dear Andrea,
You asked about the cost per kWh supplied by the new generation of batteries. I have to thank an E-Cat World reader for finding this. According to the Tesla Motors Web site, they can currently deliver electricity from their lithium-ion batteries at 0.21 Euros per kWh.
http://www.teslamotors.com/en_CA/goelectric#savings
Kind regards,
Frank Acland
D. Travchenko:
Again on the fuels: today this issue has got fuel!
I do not know what will happen to gas prices and I don’t control them, I only control, together with my wonderful Team, the products for our Customers.
Prices of fuel will change and the needs of the market will change, this is why we have an excellent business Team to watch the market and ensure we are meeting the Customer’s needs.
Warm Regards,
A.R.
IC Renoir:
I am not able to provide any information at this time and any information will be shared publicly when appropriate.
Warm Regards,
A.R.
Curiosone:
As this is a hypothetical question, my only response would be to guess, which is not something I wish to be on the record about.
Warm Regards,
A.R.
Experiment proposed for the researchers of the Martin Fleischman Memorial Project
https://www.facebook.com/MartinFleischmannMemorialProject/posts/886318188065548?comment_id=886327334731300&offset=0&total_comments=8¬if_t=feed_comment
But it seems they have no money so that to perform the experiment.
========================================================
THE ALIGNMENT OF NUCLEI AS INDISPENSABLE PRE REQUISITE FOR THE COLD FUSION OCCURRENCE CAN BE TESTED BY EXPERIMENT, as follows:
1- Fleischmann-Pons experiment will be performed in two different vessels A and B
2- In both vessels will be used an external source of magnetic field, so that to eliminate the influence of the magnetic fields of the Earth and the Sun
3 – Magnetic pulses cannot be used in any of the two vessels A and B, in order DO ONT EXCITE the nuclei in both experiments
4- The vessel A will be filled with the isotope 105Pd and deuterium. As the nucleus 105Pd has nuclear magnetic moment, the 105 Pd nuclei will be aligned toward the vector magnetic field, and the cold fusion must occur .
5- The vessel B will be filled with the isotopes 102Pd, 104Pd, 106Pd, 108Pd, 110Pd, and deuterium. As all those nuclei have null magnetic moment, and as they will not be excited (because there is not magnetic pulses applied), they cannot be aligned toward the vector magnetic field, and therefore COUD FUSION CANNOT OCCUR.
========================================================
.
REPLY BY MFMP:
Martin Fleischmann Memorial Project : Given that Pure Pd is over $5833 100g, where do you propose one gets affordable 105Pd?
Wlad Guglinski dont you think to be important in order to decypher the cold fusion mystery?
Wlad Guglinski The 105Pd natural abundance is 22, 23%. In the case my theory is correct, and MFMP decides to perform the experiment, there is need to take care, because in this version of the Fleischmann-Pons experiment the energy created by cold fusion will be 5 times greater
.
Comment by me here in JoNP:
So, anyone who is trying to explain cold fusion thinks that there is need only of proposing the nuclear reactions, and the energy produced in each of the reactions.
Nobody wishes to realize that there is need to understand the cold fusion mechanisms occurring in the phenomenon.
It is like to want to understand how the planets are attracted by the Sun, without to know the existence of the gravitation.
Let me put a hypothetical question: if the oil price will go down enough to make the E-Cat not convenient, what do you think will happen to your and Industrial Heat’s enterprise?
WG
Dr Andrea Rossi:
Gas prices are falling: how are you aware of the drop of the fuels respect the market perspectives for the E-Cat?
Warm Regards,
DT
Dear Dr Rossi,
We all would like to have more information about the 1 MW plant dlivered to the customer months ago!
Godspeed,
JCRenoir
Stefano Landi:
As you correctly write, I talked months ago of this issue and I have nothing to add to this matter, in general.
Warm Regards,
A.R.
Brandon Hurd:
I am not involved in this kind of issues, but I am sure that Industrial Heat has given all the due information where it is opportune and proper.
Warm Regards,
A.R.
Dear Andrea Rossi
Has Industrial Heat informed President Obama about the E-Cat? If so, did you get a formal response?
If not, don’t you believe the U.S. government will want to be involved with the technology when it is ready for mass roll-out in the market?
Regards
Brandon Hurd
Dear Andrea. Please give us an update on the e-cat / e-mouse system you talked months ago.
Kind regards
Stefano
Steven N Karels:
Yes, I too think that batteries could be a storage .
Warm Regards,
A.R.
Steven N. Karels:
All this issue is intriguing, even if not in the immediate future.
Warm Regards,
A.R.
Alessandro,
I recently purchased a Chevy Volt. It requires 12 Amps of 115VAC input power for 10 hours to completely charge the car’s batteries and this results in a range of about 30 miles. So a likewise continuous charge rate might yield a daily range of about 70 miles. This charging rate is equivalent to an electric output rate of 1.3 kW. Assuming typical Carnot efficiencies means a thermal generation of around 4kW. With a COP of 3, then I would guess a 10kW eCat could provide enough power for one vehicle with a daily range of 70 miles. A substantially larger eCat would be required for continuous long distance traveling. But this says the numbers are close. Getting rid of excess heat might be a problem but I suspect could be engineered to handle it.
Dear Andrea Rossi,
I think it is more than just energy cost economics. The use of batteries to store energy is more applicable to temporal energy sources such as photovoltaic or wind sources – batteries would store electrical energy during the time when the energy source is not available. Not so with the eCat which, by its nature, can run continuously for months at a time. The eCat is best suited for Baseline (continuous) electrical generation. The only way I see electricity could be used to provide the input energy for eCat operation (besides control) is if the cost of electricity becomes so low as to be much cheaper than natural gas.
Alessandro Coppi:
Interesting.
Warm Regards,
A.R.
To complete the idea of Paul, when the car is parking, it will be connected to the grid and upload energy.
Paul:
Thank you for your idea.
Warm Regards,
A.R.