zondag 6 mei 2012

Koude Fusie / E-Cat (Deel VI) :

Mijn berichten op het forum van
A. Rossi - En stand van zaken.





Zoals steeds eerst een verwijzing naar mijn vorige berichten over het onderwerp:

Alice Bailey en Nieuwe Energietechnologieën (Daniel De Caluwé)

Cold Fusion will soon be hot again (Daniel De Caluwé)

Cold Fusion/Koude fusie - Update/Vervolg (Daniel De Caluwé)

Koude Fusie / E-Cat (Deel III) : Meer informatie over de 1MW 'E-Cat'-plant van Andrea Rossi (Daniel De Caluwé)

Koude Fusie / E-Cat (Deel IV) : De test van 6 oktober. (Daniel De Caluwé)

Cold Fusion (part V) : Dr. George Miley Replicates Patterson. (Daniel De Caluwé)


En vanaf hier mijn bijdagen in The Journal of Nuclear Physics (het forum van A. Rossi) :


http://www.journal-of-nuclear-physics.com/?p=510&cpage=11#comment-73432

Daniel De Caluwé
September 16th, 2011 at 4:01 PM

Dear Mr. Rossi,

Although Mr. Krivit was very sceptical, I found the video’s of your interview very interesting. On it, you made a very good demonstration and explanation of the small E-cats, and I agree with the calculations you did (concerning the energy-balance). I also did the calculations and found that the small E-cat produced about 98.4 kWh (= 354240 kJ = 4.1 kW during 24h) per gram H2 (the amount the small E-cat consumes in 1 day), and this is 2498.2 times the energy we get if we just would burn the H2 (Ho of H2 = 141.8 kJ per gram H2). And for your information: Also Dr. Edmund Storms did a good review of this.

So, although Mr. Krivit was sceptical, for me, your demonstration and explanation of the small E-cat was very convincing.

So I wish you a lot of succes with the bigger ones and with the 1 MW plant.

Kind Regards,

Ir. Daniel De Caluwé, Belgium.


http://www.journal-of-nuclear-physics.com/?p=510&cpage=11#comment-73462

Andrea Rossi
September 16th, 2011 at 6:03 PM

Dear Daniel De Caluwè:
Thank you for your kind encouragement,
Warm Regards,
A.R.


http://www.journal-of-nuclear-physics.com/?p=510&cpage=31#comment-114307

Daniel De Caluwé
November 9th, 2011 at 7:16 AM

Dear Mr. Rossi,

1. Congratulations with the succesfull test of 28 october.

2. Reading your blog, some people already made some suggestions concerning generating electricity with steam at low temperature and low pressure. But I wonder if there’s still engineering space to increase the operating temperature of the E-cat, so that it generates steam at higher temperature and higher pressure, which makes it easier to generate electricity afterwards? Or do you operate the E-cats already at their maximum temperature (i.e. melting point of the powder, … and so on), so that there’s no engineering space anymore to increase it?

3. There appears to be a problem with 2 previous messages I posted on this blog. But when I posted the first of these, at that time there was a virus or a problem with the blog, and the second I posted when you had no time (during the 28 october test), or maybe there’s a problem with the link (to my blog) I included? Anyhow, via this message, I also want to assure you that I’m not a competitor, who wants to steal the secrets of the E-cat, and that I wish you all succes in your work and with the further development of the 1 MW plant!

Kind Regards,

Daniel De Caluwé


http://www.journal-of-nuclear-physics.com/?p=510&cpage=31#comment-114613

Andrea Rossi
November 9th, 2011 at 2:54 PM

Daniel De Caluwé:
1- Yes, we are working on this, within 2 years we will produce also electricity
2- Unfortunately, our robot spams many good comments: it should only spam comments with viruses, advertising, but many times it wrongly spams good comments. I can do nothing: it spams about 1000 comments per day, in which are dispersed 2 or three good ones… Just check that you have no viruses, or links that can be taken for ads.
Warm Regards,
A.R.


http://www.journal-of-nuclear-physics.com/?p=510&cpage=32#comment-117817

Daniel De Caluwé
November 13th, 2011 at 6:22 AM

@Mateo,

1. The energy consumption of the (peristaltic) feed water pumps (from the water reservoirs up to the container-plant) must be taken into account of the 1 MW container plant.

2. The energy consumption of the resistors, necessary to start the reaction (but not necessary during self-sustaining mode) also must be taken into account of the 1 MW plant.

3. But the energy consumption of the electrically driven air-ventilators to cool and condense the steam (during the test done with air to steam heat exchangers) may NOT be taken into account of the 1 MW plant, because in normal operation, the steam is used by the client, and does not have to be cooled in that way. (The steam is used as process heat and/or to generate electricity (later)).

4. The (eventually) energy consumption of the (eventually) condenser-pumps, that bring back the water (condensed steam) back to the water reservoirs, eventually must be taken into account of the 1 MW plant, because that also is not useful energy for the client.


http://www.journal-of-nuclear-physics.com/?p=510&cpage=48#comment-189400

Daniel De Caluwé
February 24th, 2012 at 5:34 AM

Dear Mr. Rossi,

I still follow this blog from time to time, and I still wish you a lot of succes, but reffering to the recent questions of Valeria and Piero on this blog, I remember I made similar remarks on my blog in october 2011:

My personal and first impression was that the E-cat was more suitable for industrial applications (1MW plants or more) and not so much as a small home unit. I summoned several reasons for this:

i) That there still is the need of feeding the e-cat from the grid with an energy consumption of 1/6 of the total thermal energy the e-cat produces. (This remark is the same as in the recent questions of Valeria and Piero on this blog.)

ii) the E-cat, as it operates now, cannot easily be switched on and off, because it takes (relatively much) time to start-up and also generates a lot of residual heat after it is switched off. So doesn’t that mean that the E-cat, like it operates now, is more effective when it’s used in continuous mode, generating energy to processes in a continuous way, so that you don’t loose time and energy when switching on and off all the time.

I then concluded that the E-cat was more fit for industrial applications (bigger plants, like the 1MW plant) and/or as common heating for whole appartment blocks, where it has to deliver the energy to many users, so that it operates in a more contineous mode/way.

Do you have a solution for this potential problem of (possible) energy-loss in single home-applications? (A 10 kw E-cat consumes 1.667 kW electrical energy from the grid)

(Nevertheless, a boiler can be heated for some hours at night, and central heating usually works contineously, but I’m one of these guys that just uses a gas-heater at small pace, with low energy-consumption, and compare this with a 10 kW E-cat that still consumes 1.667 kW electrical energy continuoesly, witch for me is too much).


http://www.journal-of-nuclear-physics.com/?p=580&cpage=4#comment-189505

Andrea Rossi
February 24th, 2012 at 10:38 AM

Dear Daniel De Caluwè:
Thank you for the suggestion, maybe you are right.
Warm Regards,
A.R.


http://www.journal-of-nuclear-physics.com/?p=580&cpage=4#comment-189413

Daniel De Caluwé
February 24th, 2012 at 6:25 AM

@C.Goldmann,

Could the problem (with the messages that are wrongly seen as spam), be caused by filling in the ‘Website’-space? Because I also have problems posting messages to this blog, but only when I fill in the space referred to as ‘Website’. But when I don’t refer to a website, my messages are accepted by this forum (P.S. Further information: I use ubuntu-linux)


http://www.journal-of-nuclear-physics.com/?p=580&cpage=5#comment-191279

Daniel De Caluwé
February 28th, 2012 at 6:10 AM

Dear Mr. Rossi,

On february 25th, 2012 at 12:33 PM, user ‘Rends’ mentioned a critical article in Forbes magazine:

http://www.forbes.com/sites/markgibbs/2012/02/24/dick-smith-rossi-e-cat-too-fantastic-to-be-true/

Well, this is what I quickly wrote in response to it, one day ago (You can read my answer also on page 5 of the comment-section, below the article in higher link):

“”"Daniel De Caluwé 1 day ago

1. (I think starting from january 2011) Untill 28 oktober of last year (2011), several tests has been done on the E-cats, and as a civil electromechanical engineer (from Belgium), I verified some of the previous tests, and they were very convincing to me.

2. On his blog http://www.journal-of-nuclear-physics.com/ , Andrea Rossi explained several times why he doesn’t organise any more tests:

- Several tests already have been done, with good result(s), and the latest test (of 28 oktober 2011) was done by an important customer, a company or organisation that doesn’t want to reveal its name (Mr. Rossi has a NDA-agreement with this entity), but who approved the 1MW container plant, and who buyed it. And together with this company or organisation, they improved the control system and the instrumentation, and they are also working on the issue of generating electricity with the 1 MW plant. Also, based on these tests and his coöperation with this major organisation, Mr. Rossi got a lot of investors, that support his company Leonardo Corporation.

- His competitors are very interested in his secrets, especially in the right chemical formula of the catalyser, that is used to enhance the reaction. And because pattent-approvals are still pending, Mr. Rossi doesn’t want to take the risk of doing more tests, that could reveal the secrets to his competitors.

- With his team and the companies that are choosen by his investors, he ‘s working 16 hours a day (7 day’s a week) to also make electricity (starting with the industrial 1 MW plant), and also on an automated production line for the home-ecats, that could be delivered within 18 months (worldwide), hopefully before next winter (in the northern hemisphere), but planned within 18 months. So, Mr. Rossi and his team and co-operators, have no time to do further tests. International approval of his patent(s) still is pending, so he still is not protected by it, and therefore het puts all his efforts in being the first to deliver good working home-ecats, not revealing the secrets of the E-cat to possible competitors. But as soon as pattents are approved, Mr. Rossi wrote he could give more information about the exact working of the E-cats…

Ir. Daniel De Caluwé
Belgium”"”

I quickly wrote that message (last sunday) in response (and to your defence) to the critical article in Forbes magazine.

As I believe 100% in what you do, I hope you roughly agree with above answer, and that I didn’t make too much errors?

Kind Regards,

Ir. Daniel De Caluwé
Belgium


http://www.journal-of-nuclear-physics.com/?p=580&cpage=5#comment-191629

Andrea Rossi
February 28th, 2012 at 10:37 PM

Dear Ing. Daniel De Caluwe’:
Thank you.
Warm regards,
A.R.


http://www.journal-of-nuclear-physics.com/?p=580&cpage=5#comment-191935

Daniel De Caluwé
February 29th, 2012 at 12:19 PM

Dear Mr. Rossi,

In some of your answers, you wrote that companies that want to distribute and sell e-cats, have to mail you and give a detailed description of their company. You also wrote that licenses will be given for limited geografical area’s, so here my questions:

i) Is Europe covered by one or only a few licenses, or do you give seperate licenses for every country?

ii) As I live in Belgium, how about Belgium? Do you already co-operate with a company in Belgium that will distribute the home-E-cats?

iii) If yess, could you disclose the name of that company (in Belgium)? Whenn will the list of your licensees (companies worldwide that represent you) be given?

iv) Concerning the industrial 1MW plants (and more), are these treated seperately from the home-units? (I think yess) And will industrial clients always have to contact your headquarters (Leonardo Corporation)? (I think yess)

v) Do you already co-operate with Belgian companies/clients (for industrial plants)? Of course I respect if you want to keep this secret…

vi) In autumn last year (2011), I thought ‘maybe I could ask him to represent him in Belgium’, but I am a single person, and I don’t own a distribution company, so I guess this is not possible, also because you probably work via existing and very well established distribution companies? (And don’t worry, I appreciate that, because it proves that you work very professionaly…)

Kind Regards,


http://www.journal-of-nuclear-physics.com/?p=580&cpage=6#comment-192230

Andrea Rossi
February 29th, 2012 at 8:35 PM

Dear Daniel De Caluwe’:
1- separate licenses have been given
2- Benelux has been licensed
3- When we will start the commercial strategy, the entire network of the licensees will be published, in occasion of a convention
4-Industrial plants will be treated separatedly, yes the contact will be with Leonardo Corp
5- Of course you will be entitled to ask a sub license, but only the Licensee has the right to decide if and to whom give a sub-license
Warm Regards,
A.R.


http://www.journal-of-nuclear-physics.com/?p=580&cpage=7#comment-194916

Daniel De Caluwé
March 6th, 2012 at 12:43 PM

@ Dr. Rossi,

1) You wrote: ‘Dear Gustavo:
Yes, but…how can I drive a 15 MW turbine with a 2.5 MW plant?’

I think Gustavo meant that with a 15MWth(ermal) E-cat-plant, in combination with a 15MWth Siemens turbine, and with an efficiency of 30%, you could produce 4.5MWel(ectric), from which you use 2.5MWel (= 1/6 of 15MWth) to feed your 15MWth E-cat plant. And this means that such a 15MWth E-cat plant (build with 1500 modules of 10kWth each), finally produces 2MWel netto, without the need to feed it with 2.5MWel from the grid, because this is done by the 15MWth E-cat-plant itself (via the Siemens turbine).

In this way, you could produce 2MWel netto, with a 15MWth E-cat plant, without needing electrical energy from the grid.

2) Although, in this way, a COP of 6 seems to be sufficient, I personally think that later, in a version 2.0, you better change the electrical heating (to maintain the reaction) by a thermal heating, heating the reaction chamber (with the Ni-powder inside it), with a fluid (water at higher pressure or steam or another fluid), that flows around the reaction chamber in a double-wall. You could do this by using a double-walled reaction chamber, where the Ni-powder is in the inside, and the heating fluid on the outside (flowing around the reaction chamber, between the two walls of the double-wall).

The advantage then is that you could use part of the thermal energy (that you produce with the E-cat), to sustain the reaction, and this without using electrical energy. So there probably are a lot of engineering possibilities to improve the efficiency of the E-cats?


http://www.journal-of-nuclear-physics.com/?p=580&cpage=7#comment-195104

Andrea Rossi
March 6th, 2012 at 8:43 PM

Dear Daniel De Caluwe’:
We are working in that direction too.
Warm Regards,
A.R.


http://www.journal-of-nuclear-physics.com/?p=580&cpage=7#comment-195494

Gustavo
March 7th, 2012 at 12:34 PM

and … can anybody guess how much is the COP of a 15MWth(termal) E-cat-plant, in combination with a 15MWth Siemens turbine, and with an efficiency of 30% (self-sustained)??

2.5 Mw Output (electrical) / 0 Mw Input (electrical) = INFINITE electrical perfomance!

Thank you Mr Rossi.

PD.: thanks Daniel De Caluwé for your help explaining it


http://www.journal-of-nuclear-physics.com/?p=580&cpage=9#comment-197470

Daniel De Caluwé
March 11th, 2012 at 9:35 AM

@tj,

i) You wrote: ‘The $30 million cost of a 45MW Rossi plant financed @ 4% for the 30 year life of the plant would equal $51,560,850 in capital costs.’

My answer:

I wonder how you get the amount of $51.56 million?

- Because if you only use the simple (intrest-)formula (4% /yr) * 30 years = 1.2; and then $30 million becomes $36 million and not $51.56 million

- But if you use the more complex intrest-formula, the $30 million becomes much more than $51.56:

Because (1.04)**30 = 3.24; and than $30 million becomes $97.3 million, which is much more than $51.56 million

So could you explain what formula you use and how you get these $51.56 million?

ii) Anyhow, if we work with $97.3 million, we still get $3.24 million capital costs each year for the 45 MW heat & power plant; or $0.05 (= 5 cents) per kWhe of electricity, which still is very good! ;-)

iii) Joseph Fine calculates with $30 million (E-cat only) + $15 million (for the Siemens turbine), and then we have $45 million as total investment cost, so in that case the numbers become higher.

Kind Regards,


http://www.journal-of-nuclear-physics.com/?p=580&cpage=9#comment-197804

tj
March 12th, 2012 at 2:01 AM

To — Daniel De Caluwé

The 51.5 million capital cost figure is based on borrowing 30 million @ 4%, and then paying it back over 30 years in 360 equal payments, just like a 30 year mortgage. A.R. said the e-cat is expected to have at least a 30 year life span.

When A.R. estimated a 30 million price tag for a self sustaining 45 MW electricity producing e-cat, it seemed implicit that the price included the turbine and generator.

Personally I think super critical CO2 turbine systems would offer important advantages of 40–50% efficiency, mechanical simplicity, and compact size.

Ideally the turbine coolant temperature would be engineered to be hot enough for use in a combined heating, hot water, and adsorption AC system. Large buildings are perfect candidates for combined systems.


http://www.journal-of-nuclear-physics.com/?p=462&cpage=1#comment-199311

Daniel De Caluwé
March 14th, 2012 at 7:06 PM

@ Wladimir Guglinski,

Although I have a university degree (UG, Belgium) in electro-mechanical engineering (with specialisation in energy-technology), I did not have QM-theory in my curriculum (only ‘engineers in physics’ got QM-theory in their curriculum), but I was theoretically very strong and had good results.

So, this evening, I quickly went trough this interesting introduction (as a start reading the theoretical articles on this blog ;-) , and I have following questions:

1. First see what you wrote below Fig2: (point 2)

‘Looking at the figure 2, we see that the electron is submitted to a centripetal acceleration due to its helical trajectory. Such centripetal acceleration does not play any role in the resonance between the electron and the Dirac strings (of the proton). But due to a coincidence, the radius of the electron’s helical trajectory is the same radius considered by Bohr in his calculus. Other coincidence is the fact that the force on the electron due to its helical trajectory is equal the force of attraction proton-electron, considered by Bohr. The radius of Bohr is also equal the radius of the electron’s orbit in the helical trajectory.
Thanks to this series of coincidences, the Bohr theory is able to yield those fantastic successes, in spite of his model is wrong.’

So you wrote:

Other coincidence is the fact that the force on the electron due to its helical trajectory is equal the force of attraction proton-electron

And here my question: I would expect that ‘the force of the attraction by the proton on the electron’, which is mainly due to the opposing electrical loads (Coulomb), is much stronger than ‘the force on the electron due to its helical trajectory’ (because the mass of the electron is very low), so are you sure that the above statement is right?

2. I wonder what would be the result, if we just would apply the macroscopic (and ordinary and wellknown) laws of Coulomb (the force of attraction between opposite electrical loads is proportional to the product of the electrical loads, and reverse proportional with the square of the distance) and of gravity (the same but using the masses of the particles), and this applied to the simple model of a hydrogen atom (1 proton and 1 electron)? (Electrical force being stronger in this case, because the mass of the electon is very low). Certainly, because the two forces increase at smaller distances, the speed of the electron would increase on lower shells, making the very small mass of the electron becoming important at very low distance… ;-)

Anyhow, in the past I already read part of your responses, and I found them very interesting!

Kind Regards,


http://www.journal-of-nuclear-physics.com/?p=614&cpage=2#comment-221104

Daniel De Caluwé
April 20th, 2012 at 11:16 AM

Dear Ioannis,

Although (and up to now) I have no experience in QM-theory, I quickly went trough the latest version of the pdf.file (SEPPv3.pdf), and I have following remarks:

i) On page 17 (of the pdf-file) you calculated the age of the universe (‘Tage’ = the ‘universe expansion time’) and its acceleration/deceleration (‘au(r)’), assuming that the expansion takes place at the speed of light (in the calculation of ‘Tage’), and that it happens with a constant acceleration (it looks more like you calculated the mean acceleration instead of the probably time-variabele acceleration.) [At the present level of scientific understanding, and with the knowledge we got from the measurements with the Hubble-telescope, I thought the universe is estimated to be about 15 billion years old, with, in the beginning (the first 5/6/7 billion years) at a decelerating pace (due to gravity), and after that (and upto now), with an ever increasing acceleration. So according to present science, the acceleration of the universe is not happening at a constant pace, but with a time-variable acceleration au(t)].

ii) Also on page 17, I think you made a calculation-mistake:

You wrote d(iU) = 1.4093 * 10^−15 m eq.(41), but as d(iU) = 2*re, and as re = 2.8179*10^-15 m, I thought d(iU) = 5.6358*10^-15, isn’t it?

iii) On page 38, where you discuss the very intresting Experiment #3 : (from the second line in the column to the right) you calculate TA , but am I right that you wrongly multiplied with a factor 3? [TA = (6.28*1*10^-2)/(3.484229*10^5) and NOT (6.28*3*10^-2)/(3.484229*10^5)]?

iv) But besides these details, I found your vision about neutrino’s and (the properties of) the Aether very intresting!

Kind Regards,

Ir. Daniel De Caluwé, Belgium


http://www.journal-of-nuclear-physics.com/?p=614&cpage=2#comment-221177

Ioannis
April 20th, 2012 at 3:16 PM

Dear Daniel,

Thank you very much for commenting my work. Additionally I congratulate you for this since nobody until today has ever commented my work. Although I sent my work to hundreds of Universities and Research Institutes through individual e-mailing, never have a feedback (maybe is a little bit early) and I thought that my work will be unnoticed and disintenteresting for many reasons.

I am an Electronic Engineer and I am also not very aware for most of Quantum Mechanics in mathematical perspective due to my profession. This work is based on an original idea as also it is the result of reading hundreds of patents and Scientific Papers, combined with much intuition, knowledge on different fields, comparing and partially experimenting where it was possible, the last 12 years.

Now about your comments:

a)You are right that the acceleration of the Universe must be time depended but as you may notice in my work, I calculated the current acceleration using the todays value of Gravitational constant. Even after 100 or 1000 years the acceleration will be practically almost the same, if you calculate the Gravitational constant for a future radius using the Graphic and formula of page 50. The result of deceleration and not acceleration (opposing to what is today believed) comes from the Universe Force on page 49 which becomes negative and indicates a deceleration. As you will see on page 50 this value of deceleration is equal to the calculated acceleration on page 17. Again the Universe Force (which is another form of the Unified Field Force) is the indication that the Universe decelerates. The discovered Hubble constant comes from Astrophysical observations which are not 100% objective, limited due to technological limitations of the measuring equipment.

b)This probably needs to be discussed since it is not clear in my work. The problem that arises is how we measure the distance. Since the entire work is based on the findings of the Electron-Positron pair, then about the distance we must prove (it is not very clear for me too) that Anti-matter was captured by the Aether which means it is blocked at the center of the Universe and not at his border (which does not make sense). As the todays radius of the Universe is calculated in my work, it points to the capture of Anti-matter at the momemt of Big Bang by the Aether. Now the initial Universe diameter like it is considered is probably not correct. I cannot say for sure since most theories todays claim that the diameter of the Universe is on the same value.

c)Yes you are right about it. It must be 1. I am sorry about it.

Probably there are some other minor mistakes in my work and this is due to the pressure to finish in time as also the volume of information is huge which makes most probable to make someone mistakes. Just keep in mind that on this work most of todays Quantum Physics and Cosmology is questioned. But this does not mean that I am 100% right. The reason for the challenge is to receive some feedback where it will help me to see where I stand. Again the simplest test is the Casimir Force. If it is correct then many things of my work are probable on the right way.

Another interesting semi-indication of my calculations is the Quantum Length (6.7438E-58m)where the link on my paper gives some clues about it:
http://www.sciencedaily.com/releases/2011/06/110630111540.htm

The Integral Project claims as you may read on the above link that the Quantization of space is below the 1E-48m which makes almost all Quantum Gravity theories invalid since they use the known Planck Length (1.616E-35m). You can find on my work how the Planck Units are disproved mathematically.

Some other indication is the Temperature of the Universe calculated at 2.768 Kelvin, which is very near to the 2.715 Kelvin as it is today measured.

You mention something about the assumption of Universe expansion with the speed of light, which I find your sceptisism correct. Well if you ever read the 1st version I used another expansion velocity that of matter. The correct conclusion is although I had planned to write about it, I did not (no time): Universe expands with the speed of light due to the photons of the big bang. Secondly expands with a speed 0.79c (if I remember correctly) if we accept that Anti-matter was captured and matter was that who expanded in space. So we have two horizons: One due to the photons and the second due to the matter expansion which can be calculated precisely through the Electron-Positron pair findings.

About the Aether, if you are interested to detect it(I would be very glad if someone send some feedback about it), just download some arbitrary Earth’s Magnetic Field data from the link provided on my web site and analyse them with the same way I did, using your EXCEL. You will not need a set up of an experiment to do this. Actually Experiment #3 has the same principle like Experiment #1. Of course it would be very interesting if someone could observe the frequencies on Experiment #3.

I am open to discuss everything related to my work!

Best Wishes and Thank you

Ioannis Xydous

Web Site:http://www.ioannisxydous.gr/


http://www.journal-of-nuclear-physics.com/?p=619&cpage=1#comment-225684

Daniel De Caluwé
April 30th, 2012 at 7:11 AM

Dear Wladimir,

I’ve read your article ‘How zitterbewegung contributes for cold fusion in Pamela Mosier-Boss experiment’ (see: http://peswiki.com/index.php/Article:_How_zitterbewegung_contributes_for_cold_fusion_in_Pamela_Mosier-Boss_experiment )

And like all your articles (and answers) about QRT, your explanation is very clear, and proves you have a profound knowledge of the field, and also a very interesting vision/theory to offer (QRT), that as well explains new fenomena (like LENR/Cold Fusion), corrects wrong explanations wrongly done with the old theory; but also fits well where the old theory is valid. (So it’s a very good candidate to expand the old knowledge and theories, explaining very well new fenomena, and thus going some steps further, but respecting (and building further on) the good work done by the old theories so far, doing so without throwing away the child with the bathwater).

So, about the article (link given above), I have only a minor question to ask:

In figures 8 and 10 (for figure 8, see: http://peswiki.com/index.php/Image:AAAfig8-coldFUSION-pamelaMOSIERboss.gif ; and for figure 10, see: http://peswiki.com/index.php/Image:AAAfig10-coldFUSION-pamelaMOSIERboss.gif ), you only drew 1 electron, while both deuterions (fig. 8) and both He-nuclei (fig. 10) have one electron each, so actually, they share 2 electrons.

So, as the probability that both electrons share the same position is very low (if not zero ;-) , and while the first electron is in between the two deuterions (fig. 8.B) or He-nuclei (fig. 10.B), could the second electron (in these cases probably often being on the side of one of the two surrounding Pd-atoms (like in fig. 8.AorC and fig. 10.AorC), and thus not in between the two deuterions (fig. 8.B) or He-nuclei (fig. 10.B), disturb the fusion-process (of the deuterions in fig. 8.B and the He-nuclei in fig. 10.B), reducing the fusion-force (indicated by the two green arrows in figure 8.B and figure 10.B), to only one time the green arrow (like in figures 8.A, 10.A, 8.C and 10.C), and not two times, so that the force (in that case only one green arrow in the figures) acts upon the whole chain, not trying to fuse together the nuclei (deuterions (fig. 8.B) or He (fig. 10.C))? But anyhow, in that case, maybe an ion and a neutral atom will do? Or maybe the second electron as other and non disturbing possible positions too (reducing only a little bit (and in certain positions) the probability for fusion)?


http://www.journal-of-nuclear-physics.com/?p=619&cpage=1#comment-225737

Daniel De Caluwé
April 30th, 2012 at 9:21 AM

Sorry, but although my question remains valid (for the deuterium-pair in figure 8), I need to correct my previous message (see: http://www.journal-of-nuclear-physics.com/?p=619&cpage=1#comment-225684 ) for the helium pair (in figure 10) as follows:

- As every helium atom (He: Z=2; A=4) has 2 protons, 2 neutrons and 2 electrons, the helium pair in figure 10 shares 4 electrons in total. So, in this case (of helium), we need to consider the position of 4 electrons in figure 10, and so the question in my previous message becomes even more complicated…

- But my question, as formulated in my previous message (see: http://www.journal-of-nuclear-physics.com/?p=619&cpage=1#comment-225684 ), remains valid for the deuterium-pair in figure 8, where we need to consider the position of 2 electrons (1 electron for each deuterium atom).

- And also, at the end of my previous message, where I last referred to figure 10.C, I meant (and it should have been) figure 10.B.

My apologizes for these mistakes.


http://www.journal-of-nuclear-physics.com/?p=619&cpage=1#comment-225921

Wladimir Guglinski
April 30th, 2012 at 8:50 PM

Dear Daniel De Caluwé

The electron orbits in the helium is the orbitron in the link:
http://www.webelements.com/nexus/content/orbitron-atomic-orbitals

The resonance is due to the two electrons along the z-axis (red orbit).
They move in two orthogonal planes (one in the z-y plane, and the other in the z-x plane).
When the electron of the z-y plane is in the top in the figure 10 in my paper, the other one (of the z-x plane) is the center of the distance between the two nucleons 2He4.

If you analyse their positions, you will realize that they will work together in partnership, increasing the resonance process.

The other two electrons move along the x-y plane (blue orbit).
They do not contribute for the resonance process.


http://www.journal-of-nuclear-physics.com/?p=619&cpage=1#comment-226278

Wladimir Guglinski
May 1st, 2012 at 8:39 PM

Dear Daniel De Caluwé,

I was thinking about your question, and there is another conclusion about the motion of the electron.

When I wrote the paper on Mosier-Boss experiment, published in Peswiki in 2009, I was thinking about the electron’s contribution from the electromagnetic interaction viewpoint.

However, according to Quantum Ring Theory there are laws of induction between electric massless particles of the aether and gravitons.
They are laws similar to that discovered by Faraday for the electromagnetism:

- the motion of an electric particle of the aether induces a flux of gravitons

- a flux of gravitons induces the motion of electric particles.

As the electron’s field is composed by electric particles of the aether, so an electron with relativistic speed can induce a flux of gravitons (and they can form a gravitational field).

Therefore, in the case of the electron moving in the Fig. 10 of my paper, the influence of the electron is not only electromagnetic, it is also gravitational (with the magnitude of the electromagnetism).


http://www.journal-of-nuclear-physics.com/?p=619&cpage=2#comment-227493

Wladimir Guglinski
May 4th, 2012 at 11:02 PM

P.G.Sharrow wrote in May 3rd, 2012 at 9:54 PM:

“Wladimir Guglinski
May 3rd, 2012 at 4:16 AM says
“The gravitons and the electric particles of the aether are similar.”

My examination of all that is known to me brought a similar conclusion.
Most subatomic “particles” are just quanta of Aether with different spin signatures.”

Dear Sharrow,
after the moment when one arrives to the conclusion on the existence of the massless particles of the aether, such conclusion becomes to him so obvious, evident, logic.

It becomes so obvious that he gets astonished why the physicists did not arrive to the same conclusion.

Regards
Wlad


http://www.journal-of-nuclear-physics.com/?p=619&cpage=2#comment-227487

Wladimir Guglinski
May 4th, 2012 at 10:54 PM

Dear Joe,
why have the theorists supposed the existence of the graviton, but they did not suppose the existence of the electric particles of the aether ?

In the case of gravity, the reason is obvious. There is attraction between the bodies. And so, there must be an agent responsible for such attraction,which is known as gravitational. Such agent is supposed to be the graviton.

In the case of electricity, the physicists discovered the existence of the electron. So, there is no need (as happened in the case of gravity) to consider an agent responsible for the existence of electricity, since such agent is the own electron.
Besides, to consider electric particles of the aether would be an unacceptable speculation, because from Einstein’s theory the space is supposed to be empty.

However, particles as the electron and the proton have electric fields. And so a question arises: what is the agent which composes the electric fields of proton and electron ?

This is a fundamental question, but it is not considered in current theories of Modern Physics. It is merely a philosophycal speculation, and the physicists cannot take it seriously. They know that proton and electron have an electric field. Nevertheless, it is not of their interest to speculate about the “material” structure of such fields. The concept of field in Modern Physics is an abstract concept, without physical meaning.

On another hand, they suppose the existence of the graviton, and they try to detect it. But in spite of their effort, the graviton shows itself to be undetectable.

So, the detection of the electric particles of the aether would be in worst situation. First because the physicists do not believe in the existence of the aether. And second because, as they do not suppose the existence of electric particles of the aether, they do not try to detect them.


Verder toegevoegd op 15/10/2012:


http://www.journal-of-nuclear-physics.com/?p=629&cpage=3#comment-239247

Daniel De Caluwé
May 25th, 2012 at 7:11 AM

Dear Ioannis,

Based on your first message, I contacted a Dutch group of free-energy (Zero Point Energy) experimenters, that maybe have (at least some) of the equipment you mentioned in your first message.

But I don’t know if they can do the experiment, and I leave it up to them to react here, or to contact you via your e-mail address or at your website.

Kind Regards,

Ir. Daniel De Caluwé, Belgium.


http://www.journal-of-nuclear-physics.com/?p=629&cpage=3#comment-239974

Ioannis
May 26th, 2012 at 10:56 AM

Dear Daniel De Caluwé,

Thank you for your interest as also that you forwarded my message to free experimenters in your country. The valid information is the first message of me below yours. The previous will not help. They need a Data Logger system with at least 18-Bit Amplitude resolution, otherwise they will detect nothing. I already tried something as I mentioned which seems encouraging (I saw again the Aether Signal), but I have to make more experiments with other sensors too. Probably the next couple of day, I will publish some pictures of the equipment and the recordings.

I will look forward with great interest and patience, the results from other experiments! Thank you very much for your support!

Best Wishes

Ioannis Xydous

Electronic Engineer

Web Site: http://www.ioannisxydous.gr/
E-mail: SEPP@ioannisxydous.gr


http://www.journal-of-nuclear-physics.com/?p=510&cpage=58#comment-252382

Daniel De Caluwé
June 11th, 2012 at 5:11 PM

@Joseph Fine,

You wrote: ‘Case II: Multi-Stage/Cascaded

A 4-stage design with the same 227-10 kW modules could be organized having 10-Stage 1 units, 24 Stage 2 units, 57-Stage 3 units and 136 Stage 4 units. That is, 136 Stage-4 10-kW output units produce 544 kW-Electric with a total first stage electric input power demand of only 16.66 kW and a power gain of 544/16.66 = 32.6; that is, the 544 kW of electric output are produced using only 16.66 kW of electrical input.’

My answer: Theoretically you’re right, but practically, you need a turbine and an alternator after each stage (4 stages in total), and so the investment cost probably will be too high? ;-)


http://www.journal-of-nuclear-physics.com/?p=510&cpage=58#comment-252863

Joseph Fine
June 12th, 2012 at 7:53 AM

Pekka, Daniel, A.R.

Of course, you are right.

As an alternative, use a thermal-to-electric conversion factor of .4166, with a power gain of 2.5.

So, each group of six 10 kW modules requires an “average” electrical input of 10 kW. That same set or “clowder” of 6 cats produces .4166*60 = 25 kW, of which an “average” of 10 kW can be fed back to the input. (But, it sometimes needs 20 kW!!!)

That is, a net of 25-10 = 15 kW is available for further use, with apparently zero power input. Each clowder (of 6 cats) may sometimes require twice the average power of 10 kW (and zero kW in self-sustain mode). Assuming an average of 15 kW is available from each group or “clowder” of 6 modules, it is much easier to scale up.

If 120 modules are organized in 20 “clowders of 6 cats”, the average power production is 20*15 kW = 300 kW-E. That reduces the need for a plumbers nightmare of steam turbines et cetera. Each clowder will sometimes need 20 kW at the input, rather than 10 kW, (in self sustain mode, it needs zero input). The system has to be able to provide for this. Designing in extra groups/clowders (@ 15 kW output each) can help meet peak input power demands.

Best regards,

Joseph


http://www.journal-of-nuclear-physics.com/?p=695&cpage=2#comment-302594

Daniel De Caluwé
August 15th, 2012 at 8:16 AM

From a farmer son of a small country (Belgium), with an Italian prime minister (Elio Di Rupo), I wish Dr. Rossi and all readers (especially Italian readers ;-) : Buon Ferragosto. Here in Flanders (Belgium), and in the past, the month of august was also called the ‘month of harvest’, so I guess that ‘Buon Ferragosto’ also means ‘I wish you a good Harvest’? In that case, this certainly is a good wish to Dr. Rossi, who, with his 600°C to 1200°C and also gas-driven Hot Cat (hot E-cat) will be able to produce electricity, and that certainly is a ‘good harvest’! So, I wish him and his team a ‘buon ferragosto’ again!


http://www.journal-of-nuclear-physics.com/?p=695&cpage=2#comment-302612

Andrea Rossi
August 15th, 2012 at 9:00 AM

Dear daniel De Caluwè:

Thank you for your kind wish.
Warm regards and “Buon Ferragosto” to you!
Warm Regards,
A.R.


http://www.journal-of-nuclear-physics.com/?p=695&cpage=2#comment-302605

Daniel De Caluwé
August 15th, 2012 at 8:38 AM

Oh, what i forgot to ask in my previous message:

As the start-up time and the shut-down time both will be about 1 hour, am I right then that the Hot-cat (in the 1 MW plant and more) will be able to co-operate with renewable energy sources like wind-power and solar cells? I think yess, because 1 hour delay time still is fast enough to control or stabalize (the frequency and the voltage) of the electricity-grid, when there’s no wind and no sun. So in case of no wind, we quickly could start the Hot-Cat power plant, just in the same way as they do now with the gas-powered plants, that are easier to regulate than our big nuclear plants…

P.S. And have you heard of the problems with some of our big nuclear plants? We defenitly need the Hot-cat power plants!

Kind Regards,

Ir. Daniel De Caluwé, Belgium


http://www.journal-of-nuclear-physics.com/?p=695&cpage=2#comment-302609

Andrea Rossi
August 15th, 2012 at 8:56 AM

Dear Daniel De caluwè:
Yes, our plants are fit to be connected and intergrated with the existing plants of any kind.
Warm Regards,
A.R.


http://www.journal-of-nuclear-physics.com/?p=695&cpage=2#comment-303039

Daniel De Caluwé
August 16th, 2012 at 5:11 AM

Dear Wladimir,

When you first mentioned the article in nature, I’ve read it, and I agree: it is very similar to your QRT-theory, that you published in 2006. (I remember you also mentioned a difference with them, and I agreed with your arguments about that too).

So, I agree with you that you protested, because you worked out these ideas very much earlier, and you even published a book about QRT in 2006. By the way, I like your theory, and I hope to study it more carefully, like I hope to study many other theoretical articles in this blog.

Further I hope that the scientific world will be honest enough to recognize your work, and not to attack you first, but steal your ideas later. (Unfortunately this tactics is often used by not so honest people, and I experienced this also myself, but in a totally different field).

Kind Regards,

Ir. Daniel De Caluwé, Belgium


http://www.journal-of-nuclear-physics.com/?p=695&cpage=2#comment-303437

Wladimir Guglinski
August 16th, 2012 at 4:33 PM

Dear Daniel De Caluwé,
thanks to your kind words

regards
WLAD


http://www.journal-of-nuclear-physics.com/?p=733&cpage=2#comment-322950

Daniel De Caluwé
September 14th, 2012 at 4:58 PM

So, if I understood well, the working of the E-cat is not SF (Science Fiction), but based on SF (Spin Fusion)? ;-)


http://www.journal-of-nuclear-physics.com/?p=733&cpage=3#comment-327233

Daniel De Caluwé
September 18th, 2012 at 6:14 AM

@Robert Curto,

At the moment and in Europe, renewable and variable energy sources (like wind-farms and solar cells) co-operate with gas-powered power-stations. (The gas powered stations take over the electricity production in case there’s no wind or sun). But here (and in the near future) comes in the big advantage of the (Hot) E-cats: As the E-cats can be started up in one hour, and be turned down in one hour, they certainly could co-operate with variable wind-farms and solar cells. (Based on the prediction and measurements of wind and weather, and on weather forecasts, we could quickly start up the necessary amount of Hot E-cats, to produce the electrical power that the wind and the sun fail to deliver, and this the E-cats do much quicker than gas powered plants (because they can be started up and turned down in 1 hour), and they also are more flexible (if they are switched on and of per 10kw module or even per 1MW unit) and cheaper (COP = 6, so the gas driven Hot E-cat plants produce electricity at a much lower cost than the ordinary gas-powered power-stations… ;-)

But of course, I know that the industrial E-cats (as well E-cat and Hot E-cat) also are very good for many industrial clients themselves (producing as well thermal energy (E-cat and Hot E-cat) as electricity (in the future with Hot E-cat)). And yes, compared with gas-powered power stations, the gas-driven Hot E-cats consume less primary energy (in this case gas) than the ordinary gas powered power stations, and certainly they gradually will take a growing share in the energy-mix, but I don’t think they will replace everything immediately. (There always will be an energy-mix of different energy sources, and being able to co-operate easily with renewable energy sources is a very strong point of the E-cats!)

Kind Regards,


http://www.journal-of-nuclear-physics.com/?p=733&cpage=3#comment-328375

Robert Curto
September 19th, 2012 at 6:06 AM

Dear Daniel, thanks for the excellent information.
You explained the advantage of E-cats very well.

Robert Curto


http://www.journal-of-nuclear-physics.com/?p=748&cpage=1#comment-354329

Daniel De Caluwé
October 11th, 2012 at 10:34 AM

About Ether and the Michelson-Morley-experiment

Dear Wladimir and other readers,

About two years (plus four months) ago, I wrote something about ether and the Michelson-Morley experiment on the Dutch niburu-forum:

http://niburuarchief.info/showthread.php?tid=20448&pid=317784#pid317784

My opinion was and is that the Michelson-Morley experiment only proved that the speed of light was independent of the relative motion of the sender (of the light) in respect to the receiver, so the speed of the relative motion was not transferred to the (speed of the) light itself, and so their experiment proved (far in advance) this property of propagation of the light explained by Einstein in his theory of Special Relativity.

And based on this experiment, the scientific community rejected the existence of the ether, but in fact, the Michelson-Morley experiment only rejected a kind of a ether that transfers the relative movement (of sender towards receiver) to the speed of light. But if we suppose an ether that doesn’t transfer the relative motion of sender and receiver to the speed of light, than such an ether still would be possible, and could not be rejected by the scientific community based on the Michelson-Morley experiment.

Is such an ether, (that doesn’t transfer the relative movement of sender and receiver to the speed of light), possible? I think the answer is yes, especially when we envision a kind of ether that behaves like a higher dimension that supports the propagation of light, but that doesn’t transfer the relative motion of sender and receiver, who belong to the normal/ordinary (and lower) physical dimension.

Is there an interaction between the (lower and) ordinary physical dimension and the higher dimension of such an ether? The answer is yes:

i) For instance when we heat up a piece of metal that starts to glow and sends its light (based on the well known mechanism of excitation of electrons that send fotons when they fall back in their normal orbits).

ii) And also we know by experiment(s) that the speed of light depends of the physical medium where it is passing through. The speed of light is the highest in the vacuüm, but decreases in media with higher density. In the air, the speed of light is already a little bit lower, and in water the speed is lower than in air. (Also the laws of the breaking of the light are based on this). So if we suppose that light is propagated in the higher dimension of the ether, than by experiment we know that the fysical medium that surrounds it has influence on the speed of light in the ether. (So that’s a second kind of interaction between the higher dimension of the supposed ether and the lower physical world).

iii) And also via Einsteins’ General Theory of Relativity, we know that light is bend by the interaction of heavy bodies. So gravity, that is supposed to belong to the (lower) physical dimension ;-) , also has an influence on the propagation of light through our hypothised ether.

But to conclude so far: If we suppose or hypothise such an ether (in a higher dimension than the ordinary physical world), we know from the interaction explained under i), that electrons have to do with the interaction between the two worlds (the ordinary physical world and the higher ether dimension) and because of the measured interactions under ii) and iii), we also know that there’s also an interaction because of the mass mainly composed of particles (neutrons and protons) in the nucleus.

Does this make sense and/or can we do something with this? ;-)

Kind Regards,
Daniel.


http://www.journal-of-nuclear-physics.com/?p=748&cpage=1#comment-355205

Ioannis
October 12th, 2012 at 3:07 AM

Dear Daniel De Caluwé and others,

I am tempted for once more to intervene on this discussion about the Aether without to have the intention to be annoying. Daniel, I believe that you put the subject on the right frame. Yes, the Aether as entity can be found on a higher dimension or better on a hidden dimension. The Aether is imaginary as I prove in my work: http://www.ioannisxydous.gr/

The Michelson-Morey Experiment brought null results since they were expecting to find an opposing “Aether wind” (translational motion) in regards to the motion of light. As you will see on my Web Site: http://www.ioannisxydous.gr/ , I propose an experiment (it does not need a set up) for the indirect detection of Aether: http://www.ioannisxydous.gr/AetherDetection/

which is the effect of the imaginary Aether in our world. This effect makes all matter and non matter (free space) of the Universe to rotate with the Tangential Velocity of 348.43 Km/sec. The proposed experiment simply detects the rotational motion of the Universe. The experiments which took place so far for the measurement of our absolute motion in free space are the following:

(see also the below on the link: http://www.ioannisxydous.gr/AetherDetection/ )

1925-26 David Miller: Interferometer, continuous light. Result: 208 Km /sec
1927 K.Illingworth: Interferometer, continuous light. Result: 369+/-123 Km/sec
1973 Stefan Marinov: Rotating mirrors, chopped light. Result: 130+/-100 Km/sec
1975-76 Stefan Marinov: Interferometer, Rotating mirrors, chopped light. Result: 303+/20 Km/sec
1976 Muller : Velocity towards Leo. CMB. Result: ~400 Km/sec
1977 G. Smoot : CMB Result: 390+/-60 Km/sec
1978 Wilkinson/Corey : CMB Result: 320+/-80 Km/sec
1978 Monstein/Wesley : Muon flux Anisotropy Result: 359+/-180 Km/sec
1984 Stefan Marinov: Coupled shutters, chopped light Result: 362+/-40 Km/sec
1986 E.Silvertooth : Rotating mirrors, chopped light Result: 378+/-? Km/sec
2006 M. Consoli : Analysis of rotating optical resonators Result: 276+/-71 Km/sec

2012 I.Xydous : Analysis of Earth’s Rotating Magnetic Field Result: for 348.43 Km/sec (theoretical), the Aether frequencies:
8.67mHz and 17.36mHz superpositioned on Earth’s long period Magnetic Field.

Comments and Conslusions:

i) The Aether is invisible and can be indirectly detected.
ii) Almost all of the above attempts of the past, as you will notice they have (the most of them) circular or closed loop topologies.
iii) Due to the (ii), they attempt to detect a circular motion.
iv) According to my theory and the above past experiments the conclusion is: The Aether of free space is INVISIBLE (IMAGINARY), STATIONARY (No translational motion) but ROTATING with a TANGENTIAL VELOCITY of 348.43 Km/sec.
v) All matter like: Protons, Electrons and every particle and sub-particle which comes into existence, enters our material Universe with this tangential velocity of 348.43 Km/sec.
vi) The Aether is the cause (actually in our real world is the spinning mass) of attraction between Proton-Proton in Nuclear Force. The Aether with other world has the power of a strong Gravitational Field, which shields against the Electrostatic Force. On page 14 on my paper SEPPv5, it is proved clearly in case of an Electron, the Gravitational constant is G=2.78E32 N*m2/Kgr2 on its surface. It is 1E43 times stronger than Earth’s Gravity.
vii) The new expression of Charge and Planck Constant include Aether’s Tangential Velocity. See page 37: Eq.98, Eq.99, Eq.100. It is very obvious! (I did not add something. It was always there, but wrongly interpreted due to the development of Quantum Physics all these years).

I strongly encourage everyone to read my work and I will be very glad to discuss all these. This work is completely formulated with simple Mathematics (of High School) and less philosophical. Nothing is left in the “Air”. There are complete (understandable mathematically formulated) answers on the following subjects:

Aether, Massless Neutrinos (exact imaginary mass, Upper and Lower Velocities), Dirac’s Magnetic Monopole, Complete Coulomb Force, Complete Magnetic Force, Complete Casimir Force, Nuclear Strong Force, Quantum Newtonian Gravity, Unified Field Force, Universe Properties: Age, Acceleration (actually deceleration), radius, Temperature, Aether Control, Antigravity, Space-Time Engineering and much much more….

I hope you will enjoy the ride!

Ioannis Xydous

Electronic Engineer

Web Site: http://www.ioannisxydous.gr/

Switzerland


http://www.journal-of-nuclear-physics.com/?p=748&cpage=3#comment-361751

Daniel De Caluwé
October 17th, 2012 at 9:57 AM

The strange thing about light/quanta of light/fotons is that the relative motion of the sender in respect to the receiver is not added to the velocity of light, as proven by Michelson-Morley and all other experiments done to prove that Einsteins’ theory of Special Relativity was right, and this is in contrast with all other mechanical phenomena tested en expierenced in our world. If I throw a ball or a bullet towards the reader, then that ball or bullet comes in with a higher speed when I also run towards the reader, so my (relative) speed is added to the speed of the ball or the bullet, but this is NOT so in the case of ‘bullets of light’ (= fotons = quanta of light), because they always come in at the same velocity, not dependend of the relative motion of the sender towards the receiver (proven by Michelson-Morley and many other experiments), so, if we treat fotons/quanta of light just as a form of very very very light matter, why doesn’t it behave as ordinary matter? Why is my speed not added to the speed of light if I move towards you? Does light/fotons/quanta of light belong to another higher dimension/realm, very close and with a lot of interactions with our ordinary world/matter, but because of this exceptional quality not really following the rules of the ordinary world/matter? And oh yes, in many ways, light/fotons/quanta of light behave like small and very very light bullets, because they move slower in water than in air and slower in air than in vacuüm, and these small and very light bullets (fotons) also are bend by heavy objects (like big stars and black holes), and there even is a doppler effect (frequency shift) like their is with sound in the air (think of the frequency shift of the sound when a train is passing), so from this point of view, light (fotons/quanta of light) seem and appear to behave normal, but there’s one exception: it doesn’t get the relative motion (of the sender in respect to the receiver), and this in contrast with all other things that move in this, our world. My question towards the scientific community is why? Why is the relative motion of the sender towards the receiver, and in the case of light, only transferred as a frequency-shift of the light (doppler effect; red-shift; blue-shift), and not as a change of the speed, like in the case of normal material objects/bullets, that always come in with the relative speed (of sender towards receiver) added to their own velocity?

But I agree, ‘bullets of light’ (= fotons = quanta of light) can move through space (and the vacuüm) without the existence of an aether or a carrier, we just can treat them as very very light material objects or bullets, moving through the vacuüm of space that is really empty, but why then, is the relative motion (of sender towards receiver) not added in their case, like is the case with all other material objects/bullets? Why these ‘bullets of light’ (= fotons = quanta of light) have this exception and behave different?


Verder toegevoegd op 4/11/2012:


http://www.journal-of-nuclear-physics.com/?p=748&cpage=5#comment-376827

Steven N. Karels
October 28th, 2012 at 1:59 PM

Dear Andrea Rossi,

Estimate of eCat Isoptopic Changes

Assuming an eCat (Thermal or Hot) runs continuously for 6 months, has a 10 kW thermal output, what percentage of the Nickel is converted from one isotope to another isotope?

Fuel load is 20 grams – assume 10 grams are Nickel.

Natural Nickel isotope distribution: 58Ni (68.1%), 60Ni(26.2%), 61Ni(1.1%), 62Ni(3.6%) and 64Ni(0.9%).

If we assume a Ni + H -> Cu -> Ni reaction only, then the following reactions are of interest

58Ni + H -> 59Cu (halflife of 81.5 sec) -> 59Ni (halflife of 76,500 yrs)
60Ni + H -> 61Cu (halflife 3.33 hrs) -> 61Ni (stable)
61Ni + H -> 62Cu (halflife of 9.67 sec) -> 62Ni (stable)
62Ni + H -> 63Cu (stable)
64Ni + H -> 65Cu (stable)

The mass defect for each reaction is about 1 amu. The total number of Nickel atoms is about 1.04 x 10**23. If all the Nickel atoms went through the change, about 4,260 MWh of energy would be released.

An eCat producing 10kW of power for 6 months will produce about 43.1 MWh of energy or about 1% of the total conversion energy.

Assuming the conversion probability is the same for all Nickel isotopes, we should have the following isotopic distributions:

58Ni(68.1% – 0.68% = 67.4%)
59Ni(0.68%)
60Ni(26.2% – 0.26% = 25.9%)
61Ni(1.1% – 0.01% + 0.26% = 1.35%)
62Ni(3.6% – 0.04% = 3.56%)
64Ni(0.9% – 0.009% = 0.9%)

Copper isotopes

63Cu(3.6%*0.01 = 0.036%) of 10 grams = 0.036 grams
65Cu(0.9%*0.01% = 0.009%) of 10 grams = 0.009 grams

A difficult task to measure micrograms of copper. However, the distinguishable isotope should be the presence of 59Ni at around 0.68% should be observable.

http://www.journal-of-nuclear-physics.com/?p=748&cpage=5#comment-378087

Daniel De Caluwé
October 29th, 2012 at 5:02 AM

Dear Steven N. Karels,

Roughly, I agree with your approach, but as there are no residual nuclear reactions when the E-cat is stopped, the first reaction that you mentioned (i.e. ’58Ni + H -> 59Cu (halflife of 81.5 sec) -> 59Ni (halflife of 76,500 yrs)’) will not occur, I think…

So, could it be that the first reaction does not occur because some of the (four) others are easier? (Remember that processes in nature always tend to follow the way of least resistance).

Kind Regards,

http://www.journal-of-nuclear-physics.com/?p=748&cpage=5#comment-377613

Steven N. Karels
October 28th, 2012 at 11:12 PM

WaltC,

I think the salient point is that where there was no 59Ni in the sample before the start of the eCat reaction, at the end of six months of continuous operation, there should be a detectable amount. This remains true until several times the 76,500 year half-life.

Likewise, copper will be formed but in very small amounts.

CORRECTION

Copper isotopes

63Cu(3.6%*0.01 = 0.036%) of 10 grams = 0.036 grams
65Cu(0.9%*0.01% = 0.009%) of 10 grams = 0.009 grams

SHOULD READ

Copper isotopes

63Cu(3.6%*0.01 = 0.036%) of 10 grams = 0.0036 grams = 3,600 micro grams
65Cu(0.9%*0.01% = 0.009%) of 10 grams = 0.0009 grams = 900 micro grams

If we do not see the formation of 59Ni, then I would question the belief that a Ni to Ni operation is the majority energy producing process.

http://www.journal-of-nuclear-physics.com/?p=748&cpage=5#comment-378579

Steven N. Karels
October 29th, 2012 at 10:42 AM

Daniel.

I hope there is uniform probability for all stable isotopes of Nickel. If 58Ni does not participate in eCat reactions, we have lost 68% of the Nickel on the earth as a fuel.

I do not see why the 58Ni reaction should not work. Please explain.

http://www.journal-of-nuclear-physics.com/?p=748&cpage=5#comment-378642

Daniel De Caluwé
October 29th, 2012 at 12:03 PM

@ Steven,

As there is no residual radioactivity measured in the fuel after a short time (3 hours as affirmed by A. Rossi in his recent message) after the E-cat is stopped, I wrote that reaction number one (’58Ni + H -> 59Cu (halflife of 81.5 sec) -> 59Ni (halflife of 76,500 yrs)’) does not happen, and this because of the halflife of 76,500 yrs for the last step (59Cu -> 59Ni).

http://www.journal-of-nuclear-physics.com/?p=748&cpage=5#comment-378685

Steven Karels
October 29th, 2012 at 12:33 PM

Daniel and Andrea Rossi,

The 58Ni + H -> 59CU could occur at some rate. The radioactive half-life of 59Cu is 81.5 seconds. So in terms of a 6 month run, essentailly all of the 59CU would convert to 59Ni. The 59Ni half-life of 76,500 years is long compared to the 6 month operation so essentially none of the 59Ni would be decayed.

The three hour time period — is that the radioactivity presents a half-life of three hours or that the measureable radioactivity is gone after three hours? Please clarify. If you can tell me the effective half-life of the residual readioactivity, it would help validate or discard the Ni-to-Ni reaction.

http://www.journal-of-nuclear-physics.com/?p=748&cpage=5#comment-378942

Daniel De Caluwé
October 29th, 2012 at 7:08 PM

@ Steven,

1. If the measurement of the radioactivity was ok after three hours, then you could look at reactions with a half-life of 3 hours devided by 7 = 25.7 minutes. (After 7 times half-life, the number of nucleons that still can desintegrate has fallen back to less than 1%).

2. In my previous message, I wrongly wrote: ‘that reaction does not happen, and this because of the halflife of 76,500 yrs for the last step (59Cu -> 59Ni)’. But of course I meant the long halflife of 59Ni, which remains (low-)radioactive for a very long time. (I think it becomes 59Co (which probably is stable) by electron capture, but unfortunately only after a very long time. So to see if this reaction occurs, you could look at the presence of 59Co, but of course, because the activity is very low, only after a long time). Was the radioactivity of 59Ni not seen because it’s so low? Anyhow, I agree with ‘Man’ that it’s a potential health hazard, with a low but long during activity, but because they didn’t measure any radioactivity after three hours, I concluded that the first reaction you mentioned earlier could not happen. (59Ni remains low-radioactive for a very long time).

3. Why do you treat all possible reactions equal? Thinking, for instance, on QRT of Wladimir Guglinsky, there maybe are preferrable reactions? (Affinity to/possibility of proton-capture can be different for the different isotopes). And also, if the goal would be to obtain stable isotopes of Cu, I would prefer working with 62Ni (and 64Ni to a lesser degree), because with 62Ni we could get stable 63Cu (and with 64Ni we could get stable 65Cu), but as the transmutation to Cu is not the main energy-source, it probably is not interesting anymore?

4. Remaining 60Ni and 61Ni as interesting isotopes? ;-)

http://www.journal-of-nuclear-physics.com/?p=748&cpage=6#comment-382359

Steven N. Karels
October 31st, 2012 at 10:44 AM

Daniel,

Sorry for the delayed response but Hurricane Sally — loss power, internet down plus some damage.

You are correct that after 5 half-life times, the residual radioactivity is small (~1%). But you multiply the half-life by five, not divide.

I did not include the 59Ni as a reaction because so little is produced that it would not affect the power output.

I used equal probabilities of reaction for all the stable isotopes because I have no information that one was favored over the other.

To Andrea Ross — Have you seen 59Ni detection in any of your long term (>6 months) spent fuel analyses?


http://www.journal-of-nuclear-physics.com/?p=748&cpage=6#comment-382765

Daniel De Caluwé
October 31st, 2012 at 3:38 PM

@Steven,

1. Sorry, but if you do not measure any radioactivity after 3 hours, you have to divide these 3 hours by 5 (3.13% nucleons still radioactive after 5 half-lives) or by 7 (0.78% of the nucleons still radioactive) to find a proxy (or rough estimation) for the possible half-life time of the nuclear reaction/transmutation that is involved.

2. And after 5 times the half-life time, you still have 3.13% of the original nucleons that are radioactive, and after 7 times the half-life time this is reduced to 0.78% of the original nucleons.

Kind Regards

http://www.journal-of-nuclear-physics.com/?p=748&cpage=7#comment-387302

Steven N. Karels
November 3rd, 2012 at 2:51 AM

Dear Andrea Rossi,

I did not mean to imply a 1 gram conversion to energy as that would be more than the total energy output observed. But I do appreciate the information that the loss may have due to loss of humidity. Making mass change measurements when the eCat has 4,000 times the change in mass can be very challenging.

a. Perhaps the conundrum could be explained in your instrumentation?

b. Can you see a 1% change in isotope distributions with your SEM?

c. For example, can you detect the natural Nickel isotope distributions of 61Ni and 64Ni?

d. Do you have any information on the accuracy and precision of the instrumentation used?


http://www.journal-of-nuclear-physics.com/?p=748&cpage=7#comment-387639

Andrea Rossi
November 3rd, 2012 at 7:46 AM

Dear Steven N. Karels:
a- yes
b- no
c- not outside the natural range: remember that we enrich 62 and 64 Ni, whose changement in % is made before cherging
d- yes, and the margin of error is superior to the range
Warm Regards,
A.R.

http://www.journal-of-nuclear-physics.com/?p=763&cpage=1#comment-387992

Daniel De Caluwé
November 3rd, 2012 at 1:01 PM

@Sankar Hajra,

I can be wrong, but I don’t agree that you can explain ‘the invariant proposition’by ‘classical physics’. The fact that the speed of light remains a constant in vacuüm, and that it is not affected by the relative movement of the sender in respect to the receiver, is an unusal aspect of the behaviour of light/fotons, that cannot be explained by classical physics. But beside that one exeption, I agree with the rest.

Kind Regards,

Ir. Daniel De Caluwé, Belgium

http://www.journal-of-nuclear-physics.com/?p=748&cpage=7#comment-388962

Daniel De Caluwé
November 4th, 2012 at 6:18 AM

Dear Dr. Rossi,

Although I don’t want to take part in a Q&A game, where you gradually (step by step) and eventually are forced to reveal the secret working of the E-cat (even when that maybe is against your initial will), the questions of some participants and your answers also make me curious, and therefore these questions: (but of course, if some of these questions would reveal too much of your secrets, I accept that you don’t answer them or even remove some of them)

1) Did you initially enrich Ni (to get more 62Ni and 64Ni) because you initially thought/presumed that the energy maybe was coming from the transmutation tot Cu (62Ni + proton -> stable 63Cu; and 64Ni + proton -> stable 65Cu)?

2) But, from the moment you knew that the transmutation to copper (Cu) only was a side-effect, did you try normal (not enriched) Ni back again?

3) And if yes (to question 2), did normal (not enriched) Ni also work and produces the same amount of energy?

4) If the enrichment of Ni (to get more 62Ni and 64Ni than naturally occuring) really is necessary to get the E-cat work (or to get the necessary amount of energy), could the fact that 62Ni and 64Ni, who are the most heavy of the stable isotopes of Ni, have most free (excess-) neutrons (6 neutrons in excess for 62Ni and 8 neutrons in excess for 64Ni), and that, thinking on QRT of Wladimir Guglinsky (where the structure of the nucleus is supposed/hypothised by a He-core, surrounded by deuterions (proton-neutron pairs) in hexagonal layers (with max 6 deuterons per layer)), so that more excess neutrons tend to enhance ‘proton-capture’ (to form deuterons in the outside hexagonal layers)?

Could that possibly be a mechanism? : ‘Proton-capture’ (62Ni/64Ni + proton -> (63Cu/65Cu)*, but not yet completely formed/finished, and therefore falling back (and/or via ‘pseudo electron-capture’ with beta and gamma radiaton that produces the heat) and going back and forth (because of the applied ‘frequencies’)?

But of course, I accept if you don’t want to answer the last question.

Kind Regards,

http://www.journal-of-nuclear-physics.com/?p=748&cpage=7#comment-389077

Andrea Rossi
November 4th, 2012 at 8:01 AM

Dear Daniel De Caluwe’:
1- No
2- No
3- No
4- Confidential
Warm Regards,
A.R.


Toegevoegd op 22 november 2012 :

http://www.journal-of-nuclear-physics.com/?p=748&cpage=4#comment-366849

Daniel De Caluwé
October 22nd, 2012 at 3:50 AM

Dear Dr. Rossi,

From the economical point of view, heating with gas is much cheaper than heating with electricity (gas bill will be lower than electricity bill), and as well for the Home E-cat as for the industrial Hot E-cat (also to produce electricity), replacing the electrical drive (electrical heating used when the E-cat is not operating in SSM (self sustaining mode)) by a heating with a gas burner is and would be a very big economical improvement, that will reduce the bill of the drive.

So, therefore my question: what’s the present situation of the gas-driven E-cat? Can and will it be used also in the Home E-cat? (It would be an enormous improvement, especially in the Home E-cat, where the electricity-bill would be replaced by a much lower gas-bill, so that we also economically could replace most of the smaller central heating units (like in appartements) by gas-driven Home E-cats…


http://www.journal-of-nuclear-physics.com/?p=748&cpage=4#comment-366893

Daniel De Caluwé
October 22nd, 2012 at 4:46 AM

Dear Dr. Rossi,

Is it possible to disclose the name of your Belgian Licensee (and/or the Licensee for the Benelux)? (I’m interested in who or what company that represents you in Belgium and/or the Benelux).


http://www.journal-of-nuclear-physics.com/?p=748&cpage=4#comment-367048

Andrea Rossi
October 22nd, 2012 at 8:25 AM

Dear Daniel De Caluwè:
Our Belgium Licensee will contact you.
As for your suggestions and considerations, I am studying them.
Warm Regards,
A.R.


En aangezien A. Rossi nu (sinds 2012, maar wel in tegenstelling met wat hij beweerde in 2011) beweert dat de vorming van koper waarschijnlijk maar een neveneffect is, schreef ik, een beetje ten einde raad, dit:

http://www.journal-of-nuclear-physics.com/?p=748&cpage=7#comment-390947

Daniel De Caluwé
November 5th, 2012 at 12:00 PM

@Steven N. Karels,

I think another mechanism is involved. I think the heavy isotopes loose neutrons.

Kind Regards,

Maar uiteraard kwam daar reactie op ;-)


http://www.journal-of-nuclear-physics.com/?p=748&cpage=7#comment-391443

Herb Gillis
November 5th, 2012 at 6:56 PM

Daniel:
If the heavy isotopes lose neutrons then that would show up in the isotope analysis of the spent fuel. Unless I misunderstood something I think A.R said that he did not see significant changes in the nickel isotope distribution. Is my understanding correct?
Regards; HRG.


http://www.journal-of-nuclear-physics.com/?p=748&cpage=7#comment-392222

Daniel De Caluwé
November 6th, 2012 at 6:28 AM

@Herb Gillis,

Yes, I agree, if the heavy isotopes lose neutrons, then that should show up in the isotope analysis of the spent fuel, that should be different from the fuel they started with. But as the fuel always has been enriched (to obtain more 62Ni en 64Ni than naturally occuring), and as some measurements indicated that the spent fuel showed more or less the natural concentrations of the isotopes, I presume that the heavy isotopes lose neutrons in the process. But this doesn’t mean that there’s a strong neutron-radiation/flux (like in ordinary fission reactions), because I think here it happens in a smooth way. And the protons of the loaded hydrogen then are, in a certain way, a catalysator for this…

But I don’t think that Dr. Rossi likes that we communicate a lot or in detail about this, because this probably or gradually/eventually could reveal part of the working of the E-cat.

Kind Regards.


http://www.journal-of-nuclear-physics.com/?p=748&cpage=7#comment-393192

Joe
November 6th, 2012 at 7:41 PM

Daniel De Caluwe,

How do you imagine that a proton could be a catalyst for neutron emission?

All the best,
Joe


http://www.journal-of-nuclear-physics.com/?p=748&cpage=7#comment-393993

Daniel De Caluwé
November 7th, 2012 at 7:59 AM

@Joe,

Pure speculative, and based on QRT, where the excess neutrons of the heavy isotopes (of 62Ni and 64Ni) are hypothised on the edge (and not in the center ;-) of the nucleus, and that they tend to pair with protons (also according to QRT), and combine this with an oscillating pressure wave (cfr. ‘frequencies’) of the loaded hydrogen… But just consider it as an idea coming out of brainstorming, and not as official science… ;-)

Because, if no significant amount of copper is formed, something else must happen, isn’t it? ;-)

(Maar aangezien A. Rossi in 2011 nog wél beweerde dat 'a significant amount of copper was formed', denk ik dat we hier een beetje op het verkeerde been gezet werden, en momenteel geloof ik wél dat er koper gevormd wordt, alleen wil A. Rossi daar niet veel over kwijt, immers dat beschouwt hij momenteel als één van zijn bedrijfsgeheimen.)


http://www.journal-of-nuclear-physics.com/?p=763&cpage=3#comment-410905

Daniel De Caluwé
November 18th, 2012 at 5:46 AM

@Koen Vandewalle,

You wrote: “The point is: too much money in the world, makes that people do no longer want to work for it. So money, or value is very important to make the world go round.”

I think we gradually should evolve to a society and a world with a world economy where there’s more co-operation, that should replace the deadly competition we have now.

At the moment, we live in a world with a world-economy that is too much based on deadly competition, and this makes slaves of most people, especially the lower classes, but also mid- and even ‘higher classes’. The world economy should SERVE the people on earth (and also elevate poor and underdevelloped countries, instead of exploit them), and not make us/them slaves in a inhumane system of deadly competition and economic war. Competition always is praised that it makes people work, and that ‘it makes the world go round’, but there’s also a lot of destruction/abuse/exploitation (of the lower classes/poor countries) in it, and exaggerated competition certainly will destroy us, if we remain on this path. In the aquarian age, humanity should learn to co-operate, otherwise we will destroy ourselves. We need a world-economy that is based on co-operation and NOT on deadly/exaggerated competition. We need a world economy that is eventually steered by the united nations in an attempt to let people/countries co-operate to produce what is necessary according to the needs of the people. G3, G4, G5, G7, G8, G20, Bilderberg, … and so on, should focus more on co-operation, so that people/countries are inspired to co-operate (and survive together instead of destroy each other), so that we get a better world with no war and peacefull co-operation in the end. (The other path is deadly competition to its extreme, which will make enslaved enemies of us all, who finally will destroy each other).

A year ago, during the ‘Arab Spring’ and the economic problems in the eurozone, I wrote a small text in English about it:

http://hetstilleverzet.blogspot.be/2011/06/to-young-people-of-spain-and-in-middle.html

And no, I’m not an ideologic marxist or communist, who reduced the freedom in their countries, but I’m just convinced that we need more co-operation (in the (world-)economic field) to survive with so many people on this beautifull planet Earth. (The present system of exaggerated and deadly competition also reduces the freedom of most people on this Earth).

And as a beginning, capital and labour should be more balanced, and have an equal voice in the steering of our world economy. So the ILO (‘The International Labour Organisation’, which is part of the United Nations) should have a stronger voice in the world economy and in the meetings of the G3, G4, G5, G7, G20, Bilderberg, … and so on. (Think, for instance, of our European ‘Rheinland-model’, with a balanced representation (equal number of representitives of unions (employees) and employers in the government of our companies).

By giving the ILO (‘The International Labour Organisation’, which is part of the United Nations) a stronger voice in the important world-wide meetings that steer the world-economy, I hope the world-economy eventually will become more humane and more co-operative, so that more people (and more countries) will get out of the present slavery and will enjoy more freedom. (But I repeat: this is not marxism nor communism, but just a more enlighted/balanced vision for this world and its economy).

Kind Regards,

Ir. Daniel De Caluwé, Belgium


http://www.journal-of-nuclear-physics.com/?p=763&cpage=3#comment-416082

Daniel De Caluwé
November 21st, 2012 at 12:50 AM

@ Bernie Koppenhofer,
@ Jouni Tuomela,

Hydrogen embrittlement (Wikipedia)

But the present explanation seems to be pressure build up:

The mechanism starts with lone hydrogen atoms diffusing through the metal. At high temperatures, the elevated solubility of hydrogen allows hydrogen to diffuse into the metal (or the hydrogen can diffuse in at a low temperature, assisted by a concentration gradient). When these hydrogen atoms re-combine in minuscule voids of the metal matrix to form hydrogen molecules, they create pressure from inside the cavity they are in. This pressure can increase to levels where the metal has reduced ductility and tensile strength up to the point where it cracks open (hydrogen induced cracking, or HIC). High-strength and low-alloy steels, nickel and titanium alloys are most susceptible.

1. But, if also LENR is involved, then Ti also is a candidate for Ti-H fusion? ;-)

2. Also in nuclear reactors, this phenomenom is welknown. At the moment, there even is a problem in some of our Belgian nuclear reactors I think… So, if LENR is involved, then, via these problems, they also got a key to a better way of producing energy? ;-)

3. As the E-cat works with hydrogen at high pressure, they have to choose the right metals?

Steel with an ultimate tensile strength of less than 1000 MPa or hardness of less than 30 HRC are not generally considered susceptible to hydrogen embrittlement. Jewett et al.[2] reports the results of tensile tests carried out on several structural metals under high-pressure molecular hydrogen environment. These tests have shown that austenitic stainless steels, aluminum (including alloys), copper (including alloys, e.g. beryllium copper) are not susceptible to hydrogen embrittlement along with few other metals.[3] For example of a severe embrittlement measured by Jewett, the elongation at failure of 17-4PH precipitation hardened stainless steel was measured to drop from 17% to only 1.7% when smooth specimens were exposed to high-pressure hydrogen.





En hier nog enkele van mijn berichten op andere fora over koude fusie/LENR en de E-cat :

Eerst mijn reactie op een bericht over James Patterson :

http://e-catsite.com/researchers/james-patterson/#comment-98

Daniel De Caluwé says:
October 24, 2011 at 18:21

Thank you for reporting this! The work of James Patterson, and the replication of his work by Dr. George Miley, indeed is very interesting, so that I also referred to it in my blog.

And the replication of his work by Dr. Miley, also proves that this certainly was not ‘junk science’!

Reading and (re)viewing this, I wonder why we still invest so much time, money and energy in very complex hot-fusion-research, while table-top-cold-fusion seems so simple and easy at hand?


En daarna twee van mijn reacties op het bericht Potential Benefits from Cold Fusion :

http://coldfusion3.com/blog/potential-benefits-from-cold-fusion#comment-170

Daniel De Caluwé says:
October 27, 2011 at 7:05 pm

Personally, I think the change will be slower and more gradual than we might think or expect at first. So, people working in the field of energy-production, should not be afraid of loosing their jobs immediately… The introduction of the E-cat into the market will be gradually.

Kind Regards,


http://coldfusion3.com/blog/potential-benefits-from-cold-fusion#comment-196

Daniel De Caluwé says:
November 23, 2011 at 10:10 am

Sorry for my late reaction, but I didn’t follow this thread for a long time.

To those who doubt and mention the generator:

1. The generator was necessary to drive the air-ventilators that were used to cool and condense the steam (during the test done with air to steam heat exchangers), and the energy consumption of these may NOT be taken into account of the 1 MW plant, because in normal operation, the steam is used by the client, and does not have to be cooled in that way. (The steam is used as process heat and/or to generate electricity (later))

2. To make up the energy balance in a correct way, read also my answer to Matteo on Rossi’s blog:

http://www.journal-of-nuclear-physics.com/?p=510&cpage=32#comment-117817

3. The test of the 1 MW plant was done (and organised) by Rossi’s customer, a company, firm or organisation that still does not want to disclose its name, but as you can see on Rossi’s blog, the same customer already ordered 13 additional plants:

http://www.journal-of-nuclear-physics.com/?p=510&cpage=33#comment-123165

(and in another message, Rossi confirmed that these additional plants were ordered by the same customer).

http://www.journal-of-nuclear-physics.com/?p=510&cpage=33#comment-123662

http://www.journal-of-nuclear-physics.com/?p=510&cpage=33#comment-123740

4. I verified also some of the previous tests, and personally, I’m convinced.

5. Rossi is not interested in further scientific tests, and here he gives the reasons why:

http://www.journal-of-nuclear-physics.com/?p=516&cpage=14#comment-126526

[quote="from Rossi's blog"]Andrea Rossi
November 21st, 2011 at 4:58 PM

‘Dear Ivan:
We are no more in the mode of public tests, the times of public tests are over. We are manufacturing plants for our Customers, and our Customers will test the plants they have bought.’

…[/quote]

http://www.journal-of-nuclear-physics.com/?p=516&cpage=14#comment-126869

6. Some months ago, I verified some of the previous tests, and they were very convincing to me, so I am and remain a strong believer! ;-)

Kind Regards,

Ir. Daniel De Caluwé
Belgium


En hier ook een snelle reactie op Cold Fusion Now - Past and Future :

http://coldfusionnow.org/?p=8796#comment-860

Daniel De Caluwé / Reply

Personally, I think the change will be slower and more gradual than we might think or expect to be. So, people working in the field of energy-production, for instance, should not be afraid of loosing their jobs immediately… I think the introduction of the E-cat into the market will be slowly and gradually.

Kind Regards,



En, last but not least, nog een paar recente berichten van Andrea Rossi over de laatste stand van zaken:

Lees daartoe eventueel eerst dit:

Rossi Talks of Breakthrough: Stability at ‘Very High’ Temperatures (E-Cat World)

En daarover hier nog enkele van zijn laatste berichten:

http://www.journal-of-nuclear-physics.com/?p=510&cpage=51#comment-227479

Andrea Rossi
May 4th, 2012 at 10:43 PM

Dear Fabio:
Thank you for your kind attention.
You are right: some puppett is trying to stop our work saying we cannot go on selling a producte that is not well known. Note: first they said my product was bullshit, now they have to change channel and try to raise the safety problem. But , as we always said, our product will hit the non military market only after due certification; besides, we know very well our product and how it works, so it is wrong to say we do not know how the E-Cat works. About the market: you are right again, People will vote with their money and decide if the E-Cat works or not.
By the way: we are working very hard on the temperatures, and we have reached a tremendous goal in the last week. We are making a test which endures since a week, that could make a revolution in the revolution. It will go ahead for a month.
For now, just working.

Warm Regards,
A.R.

http://www.journal-of-nuclear-physics.com/?p=510&cpage=51#comment-227683

Andrea Rossi
May 5th, 2012 at 8:38 AM

Dear Dr Joseph Fine:
I agree.
It’s Saturday, but today and tomorrow we will work 24 hours a day on the reactor we have made here in the USA: we have stabilized it at very high temperatures…and when I say very high I mean it. We understood the reason of the instability, so now the work is going on hard.
Warmest Regards,
A.R.


http://www.journal-of-nuclear-physics.com/?p=510&cpage=51#comment-227711

Francesco
May 5th, 2012 at 9:41 AM

Dear ing. Rossi
In its response to Dr. Joseph Fine you write:
“”we have stabilized it at very high temperatures”"
But … then the dream to produce steam for the traction is approaching?
F.T.

http://www.journal-of-nuclear-physics.com/?p=510&cpage=52#comment-227903

Andrea Rossi
May 5th, 2012 at 5:46 PM

Dear Francesco:
Yes. Sterling Engine.
Warm Regards,
A.R.


http://www.journal-of-nuclear-physics.com/?p=510&cpage=51#comment-227723

Robert Tanhaus
May 5th, 2012 at 10:17 AM

Dear Mr. Rossi,

Great!
Do you think that improvement could influence the 1st generation products?

BR
Robert Tanhaus

http://www.journal-of-nuclear-physics.com/?p=510&cpage=52#comment-227902

Andrea Rossi
May 5th, 2012 at 5:45 PM

Dear Robert Tanhaus: yes, for the industrial applications.
Warm Regards,
A.R.


http://www.journal-of-nuclear-physics.com/?p=510&cpage=51#comment-227751

Hergen
May 5th, 2012 at 11:47 AM

Dear Mr Rossi,

I follow your blog nearly daily since January 2011. Congrats to your new success. My question: Will it be possible to run the household e-cat with a higher temperature, too, in order to produce electricity?

Thank you for your answer in advance.

http://www.journal-of-nuclear-physics.com/?p=510&cpage=52#comment-227901

Andrea Rossi
May 5th, 2012 at 5:44 PM

Dear Hergen:
No, for the domestic the safety issues are still prohibitive, but for the industrial applications we made a gigantic step forward.
Warm Regards,
A.R.


http://www.journal-of-nuclear-physics.com/?p=510&cpage=51#comment-227796

Hergen
May 5th, 2012 at 1:37 PM

Dear Mr Rossi,

one further question i forgot to ask:

with your new achievement is it possible to increase the COP?

Thank You

http://www.journal-of-nuclear-physics.com/?p=510&cpage=52#comment-227890

Andrea Rossi
May 5th, 2012 at 5:31 PM

Dear Hergen:
No.
Warm Regards,
A.R.


http://www.journal-of-nuclear-physics.com/?p=510&cpage=51#comment-227817

P.G.Sharrow
May 5th, 2012 at 2:16 PM

Andrea Rossi: You say that you now understand the reason for the high temperature instability. This is wonderful news! And a reactor/amplifier may now be able to achieve quite high temperatures!

The little cat might become a Lion. Good news indeed. pg

http://www.journal-of-nuclear-physics.com/?p=510&cpage=52#comment-227889

Andrea Rossi
May 5th, 2012 at 5:30 PM

Dear P.G. Sharrow:
I must thank our military Customer, we are working with them and learning.
Warm Regards,
A.R.