Cold Fusion: The Cat is Finally out of the Box (French Article)

I missed this when it was published but today found a reference on the Journal of Nuclear Physics

Contrepoints.org, a popular French language self-described liberal website covering news and current affairs has published an article titled “Cold Fusion: the cat (E-Cat) is finally out of the box” by Jean-Pierre Cousty.

The article is largely positive in the coverage it gives of the Lugano test, and the author is enthusiastic regarding the future role that LENR will play in energy production.

Cousty writes:

We now have official proof that the future of global energy will not perhaps in the near future by wind or solar but the implementation of this cold fusion even if it takes decades. Several companies, such as Brillouin, are ready to market devices and futuristic dreams are good train as a Swiss company speaks to run cars continuously for months with a Stirling engine types. For example, problems of sea water processing freshwater have an off and inexpensive for Africa and countries in the developing solution …

Life of cold fusion will be able to resume its course with the 19th Conference Annual (ICCF-19) to be held this year in Padua in Italy April 13 to 17, 2015 with 500 participants from planned . . .

It is obvious that OPEC, EDF, AREVA etc. must adapt to the new conditions of cold fusion could quickly become commonplace. The coming years should be exciting and promising hopes in world energy.

Speaking of OPEC adapting, I have noticed Brent Oil futures prices dropped below $81 dollars a barrel today. I doubt that this is E-Cat related, but there are clearly macroeconomic pressures on OPEC, with non OPEC nations like the United States and Canada producing oil at levels not seen for 30 years. If LENR really does come online in the near future, it will surely be a further blow to the power of OPEC nations.

  • Gerard McEk

    Excellent,quite positive. The ball is starting to roll!

  • Gerard McEk

    Excellent,quite positive. The ball is starting to roll!

  • Buck

    Wow . . . very positive and openly accepting that LENR will be impacting the world and humankind’s use of energy.

    It is a surprise as the French are very focused on the Nuclear Power (Fission) industry.

    Another hole in the armor that prevents the world from accepting and changing.

    • french people are anyway under anti-nuclear propaganda,, as much as on climate and anti-cold fusion propaganda…

      This site is anyway a liberal/libertarian/freemarket site and have heretic position on many subjects. it is that third extreme political direction in france, the most taboo and undefendable, beside right and left which are all conservative and statist and well synthesized today by FN which aggregates both conservatism.

      The article is quite positive, but the comments are quite … standards. ie some enthusiast and most comfortably negative with “it is visibly a scam”…

      • Buck

        Thank you for the clarification.

  • Buck

    Wow . . . very positive and openly accepting that LENR will be impacting the world and humankind’s use of energy.

    It is a surprise as the French are very focused on the Nuclear Power (Fission) industry.

    Another hole in the armor that prevents the world from accepting and changing.

    • french people are anyway under anti-nuclear propaganda,, as much as on climate and anti-cold fusion propaganda…

      This site is anyway a liberal/libertarian/freemarket site, based on community articles, and have heretic position on many subjects. it is that third extreme political direction in france, the most taboo and undefendable, beside right and left which are all conservative and statist and well synthesized today by FN which aggregates both conservatism.

      The article is quite positive, but the comments are quite … standards. ie some enthusiast and most comfortably negative with “it is visibly a scam”…

      • Buck

        Thank you for the clarification.

  • bachcole

    Nice.

    But, how come I can read this translation of French but I can’t read Alain 1/2 of the time. Perhaps Alain should write in French and then use Google Translate? (:->)

    • Mark E Kitiman

      Are you sure about that? Others have found it to be l/sqrt(3) of the time!

  • jousterusa

    My article about E-Cat at the APEC summit tells the same story, but in much better English! 🙂 http://www.american-reporter/5,098/1.html and on CNN.

    • Joniale

      Hi, nice article. However, i dont see so positive that Obama has all the control. I want this technology to be used by all, to be spread all over the world and to crate a distributed energy generation system. As you see it is Obama who say when and how.
      Not good to use this tecnology to have politic power.

      • jousterusa

        Especially if you’re Russian! Read my response to Bachcole for more on that.

  • jousterusa

    My article about E-Cat at the APEC summit tells the same story, but in much better English! 🙂 http://www.american-reporter/5,098/1.html and on CNN. The CNN piece is at http://ireport.cnn.com/docs/DOC-1187686.

    • bachcole

      Joe, I think that you got too far ahead of the evidence on many of your points.

      Could you please explain why you said that the Chinese are manufacturing E-Cats when we have no evidence that they are? They have shown an interest in the E-Cat; this is a fact.

      Also, why did you say that THE professors are the ones handing out Nobel prizes?

      I know of NO ONE who thinks that Robert Godes is a third rate Rossi wannabe. Why did you say that. It does not help the cause of LENR.

      “President Obama may have lost the mid-term elections, but he has seized the ultimate power available to anyone in this world, and there does not exist a power on this earth than can take it from him.” We don’t know any of this.

      I am not proud of this your article.

      • jousterusa

        A few quick replies, and my thanks for your questions:

        1. One of the scientists who validated the E-Cat was chair of the Swedish Royal Academy of Sciences Chemistry Committee, which decides who will win the Nobel Prize. Another was on one of the two committees, Chemistry and Physics. The Nobel committees want to be sure they don’t get duped as MIT duped us into believing Fleischmann and Pons were frauds, so I believe that in their partial funding of the study, the Royal Swedish Academy was preparing the way for a Nobel Prize for Rossi, perhaps shared with the heirs of Pons and Fleischmann, knowing that only by doing so can the science of cold fusion be lifted beyond the reach of the powerful controversies and indefatigable skeptics.

        2. It was President Obama’s Executive Order in October 2013 on the cogeneration of heat and power that laid the foundation for bringing a successful E-Cat technology quickly into government acceptance, particularly by the DOE, which has been adamant in opposing it. As with all other such appointments, the director of the US Patent Office serves at the President’s pleasure. Mr. Obama can do anything he likes, at least compared to the rest of us, and as you have probably noted, he does. Now he’s got the world by the testicles – he can order a patent granted, he can deliver discretionary finds for development, he can conclude international treaties to ensure the propagation of this humanity-serving technology. He will do all that in good time.

        3. What has Robert Godes ever produced or demonstrated besides press releases? I believe he is full of crap, and a man just hoping to take financial advantage of new interest in cold fusion generated in reality exclusively by Rossi’s work. He memorized the TIP and started regurgitating it with his spin everywhere so he can get big money to back him – money that should go to Rossi. It’s a sucker’s play.

        • bachcole

          1. I fully accept ONE and I appreciate your educating me about it. This actually looks very promising.

          2. Just because Obama could know about the E-Cat does not mean that he does know about the E-Cat and does not mean that he believes it. I don’t see how co-generation has anything to do with it. I cannot give you a pass on this one, unless you can present more and better evidence.
          3. Robert Godes has build Brillioun. Mike McKubre obviously has faith in him. And Godes has is own theory. I guess I just assume the best of people unless proven otherwise. I think that you are being inappropriately harsh on him. In fact, I get the feeling that you are trying to build yourself up by tearing him down.

        • kemosabe

          jousterusa wrote> “1. One of the scientists who validated the E-Cat was chair of the Swedish Royal Academy of Sciences Chemistry Committee, which decides who will win the Nobel Prize.”

          Which one was that? Kullander, who is not an author, but did witness an ecat demo, and was sympathetic, was chair of the Energy committee, which does not award Nobel prizes.

          “Another was on one of the two committees, Chemistry and Physics. “

          Which one?

          jousterusa> “2. It was President Obama’s Executive Order in October 2013 on the cogeneration of heat and power that laid the foundation…”

          That’s reaching…

          jousterusa> “3. What has Robert Godes ever produced or demonstrated besides press releases? I believe he is full of crap, and a man just hoping to take financial advantage of new interest in cold fusion generated in reality exclusively by Rossi’s work. He memorized the TIP and started regurgitating it with his spin everywhere so he can get big money to back him – money that should go to Rossi.

          I also suspect Godes is exploiting the situation, but his spin and his company predate Rossi by several years, and he is not less credible than him.

        • ivanc

          I am waiting for an updated report, and you already declaring the ecat as real?
          The report has been challenged. the professors needs to answer the raised questions.

    • Joniale

      Hi, nice article. However, i dont see so positive that Obama has all the control. I want this technology to be used by all, to be spread all over the world and to crate a distributed energy generation system. As you see it is Obama who say when and how.
      Not good to use this tecnology to have politic power.

      • jousterusa

        Especially if you’re Russian! Read my response to Bachcole for more on that.

  • http://peswiki.com/index.php/Free_Energy_Blog:2014:11:12#SHT_Booth_at_Defense_Energy_Conference – see picture

    Solar Hydrogen Trends is making headway as well, as they received an invitation to install a booth at the Defense Energy Summit and Innovation Showcase.

    http://defenseenergy.com/

    I would guess that the Defense Department did a security check on the founders of Solar Hydrogen Trends and found, as Sterling Allan did with his own sleuthing, that they all have good to outstanding reputations. The Defense Department has their own intelligence organization plus can get information from the NSA, CIA, FBI, and the immigration department. They must have found no red flags. Solar Hydrogen Trends has a finance guy who is a Berkeley graduate. I doubt he would get involved in anything as quickly suicidal as a hoax. Their top scientist, as has been discussed earlier, is a star scientist, so unless they all have big gambling debts to the Russian mafia (a joke), they seem very serious and credible, with at least two verified independent tests already conducted.

    • bitplayer

      Unfortunately, I’m not seeing any indications that the Defense Energy Summit is in any way officially affiliated with the US Department of Defense (DOD). It looks like an industry group that is focused on selling energy technology to the DOD. Naturally people with DOD would attend. So there’s no indication of DOD vetting of SHT, so no basis to think that this lends any greater credibility to SHT. And nothing on the SHT website, or any SHT-related reference links, that provides any confidence about what they are doing.

      They really ought to change their name, too.

      • Alain Samoun

        I think that the name is right 😉

        • US_Citizen71

          Not to be confused with the grammar police, but the english rule is replace a with an when the following word begins with a vowel. I think it is the english way of getting back at the latin based languages for their masculine and feminine rules. : )

  • http://peswiki.com/index.php/Free_Energy_Blog:2014:11:12#SHT_Booth_at_Defense_Energy_Conference – see picture

    Solar Hydrogen Trends is making headway as well, as they received an invitation to install a booth at the Defense Energy Summit and Innovation Showcase.

    http://defenseenergy.com/

    I would guess that the Defense Department did a security check on the founders of Solar Hydrogen Trends and found, as Sterling Allan did with his own sleuthing, that they all have good to outstanding reputations. The Defense Department has their own intelligence organization plus can get information from the NSA, CIA, FBI, and the immigration department. They must have found no red flags. Solar Hydrogen Trends has a finance guy who is a Berkeley graduate. I doubt he would get involved in anything as quickly suicidal as a hoax. Their top scientist, as has been discussed earlier, is a star scientist, so unless they all have big gambling debts to the Russian mafia (a joke), they seem very serious and credible, with at least two verified independent tests already conducted.

    • bachcole

      This is freaking awesome. But I need much more certain (certain for me) evidence before I am going to believe such an extraordinary claim. I need Frank Acland to say that he saw it and tested it and is convinced.

      • I would not invest in the company unless I did my own tests and knew how it works. I do not think anyone is going to give them serious money to start mass production unless they have their own exhaustive tests. So how are they going to scam anyone? They are not selling stock to the general public. They say they are willing and eager for others to test their device. It makes no sense to me as a scam. So, if it is not a scam, what is it? If it is real, oil will quickly become obsolete. The *too good to be true* factor weighs on my mind, but lots of things these days are too good to be true, like cheap giant screen televisions and computers built into glasses. We are use to them now. Maybe in the future we will take dirt cheap energy for granted.

        • bachcole

          I am perfectly OK being completely confused by this, without taking sides or making a decision. I don’t have to make up my mind. I can just wait and watch open-mindedly and hope for the best. And the best in this case will be WOW!!!

        • bkrharold

          My impression is that this is not a scam, but a nuclear process which has yet to be characterized and explained by science. The ‘too good to be true’ factor is that without knowing the exact mechanism, exploiting it will be a hit and miss affair by trial and error. Once the basic science has been done and we understand how it works, it will replace all other forms of energy.

      • Ophelia Rump

        If they can produce the volumes which they claim they need to start selling the gas at what amounts to below cost for competitors to produce the same gas, and in doing so win the market themselves, then they will both turn hefty profits and earn confidence in their technology if they can do that.

        • bachcole

          I am completely unsure why that would be a problem. I would be doing that, I guarantee it.

  • bitplayer

    X number of communications reach Y number of people, of whom Z are active detractors and W are willing to look further, potentially generating a larger audience that would draw more communications.

    Depending on the values in the equation, this could create a positive feedback loop in number of people reached.

    And all I can say is…thank you Frank!

  • ivanc

    I see Rossi and the Lugano report as a threat to LENR credibility, We want LENR to be a science, is not a football game in which you take a party and believe or not.
    is not a religious faith in which you believe with all the force of your soul, and make true thing you do not see. (which I agree and accept as is impossible to try to probe or disprove the existence of a supreme being).
    but this is science, so the scientific method of trial and error, skepticism and test, has to be applied.
    The ideas and methods has to be challenged, and if stands the logic and experiment proves the phenomena is real then it becomes true.
    At this point we challenging the Lugano report,
    we found out the IL/sqrt(3) factor should have used instead of IL/2.
    The Power reported fits with a change of phase or incorrect positioning of clamp.
    A rare drop in resistance of a 3.3 factor has happened.
    Elements like FE and Calcium has evaporated at too low temperature.
    This are valid questions and not fanatical criticism.
    And we will like Professor Levi and team to address the issues.

    • Dods

      The science was there and when shown they chose not to see. Faked results and buried its validity. So with religious faith as you so well put it is what the scientific community have committed themselves. This sadly has put it in the situation it is today and its the best were going to get moving forward.

    • Axil Axil

      The professional practitioners of science today inspire a emergent secular theology. The major tenant of this theology is human banality. We lead as innocent children impressed by the eminence of our elders and embrace this doctrine, we have reached the point where this belief can undermine not only our own existence, but also holds as possible the destruction of all life on this planet. In common with all of our past theologies, this new secular theology imposes a doctrine of insignificance and shared joylessness in an attempt to keep us in our place. Scorn and ridicule awaits the backslidden unbeliever dispensed aplenty by the apologists of this new religion.

      One of the sticking point that lodges deeply in the gullets of “real” science is that LENR is just too perfect to be believed. They are wrong. In point of fact, it is beyond too perfect, it is absolutely perfect. The corruption of the mind that is our legacy inherited from the mindless primitive from which we evolved rebels against the concept of such perfection. Such perfection cannot exist in this life. Such perfection can only exist and be truly enjoyed in the next. From the pride and prejudice born deep within that primordial dark place, mankind does not deserve to drink fully this sweet ambrosia of the immortals.

      LENR goes way beyond a great way to produce energy, it is a doorway to a new science whose implications when fully appreciated and developed will lift mankind up to trod upon brave new worlds spread like dust before eternity. A door for humankind will open to savor the power and the prerogatives of the gods. When man is wise enough to step through this doorway past the impossible that LENR lays open into timeless and unending existence, mankind will spread like a rising tide throughout the universe.

      This perfection of LENR is its own threat to its credibility and its science is here 1000 years before its proper time. What aborigines from the dawn of our past corruption would rightly understand the wonders of our present civilization without quaking with fear at the reality of such wonders? The science that LENR will reveal and the future that it portends it just too awesome to contemplate.

      Carl Sagan explained the emotion behind our current science and cosmology when he wrote Pale Blue Dot: A Vision of the Human Future in Space. Sagan played for high stakes in this attempt to “de-deify” our entire species. His beautiful, secular psalm dedicated to our demotion is unsurpassed. In Psalm 8, King David described us as only a little lower than the angels while in Pale Blue Dot, Sagan takes great pains to obliterate any sense of cosmic significance.

      Sagan says of that picture taken by a spacecraft from a viewpoint far out in space: “We succeeded in taking that picture and, if you look at it, you see a dot. That’s here. That’s home. That’s us. On it everyone you know, everyone you love, everyone you’ve ever heard of, every human being who ever was, lived out their lives. The aggregate of all our joys and sufferings, thousands of confident religions, ideologies and economic doctrines. Every hunter and forager, every hero and coward, every creator and destroyer of civilizations, every king and peasant, every young couple in love, every hopeful child, every mother and father, every inventor and explorer, every teacher of morals, every corrupt politician, every superstar, every supreme leader, every saint and sinner in the history of our species, lived there–on a mote of dust suspended in a sunbeam”.

      With the help of LENR, it’s ours to loosen these earthly bonds, to ascend beyond this mortal coil, this claustrophobic view of human existence is about to change.

      • bachcole

        I guess Axil Axil is in rapture. (:->)

        I don’t need LENR to know that Sagan had simply taken the objective perspective much too far. GREAT things happen here on Earth, things that are centrally important to existence itself. I am almost completely healed. I recently helped someone by graciously expressing several loving gestures. A mother in a village in Malaysia is holding and loving her baby and singing to her. A father somewhere is forgiving his son for lying. A son somewhere is deciding to not ever lie. People are going to church in Tennessee tonight in order to sing praises to God. A politician is deciding to tell the truth about certain improprieties. An angry person has decided that perhaps his anger is the problem. A boy is shovelling snow for his neighbor. A car thief in Turkey is deciding that stealing a car does not deserve being in a Turkish jail and is deciding to escape. A criminal has told his boss that he will commit the crime, but killing someone just isn’t going to be part of the operation. A soldier somewhere is risking his life for his comrades. A scientist has decided to risk censure and is going to study cold fusion. I reach down and pour some love into my sweet doggie. Someone tosses some seeds out their back door for the birds, because it is so cold here in the Rockies. Al Gore calls his daughter just to say “hi” and see how she is doing. A witness expresses the courage to step forward and report a crime.

        I could go on and on, 7,000,000,000 times for every second of our waking day.

        Objectivity is great and appropriate for objects, like the Moon, bridges, atomic bombs, and stripped toothpaste. But is it worse than useless for subjective things.

        Sincerely,

        Roger Bird, aka bachcole

        • Brokeeper

          I couldn’t have said it any better. In tribute to your humanitarian prose I just fed the birds. Thank you Roger.

      • Billy Jackson

        This is the possibility that has me hooked on the technology. I understand the implications for cheaper energy, cars, heating, electric and so on.. but the possibilities of space flight via constant acceleration has me absolutely astounded that more scientists are not jumping on this.

      • NT

        YES!

      • Mark Szl

        With abundance which LENR will definately help provide, with automation and artificial intelligence that will definiately be able to do anything we do better, with the consequently increasing amount and degree of human unemployability across all domains … what will become of the human race?

        • Fibber McGourlick

          While you’re clouding about the glorious future, don’t forget that much of humanity consists of primitive ignorant people, many of them anti-science murderers armed with Kalashnikovs.

          • Mark Szl

            So you want to get rid of them? In any case there are many that are not and this advance effects them. This is about all humans. So you are not so special either.

            I repeat, what will humans do?

          • US_Citizen71

            Eat, sleep and reproduce. Beyond that anything they want. If energy becomes a non-issue as well as food, materials, etc… man will have time to explore philosophy and the arts. I’m not convinced that an AI will necessarily have that creative spark. Creativity is partially the result of misfires of nerve cells connecting random unrelated things. There is a reason that some of the most creative people on the world are semi to completely unstable. I’m not sure how you program logic to duplicate that effectively. Man will likely still give creative input he just might not be the one finishing the project.

          • Mark Szl

            Bots are already doing creativity. Making music and writing articled. Humans cannot tell the difference between these and ones by humans. Youtube “Humans need not apply”

            Aristotle had the vision of a society where humans do not have to work. So did the creator of Star Trek.

            If we have to compete with Robots then i think we will have to merge with technology and become more and more like robots. Homo sapiens will get teplaced by homo cyborg.

        • bachcole

          I do not fear. Necessity will force this future that you see to never exist. For example, manufacturers will find that they can’t get a sufficient price for their A.I.-LENR manufactured goods because no one will be making any money to pay for them.

    • Gerrit

      Ivanc, if science would take this seriously they could simply start with replicating the numerous reports of successful experiments showing excess heat and transmutations and work their way towards fully understanding the phenomenon which would enable increased output energies.

      There are plenty of opportunities for science to apply the scientific method of trial and error, skepticism and test in the field of LENR. There is absolutely no reason for an entrepreneur / inventor to supply science with his knowledge so that science has something to work on.

      You are barking up the wrong tree.

  • Billy Jackson

    Brick by brick we see the report dismantle the resistance of ignorance and skepticism.

    • GreenWin

      Are we talkin’ reclaimed brownsfield bricks here Billy??

  • Dods

    The science was there and when shown they chose not to see. Faked results and buried its validity. So with religious faith as you so well put it is what the scientific community have committed themselves. This sadly has put it in the situation it is today and its the best were going to get moving forward.

  • Ophelia Rump

    It sounds like E-CatWorld.com content is beginning to drive world news.

    • LilyLover

      About time! Frank is my Glenn Greenwald of Science life and things that matter. If any of us have any influence in the world of ‘awards for brave journalism’, can we please nominate Frank?

      • GreenWin

        Yes Lily. We can and should. Because real journalism is all about finding a nascent spark and following to see what it causes. The “catalyst” metaphors are accurate. Rossi and Team have not just changed our understanding of physics, they have re-enabled the study of meta-physics, which teaches we are as we see ourselves.

  • I would not invest in the company unless I did my own tests and knew how it works. I do not think anyone is going to give them serious money to start mass production unless they have their own exhaustive tests. So how are they going to scam anyone? They are not selling stock to the general public. They say they are willing and eager for others to test their device. It makes no sense to me as a scam. So, if it is not a scam, what is it? If it is real, oil will quickly become obsolete. The *too good to be true* factor weighs on my mind, but lots of things these days are too good to be true, like cheap giant screen televisions and computers built into glasses. We are use to them now. Maybe in the future we will take dirt cheap energy for granted.

    • bitplayer

      I would like to believe this is true. After all, chlorophyll works. So it seems likely that there are other combinations of techniques that will produce hydrogen from water. And maybe if they can snake the electrons efficiently into the right spots, it would not take too much input power.

      http://en.wikipedia.org/wiki/Photosynthetic_efficiency

      When SHT first came into view I believe there were some thoughts that they were converting oxygen to hydrogen, however, I can’t find anything anything that implies that. Their “independent reviews” don’t have enough information to be conclusive one way or the other. And such a claim would be a show stopper.

      One concern is that their principals, from a preliminary review, have a remarkably small internet footprint. Their investment manager is the only one that has much presence:

      https://www.linkedin.com/in/kirillgichunts

      I know startups like to be stealthy, however, this is very muddy picture for me.

  • Ophelia Rump

    If they can produce the volumes which they claim they need to start selling the gas at what amounts to below cost for competitors to produce the same gas, and in doing so win the market themselves, then they will both turn hefty profits and earn confidence in their technology if they can do that.

  • Axil Axil

    One of the sticking point that lodges deeply in the gullets of
    “real” science is that LENR is just too perfect to be
    believed. They are wrong. In point of fact, it is beyond too perfect,
    it is absolutely perfect. The corruption of the mind that is our
    legacy inherited from the mindless primitive from which we evolved
    rebels against the concept of such perfection. Such perfection cannot
    exist in this life. Such perfection can only exist and be truly
    enjoyed in the next. From the pride and prejudice born deep within
    that primordial dark place, mankind does not deserve to drink fully
    this sweet ambrosia of the immortals.

    LENR goes way beyond a great way to produce energy, it is a
    doorway to a new science whose implications when fully appreciated
    and developed will lift mankind up to trod upon brave new worlds
    spread like dust before eternity. A door for humankind will open to
    savor the power and the prerogatives of the gods. When man is wise
    enough to step through this doorway past the impossible that LENR
    lays open into timeless and unending existence, mankind will spread
    like a rising tide throughout the universe.

    This perfection of LENR is its own threat to its credibility and
    its science is here 1000 years before its proper time. What
    aborigines from the dawn of our past corruption would rightly
    understand the wonders of our present civilization without quaking
    with fear at the reality of such wonders? The science that LENR will
    reveal and the future that it portends it just too awesome to
    contemplate.

    Carl Sagan explained the emotion behind our current science and
    cosmology when he wrote Pale Blue Dot: A Vision of the Human Future
    in Space. Sagan played for high stakes in this attempt to “de-deify”
    our entire species. His beautiful, secular psalm dedicated to our
    demotion is unsurpassed. In Psalm 8, King David described us as only
    a little lower than the angels while in Pale Blue Dot, Sagan takes
    great pains to obliterate any sense of cosmic significance:

    Sagan says of that picture taken from by a spacecraft from a
    viewpoint far out in space: “We succeeded in taking that picture
    and, if you look at it, you see a dot. That’s here. That’s home.
    That’s us. On it everyone you know, everyone you love, everyone
    you’ve ever heard of, every human being who ever was, lived out their
    lives. The aggregate of all our joys and sufferings, thousands of
    confident religions, ideologies and economic doctrines. Every hunter
    and forager, every hero and coward, every creator and destroyer of
    civilizations, every king and peasant, every young couple in love,
    every hopeful child, every mother and father, every inventor and
    explorer, every teacher of morals, every corrupt politician, every
    superstar, every supreme leader, every saint and sinner in the
    history of our species, lived there–on a mote of dust suspended in a
    sunbeam”.

    With the help of LENR, this view claustrophobic view of human
    existence is about to change.

    • Billy Jackson

      This is the possibility that has me hooked on the technology. I understand the implications for cheaper energy, cars, heating, electric and so on.. but the possibilities of space flight via constant acceleration has me absolutely astounded that more scientists are not jumping on this.

    • NT

      YES!

  • Paul Smith

    From that article:
    “…And China? They are not slow people. They saw the cold fusion device called the Energy Catalyzer coming, and as they have done with everything else you can possibly imagine, they have started manufacturing it under license from an American company callled Industrial Heat LLC of Raleigh, N.C….”

    So, it seems that China has already started manufacturing the E-Cats under license of IH.
    If true, this really is a great news.

    • hempenearth

      This has not been confirmed by a second source or preferably third source.

  • Paul Smith

    From that article:
    “…And China? They are not slow people. They saw the cold fusion device called the Energy Catalyzer coming, and as they have done with everything else you can possibly imagine, they have started manufacturing it under license from an American company callled Industrial Heat LLC of Raleigh, N.C….”

    So, it seems that China has already started manufacturing the E-Cats under license of IH.
    If true, this really is a great news.

    • hempenearth

      This has not been confirmed by a second source or preferably third source.

    • bachcole

      Paul, everyone including my own son know that I am a dye-in-the-wool believer in the LENR+ and the E-Cat, and I have been following this story with a freaking microscope for 37 months, and as far as I know, there is no evidence that the Chinese are manufacturing E-Cats or that they are even contemplating it. It is certain that they have shown an interest in it and have met with Darden. Other than that, here at e-catworld it is pure speculation whether they are manufacturing E-Cats or what. I consider this to be irresponsible on Joe’s part. The skeptopaths are going to eat him alive for this.

      It seems probable that at some point, either past or future, the Chinese will start manufacturing the E-Cat or parts of the E-Cat. But we have no evidence so far.

  • Mark E Kitiman

    Are you sure about that? Others have found it to be l/sqrt(3) of the time!

  • Alain Samoun

    Good article as was a previous article in 2012,but the online journal is kind of confidential for the french public.

  • Alain Samoun

    Good article as was a previous article in 2012,but the online journal is kind of confidential for the french public.

  • ivanc

    No one have answered my questions and you keep going with retorical arguments and idealistic positions.
    It is not a matter of beliving but a matter os knowing….

    • Ophelia Rump

      Perhaps we are tired of arguing, because it is pointless.
      If you are not open to the possibility no one can prove to you the reality.
      If you are open to the possibility then there is sufficient evidence of probability.
      If you are open to the probability, then you look forward to the reality, and see the early works as real, valid, and necessary steps toward it.

      Enlightenment is a process which each individual must undertake for themselves,
      be the result positive or negative outcome.

      The days of testing are over, so the days of proving are over. The days of marketing have begun.

    • US_Citizen71

      Do you believe the measurement from the wall to the control box was incorrect?

      If no, then your “IL/sqrt(3) factor should have used instead of IL/2” issue is nothing but minutiae.

      If you believe it was incorrect why do you believe it was incorrect?

      Let’s start there.

      • ivanc
        • US_Citizen71

          You didn’t answer my question.

          • ivanc

            You did not read the link, the answer is there., your question why to use:
            IL/sqrt(3) to divide the currents.
            also see my post of D to Y transformation.
            If you use the Y Transformation then IL = IYresistance.
            so know we only have to workout the equivalent resistance.
            This equivalant resistance is independent of all factor mentioned in your arguments.
            Then if you workout the power for D and Y. you find the same value if you use the 1l/sqrt(3) what demostrates my argument is true.

      • I don’t understand the debate…

        The power-meter measure the real power transmitted… from the wall socket, and a second power-meter to the reactor…

        sqrt(3) is only valid with sinusoidal if balanced… the power-meter have it right facing any of those rules of thumbs….

        as said on other comments, it seems clear that :

        – the current is pulsed sinusoidal from triac switch

        – the policy of switching is unknown

        – tric can switch on anytime, but have to switch off at zero current/voltage

        – it seems temperature stabilized by a feedback loop

        – some data available seems to show two pais of pulses, and not 3 as expected if the switch was switching in balanced mode

        – the paragraph on current estimation for loses let think that I3 is twice I1=I2… not knowing what they mean by twice, or by average current..

        the most probable is (taken from someone else comment and checked myself) is that the controller inject one pulse (until U1=U3) from C1 to C3 , 60 degrees later one from C2 to C3 wait 120 degrees, then negative pulse from C1 to C3, then 60 degrees later negative from C2 to C3, then 120 degrees silence and again…

        the average current discussed that respect I1=I2=I3/2 is the average absolute value. the power is twice . the effective current I3rms = sqrt(2)I1rms=sqrt(2)I2rms

        remains the evidence that the resistance of the reactor seems divided by 3 after kelvin temperature double from 720K to 1520K… it is not at all classical classic alloys netagive temperature coefficient , but as said Rossi, it is a proprietary alloy, doped metal, having proprietary properties.

        I have seen reports of such resistance excursion in some metal alloys with much less temperature change, when the lattice order change.
        Since we have LENR reactor and since Rossi discussed of a “Mouse” LENR excitator, and since we have here as only excitator this doped alloy, I can propose that the doped alloy is the mouse, and this it is thus not a heating coil but a LENR device.

        Since Type II superconduction, electromigration, proton conduction, resistance change, have been observed in LENR experiments, there is a logical equivalence to assert that LENR is impossible and that the Mouse doped alloy cannot have changed so much of resistivity.

        • US_Citizen71

          “Figure 5.The PCE display downstream from the control unit. On the left, one can see the current’s waveshape (identical in both PCEs), on the right its harmonics analysis.” – http://www.elforsk.se/Global/Omv%C3%A4rld_system/filer/LuganoReportSubmit.pdf

          I believe what is being shown is the plot of 1 cycle of one phase. The wave form is being clipped in the center (peak) and possibly at the slope up and slope down ends of each half cycle, so what is plotted in a portion of the rise and a portion of the fall of the power waveform on each half cycle.

        • US_Citizen71

          I think that if all three phases were plotted together the clipping would result in a signal that essentially is six times the frequency of the power in. As one positive pulse would zero as a negative pulse from another phase would begin to increase.

        • ivanc

          Havent you seen my Post, the D to Y transformation is independent of triggering, pulse, etc, and the 1/sqrt(3) applies to all waves and forms. because all waves are the sum of harmonics of sin waves. this is way you able to use RMS meters, otherwise you need different meter for each wave form.
          Do see the data from the wall? and the data after the wall?, then tell me how much power was lost in the controller. I am talking about data that could be turn into equations to solve the circuit.

      • ivanc

        I Have found the irrefutable argument why IL/sqrt(3) should be used.

        please read carefully.

        Any Delta circuit has a Y equivalent .
        if you have a Y then Iresistor=Iline.

        so you know have to see how to transform the resistances for the Y equivalente of a delta.
        D to Y

        http://en.wikipedia.org/wiki/Y-%CE%94_transform

        for one R1, see the article for more detail

        R1=(rb rc)/(ra+rb+rc)

        now lets suppose all r abc=1 (the 3 r in the delta are 1 ohm)

        power in each case D o Y is 3 times power in individual r .

        R1=1/3

        v=R IL , P=R IL^2, P=1/3 IL^2

        now use the same values for the delta. r=1

        P=(IL/sqrt(3))^2 * 1 = 1/3 IL^2

        OHHHH!, why both results are equal.

        Now the Delta to Y transformation is independent of the current or voltage.

        or phase….. after this there is no more argument.

        • US_Citizen71

          So obviously you agree that there it no problem with the power measurement by the PCE-830 on the power in from the wall. The minutiae you continue to ramble on about is the equation used to determine the current in any of the legs (C1 on the report diagram) which is stated as 19.7A. The C1 legs never have more than two phases flowing them, the third phase flows through the other two legs. The splits C1a and C1b (C2 on the report diagram) never have more than one phase flowing through them. If current flowing in each phase is equal then the current flowing through C1a or C1b would be half of the current in the C1 leg being split.

    • Obvious

      The reactor can use ~50 A, and produce 3000 W of heat from electricity.
      But if it is only turned on 30% of the time, (duty cycle), then
      the average measured W/h is only 900.
      Not only is this a good solution, it is almost certainly the only solution.
      The COP = self-sustain time/ input time.
      Note how the time cancels. 😉

  • Daniel Maris

    It’s not a matter of believing or knowing, it’s a matter of following with interest.

  • bitplayer

    “I’ll have what he’s having.” 🙂

  • pelgrim108

    Positive story about 2nd third party test on Hollands 4th biggest infotainment opinion website. Dated 10-13-14
    http://www.geenstijl.nl/mt/archieven/2014/10/omg_cold_fusion_via_ecat_welli.html

    Translation to English by me:

    Cold Fusion via E-Cat probably still possible

    Unlimited renewable energy that hardly needs fuel. No waste. A nuclear reaction without radiation. Get a bottle of Spa Blue, fill it with fuel and it will fuel your car for the rest of your life. Too good to be true? Not according to Andrea Rossi, whose E-Cat (Energy Catalyzer) http://www.extremetech.com/extreme/191754-cold-fusion-reactor-verified-by-third-party-researchers-seems-to-have-1-million-times-the-energy-density-of-gasoline shows a process of cold fusion is taking place, meeting all the above requirements.
    At least, thats what he says. He has been ridiculed for it for years by the scientific community. Cold fusion would mean that all our energy and environmental problems have been resolved and thus we enter the age of the Energy Revolution. So it is not true. Several critical blogs shot holes in research into Rossi’s Cold Fusion. Like this one from 2013. http://scienceblogs.com/startswithabang/2013/05/21/the-e-cat-is-back-and-people-are-still-falling-for-it/
    However recently a carefull scientific report (PDF) http://www.sifferkoll.se/sifferkoll/wp-content/uploads/2014/10/LuganoReportSubmit.pdf has been published which surely hints to the notion that Andrea Rossi’s E-Cat possibly is something very revolutionary. In any case, its something mysterious in terms of the ridiculous amount of energy generation.
    Ah well, think of it this way, Copernicus was also ridiculed for his statements at the time. Who knows, maybe Andrea Rossi has saved the planet. If it is true then we will experience it in the coming years.
    The fact remains that there is no need to worry about the environment or fossil fuels because there will eventually be a technological revolution which will again help humanity along. Will it be Rossi’s cold fusion, thats the question.
    Maybe the Swedish documentary https://www.youtube.com/watch?v=LHXc7NNMiWo provides something down here. Anyway, knowing humanity, we will always invent something.
    Everything will be fine.

    • Alain Samoun

      Pelgrim you’re sure an optimist! I don’t think that the coming years are going to be so rosy,but if you are looking at the long term I hope that you are right!

      • pelgrim108

        Geenstijl wrote the article. Its hollands biggest infotainment website. Only three regular mainstream press websites are bigger in Holland.
        Now for my own opinion:
        – avoid the installation of a totalitarian world government by the super-elite / bankowners
        – avoid being hit by an asteroid
        – avoid being damaged to much by a solar flare
        – avoid being propagandized into supporting another war against people that are not that much different from us, while the real reason is that the leadership will not join the globalist robber bankerclub.
        Then humanity will be fine.

        • Alain Samoun

          Your opinion seems more reasonable 😉

  • pelgrim108

    Positive story about 2nd third party test on Hollands 4th biggest infotainment opinion website.
    Dated 10-13-14
    http://www.geenstijl.nl/mt/archieven/2014/10/omg_cold_fusion_via_ecat_welli.html

    Translation to English by me:

    Cold Fusion via E-Cat probably still possible

    Unlimited renewable energy that hardly needs fuel. No waste. A nuclear reaction without radiation. Get a bottle of Spa Blue, fill it with fuel and it will fuel your car for the rest of your life. Too good to be true? Not according to Andrea Rossi, whose E-Cat (Energy Catalyzer) http://www.extremetech.com/extreme/191754-cold-fusion-reactor-verified-by-third-party-researchers-seems-to-have-1-million-times-the-energy-density-of-gasoline shows a process of cold fusion is taking place, meeting all the above requirements.
    At least, thats what he says. He has been ridiculed for it for years by the scientific community. Cold fusion would mean that all our energy and environmental problems have been resolved and thus we enter the age of the Energy Revolution. So it is not true. Several critical blogs shot holes in research into Rossi’s Cold Fusion. Like this one from 2013. http://scienceblogs.com/startswithabang/2013/05/21/the-e-cat-is-back-and-people-are-still-falling-for-it/
    However recently a carefull scientific report (PDF) http://www.sifferkoll.se/sifferkoll/wp-content/uploads/2014/10/LuganoReportSubmit.pdf has been published which surely hints to the notion that Andrea Rossi’s E-Cat possibly is something very revolutionary. In any case, its something mysterious in terms of the ridiculous amount of energy generation.
    Ah well, think of it this way, Copernicus was also ridiculed for his statements at the time. Who knows, maybe Andrea Rossi has saved the planet. If it is true then we will experience it in the coming years.
    The fact remains that there is no need to worry about the environment or fossil fuels because there will eventually be a technological revolution which will again help humanity along. Will it be Rossi’s cold fusion, thats the question.
    Maybe the Swedish documentary https://www.youtube.com/watch?v=LHXc7NNMiWo provides something down here. Anyway, knowing humanity, we will always invent something.
    Everything will be fine.

    • Alain Samoun

      Pelgrim you’re sure an optimist! I don’t think that the coming years are going to be so rosy,but if you are looking at the long term I hope that you are right!

      • pelgrim108

        Geenstijl wrote the article. Its hollands biggest infotainment website. Only three regular mainstream press websites are bigger in Holland.
        Now for my own opinion:
        – avoid the installation of a totalitarian world government by the super-elite / bankowners
        – avoid being hit by an asteroid
        – avoid being damaged to much by a solar flare
        – avoid being propagandized into supporting another war against people that are not that much different from us, while the real reason is that the leadership will not join the globalist robber bankerclub.
        Then humanity will be fine.

        • Alain Samoun

          Your opinion seems more reasonable 😉

  • US_Citizen71

    Do you believe the measurement from the wall to the control box was incorrect?

    If not then your “IL/sqrt(3) factor should have used instead of IL/2” issue is nothing but minutiae.

    If you believe it was incorrect why do you believe it was incorrect?

    Let’s start there.

    • I don’t understand the debate…

      The power-meter measure the real power transmitted… from the wall socket, and a second power-meter to the reactor…

      sqrt(3) is only valid with sinusoidal if balanced… the power-meter have it right facing any of those rules of thumbs….

      as said on other comments, it seems clear that :

      – the current is pulsed sinusoidal from triac switch

      – the policy of switching is unknown

      – tric can switch on anytime, but have to switch off at zero current/voltage

      – it seems temperature stabilized by a feedback loop

      – some data available seems to show two pais of pulses, and not 3 as expected if the switch was switching in balanced mode

      – the paragraph on current estimation for loses let think that I3 is twice I1=I2… not knowing what they mean by twice, or by average current..

      the most probable is (taken from someone else comment and checked myself) is that the controller inject one pulse (until U1=U3) from C1 to C3 , 60 degrees later one from C2 to C3 wait 120 degrees, then negative pulse from C1 to C3, then 60 degrees later negative from C2 to C3, then 120 degrees silence and again…

      the average current discussed that respect I1=I2=I3/2 is the average absolute value. the power is twice . the effective current I3rms = sqrt(2)I1rms=sqrt(2)I2rms

      remains the evidence that the resistance of the reactor seems divided by 3 after kelvin temperature double from 720K to 1520K… it is not at all classical classic alloys netagive temperature coefficient , but as said Rossi, it is a proprietary alloy, doped metal, having proprietary properties.

      I have seen reports of such resistance excursion in some metal alloys with much less temperature change, when the lattice order change.
      Since we have LENR reactor and since Rossi discussed of a “Mouse” LENR excitator, and since we have here as only excitator this doped alloy, I can propose that the doped alloy is the mouse, and this it is thus not a heating coil but a LENR device.

      Since Type II superconduction, electromigration, proton conduction, resistance change, have been observed in LENR experiments, there is a logical equivalence to assert that LENR is impossible and that the Mouse doped alloy cannot have changed so much of resistivity.

      • US_Citizen71

        “Figure 5.The PCE display downstream from the control unit. On the left, one can see the current’s waveshape (identical in both PCEs), on the right its harmonics analysis.” – http://www.elforsk.se/Global/Omv%C3%A4rld_system/filer/LuganoReportSubmit.pdf

        I believe what is being shown is the plot of 1 cycle of one phase. The wave form is being clipped in the center (peak) and possibly at the slope up and slop down ends of each half cycle, so what is plotted in a portion of the rise and a portion of the fall of the power waveform on each half cycle.

      • US_Citizen71

        I think that if all three phases were plotted together the clipping would result in a signal that essentially is six times the frequency of the power in. As one positive pulse would zero as a negative pulse from another phase would begin to increase.

  • Obvious

    The reactor can use ~50 A, and produce 3000 W of heat from electricity.
    But if it is only on 30% of the time, (duty cycle), then the measured W/h is only 900.

  • jousterusa

    A few quick replies, and my thanks for your questions:

    1. One of the scientists who validated the E-Cat was chair of the Swedish Royal Academy of Sciences Chemistry Committee, which decides who will win the Nobel Prize. Another was on one of the two committees, Chemistry and Physics. The Nobel committees want to be sure they don’t get duped as MIT duped us into believing Fleischmann and Pons were frauds, so I believe that in their partial funding of the study, the Royal Swedish Academy was preparing the way for a Nobel Prize for Rossi, perhaps shared with the heirs of Pons and Fleischmann, knowing that only by doing so can the science of cold fusion be lifted beyond the reach of the powerful controversies and indefatigable skeptics.

    2. It was President Obama’s Executive Order in October 2013 on the cogeneration of heat and power that laid the foundation for bringing a successful E-Cat technology quickly into government acceptance, particularly by the DOE, which has been adamant in opposing it. As with all other such appointments, the director of the US Patent Office serves at the President’s pleasure. Mr. Obama can do anything he likes, at least compared to the rest of us, and as you have probably noted, he does. Now he’s got the world by the testicles – he can order a patent granted, he can deliver discretionary finds for development, he can conclude international treaties to ensure the propagation of this humanity-serving technology. He will do all that in good time.

    3. What has Robert Godes ever produced or demonstrated besides press releases? I believe he is full of crap, and a man just hoping to take financial advantage of new interest in cold fusion generated in reality exclusively by Rossi’s work. He memorized the TIP and started regurgitating it with his spin everywhere he could to get big money to back him. It’s a sucker’s play.

    • ivanc

      I am waiting for an updated report, and you already declaring the ecat as real?
      The report has been challenged. the professors needs to answer the raised questions.

  • BroKeeper

    The oil markets are further exacerbated by LENR’s endorsements today. No upturns have yet occurred during significant postings since ITPR initial release on Oct 8th. Yes, along with lower global demand, an increase in US oil production while OPEC unchanged theirs continues to force Brent Oil prices down but the news today was followed by a sharper drop of about 2.0 points to 79.84. The argument for coincidence is fading when the data suggests otherwise and the majority of the collected announcements have been consolidated quickly at one source – the ECW. (Don’t fight it Frank, enjoy it.)

    I still stand behind my prediction of last January 19th: “2015 will begin commercial awakening of its industrial application and prematurely force companies change their “Five-year-plans”. New energy stocks will begin to rise and petroleum/coal/utility stocks begin to fall. Gas prices may fall below $3.00 per gallon. Electricity bills will level off and perhaps drop some.”

    You will begin seeing within the next year corporation’s juxtaposition of their long term plans and profit forecasts alongside the realization of these new energy sources.

    Today’s Yahoo News:
    “Gas to average under $3 in 2015, government says”.
    Although LENR is not mentioned here, its underlining psychic influence it has cannot be under estimated.

    http://news.yahoo.com/gas-average-under-3-2015-government-says-195450750–finance.html

    • Alain Samoun

      Bookeeper:
      Don’t you think that the lower price of oil of today is because the low outlook of the economy?
      This type of drop, in the past, was a warning of an economic crisis, like in 2008.

      • BroKeeper

        Alain, I agree with you. I am only pointing out the short term and the long term. Surely in the short term, the three current influences mentioned above has the greatest effect on oil’s price but if you follow the charts associated with LENR announcements one can see the distinct deltas when comparing it against those without announcements.

        In the long term LENR will have lasting psychological effect knowing oil’s relatively short leash of sustained profitability per investment interests and buffer any major gains. The effects may actually decrease in time as price of oil becomes more affordable thus decelerating the urgency to sell off – always follows the money (profit).

        • Alain Samoun

          When you say:

          “the news today was followed by a sharper drop of about 2.0 points to 79.84”

          Which news are you referring to?

          • BroKeeper
          • Alain Samoun

            I see,it’s the crude oil price,I though that you were talking about an LENR news,like the endorsement of Elforsk. I still believe that LENR is still not on the radar of oil companies and investors – How many of them know what we know in your opinion? Not like us, their elites are not that well informed. They are probably more worry about the effect of Tesla Motors and the Elon Musk’s batteries in a relative near futur.

          • BroKeeper

            Just google “LENR” and a major oil company like “Shell” or “BP” and you’ll come up with enough leads. It is oil futures investor’s job to know.

          • US_Citizen71

            More than likely a large oil company would have a small team of techs constantly scanning the internet for mentioned of words like energy. Searching through articles for an informational edge is part of business. At some point enough articles containing something like LENR, ECat, Rossi, etc and energy would trigger a flag in a report causing a low level tech to review the list of articles. The tech has feeling that what he read might be important so he sends it up the chain. It makes it to a vice-president that throws some resources into more research and report is passed on up. Rinse and repeat across the energy companies.

          • LCD

            Well the single point reason underlying the current oil price is OPEC. In the past they would have closed the spigots and forced the price back up.

            So the question is why are they not.

            Their answer is that they are going after market share not price. The US and Canada interpret that as we’re going to lower the price so that your expensive methods of extracting it won’t work anymore.

            But the flip side to that is why? If demand continues for another fifty years to rise why not get the best price for the barrel.

            So there is something more. It’s almost as if they don’t believe the market is going to continue to grow. So why is that? Last time I checked more cars and more people are being made at a higher rate every year.

            If course the counter to that is battery capacities can grow and replace oil with e.g. coal/hydrogen/electrical.

            But it’s that the only reason, it’s hard to imagine.

          • Alain Samoun

            “So there is something more”
            What about an economic crisis in the near future?

          • LCD

            From?

          • Alain Samoun

            Economic: Stock market bubble,Global finance collapse,debt,unemployment…
            Politic: Ukraine,Middle East,European disintegration…

            Your choice …

          • LCD

            Oh you mean the same thing it’s been for the last fifty years.

          • Alain Samoun

            Yeep! 2008 is the last finance collapse,be ready for the next one,among other things…

          • Alain Samoun

            More or less + now climate change + 7 billions people + end of resources + robotization of economy ….

    • GreenWin

      A clear analysis Bro. And I agree, our sim is diminishing the LENR effect on oil pricing. However, it is fascinating to follow the calculated release of LENR info. At once frustrating because the data & conclusions are immediate and clear, but they must be “salted” into consciousness to avoid panic. Axil made an interesting comment that this technology is a thousand years early. I think not. If we consider the planet a “student” – when the student is ready the teacher arrives. Isn’t it mysterious the “teacher” is viewed a lowly buffoon by the high priests of science?

      • BroKeeper

        A little salt goes a long way – hides the bitter taste of panic. 🙂

  • Brokeeper

    The oil markets are further exacerbated by LENR’s endorsements today. No upturns have yet occurred during significant postings since ITPR initial release on Oct 8th. Yes, along with lower global demand, an increase in US oil production while OPEC unchanged theirs continues to force Brent Oil prices down but the news today was followed by a sharper drop of about 2.0 points to 79.84. The argument for coincidence is fading when the data suggests otherwise and the majority of the collected announcements have been consolidated quickly at one source – the ECW. (Don’t fight it Frank, enjoy it.)

    I still stand behind my prediction of last January 19th: “2015 will begin commercial awakening of its industrial application and prematurely force companies change their “Five-year-plans”. New energy stocks will begin to rise and petroleum/coal/utility stocks begin to fall. Gas prices may fall below $3.00 per gallon. Electricity bills will level off and perhaps drop some.”

    You will begin seeing within the next year corporation’s juxtaposition of their long term plans and profit forecasts alongside the realization of these new energy sources.

    Today’s Yahoo News:
    “Gas to average under $3 in 2015, government says”.
    Although LENR is not mentioned here, its underlining psychic influence cannot be under estimated.

    http://news.yahoo.com/gas-average-under-3-2015-government-says-195450750–finance.html

    • Alain Samoun

      Bookeeper:
      Don’t you think that the lower price of oil of today is because the low outlook of the economy?
      This type of drop, in the past, was a warning of an economic crisis, like in 2008.

      • Brokeeper

        Alain, I agree with you. I am only pointing out the short term and the long term. Surely in the short term, the three current influences mentioned above has the greatest effect on oil’s price but if you follow the charts associated with LENR announcements one can see the distinct deltas when comparing it against those without announcements.

        In the long term LENR will have lasting psychological effect knowing oil’s relatively short leash of sustained profitability per investment interests will buffer any major gains. The effects may actually decrease in time as price of oil becomes more affordable thus decelerating the urgency to sell off – always follows the money (profit).

        Notice the breakout trends (yellow) in the charts of http://www.sifferkoll.se/sifferkoll/

        • Alain Samoun

          When you say:

          “the news today was followed by a sharper drop of about 2.0 points to 79.84”

          Which news are you referring to?

          • Brokeeper
          • Alain Samoun

            I see,it’s the crude oil price,I though that you were talking about an LENR news,like the endorsement of Elforsk. I still believe that LENR is still not on the radar of oil companies and investors – How many of them know what we know in your opinion? Not like us, their elites are not that well informed. They are probably more worry about the effect of Tesla Motors and the Elon Musk’s batteries in a relative near futur.

          • Brokeeper

            Just google “LENR” and a major oil company like “Shell” or “BP” and you’ll come up with enough leads. It is oil futures investor’s job to know.

          • US_Citizen71

            More than likely a large oil company would have a small team of techs constantly scanning the internet for mentioned of words like energy. Searching through articles for an informational edge is part of business. At some point enough articles containing something like LENR, ECat, Rossi, etc and energy would trigger a flag in a report causing a low level tech to review the list of articles. The tech has feeling that what he read might be important so he sends it up the chain. It makes it to a vice-president that throws some resources into more research and report is passed on up. Rinse and repeat across the energy companies.

          • LCD

            Well the single point reason underlying the current oil price is OPEC. In the past they would have closed the spigots and forced the price back up.

            So the question is why are they not.

            Their answer is that they are going after market share not price. The US and Canada interpret that as we’re going to lower the price so that your expensive methods of extracting it won’t work anymore.

            But the flip side to that is why? If demand continues for another fifty years to rise why not get the best price for the barrel.

            So there is something more. It’s almost as if they don’t believe the market is going to continue to grow. So why is that? Last time I checked more cars and more people are being made at a higher rate every year.

            If course the counter to that is battery capacities can grow and replace oil with e.g. coal/hydrogen/electrical.

            But it’s that the only reason, it’s hard to imagine.

          • Alain Samoun

            “So there is something more”
            What about an economic crisis in the near future?

          • LCD

            From?

          • Alain Samoun

            Economic: Stock market bubble,Global finance collapse,debt,unemployment…
            Politic: Ukraine,Middle East,European disintegration…

            Your choice …

          • LCD

            Oh you mean the same thing it’s been for the last fifty years.

          • Alain Samoun

            Yeep! 2008 is the last finance collapse,be ready for the next one,among other things…

          • Alain Samoun

            More or less + now climate change + 7 billions people + end of resources + robotization of economy ….

    • GreenWin

      A clear analysis Bro. And I agree, our sim is diminishing the LENR effect on oil pricing. However, it is fascinating to follow the calculated release of LENR info. At once frustrating because the data & conclusions are immediate and clear, but they must be “salted” into consciousness to avoid panic. Axil made an interesting comment that this technology is a thousand years early. I think not. If we consider the planet a “student” – when the student is ready the teacher arrives. Isn’t it mysterious the “teacher” is viewed a lowly buffoon by the high priests of science?

      • Brokeeper

        A little salt goes a long way – hides the bitter taste of panic. 🙂

  • US_Citizen71

    You didn’t answer my question.

  • Gerrit

    Ivanc, if science would take this seriously they could simply start with replicating the numerous reports of successful experiments showing excess heat and transmutations and work their way towards fully understanding the phenomenon which would enable increased output energies.

    There are plenty of opportunities for science to apply the scientific method of trial and error, skepticism and test in the field of LENR. There is absolutely no reason for an entrepreneur / inventor to supply science with his knowledge so that science has something to work on.

    You are barking up the wrong tree.

  • georgehants

    A lot of Truth on this page, that future generations will read regarding the the crimes of science on Cold Fusion and many other important subjects.
    We are forgetting to give the Russians due credit for being the first peer reviewed premier journal to cover the TPIR.
    I hope this is not childish sour grapes.

  • georgehants

    A lot of Truth on this page, that future generations will read regarding the crimes of science on Cold Fusion and many other important subjects.
    We are forgetting to give the Russians due credit for being the first peer reviewed premier journal to cover the TPIR.
    I hope this is not childish sour grapes.

    • bachcole

      I think that you may be casting too broad a net with the word “crime”. Yes, there definitely has been malfeasance, at MIT and elsewhere. And there have been some really ugly and rude behavior. But most people shy away from cold fusion because of fear, and although shying away because of fear shows a lack of inner strength and character, it is not a crime. It is gutlessness; it is shameful; it is not a crime.

      • US_Citizen71

        I believe he is accusing them of crimes against humanity because the suppression of the technology prevented improvements in the status quo. Which likely lead to needless suffering and death. I think he means it more in a philosophical sense than a legal one. But I could be wrong.

  • ivanc

    I would like to repeat this post, because here is the irrefutable reason they should have used IL/sqrt(3)

    Any delta have a Y equivalent.

    if you have a Y then Iresistor=Iline.

    so you know have to see how to transform the resistances for the Y equivalente of a delta.
    D to Y

    http://en.wikipedia.org/wiki/Y-%CE%94_transform

    R1=(rb rc)/(ra+rb+rc) for one , the others are similar. see the link.

    now lets suppose all r abc=1 (each resistor in the delta = 1 ohm)

    power in each case D o Y is 3 times power in individual r delta or r Y.

    R1=1/3

    v=R IL , P=R IL^2, P=1/3 IL^2

    now, do the same calculation for the delta . r=1

    P=(IL/sqrt(3))^2 * 1 = 1/3 IL^2

    OHHHH!, both results are equal !!!!!!!!!!!!!!!!!.

    Now the Delta to Y transformation is independent of the current or voltage.

    or phase….. after this there is no more argument.

    • US_Citizen71

      The minutiae you continue to ramble on about is the equation used to determine the current in any of the legs (C1 on the report diagram) which is stated as 19.7A. The C1 legs never have more than two phases flowing them, the third phase flows through the other two legs. The splits C1a and C1b (C2 on the report diagram) never have more than one phase flowing through them. If current flowing in each phase is equal then the current flowing through C1a or C1b would be half of the current in the C1 leg being split.

      • ivanc

        There is no more blind that the one who does not want to see.
        This is important because a electric circuit with the data given power, currents, resistance in cables, is fully solvable, so if we have the correct equations we able to find out the bits not published because they follow natural laws.
        The point that make my argument irrefutable is. the transformation from D to Y is fully independent of phase, current, voltage, angle of triggering etc. only depends in the value or the resistors in the Delta. In a Y Iline = Iresistor. This Invalidates your argument about timing and only one phase etc.
        Also please note is impossible to have only one line active at a given time, you need at least two, otherwise there is no current flow.

        • US_Citizen71

          No it is not impossible for only one phase to be active at a time that is the purpose of the control box. Have you looked at the wiring diagram on the report and the wave form being used? Only a portion of the leading and trailing part of each half wave is being used. Causing two pulses positive and two pulses negative for each phase on each cycle. They appear to be timed so that only one pulse from any phase is active at any given time.

          • ivanc

            ok, draw a delta with 3 cables feeding it. and put 3 switches in off position in the middle of the feeding cables , (there is no neutral, is only 3 lines like in the report)
            now turn on one line , now tell me is there any path for current flow? Get a piece of paper and a pencil before you answer . I am starting to loose my patience, is no easy to teach electricity to people with little technical insight.

          • US_Citizen71

            What about a three phase transformer with three separate output windings? That would take the three phase power in and give three separate single phase lines out.

            I think option two on the PDF below would fit the wiring diagram.

            http://www.ft-transformers.co.uk/wp-content/uploads/2012/06/Supplying-Single-Phase-Loads-from-a-Three-Phase-Supply-Tra.pdf

            edit:I believe it is called a Delta-wye transformer with a floating neutral or a three phase transformer with L-L star connected secondary.

          • US_Citizen71

            After a long conversation with a power systems engineer I will concede the current wouldn’t be half, but the power would be.

          • ivanc

            the power has to be 1/3 of the total input power, There is 3 resistor, and each one uses power.

          • US_Citizen71

            On average yes 1/3 of the total power is dissipated in each resistor. But during any instant along a full cycle each of the resistors will be dissipating a different amount of power. Peak power will not occur during peak voltage or peak current since the current will trail the voltage. Also the circuit would need to be powered by a Y configuration with a floating neutral likely from a secondary on a transformer. This explains the increase of current and drop in voltage when compared with the line in from the main. So effectively only one phase is being dissipated by each resistor. I have to admit that you have had extreme patience in dealing with those of us who are lacking in knowledge in this area. But obviously something is missing to do a complete and correct analysis as the first meter on the line in shows a different total power than the one you derive from the C1 current measurement.

            …and you still haven’t answered my question of if you disagree with the measurement done on the input power from the wall.

    • Obvious

      ****This post is under severe editing to fix up at least one good version – O***
      (I’ll mark it updated once it is not such a mess)
      I’m having a brain f.art with some stuff …need a break…
      ——————————
      Late note:Forget it. This post is a mess and is mostly messed up. Ignore it.
      —————————————————————————
      I repeat my answer here, since the other one is buried in pages of posts.

      Rxy=(ra+rb)rc /(ra+rb+rc)
      You measure the resistance between two delta corners (x,y), so Rxy
      Ignore C2 cables for now.
      One resistor is directly across the two corners (rc).
      Two resistors are across the one resistor, connected to the same corners. These are in series (ra+rb), and both are in parallel with the one resistor.(Draw the delta with one flat side up, this is the resistor side we are working with).
      Since all r’s are equal then:
      …….Rxy = (ra+rb)rc /(ra+rb+rc)
      …….Rxy = (r+r)r/r+r+r
      …….Rxy = (2r)r/3r ….. using 1 for r…
      …….Rxy = 2(1)1/3(1)
      …….Rxy = 2/3 r
      Solving for r…
      ………..Rxy = 2/3r
      ……3/6Rxy = r
      …….. …….r = 1/2Rxy

      This describes resistance when current is forced to flow across the entire delta, but through two legs only. (In one and out the other, series-parallel.) This could look like this:
      .
      …..Rxy = P/(I^2)
      …..Rxy = 479/(388.09)
      …..2/3r = 1.234………………get rid of denominator….
      **(Rxy) = 1.234/0.66667 = 0.8827 .****this value is what the ohmmeter will read.*
      ……Rxy = 0.8827
      then..
      …..r = 1/2Rxy
      …..r = 1/2(0.8827)
      …..r = 0.411

      ……..

      I repeat: If you insist on doing the math using the square root of three, you are only calculating one phase at a time. If this is the case, you must choose a fraction of the 479 W, for one phase, and then the remainder of that fraction that makes the total 479 W for the other phase. (The third phase is the inverse of the other two phases that reduces the circuit to zero.) The results of the two phases are still in parallel with each other, so in order to calculate the whole they must be combined using parallel circuit rules. If you insist on using the entire 479 W in one equation, then you are solving the entire equation at once, then the result is neither multiplied nor divided by any factor. It is the answer for the total circuit.
      ——————————————————————————–
      Edit: I found after much work that using I/sqrt(3) DOES tangle with V, even though V is unknown. It is unavoidable that you must modify V in the Delta (even though we have no V measurement), if you are converting to phase current. What happens is that P is not conserved otherwise. V must increase by sqrt(3) in order to use the phase current which is divided by sqrt(3). V increases by the square in P = V^2/R so the increase can be substantial and is equal to the magnitude of the sqrt(3) division on the current side. Since V is Squared, and is an unknown, lets use Vu = 1 to allow comparison.
      Sqrt(3)^2 =3. This means that a factor or quotient of 3(Vu) is needed to reconcile the power.
      IE: If you use phase current, then you must know and use phase voltage or incorporate it’s relative ratio of change to effective power in order to use power calculations or derive R.

      P=I^2R………………………P=I^2R………….substitute with 1…..P = 1R……3…….P = 3R
      …………………………………P=V^2/R…………………………………..P = 1/R…………..P = 3/R
      ——————————————————————–

      So then:
      Rxy = W/I^2
      Rxy = 479/388.09
      Rxy = 1.234
      r = 3Rxy/2
      r = (3)1.234/2
      r = 1.851

      Edit: the calcs immediately above are wrong.
      The right version is:
      Rxy = W/I^2
      Rxy = 479/388.09
      Rxy = 1.234
      ————————————–this part is OK—————-BUT:
      Its describes when all power is forced to be made using one resistor, so it is equivalent to three real resistors in series so:

      r = Rxy/3
      r = 1.234/3
      r = 0.411

      Third Proof of r=0.411

      In this case, we force all current through two reactor resistors.
      Assume maybe one resistor is open circuit, but somehow the current makes it to two the far C1 cables, where they are joined. This is a parallel circuit of two resistors. Power goes in, splits, goes through two resistors, and exits the resistors into a common return wire. This does not happen in real life, but due to the “unreasonable effectiveness of mathematics”…
      First we calculate the parallel resistance of two r. Rp = r in parallel.
      ..Rp = (ra*rb)/(ra+rb)
      ..Rp = r^2/r……………….use 1 for r…..
      ..Rp = 1/2 r
      2Rp.= r
      …..r = 2Rp
      then
      ….P = (I^2)2Rp
      2Rp = P/(I^2)
      2Rp = 479/388.09
      2Rp = 1.234
      Rp = 1.234/2
      Rp = 0.617
      This is the resistance of two resistors, so now we need to make same as three.
      ..r = 2/3Rp
      ..r = 0.66667(0.617)
      ..r = 0.411

      Fourth Proof of r = 0.411
      This one is very tricky. This is the phase current version. All previous iterations
      of r = 0.411 used Line current (measured).
      Converting the 19.7 A Line current (IL) into Phase current (Ip) is easy enough.
      Ip = IL/sqrt(3)
      Ip = 19.7/1.73~
      Ip = 11.37~
      Ip is considered to be the same in all three phases, and is known to be the above ratio to IL in a delta resistor configuration.
      P “into: the total circuit must be distributed somehow, the easiest is evenly distributed between all three phases. This is a balanced circuit with balanced resistors and current. V is unknown, both for line and phases. Vx can be represented as factor to P, since P=V^2/R….. so P=Vx^2/R…..and…..V=IR……Then Vxl is the line power factor for V, and Vxp is the phase power factor. Pp = P phase, PL = P line (“I” looks too much like “l”).

      Using sqrt(3) current, which is ~11.37 A (19.7/sqrt(3)for the dummy run, calculating the power in one resistor can be attempted. This is much more complicated that it appears.
      Ptotal = 479 W. So if 11.37 A (Ip) average is in each resistor then intuitively one might believe that total Power should be equally divided in all three resistors, and it is a simple
      1/3P relation. Lets try that as a test:
      1/3P = 159.66667~.
      P = I^2R
      R = P/I^2
      R = 159.66667/11.37^2
      R = 159.66667/129.5
      R = 1.234 Ohms (this is three time higher than the previous three proofs of R = 0.411)
      Then we try Ptotal = I^2R
      …………………..479 = 3(11.38^2)(1.234)
      …………………..479 = 3(129.5)(1.234)
      …………………..479 = 479 ……………………..seems to work…..
      But what happens when all the power is forced though one resistor? Line current should be three times higher than the current in one phase so 3(11.37)= 34.11 A (!!). This is much larger than 19.7A going in to the circuit! Already there is a problem.
      Well, lets continue on this wrong path….. P in the single resistor case then is Ptotal, and current will be for this example 3(Ip).
      ….P = I^2R
      479 = (34.11^2)1.234
      479 = (1163.5)1.234
      479 = 1435.75……………..definitely not right!
      Just to work out what might be wrong…
      1435.75/479 = 2.9973~ or 3 times greater than the right answer. Why 3? Good question.
      Well, we could say that R is 1/3 less in this case…. but that is non-physical. We cannot just chop the resistor in three because we just calculated it to be three times higher. We also just added three phases together, so the three resistors should actually be multiplied by three also. This makes things more than three times worse (3*1.23) than the already 3 times too high! So the resistors now have a problem of being a factor 9 of nine times greater than they should be! Looks like a squaring error, doesn’t it? Three squared wrong (and a bit).
      ——–The error is caused by not adjusting V to match. P = V^2/R. Phase current seems to be operating on phase voltage. And there is still more to it than that. Even dividing phase voltage by three, and thereby changing phase power, there are still big gremlins in the assumption that using three times phase current is OK for (besides messing with the un-measured volts). Not the least of which is explaining why phase voltage must be divided by three, when all delta corners are connected directly to the Line voltage. Or explaining how 11.37 A travels down each branch, when again the corners are all connected to that line voltage and resistance is constant (or should be).
      The real reason that R ends being 1.23 in this case is that using sqrt(3) means that we are making a special kind of series circuit. “Circuit” is the key word.

      **Lets just do it the right way, and then it is easier to see where the other wrong version for phase current math is being used incorrectly.
      First, sqrt(3) is a vector sum of two phases’ current flowing in one line. When measured as RMS it is the average vector of two directly opposed currents that can possibly flow in one side (resistor) of the delta at a time, being sourced from three possible locations (the delta corners). If all the current coming “from” one side of the resistor is drawn as a line of length I1, then that line is sourced from two other phases that can possibly “push” current that way. A second line is drawn now, to represent the other side, and current flowing the opposite direction. This is important. Since each corner’s current is sourced at an 120° phase angle from the resistor we are monitoring, then these lines are drawn at a 240° angle “outward” from each other when one line is started at the end of the other. (These are vector currents.) This makes a 120° inside angle, which is an Iscosceles triangle, when closed with a third line. The angle that is halfway between the tips of the two lines opposite where the original two lines join is the average of the two angles. It is obviously 60°. This 60° angle then splits the Isosceles triangle into two right angles. Note that this latter triangle is made from a length of 1, an length of 1/2 (since the new line cuts one line exactly in half) and an unknown length. The length of this line is Sqrt(3) when the hypotenuse is 2 units long, and the remaining leg is 1 unit long (double the 1 and 1/2 lengths). The sqrt(3) is then the average current (ie: length) of of the vector which describes the average of the two opposed currents, for this one phase.

      In the above example, the opposed currents are that form the first pair of vector long lines are 19.7 units long. the sqrt(3) last line is the 11.37 units long line.
      Now repeat the example with a new delta side current vectors, starting at the origin of the first line 19.7 unit long line from the above, but offset by 120°, since it is a new delta side that is 120° to the first. Completing the operation, and erasing all the messy triangles and 19.7 unit lines, we now have two lines 11.37 units long that are 120° apart.

      If we repeat one more time for the third delta side, we now have three lines 120° apart from each other that are 11.37 units long with a common origin. The vector sum of these three lines is ZERO. So multiplying 11.37 three times looks good on paper, but the sum is zero, due to their respective vector angles, not 34.11. Even adding 11.37 plus 11.37 is wrong if their vector angles (which include a relative +/- sign) are not respected. If the three 11.37 unit long lines are rearranged so that each end of the lines are connected to one of the others, they form a delta (the three lines are the same length, so they form an equilateral triangle), the picture begins to form of what is actually going on when using sqrt(3).

      The actual circuit using sqrt(3) for current is current flowing from one corner, around the delta, and returning to the origin. It is easier to consider that at the “start” corner, the opposite end of the current traverse is “not quite” touching the origin. The three resistors are effectively in series, and the phase current is the same current all the way around. this is why the current is equal all the way around.
      BUT: we do not want calculate a zero for an answer, so we can’t use the entire new mini vector delta we just built to supply answers directly for P or R (although this explains why R looks three times bigger).

      ***To be continued….

      ********
      ………
      But that doesn’t matter, unless you want to build a reactor or do even more unnecessarily complex math. Rxy is good enough to work out the entire reactor circuit as a whole, which is much simpler than vector math. If we are attempting to decipher the “Joule heat problem”, we only need the entire reactor values.
      I have already solved the entire circuit. The solution is so simple I can’t believe we spent all this time fretting over it.

      And… If all the W are used up in a combined two phase equivalent calculation, then the average current in the one pair of C2 must be 1/2 of C1. There is no other answer.
      A second proof of this is that if one were to put the amp probes around both of the C2 cables (at the same time), instead of C1, they would all read the same current as at the C1 cables.
      Proof 3: The average vector sum of the two C2 cable currents for two combined, simultaneous, unbalanced, complimentary, current-carrying phases equals one half of the average total current for each cable in a balanced three phase circuit. The average total current is a conserved value, and so is the same as the current measured at C1. This is easier to see when it is realized that one corner of the delta is the simultaneous return for both of the other corners from which the vector currents are calculated. This third corner is the one that the professors used.

      Most of the confusion comes from considering this third corner to be the “input” side. It can be the input side, but then the total power (conserved) and the total current (conserved) are 100% at this corner, and either current or power can be divided between the two other legs, but the division along each leg must be complimentary fractions of the total value for power and current. The simplest complimentary fraction to use is 1/2. This divides the equilateral triangle of the delta into two equal isosceles triangles (this makes vector addition easy). Dividing the current or power unevenly (1/12 and 11/12ths for example) makes for complicated complimentary calculations requiring complex trigonometry to solve each branch. (1/3 and 2/3 branches might not be too bad to do).

      • AlbertNN

        It makes no sense at all to divide the power between the phases.

        • Obvious

          The power of a three phase circuit is the sum of the power of (any) two phases at any given time if the delta is balanced.
          The total power summed for all three phases is zero.
          The inverse circuit of two phases flowing current through two resistors (power) to a single return phase is a single phase supplying all the power complimentarily distributed between two phases.
          Current cannot flow through the wires in two directions at once.
          When you measure the average power in a three phase circuit with only one clamp, you are reading the inverse power delivered by the other two phases (towards the one clamp). The division of power between those two phases towards the one clamp at any one time are unknown in an unbalanced system, but must be 1/2, on average, in a balanced system.

          Edit: Maybe I answered something that wasn’t a question… ?
          I wouldn’t divide the power between phases if I could avoid it.
          It was part of a longer discussion, but it demonstrates some physical fundamentals that can be lost in the math.

          • Mark Szl

            Can yourself, ivanc, thomas kaminski, freethinker etc come together and review this whole thing in one place. Separate what is in the scope of the report from what is just speculation on what the applied waveforms could been that are be consistent with the findings. As you said, this is scattered all over the place and one cannot follow this stuff. Maybe if it is clear enough then the admin/Frank can put up a page … with a circuit diagram.

          • ivanc

            Frank has the Key to enable this, The Lugano report is about thecnical data any way. is not just to have faith and cross fingers

          • Obvious

            That is a good idea.
            The other wiring discussions ended up getting buried under on-topic posts…..

      • Obvious

        Now, once we have our heads wrapped around that, lets divide the equilateral delta circuit into two isosceles triangles. We can consider the C2 cables to be a separate delta, (minus the reactor resistors), and deal with that later.
        Lets call the delta corners respectively x, y, z.
        Total power and current are at one corner, x.
        Each 1/2 current goes down each of the xy and xz resistors.
        We have a problem now: We cannot simply split the third resistor in half. What to do? (**If you are a quick study you can see that there is actually no problem. I’ll bring this up in a moment).
        The best solution is to solve for the current value of the third resistor as if it were part of each branch. Current from one branch, say xy, will now flow into both the y corner and the z corner. This current is split in half when this happens (equal sides, remember. It is actually more complicated than this, but the net effect is the same). So now we have 1/4 the current arriving at y, and 1/4 the current arriving at z.
        The same thing happens with the xz side, so that 1/4 of the current in that side splits into 1/4 each at the z and y corners.
        BUT, the current flowing from zy and yz from each side are in opposite directions. So they cancel. The net current is zero.
        (**If you know your V = IR and P rules well, you would have noticed that the voltage across the zy resistor would be zero, since the currents are equal on both the xy and xz sides, and the V is the same in each branch because P= (V^2)/I so no current flows anyways. If the currents were unbalanced, the zy resistor would flow the excess from one side to the other (as when the current is split unevenly between xy and xz). This is in fact why deltas are used in the first place: to balance the loads).
        So:
        1 x I goes in one corner, and comes out 1/2 I at each other corner. One resistor across the two 1/2 I corners does not participate.

        Now: The C2 cables. They are a larger delta, connected so that two C2 cables are on each side (2(C2) each). The C1 cables become the corners of the delta. Note from the example above, that two of those series-ed C2 cables (on the zy side) will not participate in the current. So the resistance of the C2 cables are 2 x C2 resistance on each side.
        BUT the sums of the 1/4 currents from each side still show up at z and y. So they have 1/2 each of the total input current. (actually there is no 1/4 current, since, as demonstrated above, the zy link does not conduct current at all). The current in each of the two series-ed C2 cables (one from each side) is 1/2 I.
        But don’t get carried away. The two sets of two C2 series cables are in parallel with each other. If you want to run all the I current through them at once:
        then their total effective resistance is (2(C2)) * (2(C2)/(2(C2)) + (2(C2)).
        This goes for the reactor resistors also.
        Thank you for your time.

        • Obvious

          So what does all the above mean?
          It means that from the perspective of a current split equally from one corner of a delta, the reactor resistors act as though there are only two parallel resistors.
          So the correct effective resistance of r in this case is r*r/r or just r.
          So the effective resistance of the whole reactor circuit is 1.234 ohms.
          The resistance of the C2 cables in the entire circuit is (using 0.002811 Ohms for each C2 length) is (0.005622)(0.005622)/0.005622+0.005622
          = 0.002811 (!!)
          And the C1 (0.004375 Ohms) circuit is C1+((C1*C1)/(C1+C1)) =
          0.004375 + 0.0021875 = 0.0065625 Ohms

          Therefore the total effective resistance of all the C1 plus CF2 cables is
          0.00065625 + 0.002811 = 0.0093735 Ohms
          Therefore: 0.0093735 times 19.7^2 = 3.637761615 Watts. This is the proper Joule heat value for the reactor. NOT 6.73 Watts, as written in the Lugano report.

          Since this error propagates through all the values calculated by the professors, it has almost no effect on the function derived for the ratio of the dummy run Joule heat to the Active current. So active run currents calculated (by several of us) from the ratio of relative Joule heat values remain essentially valid. (a Watt or two difference?)

          • Obvious

            Now for the factor of dummy run Joule heat to active run Joule heat.

            You could re-calculate the active run Joule heat using the above values. They are so close to half, that for a first approximation 1/2 is close enough.
            Joule heat is reported at 37.77 W, so a (nearly) corrected 1/2 is 18.9 Watts and is close enough (I am using 20.42~, but we can get to that later for the rest of the values. Because of I^2, the difference is a factor of 0.96 times R in this case).

            Lets use the first row of values for the reactor in the report.

            If we multiply the reported Joule heat for row one, 37.77 W, by the new value for the dummy run of 3.64 W, we get 137.4828. Divide this by the old value of 6.73 W and we get 20.43 W. In this way we arrive at a corrected active run Joule heat.

            Divide the active run Joule value by the dummy run. This gives us a factor of the increase in Watts to the dummy run. Using 20.43 divided by 3.64, we get a factor of 5.61. This is a Watt factor, so by using the square root, we obtain a current factor for calculating the active run. (SqrRt(5.61) = 2.3690. (You may get a slightly different number, since my spreadsheet retains all the many decimal places that I have rounded here).

            2.37 is the factor to multiply the dummy run current of 19.7 A with. This gives us 46.6694 A. [Ivan, you will agree this is pretty darn close to your value for this row].
            46.6694 A , squared, times 1.23 Ohms gives us the active run Watts using W = I^R.
            This gives us an active run Watts of 2678.98 Watts.

            [Ivan, you had 2688.21 W for this value. Certainly close enough. Nicely done. Note that my factoring is different, but essentially the same values came out. My different total Watt value is due to the dummy run Joule heat adjustment, which (as can be seen) is insignificant in terms of total power developed. My rounding the total reactor resistance also came in here, but even more insignificantly. This also means we are both on the right path at this point in our journey. Our wildly different, wandering paths are ending up at the same destination.]

          • Obvious

            So why does the calculated active run power deviate so much from the reported value in the report? Our active run calculations must be close to reality. We have arrived at essentially the same power, which is still three times higher than in the Lugano report.
            The answer, my friend, is that the report values are Joules. That is, Watts per second.
            And if the reactor was turned off for 70% of the time, then the average Joules are then 70% less, but the circuit Watts are the same (or zero).
            We cannot calculate the circuit when power = zero and current = zero.
            But the clock counts mercilessly forwards, reactor on or off, and we divide the active power over time.
            Thanks again everyone, for your attention..

          • Obvious

            And… I think that I can explain the OL and goofy waveform on the Lugano report PCE-830 photo in (mercifully) probably two or three sentences.

      • ivanc

        “I repeat: If you insist on doing the math using the square root of three, you are only calculating one phase at a time.”

        Were you got that idea?

        I am calculating the I in the full cycle for One phase, the other phases are the same.
        and I do divide the total power/3 has there are 3 equal resistors.

        you need to undestand better RMS, Your data is in RMS

        • Obvious

          RMS means that the circuit has equivalent values to a DC circuit.
          But a DC circuit cannot have current flowing from three “positive” leads at the same time to a non-existent “negative”. No current would flow.
          You must either split the power from one lead, and return it via two leads, effectively connected together at far end of the circuit at some “negative” place.
          Or you must use two “positive” leads that arrive at one “negative” lead.
          If you split the power unequally between two leads, then these must be complimentarily proportioned so that the sum of power measured at their “negative” ends is equal to the input power.

          Edit: several comments are coming in (to me anyways) several hours after being sent. I’ll have another look at your idea, Ivan, and see if I can figure out the specific step where we disagree.

          Edit 2: Ivan, the formula you have above converts one wye resistor (R1) to its delta equivalent. So using 1 for r, R1 (and R2, R3) = 1/3 R1
          The reverse is also true. The delta formula is ra= (R1*R2)+(R2*R3)+(R3*R1)/R1
          So power in each delta resistor, using 1 again, is 3/1. Three of these is 9/3 or 3 times the Wye resistor.
          One delta resistor is equal to 3 wye resistors.
          The resistors are not equivalent.
          Using the Wye with 1 ohms resistors…
          ……P = 1/3 IL^2
          3P = IL^2
          But using equivalent resistors for the delta, then trying to use
          ..P = (IL/sqrt(3))^2 * 3
          3P = (IL/sqrt(3)^2
          ..P = (IL^2)

          The reason for this is because power is also the product of voltage and current. The relationship between voltage in the phase of a delta connected load compared to a wye connected load is sqrt(3) times greater. Since current is dependent on the voltage applied, the current will also be sqrt(3) times higher per phase (compared to a wye) in a delta, which creates a value 3 times higher in a delta than a wye. We do not have a voltage measurement so this is not immediately obvious.

          ….something maybe not quite right down below here….I think it is because we don’t know V we might not be able to use sqrt(3) at all?

          Each of the three reactor resistors may see a phase current of IL/sqrt(3), but multiplying them by three just gives the total reactor resistance. The reactor resistors “see” sqrt(3) times the voltage (which we do not know).
          P = (V^2)I.
          This means that the effective power at each resistor must be modified by (sqrt(3))^2 (= three) before doing the resistor calculations
          with IL/sqrt(3).

          Sorry for the constant edits, if you are following along.

          • ivanc

            Now you are close.
            you got Ry=Rdelta/3
            or Rdelta=3Ry
            The power in one resistor = IY ^ 2 * Ry
            replacing.
            p=IL*IL * Rdelta/3
            p=IL*IL*Rdelta/(sqrt(3)*sqrt(3))
            p=(IL/sqr(3))^2 * Rdelta
            And this is what I wanted to prove.
            The Key is the conversion to delta to Y is independent of voltage, current, phase, etc etc…… (independent!!!!!!!!!!!)

            The total power is 3 times the power in one resistor.

          • ivanc

            Some people may not have clear why this is important, if you have a Y 3phase, there is no question on what is the I on the resistor, is just ILine. because the line and the resistor are in series. but the power in each resistor (delta or Y) has to be the same.
            only the size of the resistor will change to accommodate the conversion, then is easy to find out the phase current if the configurantion is a delta, because there is formulas for the conversion. see my other posts, there is a Link, or google delta to y 3phase equivalent.

          • Obvious

            Good day, Ivan. I’m back at this again. Lets see if we can put this to bed today.
            I had dreams of P R V etc floating around all night….
            I’m just going over your newer comments.
            For the sake of both our sanity, I won’t re-edit my earlier comments.

          • Obvious

            The power in a delta is higher than an equivalent wye. This is why there is delta-wye starters for some motors. You start the motor on wye so they don’t burn out at low starting speeds, but low speed torque is sacrificed. Current is 1/3 of a delta in a wye. (Electric motor torque is maximum at 0 rpm, but so is current. Dropping the current by a third keeps the windings cooler on start, as long it can still start turning.)

            The wye has less line voltage by V/sqrt(3), but 3 times the effective resistance (total impedance) comparable to a delta.
            The delta has line voltage, and 1/3 the effective resistance compared to a wye.
            If I can remember correctly, the total effective power difference (line V being the same) for a delta is about 1.4(?) times higher than a wye.

            Edit: Where get thrown a for loop is that V is unknown, and even though VLine is constant both circuits the unknown V is higher in the delta phases, “sneaking” in extra power that is has to be somehow accounted for by only manipulating the R and I equations for P.

          • ivanc

            in a motor the impedances are constant, so is not the kind of equivalence we talking here. if you get 3 equal resistor and put in d o Y the power is different.

            Here we talking about a circuit that is equivalent between a D and Y, with means the load or power is equivalent.
            between the D and Y.
            the IL is same
            the Vl is same.

            be x an element of the delta.
            but the voltage in Rx is different
            the Ix are different
            and Rxd different of Rxy.

            but this values are related by formulas.

            From the point of view of the generator an equivalent load means same power.
            so power in delta should be equal to power in Y.
            otherwise …..what is the equivalence we talking about.

          • Obvious

            The equivalence of the Y to a delta allows the delta to be converted to a circuit that is intuitively easier to imagine as a DC circuit. It opens the delta up, eliminating the “cross-current” or “cancelling current” section across one side (from the viewpoint of one corner of the delta) that I described earlier, by turning it into a parallel branch.

          • ivanc

            you are really close: ( I have put some comments in your expressions)
            “Using the Wye with 1 ohms resistors…
            ……P = 1/3 IL^2 (power in 1 Y resistor)
            3P = IL^2 —————–(expresion to match—-(a)
            But using equivalent resistors for the delta, then trying to use
            ..P = (IL/sqrt(3))^2 * 3 ———–(b) (power in 3 delta resistors?)
            3P = (IL/sqrt(3)^2 ———–(c) wrong expression derived from b
            ..P = (IL^2)” ———–wrong conclusion because equations above are wrong

            There is an error in moving the 3 between expressions b and c
            but also is an error to multiply right side of b by 3
            because we calculating the power in 1 resistor in (a)
            then we need the power in 1resistor in (b)

            it should be P = (IL/sqrt(3))^2 =IL^2 /3=1/3 IL^2
            3P=IL^2

            So the both Delta and Y Power are equal.
            🙂

          • Obvious

            For (b) I had 3 (bold) because the equivalent resistor in a delta is 3 times one in an equivalent wye. This is just the equation for one r of the delta so far. Obviously the r doesn’t really change, but this is the larger impedance that the total circuit sees because the current increases due to increased V in the sides of the delta relative to a wye. Three times the r, times three again for each phase ends up with 9 times the power from which to derive the resistance value from.
            Where my brain is stalling for some reason is when I try to force the r to stay constant, what happens to power. It seems to end up as divided by 9 for the complete circuit, which should work out OK since the 9 times (3^2) cancels the 9 times power in the r calculations. Somehow.
            Edit: Where I was getting stuck was foolishly inserting a V equation into the mix, hoping V would get cancelled somewhere, but since it is unknown it gets “trapped” as a useless constant.

          • ivanc

            R is not constant between D and Y. the voltage also changes.
            but the power do not need voltage if you use p=I^2 R

            Ry1=rd1 rd2/ (rd1+rd2+rd2), if all Rds equal. (lets not asume rd=1, just any value)

            then.
            Ry=rd rd / 3 rd = rd/3 ………………(a)
            power in one Ry:
            Pry=Iry Iry Ry ——————-(b)
            power in one rd:
            Prd=Ird Ird Rd ——————(c)

            IL= I line or I in feeding cables. (external to Y or D)

            Iry=IL ————-because is a y and cable in series with Yresistor.

            Ird=IL / sqrt(3) —- expression I want to prove. current i one delta resistor.

            3Pry=3Prd ——total power in delta or Y, total input power in this case.
            Pry=Prd ——- (d) Power in ry equal power in rd

            now is just algebra.

            Pry=IL IL Ry ———————(e)

            because (d) we could say:

            lrd lrd Rd = IL IL Ry — from (a) we know Ry=Rd/3

            lrd lrd Rd = IL^2 Rd/3 —- Rd are in both sides of equation
            so cancel each other.
            Ird ^2 = IL^2/3 ———— finally taking sqrt() in both sides.

            Ird=IL/sqrt(3) ===== equal to expression I wanted to prove…….
            ————–proved——————-

            If you could find one wrong expression or equation, please point out.
            one error is sufficient to make my argument wrong.

            In other post I will comment the errors in your argument.

          • ivanc

            In one to say the equivalent Ry to Rd
            in the other the equivalent Rd to Ry, is like retrofeeding, you will end in an infinite loop, this is were your argument is wrong.

          • Obvious

            Hold the correct constants.
            Wye has a higher V, Delta has a higher I. Holding P and R means that due to R in the denominator for P/1=V^2/R and as a factor in P/1=I^2R/1 that
            V and I are complimentary fractions of R.

  • US_Citizen71

    So obviously you agree that there it no problem with the power measurement by the PCE-830 on the power in from the wall. The minutiae you continue to ramble on about can’t be proved for the circuit under discussion as there are to many missing values and unknowns about the circuit. You are correct discussion is over as one sided ramblings are not a discussion.

  • US_Citizen71

    see my reply to you same exact post below.

  • BroKeeper

    I couldn’t have said it any better. In tribute to your humanitarian prose I just fed the birds. Thank you Roger.

  • Obvious

    I repeat my answer here, since the other one is buried in pages of posts.

    Rxy=(ra+rb)rc /(ra+rb+rc)
    You measure the resistance between two delta corners (x,y), so Rxy
    Ignore C2 cables for now.
    One resistor is directly across the two corners (rc).
    Two resistors are across the one resistor, connected to the same corners. These are in series (ra+rb), and both are in parallel with the one resistor.(Draw the delta with one flat side up, this is the resistor side we are working with).
    Since all r’s are equal then
    Rxy = (r+r)r/r+r+r
    Rxy = (2r)r/3r
    Rxy = 2r/3
    3(Rxy) = 2r
    r =3(Rxy)/2

    I repeat: If you insist on doing the math using the square root of three, you are only calculating one phase at a time. If this is the case, you must choose a fraction of the 479 W, for one phase, and then the remainder of that fraction that makes the total 479 W for the other phase. (The third phase is the inverse of the other two phases that reduces the circuit to zero.) The results of the two phases are still in parallel with each other, so in order to calculate the whole they must be combined using parallel circuit rules. If you insist on using the entire 479 W in one equation, then you are solving the entire equation at once, then the result is neither multiplied nor divided by any factor. It is the answer for the total circuit.
    So then:
    Rxy = W/I^2
    Rxy = 479/388.09
    Rxy = 1.234
    r = 3Rxy/2
    r = (3)1.234/2
    r = 1.851
    But that doesn’t matter, unless you want to build a reactor or do even more unnecessarily complex math. Rxy is good enough to work out the entire reactor circuit as a whole, which is much simpler than vector math. If we are attempting to decipher the “Joule heat problem”, we only need the entire reactor values.
    I have already solved the entire circuit. The solution is so simple I can’t believe we spent all this time fretting over it.

    • Obvious

      Now, once we have our heads wrapped around that, lets divide the equilateral delta circuit into two isosceles triangles. We can consider the C2 cables to be a separate delta, (minus the reactor resistors), and deal with that later.
      Lets call the delta corners respectively x, y, z.
      Total power and current are at one corner, x.
      Each 1/2 current goes down each of the xy and xz resistors.
      We have a problem now: We cannot simply split the third resistor in half. What to do? (**If you are a quick study you can see that there is actually no problem. I’ll bring this up in a moment).
      The best solution is to solve for the current value of the third resistor as if it were part of each branch. Current from one branch, say xy, will now flow into both the y corner and the z corner. This current is spit in half when this happens (equal sides, remember). So now we have 1/4 the current arriving at y, and 1/4 the current arriving at z.
      The same thing happens with the xz side, so that 1/4 of the current in that side splits into 1/4 each at the z and y corners.
      BUT, the current flowing from zy and yz from each side are in opposite directions. So they cancel. The net current is zero.
      (**If you know your V = IR and P rules well, you would have noticed that the voltage across the zy resistor would be zero, since the currents are equal on both the xy and xz sides, and the V is the same in each branch because P= (V^2)I so no current flows anyways. If the currents were unbalanced, the zy resistor would flow the excess from one side to the other (as when the current is split unevenly between xy and xz). This is in fact why deltas are used in the first place: to balance the loads).
      So:
      1 x I goes in one corner, and comes out 1/2 I at each other corner. One resistor across the two 1/2 I corners does not participate.
      Now: The C2 cables. They are a larger delta, connected so that two C2 cables are on each side (2(C2) each). The C1 cables become the corners of the delta. Note from the example above, that two of those series-ed C2 cables (on the zy side) will not participate in the current. So the resistance of the C2 cables are 2 x C2 resistance on each side.
      BUT the sums of the 1/4 currents from each side still show up at z and y. So they have 1/2 each of the total input current. The current in each of the two series-ed C2 cables (one from each side) is 1/2 I.
      But don’t get carried away. The two sets of two C2 series cables are in parallel with each other. If you want to run all the I current through them at once:
      then their total effective resistance is (2(C2)) * (2(C2)/((C2)) + (2(C2)).
      This goes for the reactor resistors also.
      Thank you for your time.

      • Obvious

        So what does all the above mean?
        It means that from the perspective of a current split equally from one corner of a delta, the reactor resistors act as though they are two parallel resistors.
        So the correct resistance of r in this case is r*r/r or just r.
        So the effective resistance of the whole reactor circuit is 1.234 ohms.
        The resistance of the C2 cables in the entire circuit is (using 0.002811 Ohms for each C2 length) is (0.005622)(0.005622)/0.005622+0.005622
        = 0.002811 (!!)
        And the C1 (0.004375 Ohms) circuit is C1+((C1*C1)/(C1+C1)) = .

        …..still working…..

        • Obvious

          Now for the factor of dummy run Joule heat to active run Joule heat.

          You could re-calculate the active run Joule heat using the above values. They are so close to half, that for a first approximation 1/2 is close enough.
          Joule heat is reported at 37.77 W, so a (nearly) corrected 1/2 is 18.9 Watts and is close enough (I am using 20.42~, but we can get to that later for the rest of the values. Because of I^2, the difference is a factor of 0.96 times R in this case).

          Lets use the first row of values for the reactor in the report.

          If we multiply the reported Joule heat for row one, 37.77 W, by the new value for the dummy run of 3.64 W, we get 137.4828. Divide this by the old value of 6.73 W and we get 20.43 W. In this way we arrive at a corrected active run Joule heat.

          Divide the active run Joule value by the dummy run. This gives us a factor of the increase in Watts to the dummy run. Using 20.43 divided by 3.64, we get a factor of 5.61. This is a Watt factor, so by using the square root, we obtain a current factor for calculating the active run. (SqrRt(5.61) = 2.3690. (You may get a slightly different number, since my spreadsheet retains all the many decimal places that I have rounded here).

          2.37 is the factor to multiply the dummy run current of 19.7 A with. This gives us 46.6694 A. [Ivan, you will agree this is pretty darn close to your value for this row].
          46.6694 A , squared, times 1.23 Ohms gives us the active run Watts using W = I^R.
          This gives us an active run Watts of 2678.98 Watts.

          [Ivan, you had 2688.21 W for this value. Certainly close enough. Nicely done. Note that my factoring is different, but essentially the same values came out. My different total Watt value is due to the dummy run Joule heat adjustment, which (as can be seen) is insignificant in terms of total power developed. This also means we are both on the right path at this point in our journey.]

          ……still working……

          • Obvious

            So why does the calculated active run power deviate so much from the reported value in the report? Our active run calculations must be close to reality. We have arrived at essentially the same power, which is still three times higher than in the Lugano report.
            The answer, my friend, is that the report values are Joules. That is, Watts per second.
            And if the reactor was turned off for 70% of the time, then the average Joules are then 70% less, but the circuit Watts are the same (or zero).
            We cannot calculate the circuit when power = zero and current = zero.
            But the clock counts mercilessly forwards, reactor on or off, and we divide the active power over time.
            Thanks again everyone, for your attention..

          • Obvious

            And… I think (mercifully) that I can explain the OL and goofy waveform on the Lugano report PCE-830 photo in probably two or three sentences.

  • US_Citizen71

    No it is not impossible for only one phase to be active at a time that is the purpose of the control box. Have you looked at the wiring diagram on the report?

  • Obvious

    The power of a three phase circuit is the sum of the power of (any) two phases at any given time if the delta is balanced.
    The total power summed for all three phases is zero.
    The inverse circuit of two phases flowing current through two resistors (power) to a single return phase is a single phase supplying all the power complimentarily distributed between two phases.
    Current cannot flow through the wires in two directions at once.
    When you measure the average power in a three phase circuit with only one clamp, you are reading the inverse power delivered by the other two phases (towards the one clamp). The division of power between those two phases towards the one clamp at any one time are unknown in an unbalanced system, but must be 1/2, on average, in a balanced system.

    • Mark Szl

      Can yourself, ivanc, thomas kaminski, freethinker etc come together and review this whole thing in one place. Separate what is in the scope of the report from what is just speculation on what the applied waveforms could been that are be consistent with the findings. As you said, this is scattered all over the place and one cannot follow this stuff. Maybe if it is clear enough then the admin/Frank can put up a page … with a circuit diagram.

      • Obvious

        That is a good idea.
        The other wiring discussions ended up getting buried under on-topic posts…..

  • Obvious

    RMS means that the circuit has equivalent values to a DC circuit.
    But a DC circuit cannot have current flowing from three “positive” leads at the same time to a non-existent “negative”. No current would flow.
    You must either split the power from one lead, and return it via two leads, effectively connected together at far end of the circuit at some “negative” place.
    Or you must use two “positive” leads that arrive at one “negative” lead.
    If you split the power unequally between two leads, then these must be complimentarily proportioned so that the sum of power measured at their “negative” ends is equal to the input power.

  • US_Citizen71

    I believe he is accusing them of crimes against humanity because the suppression of the technology prevented improvements in the status quo. Which likely lead to needless suffering and death. I think he means it more in a philosophical sense than a legal one. But I could be wrong.

  • US_Citizen71

    Eat, sleep and reproduce. Beyond that anything they want. If energy becomes a non-issue as well as food, materials, etc… man will have time to explore philosophy and the arts. I’m not convinced that an AI will necessarily have that creative spark. Creativity is partially the result of misfires of nerve cells connecting random unrelated things. There is a reason that some of the most creative people on the world are semi to completely unstable. I’m not sure how you program logic to duplicate that effectively. Man will likely still give creative input he just might not be the one finishing the project.

  • GreenWin

    As Algore would say (about electric measurements) “The debate is over.” Trying to poke holes in the Lugano Report with Ohm’s Law is futile. We’ve moved far beyond doubt to action.

    • ivanc

      Who has moved? You? Rossi? Levi? , the 1mw reactor?, the customer? , is all rhetoric an hopes. and speculation. the hard evidence is the report, and I am challenging it. We waiting an update from Levi, the report if full of inconsistent data, we need the voltage readings to confirm the observation.

      • GreenWin

        Well, good luck with that ivan. For the rest of the planet, “The debate is over.”

  • GreenWin

    As Algore would say (about electric measurements) “The debate is over.” Trying to poke holes in the Lugano Report with Ohm’s Law is futile. We’ve moved far beyond doubt to action.

    • ivanc

      Who has moved? You? Rossi? Levi? , the 1mw reactor?, the customer? , is all rhetoric an hopes. and speculation. the hard evidence is the report, and I am challenging it. We waiting an update from Levi, the report if full of inconsistent data, we need the voltage readings to confirm the observation.

      • GreenWin

        Well, good luck with that ivan. For the rest of the planet, “The debate is over.”

  • Obvious

    Good day, Ivan. I’m back at this again. Lets see if we can put this to bed today.
    I had dreams of P R V etc floating around all night….
    I’m just going over your newer comments.
    For the sake of both our sanity, I won’t re-edit my earlier comments.

  • Obvious

    For (b) I had 3 (bold) because the equivalent resistor in a delta is 3 times one in an equivalent wye. This is just the equation for one r of the delta so far. Obviously the r doesn’t really change, but this is the larger impedance that the total circuit sees because the current increases due to increased V in the sides of the delta relative to a wye. Three times the r, times three again for each phase ends up with 9 times the power from which to derive the resistance value from.
    Where my brain is stalling for some reason is when I try to force the r to stay constant, what happens to power. It seems to end up as divided by 9 for the complete circuit, which should work out OK since the 9 times (3^2) cancels the 9 times power in the r calculations. Somehow.

  • Obvious

    The power in a delta is higher than an equivalent wye. This is why there is delta-wye starters for some motors. You start the motor on wye so they don’t burn out at low starting speeds, but low speed torque is sacrificed. Current is 1/3 of a delta in a wye. (Electric motor torque is maximum at 0 rpm, but so is current. Dropping the current by a third keeps the windings cooler on start, as long it can still start turning.)

    The wye has less line voltage by V/sqrt(3), but 3 times the effective resistance (total impedance) comparable to a delta.
    The delta has line voltage, and 1/3 the effective resistance compared to a wye.
    If I can remember correctly, the total effective power difference (line V being the same) for a delta is about 1.4(?) times higher than a wye.

  • Obvious

    Ivan,

    Lets quickly go back to step one: basic assumptions and first principles.

    The line current divided by square root of three for one phase is a vector sum of the currents coming from the other phases. The line current in a delta is composed of two
    currents that are out of phase by 120 degrees. This is where sqrt(3) comes from.
    The vector sum of the two phase current components is the square root of
    three times the current in any phase.
    So if you know one vector current in one phase, the other two phases must total to sqrt(3) times the current in the one phase. But since a third phase can’t also conduct through a wire at the same time as two others are conducting in one phase (it would short out), the current multiplied by the sqrt(3) is only in one other phase at a time. In this way power stays constant in the equation.

    • Obvious

      The math above is a bit whacked.
      19.7*sqrt(3) = 34.081 (two remaining delta sides, if one has 11.37 A)
      19.7/sqrt(3) = 11.37 (one delta side, C2 cable)
      total………….= 45.451
      All the above values are magnitudes, not respecting their sign (+/-). Their signs are assigned using the correct vector polarity, which is relative to an arbitrary choice of frame of reference.
      If you choose the start vector point at one corner of a delta (say x) leading to another (y), then the magnitudes of the values between xz and zy will resolve to relative vector quantities with appropriate sign compared to the reference point.
      If both C2 cables coming from a C1 have each 11.37 A, (total 22.74 A) then the last side has -22.74 to make the total zero.

      • Obvious

        The “short” version is this: If you use sqrt(3), you are using a vector quantity. The average RMS is integrated over time. Time cannot be ignored, but can be “halted” at an instant in time in order to simply things. But we must hold the time constant consistently in one location, using one description that describes what is happening at that one instant.
        So at the instant you freeze time at a point when one delta side sees 11.37 amps, then the rest of the equation must follow the consequences of that “measurement” at this instant. The average RMS values of the other sides are averages of time-integrated values. So they cannot be used when compared to a frozen instant of time in another side of the equation.
        Moving the frozen viewpoint of time to the C1 cable at RMS 19.7 A means that measured (or arbitrarily calculated RMS) values cannot be used for any other of the following parts of the circuit. You must use whatever math is appropriate based on only one RMS measurement per averaged factor (V and I). Resistance is not a time-sensitive RMS value, so it remains independent (time invariant). It is already “frozen in time” for the purposes of the relevant equations.
        We must also freeze P at some point. It must be tied to the same time frame as the other factors. Since P is a function of V and I, this follows that P is frozen in time with these two functions, and must be frozen from the same viewpoint.
        Since V is unknown, we must hold it constant, leaving only I as the variant to be held still from the viewpoint of any math applied to the circuit.

        • Obvious

          Just before I head out:
          The problem we keep tripping over in this power story is time. When measured in seconds, and t=1 in the denominator or as a factor, or both, and gets cancelled somewhere, then it is ignored for convenience sake and then we forget that it is still part of the original equations from which later calculations are made.

    • US_Citizen71

      What do you think about option 2 in the below PDF being inside of the control box? Wouldn’t that solve the problem as well?

      http://www.ft-transformers.co.uk/wp-content/uploads/2012/06/Supplying-Single-Phase-Loads-from-a-Three-Phase-Supply-Tra.pdf

      • Obvious

        Yep. I thought that switching to a wye and delta, or vice versa with the control might throw wrench in the works, and add a new dimension of electrical ugliness for discussion. But it seems that the delta uses the most power anyways. If the box ran it wye for dummy, then delta for the run, then the proportion of Joule heating between active and dummy might be non-linear or at least have a sudden “phase change”. lol

  • Obvious

    Ivan,

    Lets quickly go back to step one: basic assumptions and first principles.

    The line current divided by square root of three for one phase is a vector sum of the currents coming from the other phases. The line current in a delta is composed of two
    currents that are out of phase by 120 degrees. This is where sqrt(3) comes from.
    The vector sum of the two phase current components is the square root of
    three times the current in any phase.
    So if you know one vector current in one phase, the other two phases must total to sqrt(3) times the current in the known phase. But since a third phase can’t also conduct through a wire at the same time as two others phases are conducting in one wire (it would short out, changing the sqrt(3) equation in infinite regression), the current multiplied by the sqrt(3) is only in one other delta side at a time. In this way power stays constant in the equation.

    So if one delta edge current is held constant at IL[19.7]/sqrt(3) (always know what you hold), then one other side must have 19.7 times sqrt(3) IL at the same time. The last side has the inverse sum of the vectors of the first two sides, to make the circuit a circle with a zero total vector sum.
    Edit: some scrambling of phase and delta sides is going on above, but I hope the general meaning is clear) oops voted myself down by mistake….

    • ivanc

      RMS is not a vector is an absolute value
      so Ir = Iline/sqrt(3) is not a vector. is an absolute value.

      Have you read my last responses to Y to D ?
      my post needs to be aproved…. so please be patient

      • Obvious

        The sqrt(3) is the vector angle descriptor. When used to factor I, it describes the relative length of the base of an Isosceles triangle to its sides. The Isosceles triangle with a 120° angle point represents one phase of three phases of power. Connecting three of these triangles at the 120° point makes a set of three phases enclosed in the inside of a circle so that all 360° are accounted for. This is the basis of all 3 phase-related mathematics.

        If you bisect the circle with a line through the exact centre, everything on one side of the is “plus” and everything on the other is “minus”, regardless of the angle the line makes to three inscribed triangles. If you choose to bisect the circle superimposed on one side of one of the triangles, you will bisect the opposite triangle exactly.

        I think that the three lines that split a circle to make three equal Isosceles triangles inside a circle is the wye.
        And the three bases of the three triangles are one equilateral triangle inside the same circle and that describes the delta.

        Try this interactive triangle. Hold the angle at 120° and see how the numbers work out on the sides and bottom (19.7, 11.37, 34.1, 46.7). Less than 120° is a reduced conduction angle/period. I think the area of the triangle is proportional to the power, and maximum potential power per phase is 100% of the 120° conduction angle.
        http://www.mathopenref.com/isosceles.html

        • ivanc

          No my friend I am not using vectors at all. RMS is absolute value.
          and all the resistances has a IL/sqrt(3) current at the same time. (in RMS)

          and all resistor burn the same power .
          Is “absolute values” so no vector of phase has place in the discussion.
          Please see my demos and try to find an error in the equations.
          also
          D to Y equivalent, means you put the circuit in delta o Y to the power supply and the power supply see no difference. this is why is called equivalent.
          Each Idea I have tried to explain you make it more complex, because you trying to do an instantaneous regime. but you have no data for that, you just have RMS.
          so RMS is the same in all 3 resistors.
          RMS and angle analysis are incompatible!
          you can not mix an escalar with a vector. (a vector moves evolves on time) RMS is just a fix value, an equivalent for the full cycle.
          as you said it does the work DC does. but obviously does not behave like DC.
          This is why the 1/sqrt(3) cames in the picture. in a 3 phase circuit.
          Instead of trying to invent the wheel get a book in 3 phase and see where is your confusion. all this has been analysed and is well documented for more than a century.

          • Mark Szl

            OK, I’ve gotten lost as to the relevance of all this sqrt(3) versus 1/2 stuff to the conclusions of the study. Please remind me again.

          • Obvious

            We are very complicatedly figuring out the resistance of the reactor resistor coils. This is needed to work out the correct current in the active run.

        • Obvious

          You will like this. Bisect one 120° triangle so that the base is divided in exactly 1/2, starting the line at the 120° corner (center of the circle).

          You will have two Right Angle triangles. Using Pythagoras’ Theorem:

          For one of these smaller triangles, the base is sqrt(3) long if the new 90° angle side is 1 long and the radius-length side is 2 long. These ratios are forced since the angles of the triangle are 90, 60 and 30.
          Dividing 19.7 by the length of these sides gives values of 11.37, 9.85, and 19.7 .
          Very familiar numbers.

          Considering that we are on the verge of deriving Ohms Law and Kirchhoff’s Laws from first principles with only a very limited set of data still missing V, I am amazed at what can be done in a few days.

          V is like dark energy.
          We can’t measure it directly (in this circuit), it’s invisible, but we can feel its effect in the circuit and quantify that as a constant.
          What is the EMF of the universe anyways? Infinite? Cancels out? Local galaxy rotation may generate a current?….gravity and mass vs Higgs field friction?
          Friday night.

    • Obvious

      The math above is a bit whacked.
      19.7*sqrt(3) = 34.081 (two remaining delta sides, if one has 11.37 A)
      19.7/sqrt(3) = 11.37 (one delta side, C2 cable)
      total………….= 45.451
      All the above values are magnitudes, not respecting their sign (+/-). Their signs are assigned using the correct vector polarity, which is relative to an arbitrary choice of frame of reference.
      If you choose the start vector point at one corner of a delta (say x) leading to another (y), then the magnitudes of the values between xz and zy will resolve to relative vector quantities with appropriate sign compared to the reference point.
      If both C2 cables coming from a C1 have each 11.37 A, (and therefore also their respective resistors) (sum 22.74 A) then the last side has -22.74 A to make the total zero.

      • Obvious

        The “short” version is this: If you use sqrt(3), you are using a vector quantity. The average RMS is integrated over time. Time cannot be ignored, but can be “halted” at an instant in time in order to simply things. But we must hold the time constant consistently in one location, using one description that describes what is happening at that one instant.
        So at the instant you freeze time at a point when one delta side sees 11.37 amps, then the rest of the equation must follow the consequences of that “measurement” at this instant. The average RMS values of the other sides are averages of time-integrated values. So they cannot be used when compared to a frozen instant of time in another side of the equation.
        Moving the frozen viewpoint of time to the C1 cable at RMS 19.7 A means that measured (or arbitrarily calculated RMS) values for current cannot be used for any other of the following parts of the circuit. You must use whatever math is appropriate based on only one RMS measurement per averaged factor (V and I). Resistance is not a time-sensitive RMS value, so it remains independent (time invariant). It is already “frozen in time” for the purposes of the relevant equations.
        We must also freeze P at some point. It must be tied to the same time frame as the other factors. Since P is a function of V and I, this follows that P is frozen in time with these two functions, and must be frozen from the same viewpoint.
        Since V is unknown, we must hold it constant, leaving only the circuit distribution of I as the variant to be forced to be held still from the viewpoint of any math applied to the circuit.
        Edit: As I mentioned a day or two ago, since V is unknown and therefore must be held constant, then P is automatically held constant, since P = V^2/R and both V and R are both constants in this case.
        (This means, per my strange conversation a couple of days ago, that in 3-D Wattspace, made of dimensions V, I, and R there is only one solution, and I^2R is fixed in a specific point on a plane described by I^2R within 3-D Wattspace when V is a constant).

        • Obvious

          Just before I head out:
          The problem we keep tripping over in this power story is time. When measured in seconds, and t=1 in the denominator or as a factor, or both, and gets cancelled somewhere, then it is ignored for convenience sake and then we forget that it is still part of the original equations from which later calculations are made.

    • US_Citizen71

      What do you think about option 2 in the below PDF being inside of the control box? Wouldn’t that solve the problem as well?

      http://www.ft-transformers.co.uk/wp-content/uploads/2012/06/Supplying-Single-Phase-Loads-from-a-Three-Phase-Supply-Tra.pdf

      edit:I believe it is called a Delta-wye transformer with a floating neutral or a three phase transformer with L-L star connected secondary.

      • Obvious

        Yep. I thought that switching to a wye and delta, or vice versa with the control box might throw a wrench in the works, and add a new dimension of electrical ugliness for discussion. But it seems that the delta uses the most power anyways. If the box ran it wye for dummy, then delta for the run, then the proportion of Joule heating between active and dummy might be non-linear or at least have a sudden “phase change”. lol

  • Obvious

    The equivalence of the Y to a delta allows the delta to be converted to a circuit that is intuitively easier to imagine as a DC circuit. It opens the delta up, eliminating the “cross-current” or “cancelling current” section across one side (from the viewpoint of one corner of the delta) that I described earlier, by turning it into a parallel branch.

  • Obvious

    The sqrt(3) is the vector angle descriptor. When used to factor I, it describes the relative length of the base of an Isosceles triangle to its sides. The Isosceles triangle with a 120° angle point represents one phase of three phases of power. Connecting three of these triangles at the 120° point makes a set of three phases describing the inside of a circle so that all 360° are accounted for. This is the basis of all 3 phase-related mathematics.

    • Obvious

      You will like this. Bisect the triangle so that the base is divided in exactly 1/2, starting the line at the 120° corner (center of the circle).

      You will have two Right Angle triangles. Using Pythagoras’ Theorem:

      For one of these smaller triangles, the 90° angle’s side is sqrt(3) long, if the opposite side is 2 long and the base is 1 long. These ratios are forced since the angles of the triangle are 90, 60 and 30.
      Dividing 19.7 by the length of the sides by gives sides of 11.37, 9.85, and 19.7 .
      Very familiar numbers.

  • US_Citizen71

    After a long conversation with a power systems engineer I will concede the current wouldn’t be half, but the power would be.

  • Obvious

    Hold the correct constants.
    Wye has a higher V, Delta has a higher I. Holding P and R means that due to R in the denominator for P/1=V^2/R and as a factor in P/1=I^2R/1 that
    V and I are complimentary fractions of R.

  • US_Citizen71

    On average yes 1/3 of the total power is dissipated in each resistor. But during any instant along a full cycle each of the the resistors will be dissipating a different amount of power. Also the circuit would need to be powered by a wye configuration with a floating neutral likely from a secondary on a transformer. This explains the increase of current and drop in voltage when compared with the line in from the main. So effectively only one phrase is being dissipated by each resistor. They simplified the math and the situation which does induce an error but this error reduces the calculated COP not increases it. So other than being the electrical equivalent of the grammar police you are pointing out that the reported COP in reality would be slightly higher.

  • Obvious

    Ivan,
    Let me first apologize for the torture I have put you through.
    You are very patient, and for that I thank you.
    I have figured out what no textbook or website seems to be able to explain clearly.
    I hope that we can agree on this, and then move forward.
    The square root of three times line current is the vector sum of two 120° separated current flows for one phase.
    In other words, it already incorporates the vector math, and is the final result. (that is why it is so much smaller) than line current.
    The reason that sqrt(3) is the side of a triangle is due to: that for product of the vector of two equal length sides, the short cut to the same point is described by an a triangle with a sqrt(3)times the length sides.
    So, indeed, sqrt(3) is a vector, but is a completed vector operation for two of three equal vectors, (the third of which would return the vector solution to the origin).

    • Obvious

      And I’m not sure what this means, but I worked it out so I may as well save it for posterity.

      Using IL as the base of each equal 120° triangle formed inside a circle, and a using a line that exactly dissects one of these 120° triangles so that the line continues, superimposed on the common line for the other two triangles (same one as the circle radius), then the sum of the square of one complete base length plus the square one half of another (the one that is dissected) = Power
      (The three bases of the 120° triangles make an equilateral triangle within the circle, so the line dissects the equilateral triangle, and so we are using 1/2 an equilateral triangle).

      IE: 19.7^2 + 9.85^2 = 485.1125

      This means P = IL^2 + (1/2IL)^2

  • Obvious

    Ivan,
    Let me first apologize for the torture I have put you through.
    You are very patient, and for that I thank you.
    I have figured out what no textbook or website seems to be able to explain clearly.
    I hope that we can agree on this, and then move forward.

    The square root of three times line current is the vector sum of two 120° separated current flows for one phase.

    The reason that sqrt(3) is the side of a triangle is due to: that for sum of the vector of two equal length vector components, offset by 120°, the length of the short cut to the end point from the origin is described by a triangle with sqrt(3) times the length of one component.

    In other words, it already incorporates the vector math, and is the final result. (That is why it is so much smaller than line current).

    So, indeed, sqrt(3) is a vector, but is a completed vector operation for two of three equal but 120° offset vectors, for each phase (the third of which would return the vector solution to the origin).

    • Obvious

      And I’m not sure what this means, but I worked it out so I may as well save it for posterity.

      Using IL as the base of each equal 120° triangle formed inside a circle, and a using a line that exactly dissects one of these 120° triangles so that the line continues, superimposed on the common line for the other two triangles (same one as the circle radius), then the sum of the square of one complete base length (becomes the hypotenuse) plus the square one half of another (the one that is dissected) = Power (total)
      (The three bases of the 120° triangles make an equilateral triangle within the circle, so the line dissects the equilateral triangle, and so we are using 1/2 an equilateral triangle which is a right angle triangle).

      IE: 19.7^2 + 9.85^2 = 485.1125 (The professors used 486 W)

      This means P = IL^2 + (1/2IL)^2

      This seems to be a unique solution I have never encountered before.
      It indicates, unless it is a fluke (and I am still uncertain), that almost unbelievably in a balanced 3 phase delta, we can calculate the total power (before line losses) knowing only the Line current.
      Is this possible?

  • Obvious

    We are very complicatedly figuring out the resistance of the reactor resistor coils. This is needed to work out the correct current in the active run.