Mats Lewan: Testers Rule Out Inverted Clamp Hypothesis, Rossi Comments (Mark E. Kitiman)

This comment was posted on the Always on Thread by Mark E. Kitiman

Mats Lewan has been in contact with the professors and answered a question about the clamps that was put forward by DickieFix:

Question
2. If one clamp is inverted on both power analyzers, they would measure a
third of the effective power but still show the correct current value
(but negative).

Answer:

No 2. I have discussed this with the researchers and they tell me that
they have excluded this hypothesis, looking at log files and photos.

Source: https://matslew.wordpress.com/2014/10/09/interview-on-radio-show-free-energy-quest-tonight/#comment-3959

Admin edit to add:

Andrea Rossi responded to a question about this on the Journal of Nuclear Physics this week, saying:

Andrea Rossi
November 13th, 2014 at 1:59 PM
Silvio Caggia:
The photo #5 of the Report of the Independent Third Party is very important and has been made on purpose from the Professors. They explained to me that the photo has been taken during the set up of the measurement stuff and they were controlling that the PCE830 was surely able to read perfectly the waves also in extreme conditions: for this reason , as surely have understood the experts and the reviewers to whom the Professors have given the report before the publication, the photo shows the wave also when the system has been put in overload; you can understand it from the acronym “OL” that you can read on the display, while the wave is perfectly described by the instrument.
Thank you for the intelligent question.

  • Signs of life from the testing team! I am so glad to hear that they are taking an open-minded approach to potential flaws in their analysis/measurements and triple-checking.

    More of this, please.

    • psi2u2

      What a good response.

    • LuFong

      It’s a bit disconcerting to me that it’s been well over a month since the (presumably peer reviewed) report has been released and these power issues identified and there has been no official response by the testing team. Instead we’ve gotten some indirect statements from Mats Lewan and a number of quite belligerent responses from Rossi who has nothing to do with the report or the measurements (his words). Rossi’s most recent statement seems to indicate the matter is closed:

      vvvvvv

      Andrea Rossi
      November 13th, 2014 at 6:46 AM

      JC Renoir:

      I spammed your comment for reasons you easily can understand, but I will respond to the part concerning the questions emerged on some blog related to the Report of the ITP: you ask when I will answer to the questions put here and there. As a matter of fact, all the questions have been answered in this blog, directly from me or from other expert Readers. It is true that notwithstanding this fact, somebody continues to put again and again the same questions, but the intention of these guys is not to make clear obscure points, but to try to pull us in a discussion where they get confidential information; obviously there are also the agenda-motivated guys: our policy with them is just to ignore them, after the answer has been already given regarding the issue they raise.

      To all the intelligent and honest questions we have answered .

      Warm Regards,

      A.R.
      ^^^^^^^

      I do hope the testers answer the questions in a timely manner with actual supportive data. The people raising these questions are knowledgeable people and have spent a great deal of time analyzing the data, such as it is. A complete response will only strengthen the report. Rossi again is attacking those that are looking at the data and trying to make sense of it but frankly I don’t care what Rossi says since it’s not his report or measurements.

      • ivanc

        is OK, You can not really expect nothing better from Rossi, Hi hides under the cover of secrecy etc etc, if the reported ecat works as it say it works, we already have all currents and voltages, resistances, etc…. The report did gave the data to calculate all this.
        but…
        We just trying to confirm the drop in resistance was real (because this is an almost impossible event)….
        So Please confirm the info you have already given to us by releasing the voltage readings.

      • Freethinker

        But LuFong,

        you are not looking at the data in the perspective of the scope, but demand MORE data to conclude that the claims made in the scope are acceptable. This is because you question the measurements of the power in. And by getting more data you think you will be able to point out errors, although that extra data does not cover the scope you wish to extend to. This is not logical.

        I don’t pretend that I do not want more data, I do, because I would like to learn more of the device. But that aside, the scope is clear, that data has been readily described. Two observables, power in, temperature of the reactor surface. Hence the claims within the scope can treated as valid.

        Also, your assertion that it must be important to the testers to publicly “strengthen” the report is naive. If and when there will be a publication of the report in some journal or other venue, then there might be room for such improvements. If you are lucky they might update the current report as is, or release an addendum.

        I doubt that the scope of the test will be any different, and thus only clarifications that relates to things outside the black box will be addressed. Not those that you, Thomas Clarke, ivanc, and DickeFix are so tenaciously chasing.

        I think you need to come to terms with that, and simply choose whether or not the data presented is strong enough to convince you personally. If you do not trust the power-in measurements, you don’t. If you don’t trust the thermal measurements, you don’t.

        It is your choice, logical or not.

        • LuFong

          You seem to have completely overlooked the point of a scientific test. The point of this scientific test is to get experimental data to demonstrate an effect. If there is not enough data then the scientific test is not sufficient to demonstrate the effect. And to this end the more data the better especially if the data indicates an anomalous effect. This is the case here and you should know better.

          You and others seem to be content to assume that the researchers are a) competent enough not to make measurement errors b) unbiased enough not to make measurement errors c) accepting of any explanation without supporting data in order to support other data in the test. But this is not how science works. Science relies on the data alone to ascertain the adequacy of the test. More data is required to do this.

          Clearly the testers are not independent testers. Clearly the testers are not experts in power measurements. And clearly what is going on in this test is anomalous, in more ways than one, and that more data and replication is necessary to show the claimed results.

          This is how science works. The bottom line is that this report will never be accepted in any scientific journal. It doesn’t mean that the information contain therein is not important or interesting. It just means that the report is not competent science.

          • Freethinker

            Good to know from you how science works. I did not know. You are equally lost as your compadres Thomas Clarke, Mike, Dickefix and Ivan. You do really not know what a scope is? What a black box test is?

            Yes, the outset is that they have done their measurements properly. Inspite the pathoskeptic viewpoint that these measurements are more elaborate than rocket science, that is not the case.

            But ofcourse you should question the independence of the testers. Naturally. Thanks. It helps me.

            No LuFong, science does not work like so that you use that for witch you have no real data to make conclusions.

            But thanks for the lesson in science. It was unreal.

  • Signs of life from the testing team! I am so glad to hear that they are taking an open-minded approach to potential flaws in their analysis/measurements and triple-checking.

    More of this, please.

    • psi2u2

      What a good response.

    • Thomas Clarke

      Checking would be resolving the current/power anomaly by seeing how heater resistance varies with temperature in the active test. They have not yet done this.

      If they did it would confirm or deny the anomaly. So why has it not been done?

      • US_Citizen71

        It is being done by committee.

        • Thomas Clarke

          That is , I agree, a likely reason for delay. But it is no excuse!

  • Obvious

    My idea for of the OL photo was that only one amp clamp, the I3, was connected, and no voltage lines.
    I was at one point working out whether the amps would still show. (Probably?)
    The reference line should then jump to 0° instead of 30°, as Andreas S had shown in that waveform analysis, since there are no other phases connected to the meter to reference the voltage crossing.
    Hearing some responses from the test team is welcome.
    I would prefer to hear from them directly, rather than second hand, though.

  • Obvious

    My idea for of the OL photo was that only one amp clamp, the I3, was connected, and no voltage lines.
    I was at one point working out whether the amps would still show. (Probably?)
    The reference line should then jump to 0° instead of 30°, as Andreas S had shown in that waveform analysis, since there are no Volt leads connected to the meter to reference the voltage crossing. (No V measurements at all, so no sines to cross).

    Hearing some responses from the test team is welcome.
    I would prefer to hear from them directly, rather than second hand, though.

  • Freethinker

    Well,

    it means that the input power was in all aspects correctly measured, and as the temperature can be said to be reasonably well measured, the claims made in the report stand.

    Not that this will scare away the patoskeps since they tend to be quite insatiable when it comes to evidence. The more evidence, data and other information they get the higher they raise the bar, and claim fraud or gullible flocks of scientists, and analyze things still outside the scope for which the data and information is not mean to reflect, and draw their own conclusions.

    But still. It’s very good to get that confirmation.

    • Thomas Clarke

      The anomaly remains. The testers claim (without giving their reasons) that a reversed clamp does not explain it.

      So Freethinker’s comment is clearly wrong. Note that they have not addressed the anomaly, nor stated they have data to show miracle variable non-Inconel resistance.

      • Freethinker

        No, Thomas Clarke,

        as has been explained to you and ivanc many times before, it is you who are wrong, as you assume too many things, do not have the relevant information as you look at things outside of the scope of the test, consider uncorrelated information to be correlated, then confabulate based on your preset bias that the ECAT does not work and build up your conjectures that the claims in the report are void.

        Your conjectures are rubbish and disingenuous, and does not hold water.

        • Name calling, ridicule and ‘ad hominem’ attacks no longer cut it, so a more intelligent form of introducing uncertainty and doubt are called for. These relatively subtle attacks may be based on an active assumption of either fraud or incompetence without any real evidence to justify such an assumption, but constant repetition of the false premise that there are ‘anomalies’ is apparently seen as an adequate substitute.

          Presumably the intention is to minimise the effect of the positive outcome of the tests and to attempt to introduce further delays in steadily widening acceptance. However such trolling is coming very late in the day, and as industry ‘movers and shakers’ become increasingly involved (seen and unseen), its effects can only be marginal at most.

          Still, it may keep the wolf from the door, for a while at least.

          • Thomas Clarke

            I apologise, in my reply above I misunderstood the thrust of your comment. You are saying that I am not doing ad hom attacks, but a “relatively subtle attack that may be based on an active assumption of either fraud or incompetence”.

            I see you are making a general point. Perhaps you would like to consider the facts of this case to see whether any such assumption is part of my argument?

            There is no denying the anomaly in the electrical measurements. It can be explained
            (1) by a change in heater resistance that appears impossible, and certainly is inconsistent with the tester’s statement that the heater is Inconel wire.

            (2) By a mismeasurement of input power (such that the real input would be enough to account for the output with COP ~ 1) or mismeasurement of the current.

            I’m saying that a lack of reply to this matter, given that the testers state in their report that they have the data stored which would allow them to do so, is most surprising. I’m also saying that the report’s conclusions cannot be trusted when such an anomaly exists.

            It is surprising to me that the testers have not yet replied to this anomaly, because it is so serious and any scientist would want to put the record straight as soon as possible. If I assumed fraud or incompetence then I would not be so surprised.

            I can assure you that if the testers resolve this anomaly in a way consistent with their data and that preserves the low power in measurement the report will be significantly more convincing to “movers and shakers” than it is now.

            Best wishes, Tom

      • Thomas Clarke

        @Fortyniner & Freethinker

        I’m not sure who you accuse of such attacks? I’m pointing out that an anomaly in the data that could be explained by a mismeasurement of power in has not been resolved. I’m further saying that this anomaly can be resolved (giving as explanation a variable resistance heater coil or the certainty of an error in one or other set of measurements shown in the report) by data the testers say they have safely stored on a USB stick for exactly such purpose.

        Now – which of those two statements do you believe either “false” or “subtle” or “outside the scope of the test”?

        Given you agree both are true, you may think of some other reason why the testers have not yet addressed this anomaly. In science it is usual, when somone detects an error in a paper, to issue a correction. There is no ad-hom attack, nor personal animosity, in this process.

        I am expressing surprise at the length of time it has taken the testers to respond, but hope that soon they will resolve the matter and either retract their claims as based on some not understood measurement error or modify the report stating exactly what was the error. This (address errors and concerns) is what they said they would do when pre-publishing the paper.

        • Freethinker

          Thomas Clarke, you are slick. But you have already shown your true face.

          You can question these things all you want but the fact remains:

          * The details of the output of the control box is part of the black box.

          * The details of the impedance of the active reactor is within the black box.

          * The details of the alloys used or any other details of the reactor innards is in the black box.

          The scope if to measure power in and out. Power out proxied by the temperature of the reactor surface. You have those data. Those data give at hand that the claims in the report are reasonable and thus stand.

          Whatever you think you are expressing, it is plain to people who has followed the ECAT on this site, that you are biased. You make that clear, either consciously or unconsciously, in this sentence:

          “I am expressing surprise at the length of time it has taken the testers
          to respond, but hope that soon they will resolve the matter and either
          retract their claims as based on some not understood measurement error
          or modify the report stating exactly what was the error.”

          Like it is either wrong or it is wrong. You do not consider that there is NO error, because you already KNOW there is an error.

          You are a true pathoskeptic.

          • Thomas Clarke

            I know that there is an error because it is possible to work out the heater resistance from the dummy and active test data and the two values do not agree, as they should for an Inconel heater element. Perhaps I have made a mistake in this: but everyone I know competent to evaluate the matter agrees.

            The point is that this error can be resolved by the report authors, after which there would be no error, or some measurement issue that they cannot identify and agree is an error.

            I’m not sure why being able to identify a clear error in a report makes me a pathoskeptic?

            As for details: the testers stated that the heater element was inconel – so they obviously felt competent to evaluate this. The resistance of the heater under any test condition is contained in the stored experimental data the testers have collected – so clearly it is “out of the box”. In fact the anomaly has come to light exactly because this resistance was published in the report.

            The impedance of the heater is not relevant to this question except for its series equivalent resistive component.

          • Freethinker

            “I’m not sure why being able to identify a clear error in a report makes me a pathoskeptic?”

            As said so many times before: because you base your argument on information that is not relevant, not there, and you know that you are right that the ECAT cannot work.

            Had you been just a tiny spec more humble, realizing that what you have is conjecture, it would be another thing, but you don’t, so you are a pathological skeptic, as that kind of behavior is the hallmark of such an individual.

            And you will not rest with the information you get, were it to show NO error at all. Pathoskeps are insatiable, and will never accept any level of evidence, but constantly raise the bar of what is evidence, regardless of new information that is presented.

            So, yes Thomas Clarke, you are a pathological skeptic.

          • Thomas Clarke

            Just one new point to answer. Were the electrical input anomaly to be resolved there would, it is true, be many other issues about this report. They have all been raised.

            You are free to think that my seeing these issues as valid indicates bias – just as I am equally free to think that your dismissing them as invalid indicates your bias. You will see that such argument is entirely unhelpful. It is only by identifying the detailed evidence, so that others can judge it for themselves, that progress can be made.

            The electrical current issue is a little different from all the others because it is definite, and indicates an error that can be further investigated and corrected by the report authors now. That is surely worth doing.

          • bachcole

            I presume that he has not been abusive (since he is still here), so that is better than being a maryyugo.

            Thomas, if Colombus or Edison or Magellan or a 100 other discoverers and explorers had your attitude, the world would be a very different place.

          • Freethinker, this is what Thomas Clarke just wrote on theeestory.com

            “I’m a bit surprised that Fibb and parallel choose to defend the indefensible on this thread. The case for LENR is not strong, but as it is so much less weak than the case for Rossi it bemuses me that they defend him.

            This latest test, as others have said, is not the first time Rossi has been shown to make unsubstantiated technical claims. He has a long history of this. He has also repeatedly lied on his blog. All of that behaviour is quite different from typical LENR researchers. Defending Rossi does them no justice.”

            http://theeestory.ning.com/xn/detail/6495062:Comment:124236

            note his strong anti Rossi bias, that he tried to soften here. I’m glad you saw right through his BS. Well done

          • Thomas Clarke

            Fibb – you seem personally interested in my posting here. Do not however confuse me (I have a clear public identity) with any anonymous posters on other sites. Thanks.

            For your idea that I have a strong anti-Rossi bias. That must be opinion, but it is true (and i would not hide) that I find many aspects of Rossi’s statements and the tests over the last 4 years to be problematic. In particular it is surprising that a device that works with “LENR+” capability should prove so difficult to test in a way that gives clear results. I expect the testers, having now conducted 3 tests, all criticised, share my frustration with this.

            Perhaps more than you, I try to keep ad homs and feelings out of comments on technical stuff. It helps. In this case the latest report has one clear inconsistency which can and should be resolved by the testers. You could agree with that, expecting the resolution to be one that makes the report more convincing and preserves the stated COP of >3.

            I cannot see what you find objectionable about my statements on this thread.

            Moderators – the anonymous poster Fibb thinks that I am the same as some anonymous poster on another site he knows. Whether this is true or not it would seem only relevant as an ad hom. Please, if your blog policy forbids ad homs, delete his post.

          • Frank Acland

            approved

          • Thomas Clarke

            It is your site, but I’d like to comment here on your policy.

            I post here under my true name, Anyone can find out who I am easily, and I have a presence on the web related to my work. I have never posted on this site or related ones under any other ID.

            It is thus not relevant, and clearly an ad hom, to claim I am some anonymous poster.

            It is not my way to troll and while I differ with you guys on how likely I consider Rossi’s stuff to be real, and doubt I would post here much, I’m here now for one specific reason. I’m not flaunting my general views (which would become repetitive and boring, as well as unwelcome here) because I see no point in that.

            The current independent test report contains an error that can be addressed by the authors. If the measured COP is real they will be able, with additional data that they certainly possess, to clear up the error in a way that greatly strengthens the report, which I’m sure all here would like. If the measured COP is in fact not as stated in the report then it is equally better this be made clear rather than left uncertain.

            The authors of the report have said they will read and respond to internet comment, hence my comment.

            I’m not alone, or original, in doing this. A number of others here, as skilled in the electrical issues here as me or more, have a similar view to mine.

            The specific matter here is that the authors of the report have stated informally that they reckon clamps are not wrongly connected. That would have explained the error. What those unskilled in the electrical stuff may not realise is that if that explanation does not work then there must be some other explanation.

            The report authors have the ability to clear this matter up. I very much hope that they will choose to do this. If the report is to be considered a scientific document (and some here have argued that it is not) the authors have a duty to follow up this matter.

            I hope very much they will do this. I’ll continue to post on this specific matter so that the testers, in particular, realise what is the issue here. I have not heard from them that they actually realise the precise problem, though they are obviously responding to some extent to concerns.

            They may, for example, think that the reversed clamp issue was the main problem. It is not. It is one possible solution to a problem that needs a solution.

  • bachcole

    ZZZZZZZ Only of interest to technophiles who aren’t paying and/or who can’t pay attention to the social evidence.

    • psi2u2

      Now now, let’s be merciful. 😉

    • ivanc

      what kind of scientific evidence is “social evidence” ?

      • psi2u2

        Significant evidence.

      • georgehants

        Socially it has been known for thousands of years by ordinary people, thousands of obvious, commonsense Facts that have been and still are, completely ignored by our scientific experts.
        For example animals are conscious and have feelings.
        Millions of animals tortured and abused because science has been so dumb in believing they are just unfeeling fodder for them to do with as they like.

      • bachcole

        Social evidence is seeing that the professors are competent since they have been doing critical thinking and science 16 hours a day since they got to their 15th birthday. They also have their entire careers to lose (which most scientists love) if they are lying or incompetent. They also have to put up with castigation by so many people who disbelieve. Why would they do this if what they are seeing is not true? That is social evidence.

        Industrial Heat’s going out on a limb and investing real money in this project. They stand to lose the money and their reputation. Darden has an exceptional reputation; do you think that he would throw all of this away on a lark? There are scores of researchers who say that they have gotten excess heat. Are all of these people stupid, corrupt, or incompetent? That is social evidence, and frankly it is overwhelming.

        There are many, many other researchers who also stand to look like morons if LENR is not true. There is no explanation for their behavior other than the fact that LENR IS true.

        Remember that virtually all of the so-called laws of elementary particles that you and so many others believe in and are trying to defend were established using “tools” that have velocities that are approaching the speed of light. Resonance, vibration, and slow elementary particles play no role in the formation of these so-called laws. So there is plenty of wiggle room for the grandly well established laws of elementary particles to be only a subset of how things work.

        • Thomas Clarke

          This is a very broad post and to reply in full would take time.

          I can understand why what you call “social evidence” seems to you overwhelming. It is needed, since the scientific evidence is currently very weak.

          I would recommend that you consider these matters without personal attacks or animosity. Mostly, scientists are motivated by a search for truth, not glory, nor animosity. It works best if you assume everyone is honest and concentrate on the scientific facts, since then they can be debated fairly without people losing their temper.

          I think you misunderstand the nature of science. Most scientists never make great discoveries. A scientist going out on a limb and hoping for proof of a great discovery is not considered a moron. A scientist who makes a mistake in interpretation and chases a theory which is in fact wrong is also not considered a moron, such things happen all the time to the best of scientists.

          Rather, scientists who do not provide good evidence for their statements tend not to get published. It is OK to make outlandish speculations as long as you do not claim you have evidence for them which does not exist.

          As for IH, you will see that many high tech companies have proposed to make money from ideas that frankly do not work at all. Obviously if Rossi’s device worked as billed it would be immensely valuable. Perhaps enough to convince a VC that a big risk is justified. Or maybe also that the story has enough traction to attract other investors and therefore the risk for an initial enabling investor could be minimised. Really there are so many possibilities here that speculation is fruitless.

          The evidence for nuclear theory comes from many many different things, not just particle accelerators. Resonances, vibrations, and slow elementary particles play a strong part in both the theoretical and experimental work that supports existing theory.

          One final point. “LENR is true”. That statement may seem clear to you, but as a scientific hyponthesis it is very weak, Let me ask you, how would you disprove “LENR is true”. If you cannot think of an experiment which would do so, by invalidating some prediction made, then the statement does not yet correspond to anything in science.

    • Freethinker

      They are not paying attention to the technical evidence either. They rather use information they do not have and multiply with their preset notion that it cannot work, and create conjecture.

    • Dods

      I think this Bruce Lee quote summarises the situation perfectly.

      “Its like a finger pointing away to the moon. Don’t concentrate on the finger or you will miss all that heavenly glory.”

  • ivanc

    The question is wrong. there is no negative currents in AC. so if the testers say No they Right they are Measuring absolute values RMS. The order of the clamps is important, or the internal order of the phases inside the controller.
    The real question to the professor is : “Do you have the voltage Readings?” “Could you Please release them?”.
    The current is no issue it given by the joule heating, as Joule_heating=I^2 R and R is given in the report.
    But maybe they could release the reading of the current too.

    The issue here is the drop in value of the resistor by a factor of 3.3.

    • Gerrit

      Ivanc, what are you planning to do when you will finally have all the answers for everything you are currently doubtful about?

      What are you planning to do when it turns out things are really the way they appear and the ecat is working?

      • ivanc

        I will be happy we have an ecat that works.
        At the moment we do not know that.

        • Gerrit

          I agree that at the moment we cannot scientifically exclude the possibility that the ecat doesn’t work, even when we have solid indications that it does.

          But even if the ecat really works today, we will have final confirmation only when (if) it has been in operation at a customer site for several months and we are all able to verify that information.

          The small scientific work that has been performed in the last 25 years in the field of LENR strongly indicates that LENRs are real.

          We could have had a scientific consensus several years ago had the larger science community investigated this topic sincerely instead of dismissing and ridiculing it.

          • Thomas Clarke

            The current report, if convincing, would surely be confirmation, or at least evidence, that e-cats work as billed. And while e-cats working as billed would probably mean LENR occurring – LENR occurring certainly does not mean that e-cats work.

            Rossi has shown a remarkable interest in obtaining scientific proof for his device. And proving that an electric heater that provides 3X as much heat out as electric power in is surely not so difficult.

          • Gerrit

            I have explained in previous posts why Rossi had the public demonstration back in 2011. He did not initiate the third party test, the third party did.

            Apparently it is much more difficult to prove a COP of 3 to a bunch of nitpickers than any scientists would have imagined.

  • psi2u2

    Now now, let’s be merciful. 😉

  • psi2u2

    As LENR G says in the first post to the thread, its good to hear from the testers and many of us look forward to them explaining what they did and experienced.

  • psi2u2

    As LENR G says in the first post to the thread, its good to hear from the testers and many of us look forward to them explaining what they did and experienced.

  • Thomas Clarke

    The testers could easily confirm or deny the drop in resistance by 3, using the active run data they have collected. It seems they pay attention to the matter so I would expect them to do this?

    I agree with ivanc – the question is wrong, and if they did not realise this they are misunderstanding.

    The RMS line currents will always be positive. The line powers, not normally displayed, but available, will have one negative.

    • Freethinker

      Thomas Clarke,

      the instrument would show whatever is fed into it and what it consequently was ordered to show. If it was attached to the control box output, then that is what the control box output is generating at the moment of measurement.

      Further, your constant trolling on the subject of the aka “factor 3” – which is in fact rather a factor of 2.5 and can be explained either by impedance differences in the active vs. dummy reactor, or simply by making a few simple calculations where in fact the voltage is taken into consideration – will lead you nowhere. It is in the black box, and you will likely not get that information.

      • Thomas Clarke

        I wonder where you get 2.5 from? The anomaly is actually 3.3X change in power measurement or resistance between dummy and active tests.

        I agree it could be explained by supposing that the active reactor used a different (and lower resistance) heater element. But that is surely not what the testers were told: they say the active reactor is the same as the dummy but with powder added.

        I don’t understand your “simple calculations where voltage is taken into consideration”. If you could elaborate I’d be able to comment.

        The information that will resolve this is, fortunately, outside the black box. It is the power and current measurements already taken and now on a securely stored USB stick, we are told.

        • Freethinker

          And why would I tie up myself a nice Sunday, to repeatedly tell you that you are WRONG. I have given DickeFix a suggestion on one way to look at it with numbers. And the 2.5 you get by dividing the P=R*I² formulas for the joule heating. Your 3.3 bs must be from the clamp reasoning which is clearly bogus.

          Are you getting paid for weekend work?

          • Thomas Clarke

            I did this quite a while ago. It explains the 3.3 factor. The columns of a spreadsheet come out one after another – sorry for the lousy formatting.

            Wire Joule Heating/W
            Total Input Power/W
            ratio
            real/dummy

            dummy (500C)
            6.7
            486
            72.53731343
            3.332784711

            real (1250C)
            36.8
            796.7
            21.64945652

            real (1400C)
            41.7
            912.4
            21.88009592

          • Freethinker

            And you flaunt it as were it the truth.

            The only place you see 3.3 is the ratio between the joule heating and the power-in for the dummy case.

            I assume you intend to apply the P=R*I² formula and divide the joule heat case for the dummy and the active reactor? That give you a current ratio of about 2.5. Now take the case for the 1400C vs the dummy and take the ratio of the input power, you get 912/486=1.9. Now remember that number.

            Take into account the voltage RMS if we take 50A per line as the RMS value for the 1400C case and we know we have 19.7 A per line for the dummy, we have Ua=912/150= 6.1 V and Ud=486/59.1=8.20 V. The ratio in voltage Ua/Ud=0.75. Apply that to negotiate the in-power using the current ratio and the voltage ratio, P=U*I, Pa/Pd=Ua/Ud*Ia/Id -> Pa/Pd=0.75*2.5 = 1.9. And that is the same as 912/486=1.9. Hence there is no monkey business going on.

            But it is conjecture, as there is too much assuming, same as in the things you are doing. Also it is totally unnecessary, as it is outside the scope, and the data we need to check the claims are there, and are viable.

            I trust this does not suffice to convince you?

          • Thomas Clarke

            There is no conjecture. And what I do is not what you suppose above.

            Pj = I^2Rj
            Ptot = I^2Rtot
            Rj = wire resistance (constant)
            Rtot = wire + heater resistance ~ heater resistance

            If we take the ratio Ptot/Pj you will see that I (and V) drops out and we look at the ratio Rtot/Rj. Which should be constant.

            We calculate the Ptot/Pj ratio for the three tests and get:
            dummy: 72.5
            active (1250C): 21.65
            active (1400C): 21.88

            The two active ratios are the same, as you’d expect. The slight difference would be from the small PTC of inconel wire between 1250C and 1400C.

            However the dummy ratio is 3.3 X larger than the active ratio.

            The 3.3 comes from (Ptot/Pj)dummy / (Ptot/Pj)active. Not what you state above.

            This indicates Rtot, and hence the heater resistance, some 3.3 X larger or else a mistake somewhere between the dummy and active tests.

          • Freethinker

            “There is no conjecture”

            But you are blind. You conjecture that the R is the same, you conjecture the voltage is same – it does not “drop out”, it is implicit in the values you use. And you compare two completely different situations.

            You have no problem rejecting the claims that are within he scope, but you have no problem building arguments based on conjectures and the confabulations that stems from the fact that you are already convinced that LENR is pseudo science and ECAT is in extension not capable of working.

          • Thomas Clarke

            You will see from the equations I use that I make no assumption, or conjecture, about the voltage.

            I use the numbers in the report to conclude that either there is an error or the heater resistance varies by a factor of 3.3. That is all.

            I am not “conjecturing R is the same”. But a large difference needs to be explained. In this case there is no easy way to that – hence the problem. You might want to refer to others, not me, for why a heater with the characteristics needed is unlikely to exist. You can refer to the Inconel material data sheets for why all Inconel wires do not have the required behevaviour. That is why I am confident saying there is an error

          • Freethinker

            Thomas Clarke,

            You are making implicit assumptions of the voltage as you use the values of the joule heating and the in powers to in the base of your argument and saying that the heater resistance varies by a factor of 3.3.

            But you want to bring this line of argument into the out of scope domain, and you keep repeating, over an over again, the same conjectures for which you have no base. You do not have no information, no correlated information and only your own confabulations, thats all. It is all conjecture.

            But this does not matter to you as though the repeating of the same infinitum will make it a reality.

            You are out of scope, and your conjectures are baseless, you have a strong conviction before this argument that Rossi’s ECAT cannot work. You have even admitted it indirectly as you claim you KNOW the report is wrong, and has errors.

            You refuse look at the scope and the data presented. You have no problem rejecting that which is very well described and quite uncontroversial, but at the same time embrace notions that are very much ill founded, based on data that the scope do not cover.

          • Dr. Mike

            Freethinker,
            Please Goggle “3 phase power equations”. Perhaps after checking out these equations you will not calculate the RMS voltage as Power/(3*RMS line current) for a 3-phase system. However, you are correct in that for the power numbers to be correct in the Lugano report, the active run voltages must have been lower than the dummy run voltage. Were the active run voltages really lower than the dummy run voltage? Was the potentiometer setting on the TRIAC controller lower for active runs than the dummy run? Was the potentiometer turned up initially past the dummy setting and then turned down as the LENR reactions started producing power to reach the first operating point? If the authors tell us this is how the active run was ramped up in power, then we will know that something strange is going on in the active line current, since there is no known material that can have an initial resistivity of about 100 microhm-cm at 450C and then have the resistivity drop by a factor of 3.3X at 1260C, but doesn’t drop further going to 1400C. The authors need to put an explanation in the revised report for why the higher power active run voltage was lower than the dummy run voltage, if this actually happened. If they don’t know why the active run voltage was lower than the dummy run voltage they should state that in the report. If the potentiometer setting for the active runs was higher than for the dummy run, the authors should realize there is a problem with their input power numbers for the active runs. A simple question: What were the potentiometer settings for the dummy run and the two portions of the active run?
            Dr. Mike

          • Thomas Clarke

            I agree mostly with this, but I’m not sure the potentiometer setting gives useful information. It will (I imagine) set the temperature to which the device is heated. But the heat could come from the input power or some putative LENR effect. Because of feedback the pot setting would likely be correlated to the temperature and nothing else.

            The crucial evidence for varying resistance (or not) comes from the currents and powers measured as the device heats up on the active test.

          • Dr. Mike

            Thomas,
            The data from the ramp-up portion of the active run is crucial for determining what is going on. Maybe one day we will see that data!
            Dr. Mike

          • Freethinker

            Mike,

            You are as lost as are Thomas Clarke. You also fail to grasp what is the scope. You do not have a clue as to what the control box actually deliver in terms of current and voltage, except that you know the RMS values of the current for the dummy, and a hand waving number for the active run. What you do have is the input power. The total input power. And that is the important number.

            Your claim that the authors “need” to explain anything is ridiculous. Again: the scope is to measure input power, and the temperature of the reactor surface. Those data has been adequately explained and as such the claims are reasonable.

            All your questions you pose only serve to illustrate my exact thesis – there is a lot of no information, irrelevant information, uncorrelated information, and pure confabulations that goes into conjectures there the claims in the report are invalid.

          • DickeFix

            Freethinker, as I tried to explain to you, you fool yourself with your calculation.

            1. You start with calculating the current ratio from Joule heating Ia/Id=2.5. This agrees well with what is stated in the report Ia=50A and Id=19.7A

            2. You calculate the voltage ratio Ua/Ud from Pa/Ia and Pd/Id assuming Pa and Pd are correct. You get Ua/Ud=0.75

            3. You multiply the result in 2 with the result in 1 and get Pa/Pd=Ua/Ud*Ia/Id

            Of course this agrees with the measured Pa/Pd that you used in 2. It is a circular proof!

            4. You should instead compare the Ua/Ud you get in 2 with what it should be according to Ohms law: Ua/Ud=Ra/Rd*Ia/Id

            Two possibilities:
            a) If Ra=Rd you get Ua/Ud=Ia/Id=2.5. Hence the Ua/Ud you get in 2 is wrong and then Pa or Pd must be wrong
            b) To get agreement between 4 and 2 you need Ra/Rd=0.75/2.5=0.3 which is impossible in an Inconel wire and probably also in a semiconductor heating element unless you have a very large bandgap.

          • Freethinker

            It is no more circular than your numeric nonsense. See that is my point, you do not have ample data. You claim the input power to be wrong. Then explain that without the conjecture and other confabulatory bs you are doing now. You have nothing to show, but you hot desire to disprove this.

          • Obvious

            R does not vary (much). The Watts are Joule Watts in the report (Watts/s).
            Turning the device on only 30% of the time fixes the math perfectly. It can be 30% of a day, an hour, a minute, a second, 100’s of a second. Whatever. As long as it is constantly 30% of the 100% total time, the electrical Watts stay the same, but total Power will be 30% of the possible 100% suggested by the electric Watts. Because we don’t measure the electric circuit when it is off, but time still passes.

            Edit: What I mean is that if you multiply the RMS current by three and current is a part of Power in Joules, and the three times multiple is a ratio, then you must multiply time by three also to have a consistent result in Joules.
            IE: 100 W/s with 1 Ohm, 1 Volt, 100 A. With R and V constant:
            100 W/s times three is 300 W/ 3 seconds. Not 300 W in one second.
            —– 100W/s(3/3), not 100W/s(3/1) —-

          • Thomas Clarke

            You repeated this post below (some way!) I refer those interested to my reply and further comments there.

          • bachcole

            What if muons play a role in LENR? How can muons NOT play a role in LENR, given that there is no attempt to shield them. Perhaps the role is exceedingly insignificant. But I see no way that the muons can’t effect LENR. A very ignorant but curious mind wants to know.

          • ivanc

            You have not check the equations, I been waiting for your feedback of what is wrong with them!, do I have to assume you could not find any thing wrong?

          • Obvious

            Sorry, Ivan. I have been working out a concise reply, without bad math (this time).
            The delta-wye transformation equivalence transformation equations?
            1/3—>zero<— 3/1?

          • ivanc

            Please take in account that the power supply has to see this two different circuits (D to Y) (Y to D) equivalent.!

  • Deepak Saharan

    I have been following ecat from almost 4 years. I have slowly started kind of hating theorist of physics. Yes , I accept that theory is must for replicating and understanding . But look at the times of today where physicist have gone to membrane theory and origin of universe. All these things are creating mindsets which I feel is major road block for something new. From half century we have spent trillions of dollars to understand the nature of universe and what is the output??. I am not questioning the cognitive abilities of physicists but practical importance of these theories have been negligible or even negative for some cases.

    • georgehants

      Deepak, Many Wonderful theorists and they I think are very important in the Understanding of our reality, they cost very little.
      I think it is the administration that then allows billions to be wasted on the pursuit of politically motivated schemes, far to much spent on particle physics that in past years made great discoveries at very little cost, Hot Fusion is the craziest scam and of course silly grandiose ideas in space etc, billions on unproven global warming. etc. etc.
      Many of the working scientists could feel good about their lives if they where doing something to help the World, such as Cold Fusion etc.
      In the end, it is I think up to the working scientists to form a union and demand that they determine the direction of science and not the religious like priests that they now allow to make them all look like wasteful automatons.
      The Wonderful theory below has cost very little but adds much to the excitement and fun of science.
      https://medium.com/the-physics-arxiv-blog/deeper-than-quantum-mechanics-david-deutschs-new-theory-of-reality-9b8281bc793a

    • Over the last few decades, more money and ingenuity seems to have gone into shoring up the ailing Standard Model/Big Bang theory, than into any effort to start again, this time to try to understand what the data is actually telling us. Just as for fake ‘climate science’, careers and reputations depend on not rocking the boat, and whole branches of science research become self-perpetuating, inward looking and essentially irrelevant.

      Until the observations of dissenters such as Halton Arp are taken into consideration and currently taboo theories such as ‘Electric Universe’ are considered objectively, we will stay locked into pointless ‘big physics’ with its continual attempts to find evidence that supports established ideas, rather than seeking to formulate new ones that actually fit the data (the incredibly expensive LHC experiment to find a ‘Higgs boson’ is a classic case in point).

    • Warthog

      No, theory is NOT a “must for replicating”. That is solely the purpose of experimental measurements. Theory CAN be a tool for understanding and possibly for prediction, if it posits NEW experiments that can be run. This idea that no phenomenon is “real” unless it agrees with theory is a gigantic bastardization of the scientific process.

      • bachcole

        Outstanding. Theory would be great for optimizing, but not replicating. Gravity and the mathematical modelling of gravity’s behavior has been replicated hundreds of thousands of times, perhaps millions of times, but we still don’t have a theory of why and how it works.

        • Thomas Clarke

          No, but it makes clear and very precise predictions which can be checked. If these turned out contrary to GR, we would need some other theory. There are various other theories which deliver different experimental results. However, so far, I believe, GR is still winning this one on experimental accuracy. It would be exciting to find strong evidence for some alternate theory.

      • Thomas Clarke

        It is rather that no phenomenon is understood unless it agrees with a theory.

        For example, replicable excess heat in Ni-H systems could be:
        experimental error
        lattice-based chemical effects
        aliens
        God
        LENR

        Each of these hypotheses might, if suitably elaborated, explain the (real) results.

        • Warthog

          Sorry, but no. Simply wrong, and wrong-headed. Science is about replicated experiments….NOT theory. ANY theory can be pitched out at any time if even a single well-executed experiment contradicts it. The idea that an experiment has to be discarded if it doesn’t agree with “a theory” is turning science 180 degrees. And any supposed “scientist” who claims the latter should have any and all degrees in science rescinded, and never be allowed to enter a laboratory again.

          The pathological skeptics, though, constantly do precisely that. And they also TRY, with more or less success, to execute the punishment they themselves deserve, on researchers who report positive results on LENR.

          • ivanc

            I fully agree with you

          • Thomas Clarke

            What I say is npt contradicted by what you say. I agree that experiments determine which theories are correct.

            That is why a theory that cannot be disproven by any experiment should be looked at with great suspicion.

            In this case if experiments show heat excess in Ni-H systems that is all good, and needs explanation. But it does not show LENR, unless the LENR theory wins over all the other theories that predict the same excess heat.

            So i don’t know if i’m a pathoskeptic – but i certainly don’t recognise what you say they do. The issue is which theory best predicts the experimental results. All of them.

            EDIT NB ivanc is fully agreeing with warthog not me!

          • ivanc

            The theory does not matter, the data is what matters, once you have the data right you could work out the theory, the theory could change, the data no.
            So I fully agree that if a scientist refuses to see the data, He should be forbidden to enter in a lab.
            But do not expect a scientist to believe you if you do not give the data!!!!!!

    • Joniale

      I agree with your comment.Besides, they try to break the progress of these practical experiments putting the scientific method as excuse.As said before that is a missinterpretation of the scientific method that shows that the people is also corrupt in science. The facts,proofs and evidence should be enough to trigger curiosity and avoid black mailing in science for other personal reasons. The science is at the end done by persons who have their own agendas.
      What i want to trigger an really see is an exercise of autocritic in science about what has happenned with cold fussion since 1989. I think it is evident how the facts and proofs has not been taken serious and how the area has been blocked. That is embarrassing!!.history should put this well clear how this happenned as already happenned many times before.But that’s not enough, it’s needed a way of control so that lobbies and other personal mindsets are excluded from the acceptance process. What’s important are the facts and not the authorities it’s important to force an spirit of open mind in science for the facts. I don’t want the people to believe but to listen and study the announced facts.
      Science must study the announced facts and don’t get lazy with the excuse that has been proven wrong before or the general consensus is other.
      I really want to see that after LENR is out. This immobilized science has costed us a lot this corrupt behavior and lazyness has costed us maybe the contamination of the planet by letting oil to stay.

  • US_Citizen71

    It is being done by committee.

    • Thomas Clarke

      That is , I agree, a likely reason for delay. But it is no excuse!

  • psi2u2

    Significant evidence.

  • Gerrit

    Ivanc, what are you planning to do when you will finally have all the answers for everything you are currently doubtful about?

    What are you planning to do when it turns out things are really the way they appear and the ecat is working?

  • georgehants

    Deepak, Many Wonderful theorists and they I think are very important in the Understanding of our reality, they cost very little.
    I think it is the administration that then allows billions to be wasted on the pursuit of politically motivated schemes, far to much spent on particle physics that in past years made great discoveries at very little cost, Hot Fusion is the craziest scam and of course silly grandiose ideas in space etc, billions on unproven global warming. etc. etc.
    Many of the working scientists could feel good about their lives if they where doing something to help the World, such as Cold Fusion etc.
    In the end, it is I think up to the working scientists to form a union and demand that they determine the direction of science and not the religious like priests that they now allow to make them all look like wasteful automatons.
    The Wonderful theory below has cost very little but adds much to the excitement and fun of science.
    https://medium.com/the-physics-arxiv-blog/deeper-than-quantum-mechanics-david-deutschs-new-theory-of-reality-9b8281bc793a

  • georgehants

    Socially it has been known for thousands of years by ordinary people, thousands of obvious, commonsense Facts that have been and still are, completely ignored by our scientific experts.
    For example animals are conscious and have feelings.
    Millions of animals tortured and abused because science has been so dumb in believing they are just unfeeling fodder for them to do with as they like.

  • Freethinker

    No, Thomas Clarke,

    as has been explained to you and ivanc many times before, it is you who are wrong, as you assume too many things, do not have the relevant information as you look at things outside of the scope of the test, consider uncorrelated information to be correlated, then confabulate based on your preset bias that the ECAT does not work and build up your conjectures that the claims in the report are void.

    Your conjectures are rubbish and disingenuous, and does not hold water.

    • Fortyniner

      Name calling, ridicule and ‘ad hominem’ attacks no longer cut it, so a more intelligent form of introducing uncertainty and doubt are called for. These relatively subtle attacks may be based on an active assumption of either fraud or incompetence without a shred of real evidence, but constant repetition is seen as an adequate substitute.

      Presumably the intention must be an attempt to minimise the effect of the positive outcome of the tests and to introduce further delays in acceptance. However such trolling is coming very late in the day, and as industry ‘movers and shakers’ become increasingly involved (seen and unseen), more and more irrelevant.

    • Freethinker

      Thomas Clarke,

      the instrument would show whatever is fed into it and what it consequently was ordered to show. If it was attached to the control box output, then that is what the control box output is generating at the moment of measurement.

      Further, your constant trolling on the subject of the aka “factor 3” – which is in fact rather a factor of 2.5 and can be explained either by impedance differences in the active vs. dummy reactor, or simply by making a few simple calculations where in fact the voltage is taken into consideration – will lead you nowhere. It is in the black box, and you will likely not get that information.

      • bachcole

        Social evidence is seeing that the professors are competent since they have been doing critical thinking and science 16 hours a day since they got to their 15th birthday. They also have their entire careers to lose (which most scientists love) if they are lying or incompetent. They also have to put up with castigation by so many people who disbelieve. Why would they do this if what they are seeing is not true? That is social evidence.

        Industrial Heat’s going out on a limb and investing real money in this project. They stand to lose the money and their reputation. Darden has an exceptional reputation; do you think that he would throw all of this away on a lark? There are scores of researchers who say that they have gotten excess heat. Are all of these people stupid, corrupt, or incompetent? That is social evidence, and frankly it is overwhelming.

        There are many, many other researchers who also stand to look like morons if LENR is not true. There is no explanation for their behavior other than the fact that LENR IS true.

        Remember that virtually all of the so-called laws of elementary particles that you and so many others believe in and are trying to defend were established using “tools” that have velocities that are approaching the speed of light. Resonance, vibration, and slow elementary particles play no role in the formation of these so-called laws. So there is plenty of wiggle room for the grandly well established laws of elementary particles to be only a subset of how things work.

        • Freethinker

          You simply do not grip the concept of scope, Thomas Clarke.

          The scope is to compare the power-in to that of power-out as proxied by the thermal measurements of the reactor surface.

          The joule heating was there to address the component from a thermal measuring perspective, for the output power. Hence it was measured within the context of the scope of the report, with presented with relevant detail for that purpose. It is so shown that the joule heating is of little or no consequence for the outcome.

          The fact that you take two different situations and care to divide the power as a function of current for those two, uncorrelated, situations, without even considering the voltage, show that you are really using non information in your conjectures. Both you and ivanc are formula orthodox people that think you just have to apply the schoolbook formulas without asking yourself what data you have at hand.

          You are both meaning that the entire power diff emanates from the P=R*I² formula. For this you have no information that will substantiate your claim but you have only conjectures. I have given you and the other “factor 3” comrades two alternative explanations that are equally conjectures, but to show you that there are alternative explanations to what you see.

          But it does not matter what you think is going on in the black box. Because: if the output power is more than three times larger that what you put in, then the claims in the report stand. As it has been confirmed that no clamps were inverted, the input power can be relied upon. Note, that this result is there regardless of the pixies and unicorns you think you see in the black box.

          • Thomas Clarke

            I can see that we may have no meeting of minds here, but to address the new points you make:

            You are right that the Joule heating was not directly relevant to the intended result – it was supposed to be a slight power correction. Nevertheless its intention does not change its fact. The numbers given, which are directly a proxy for the heater current, do not match the power given unless the heater changes resistance. Voltage is irrelevant in this calculation, I can assure you. This extra measurement shows there is an anomaly.

            P = I*R^2 applies in this case. It would only be untrue if the heater was a nonlinear element. No-one has hypothesised this. The SCR-controlled waveform and the possibility of reactive components do not change this – it is a very robust conclusion, not a conjecture.

            The alternatives (if these are yours) would be R varying with temperature, or R different between the two tests. I agree both would explain the data. R varying with temperature seems very unlikely on material science grounds but in any case can be checked by the testers. They would want to confirm this if it is true since it would strengthen the report. R different between the two tests indicates some hidden change in the reactor that could invalidate the other assumptions and is certainly contrary to what the testers have stated. Again, they would need to clarify that in the report for it to be taken seriously by others.

            Finally you say it has been confirmed the clamps were not reversed. This is not relevant to the argument. If the error does not come from a reversed clamp then it much come from something else. An explanation is still needed. Your argument is thus wrong.

          • Obvious

            R does not vary (much). The Watts are Joule Watts in the report (Watts/s).
            Turning the device on only 30% of the time fixes the math perfectly. It can be 30% of a day, an hour, a minute, a second, 100’s of a second. Whatever. As long as it is constantly 30% of the 100% total time, the electrical Watts stay the same, but total Power will be 30% of the possible 100% suggested by the electric Watts. Because we don’t measure the electric circuit when it is off, but time still passes.

          • Thomas Clarke

            That is not true. The PCE-830 measures power correctly, and current as true RMS. In this case regardless of duty cycle, or how that changes:
            P = IRMS^2R.

            The power is an average ove rtime, and the RMS current is the same. So in both cases the off period is taken into account.

            You can also see this in the report where the authors correctly use this equation, without needing to consider duty cycle.

          • Obvious

            Exactly. And the report show the actual measured Power used.
            But when we extract a pure electrical ratio from Joule heating, we are not factoring the time period. We also do not know the duty cycle of the dummy run.

      • Thomas Clarke

        I wonder where you get 2.5 from? The anomaly is actually 3.3X change in power measurement or resistance between dummy and active tests.

        I agree it could be explained by supposing that the active reactor used a different (and lower resistance) heater element. But that is surely not what the testers were told: they say the active reactor is the same as the dummy but with powder added.

        I don’t understand your “simple calculations where voltage is taken into consideration”. If you could elaborate I’d be able to comment.

        The information that will resolve this is, fortunately, outside the black box. It is the power and current measurements already taken and now on a securely stored USB stick, we are told.

        • Freethinker

          And why would I tie up myself a nice Sunday, to repeatable tell you that you are WRONG. I have given DickeFix a suggestion on one way to look at it with numbers. And the 2.5 you get by dividing the P=R*I² formulas for the joule heating. Your 3.3 bs must be from the clamp reasoning which is clearly bogus.

          Are you getting paid for weekend work?

          • Thomas Clarke

            I did this quite a while ago. It explains the 3.3 factor. The columns of a spreadsheet come out one after another – sorry for the lousy formatting.

            Wire Joule Heating/W
            Total Input Power/W
            ratio
            real/dummy

            dummy (500C)
            6.7
            486
            72.53731343
            3.332784711

            real (1250C)
            36.8
            796.7
            21.64945652

            real (1400C)
            41.7
            912.4
            21.88009592

          • Freethinker

            And you flaunt it as were it the truth.

            The only place you see 3.3 is the ratio between the joule heating and the power-in for the dummy case.

            I assume you intend to apply the P=R*I² formula and divide the joule heat case for the dummy and the active reactor? That give you a current ratio of about 2.5. Now take the case for the 1400C vs the dummy and take the ratio of the input power, you get 912/486=1.9. Now remember that number.

            Take into account the voltage RMS if we take 50A per line as the RMS value for the 1400C case and we know we have 19.7 A per line for the dummy, we have Ua=912/150= 6.1 V and Ud=486/59.1=8.20 V. The ratio in voltage Ua/Ud=0.75. Apply that to negotiate the in-power using the current ratio and the voltage ratio, P=U*I, Pa/Pd=Ua/Ud*Ia/Id -> Pa/Pd=0.75*2.5 = 1.9. And that is the same as 912/486=1.9. Hence there is no monkey business going on.

            But it is conjecture, as there is too much assuming, same as in the things you are doing. Also it is totally unnecessary, as it is outside the scope, and the data we need to check the claims are there, and are viable.

            I trust this does not suffice to convince you?

          • Thomas Clarke

            There is no conjecture. And what I do is not what you suppose above.

            Pj = I^2Rj
            Ptot = I^2Rtot
            Rj = wire resistance (constant)
            Rtot = wire + heater resistance ~ heater resistance

            If we take the ratio Ptot/Pj you will see that I (and V) drops out and we look at the ratio Rtot/Rj. Which should be constant.

            We calculate the Ptot/Pj ratio for the three tests and get:
            dummy: 72.5
            active (1250C): 21.65
            active (1400C): 21.88

            The two active ratios are the same, as you’d expect. The slight difference would be from the small PTC of inconel wire between 1250C and 1400C.

            However the dummy ratio is 3.3 X larger than the active ratio.

            The 3.3 comes from (Ptot/Pj)dummy / (Ptot/Pj)active. Not what you state above.

            This indicates Rtot, and hence the heater resistance, some 3.3 X larger or else a mistake somewhere between the dummy and active tests.

          • Freethinker

            “There is no conjecture”

            But you are blind. You conjecture that the R is the same, you conjecture the voltage is same – it does not “drop out”, it is implicit in the values you use. And you compare two completely different situations.

            You have no problem rejecting the claims that are within he scope, but you have no problem building arguments based on conjectures and the confabulations that stems from the fact that you are already convinced that LENR is pseudo science and ECAT is in extension not capable of working.

          • Dr. Mike

            Freethinker,
            Please Goggle “3 phase power equations”. Perhaps after checking out these equations you will not calculate the RMS voltage as Power/(3*RMS line current) for a 3-phase system. However, you are correct in that for the power numbers to be correct in the Lugano report, the active run voltages must have been lower than the dummy run voltage. Were the active run voltages really lower than the dummy run voltage? Was the potentiometer setting on the TRIAC controller lower for active runs than the dummy run? Was the potentiometer turned up initially past the dummy setting and then turned down as the LENR reactions started producing power to reach the first operating point? If the authors tell us this is how the active run was ramped up in power, then we will know that something strange is going on in the active line current, since there is no known material that can have an initial resistivity of about 100 microhm-cm at 450C and then have the resistivity drop by a factor of 3.3X at 1260C, but doesn’t drop further going to 1400C. The authors need to put an explanation in the revised report for why the higher power active run voltage was lower than the dummy run voltage, if this actually happened. If they don’t know why the active run voltage was lower than the dummy run voltage they should state that in the report. If the potentiometer setting for the active runs was higher than for the dummy run, the authors should realize there is a problem with their input power numbers for the active runs. A simple question: What were the potentiometer settings for the dummy run and the two portions of the active run?
            Dr. Mike

          • Freethinker

            Mike,

            You are as lost as are Thomas Clarke. You also fail to grasp what is the scope. You do not have a clue as to what the control box actually deliver in terms of current and voltage, except that you know the RMS values of the current for the dummy, and a hand waving number for the active run. What you do have is the input power. The total input power. And that is the important number.

            Your claim that the authors “need” to explain anything is ridiculous. Again: the scope is to measure input power, and the temperature of the reactor surface. Those data has been adequately explained and as such the claims are reasonable.

            All your questions you pose only serve to illustrate my exact thesis – there is a lot of no information, irrelevant information, uncorrelated information, and pure confabulations that goes into conjectures there the claims in the report are invalid.

          • Obvious

            R does not vary (much). The Watts are Joule Watts in the report (Watts/s).
            Turning the device on only 30% of the time fixes the math perfectly. It can be 30% of a day, an hour, a minute, a second, 100’s of a second. Whatever. As long as it is constantly 30% of the 100% total time, the electrical Watts stay the same, but total Power will be 30% of the possible 100% suggested by the electric Watts. Because we don’t measure the electric circuit when it is off, but time still passes.

          • ivanc

            You have not check the equations, I been waiting for your feedback of what is wrong with them!, do I have to assume you could not find any thing wrong?

          • Obvious

            Sorry, Ivan. I have been working out a concise reply, without bad math (this time).
            The delta-wye transformation equivalence transformation equations?
            1/3—>zero<— 3/1?

  • Fortyniner

    Over the last few decades, more money and ingenuity seems to have gone into shoring up the ailing Standard Model/Big Bang theory, than into any effort to start again, this time to try to understand what the data is actually telling us. Just as for fake ‘climate science’, careers and reputations depend on not rocking the boat, and whole branches of science research become self-perpetuating, inward looking and essentially irrelevant.

    Until the observations of dissenters such as Halton Arp are taken into consideration and currently taboo theories such as ‘Electric Universe’ are considered objectively, we will stay locked into pointless ‘big physics’ with its continual attempts to find evidence that supports established ideas, rather than seeking to formulate new ones that actually fit the data (the incredibly expensive LHC experiment to find a ‘Higgs boson’ is a classic case in point).

  • Freethinker

    They are not paying attention to the technical evidence either. They rather use information they do not have and multiply with their preset notion that it cannot work, and create conjecture.

  • Dods

    I think this Bruce Lee quote summarises the situation perfectly.

    “Its like a finger pointing away to the moon. Don’t concentrate on the finger or you will miss all that heavenly glory.”

  • Warthog

    No, theory is NOT a “must for replicating”. That is solely the purpose of experimental measurements. Theory CAN be a tool for understanding and possibly for prediction, if it posits NEW experiments that can be run. This idea that no phenomenon is “real” unless it agrees with theory is a gigantic bastardization of the scientific process.

  • Gerrit

    I agree that at the moment we cannot scientifically exclude the possibility that the ecat doesn’t work, even when we have solid indications that it does.

    But even if the ecat really works today, we will have final confirmation only when (if) it has been in operation at a customer site for several months and we are all able to verify that information.

    The small scientific work that has been performed in the last 25 years in the field of LENR strongly indicates that LENRs are real.

    We could have had a scientific consensus several years ago had the larger science community investigated this topic sincerely instead of dismissing and ridiculing it.

  • Joniale

    I agree with your comment.Besides, they try to break the progress of these practical experiments putting the scientific method as excuse.As said before that is a missinterpretation of the scientific method that shows that the people is also corrupt in science. The facts,proofs and evidence should be enough to trigger curiosity and avoid black mailing in science for other personal reasons. The science is at the end done by persons who have their own agendas.
    What i want to trigger an really see is an exercise of autocritic in science about what has happenned with cold fussion since 1989. I think it is evident how the facts and proofs has not been taken serious and how the area has been blocked. That is embarrassing!!.history should put this well clear how this happenned as already happenned many times before.But that’s not enough, it’s needed a way of control so that lobbies and other personal mindsets are excluded from the acceptance process. What’s important are the facts and not the authorities it’s important to force an spirit of open mind in science for the facts. I don’t want the people to believe but to listen and study the announced facts.
    Science must study the announced facts and don’t get lazy with the excuse that has been proven wrong before or the general consensus is other.
    I really want to see that after LENR is out. This immobilized science has costed us a lot this corrupt behavior and lazyness has costed us maybe the contamination of the planet by letting oil to stay.

  • Freethinker

    Thomas Clarke, you are slick. But you have already shown your true face.

    You can question these things all you want but the fact remains:

    * The details of the output of the control box is part of the black box.

    * The details of the impedance of the active reactor is within the black box.

    * The details of the alloys used or any other details of the reactor innards is in the black box.

    The scope if to measure power in and out. Power out proxied by the temperature of the reactor surface. You have those data. Those data give at hand that the claims in the report are reasonable and thus stand.

    Whatever you think you are expressing, it is plain to people who has followed the ECAT on this site, that you are biased. You make that clear, either consciously or unconsciously, in this sentence:

    “I am expressing surprise at the length of time it has taken the testers
    to respond, but hope that soon they will resolve the matter and either
    retract their claims as based on some not understood measurement error
    or modify the report stating exactly what was the error.”

    Like it is either wrong or it is wrong. You do not consider that there is NO error, because you already KNOW there is an error.

    You are a true pathoskeptic.

    • Thomas Clarke

      I know that there is an error because it is possible to work out the heater resistance from the dummy and active test data and the two values do not agree, as they should for an Inconel heater element. Perhaps I have made a mistake in this: but everyone I know competent to evaluate the matter agrees.

      The point is that this error can be resolved by the report authors, after which there would be no error, or some measurement issue that they cannot identify and agree is an error.

      I’m not sure why being able to identify a clear error in a report makes me a pathoskeptic?

      As for details: the testers stated that the heater element was inconel – so they obviously felt competent to evaluate this. The resistance of the heater under any test condition is contained in the stored experimental data the testers have collected – so clearly it is “out of the box”. In fact the anomaly has come to light exactly because this resistance was published in the report.

      The impedance of the heater is not relevant to this question except for its series equivalent resistive component.

      • Freethinker

        “I’m not sure why being able to identify a clear error in a report makes me a pathoskeptic?”

        As said so many times before: because you base your argument on information that is not relevant, not there, and you know that you are right that the ECAT cannot work.

        Had you been just a tiny spec more humble, realizing that what you have is conjecture, it would be another thing, but you don’t, so you are a pathological skeptic, as that kind of behavior is the hallmark of such an individual.

        And you will not rest with the information you get, were it to show NO error at all. Pathoskeps are insatiable, and will never accept any level of evidence, but constantly raise the bar of what is evidence, regardless of new information that is presented.

        So, yes Thomas Clarke, you are a pathological skeptic.

        • Thomas Clarke

          Just one new point to answer. Were the electrical input anomaly to be resolved there would, it is true, be many other issues about this report. They have all been raised.

          You are free to think that my seeing these issues as valid indicates bias – just as I am equally free to think that your dismissing them as invalid indicates your bias. You will see that such argument is entirely unhelpful. It is only by identifying the detailed evidence, so that others can judge it for themselves, that progress can be made.

          The electrical current issue is a little different from all the others because it is definite, and indicates an error that can be further investigated and corrected by the report authors now. That is surely worth doing.

        • Freethinker

          Wow Ivan.

          I feel so much better now when you have used all your considerable skillz in conjecture and confabulations to calculate the COP to less than one unless a miraculous drop in resistance.

          You are remarkably ignorant in your approach as you have NO clue as to the setup of the control box in the dummy run compared to the active run, and assume that it is the same in both. Further you assume same components, ignoring the fact that in the dummy you have an idle reactor and in the active situation, you have exactly that, an active reactor in where novel nuclear processes are at work.

          Exactly what in the input power measurement is then wrong? No bullshit about some resistors or joule heating, but why has the PCE-830 not measured the input power correctly? Explain that to me.

          Or do you have some opinions of the measurements of the temperature of the reactor surface that will have a negative impact on the output power computation?

          If you fail to explain any of those two things satisfactory and without conjectures and confabulations, then the claims in the report stand.

    • EEStorFanFibb

      Freethinker, this is what Thomas Clarke just wrote on theeestory.com

      “I’m a bit surprised that Fibb and parallel choose to defend the indefensible on this thread. The case for LENR is not strong, but as it is so much less weak than the case for Rossi it bemuses me that they defend him.

      This latest test, as others have said, is not the first time Rossi has been shown to make unsubstantiated technical claims. He has a long history of this. He has also repeatedly lied on his blog. All of that behaviour is quite different from typical LENR researchers. Defending Rossi does them no justice.”

      http://theeestory.ning.com/xn/detail/6495062:Comment:124236

      not his strong anti Rossi bias, that he tried to soften here. I’m glad you saw right through his BS. Well done

  • Freethinker

    To be honest, you do not deserev any more focus from my side so I simply repost a comment from above:

    Thomas Clarke, you are slick. But you have already shown your true face.

    You can question these things all you want but the fact remains:

    * The details of the output of the control box is part of the black box.

    * The details of the impedance of the active reactor is within the black box.

    * The details of the alloys used or any other details of the reactor innards is in the black box.

    The scope if to measure power in and out. Power out proxied by the temperature of the reactor surface. You have those data. Those data give at hand that the claims in the report are reasonable and thus stand.

    Whatever you think you are expressing, it is plain to people who has followed the ECAT on this site, that you are biased. You make that clear, either consciously or unconsciously, in this sentence:

    “I am expressing surprise at the length of time it has taken the testers to respond, but hope that soon they will resolve the matter and either retract their claims as based on some not understood measurement error or modify the report stating exactly what was the error.”

    Like it is either wrong or it is wrong. You do not consider that there is NO error, because you already KNOW there is an error.

    You are a true pathoskeptic.

  • EEStorFanFibb

    Freethinker has pegged Thomas Clarke EXACTLY right. Well done.

  • ivanc

    I see people arguing again about the 3.3 factor.
    The Joule Heating, currents, saying is irrelevant…..etc etc.
    The circuit is simple, and is fully solvable, we have a dummy run with the same components of the active run. the same controller. the same meters. and all voltages and currents and powers in each element are interrelated by know functions. so there is no uncertainty/speculation/conjeture. This is why the testers (professors) are silent. they know the calculation and formulas we have put in front of them are right and they are not easily able to answer.
    The math required is not of very high level, just a bit of equations, all numbers are absolute value. RMS.
    There is no information of duty cycle, and is not needed, as they using RMS.
    I have fully calculated the circuit and published it.
    I have prove Iresistor = IL /sqr(3)
    The calculation were done using Recat constant. and Recat variable.
    For the given Joule heating data, cable resistors, if Recat constant, then COP less than 1.
    For the variable Recat calculation for the report to be true, Recat has to drop value 3.3 times, but in the range of power in the active run it remains quasi constant.
    The report points to really extraordinary events, o just errors in measurements.
    This is why a scientists, take double set of measurements, and they did, we just want to see the measurements of this meters, voltage,current,power. we have been given Power, and current to a reasonable accuracy, they say about 50A, and the calculations did confirm that for the active run.
    so the only possible explanation is a miraculous drop in the resistance.

    • Mark Szl

      Would that dosclosure give away any secrets, any IP issues valuable to IH?

      • ivanc

        NO, We already have the calculated voltages. we just want to confirm they are similar to what they have measured.

        • Mark Szl

          That issue about the lowered resistance. Could it be from melting at higher temperature.

          • Thomas Clarke

            I can’t think of any mechanism.

            But anyway the point is that we know the testers have the data to say whether resistance changed during the active test. It would explain the anomaly if it did. The fact that they are silent tells me that most likely it did not change.

    • Thomas Clarke

      Yes. And the Profs are able to answer. They have definitive data to determine whether the heater resistance is constant. They may not to be able to pin down the precise error, but they can get it down to one of:

      (1) the heater resistance changes with temperature as Rossi has said. The report’s conclusions are correct, there is no electrical measurement issue.

      (2) One of the following:
      the heater resistance is 3.3X lower throughout the active test then it is in the dummy test
      the power was mis-measured on input during the active test, and is in fact 3.3X higher
      the current was mis-measured by sqrt(3) during either the active or the dummy test

      Whichever of these is in fact true, it deserves comment in the report, or in an additional document replying to criticism.

      I am confident the Swedish independent testers want to get to the truth, so I hope this additional comment will be forthcoming.

      • US_Citizen71

        Of course the resistance changes with temperature that is expected. Take several feet of bare wire make a non-overlapping coil. Attach the ends to an Ohm meter. Then take a propane torch and heat the coil and watch the resistance measured change before your eyes. The opposite works as well. Dip the coil in liquid nitrogen or old style copier fluid for mimeographs with dry ice dissolved in it and the same thing will occur but in reverse.

        edit; The total power in all runs was calculated by the PCE 830 units not via equations done by the professors. So the error on the joule heating has nothing to do with the measured power, just with what they subtracted from it to develop their COP numbers.

        • Thomas Clarke

          Yes, but that is a small positive TC. The required change is a large negative TC (reduction to 1/3 of original value). Only possible for semiconductors and the temperature at which this happens is too high for any semiconductor known. Anyone with a candidate please post it in reply. Diamond?

          • US_Citizen71

            No change in resistance is required to explain the change in current. I=V/R

            V=1 R=1 then I=1

            If V=3 and R=1 then I=3

            There you go 3 times the current no change in resistance.

            The Vrms of a TRIAC controled system can be calculated fairly close with the following:

            Vrms(switched) = √Duty cycle x Vrms(pure sine)

            Vrms(pure sine) or full wave = 100 volts rms
            Duty cycle = 1% or .01
            √.01 x 100 volts = 10 volts rms

            Vrms(pure sine) or full wave = 100 volts rms
            Duty cycle = 10% or .1
            √.1 x 100 volts = 31.62 volts rms

            Vrms(pure sine) or full wave = 100 volts rms
            Duty cycle = 100% or 1
            √1 x 100 volts = 100 volts rms

            A TrueRMS multimeter like the PCE 830 wil get a more correct value as where the clipping happens matters a little bit . But this should give you an understanding of why the Vrms value is not constant in a TRIAC control power system.

    • Freethinker

      Wow Ivan.

      I feel so much better now when you have used all your considerable skillz in conjecture and confabulations to calculate the COP to less than one unless a miraculous drop in resistance.

      You are remarkably ignorant in your approach as you have NO clue as to the setup of the control box in the dummy run compared to the active run, and assume that it is the same in both. Further you assume same components, ignoring the fact that in the dummy you have an idle reactor and in the active situation, you have exactly that, an active reactor in where novel nuclear processes are at work.

      Exactly what in the input power measurement is then wrong? No bullshit about some resistors or joule heating, but why has the PCE-830 not measured the input power correctly? Explain that to me.

      Or do you have some opinions of the measurements of the temperature of the reactor surface that will have a negative impact on the output power computation?

      If you fail to explain any of those two things satisfactory and without conjectures and confabulations, then the claims in the report stand.

      • ivanc

        So You want me to ignore, the conclusions that cames from the hard data in the report. and start to speculate.
        What setup of the control box? it has 3 wires out and a potentiometer to control.
        the only difference will be the postion of the potenciometer, as this is how they change the feeding ( voltage or current, they are proportional).

        now what you want me to speculate ignoring the hard data.?

        Well the measurents could be wrong for two reasons.
        one: The clamps were conected wrong.
        two: The controler changes the sequence of phases internally, this could be achieved with some electronic and a remote control.

        They could check all this will the two meters, but they have not given data except partially for the one connected to the ecat.

        You want me to suppose they did all correctly.!!!!!!!!!!!!!!!!!!
        (SUPOSE = speculate)

        This is why the data is publised in anexes, so people do not have to suppose but see.

        What has to do temperature with amps and volts and watts. temperature depends of area, volume….etc etc…. nothing to do in this argument

        • Freethinker

          As hard as it may seem to you, I want you to stop speculate and stop using conjectures to make conclusions that the data do not support.

          • ivanc

            You are really funny the the one speculating here Are you.
            You are the one making conjectures…. etc etc etc.

            The data is in the report and the circuit can be calculated

          • Mark Szl

            Why are you being so hard on ivanc if all he is asking for is really just double checking the data published in the report. Does not sound like ivanc is going outside of the scope of the report to me.

          • Freethinker

            Mark, you, just as everybody else, are free to have an opinion. I just don’t happen to agree.

        • US_Citizen71

          The potentiometer is on the trigger to the triac. Changing the potentiometer changes the chop of the waveform.

      • Thomas Clarke

        Freethinker,

        We don’t know the input power is measured wrong. All we know is that the input power or the current is measured wrong, or the resistance miraculously changes.

        In an experiment supposed to prove COP > 1 – something quite unusual and undoubtedly worth a Nobel prize if true – it is no good to say – well – we might have extraordinary results and made a mistake with current, or we might have made a mistake with power,

        Do you see? No-one can prove what the mistake is (except maybe the testers) but we can prove there is a mistake somewhere and it could be the power.

        The above is fact, not conjecture. For you to say somehow the power must be the correct measurement and the current the wrong one is conjecture.

        • Freethinker

          Thomas Clarke, it is conjecture because you build it up from information that does not go together.

          I take the input power measurements at face value, as the are measured and presented as intended within the scope. They are sets of information that are coherent, and relevant, supported by the data from two PCE-830, and analyzed by intelligent and competent people. Hence I do not conjecture. I don’t pick data intended for one purpose, and use it in another purpose where the data is no longer relevant or correlated with other data used.

          You are constantly claiming that you are correct in your assertion, but you are not. You fight fiercely to bring this discussion into the domain that is outside the scope, where nobody can argue from a standpoint of fact.

          But it is befitting your already preconceived notion that LENR is quackery and the ECAT cannot work anyway. You are a pathological skeptic who are engaged in trolling on this website.

          I far as I go, you have no credibility.

  • Warthog

    Sorry, but no. Simply wrong, and wrong-headed. Science is about replicated experiments….NOT theory. ANY theory can be pitched out at any time if even a single well-executed experiment contradicts it. The idea that an experiment has to be discarded if it doesn’t agree with “a theory” is turning science 180 degrees. And any supposed “scientist” who claims the latter should have any and all degrees in science rescinded, and never be allowed to enter a laboratory again.

    The pathological skeptics, though, constantly do precisely that. And they also TRY, with more or less success, to execute the punishment they themselves deserve, on researchers who report positive results on LENR.

  • Freethinker

    Thomas Clarke,

    You are making implicit assumptions of the voltage as you use the values of the joule heating and the in powers to in the base of your argument and saying that the heater resistance varies by a factor of 3.3.

    But you want to bring this line of argument into the out of scope domain, and you keep repeating, over an over again, the same conjectures for which you have no base. You do not have no information, no correlated information and only your own confabulations, thats all. It is all conjecture.

    But this does not matter to you as though the repeating of the same infinitum will make it a reality.

    You are out of scope, and your conjectures are baseless, you have a strong conviction before this argument that Rossi’s ECAT cannot work. You have even admitted it indirectly as you claim you KNOW the report is wrong, and has errors.

    You refuse look at the scope and the data presented. You have no problem rejecting that which is very well described and quite uncontroversial, but at the same time embrace notions that are very much ill founded, based on data that the scope do not cover.

  • Obvious

    Your point is a good one.
    I will go through the math again and double-check that time is conserved properly when the Joule factoring is done to obtain active run current. Just to ensure there are no loopholes.

  • Freethinker

    It is no more circular than your numeric nonsense. See that is my point, you do not have ample data. You claim the input power to be wrong. Then explain that without the conjecture and other confabulatory bs you are doing now. You have nothing to show, but you hot desire to disprove this.

  • kdk

    It seemed like one of those long-shots that people latch onto like the hidden 3rd wire… We know cold fusion is impossible, despite all the other experiments, so it just has to be wrong…

  • kdk

    It seemed like one of those long-shots that people latch onto like the hidden 3rd wire… We know cold fusion is impossible, despite all the other experiments, so it just has to be wrong…

  • US_Citizen71

    Currently there is a debate raging on about the current reading of 19.7A. I want to examine how that number applies to the report. All page references refer to the report located at http://www.elforsk.se/Global/Omv%C3%A4rld_system/filer/LuganoReportSubmit.pdf

    ‘The E-Cat’s control apparatus consists of a three-phase TRIAC power regulator, driven by a programmable microcontroller; its maximum nominal power consumption is 360 W.’ – page 3

    This is the power requirement stated by IH/Rossi to the testers for the control box.

    ‘The power analyzers were two PCE 830 units from PCE Instruments, capable of measuring, and displaying on an LCD display, electric current, voltage and power values, as well as the corresponding waveforms. These instruments are capable of reading voltage and AC current values up to 5 kHz.

    The choice of instruments was warranted both by the straightforwardness of the experimental setup and the precision of the instruments themselves. Designing a calorimetric measurement by means of a cooling fluid would have been more complex, especially in the light of the high temperatures reached by the E-Cat.

    All the instruments used during the test are property of the authors of the present paper, and were calibrated in their respective manufacturers’ laboratories. Moreover, once in Lugano, a further check was made to ensure that the PCEs and the IR cameras were not yielding anomalous readings. For this purpose, before the official commencement of the test, both PCEs were individually connected to the power mains selected for powering the reactor. For each of the three phases, readings returned a value of 230 ± 2V, which is appropriate for an industrial establishment power network.’ – page 4

    Both PCE 830 meters showed standard industrial three phase power and appear to be similiarly calibrated. We also see that the authors were able to correctly setup and measure the main input twice.

    ‘Figure 4 details the electrical connections of all elements of the experimental setup. The two PCEs were inserted one upstream and one downstream of the control unit: the first allowed us to measure the current, voltage and power supplied to the system by the power mains; the second measured these same quantities as input to the reactor. Readings were consistent, showing the same current waveform; furthermore, they enabled us to measure the power consumption of the control system, which, at full capacity, was seen to be

    the same as the nominal value declared by the manufacturer.

    Special attention was given to measuring the current and voltage input to the system: the absence of any DC component in the power supply was verified in various occasions in the course of the test, by means of digital multimeters and supplementary clamp ammeters. We also verified that all the harmonics of the waveforms input to the system were amply included in the range measurable by the PCEs (Figure 5). The three-phase current line supplying all the energy used for the test came from an electrical panel belonging to the establishment hosting our laboratory, to which further unrelated three-phase current equipment was connected.’ – page 5

    The Rossi/IH 360w max power usage for the control box is confirmed and no hidden DC current was detected. We also can conclude that the power values on both PCE 830 units during all runs never were separated by more than 360w.

    Now we get to the meat of the matter.

    ‘We may calculate the dissipated heat to the limited extent of the dummy reactor: the results relevant to the E-Cat will be given in Table 7, due to the fact that the average current values changed from day to day.

    Measurements performed during the dummy run with the PCE and ammeter clamps allowed us to measure an average current, for each of the three C1 cables, of I1 = 19.7A, and, for each C2 cable, a current of I1 / 2 = I2 = 9.85 A. The evaluation of heat dissipated by the first circuit is:

    WC1 = 3(R1I1²) = 3(4.375 ∙ 10–3 ∙ (19.7)²) = 5.1 [W] (9)

    For the second circuit we have:

    WC2 = 6(R2I2²) = 6(2.811 ∙ 10–3 ∙ (9.85)²) = 1.6 [W] (10)

    By adding the results, we have the total thermal power dissipated by the entire wiring of the dummy.

    Wtot.dummy = 5.1 + 1.6 = 6.7 ≈ 7[W] (11)

    In the calculations that follow, relevant to the dummy reactor and the E-Cat’s power production and consumption, the watts dissipated by Joule heating will be subtracted from the power supply values.

    ‘ -page 14

    The above section shows the values and equations in question. The resistance values were calculated on pg13-14 if one has questions on how they were determined. It should be noted no voltages are give at all and no resistance values are given for the reactor itself anywhere in the report.

    The entire debate is over the current used in this equation:

    WC2 = 6(R2I2²) = 6(2.811 ∙ 10–3 ∙ (9.85)²) = 1.6 [W] (10)

    Instead of continuing the debate we will work forward with the value being off by a factor of 3.3. So 1.6 watts becomes 5.28 watts. The total Joule heat becomes 10.38 watts instead of the 7 watts from the report. The reported value was 3.38 watts low according to this math.

    10.38 / 7 = 1.483 is the ratio of the argued correct value to the report value.

    Now we will plug these numbers into the rest of the report

    For the dummy reactor:

    ‘Let us now compare this dissipated power with the power supply, the average of which over 23 hours of test

    is = (486 ± 24) W (uncertainty here is 5% of average, calculated as standard deviation). Keeping in mind the

    Joule heating of the power cables discussed in paragraph 4.3, we have the following results:

    Power supply (W) Joule heating (W) Actual input (W) Output (W)

    486 ± 24 7 486 – 7 = 479 ± 24 446 ± 10’ – page 20

    Subtracting out 10.38 watts instead of 7 watts for joule heating makes the actual input value 475.62 watts still within 5% standard deviation. Table 7 holds all the values for the active run. Multipling the listed joule heating value times the 1.483 ratio calculated earlier will give the updated values for the active runs as the same equations were used throughout the report. The difference between the updated value and the report value must be subtracted from the ‘Net Production’ column as well as the ‘Consumption’ column to give an updated values. The new Net Production value plus the new Consuption value must then be divided by the new Consuption column to calculate the new COP. I will calculate the new values for the first and last rows of table 7 below.

    First column:

    Joule heating: 37.77 watts x 1.483 = 56.01 watts

    Difference: 56.01 watts – 37.77 watts = 18.24 watts

    Net Production: 1658.21 watts – 18.24 watts = 1639.97 watts

    Consumption: 815.86 watts – 18.24 watts = 797.62 watts

    New COP: (1639.97 watts + 797.62 watts)/797.62 watts = 3.06

    Last column:

    Joule heating: 41.25 watts x 1.483 = 61.17 watts

    Difference: 61.17 watts – 41.25 watts = 19.92 watts

    Net Production: 2393.74 watts – 19.92 watts = 2373.82 watts

    Consumption: 906.31 watts – 19.92 watts = 886.39 watts

    New COP: (2373.94 watts + 886.39 watts)/886.39 watts = 3.68

    • Obvious

      The active run current increases by the multiplication of dummy run current by the square root of the ratio of change from the dummy run to the active run, if resistance of the reactor stays the same. I cut the Joule heating in half, and it made no difference.
      Any linear adjustment to the all the Joule heat values has zero effect on the factor derived from it’s rate of change.
      Changing the dummy Joule heat only, and leaving the Joule heat alone for the active could work.

      • US_Citizen71

        The ratio of the error would remain the same in the active run because the same equations were used through only the values were changed. The joule heating value was subtracted from the consumption value and the net power created by the reactor so to maintain parity it must be done on the active run calculations.

        • Obvious

          Yes. And it affects COP slightly, but does nothing to address the rate of current increase.
          Despite my differences with several detractors of the report COP, my active run current calculations, using two entirely separate routes to arrive at the values, agree within 2 decimal places. These are identical to what Ivan has posted, and I am sure they are the same as everyone else’s that have done the math in a similar way.

          • US_Citizen71

            An increase in voltage from the control box also explains the increase in current. The triac could chop the wave form more at lower power than it is at higher power causing the voltage to increase. Correct?

          • Obvious

            No, or not exactly.
            V is totally dependent on I and R, and it cannot be moved without moving at least one (and I am almost 100% certain) both of I and R.
            I do have a suspicion that V is involved in some other way, but I can’t find a way to involve V without the I and R un-doing any changes.

          • US_Citizen71

            Vrms in AC is determined by the wave form. If only the first 10% of the upslope is allowed through at low power the voltage would be lower than the full wave I believe. I=V/R If V increases while R remains the same I increases.

          • Obvious

            I’m still fighting with it.

            Here’s a new formula I derived myself from doing the trigonometry with real triangles in a delta configuration, using a right angle triangle and report currents…long story.
            It must only apply to balanced delta configuration:

            Ptotal = IL^2 + (1/2IL)^2

            I don’t know exactly the full repercussions of this formula, but it works.
            It might actually derive better W values than in the report, but the formula is extremely finicky when it comes to rounding. Keep as many decimal places as you can. It can also derive the current consumed by line loss by inputting the correct W for Joule heat loss, and running in reverse, or subtracting the line loss W from the total P and running the values.

          • Dr. Mike

            Obvious,
            An interesting equation! Sure enough if you use this equation you get fairly close to the power delivered to the load. But how can this be- power only dependent on the line current? I believe I have found out why your equation works. Your equation from above: Ptotal = IL^2 + (1/2IL)^2 can be simplified to Ptotal = 1.25*IL^2. However, the power to the 3 heater coils is Pcoils = 3*Rcoil*Icoil^2, but Icoil = IL/SQRT(3). Therefore, Pcoil = Rcoil*IL^2.. Assuming the Cu wire Joule heating is small relative to the heater wire Joule heating, Pcoil approximately equals Ptotal. Therefore, Ptotal = Pcoil = Rcoil*IL^2. You derived an equation Ptotal = 1.25*IL^2. Why does your equation work? Because the coil resistance can be calculated as 1.235 ohms for the Hot-Cat. The 1.25 factor “works” only because the actual heater wire resistance is 1.235 ohms! Try your formula on a an example where a 3-phase line current of 10A is delivered to a 5 ohm delta load. The actual total power delivered to the load is 500W, whereas your calculation would give 125W.
            Dr. Mike

          • Obvious

            Delivering 10 A through a delta with 5 ohms inductance is impossible to make 500 W. 5 ohms inductance is not the same a as 5 ohms resistance in each delta side.

            Using your example, it can be shown that 50 V would be required.

            P = VI, V= 500/10 therefore V = 50.

            In a delta, the line voltage is the same as phase voltage.
            Try and calculate the power per 5 ohm resistor at 50V…..
            P = V^2/R, P = 2500/5 therefore P = 500 W. But there are 3 R’s!!
            so try dividing power three ways, one for each resistor…..
            …….1/3P = V^2/R
            …….1/3P = 2500/R
            166.6667 = 2500/R
            ………….R = 2500/166.6667
            ………….R = 15 !!! [Use R = (R1+R2+R3) and this fixes this part?]. But this is weird, since now we have 1/3 the power and three times the resistors!
            There is a hint of something going on here….
            Actually the three R’s in the delta, in order make 500 W, and have a 5 ohm inductance would have to be (R1+R2)R3/(R1+R2+R3) = 5
            This is due to “forcing” current IN one leg, and OUT one other, so that the resistors are series-parallel (one in parallel with two in series)
            so…since the r’s are all the same….
            This is what happens when all the resistance is calculated by diving the total power by current. You get the inductance of the whole delta from the perspective of two out of three phases, not resistance of each resistor. Mathematically you have disconnected one delta cable.
            Using 5 ohm resistors…..
            Rdelta inductance = 2R(R)/3R
            ………………………5 = 2R(R)/3R…..use some 1’s for R to get a factor..
            ……………………….. = 2/3 (5)
            Rdelta inductance = 3.33333~ ohms NOT 5
            Now:
            P = I^2R
            P = 10^R
            P = 100(3.3333333~)
            P = 333.33333~
            So to get 500W, you need 3/2 (5) ohms, or boost I^2 to make the difference. This gets really complicated, since this messes with V. In effect, you must “steal” from V (was 50V, what is it doing now?) to make this work. If you steal from V, then I changes and P changes if you hold R. ad infinitum, nearly
            Best to do some test numbers and work it out by estimating. Lots of work.
            Reversing my equation gives an answer of 20A Line current to satisfy the 500W.
            Well, might as well try it.
            …..P = I^2R
            …..P = 400(R)
            …..R = 500/400
            …..R = 1.25…….R is very different. This would be inductance of the delta required to dissipate 500W. And 20 A are required.
            Then the old way…using 10A and 1.25 ohms inductance:
            P = I^2(R)
            P = 100(1.25)
            P = 125 W………There is my 125 W!!

          • Freethinker

            Mark, you, just as everybody else, are free to have an opinion. I just don’t happen to agree.

          • Obvious

            Good morning. I had a restless sleep filled with sqrt(3)’s floating all around.
            I can see why ivanc likes the sqrt(3) so much, other than that it is the line phase modifier.
            The extrapolation equations are absolutely soaked in sqrt(3)’s.
            Some quick notes, just so I get them out of my system:
            (not picking on anyone, just accrediting the quotes. The math seems fine.

            “Lets call Jof=57.66513854” -ivanc
            10^2/sqrt(3) = 57.7350269……very close. Why?

            “I have updated the table to show input power = consumed power – joule heating.
            The factor is now 3.4 to 3.5 instead to 3.3” – ivanc
            and
            “….for the given figures to be correct the heater resistance must decrease by 3.3 (sorry, in fact as Mikepoints out, 3.45) in the active test.” -Thomas Clarke
            2*sqrt(3) = 3.464………very close again. Why?

    • Dr. Mike

      US_Citizen71,
      I believe you are confused about the 3.3X factor. The error in the current in the C2 lines is that the C1 current should be divided by SQRT(3) rather than divided by 2 to calculate C2 currents. This would make the Joule heating in all 6 C2 lines total 2.18W, rather than the 1.6W stated in the report The factor of 3.3 is the relative fraction that the resistance of the Inconel heater wire must decrease in the active run to make the Input power numbers agree with the Inconel heater wire currents which can be calculated fro the Joule heating numbers.
      Dr. Mike

      • US_Citizen71

        Yes I understand that current that would come from the updated value is better than cold fusion itself as it creates current out of thin air. I used that values that I did to show that effect of changing the joule heating is almost meaningless.

        • Dr. Mike

          US_Citizen71,
          I don’t know what you mean by “I understand that current that would come from the updated value is
          better than cold fusion itself as it creates current out of thin air”. Don’t you realize that each line current is the superposition of 2 currents that are 120 deg out of phase? The RMS value of this superposition of 2 phase currents is not equal to 2 times the RMS current going to each phase! However, the instantaneous line current is equal to the sum of the two phase currents- no current is being created out of thin air at any point in time. If you would Google “3 phase delta current equations”, you might get a better understanding of where the 1/SQRT(3) factor comes in. I certainly did not remember this 1/SQRT(3) factor from my power and motors course from over 40 years ago. However, after this error was pointed out in the report, I was able to understand why you can not sum out of phase RMS currents. Take a little time to study what information is readily available on the internet, or check a textbook on 3 phase power systems, and see if you don’t agree that the authors have made an error in their calculation of the C2 heater wire currents. Wouldn’t you like to see this error fixed when the report gets revised?
          Dr. Mike

          • US_Citizen71

            Work the equation backwards and you will find the current in C2 would be higher than the current in C2 should be by approximately 40% if the 5.28 watts was the answer.

          • Dr. Mike

            US_Citizen71,
            The C2 RMS current should be 19.7/SQRT(3) = 11.27A for the dummy run if the C1 current was measured to be 19.7A. Would you like to see this error fixed in the report? Were you able to figure out why the C1 RMS current is not equal to the sum of two C2 RMS currents?
            Dr. Mike

          • US_Citizen71

            They obviously used the wrong equation and yes I would like to see them fix it. The three phase power threw them and many commenters as well including myself. It took a conversation with a power systems engineer for me to knock the rust out and remember the correct way to figure the current.

      • Obvious

        That is incorrect, in my opinion.
        If you use Lcurrent/sqrt(3) then you make a vector angle sum of current for one phase. Three times that vector is zero net current, and they have sign (+/-), relative to each other. The three vectors point away from the origin at 120° angles to each other.
        The proper, positive, vector sum is then the vector sum of TWO sqrt(3) current vectors.
        This latter sum is the DC equivalent, “forward, (+)” current in the circuit.

        • Dr. Mike

          Obvious,
          When ivanc first brought up this error, I had to Google “3-phase delta current equations” to verify that he was correct. Why don’t you do the same? You appear to have a basic understanding of electrical equations and should be able to understand the textbook explanation of why the RMS current in the C1 line is not the sum of 2 RMS currents in the C2 lines.
          Dr. Mike

          • Obvious

            The sum of two RMS currents, using sqrt(3), in one C2 cable is greater than 1/2 of the C1 cable for a very good reason. The sum of two identical RMS sqrt(3) values is greater than 1/2 of the forward current flowing through the entire circuit at any one time.
            In fact, one RMS sqrt(3) “measurement” is exactly equal to the
            RMS sqrt(3) current flowing in the opposite direction in the other two phases at any one time.

          • Obvious

            Hmmm.
            If two times IL(sqrt(3) is 240 degrees’ worth, then the maximum forward current of 180° out of 360° is 240/1.333~ (240/180 = 1.33333~)
            2(11.37) = 22.74 A, 22.74/1.3333~ = 17.055 A.
            My triangle figuring came up with this value several times, and I was wondering why that was, and not 19.7.
            So there you have it.
            1/2 total forward delta I must be 17.055 A at any one time.

  • US_Citizen71

    Currently there is a debate raging on about the current reading of 19.7A and dividing it by 2 to determine the current used to determine the joule heating of the shorter C2 cables. I want to examine how that number applies to the report. All page references refer to the report located at http://www.elforsk.se/Global/Omv%C3%A4rld_system/filer/LuganoReportSubmit.pdf

    ‘The E-Cat’s control apparatus consists of a three-phase TRIAC power regulator, driven by a programmable microcontroller; its maximum nominal power consumption is 360 W.’ – page 3

    This is the power requirement stated by IH/Rossi to the testers for the control box.

    ‘The power analyzers were two PCE 830 units from PCE Instruments, capable of measuring, and displaying on an LCD display, electric current, voltage and power values, as well as the corresponding waveforms. These instruments are capable of reading voltage and AC current values up to 5 kHz.

    The choice of instruments was warranted both by the straightforwardness of the experimental setup and the precision of the instruments themselves. Designing a calorimetric measurement by means of a cooling fluid would have been more complex, especially in the light of the high temperatures reached by the E-Cat.

    All the instruments used during the test are property of the authors of the present paper, and were calibrated in their respective manufacturers’ laboratories. Moreover, once in Lugano, a further check was made to ensure that the PCEs and the IR cameras were not yielding anomalous readings. For this purpose, before the official commencement of the test, both PCEs were individually connected to the power mains selected for powering the reactor. For each of the three phases, readings returned a value of 230 ± 2V, which is appropriate for an industrial establishment power network.’ – page 4

    Both PCE 830 meters showed standard industrial three phase power and appear to be similiarly calibrated. We also see that the authors were able to correctly setup and measure the main input twice.

    ‘Figure 4 details the electrical connections of all elements of the experimental setup. The two PCEs were inserted one upstream and one downstream of the control unit: the first allowed us to measure the current, voltage and power supplied to the system by the power mains; the second measured these same quantities as input to the reactor. Readings were consistent, showing the same current waveform; furthermore, they enabled us to measure the power consumption of the control system, which, at full capacity, was seen to be

    the same as the nominal value declared by the manufacturer.

    Special attention was given to measuring the current and voltage input to the system: the absence of any DC component in the power supply was verified in various occasions in the course of the test, by means of digital multimeters and supplementary clamp ammeters. We also verified that all the harmonics of the waveforms input to the system were amply included in the range measurable by the PCEs (Figure 5). The three-phase current line supplying all the energy used for the test came from an electrical panel belonging to the establishment hosting our laboratory, to which further unrelated three-phase current equipment was connected.’ – page 5

    The Rossi/IH 360w max power usage for the control box is confirmed and no hidden DC current was detected. We also can conclude that the power values on both PCE 830 units during all runs never were separated by more than 360w.

    Now we get to the meat of the matter.

    ‘We may calculate the dissipated heat to the limited extent of the dummy reactor: the results relevant to the E-Cat will be given in Table 7, due to the fact that the average current values changed from day to day.

    Measurements performed during the dummy run with the PCE and ammeter clamps allowed us to measure an average current, for each of the three C1 cables, of I1 = 19.7A, and, for each C2 cable, a current of I1 / 2 = I2 = 9.85 A. The evaluation of heat dissipated by the first circuit is:

    WC1 = 3(R1I1²) = 3(4.375 ∙ 10–3 ∙ (19.7)²) = 5.1 [W] (9)

    For the second circuit we have:

    WC2 = 6(R2I2²) = 6(2.811 ∙ 10–3 ∙ (9.85)²) = 1.6 [W] (10)

    By adding the results, we have the total thermal power dissipated by the entire wiring of the dummy.

    Wtot.dummy = 5.1 + 1.6 = 6.7 ≈ 7[W] (11)

    In the calculations that follow, relevant to the dummy reactor and the E-Cat’s power production and consumption, the watts dissipated by Joule heating will be subtracted from the power supply values.

    ‘ -page 14

    The above section shows the values and equations in question. The resistance values were calculated on pg13-14 if one has questions on how they were determined. It should be noted no voltages are give at all and no resistance values are given for the reactor itself anywhere in the report.

    The entire debate is over the current used in this equation:

    WC2 = 6(R2I2²) = 6(2.811 ∙ 10–3 ∙ (9.85)²) = 1.6 [W] (10)

    Instead of continuing the debate we will work forward with the value being off by a factor of 3.3. So 1.6 watts becomes 5.28 watts. The total Joule heat becomes 10.38 watts instead of the 7 watts from the report. The reported value was 3.38 watts low according to this math.

    10.38 / 7 = 1.483 is the ratio of the argued correct value to the report value.

    Now we will plug these numbers into the rest of the report

    For the dummy reactor:

    ‘Let us now compare this dissipated power with the power supply, the average of which over 23 hours of test

    is = (486 ± 24) W (uncertainty here is 5% of average, calculated as standard deviation). Keeping in mind the

    Joule heating of the power cables discussed in paragraph 4.3, we have the following results:

    Power supply (W) Joule heating (W) Actual input (W) Output (W)

    486 ± 24 7 486 – 7 = 479 ± 24 446 ± 10’ – page 20

    Subtracting out 10.38 watts instead of 7 watts for joule heating makes the actual input value 475.62 watts still within 5% standard deviation.

    Table 7 holds all the values for the active run. Multiplying the listed joule heating value times the 1.483 ratio calculated earlier will give the updated values for the active runs as the same equations were used throughout the report. The difference between the updated value and the report value must be subtracted from the ‘Net Production’ column as well as the ‘Consumption’ column to give updated values. The new Net Production value plus the new Consumption value must then be divided by the new Consumption value to calculate the new COP. I will calculate the new values for the first and last rows of table 7 below.

    First row:

    Joule heating: 37.77 watts x 1.483 = 56.01 watts

    Difference: 56.01 watts – 37.77 watts = 18.24 watts

    Net Production: 1658.21 watts – 18.24 watts = 1639.97 watts

    Consumption: 815.86 watts – 18.24 watts = 797.62 watts

    New COP: (1639.97 watts + 797.62 watts)/797.62 watts = 3.06

    Last row:

    Joule heating: 41.25 watts x 1.483 = 61.17 watts

    Difference: 61.17 watts – 41.25 watts = 19.92 watts

    Net Production: 2393.74 watts – 19.92 watts = 2373.82 watts

    Consumption: 906.31 watts – 19.92 watts = 886.39 watts

    New COP: (2373.94 watts + 886.39 watts)/886.39 watts = 3.68

    EDIT: Multiplying the C2 joule heating value times 3.3 produces a value beyond possibility. The exercise above was not to determine the actual value for the joule heating of C2. I did it show how little the difference being debated matters to the values in the report. The conjured 3.3 controversy can be explained by the simple fact that the circuit is being controlled by a TRIAC. Chopping the waveform to different degrees changes the voltage. A voltage increase explains the current increase no near miracle drop to the resistance is needed.

    • Obvious

      The active run current increases by the multiplication of dummy run current by the square root of the ratio of change from the dummy run to the active run, if resistance of the reactor stays the same. I cut the Joule heating in half, and it made no difference.
      Any linear adjustment to the all the Joule heat values has zero effect on the factor derived from it’s rate of change.
      Changing the dummy Joule heat only, and leaving the Joule heat alone for the active run could work.

      • US_Citizen71

        The ratio of the error would remain the same in the active run because the same equations were used throughout only the values were changed. The joule heating value was subtracted from the consumption value and the net power created by the reactor so to maintain parity it must be done on the active run calculations.

        • Obvious

          Yes. And it affects COP slightly, but does nothing to address the rate of current increase.
          Despite my differences with several detractors of the report COP, my active run current calculations, using two entirely separate routes to arrive at the values, agree within 2 decimal places. These are identical to what Ivan has posted, and I am sure they are the same as everyone else’s that have done the math in a similar way.
          Edit: My best simulation results in about ~300 more Watts input than reported for the report on the active run, which isn’t bad. It chops the COP down a fair bit, to 2.37 to 2.48 COP, but all considering it might be close.I’m looking at this to see that it does work out in another route of calculations, so that I am satisfied that it approaches reality.
          (I am using 3.64 W Joule heat for the dummy)

          • US_Citizen71

            An increase in voltage from the control box also explains the increase in current. The triac could chop the wave form more at lower power than it is at higher power causing the voltage to increase. Correct?

          • Obvious

            No, or not exactly.
            V is totally dependent on I and R, and it cannot be moved without moving at least one (and I am almost 100% certain) both of I and R.
            I do have a suspicion that V is involved in some other way, but I can’t find a way to involve V without the I and R un-doing any changes.

          • US_Citizen71

            Vrms in AC is determined by the wave form. If only the first 10% of the upslope is allowed through at low power the voltage would be lower than the full wave. I=V/R If V increases while R remains the same I increases.

          • Obvious

            I’m still fighting with it.

            Here’s a new formula I derived myself from doing the trigonometry with real triangles in a delta configuration, using a right angle triangle and report currents…long story.
            It must only apply to a balanced delta configuration:

            Ptotal = IL^2 + (1/2IL)^2

            (just plug in the IL current to start with)

            ie: Ptotal = 46.7^2 + 23.35^2………predicted row one current…..
            …..Ptotal = 2726.1225
            no questions asked, no V, no R…..no t…..

            or Ptotal = 19.7^2 + 9.85^2
            ….Ptotal = 388.09 + 97.0225
            ….Ptotal = 485.1125

            I don’t know exactly the full repercussions of this formula, but it works.
            It might actually derive better W values than in the report, but the formula is extremely finicky when it comes to rounding. Keep as many decimal places as you can. It can also derive the current consumed by line loss by inputting the correct W for Joule heat loss, and running in reverse, or subtracting the line loss W from the total P and running the values.
            I have also derived the resistor values using this more intensively, but I haven’t quite got that (the proofs) fully worked out. Considering that this formula is V and R independent, it is pretty far-out.
            Edit: I think I can make a variant of this formula for the wye, using the same idea that generated this formula, also.

          • Dr. Mike

            Obvious,
            An interesting equation! Sure enough if you use this equation you get fairly close to the power delivered to the load. But how can this be- power only dependent on the line current? I believe I have found out why your equation works. Your equation from above: Ptotal = IL^2 + (1/2IL)^2 can be simplified to Ptotal = 1.25*IL^2. However, the power to the 3 heater coils is Pcoils = 3*Rcoil*Icoil^2, but Icoil = IL/SQRT(3). Therefore, Pcoil = Rcoil*IL^2.. Assuming the Cu wire Joule heating is small relative to the heater wire Joule heating, Pcoil approximately equals Ptotal. Therefore, Ptotal = Pcoil = Rcoil*IL^2. You derived an equation Ptotal = 1.25*IL^2. Why does your equation work? Because the coil resistance can be calculated as 1.235 ohms for the Hot-Cat. The 1.25 factor “works” only because the actual heater wire resistance is 1.235 ohms! Try your formula on a an example where a 3-phase line current of 10A is delivered to a 5 ohm delta load. The actual total power delivered to the load is 500W, whereas your calculation would give 125W.
            Dr. Mike

          • Obvious

            Delivering 10 A through a delta with 5 ohms inductance is impossible to make 500 W. 5 ohms inductance is not the same a as 5 ohms resistance in each delta side.

            Using your example, it can be shown that 50 V would be required.

            P = VI, V= 500/10 therefore V = 50.

            In a delta, the line voltage is the same as phase voltage.
            Try and calculate the power per 5 ohm resistor at 50V…..
            P = V^2/R, P = 2500/5 therefore P = 500 W. But there are 3 R’s!!
            so try dividing power three ways, one for each resistor…..
            …….1/3P = V^2/R
            …….1/3P = 2500/R
            166.6667 = 2500/R
            ………….R = 2500/166.6667
            ………….R = 15 !!! [Use R = (R1+R2+R3) and this fixes this part?]. But this is weird, since now we have 1/3 the power and three times the resistors!
            There is a hint of something going on here….
            Actually the three R’s in the delta, in order make 500 W, and have a 5 ohm inductance would have to be (R1+R2)R3/(R1+R2+R3) = 5
            This is due to “forcing” current IN one leg, and OUT one other, so that the resistors are series-parallel (one in parallel with two in series).

            This is what happens when all the resistance is calculated by diving the total power by current. You get the inductance of the whole delta from the perspective of two out of three phases, not resistance of each resistor. Mathematically you have disconnected one delta cable.
            Using 5 ohm resistors…..
            and…since the r’s are all the same….

            Rdelta inductance = 2R(R)/3R
            ………………………5 = 2R(R)/3R…..use some 1’s for R to get a factor..
            ……………………….. = 2/3 (5)
            Rdelta inductance = 3.33333~ ohms NOT 5
            Now:
            P = I^2R
            P = 10^R
            P = 100(3.3333333~)
            P = 333.33333~
            So to get 500W, you need 3/2 (5) ohms, or boost I^2 to make the difference. This gets really complicated, since this messes with V. In effect, you must “steal” from V (was 50V, what is it doing now?) to make this work. If you steal from V, then I changes and P changes if you hold R. ad infinitum, nearly
            Best to do some test numbers and work it out by estimating. Lots of work.
            Reversing my equation gives an answer of 20A Line current to satisfy the 500W.
            Well, might as well try it.
            …..P = I^2R
            …..P = 400(R)
            …..R = 500/400
            …..R = 1.25…….R is very different. This would be inductance of the delta required to dissipate 500W. And 20 A are required.
            Then the old way…using 10A and 1.25 ohms inductance:
            P = I^2(R)
            P = 100(1.25)
            P = 125 W………There is my 125 W!!

    • Dr. Mike

      US_Citizen71,
      I believe you are confused about the 3.3X factor. The error in the current in the C2 lines is that the C1 current should be divided by SQRT(3) rather than divided by 2 to calculate C2 currents. This would make the Joule heating in all 6 C2 lines total 2.18W, rather than the 1.6W stated in the report The factor of 3.3 is the relative fraction that the resistance of the Inconel heater wire must decrease in the active run to make the Input power numbers agree with the Inconel heater wire currents which can be calculated fro the Joule heating numbers.
      Dr. Mike

      • US_Citizen71

        Yes I understand that current that would come from the updated value is better than cold fusion itself as it creates current out of thin air. I used the values that I did to show that the effect of changing the joule heating is almost meaningless.

        • Dr. Mike

          US_Citizen71,
          I don’t know what you mean by “I understand that current that would come from the updated value is
          better than cold fusion itself as it creates current out of thin air”. Don’t you realize that each line current is the superposition of 2 currents that are 120 deg out of phase? The RMS value of this superposition of 2 phase currents is not equal to 2 times the RMS current going to each phase! However, the instantaneous line current is equal to the sum of the two phase currents- no current is being created out of thin air at any point in time. If you would Google “3 phase delta current equations”, you might get a better understanding of where the 1/SQRT(3) factor comes in. I certainly did not remember this 1/SQRT(3) factor from my power and motors course from over 40 years ago. However, after this error was pointed out in the report, I was able to understand why you can not sum out of phase RMS currents. Take a little time to study what information is readily available on the internet, or check a textbook on 3 phase power systems, and see if you don’t agree that the authors have made an error in their calculation of the C2 heater wire currents. Wouldn’t you like to see this error fixed when the report gets revised?
          Dr. Mike

          • Thomas Clarke

            What definitely takes time is to see the sqrt(3) applies even if the duty cycle is small. But it does!

            Peer review has the great merit that errors like this, which everyone makes occasionally, get caught more often. In this case the error does not affect the results much but it still looks bad.

          • Dr. Mike

            Thomas,
            I agree totally with both statements. As you remember, I thought SQRT(3) factor perhaps didn’t apply to a short duty cycle, but I found where I had made a very fundamental error in my calculation. Frank’s forum here on his “e-catworld” website is giving the Lugano report a much better peer review than many published papers receive. I wish there were a couple of good thermal engineers giving input to the review of the report so that some my questions about the thermal issues that I brought up could be answered by someone with expertise in the field.
            Dr. Mike

          • US_Citizen71

            Work the equation backwards and you will find the current in C2 on the 5.28w equation would be higher than the current in C2 should be if calculated correctly by approximately 60%.

          • Dr. Mike

            US_Citizen71,
            The C2 RMS current should be 19.7/SQRT(3) = 11.27A for the dummy run if the C1 current was measured to be 19.7A. Would you like to see this error fixed in the report? Were you able to figure out why the C1 RMS current is not equal to the sum of two C2 RMS currents?
            Dr. Mike

          • US_Citizen71

            They obviously used the wrong equation and yes I would like to see them fix it. The three phase power threw them and many commenters as well including myself. It took a conversation with a power systems engineer for me to knock the rust out and remember the correct way to figure the current.

      • Obvious

        That is incorrect, in my opinion.
        If you use Lcurrent/sqrt(3) then you make a vector angle sum of current for one phase. Three times that vector is zero net current, since each vector has
        sign (+/-), relative to each other. The three vectors point away from the origin at 120° angles to each other.
        The proper, positive, vector sum is then the vector sum of TWO sqrt(3) current vectors.
        This latter sum is the DC equivalent, “forward, (+)” current in the circuit.

        • Dr. Mike

          Obvious,
          When ivanc first brought up this error, I had to Google “3-phase delta current equations” to verify that he was correct. Why don’t you do the same? You appear to have a basic understanding of electrical equations and should be able to understand the textbook explanation of why the RMS current in the C1 line is not the sum of 2 RMS currents in the C2 lines.
          Dr. Mike

          • Obvious

            The sum of two RMS currents, using sqrt(3), in one C2 cable is greater than 1/2 of the C1 cable for a very good reason. The sum of two identical (but 120° separated) RMS sqrt(3) values is greater than 1/2 of the forward current flowing through the entire circuit at any one time.
            In fact, one RMS sqrt(3) “measurement” is exactly equal to the
            sum of the RMS sqrt(3) currents flowing in the opposite direction in the other two phases at any one time.

            Edit:
            One sqrt(3) is the halfway point between the final vector distance travelled by two opposed, equal, length lines of length IL, separated by 120°, in one phase. So two times sqrt(3) is the vector sum of the two vector halfway points of two opposed IL, each separated by 120°, for two phases. But since the two phases total 240°, they are greater than 180°, and therefore the vector sum of the two sqrt(3) current values make a total vector that exceeds one half of the total possible “forward” current.

          • Obvious

            Hmmm.
            If two times IL(sqrt(3) is 240 degrees’ worth, then the maximum forward current of 180° out of 360° is 240/1.333~ (240/180 = 1.33333~)
            2(11.37) = 22.74 A, 22.74/1.3333~ = 17.055 A.
            My triangle figuring came up with this value several times, and I was wondering why that was, and not 19.7.
            So there you have it.
            1/2 total forward delta I must be 17.055 A at any one time, if IL = 19.7A

            So everyone was wrong.

    • AlbertNN

      The issue with the joule heating and the error in the formula is not the impact they have on the calculated COP in the article. Instead it is remarkable to find such a basic error of Maxwells equations in this article, and second the tabulated values of the joule heating currents does support the hypothesis that one current clamp was reversed in the actual test.

      • Thomas Clarke

        I don’t understand what you are saying here.

        The sqrt(3) error is quite separate from the X3.3 issue. The X3.3 calculation is independent of the sqrt(3) error because it is based on ratios. Any error in wire resistance just cancels out.

      • US_Citizen71

        The math for the reversed clamp theory requires the voltage to remain constant. Something that there should be no expectation of with a triac in the circuit. The power to the coils is being regulated by chopping the waveform which would vary the voltage.

    • ivanc

      If I am lucky and this is published….

      There is no data for the referred readings.

      The joule should be small, but they gave us all the data to calculate the IL current when they did that calculation.

      you saying the max diff should be 360w. who knows 360w is a number calculated under maximum stress, or normal stress??? (probably) in this case is just an arbitrary number you need to see the data to be able to comment.

      but:

      “The exercise above was not to determine the actual value for the joule
      heating of C2. I did it show how little the difference being debated
      matters to the values in the report.”

      I repeat to you the problem is not the joule value, but this value encodes the IL the current supplied in the feeding lines. and this current allow us to calculate the circuit.
      the Joule value is just the carrier of this info, and to do the calculation you have to use the same method as they did, using Ip=1/2 IL. so the current you find is the current they read.

      Does did answer your question? if not Why?

  • Dr. Mike

    Thomas,
    The data from the ramp-up portion of the active run is crucial for determining what is going on. Maybe one day we will see that data!
    Dr. Mike

  • Thomas Clarke

    This looks to me like a great distraction!

    While it is true that the report miscalculates the the wire currents slightly, and hence the calculated wire resistance, that has no real effect:

    (1) the adjustment to the Joule heating values has a few % effect on COP. Irrelevant.

    (2) The power anomaly we all care about depends on the ratio of Joule heating and total power in one case divided by the ratio of Joule heating and total power in another case. Hence a constant multiplier of Joule heating power has no effect on it at all.

    The only issue would be if the authors had calculated Joule heating for the active run different from how the calculated it for the dummy run (contrary to what they state). That would maybe explain a difference.

  • Thomas Clarke

    -> kdk
    I don’t really understand that. There is an error in the report. It is a major error, because it could be explained by power under-measurement X3 on the active test. It can easily be cleared up by the report’s authors explaining how they got it, and proving or denying the idea of a variable resistance heating element.

    The only mystery is why they are silent on this matter when they are obviously responding to criticism (the clamp hypothesis).

    • Freethinker

      No, Thomas Clarke, there is no error in the report of the sorts you imply. You have an opinion of there being an error, and that opnion you base on the preconceived notion that LENR is quackery and the ECAT cannot work. You base your conclusions on limited data taken out of its context, apply said data for uncorrelated situations, and add a fair amount of confabulations to build your conjectures.

      You have the desire to feed a dispute on the claims in the report, to be debated in the domain outside the scope of the test, for which you have no or little relevant data. Nobody can settle the dispute in that domain with the data at hand, but of course they do not have to. The report is clear enough, as the input power and the reactor surface temperature has been measured well enough to support the claims in the report.

      You are trolling, and your reasons for doing this I can only imagine.

      • Thomas Clarke

        Freethinker.

        Answering the substantive points you make: you are saying that because the report has a clear measurement of input power it is therefore correct.

        That standard of evidence (clarity => correctness) is of course open to you to adopt.

        However most people know that mistakes can happen and therefore they look for all related and relevant measurements to be correct.

        Let me give some examples:
        The testers got the sqrt(3) factor wrong – thought it was 2. That is a real error, but when you follow the numbers through you find that it makes no significant change to the headline COP. It shows the testers made a mistake, but that can be corrected and it was anyway not one that matters. So this is not a criticism that invalidates the report.

        However, the testers also provide measured currents that don’t match the measured power. Your logic says that the measured power, because it is clear, must be correct. Therefore either Rossi has invested a material with miracle properties, or the measured currents must be wrongly measured in either the dummy or the active test. That statement is fact, not conjecture. You cannot get away from it.

        Now we have a problem. The measured currents are also clearly measured, given (as Joule heating powers) in the same table as the measured input powers. Why do you conjecture that the measured currents are wrong and the measured powers are correct?

        This is a significant error – because if it is the currents wrong then maybe the COP > 3 stands (depending on which current is wrong). Whereas if the measured power is wrong, since we know the dummy power measurement was correct, we are forced to recalculate COP at ~ 1. So unlike the sqrt(3) error this one must be resolved, else the report has no validity.

        I fail to understand why you say no-one can settle the dispute with the data at hand. The testers very likely can. They can:
        (1) Check resistance to see if it changes during the active test.
        (2) Check all the PCE-830 measurements (specifically line powers) to make sure that the power measurement taken is correct. Doing this step they can ask for a second opinion from someone completely independent of the tests and skillfull in three phase power measurements. The testers themselves are clearly not so skilful because of the known mistakes in the published report.

        These steps are the minimum that would be required in a report which is to be taken seriously by scientists, and they are not unreasonable.

        The beauty of this test, and why I am posting here to make sure this is understood, is that the testers probably do have enough data to resolve this issue completely and solve the mystery.

        If they choose to do this.

      • DickeFix

        Freethinker, your thinking seem not to be as free as your name indicates. You refuse to admit the inconsistencies in the reported results. You are suspicious towards all people that try to logically analyze the measured results and you reduce their potential explanations to “conjectures”. In short, you are trolling, and your reasons for doing this I can only imagine.

      • kemosabe

        “You have an opinion of there being an error, and that opnion you base on the preconceived notion that LENR is quackery and the ECAT cannot work.

        The speculation of an error does not require any preconceived notion about LENR. It only requires a very justified notion that the heater coils do not decrease in resistance by a factor of 3 between the dummy run and the live run. And this is justified by the report itself which states (emphasis, mine): “Three braided high-temperature grade Inconel cables exit from each of the two caps: these are the resistors wound in parallel non-overlapping coils inside the reactor .

        If the heating coils are not inconel, as described, then the report is misleading. And in any case, a material that decreases multifold in resistance in that temperature range and then becomes constant would be as exotic as cold fusion itself, and it would be additionally surprising that the factor coincidentally equals the claimed COP. This represents a serious multiplication of miracles problem.

        The report was intended to show plausible evidence for an excess heat effect (regardless of the plausibility of cold fusion itself). Instead, it requires acceptance of highly anomalous behavior of heater coils, which have nothing to do with LENR itself. That makes the evidence implausible, and the authors presumably have the information needed to resolve this anomaly. Why don’t they?

      • Thomas Clarke

        I’ll leave it at DickieFix’s reply below, unless Freethinker wants his two substantive points shown false again, in which case my moderated out reply I think does it?

      • Dr. Mike

        Freethinker,
        What Thomas Clarke is doing is peer reviewing the report based on his knowledge of electrical engineering. This has nothing to do with his opinion on whether or not he thinks LENR works. I’m sure both he and I were quite happy that the analysis of the “ash” clearly showed that nuclear reactions were taking place in the reactor in the Lugano test. I haven’t seen anyone question these “ash” results, other than even the authors wish they had been able to retrieve a larger quantity of the “ash”.
        Your statement “You base your conclusions on limited data taken out of its context,
        apply said data for uncorrelated situations, and add a fair amount of
        confabulations to build your conjectures” seems to imply that an electrical engineer can look at data saying the Joule heating in Cu wires went up by a factor of 6.25, but then can not conclude that the current through those wires had to go up by a factor of 2.5. What other conclusion could be derived from this data? Since the heater wires are connected in series with the Cu wires, is it not possible for an electrical engineer to conclude that that the heater wire current must also go up by the same factor of 2.5?
        If you would like to argue that the Lugano scientists completely changed the set-up for the active run, it would mean that the dummy run was not a legitimate control for the experiment. This is possible, but it would mean that the authors don’t know how to run a controlled experiment. If you look at the results of the first independent report, you will see that it is clear the authors understand the need for having a good control in their experiment.
        Dr. Mike

        • Freethinker

          Thomas Clarke is not peer reviewing anything.

          He is trolling, and it is quite a hindrance for an individual, claiming to peer review in all honesty, that he is biased to assume that things cannot work, when looking into data that is not relevant and not correlated. To assume that a few single datums are relevant in a another scope, not covered by the analysis or the intention by the testers, will not make his conclusions right, as they are based on conjecture and, his opinions of the LENR field, and the clear notion that the ECAT cannot work.

          See, Mike I can repeat all I have written about this, as I am fully committed in my view on this, but it will make no difference. Thomas Clarke is a troll, and will continue to chase this, and monotonically drive the discussion in this forum, with you, Ivan and DickeFix to cheer him on.

          And that is exactly what he wants, to corner this discussion, so the outcome of the report is ambivalent. To corner it into the out of scope discussion where there is not enough data to say anything, but deliver conjectures.

          You know my take, this discussion is irrelevant, as the input power is correctly measured, using the two PCE-830 and the sane minds of the testers.

          Understand that you will never get any data on what is in the black box. The detailed output of the control box is part of the black box, the details of the reactor innards is part of the black box.

          But that is of course only my opinion.

          • bachcole

            Year: 1494

            CHURCH ELDER: Everyone knows that the Earth is 25,000 miles in circumference. So Columbus must either be a fool or a crook or a devil.

            PEASANT: But Father, I thought that the world was flat.

            CHURCH ELDER: No, my son. The world is round and we have calculated that it is 25,000 miles in circumference. That fool Columbus explains his theories by saying that the world is only 17,000 miles in circumference.

            PEASANT: But Father, what if there is other land between the East Indies and Spain?

            CHURCH ELDER: Don’t be such a fool. You are speculating. You need to understand science and logic and don’t be such an ignoramus.

          • Thomas Clarke

            I think the difference here is that Freethinker is making judgements based on an approximate understanding of the report, and expects others to do the same. In that case expectations (bias if you like) are likely to influence what is considered likely.

            Whereas Dr. Mike, ivanc, I and DickeFix are looking at one aspect of the report about which precise judgements can be made and seeing where they lead. You get the same answers whatever your expectations.

            I agree there are other aspects of the report where the judgement of probabilities is more difficult – such as how accurate do you consider the IR calorimetry – and in that case expectations might, however hard you try, influence judgement. That cuts both ways, with your expectations just as likely to influence you as anyone else’s.

            And what I want is a comment from the testers that will resolve this anomaly and make the report outcome clearer, whatever it is. They have the data to enable them to do this.

            I’d expect that to be shared by all here. And I think it is quite possible.

      • kemosabe

        To emphasize the point: It is a far greater conjecture or confabulation to assume that the heater coil magically changes resistance by a factor equal to the claimed COP, than it is to assume an error was made, deliberately or otherwise. Because if the coil is not magical, then there’s an inconsistency in the reported data — you know *in the report* which makes it part of the scope.

    • bachcole

      Thomas Clarke, if I were as 100% certain as you are that there was an error in the calculations, and if I was in a forum that simply refused to believe me, no matter how civil I was or how good my math was, I would leave that forum, with a little bit of frustration, thinking what a bunch of moron they were. (I am NOT suggesting that you leave.) I AM suggesting that you are not 100% certain, even if you think that you are. NO ONE is believing you. (:->)

      • Thomas Clarke

        Well I think quite a lot are believing me, Dr. Mike, and invanc. We are all saying the same thing!

        I find this mystery fascinating because the testers probably have recorded enough data to resolve it. I’m hoping that they will do this. I think making precisely what the issue is as clear as possible is the best way to encourage them to do this.

  • Freethinker

    Thomas Clarke, it is conjecture because you build it up from information that does not go together.

    I take the input power measurements at face value, as the are measured and presented as intended within the scope. They are sets of information that are coherent, and relevant, supported by the data from two PCE-830, and analyzed by intelligent and competent people. Hence I do not conjecture. I don’t pick data intended for one purpose, and use it in another purpose where the data is no longer relevant or correlated with other data used.

    You are constantly claiming that you are correct in your assertion, but you are not. You fight fiercely to bring this discussion into the domain that is outside the scope, where nobody can argue from a standpoint of fact.

    But it is befitting your already preconceived notion that LENR is quackery and the ECAT cannot work anyway. You are a pathological skeptic who are engaged in trolling on this website.

    I far as I go, you have no credibility.

  • Freethinker

    No, Thomas Clarke, there is no error in the report of the sorts you imply. You have an opinion of there being an error, and that opnion you base on the preconceived notion that LENR is quackery and the ECAT cannot work. You base your conclusions on limited data taken out of its context, apply said data for uncorrelated situations, and add a fair amount of confabulations to build your conjectures.

    You have the desire to feed a dispute on the claims in the report, to be debated in the domain outside the scope of the test, for which you have no or little relevant data. Nobody can settle the dispute in that domain with the data at hand, but of course they do not have to. The report is clear enough, as the input power and the reactor surface temperature has been measured well enough to support the claims in the report.

    You are trolling, and your reasons for doing this I can only imagine.

    • Thomas Clarke

      Freethinker.

      Answering the substantive points you make: you are saying that because the report has a clear measurement of input power it is therefore correct.

      That standard of evidence (clarity => correctness) is of course open to you to adopt.

      However most people know that mistakes can happen and therefore they look for all related and relevant measurements to be correct.

      Let me give some examples:
      The testers got the sqrt(3) factor wrong – thought it was 2. That is a real error, but when you follow the numbers through you find that it makes no significant change to the headline COP. It shows the testers made a mistake, but that can be corrected and it was anyway not one that matters.

      However, the testers also provide measured currents that don’t macth the measured power. Your logic says that the measured power, because iy is clear must be correct. Therefore either Rossi has invested a material with miracle properties, or the measured currents must be wrongly measured in either the dummy of the active test.

      Now we have a problem. The measured currents are also clearly measured, given in the same table as the measured powers. Why do you conjecture that the measured currents are wrong and the measured powers are correct?

      This is a real error – because if it is the currents wrong then maybe the COP > 3 stands. Whereas if the measured power is wrong, since we know the dummy power measurement was correct, we are forced to recalculate COP at ~ 1.

      I fail to understand why you say no-one can settle the dispute with the data at hand. the testers very likely can. They can:
      (1) Check resistance to see if it changes during the active test.
      (2) Check all the PCE-830 measurements (specifically line powers) to make sure that the power measurement taken is correct. Doing this step they can ask for a second opinion from somone completely independent of the tests and skillfull in three phase power measurements.

      The beauty of this test, and why I am posting here to make sure this is understood, is that the testers probably do have enough data to resolve this issue completely.

      If they choose to do this.

    • Dr. Mike

      Freethinker,
      What Thomas Clarke is doing is peer reviewing the report based on his knowledge of electrical engineering. This has nothing to do with his opinion on whether or not he thinks LENR works. I’m sure both he and I were quite happy that the analysis of the “ash” clearly showed that nuclear reactions were taking place in the reactor in the Lugano test. I haven’t seen anyone question these “ash” results, other than even the authors wish they had been able to retrieve a larger quantity of the “ash”.
      Your statement “You base your conclusions on limited data taken out of its context,
      apply said data for uncorrelated situations, and add a fair amount of
      confabulations to build your conjectures” seems to imply that an electrical engineer can look at data saying the Joule heating in Cu wires went up by a factor of 6.25, but then can not conclude that the current through those wires had to go up by a factor of 2.5. What other conclusion could be derived from this data? Since the heater wires are connected in series with the Cu wires, is it not possible for an electrical engineer to conclude that that the heater wire current must also go up by the same factor of 2.5?
      If you would like to argue that the Lugano scientists completely changed the set-up for the active run, it would mean that the dummy run was not a legitimate control for the experiment. This is possible, but it would mean that the authors don’t know how to run a controlled experiment. If you look at the results of the first independent report, you will see that it is clear the authors understand the need for having a good control in their experiment.
      Dr. Mike

      • Freethinker

        Thomas Clarke is not peer reviewing anything.

        He is trolling, and it is quite a hindrance for an individual, claiming to peer review in all honesty, that he is biased to assume that things cannot work, when looking into data that is not relevant and not correlated. To assume that a few single datums are relevant in a another scope, not covered by the analysis or the intention by the testers, will not make his conclusions right, as they are based on conjecture and, his opinions of the LENR field, and the clear notion that the ECAT cannot work.

        See, Mike I can repeat all I have written about this, as I am fully committed in my view on this, but it will make no difference. Thomas Clarke is a troll, and will continue to chase this, and monotonically drive the discussion in this forum, with you, Ivan and DickeFix to cheer him on.

        And that is exactly what he wants, to corner this discussion, so the outcome of the report is ambivalent. To corner it into the out of scope discussion where there is not enough data to say anything, but deliver conjectures.

        You know my take, this discussion is irrelevant, as the input power is correctly measured, using the two PCE-830 and the sane minds of the testers.

        Understand that you will never get any data on what is in the black box. The detailed output of the control box is part of the black box, the details of the reactor innards is part of the black box.

        But that is of course only my opinion.

  • US_Citizen71

    The math for the reversed clamp theory requires the voltage to remain constant. Something that there should be no expectation of with a triac in the circuit. The power to the coils is being regulated by chopping the waveform which would vary the voltage.

    • Obvious

      I think the 3 and 6 are just how many lengths of cable of each type there are.
      No special inference.

  • US_Citizen71

    The potentiometer is on the trigger to the triac. Changing the potentiometer changes the chop of the waveform.

  • Thomas Clarke

    @ Freethinker

    You have repeated a number of times that: “there is no error, I am trolling, etc”.

    For the facts of the matter I’d refer you on this thread to Dr. Mike and ivanc, both of whom have worked through the electrical issues and I think agree with me. Perhaps you consider them trolls too? Or is it my impoliteness in explaining to anyone who mistakenly considers this error to be insignificant why it in fact is significant?

    Or, if you like, we could go through this together. We start with the 3.3X difference between ratio of total and Joule heating powers between the dummy and active tests and go through the possible things that could cause it. For each such thing we then work out whether it is an error, and if so how does it change the headline COP.

    If you care about scientific validation of LENR that will make the world take notice and release development funds, you surely want the report to be improved and anomalies explained? So although my view as to what is likely here is different from yours, we both want the same outcome.

    You say that the data to resolve this does not exist. Again, I refer you to Dr. Mike or ivanc. The data exists which will clear up:
    (1) is this some extraordinary variable resistor?
    (2) is this some heating element that switches to a lower resistance between the two tests?
    (3) is this an error in current measurements?
    (4) is this an error in power measurements?
    (5) Is this a reversed clamp (this is actually just one possible version of 4)?

    The PCE-830 took all this data, and we can assume it is correct. So by looking carefully the mystery can very likely be understood. After that, we can see whether the error corrected keeps the headline COP, or changes it.

    I can’t guarantee there will not remain some mystery – but surely, given the stakes here, the testers owe it to themselves and the scientific community to do what they can to resolve this issue? The report will be stronger with this matter resolved.

    • Mark Szl

      I agree, more needs to said and given about these issues by the Profs involved in the test. However has a good contact with them, please ask them for this data or ask if they are working on getting this data and when we can expect it released.

      Maybe that they are working on it by re-doing the experiment all over again and that is why we have not heard much lately.

    • Freethinker

      See Thomas Clarke, there is no need for you to regurgitate you conjectures again. I can only repeat my previous statements on you and your efforts.

    • ivanc

      Actually freethinker is acting like a troll, close minded, and with out capacity to defend his arguments, he just goes to offend, I Fully agree with your analysis of the 3.3 factor, I have even made a table and calculated I, R ,V and the 3.3 factor as last column in my table.
      The numbers are there.
      People who know about electricity will agree with us.
      The others think in mysterious ways they do not understand the concepts, How they will calculate the input power if they do not know electricity????

  • US_Citizen71

    Of course the resistance changes with temperature that is expected. Take several feet of bare wire make a non-overlapping coil. Attach the ends to an Ohm meter. Then take a propane torch and heat the coil and watch the resistance measured change before your eyes.

  • ivanc

    What is importat about IL/sqrt(3)?.

    IL/sqrt(3) is the current that is circulating in each of
    the ecat resistors:

    …………………………………| IL
    …………………………………| cu Line
    …………………………………|
    ………………………………/…… cu Phase (6 of them,no space to label all)
    ……………………………./………..
    …………………….ecat-R…………ecat-R.
    ……………IL/sqrt(3) /………………… IL/sqrt(3)
    ………………………./…………………….
    cu_line________———-ecat-R.———__________cu Line
    IL…………………………… IL/sqrt(3)…………………………………IL

    The closest posible circuit under the editor capabilities.
    ignore the dots, they are needed otherwise the editor will eliminate the formating

    Calculation of Power, voltage, current

    From report:
    WC1 = 3(R1I12) = 3(4.375 ∙ 10–3 ∙ (19.7)2) = 5.1 [W]
    WC2 = 6(R2I22) = 6(2.811 ∙ 10–3 ∙ (9.85)2) = 1.6 [W]
    Wtot.dummy = 5.1 + 1.6 = 6.7 ≈ 7[W]

    Note the factors (3) and (6) this shows the testers are using a balanced 3phase voltage and Current.
    so ideas about open deltas and unbalanced voltages have no base.

    Joule heat in Dummy reactor: Jh=3 iL^2 Rcu1+ 6 (iL/2)^2 Rcu2 (a)

    (using
    i/2 because this is how the testers did the calculation of joule
    heating, and I am using their calculated joule data. but they should
    have used i/sqrt(3) )

    …calculating the current and resistor of
    ecat: solving iL from (a) iL=sqrt(Jh*4/(12Rcu1+6Rcu2))

    Rcu1=0.004375
    Rcu2=0.002811
    iL=sqrt(jh*57.66513854) (b) , we get this by substituting Rcu1 and Rcu2 in above equation.

    Lets call Jof=57.66513854

    now the resistor value in ecat. using sqrt(3) Because is the correct way
    input power 479=3R (IL/sqrt(3))^2 (479 is net imput power or power in R-ecat)

    IL=19.7 amps
    R=479/( IL^2)=1.234249787 ohms.

    Assuming the ecat resistor is constant. (as it should be):

    R e-cat: 1.234249787

    Lugano Iphase has to be equal to calculated Iphase, because Rcu2 in series with ecat R—
    -Constant R—————————————————Variable R—————————–
    ——————————————–Input—————–Lugano———————————
    No——Joule—–IL——-Iphase—power——-V——–data——data——————V——Factor
    ———-From—-sqrt——IL/——-3ip^2*R—–I*R—–Total——Input———R——-I*R—–of R
    ———-Data—(Jo*jof)– sqrt(3)—————————Pwr——–Pwr—iPwr/(3*Ip^2)—-Change
    1.00—-37.77—46.67—26.94—-2688.21—33.26—2128.32—815.86—0.375—10.09—3.3
    2.00—-36.98—46.18—26.66—-2631.98—32.91—2119.76—799.84—0.375—10.00—3.3
    3.00—-36.49—45.87—26.48—-2597.11—32.69—2110.18—791.48—0.376—-9.96—3.3
    4.00—-36.41—45.82—26.45—-2591.42—32.65—2114.79—790.69—0.377—-9.96—3.3
    5.00—-36.13—45.64—26.35—-2571.49—32.53—2058.32—785.79—0.377—-9.94—3.3
    6.00—-42.43—49.46—28.56—-3019.88—35.25—2809.28—923.71—0.378—10.78—3.3
    7.00—-42.18—49.32—28.47—-3002.08—35.14—2846.32—921.91—0.379—10.79—3.3
    8.00—-41.89—49.15—28.38—-2981.44—35.02—2836.42—918.24—0.380—10.79—3.2
    9.00—-41.75—49.07—28.33—-2971.48—34.96—2820.11—917.9—-0.381—10.80—3.2
    10.00—41.93—49.17—28.39—-2984.29—35.04—2774.07—913.4—-0.378—10.72—3.3
    11.00—41.52—48.93—28.25—-2955.11—34.87—2800.31—904.77—0.378—10.68—3.3
    12.00—41.60—48.98—28.28—-2960.80—34.90—2826.51—906.98—0.378—10.69—3.3
    13.00—41.62—48.99—28.28—-2962.23—34.91—2831.67—910.47—0.379—10.73—3.3
    14.00—41.55—48.95—28.26—-2957.25—34.88—2823.63—908.13—0.379—10.71—3.3
    15.00—41.46—48.90—28.23—-2950.84—34.84—2883.12—905.01—0.379—10.69—3.3
    16.00—41.25—48.77—28.16—-2935.89—34.75—2886.18—906.31—0.381—10.73—3.2
    Dum round-7.00–20.09—11.60—498.21—-14.32–~486——-498.21—-1.234—14.32—1.0
    Dum real–6.73—19.70—11.37—-479.00—-14.04—486——–479.00—-1.234—14.04—1.0

    https://docs.google.com/spreadsheets/d/1tuKwWUxC2Gq_MtEED4_XYYIqehpOv47mGVx9yio-GKs/pubhtml

    Lugano Data for this table, page 22 report, section 4.3 report.
    Calcuated columns using formulas in headings. you could check formulas in any 3phase book.
    If the formating wrong in you browser, copy and paste to notepad..
    If you do not understand something please ask!
    Do not mix with phase, duty cycle and other incompatible ideas, all data is RMS. so let use the RMS way.
    To verify the calculations see the last two columns, where the formulas correctly reproduce the dummy data.

    • US_Citizen71

      “Do not mix with phase, duty cycle and other incompatible ideas, all data is RMS. so let use the RMS way.” – Stated another way please do the calculation incorrectly.

      You are assuming constant Vrms voltage I believe since you are ignoring the duty cycle. You can not ignore the duty cycle with a triac in the circuit. Vrms is different for 10% duty cycle vs Vrms for 100% duty cycle. Constant voltage is not an option when the waveform is chopped to control the power.

      • Thomas Clarke

        NB – I’ve edited this post to correct it as detailed below under EDIT.

        The ratio can be no more than sqrt(3) for any switched 3 phase waveform. Andreas.s has posted a proof on Mats thread. I have not checked this, but here is a way to see it is most likely true.

        Consider the possible instantaneous values of voltages on the three lines. Work out the 3 currents in the delta load. Work out the ratio between two delta load branches that join. See whether you can ever get a ratio larger than sqrt(3). You will not be able to.

        For example, you can get one of the lines with current divided equally, but only of if the other two lines the current is divided 0/100%. In this specific case the RMS values are:
        sqrt((1/4 +1/4 + 0 + 1 + 0 + 1)/6)= sqrt(5/12)

        and

        sqrt((1+1+1)/3)=1

        In this example the ratio is sqrt(12/5) < sqrt(3)

        This is an intuitive way to see why there has to be a limit b
        set x = a-b (the only thing that matters)

        inner resistor delta currents are:
        a-b, a-b, 0. Each current is counted twice (one at each end of its resistor)

        outer line currents are:
        L1: (a-b)+0= x
        L2: (a-b) + 0 = x
        L3: (a-b) + (a-b) =2x

        The complicated issue is how do we work out a total RMS current from these separate RMS currents on different wires. We need:
        Ptotal = Irms_total^2*R
        In other words the power is the sum of the square currents. So we must do a squared addition of the individual currents, and then divide by the number of wires, before finally doing a square root to get from power back to equivalent current.
        For inner, we have sqrt(4*x^2/6) = x* sqrt(2/3)
        For outer, we have sqrt((x^2 + x^2 + 4x^2)/3)= sqrt(2)*
        ratio outer/inner = sqrt(3) as expected!

        • US_Citizen71

          I=V/R

          V=1 R=1 then I=1

          If V=3 and R=1 then I=3

          There you go 3 times the current no change in resistance.

          No change in resistance is required to explain the change in current.

          The Vrms of a chopped wave can be calculated fairly close with the following:

          Vrms(switched) = √Duty cycle x Vrms(pure sine)

          Vrms(pure sine) or full wave = 100 volts rms
          Duty cycle = 1% or .01
          √.01 x 100 volts = 10 volts rms

          Vrms(pure sine) or full wave = 100 volts rms
          Duty cycle = 10% or .1
          √.1 x 100 volts = 31.62 volts rms

          Vrms(pure sine) or full wave = 100 volts rms
          Duty cycle = 100% or 1
          √1 x 100 volts = 100 volts rms

          • Dr. Mike

            US_Citizen71,
            Your calculation of getting the current to go up by a factor of 3 when the voltage goes up by a factor of 3 is absolutely correct. However, look at ivanc’s Table a few comments up. Although the current can be calculated to go up by a factor of 2.35 for the first part of the active run and 2.5 for the second part, relative to the dummy runs, the voltage as calculated from the power measurements is lower for the active runs than the dummy run! This is the discrepancy in the reported data that needs to be addressed by the Lugano scientists. Their computer logged data from the PCE-830’s should enable them to figure out why the data is in conflict.
            Dr. Mike

          • US_Citizen71

            To be honest I have no idea what he is attempting to show there. Back in my Navy days we would have called that a shotgun answer. It is such a mess due to formatting I am not exactly sure what headers go with what column or what the headers are intended to mean. Assuming absolute correct values in that is a large stretch, we do not know impedance, inductance, power factor or capacitance values of the circuit so exact values cannot be assured. Then you add to that that all of the resistances will change due to heating and not equally.

          • Dr. Mike

            US_Citizen71,
            Ivanc is trying to put spreadsheet results into a format that can be read in these comments. It was easier for me to understand his spreadsheet since I have made a similar spreadsheet in EXCEL. (He really went to a lot of work to get the results of his spreadsheet posted in these comments). The next to the last column are the calculated heater coil voltages, based of the reported power measurements. You can see the active run voltages at 10-11 volts are lower than the dummy run voltage at about 14V. The 6th column gives what the voltage should have been based on the calculated currents from Cu wire Joule heating and the calculated heater wire resistance of 1.234 ohms. These voltages are 33-35V.
            We do know the circuit. It is essentially a pure resistive load made up of the heater wire in a 3-phase delta configuration so the power factor should be really close to 1.000 with no capacitance and close to no inductance.
            Thanks for your service in the Navy! My brother spent 30+ years in the Navy in emergency medicine.
            Dr. Mike

          • US_Citizen71

            Thanks, I was a knuckle banger, machinist mate. I’ll wait for his new spread sheet before attempting to analyze it further I would recommend https://docs.google.com as an easy way to share it to the world.

          • ivanc

            Thanks for the suggestion.
            I do it tonigth after work.
            ….. and I just used to work as an Electrical Engineer.
            maybe the last bit will not help!

          • ivanc
          • ivanc

            I you use a true RMS meter, you do not have to do anything. the reading is the RMS, and is the value.
            Your Idea is a blasphemy in the field of electricity and alternate currents.
            This kind of analysis is done when you are given a wave form, a max and then they ask you to calculate the RMS.
            But in this case the Meter is given the RMS……..
            !!!!!!!!!!!!!OHHHHHHHHHHHHHHH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
            Is so frustrating….. Were are you getting all this fantasies from??????

        • ivanc

          What he says is wrong, RMS and Instantaneous current are different.
          RMS is the equivalent to DC, or said differently, causes the effect of DC, so the duty cycle is implicit in. the sqrt(3) has nothing to do with RMS, the sqrt(3) is a consequence of using a delta.
          !!!Why to speak and comment and fight against things you do not know!!!!!, get a book.

          • Thomas Clarke

            Ivan,

            I was not saying that RMS & instantaneous current are the same. But I did make a mistake now corrected – if your post referred to me. it is hard to tell with this thread who is replying to who.

            For this problem the instantaneous powers are always in ratio 3 the same as the RMS effective current ratio of sqrt(3). The individual instantaneous currents change of course, but when you sum the squares of the currents it turns out the power is identical.

      • ivanc

        The duty cycle is already expressed in the RMS, at 100% duty cycle the RMS will be maximum (I did not said equal to the max of the wave), as the duty cycle is reduced the RMS also is reduced, when duty cycle is 0 RMS is also 0. this is why you do not have to worry about duty cycle in RMS, The RMS meter has worked out this issue already.

        Please note this apply to RMS amps or volts.

        “In mathematics, the root mean square (abbreviated RMS or rms), also known as the quadratic mean, is a statistical measure of the magnitude of a varying quantity. It is especially useful when variates are positive and negative, e.g., sinusoids. In the field of electrical engineering, the RMS value of a periodic current is equal to the DC current that delivers the same average power to a resistor as the periodic current.”

        IS A STATISTICAL MEASURE.

        • US_Citizen71

          “IS A STATISTICAL MEASURE.” Of the current waveform / duty cycle. We do not know the duty cycle used for the dummy run we only know what the RMS current was, a little educated guessing and plugging of numbers can get some approximate values for some of the unknowns at that duty cycle. But we do not know impedance, inductance, power factor or capacitance values of the circuit so exact values cannot be assured. When the duty cycle changes the current will change no change in resistance is needed as the voltage will change. (see my answer to Thomas Clarke below)

          • ivanc

            you do not need to know the duty cycle this is the benefit of using RMS, and in this case even more as the load is resistive.
            RMS is the a kind of average (not real average) but is the effective value that will compare to DC for the cycle.

            Gentleman. How difficult is to understand this.???

      • Obvious

        I have examined the extrapolated current issue a bit more.
        The extrapolated currents I have are identical to Ivans.
        The 3.3 difference be mostly reconciled in the active run by dividing the extrapolated current by sqrt(3). This will result in a bit higher input current (possibly fixed with my new current of 17.055 A, but this may also simply fade away since it will apply equally to all values). The new sqrt(3)-corrected values will also include cable Joule heating.
        Importantly, the sqrt(3) denominator for the active run current correction is not a fluke.
        It is a direct consequence of un-monitored V changes.
        We cannot actually extrapolate the real active run current, only the active run power increase that causes the increased Joule heating of the cables. It may look like we can extrapolate for current, but only because we are ignoring the effect of V. For each constant R and “mobile” I value (squared) and a known Power, there is only one acceptable (real) V (also squared).
        P = I^2R and P = V^2/R
        The numerical (nominal) rate of exchange of V for I is 1:1, but one is a quotient (I), and the other (V) is a denominator to both R and P, so the exchange rate is not a “fair” one. They are inversely proportional, but not linearly so. These exchanges are numerically equivalent but are not equal in the real world. This inverse proportionality is the reason there is a fixed value for V for every set of “known” I and P and R values.
        Since the numerical equivalents of I and V are effectively vectors of “opposite polarity” to each other, sqrt(3) is the vector-summed position of the numerical rate of change from V to I or the inverse.

        Edit: and the reason for sqrt(3) (in my above discussion) is a direct consequence of three phase power.
        The physical example of this sqrt(3) exchange is the delta to wye transformation.
        In a wye, there is phase Voltage, and it is sqrt(3) Line Voltage. This is the angle of V to I in a wye. Current is Line Current in a wye.
        In a delta, there is phase current and it is sqrt(3) of Line Current. Phase voltage is Line Voltage. Phase current is at a sqr(3) angle to Line V (or vice-versa).

        • Obvious

          To watch the very finicky exchange rate at work, try this web-based calculator.

          http://www.rapidtables.com/calc/electric/Watt_to_Amp_Calculator.htm

          First check the first box to 3 phase.
          Put a power factor of 1 in (this is normal resistive Joule heat: motors, etc. are less).
          Put in 485 W.
          Calculate by clicking the button.
          INFINITY. Whoops no volts. With infinity, you could make any amps you wanted…….[ahem]….
          Well, put in a volt value. Best to start low. Try and guess the V to make 19.7A in the result.
          (Maybe you have worked this out already, or at least think you have.) ****Hint: less than 15 V is a good start.**** Don’t go crazy with the Volt values or this will take all day.
          Move the Volts up and down, maybe 10 volts at a time until you get close to the desired 19.7A.
          Note something interesting after a few tries?
          Whoa, now you are really close, using 0.1 volt increments. But it is quite sensitive. See how many zeros you can put on 19.7 A.
          Just for those that don’t have all night:
          ……..485 W……………Power Factor 1 …………….
          Volts…..Amps Calc………Decimal Volts…….Amp Calc
          1………..280.01………………………….0.2 —…….1400.07
          2.8……..100.000
          10………..28.001………………………..0.1 —…….2800.15
          14………..20.00
          20………..14.00………………………….0.02 …….14000.7
          30………….9.33………………………….0.01……..28000.5
          40………….7.00………………………….0.002….140007.4
          50…………..5.6
          60…………..4.67
          100…………2.8
          Toss a couple sqr(3) in these values, (which already incorporate squares for V and I), and the logarithmic exchange rate change can
          be, er…, shocking.
          .
          Better still, check out one of the power equations below the calculator on the same website, using P/sqrt(3) in part of the equation. The three phase calculation there is a cousin-brother of the one I derived, earlier, but still has V attached as well being a phase equation. It is an inside-out version of mine.
          (And works without a balanced delta. A balanced, three phase delta forces an extreme level of geometric control on the range of possible real solutions for Power, depending especially on the accuracy of the values. And Current, which is the actual carrier of force in an electrical power equation).
          .

          • Mark Szl

            I seems like everyone or almost everyone is now agreeing that there is an issue with the data even though it _may_ not effect the qualitative conclusions of the report.

            Admin/Frank! Can these individuals summarize the issue in a new thread?

          • Frank Acland

            I would prefer this discussion to stay in this thread.

      • ivanc

        Yes, you could ignore duty cycle if using RMS

    • US_Citizen71

      First question.
      “…calculating the current and resistor of
      ecat: solving iL from (a) iL=sqrt(Jh*4/(12Rcu1+6Rcu2))”

      This appears to be based off of I=sqrt(P/R)

      but you’ve changed it to I=sqrt(4P/2.3128938681604481344403320996299R) where P is the joule heating and R is the total of C1 and C2 copper cable resistance.

      What is going on here.

      • Obvious

        We could spend all day on the cables if we do not agree on their effective resistance.
        For solving how much the cables “see” power, we can (temporarily) ignore the reactor resistors (call them either superconductors, bolted to a large gold bar, or absent for now), and build a delta entirely from C2 cables, each side with two C2 cables in series. The C1’s can stick out the ends of the new delta and we will have to work on them a bit later, since they depend on the interpretation of the C2 wires’ current flow.
        Does that make sense so far?

        • US_Citizen71

          I’m not trying to be obtuse, I’m simply trying to understand what he put up. I see the he is using the basic equation to determine current from power and resistance but I don’t understand what he is substituted into the equation and why.

          edit:removed incorrect equation

          The current has to be figured different because of the 3phase I was doing DC. That likely explains his part of what he is doing in his equation.

          • Obvious

            Sorry, my post wasn’t specifically directed at you (or anyone) in particular.
            It was just a suggestion, in case the subject gets more complicated.

          • US_Citizen71

            no worries ( :

        • Dr. Mike

          Obvious,
          The authors have calculated the Joule heating summed in all Cu wires correctly except the C2 currents are equal to the C1 currents divided by SQRT(3), rather than divided by 2. When they make this correction they will get the Cu wire Joule heating for the dummy run to be 7.3W rather than 6.7W. Their implied assumptions in their Joule heating calculation include:
          1. The 3 Ci wire lengths are equal.
          2. The 6 C2 wire lengths are equal.
          3. All C1 currents are equal.
          4. All C2 currents are equal.
          5. The effects of heating in the Cu wire that is near the connection to the Inconel heater wire is ignored. (The resistivity will be higher in the heated wire.)

          These assumptions are all quite valid for getting a measure of the Cu wire Joule heating within a few tenths of a Watt.
          Dr. Mike

          • Obvious

            Looking on some wire companies websites, I see that a roll of 100′ of 0 gauge AWG wire has about .001 ohm resistance. We are not making a piano here or wiring the Griswalds Christmas lights. These are big wires. They have been overestimated, IMO, but what a are few W in hundreds? One bad crimp and the resistance could triple that. Car taillight bulbs use, what 15, 20 W? And we are trying to get a ratio from that? Could anyone calculate the charge in a battery or Joule heat of a harness in a car by measuring the heat of a tail light?

          • Dr. Mike

            Obvious,
            The Joule heating in the wire is small, and is a small factor in the results. I can’t tell you if the authors correctly measured the wire diameters to get an accurate calculation of the Cu wire resistance (the wire cross sections given in the report do not correspond to common Cu wire gauges). It doesn’t matter if they calculated the Cu wire resistance or the Joule heating accurately. All that matters is that they always calculated Joule heating the same way. The current in the C1 lines was reported to be 19.7A for the dummy run. Although they did not have a column in Table 7 for the C1 current for the 16 active run files , we can calculate those current as 19.7 times the square root of the Cu wire Joule heating power ratios. All errors in calculating the Cu wire resistance and even the fact that the authors miscalculated the C2 currents do not affect the calculated C1 currents. It’s these high C1 currents that don’t agree with the measured TRIAC power levels for the active runs.
            Your point of the wire crimp is quite valid. It would have been good experimental procedure to have a means of measuring the voltage directly across the heater wires to verify there were no bad connections in the set-up. The professors may have done this as part of their set-up after the reactor was reconnected for the active run. If they checked the connections in this manner, they should add one sentence in the revised report stating this.
            Dr. Mike

          • Obvious

            Please see my post to Ivan where I demonstrate the actual DC equivalent of the circuit is 17.06 V.
            I agree that one can disassemble the author’s use of math and related assumptions by de-integrating their reported values.

        • ivanc

          The resistance is given in the report

          • Obvious

            By effective resistance, I mean whether or not the C2 cables see more current on average than do the C1 cables.

          • ivanc

            The cables see RMS, there is no more or less just RMS

          • Obvious

            The RMS measurement is on the C1, not C2 cables.
            If the professors slid the amp probe down the C1 wire, and measured the both C2 cables together, side by side, as they exit the C1, then this measurement would also read 19.7 A for the dummy run.
            This indicates two things:
            1. That all the currents coming from all cables towards or away from any one C1 cable average out in a way that the probe considers to be the RMS equivalent.
            2. That this equivalence continues to the C2 cables, as long as they act as a pair in the same orientation. Individually, the C2 cables will act differently. The current in one C2 cable is phase current. Some currents in one C2 cable are out of phase with currents in it’s “twin”, which reach a balance in their common C1 cable.
            3. Folding any one these cables back on itself, and measuring the current of the folded-over wire (both wire halfs at once) as though it is one double wire, (whether C2 or C1) will result in a zero reading. The RMS of one half of the folded wire is exactly the same RMS in the other before folding. Together, the average instant current will be zero, so the RMS is zero of a folded-on-itself wire. This is the basis of the reason that the two C2 cables in (1.) above will read the same as the C1 cable.

          • ivanc

            I in

            ic2=ic1/qsrt(3)

          • Obvious

            sqrt(3) is a vector.
            Actually, the 19.7 A is also a vector, since it describes only the activity of one of three cables (or sets of cables) in an equilateral triangle of cables.
            19.7 A is by itself is a magnitude without sign. This is easily proven.
            3 x 19.7 A is not the correct total current in the system.
            We must assign a sign to 19.7 A to make a DC circuit equivalent, if we adding the contributions from other cables.
            In a balanced triangle with 19.7 A going “in” each line, then the sum is actually zero.
            This is easily shown, since 19.7^2 times 1.23 is 477 (W).
            The sum of 477 + 477 +477 is also not a correct answer.
            Therefore, we choose a sign for one of the 19.7 A (say +19.7) to make a DC-type value. Everything else must be given the sign which is relative to this + value. A point of reference must be chosen, or the resulting math is meaningless.
            They cannot all be +19.7, either. The total sum of all the +currents must be equal to the sum of all the -currents.
            Same goes for power. The sum of all the +currents, squared, times resistance equals the total measured +power. The total measured power from doing the sum of -currents squared times resistance is equal to -total power. The sum of total -power plus total +power is zero.
            The magnitudes of the 19.7 A, or 11.38 A (if you must) can be used to calculate the power in one third of the delta.
            But you can’t just add them up any old way. If the magnitudes are without sign, then the sum answer has no sign. The correct way for adding all magnitudes together is to add the RMS values, and then divide by the number of RMS magnitudes used in the summing, for a total RMS average. (Another RMS-ing of the RMS values would be best, but since the values are same for each side of the delta, this is pointless).
            Having now returned to the original RMS value, this value is still only the one that is from this reference point.
            Since the triangle has three sides, this RMS value is actually incorrect for the delta as a whole. It only represents 1/3 of the triangle, and we need 1/2 of the triangle forward current to actually describe the DC equivalent properly.
            But most of this is a waste of time.
            We already know that the RMS current in one set of cables in one corner if the delta sees, on average, 19.7A. So the other three must also, on average. One phase in, two phases back out for each cable.
            The RMS for each C1 is then the return current average, not the input current average. If this was two phase, then one in for one out would occur, and the math would simplify immensely. The same RMS would be the result on either end of the circuit (and we would not add then up).
            Since the real RMS measured at each C1 cable actually is the power returning from the other two phases, then this current is higher than the actual “input” power on this one line.
            The real input power on each on C1 line is 1/2 of the summed vector RMS average of the other two C1 cables. This looks like it would be exactly the same number as 19.7 we started with, BUT these cables “add power” for a total of 240°, and the only allowable DC equivalent is 180°. Here come those nasty vectors again.
            The inside angle of the overlarge 240° angle is 120°. This makes a special triangle, the Isosceles, just like my other triangle forays.
            The vector sum of the two opposing cables is 2(sqrt(3)*19.7A = 22.74 A. This is higher than the 19.7 A because this is the 180° equivalent of the return currents from the opposite side of the circuit from one side.
            Now, consider the first 19.7 A in the cable we began with. It is too high, since it is only 120° of 180°. It is also facing the wrong way compared to the others. We must divide the 120° value into a wider 180° equivalent. It is 11.37 A.
            This is unbalanced, so the route to fix it to make the real DC equivalent amperage is to add the 22.74 to the 11.37 and average them.
            This gives a real, correct, proper, DC equivalent forward amperage of 17.06 A.
            My geometry forays have already confirmed this value.

      • ivanc

        Jh=3 iL^2 Rcu1+ 6 (iL/2)^2 Rcu2 (a) this is the equation Levi team used.
        I want to know IL so:
        now getting rid of denominators
        4jh=12 IL^2 Rcu1 + 6 IL^2 Rcu2
        factorizing IL^2
        4jh=IL^2 (12Rcu1+6Rcu2)
        then
        IL^2=4jh/(12Rcu1+6Rcu2)
        then
        IL = sqrt(4jh/(12Rcu1+6Rcu2))
        Now you could replace values jh is know, also Rcu1 and Rcu2, Now you know IL for every posible joule heating value. (the thing will change is Jh)

        and after this you know the I in ecat resistor Iph=IL/sqrt(3). and so on.

        Now I know why you can not really undestand electricity, you need to learn more algebra.!!!

        But thanks for asking!!!!!!!

        • US_Citizen71

          I understand algebra just fine, my question was partly due to tired eyes and partly due to you needing a set of parentheses added. Otherwise you cannot end up with iL=sqrt(jh*57.66513854) was the thought that came into my mind.

          iL=sqrt(Jh*4/(12Rcu1+6Rcu2)) as stated Jh*4 would be done before division by the rest of the equation, standard order of operations. I deal with equations in databases, computers are completely unforgiving if you forget to break up an equation correctly, it tends to affect how I look at equations. It works out using your derived equation or your simplified version of iL=sqrt(jh*57.66513854). I didn’t do the math when I asked the question.

          • ivanc

            I invite you to put it in a computer, …. I will work! I already did it.

          • US_Citizen71

            I just did will post it on Google Docs in a bit.

    • Obvious

      I think the 3 and 6 are just how many lengths of cable of each type there are.
      No special inference.
      Perhaps we can assume they are demonstrating that they are RMS average equal carriers of current.

      • ivanc

        They calculating the joule in each cable, then multiply one by 3 the other by 6, the number of cables producing the total joule heating, and they did not use duty factors or phases etc….

        • Obvious

          Yes, They used as much as an average as possible.
          Finding a good phase modulation rate that works by math alone is like trying to spot a planet crossing a star from a sea of data. I’ll leave that to those that can.
          The RMS, I agree, is good enough as a proxy for time. There is no way to determine on/of, but RMS should catch it all and average it out, over many hours. The meters are good.
          But if the RMS is good, then all three Lines are the same average current, then how can any more get in or out than any one line says, on average?

          • ivanc

            RMS is measuring absolute values, in reality the currents are out of phase this is why you can not add up then directly, but you could work with RMS because is the equivalent to DC

          • Obvious

            Yes.

  • US_Citizen71

    You are assuming constant Vrms voltage I believe since you are ignoring the duty cycle. You can not ignore the duty cycle with a triac in the circuit. Vrms is different for 10% duty cycle vs Vrms for 100% duty cycle. Constant voltage is not an option when the waveform is chopped to control the power.

    • Obvious

      I have examined the extrapolated current issue a bit more.
      The extrapolated currents I have are identical to Ivans.
      The 3.3 difference be mostly reconciled in the active run by dividing the extrapolated current by sqrt(3). This will result in a bit higher input current (possibly fixed with my new current of 17.055 A, but this may also simply fade away since it will apply equally to all values). The new sqrt(3)-corrected values will also include cable Joule heating.
      Importantly, the sqrt(3) denominator for the active run is not a fluke.
      It is a direct consequence of un-monitored V changes.
      We cannot actually extrapolate the real active run current, only the active run power increase that causes the increased Joule heating of the cables. It may look like we can extrapolate for current, but only because we are ignoring the effect of V. For each constant R and “mobile” I value (squared) and a known Power, there is only one acceptable (real) V (also squared).
      P = I^2R and P = V^2/R
      The numerical (nominal) rate of exchange of V for I is 1:1, but one is a quotient (I), and the other (V) is a denominator to both R and P, so the exchange rate is not a “fair” one. They are inversely proportional, but not linearly so. These exchanges are numerically equivalent but are not equal in the real world. This inverse proportionality is the reason there is a fixed value for V for every set of “known” I and P and R values.
      Since the numerical equivalents of I and V are effectively vectors of “opposite polarity” to each other, sqrt(3) is the vector-summed position of the numerical rate of change from V to I or the inverse.

      • Obvious

        To watch the very finicky exchange rate at work, try this web-based calculator.

        http://www.rapidtables.com/calc/electric/Watt_to_Amp_Calculator.htm

        First check the first box to 3 phase.
        Put a power factor of 1 in (this is normal resistive Joule heat: motors, etc. are less).
        Put in 485 W.
        Calculate by clicking the button.
        INFINITY. Whoops no volts. With infinity, you could make any amps you wanted….
        Well, put in a volt value. Best to start low. Try and guess the V to make 19.7A in the result.
        (Maybe you have worked this out already, or at least think you have.) ****Hint: less than 15 V is a good start.**** Don’t go crazy with the Volt values or this will take all day.
        Move the Volts up and down, maybe 10 volts at a time until you get close to the desired 19.7A.
        Note something interesting after a few tries?
        Whoa, now you are really close, using 0.1 volt increments. But it is quite sensitive.
        Just for those that don’t have all night:
        ……..485 W……………Power Factor 1 …………….
        Volts…..Amps Calc………Decimal Volts…….Amp Calc
        1………..280.01………………………….0.2 —…….1400.07
        2.8……..100.000
        10………..28.001………………………..0.1 —…….2800.15
        14………..20.00
        20………..14.00………………………….0.02 …….14000.7
        30………….9.33………………………….0.01……..28000.5
        40………….7.00………………………….0.002….140007.4
        50…………..5.6
        60…………..4.67
        100…………2.8
        .
        Better still, check out one of the power equations below the calculator on the same website, using P/sqrt(3) in part of the equation. The three phase calculation there is a cousin-brother of the one I derived, earlier, but still has V attached as well being a phase equation. It is an inside-out version of mine.
        .

    • ivanc

      Yes, you could ignore duty cycle if using RMS

  • Dr. Mike

    Thomas,
    I agree totally with both statements. As you remember, I thought SQRT(3) factor perhaps didn’t apply to a short duty cycle, but I found where I had made a very fundamental error in my calculation. Frank’s forum here on his “e-catworld” website is giving the Lugano report a much better peer review than many published papers receive. I wish there were a couple of good thermal engineers giving input to the review of the report so that some my questions about the thermal issues that I brought up could be answered by someone with expertise in the field.
    Dr. Mike

  • US_Citizen71

    No change in resistance is required to explain the change in current. I=V/R

    V=1 R=1 then I=1

    If V=3 and R=1 then I=3

    There you go 3 times the current no change in resistance.

    The Vrms of a TRIAC controled system can be calculated fairly close with the following:

    Vrms(switched) = √Duty cycle x Vrms(pure sine)

    Vrms(pure sine) or full wave = 100 volts rms
    Duty cycle = 1% or .01
    √.01 x 100 volts = 10 volts rms

    Vrms(pure sine) or full wave = 100 volts rms
    Duty cycle = 10% or .1
    √.1 x 100 volts = 31.62 volts rms

    Vrms(pure sine) or full wave = 100 volts rms
    Duty cycle = 100% or 1
    √1 x 100 volts = 100 volts rms

    A TrueRMS multimeter like the PCE 830 wil get a more correct value as where the clipping happens matters a little bit . But this should give you an understanding of why the Vrms value is not constant in a TRIAC control power system.

  • US_Citizen71

    I=V/R

    V=1 R=1 then I=1

    If V=3 and R=1 then I=3

    There you go 3 times the current no change in resistance.

    No change in resistance is required to explain the change in current.

    The Vrms of a chopped wave can be calculated fairly close with the following:

    Vrms(switched) = √Duty cycle x Vrms(pure sine)

    Vrms(pure sine) or full wave = 100 volts rms
    Duty cycle = 1% or .01
    √.01 x 100 volts = 10 volts rms

    Vrms(pure sine) or full wave = 100 volts rms
    Duty cycle = 10% or .1
    √.1 x 100 volts = 31.62 volts rms

    Vrms(pure sine) or full wave = 100 volts rms
    Duty cycle = 100% or 1
    √1 x 100 volts = 100 volts rms

    • Dr. Mike

      US_Citizen71,
      Your calculation of getting the current to go up by a factor of 3 when the voltage goes up by a factor of 3 is absolutely correct. However, look at ivanc’s Table a few comments up. Although the current can be calculated to go up by a factor of 2.35 for the first part of the active run and 2.5 for the second part, relative to the dummy runs, the voltage as calculated from the power measurements is lower for the active runs than the dummy run! This is the discrepancy in the reported data that needs to be addressed by the Lugano scientists. Their computer logged data from the PCE-830’s should enable them to figure out why the data is in conflict.
      Dr. Mike

      • US_Citizen71

        To be honest I have no idea what he is attempting to show there. Back in my Navy days we would have called that a shotgun answer. It is such a mess due to formatting I am not exactly sure what headers go with what column or what the headers are intended to mean. Assuming absolute correct values in that is a large stretch, we do not know impedance, inductance, power factor or capacitance values of the circuit so exact values cannot be assured.

        • Dr. Mike

          US_Citizen71,
          Ivanc is trying to put spreadsheet results into a format that can be read in these comments. It was easier for me to understand his spreadsheet since I have made a similar spreadsheet in EXCEL. (He really went to a lot of work to get the results of his spreadsheet posted in these comments). The next to the last column are the calculated heater coil voltages, based of the reported power measurements. You can see the active run voltages at 10-11 volts are lower than the dummy run voltage at about 14V. The 6th column gives what the voltage should have been based on the calculated currents from Cu wire Joule heating and the calculated heater wire resistance of 1.234 ohms. These voltages are 33-35V.
          We do know the circuit. It is essentially a pure resistive load made up of the heater wire in a 3-phase delta configuration so the power factor should be really close to 1.000 with no capacitance and close to no inductance.
          Thanks for your service in the Navy! My brother spent 30+ years in the Navy in emergency medicine.
          Dr. Mike

  • Freethinker

    See Thomas Clarke, there is no need for you to regurgitate you conjectures again. I can only repeat my previous statements on you and your efforts.

  • US_Citizen71

    “IS A STATISTICAL MEASURE.” Of the current waveform / duty cycle. We do not know the duty cycle used for the dummy run we only know what the RMS current was, a little educated guessing and plugging of numbers can get some approximate values for some of the unknowns at that duty cycle. But we do not know impedance, inductance, power factor or capacitance values of the circuit so exact values cannot be assured. When the duty cycle changes the current will change no change in resistance is needed as the voltage will change.

  • US_Citizen71

    First question.
    “…calculating the current and resistor of
    ecat: solving iL from (a) iL=sqrt(Jh*4/(12Rcu1+6Rcu2))”

    This appears to be based off of I=sqrt(P/R)

    but you’ve changed it to I=sqrt(4P/2.3128938681604481344403320996299R) where P is the joule heating and R is the total of C1 and C2 copper cable resistance.

    What is going on here.

    • Obvious

      We could spend all day on the cables if we do not agree on their effective resistance.
      For solving how much the cables “see” power, we can (temporarily) ignore the reactor resistors (call them either superconductors, bolted to a large gold bar, or absent for now), and build a delta entirely from C2 cables, each side with two C2 cables in series. The C1’s can stick out the ends of the new delta and we will have to work on them a bit later, since they depend on the interpretation of the C2 wires’ current flow.
      Does that make sense so far?

      • US_Citizen71

        I’m not trying to be obtuse, I’m simply trying to understand what he put up. I see the he is using the basic equation to determine current from power and resistance but I don’t understand what he is substituted into the equation and why. To get the current in the cables using the report values
        I=sqrt(joule heating or 6.7watts/((3C1 or 3 x .004375ohms) + (6C2 or 6 x .002811ohms)))

        • Obvious

          Sorry, my post wasn’t specifically directed at you (or anyone) in particular.
          It was just a suggestion, in case the subject gets more complicated.

          • US_Citizen71

            no worries ( :

      • Dr. Mike

        Obvious,
        The authors have calculated the Joule heating summed in all Cu wires correctly except the C2 currents are equal to the C1 currents divided by SQRT(3), rather than divided by 2. When they make this correction they will get the Cu wire Joule heating for the dummy run to be 7.3W rather than 6.7W. Their implied assumptions in their Joule heating calculation include:
        1. The 3 Ci wire lengths are equal.
        2. The 6 C2 wire lengths are equal.
        3. All C1 currents are equal.
        4. All C2 currents are equal.
        5. The effects of heating in the Cu wire that is near the connection to the Inconel heater wire is ignored. (The resistivity will be higher in the heated wire.)

        These assumptions are all quite valid for getting a measure of the Cu wire Joule heating within a few tenths of a Watt.
        Dr. Mike

        • Obvious

          Looking on some wire companies websites, I see that a roll of 100′ of 0 gauge AWG wire has about .001 ohm resistance. We are not making a piano here or wiring the Griswalds Christmas lights. These are big wires. They have been overestimated, IMO, but what a are few W in hundreds? One bad crimp and the resistance could triple that. Car taillight bulbs use, what 15, 20 W? And we are trying to get a ratio from that? Could anyone calculate the charge in a battery or Joule heat of a harness in a car by measuring the heat of a tail light?

          • Dr. Mike

            Obvious,
            The Joule heating in the wire is small, and is a small factor in the results. I can’t tell you if the authors correctly measured the wire diameters to get an accurate calculation of the Cu wire resistance (the wire cross sections given in the report do not correspond to common Cu wire gauges). It doesn’t matter if they calculated the Cu wire resistance or the Joule heating accurately. All that matters is that they always calculated Joule heating the same way. The current in the C1 lines was reported to be 19.7A for the dummy run. Although they did not have a column in Table 7 for the C1 current for the 16 active run files , we can calculate those current as 19.7 times the square root of the Cu wire Joule heating power ratios. All errors in calculating the Cu wire resistance and even the fact that the authors miscalculated the C2 currents do not affect the calculated C1 currents. It’s these high C1 currents that don’t agree with the measured TRIAC power levels for the active runs.
            Your point of the wire crimp is quite valid. It would have been good experimental procedure to have a means of measuring the voltage directly across the heater wires to verify there were no bad connections in the set-up. The professors may have done this as part of their set-up after the reactor was reconnected for the active run. If they checked the connections in this manner, they should add one sentence in the revised report stating this.
            Dr. Mike

          • Obvious

            Pleas see my post to Ivan where I demonstrate the actual DC equivalent of the circuit is 17.06 V.
            I agree that one can disassemble the author’s use of math and related assumptions by de-integrating their reported values.

          • Obvious

            Ivan,
            I am definitely on-side with the phase current for the dummy run. (~11.37 A)
            My updated DC equivalent equations require them to work out correctly.
            However, they must be used carefully in a DC equivalent circuit or weird things happen.

          • Obvious

            The resistor value problem is a minefield of wrong turns if not considered carefully.

      • ivanc

        The resistance is given in the report

        • Obvious

          By effective resistance, I mean whether or not the C2 cables see more current on average than do the C1 cables.

  • Obvious

    The measurement of Watts in a pure resistive electrical system, (disregarding line-loss or minor negligible losses), or the measurement of current (Amps) in same, averaged over time, is a measurement of Watt*seconds, aka the coulomb. The coulomb has a physical basis in the charge of electrons (or protons). The coulomb is equal to the charge of approximately 6.241×10^18 electrons (or protons).
    Any and all (real) electrical Watt*second measurements are directly proportional to this physical embodiment of the carrier of work. This is why when knowledge of the W or A is known, and can be described in a correct geometric relation, that all aspects such as V, I, R and W are then derivable quantities, when either only W or I are known.

  • Obvious

    The measurement of Watts in a pure resistive electrical system, (disregarding line-loss or minor negligible losses), or the measurement of current (Amps) in same, averaged over time, is a measurement of Watt*seconds, aka the coulomb. The coulomb has a physical basis in the charge of electrons (or protons). The coulomb is equal to the charge of approximately 6.241×10^18 electrons (or protons).
    Any and all (real) electrical Watt*second measurements or amp*seconds measurements are directly proportional to this physical embodiment of the carrier of work. This is why when knowledge of the W or I is known, and can be described in a correct geometric relation, that all aspects such as V, I, R and W are then derivable with only one correct quantity for each, when either only W or I are known.

    This is the physical basis for the Lugano Theorem which sates that P = IL^2 +((1/2)IL)^2 when IL is the average line current in a perfectly balanced delta resistor, three phase power circuit. The geometric basis is the case where IL= the length of a side of a triangle with a 120° corner that is common with two identical others, that are all inscribed in a perfect circle so that the 120° corners of the triangles are the same as the center of the circle, and the three equal length bases of the 120° triangles (which are connected to the outer edge of the circle) form an equilateral triangle. A line drawn with any arbitrary single angle (with respect to the inscribed triangles) that dissects the circle through the center will separate perfectly the average “positive” or “forward” current side from the average “negative” or “return” side, which must be equal in a balanced electrical circuit.

    Edit: sort of like the Deathly Hallows sign, but with the circle on the outside, and a wye added on the inside.

  • ivanc
    • US_Citizen71

      Much appreciated. So much easier to read.

  • ivanc
    • US_Citizen71

      Much appreciated. So much easier to read.

  • ecatworld

    I would prefer this discussion to stay in this thread.

  • Obvious

    Yes, They used as much as an average as possible.
    Finding a good phase modulation rate that works by math alone is like trying to spot a planet crossing a star from a sea of data. I’ll leave that to those that can.
    The RMS, I agree, is good enough as a proxy for time. There is no way to determine on/of, but RMS should catch it all and average it out, over many hours. The meters are good.
    But if the RMS is good, then all three Lines are the same average current, then how can any more get in or out than any one line says, on average?

  • Thomas Clarke

    The exact numbers in the report provide some insight.

    _First_, note that the sqrt(3) correction for the ratio of C1/C2 RMS currents is exactly true regardless of duty cycle. That is because the current ratio is exactly true at any instantaneous point in time. So Dr. Mike’s corrections below for the Joule heating power are exactly correct. It should be noted that this applies for three-leg and two-leg SCR drive.

    python script showing this from first principles, for those not believing AC theory:

    http://gyazo.com/ac1dcebd9cb46ee7ef06c267300a98db

    result:
    angle=0.000, inner RMS = 1.225, outer RMS = 2.121, ratio = 1.7321
    angle=0.209, inner RMS = 1.225, outer RMS = 2.121, ratio = 1.7321
    angle=0.419, inner RMS = 1.225, outer RMS = 2.121, ratio = 1.7321
    angle=0.628, inner RMS = 1.225, outer RMS = 2.121, ratio = 1.7321
    angle=0.838, inner RMS = 1.225, outer RMS = 2.121, ratio = 1.7321
    angle=1.047, inner RMS = 1.225, outer RMS = 2.121, ratio = 1.7321
    angle=1.257, inner RMS = 1.225, outer RMS = 2.121, ratio = 1.7321
    angle=1.466, inner RMS = 1.225, outer RMS = 2.121, ratio = 1.7321
    angle=1.676, inner RMS = 1.225, outer RMS = 2.121, ratio = 1.7321
    angle=1.885, inner RMS = 1.225, outer RMS = 2.121, ratio = 1.7321
    angle=2.094, inner RMS = 1.225, outer RMS = 2.121, ratio = 1.7321
    angle=2.304, inner RMS = 1.225, outer RMS = 2.121, ratio = 1.7321
    angle=2.513, inner RMS = 1.225, outer RMS = 2.121, ratio = 1.7321
    angle=2.723, inner RMS = 1.225, outer RMS = 2.121, ratio = 1.7321
    angle=2.932, inner RMS = 1.225, outer RMS = 2.121, ratio = 1.7321
    angle=3.142, inner RMS = 1.225, outer RMS = 2.121, ratio = 1.7321
    angle=3.351, inner RMS = 1.225, outer RMS = 2.121, ratio = 1.7321
    angle=3.560, inner RMS = 1.225, outer RMS = 2.121, ratio = 1.7321
    angle=3.770, inner RMS = 1.225, outer RMS = 2.121, ratio = 1.7321
    angle=3.979, inner RMS = 1.225, outer RMS = 2.121, ratio = 1.7321

    _Second_ note that the factor of 3.3. is in fact 3.3329. This does not depend on any approximations or output power measurement – which has a number of possible inaccuracies. Nor does it depend on the Joule heating power (now corrected). It is an internal error with the figures posted in the report and must have some explanation.

    _Third_. Normally the report authors would respond with the explanation but since over 5 weeks this has not happened the best that can be done is to try to reconstruct the error.

    Possible causes

    _A clamp reversal_
    would give X3.0, if heater resistances are exactly balanced. If there is an imbalance (possibly caused by variable resistance joints) then the three line power would be slightly different and clamp reversal could give X3.3 This error would give COP = 1 (approximately).

    _Single-line measurement for active test power_
    Similarly, if the active test power was recorded as one line only, whereas the dummy test power recorded as total, the same could happen. This error would give COP = 1 (approximately).

    _Single-line measurement of dummy current_
    Similarly, if the dummy test power was recorded as Joule power from one line only, but the active test current recorded as total Joule power, the same could happen. This is contrary to the report which shows how the dummy calculation is done. Although it does not show how the active calculation is done it is not easy to see how we get an extra X3.3 on power. this error would however preserve the power in values and COP.

    There may be other error mechanisms I cannot think of?

    • Mark Szl

      Yup if the Ecat is a dude then the sooner we know the better. People here may not want to hear that but it is much, much better to move on sooner than later.

      On the other hand, it could be that Ecat works and cleaning up this inconsistency will make more jump off the fence.

      As it stands, this inconsistency is making more jump back on the fence. The reason i know this is because i am now one of those fence sitters but before i supported Ecat.

      • Dr. Mike

        Mark Szi,
        There really shouldn’t be any question on whether E-Cats work. The data from the first independent report, the “ash” results from the second test, and soon the results of 1MW plant should be sufficient to prove E-Cats work. The only question is the data inconsistency in the Lugano report (and an error in the calculation of the C2 current which will surely be corrected in a revised report). We will have to wait for a revision of the report on the Hot-Cat to see if it worked as good as claimed in the initial Lugano report.
        Dr. Mike

    • Obvious

      1A * 1V = 1W………..R = P/1^2………1W/1^2 = 1
      2A * 1V = 2W………………………………2W/2^2 = 1/2
      1A * 2V = 2W………………………………2W/1^2 = 2/1
      1A * 3V = 3W………………………………3W/1^2 = 3/1
      3A * 1V = 3W………………………………3W/3^2 = 1/3
      10A * 1V = 10 W………………………..10W/1^2 = 10/1
      1A * 10V = 10 W………………………10W/10^2 = 1/10
      10A * 10V = 100W………………….100W/10^2 = 1
      When current is 3 times higher, then R drops by 1/3 for the same Watts.
      This is a non-physical answer.
      To hold R at 1 for this example, requires some kind of trade-off between V and I.
      V and I are not independent. Neither V nor I can be correctly extrapolated in a straight line when dealing with real power.
      V and I must relate to each other. They are inseparable in a power equation.
      Geometry can demonstrate that the translation relation between I and V that keeps R at 1, is the sqrt(3) as the denominator or quotient of the either V or I to obtain the correct answer.
      Dividing the projected active run I by sqrt(3) fixes the current value, when V is multiplied by sqrt(3) in order to maintain R at 1*Resistor (or effective Rtotal).
      (I’m not certain, but dividing P by sqrt(3) may give a straight 1:1 V or I correlation. Note that 100/sqrt(3) = 57.735 which is almost exactly the Jh factor Ivan has.)

      • Thomas Clarke

        I don’t understand your point here.

        Normally R does not change, V & I do change and V = IR.

        This relationship leads to P = I^2*R or P = V^2/R

        In this system we have P and (via Joule heating current) I. So indeed the “tradeoff” between V & I or equivanetly P and I, since P = VI, is what we use.

        The equation P = I^2*R is always true (if we have RMS I) and that is why for the given figures to be correct the heater resistance must decrease by 3.3 (sorry, in fact as Mikepoints out, 3.45) in the active test.

        The sqrt(3) is nothing to do with this and comes from the way that 3 phase currents and volatges relate to each other in a delta configuration.

        • Obvious

          What I am trying to demonstrate is that if current only is extrapolated from a data point, the invisible function of V cannot be determined. If V is not determined, and only I is considered in a multiplication of a data point, then V can reach infinity, or an invisible sqrt(3) multiple of I, if V is pretended to be 1 at the start, so it won’t be a problem in the math. Otherwise, by holding the original data point V at 1, the other visible “variable”, R, gets divided by (sqrt(3))^2, which is also untenable.
          V, even though unknown, must be incorporated in the original data as a “1V” placeholder. Then as complex mathematical inter-relations are performed, the status of V, (or a new factor of it, based on the results of mathematical manipulations) must be assessed. If this is not done, then R is affected by the invisible influence of the un-tracked V alterations.

          Edit: maybe it is actually I/(2(sqrt(3)), not I/(sqrt(3)^2). That makes more sense considering what I wrote (above) on the fly.
          That equals a denominator of 3.46, which is very close to 3.3, or better yet 3.45 mentioned by Thomas, above. This demonstrates great overall accuracy of the values reported by the professors. (And that the partial formulae answers, (although missing the V component), extrapolated from the data are on the right path for a two-dimensional solution. Add in the V modifier to rebalance the extrapolations to be consistent with the full power equations, which adds the missing power dimension, and case closed.)
          The power equation is a 3-D equation. It consists of dimensions I, R, and V. All of them are important. The power equations cannot function properly in only two dimensions, where one of the dimensions is allowed to “float”. Either infinity, zero, or an incorrect or incomplete 2-D answer will be the result of ignoring one of the dimensions. Time is also important, but can be corralled to a single location relative to the other dimensions by RMS averaging of all of the other values.

          Scotty to First Spock, after receiving the Warp Speed Transporter equations he developed in his alternate future: “Imagine that! It never occurred to me to think of SPACE as the thing that was moving!”

      • Obvious

        With a protractor and ruler, draw a line 10 units long (this can be V).
        Now using the protractor, draw a line starting at the end of the first line at a 120° angle that is 10 units long (this can be I).
        Draw a line that connects the two open ends.
        Any arbitrary angle that is created by a line that starts at the 120° corner to any point on the last line drawn will separate all possible reciprocal fractions of I/V and/or V/I.
        A line drawn from the 120° corner that goes to the exact center of the last line is the point where V/I and I/V are 1/1.
        This 120° equal bisection line (splitting the 120° angle into two equal 60° angles)creates an equilateral triangle with the 90° angle where the bisection line meets the last line drawn.
        The length of this bisection line can be measured, and will be 5 units long.
        The length of the last line drawn, that connects the two 10 unit long ends will be 17.32 units long. This line is sqrt(3) times the length of the 10 unit long lines.
        The 1/2 way point along the 17.32 unit long line will be 8.66 units long. Dividing 8.66 units by sqrt(3) equals 5 units.
        From this can be derived the translation factors and quotients required to solve for I and V when R = 1.
        Sqrt(3) times either I or V is the numeric vector sum of the two numerically equal but opposed lines, one of V length and the other of I length, but from the reference point of either one or the other, which are 120° apart. 1/2 sqrt(3) of this longer vector length is the point where R = 1, since V/I here equals one.
        Therefore ((1/2)sqrt(3))((I^2)/(V^2))*((V^2)/ I^2)) is the formula for determining the rate of increase that maintains R =1.
        This simplifies to 1/2(sqrt(3) or 0.866.
        So if power doubles, then the I and V must each increase at a rate of V/(1/2sqrt(3) and I/1/2sqrt(3) if R is to =1.
        If apparent current increases by double, but V is unknown, then the apparent increase is by 2(1/2sqrt(3)) or sqrt(3), but R is affected by R/(sqrt(3)^2) or R/3.
        Dividing the apparent current by sqrt(3) fixes the relationship of P to R and I to V, but leaves V=1, (or infinity) which is not physically correct.
        The correct usage is increased apparent current /1/2(sqrt(3), but then V (which was un-reported in the Lugano report) becomes incorporated in the other part of the relationship equations, as a permanent figure in the power equations, and is nearly impossible to cancel back out somewhere else. The only place V=1 is in the original source for the data from which correlations are extracted.
        Incorporating a V placeholder in a parallel set of equations for the Joule heat extrapolations, that is a back-check on V to P to R relations’ values makes this very apparent.

        • Dr. Mike

          Obvious,
          I’m with Thomas, I don’t understand the point you are making. Your drawing of I at a120 degree angle to V does not make sense since there is no phase difference between I and V in a resistive load. At any give instant of time the voltages across each of the three heater coils are 120 deg out of phase, but the currents in each heater wire are in-phase with the voltage.
          Dr. Mike

          • Obvious

            The case where the 120° angle is made in my discussion above is totally independent of the 3 phase power. It is a precise mathematical relation hidden in the various iterations of the Power formulae. (Although it is not immediately obvious.)
            It is forced by the only geometry that allows both V and I grow in size to allow a power increase while R is maintained steady in a real circuit, with limits imposed on the geometry by the rate of increase in Power and a constant R. Any translation of the value I requires the appropriate reciprocal translation to the value for V, or the result is incorrect in a real system. The balance point is R.
            V cannot be removed from the power equation arbitrarily.
            I^2R = V^2/R
            R/R = V^2/I^2

          • Dr. Mike

            Obvious,
            Your last equation

            R/R = V^2/I^2

            should be R*R = R^2=V^2/I^2 or R = V/I. I believe you have made a fundamental math error.

            Dr. Mike

          • Obvious

            Yes, I would say I did. I was typing faster than thinking.
            R = V/I of course.

  • ivanc

    To Frank Aukland,
    Dear Frank, you see for the number of post in this publication that this is the subject people are interested in.
    I want to ask you in search of the Truth.
    That you contact a electric engineer and ask him to peer review my calculation.
    I firmly believe people will appreciator to know the truth even if is not what they were expecting.
    For the post you see a so dissimilar theories that shows that some people does not have the understanding of electricity to evaluate this circuit.
    Thanks

    • Mark Szl

      Ivanc, you are not an EE?

      • ivanc

        I am an EE.

      • Thomas Clarke

        ivanc is doing what any wise person does, and wanting his stuff to be checked. No-one is so perfect that they never make mistakes!

    • Thomas Clarke

      ivan,

      As far as I know there is no difference between you, me, Dr. Mike over this? If you think there is I will go through your calculations carefully and critique.

      Looking quickly at them, they seem fine. One change gthat would be useful. If you gave the ratio figure to 3 decimal places we could work out what figures the exact value correlates with – it varies a little between one sample and the next and seeing this might be interesting?

      • ivanc

        Your approach was different you calculated ratios. I solved the circuit and calculated all the unknowns, Please search for errors and critique.

    • Dr. Mike

      ivanc,
      Your data looks good with one possible error. The “Consumption Power” listed in Table 7 actually appears to be the power out of the TRIAC power supply. You can see from Equation 26 on page 21, the Cu wire Joule heating is subtracted from what they call the “Consumption Power” to get the net power going to the reactor in their COP calculation. Therefore, to calculate the new variable resistance of the heater wire you need to subtract the Cu wire Joule heating numbers from the “consumption power” to get the actual power going to the reactor. Your voltage calculation actually represents the voltage at the TRIAC output. If you want to calculate the voltage across the reactor, you need to calculate it from the net power going to the reactor. The voltage across the reactor less less than the supplied voltage due to the IR drop in the Cu wires.
      Dr. Mike

      • ivanc

        Dr. Mike, You could be right on that (input power=consumption – joule heating), have you worked out how they calculated the cop?, all my combinations does not give 3.13 for item 1. but If could workout exactly how they find the cop then we know what they used for input power. I did the change and the result was identical, to see a difference you need more decimal places.
        But still is good to clarify this for completeness.

      • Thomas Clarke

        OK – I agree with this, but it is a small correction.

        We have [Ptot/Pjoule]active /[Ptot/Pjoule]dummy = Rtotactive/Rtotdummy.

        So the heater resistance change is not quite X3.3329. Instead it is X3.45

        • Dr. Mike

          Thomas,
          3.45 is what I calculated.
          Dr. Mike

  • Mark Szl

    Ivanc, you are not an EE?

    • ivanc

      I am an EE.

    • Thomas Clarke

      ivanc is doing what any wise person does, and wanting his stuff to be checked. No-one is so perfect that they never make mistakes!

  • Mark Szl

    Yup if the Ecat is a dude then the sooner we know the better. People here may not want to hear that but it is much, much better to move on sooner than later.

    On the other hand, it could be that Ecat works and cleaning up this inconsistency will make more jump off the fence.

    As it stands, this inconsistency is making more jump back on the fence. The reason i know this is because i am now one of those fence sitters but before i supported Ecat.

    • Dr. Mike

      Mark Szi,
      There really shouldn’t be any question on whether E-Cats work. The data from the first independent report, the “ash” results from the second test, and soon the results of 1MW plant should be sufficient to prove E-Cats work. The only question is the data inconsistency in the Lugano report (and an error in the calculation of the C2 current which will surely be corrected in a revised report). We will have to wait for a revision of the report on the Hot-Cat to see if it worked as good as claimed in the initial Lugano report.
      Dr. Mike

  • Thomas Clarke

    ivan,

    As far as I know there is no difference between you, me, Dr. Mike over this? If you think there is I will go through your calculations carefully and critique.

  • Thomas Clarke

    I’ve had quite a number of my comments on this thread just disappear into thin air. They don’t all do this, and I’m at a loss to know when this will happen. For example my reply to ivanc below, where I say I don’t see any difference in analysis of the figures between him, me, and Dr. Mike, which has disappeared.

  • kasom

    Ask the testers or Officine Ghidoni SA. for readings/snapshots of the grids powermeter at the lab before and after the e-cat test.
    If the total power consumption fits the report, scepts will say the e-cat fooled the grid.

  • Thomas Clarke

    Note to moderators.

    OK, goodbye all.

    It seems that there is a policy here to delete comments I make which are substantive and relate to an area (3 phase EE) where while not a great expert I have Uni level expertise. For example my post showing a python script that illustrates how the sqrt(3) factor applies instantaneously, and hence will be precise for any duty cycle. Personally, I find that interesting, and not obvious.

    I put some time into such posts, working out figures is not always quick. Cross-checking is tedious.

    I realise this is a site mostly for people with settled opinions. That is fair enough, the site can be what it wants. However the issue in this report discussed on this thread should be the same for everyone – it is anomalous, in a way that makes the COP results worthless – and that anomaly should if possible be cleared up by the testers.

    The dryest skeptic and the most fervent believer here should have the same motivation – a wish to see the report strengthened by explaining an obvious anomaly.

    There is been a lot of comment here and much of it is irrelevant to the main point. I only hope that Dr. Mike and Ivanc and DickeFix – who it seems do not so often have posts removed as me – can ensure that the testers understand the issue and respond accordingly.

    Best wishes, Tom

    • Mark Szl

      Isn’t your post with the link to the script below? It was up very early this morning.
      If others don’t make it the you could post on Matt Lewin’s blog as you have done and we can go read them over there.

      • Thomas Clarke

        Yes. It was up. Then it vanished for a long time. Now its back!

    • Frank Acland

      Sometimes I am away from the Internet for periods of time, and am not able to approve posts as fast as I would like. I have just approved many posts that were awaiting approval.

    • DickeFix

      Tom, all your comments have high quality and are often what I would have written myself if I had more time and my english was better. Hence, please stay and continue to defend the science and the objective, unbiased search for truth regarding the E-Cat. I also want to say the same to Giancarlo, Andrea S., Dr. Mike and Ivanc who all have shown a solid background in electrical engineering. It is good that several competent people are active in this forum since power calculations on switched three phase circuits are not trivial, even for an assistant professor in EE like myself.

      Regarding the temporary disappearance of some of your comments, I think it may be due to that you have edited them afterwards. If I understand correctly, every time one edits a comment, Frank Acland has to approve it again. I want to take the opportunity to thank Frank for his good work maintaining this site and for updating it so frequently with fresh news in the LENR field. I have not myself observed any biased censorship from his side. I think that also a confessed believer as Frank values an open debate in search for the truth. Even if it turns out in the end that the E-Cat was just an illusion; his work has not been in vain since the site has been a valuable meeting point for the discussion around both the E-Cat and LENR in general.

  • Dr. Mike

    ivanc,
    Your data looks good with one possible error. The “Consumption Power” listed in Table 7 actually appears to be the power out of the TRIAC power supply. You can see from Equation 26 on page 21, the Cu wire Joule heating is subtracted from what they call the “Consumption Power” to get the net power going to the reactor in their COP calculation. Therefore, to calculate the new variable resistance of the heater wire you need to subtract the Cu wire Joule heating numbers from the “consumption power” to get the actual power going to the reactor. Your voltage calculation actually represents the voltage at the TRIAC output. If you want to calculate the voltage across the reactor, you need to calculate it from the net power going to the reactor. The voltage across the reactor less less than the supplied voltage due to the IR drop in the Cu wires.
    Dr. Mike

  • Obvious

    The RMS measurement is on the C1, not C2 cables.
    If the professors slid the amp probe down the C1 wire, and measured the both C2 cables together, side by side, as they exit the C1, then this measurement would also read 19.7 A for the dummy run.
    This indicates two things:
    1. That all the currents coming from all cables towards or away from any one C1 cable average out in a way that the probe considers to be the RMS equivalent.
    2. That this equivalence continues to the C2 cables, as long as they act as a pair in the same orientation. Individually, the C2 cables will act differently. The current in one C2 cable is phase current. Some currents in one C2 cable are out of phase with currents in it’s “twin”, which reach a balance in their common C1 cable.
    3. Folding any one these cables back on itself, and measuring the current of the folded-over wire (both wires at once) as though it is one double wire, (whether C2 or C1) will result in a zero reading. The RMS of one half of the folded wire is exactly the same RMS in the other before folding. Together, the average instant current will be zero, so the RMS is zero of a folded-on-itself wire. This is the basis of the reason that the two C2 cables in (1.) above will read the same as the C1 cable.

    • ivanc

      c2=c1/qsrt(3)

      • Obvious

        sqrt(3) is a vector.
        Actually, the 19.7 A is also a vector, since it describes only the activity of one of three cables (or sets of cables) in an equilateral triangle of cables.
        19.7 A is by itself is a magnitude without sign. This is easily proven.
        3 x 19.7 A is not the correct total current in the system.
        We must assign a sign to 19.7 A to make a DC circuit equivalent, if we adding the contributions from other cables.
        In a balanced triangle with 19.7 A going “in” each line, then the sum is actually zero.
        This is easily shown, since 19.7^2 times 1.23 is 477 (W).
        The sum of 477 + 477 +477 is also not a correct answer.
        Therefore, we choose a sign for one of the 19.7 A (say +19.7) to make a DC-type value. Everything else must be given the sign which is relative to this + value. A point of reference must be chosen, or the resulting math is meaningless.
        They cannot all be +19.7, either. The total sum of all the +currents must be equal to the sum of all the -currents.
        Same goes for power. The sum of all the +currents, squared, times resistance equals the total measured +power. The total measured power from doing the sum of -currents squared times resistance is equal to -total power. The sum of total -power plus total +power is zero.
        The magnitudes of the 19.7 A, or 11.38 A (if you must) can be used to calculate the power in one third of the delta.
        But you can’t just add them up any old way. If the magnitudes are without sign, then the sum answer has no sign. The correct way for adding all magnitudes together is to add the RMS values, and then divide by the number of RMS magnitudes used in the summing, for a total RMS average. (Another RMS-ing of the RMS values would be best, but since the values are same for each side of the delta, this is pointless).
        Having now returned to the original RMS value, this value is still only the one that is from this reference point.
        Since the triangle has three sides, this RMS value is actually incorrect for the delta as a whole. It only represents 1/3 of the triangle, and we need 1/2 of the triangle forward current to actually describe the DC equivalent properly.
        But most of this is a waste of time.
        We already know that the RMS current in one set of cables in one corner if the delta sees, on average, 19.7A. So the other three must also, on average. One phase in, two phases back out for each cable.
        The RMS for each C1 is then the return current average, not the input current average. If this was two phase, then one in for one out would occur, and the math would simplify immensely. The same RMS would be the result on either end of the circuit (and we would not add then up).
        Since the real RMS measured at each C1 cable actually is the power returning from the other two phases, then this current is higher than the actual “input” power on this one line.
        The real input power on each on C1 line is 1/2 of the summed vector RMS average of the other two C1 cables. This looks like it would be exactly the same number as 19.7 we started with, BUT these cables “add power” for a total of 240°, and the only allowable DC equivalent is 180°. Here come those nasty vectors again.
        The inside angle of the overlarge 240° angle is 120°. This makes a special triangle, the Isosceles, just like my other triangle forays.
        The vector sum of the two opposing cables is 2(sqrt(3)*19.7A = 22.74 A. This is higher than the 19.7 A because this is the 180° equivalent of the return currents from the opposite side of the circuit from one side.
        Now, consider the first 19.7 A in the cable we began with. It is too high, since it is only 120° of 180°. It is also facing the wrong way compared to the others. We must divide the 120° value into a wider 180° equivalent. It is 11.37 A.
        This is unbalanced, so the route to fix it to make the real DC equivalent amperage is to add the 22.74 to the 11.37 and average them.
        This gives a real, correct, proper, DC equivalent forward amperage of 17.06 A.
        My geometry forays have already confirmed this value.

  • Obvious

    Yes.

  • LuFong

    Surely the testers must realize that trying to demonstrate an anomalous heat gain by relying on an anomalous, unidentified, unverified Inconel resistance performance is just not credible.

    Given that the testers have not yet responded it’s beginning to appear to me as a) they are not going to (for any number of reasons) b) they are working very hard to come up with an explanation and make the changes to their report. I suspect now that it’s the former rather than the latter but I hope I am wrong and that they are just being very careful.

    • Dr. Mike

      LuFong,
      I also hope they are being careful. They may have needed to bring in some outside expertise to review the saved electrical data. Also, they may already have an answer to the high currents/ low power issue in the active test, but may be trying to answer one of the other questions brought up by the many people that have commented on the report. I don’t think we will see an answer until they have a revised report that addresses a majority of the issues brought up by reviewers.
      Dr. Mike

  • LuFong

    Sorry to see you go. I don’t think your posts are being deleted intentionally by the moderator unless they contain personal attacks (which I know isn’t true). Disqus sometimes rejects posts for no apparent reason. I hope they reappear and that you continue posting your conclusions and thoughts at least elsewhere.

    • Obvious

      I have what looked like my comments disappearing a few times. They show up later, or sometimes a full refresh of the page is required to make them show up if several other pages are also open at the same time (especially if there is Disqus comment section on the bottom of any of the other pages).

  • Thomas Clarke

    @LuFong,

    My posts seem to have reappeared, which is good. I am happy to continue posting, as long as the chance of substantive posts being permanently deleted is low. The site is weird the way it removes and then resurrects posts!

    But I’m not sure there is much more to say now except to hope that the testers respond!

    Best wishes, Tom

    • Dr. Mike

      Tom,
      Glad you are staying with us. I know you are putting a lot of effort into seeing the Lugano report will eventually be revised to clarify all or most of the technical issues that have been brought up by all of us “peer reviewers”. I know it takes a long time for Frank to
      review posts, but he is trying his best to see that personal attacks
      stay off his website (guess he missed some of Freethinker’s replies to
      you). I spent more than a day working on one of my guest posts and was
      very disappointed that it did not get immediately approved. It turns
      out Frank just had just overlooked it. We need people like you that
      carefully reviews data and can respond with a technical answer. I think
      it’s great that many non-technical people are also following LENR
      development on this website, but it is people like you that are making a
      real positive difference in seeing that other scientists can review the
      data from the E-Cat tests and believe what they read. I certainly
      would like to get the second report up to a quality level of the first
      report.
      I did get to see your post on how the sqrt(3) factor
      applies instantaneously, I thought it was very good of you to provide
      this data for those few that still might believe the authors did not make a
      mistake by not using the sqrt(3) factor in their C2 current
      calculations.
      Dr. Mike

      • Mark E Kitiman

        If Thomas Clarke disappears from here, you can find his posts on plenty of other sites – an excerpt from his latest post:

        Rossi I think finds 3 phase gives him a lot more ways to get apparent
        high COP. He needs these now that presumably the obviously flakey
        calorimetry of the earlier tests no longer cuts it.

        It must be dispiriting for him to find COP going down over time.

        https://matslew.wordpress.com/2014/10/09/interview-on-radio-show-free-energy-quest-tonight/#comments

        The above post disregards the statements by the authors of the Lugano report when they point out:

        We also chose not to induce the ON/OFF power input mode used in the March 2013 test, despite the fact that we had been informed that the reactor was capable of operating under such conditions for as long a time as necessary

        and:

        It must be remarked that the COP values quoted here refer only to the performance of the reactor running at the capacity selected by us, not at its maximum potential, any evaluation of which lies beyond the purposes
        for which this test was designed. Awareness of the fact that the test would have lasted a considerable length of time prompted us to keep the reactor running at a level of operation capable of warranting both the stability and the safety of the test. Therefore, we do not know what the limits of the current technology are,
        in terms of performance and life span of the charges.

        The above conditions were set by the Professors… not by Rossi.
        The latest patent shows a COP in excess of 11.

        • Dr. Mike

          Mark,
          Thomas is entitled to his opinion on the calorimetry. I certainly am not an expert on calorimetry, but I thought the basic methodology used in the Lugano report was at least good enough to give a reasonable estimation of the total output power. Of course, If the reactor had been run at 1100C-1300C in the dummy run, it would have been hard for anyone to question the calorimetry.
          I believe Rossi can get a COP of 11 in his earlier version of the E-cat. I also believe he will achieve much higher COP’s in the future when he and everyone else better understands the physics of LENR. However, I’ll have to wait to see the revised Lugano report to see if they really did achieve a COP of >3 in the Hot-Cat. I consider the Lugano test a success just from the results of the “ash” analysis. I think these results will add more to knowledge of LENR than any COP number because any time a theory is proposed to explain LENR, that theory must be able to explain the Lugano “ash” results.
          Dr. Mike

  • ecatworld

    Sometimes I am away from the Internet for periods of time, and am not able to approve posts as fast as I would like. I have just approved many posts that were awaiting approval.

  • Obvious

    1A * 1V = 1W………..R = P/1^2………1W/1^2 = 1
    2A * 1V = 2W………………………………2W/2^2 = 1/2
    1A * 2V = 2W………………………………2W/1^2 = 2/1
    1A * 3V = 3W………………………………3W/1^2 = 3/1
    3A * 1V = 3W………………………………3W/3^2 = 1/3
    10A * 1V = 10 W………………………..10W/1^2 = 10/1
    1A * 10V = 10 W………………………10W/10^2 = 1/10
    10A * 10V = 100W………………….100W/10^2 = 1
    When current is 3 times higher, then R drops by 1/3 for the same Watts.
    This is a non-physical answer.
    To hold R at 1 for this example, requires some kind of trade-off between V and I.
    V and I are not independent. Neither V nor I can be correctly extrapolated in a straight line.
    V and I must relate to each other. They are inseparable in a power equation.

    • Thomas Clarke

      I don’t understand your point here.

      Normally R does not change, V & I do change and V = IR.

      This relationship leads to P = I^2*R or P = V^2/R

      In this system we have P and (via Joule heating current) I. So indeed the “tradeoff” between V & I or equivanetly P and I, since P = VI, is what we use.

      The equation P = I^2*R is always true (if we have RMS I) and that is why for the given figures to be correct the heater resistance must decrease by 3.3 (sorry, in fact as Mikepoints out, 3.45) in the active test.

      The sqrt(3) is nothing to do with this and comes from the way that 3 phase currents and volatges relate to each other in a delta configuration.

      • Obvious

        What I am trying to demonstrate is that if current only is extrapolated from a data point, the invisible function of V cannot be determined. If V is not determined, and only I is considered in a multiplication of a data point, then V can reach infinity, or an invisible sqrt(3) multiple of I, if V is pretended to be 1 at the start, so it won’t be a problem in the math. Otherwise, by holding the original data point V at 1, the other visible “variable”, R, gets divided by sqrt(3), which is also untenable.
        V, even though unknown, must be incorporated in the original data as a “1V” placeholder. Then as complex mathematical inter-relations are performed, the status of V, (or a new factor of it, based on the results of mathematical manipulations) must be assessed. If this is not done, then R is affected by the invisible influence of the un-tracked V alterations.

    • Obvious

      With a protractor and ruler, draw a line 10 units long (this can be V).
      Now using the protractor, draw a line starting at the end of the first line at a 120° angle that is 10 units long (this can be I).
      Draw a line that connects the two open ends.
      Any angle that is created by a line that starts at the 120° corner to any point on the last line drawn will separate all possible fractions of I/V and/or V/I.
      A line drawn from the 120° corner that goes to the exact center of the last line is the point where V/I and I/V are 1/1.
      This 120° equal bisection line (splitting the 120° angle into two equal 60° angles)creates an equilateral triangle with the 90° angle where the bisection line meets the last line drawn.
      The length of this bisection line can be measured, and will be 5 units long.
      The length of the last line drawn, that connects the two 10 unit long ends will be 17.32 units long. This line is sqrt(3) times the length of the 10 unit long lines.
      The 1/2 way point along the 17.32 unit long line will be 8.66 units long. Dividing 8.66 units by sqrt(3) equals 5 units.
      From this can be derived the translation factors and quotients required to solve for I and V when R = 1.
      Sqrt(3) times either I or V is the numeric vector sum of the two numerically equal but opposed lines, one of V length and the other of I length, but from the reference point of either one or the other, which are 120° apart. 1/2 sqrt(3) of this longer vector length is the point where R = 1, since V/I here equals one.
      Therefore ((1/2)sqrt(3))((I^2)/(V^2))*((V^2)/ I^2)) is the formula for determining the rate of increase that maintains R =1.
      This simplifies to 1/2(sqrt(3) or 0.866.
      So if power doubles, then the I and V must increase at a rate of V/(1/2sqrt(3) and I/1/2sqrt(3).
      If apparent current increases by double, but V is unknown, then the apparent increase is by 2(1/2sqrt(3)) or sqrt(3), but R is affected by R/(sqrt(3)^2) or R/3.
      Dividing the apparent current by sqrt(3) fixes the relationship of P to R and I to V, but leaves V=1, which is not physically correct.
      The correct usage is apparent current /1/2(sqrt(3), but then V (which was un-reported in the Lugano report) becomes incorporated in the equation as a permanent figure in the power equations, and is nearly impossible to cancel back out somewhere else.
      Incorporating a V placeholder in a parallel set of equations for the Joule heat extrapolations, that is a back-check on V to P relations, values makes this very apparent.

      • Dr. Mike

        Obvious,
        I’m with Thomas, I don’t understand the point you are making. Your drawing of I at a120 degree angle to V does not make sense since there is no phase difference between I and V in a resistive load. At any give instant of time the voltages across each of the three heater coils are 120 deg out of phase, but the currents in each heater wire are in-phase with the voltage.
        Dr. Mike

        • Obvious

          The case where the 120° angle is made in my discussion above is totally independent of the 3 phase power. It is a precise mathematical relation hidden in the various iterations of the Power formulae. (Although it is not immediately obvious.)
          It is forced by the only geometry that allows both V and I grow in size to allow a power increase while R is maintained steady in a real circuit, with limits imposed on the geometry by the rate of increase in Power and a constant R. Any translation of the value I requires the appropriate reciprocal translation to the value for V, or the result is incorrect in a real system. The balance point is R.
          V cannot be removed from the power equation arbitrarily.
          I^2R = V^2/R
          R/R = V^2/I^2

          • Dr. Mike

            Obvious,
            Your last equation

            R/R = V^2/I^2

            should be R*R = R^2=V^2/I^2 or R = V/I. I believe you have made a fundamental math error.

            Dr. Mike

          • Obvious

            Yes, I would say I did. I was typing faster than thinking.
            R = V/I of course.

        • LuFong

          If the testers chose to run SSM, say at the end of the test, and for say 4-8 hours, and achieved a COP significantly higher than 3 we would not be having any discussion about power calculations.

      • US_Citizen71

        Partly due to tired eyes and partly due to you needing a set of parentheses added. Otherwise you cannot end up with iL=sqrt(jh*57.66513854)

        iL=sqrt(Jh*4/(12Rcu1+6Rcu2)) as stated jh*4 would be done before division by the rest of the equation, standard order of operations.

  • Dr. Mike

    Tom,
    Glad you are staying with us. I know you are putting a lot of effort into seeing the Lugano report will eventually be revised to clarify all or most of the technical issues that have been brought up by all of us “peer reviewers”. I know it takes a long time for Frank to
    review posts, but he is trying his best to see that personal attacks
    stay off his website (guess he missed some of Freethinker’s replies to
    you). I spent more than a day working on one of my guest posts and was
    very disappointed that it did not get immediately approved. It turns
    out Frank just had just overlooked it. We need people like you that
    carefully reviews data and can respond with a technical answer. I think
    it’s great that many non-technical people are also following LENR
    development on this website, but it is people like you that are making a
    real positive difference in seeing that other scientists can review the
    data from the E-Cat tests and believe what they read. I certainly
    would like to get the second report up to a quality level of the first
    report.
    I did get to see your post on how the sqrt(3) factor
    applies instantaneously, I thought it was very good of you to provide
    this data for those few that still might believe the authors did not make a
    mistake by not using the sqrt(3) factor in their C2 current
    calculations.
    Dr. Mike

  • Dr. Mike

    Thomas,
    3.45 is what I calculated.
    Dr. Mike

  • ivanc

    https://docs.google.com/spreadsheets/d/1tuKwWUxC2Gq_MtEED4_XYYIqehpOv47mGVx9yio-GKs/pubhtml
    I have updated the table to show input power = consumed power – joule heating.
    The factor is now 3.4 to 3.5 instead to 3.3
    Thanks Dr Mike.!

    • Obvious

      2(sqrt(3)) = 3.464

  • ivanc

    I have updated the table to show input power = consumed power – joule heating

    now the factor is about 3.4

    https://docs.google.com/spread

    Thanks to DR Mike for the feedback.

  • ivanc

    I have updated the table to show input power = consumed power – joule heating

    now the factor is about 3.4

    https://docs.google.com/spreadsheets/d/1tuKwWUxC2Gq_MtEED4_XYYIqehpOv47mGVx9yio-GKs/pubhtml

    Thanks to DR Mike for the feedback.

  • Obvious

    2(sqrt(3) = 3.464

  • Dr. Mike

    Mark,
    Thomas is entitled to his opinion on the calorimetry. I certainly am not an expert on calorimetry, but I thought the basic methodology used in the Lugano report was at least good enough to give a reasonable estimation of the total output power. Of course, If the reactor had been run at 1100C-1300C in the dummy run, it would have been hard for anyone to question the calorimetry.
    I believe Rossi can get a COP of 11 in his earlier version of the E-cat. I also believe he will achieve much higher COP’s in the future when he and everyone else better understands the physics of LENR. However, I’ll have to wait to see the revised Lugano report to see if they really did achieve a COP of >3 in the Hot-Cat. I consider the Lugano test a success just from the results of the “ash” analysis. I think these results will add more to knowledge of LENR than any COP number because any time a theory is proposed to explain LENR, that theory must be able to explain the Lugano “ash” results.
    Dr. Mike

  • US_Citizen71

    I just did will post it on Google Docs in a bit.

  • US_Citizen71

    Since ivanc doesn’t show much of his work even on a spread sheet I decided to do one of my own to verify the math. Doing so I discovered what I believe to be a large basic error in his calculations that accounts for the factor of 3+. To be sure I put my sheet in front of a power systems engineer I know and he verified my equations. He was concerned with the wire in the test being too small a gauge resulting in unnecessary line loss and nothing else. The error is caused by neglecting to divide the power-in from the supply listed in table 7 of the report by 3. This is the correct thing to do as the values being calculated are from one third of the circuit and in a balanced load one third of the power is dissipated by each piece. All my work is shown, clicking on a cell will show the equation used. My Dummy Run calculations match his and that is where the similarity ends.

    https://docs.google.com/spreadsheets/d/1b_bXWnPXdsvUQT0ng1F2vATFwfQK_mXQw4nw4y4Safo/edit#gid=890186181

    • Obvious

      I ran P = I^2/R against the numbers to solve back for R, and they all come back 3.700 R (3x Rph). If the phases are considered to be in series maybe this is ok….
      I’m done with this for tonight.
      Cheers

      • US_Citizen71

        Yes because the sheet solves for for one third of the power. You did a manual check of the spreadsheet, thanks!

        • Obvious

          I have downloaded almost all the spreadsheets presented in this topic.
          I have my own also, which is now getting quite out of hand.

        • Obvious

          I am certain that the individual reactor resistances are 1.84 ohms each. The inductance of the delta as a whole is 1.234 ohms. 1.234 ohms is the resistance that would be measured by an ohmmeter across any two terminals of the delta.

    • ivanc

      Looks that my posts are delayed, I hope not banned.
      your calculation and mine are equal for the dummy run.
      This show that there is some error in the data for the active run.
      The current you have calculated is to low. they said about 50 amps has been used.
      your data is also showing the anomaly the data should match with the joule heating data.
      you starting from the input power, but this is the parameter we trying to test.
      better start calculating IL, the equation we spoke and derived step by step gives IL , from there you could calculate Ip, then the rest is easy.
      There is no error in my calculations, the error is in the data, and the match on the dummy data shows your calculation and mine are correct, but there is incoherent data.

      • US_Citizen71

        Read my reply to your earlier post.

    • Dr. Mike

      US_Citizen71,
      Your calculations look good. Please add one more column to your Table. Calculate the Cu wire Joule heating using your calculated currents and the resistances and formulas shown in the report. Perhaps, the Lugano scientists will get back to us and say they just miscalculated the active run Cu wire Joule heating, and they were in error when they said on page 3 “current through the resistor coils, normally 40-50 Amps”. Remember they really didn’t measure any resistor coil currents or they would have caught the 1/SQRT(3) factor in the C2 lines for the Cu wire Joule heating calculation. The 40-50 Amp reference has to mean something they measured, that is, the C1 line currents. Ci line currents of 40-50A are in agreement with the Cu wire Joule heating numbers in Table 7.
      Dr. Mike

      • US_Citizen71

        I will add your requests to my to do list for the sheet. I likely won’t get to it until this evening.

        • US_Citizen71

          I agree that there is a problem with the joule heating data and according to the power systems engineer I consulted we really shouldn’t even have that data to play with since the power is less than one kilowatt. They appear to have used 6 gauge (awg) and that gauge caused a drop of 1.5% of the power over the course of about 5 meters total from my calculations. Relating it to home use the same rate of drop over the wires leading from the pole to the average outlet in a home would be a 30% drop or better. They should have gone with 4 or 2 gauge wires to make the drop completely insignificant. I have yet found a reason to not believe the total power number being close other than the difference in the joule heating subtracted, since there was two True-RMS power meters one at the wall and one after the control box. The report claims that the two were insync and the difference between the two were within 360W at all times. The 360W is the drop of the control box so at most the real life COP measure would be heat observed minus the power measured coming from the wall divided by the power coming from the wall. Which still is way over unity.

          • ivanc

            You missing the point, I agree the joule doest not have significant value.
            but let us know the other measurement they did.
            and is quiet important because is the IL that was feeding the system.
            so there is not really only one measurement (power like you think)
            is two (power and current), both must be in harmony.
            And remember that power has been contested. but luckily or unluckily depends of your point of view we have the IL to double check.

          • US_Citizen71

            And you still haven’t answered my original question. What is wrong with the measurements made by the PCE 830 connected from the wall? If you do not have an answer for that then I have to this whole exchange by you is nothing but FUD. I’m sure you will ignore this again like you have the last 4 times I have asked but it is worth a fifth attempt.

  • US_Citizen71

    Since ivanc doesn’t show much of his work even on a spread sheet I decided to do one of my own to verify the math. Doing so I discovered what I believe to be a large basic error in his calculations that accounts for the factor of 3+. To be sure I put my sheet in front of a power systems engineer I know and he verified my equations. He was concerned with the wire in the test being too small a gauge resulting in unnecessary line loss and nothing else. The error is caused by neglecting to divide the power-in from the supply listed in table 7 of the report by 3. This is the correct thing to do as the values being calculated are from one third of the circuit and in a balanced load one third of the power is dissipated by each piece. All my work is shown, clicking on a cell will show the equation used. My Dummy Run calculations match his and that is where the similarity ends.

    https://docs.google.com/spreadsheets/d/1b_bXWnPXdsvUQT0ng1F2vATFwfQK_mXQw4nw4y4Safo/edit#gid=890186181

    • ivanc

      Good job,

      but you starting with power consumed,
      I am trying to calculate IL because they read that parameter to calculate joule heating.
      Please note your max iphase current is about 15.7 amps, and they reported about 50amps for the line this should translate to 50/sqrt(3) about 29 amps.
      My calculation correlate with the amp readings.

      Now please calculate the IL first, then Iphase from the found IL .
      then p=Ip^2 * R
      see were you get.

      You not using the resistance values in cu at all, but they did.

      If all calculation were correct you method and mine should match.
      Your calculation are correct for the data you have used.

      Now do the calculations finding IL first from the joule heating formula they used.

      • US_Citizen71

        Power consumed is the only value given based on measurement for each run that we can extrapolate. The resistance can be considered constant to allow some approximation of the other values, you must take into consideration that it would change as I explained to Thomas Clarke above. But you still need a 2nd variable in order to solve the equations. Without another measured current or voltage the power is the only other variable available. The report stated the max current was about 40A not 50A stop moving the goal posts. Ic1 + Ic2 for the highest power run would be approximately 40A. Im not sure that is correct equation to determine the total current in a 3 phase circuit so I won’t stand on that, but I will look into what is the correct equation this evening and alter my sheet to include it as well as the c2 currents.

        • ivanc

          You miss the point, we need to calculate the IL to verify the Input power, because the input power in the subject in question.

          “The report says:

          Measurements performed during the dummy run with the PCE and ammeter clamps allowed us to measure an
          average current, for each of the three C1 cables, of I1 = 19.7A, and, for each C2 cable, a current of I1 / 2 = I2 =
          9.85 A. The evaluation of heat dissipated by the first circuit is:
          WC1 = 3(R1I12) = 3(4.375 ∙ 10–3 ∙ (19.7)2) = 5.1 [W]
          (9)
          For the second circuit we have:
          WC2 = 6(R2I22) = 6(2.811 ∙ 10–3 ∙ (9.85)2) = 1.6 [W]
          (10)
          By adding the results, we have the total thermal power dissipated by the entire wiring of the dummy.
          Wtot.dummy = 5.1 + 1.6 = 6.7 ≈ 7[W]

          this is from where equation (a) in my calculation came from.
          in they add (9) and (10) to get total joule.
          they stated they read IL, with clamp and PCE, and averaged the readings.
          if you generalize his equation (use letters instead on numbers).
          you see:
          wtot.dummy=3(Rcu1 IL^2) + 6(Rcu2 (IL/2)^2) and this is my equation (a)
          From this equation you could find IL and solve the circuit.

          IL will give you a second way to verify the circuit. (if you see the data I calculated the lowest IL is about 45 and the largest 49 so is between 40 and 50).

          The IL I calculated have all certainty because I am just solving (IL) in the equations they used to calculate the Joule.

          both IL and Power measurements should correlate like in the dummy run, but they do not.

          Why you say my calculations are wrong, if you have not proven any of my equations wrong?, You have completely ignored the possibility that the data could be incoherent. (why the dummy calculations match?)

    • Thomas Clarke

      US_Citizen. You need to treat the dummy and active figures the same way.

      If you divide the active power by 3 to get power per phase, as you have done, and work out resistance per phase equivalent that is fine. But then you need to do the same with the dummy figures, so your dummy resistance also changes by 3.
      Your Iph for the active run (14A) is less than your Iph for the dummy run (19A – it is what you call Ic1). That vcan’t be, because the dummy run current must be smaller than the active run current.

      A simpler way to see this is to note that the ratio of powers:

      Pin/Pjoule

      must equal the ratio of resistances. If this changes (which it does) the resistance must change.

      • US_Citizen71

        I did treat the dummy run and the active runs the same in my table. Look again. I used the same Ic2=Ic1/sqrt(3) calculation used by ivanc up top where I calculate the values of the dummy run by itself. In the table I use the sqrt((1/3 x P)/R) to calculate all of the currents. The numbers match for the Dummy run. To find the c1 current in the runs, Ic2=Iph on my table Ic1=sqrt(3) x Ic2. Ic1 contains the current for more than one phase. The resistance like ivanc has said many times should be equal for all runs.

        edit:”The resistance like ivanc has said many times should be equal for all runs.” It should be mentioned that in reality the resistance would increase in the cables or in the coil as the temperature in either increases. Keeping the resistance constant is only to allow calculations that approximate what is going on. As any of the resistances increase the corresponding voltage would increase and the corresponding current would decrease. This leads to a non-linear function as the power increases.

        • ivanc

          NO the input power is not the only solid data we have, they read the current and used to calculate the joule heat in the cables, They measured current to work out joule heating, this is the way we could double check their calculation. IL is encoded in that joule data, because they giving you the rcu1 and rcu2 they used. so finding IL is simple.

    • Obvious

      I ran P = I^2/R against the numbers to solve back for R, and they all come back 3.700 R (3x Rph). If the phases are considered to be in series maybe this is ok….
      I’m done with this for tonight.
      Cheers

      • US_Citizen71

        Yes because the sheet solves for for one third of the power. You did a manual check of the spreadsheet, thanks!

        • Obvious

          I have downloaded almost all the spreadsheets presented in this topic.
          I have my own also, which is now getting quite out of hand.

        • Obvious

          I am certain that the individual reactor resistances are 1.84 ohms each. The inductance of the delta as a whole is 1.234 ohms. 1.234 ohms is the resistance that would be measured by an ohmmeter across any two terminals of the delta.
          Edit: Well, almost certain. Something is weird about calculating the right resistance. Most of my calculations come up with that number, but I’m still leery of the actual DC voltage equivalent. I suppose it is the effective DC equivalent resistance at least.

    • ivanc

      Looks that my posts are delayed, I hope not banned.
      your calculation and mine are equal for the dummy run.
      This show that there is some error in the data for the active run.
      The current you have calculated is to low. they said about 50 amps has been used.
      your data is also showing the anomaly the data should match with the joule heating data.
      you starting from the input power, but this is the parameter we trying to test.
      better start calculating IL, the equation we spoke and derived step by step gives IL , from there you could calculate Ip, then the rest is easy.
      There is no error in my calculations, the error is in the data, and the match on the dummy data shows your calculation and mine are correct, but there is incoherent data.

      • US_Citizen71

        Read my reply to your earlier post.

    • Dr. Mike

      US_Citizen71,
      Your calculations look good. Please add one more column to your Table. Calculate the Cu wire Joule heating using your calculated currents and the resistances and formulas shown in the report. Perhaps, the Lugano scientists will get back to us and say they just miscalculated the active run Cu wire Joule heating, and they were in error when they said on page 3 “current through the resistor coils, normally 40-50 Amps”. Remember they really didn’t measure any resistor coil currents or they would have caught the 1/SQRT(3) factor in the C2 lines for the Cu wire Joule heating calculation. The 40-50 Amp reference has to mean something they measured, that is, the C1 line currents. Ci line currents of 40-50A are in agreement with the Cu wire Joule heating numbers in Table 7.
      Dr. Mike

      • US_Citizen71

        I will add your requests to my to do list for the sheet. I likely won’t get to it until this evening.

  • ivanc

    Frank…. Why are you stopping my posts?????
    If you not interested in the truth…is ok …. I will understand…..

    • US_Citizen71

      Frank sleeps and does other things than approve posts at times, be patient.

      • ecatworld

        True, USC — and I try to keep up.

  • ivanc

    Frank…. Why are you stopping my posts?????
    If you not interested in the truth…is ok …. I will understand…..

    • US_Citizen71

      Frank sleeps and does other things than approve posts at times, be patient.

      • Frank Acland

        True, USC — and I try to keep up.

  • US_Citizen71

    I did treat the dummy run and the active runs the same in my table. Look again. I used the same Ic2=Ic1/sqrt(3) calculation used by ivanc up top where I calculate the values of the dummy run by itself. In the table I use the sqrt(1/3P/R) to calculate all of the currents. The numbers match for the Dummy run. To find the c1 current in the runs, Ic2=Iph on my table Ic1=sqrt(3) x Ic2. Ic1 contains the current for two phases. The resistance like ivanc has said many times should be equal for all runs.

    • Obvious

      The 57 one is probably a red herring…
      3.45 is almost certainly a function of the resistance problem, which will be found to be an error somewhere involving incorrect vector math.
      I do not believe the resistance changes in a meaningful way. This is just a heater, and has nothing (besides heat) to do with the reaction.

      Part one of my solution is that 9.85 A [ie:1/2(19.7A)] is the vector sum of two 120° separated 11.37 A phase currents heading towards a common return wire that are also each 120° on either side of a single wire lead.

      Part two is that the majority of the 19.7A RMS is contributed by the return side of two other 120° separated lines, that are each fighting the one measured wire’s “forward” current.
      There are more parts, but is good for now to consider.

    • Obvious

      All of the signs are eliminated by RMS. The RMS for one line, over time, is identical to the others.
      Square, then mean, then root is the proper order.
      The diametrically opposed version of the cheese trick is 3 times current (or infinity).

      • Andreas Moraitis

        Do you mean the PCE before the control box? From where do you take the data? The calculations of the COP are most likely based on the readings of the second meter.

  • Obvious

    Good morning. I had a restless sleep filled with sqrt(3)’s floating all around in my dreams.
    I can see why ivanc likes the sqrt(3) so much, other than that it is the line phase modifier.
    The extrapolation equations are absolutely soaked in sqrt(3)’s.
    Some quick notes, just so I get them out of my system:
    (not picking on anyone, just accrediting the quotes. The math seems fine….)

    “Lets call Jof=57.66513854” -ivanc
    10^2/sqrt(3) = 57.7350269……very close. Why?

    “I have updated the table to show input power = consumed power – joule heating.
    The factor is now 3.4 to 3.5 instead to 3.3” – ivanc
    and
    “….for the given figures to be correct the heater resistance must decrease by 3.3 (sorry, in fact as Mikepoints out, 3.45) in the active test.” -Thomas Clarke
    2*sqrt(3) = 3.464………very close again. Why?

    There are some others, also. I’ll add them above when I find them again. These are not random coincidences, IMO.

  • US_Citizen71

    Power consumed is the only value given based on measurement for each run that we can extrapolate. The resistance can be considered constant to allow some approximation of the other values, you must take into consideration that it would change as I explained to Thomas Clarke above. But you still need a 2nd variable in order to solve the equations. Without another measured current or voltage the power is the only other variable available. The report stated the max current was about 40A not 50A stop moving the goal posts. Ic1 + Ic2 for the highest power run would be approximately 40A. Im not sure that is correct equation to determine the total current in a 3 phase circuit so I won’t stand on that, but I will look into what is the correct equation this evening and alter my sheet to include it as well as the c2 currents.

  • Thomas Clarke

    Finding patterns in random numbers is what people tend to do. Mostly they are rubbish.

    If you can think of an error mechanism that gives this factor difference between active & dummy tests that will be interesting, but really we are at the moment stuck with not many error mechanisms full stop.

    57.73 depends only on the copper wire resistance (essentially arbitrary) and also has units. Hence any relationship to a specific round etc number of this quantity is purely coincidental.

    3.3329 is the power ratio characteristic of the specific error – so maybe it has some meaning. It works as 3X + 10% where the 10% is easily got from phase imbalance in the heater resistors and/or connections. Three is a number that crops up a lot in 3 phase mistakes, and I’ve given two mechanisms that lead to it below. I don’t know any error mechanism that leads to 2*sqrt(3) – that does not mean that none such exists.

    3.45 is the equivalent resistance change, relevant if you think the resistance really does change.

    • Obvious

      The 57 one is probably a red herring…
      3.45 is almost certainly a function of the resistance problem, which will be found to be an error somewhere involving incorrect vector math.
      I do not believe the resistance changes in a meaningful way. This is just a heater, and has nothing (besides heat) to do with the reaction.

      Part one of my solution is that 9.85 A [ie:1/2(19.7A)] is the vector sum of two 120° separated 11.37 A phase currents heading towards a common return wire that are also each 120° on either side of a single wire lead. (Edit: or something like that. It is more likely the vector sum of the “negative” portions.)

      Part two is that the majority of the 19.7A RMS is contributed by the return side of two other 120° separated lines, that are each fighting the one measured wire’s “forward” current.

      Edit: So, if the 19.7 A is mostly from two other wires at vectors to the one measured, then multiplying this value will inadvertently multiply two vectors that are “opposite” by
      sqrt(3) each of the one intended.

      There are more parts, but this is good for now to consider.

      • Thomas Clarke

        The math, correct or incorrect, has to be different for the dummy and active tests: that is the challenge. Consistently incorrect values would not lead to this ratio – and would be detected because the dummy in and out power would not balance.

        • Obvious

          The difference is probably that the professors used the recorded V and I to measure power. And we are trying to make a voltmeter from a spreadsheet.

  • Obvious

    Ivan,
    I am definitely on-side with the phase current for the dummy run. (~11.37 A)
    My updated DC equivalent equations require them to work out correctly.
    However, they must be used carefully in a DC equivalent circuit or weird things happen.
    (It is more complicated than it looks.)

    • ivanc

      The only thing you have to be careful is to use Ip=IL/sqrt(3),
      Then for each element you could use the DC method, repeat for each element, not for the whole.

      • Obvious

        That seems almost right to me. (just cleaned this up a bit)
        But whole DC equivalent circuits can be made, if one is very careful.
        There are (at least) three DC versions possible.
        1. Do it in pieces, (one resistor at a time) then add or multiply the parts properly. I think this requires phase current.
        2. Current flows in one leg, and out only one other, making a series-parallel circuit of resistors. Then one must be certain of the value for each resistor (which is not that hard). This would only happen for shades of a microsecond in real life, or if the third leg was open. This version seems to need phase current to make the math work out. The total Power answer for P = I^2R is three times this single solution.
        3. Current is split (1/2), and enters two legs equally, simultaneously, and exits through one leg only. In this case the resistor between the two “in” legs would not conduct. This version requires line current. The total Power answer for P = I^2R is two times this single solution. (ie: open the delta on the “out” side).
        **** This version is a nightmare using vector (phase) current and a two-into-one connected “out” line.**
        4. Is the same as 3. above, but all current is going IN one leg, and is split equally between two delta sides, and flows equally out the remaining two legs. We came to severe disagreement over this last version, earlier. But I insist it is a proper DC equivalent circuit, if you use line current in the single leg, and therefore 1/2 line current in the legs.
        5. Two delta legs are connected together by a DC test shorting wire (that has effectively zero resistance, (ie 1 inch solid gold or silver wire.).
        This shorts out the one resistor that is between these legs. Line current (from one line) then passes equally through two resistors, that are now in parallel, and goes out the remaining leg. This gives funny resistor values if you don’t pay attention to the parallel part, and that there are only two participating resistors, not three. Get that right and the power equation should work. This basically only useful for cross-checking your resistance values in the other versions, since so much gets “disconnected”.

        • Obvious

          And a serious problem exists with ALL of the above circuits.
          NONE of them will give the correct resistance for the resistors in real life, if power was turned off, power leads disconnected, and a probe connected across any two delta leads.
          All of the above versions only derive the effective DC resistance that the AC power sees in a delta, due two three phases mixing up the power delivery.
          NONE of these approximations takes the average 180° in-out, at-one-time, actual DC values into account for current.
          The actual DC current equivalent is Sqrt(3)*9.85 A.

        • Thomas Clarke

          My python script, below somewhere, shows what the delta currents are given the three 120 degree separated line voltages.

          The clever thing is that the delta mean square delta current is constant. The delta currents are 1/sqrt(3) of the line currents, both of which (RMS) are constant over time.

          That fact simplifies everything. The individual currents change of course, but the power (proportional to the sum of squares) stays constant.

          • Obvious

            I will scroll down for another look at your script. I have another bigger version also.
            I need a break from this to start fresh, right from square one with all assumptions examined one by one. Methodically.
            I have pages of crap scribbled out, each one with probably one important piece of the puzzle on it.
            The phase amps fix some things and complicates others.
            The problem, I think, begins when we assume that 19.7 A is a meaningful number that represents all the power in the circuit at the same time for the dummy, and can be simply extrapolated from. It can be useful, but how useful is something of an enigma.
            I am also going to attempt to de-integrate the reported input Watts and see what terrible path that leads down. If extrapolated resistance climbs by 2sqrt(3), then I might get somewhere.

  • ivanc

    I have updated my spreadsheet and put in the spread sheet of Us_citizen71 and calculated joule heating. and as expected the value is different to the report by a factor of 3.
    Is the same issue, it just depends from which side you look at it.
    to me the reading of IL is more likely to be correct, just because is a simple measurement.

    But the lugano testers have logged voltage and current and input power. so releasing this data will solve the issue.
    https://docs.google.com/spreadsheets/d/1tuKwWUxC2Gq_MtEED4_XYYIqehpOv47mGVx9yio-GKs/pubhtml

    • Obvious

      Really nice, from the parts I that can see.
      Can you re-post that as a downloadable Excel equivalent somehow?
      USC seems to have been able to.
      For whatever reason I’m not getting all of it.

      • ivanc

        I think you could just copy and paste into excel, I use open office and it worked ok.

        • Obvious

          I have some funny formatting issues, mostly with comments, but copy-paste worked overall.

    • Mark Szl

      So now we wait again but it is better than going on a wild goose case.

    • Dr. Mike

      ivanc,
      Everything looks good in your Tables to me. About the only thing you didn’t calculate is the real Cu wire Joule heating numbers when the correct 1/SQRT(3) factor is used for the C2 currents. I’m sure the Lugano authors will make this correction in their revised report. This correction doesn’t affect any of the results where ratios of Joule heating are used to calculate currents. It will slightly change the COP numbers.
      Dr. Mike

    • Obvious

      Some big problem here.

      I/sqrt(3) = 3(I/sqrt(3))
      Looks ugly?, but then:

      ip = IL/sqrt(3)
      ipwr = 3ip^2*ecat_R ……………this is OK?
      ipwr = 3(IL/sqrt(3))^2*R….substitute the ip equation back in…..

      divide by 1.73, then multiply 3 times…= 1.73 times IL, squared…times R..

  • Obvious

    That seems almost right to me.
    But whole DC equivalent circuits can be made, if one is very careful.
    There are (at least) three DC versions possible.
    1. Do it in pieces, then add or multiply the parts properly. I think this requires phase current.
    2. Current flows in one leg, and out only one other, making a series-parallel circuit of resistors. Then one must be certain of the value for each resistor (which is not that hard). This would only happen for shades of a microsecond in real life, or if the third leg was open. This version seems to need phase current to make the math work out.
    3. Current is split, and enters two legs equally, simultaneously, and exits through one leg only. In this case the resistor between the two “in” legs would not conduct. This version requires line current.
    4. Is the same as 3. above, but all current is going IN one leg, and is split equally between two delta sides, and flows equally out the remaining two legs. We came to severe disagreement over this last version, earlier. But I insist it is a proper DC equivalent circuit, if you use line current in the single leg, and therefore 1/2 line current in the legs.

    • Obvious

      And a serious problem exists with ALL of the above circuits.
      NONE of them will give the correct resistance for the resistors in real life, if power was turned off, power leads disconnected, and a probe connected across any two delta leads.
      All of the above versions only derive the effective DC resistance that the AC power sees in a delta, due two three phases mixing up the power delivery.
      NONE of these approximations takes the average 180° in-out, at-one-time, actual DC values into account for current.

  • Thomas Clarke

    Yes. We can speculate, but the testers can undoubtedly resolve this – they have stored data which will do it. I don’t understand why they are not doing this. Whatever the results, they are better off clearing up the anomaly than leaving it.

    I guess it may be embarassing to admit to a mistake, but in teh long run it is much worse just keepimng quiet.

  • Obvious

    Really nice, from the parts I that can see.
    Can you re-post that as a downloadable Excel equivalent somehow?
    USC seems to have been able to.
    For whatever reason I’m not getting all of it.

  • andrea.s

    Ivanc, Obvious et al:

    I have posted an updated analysis of the input power on this link.

    http://www.cobraf.com/forum/immagini/R_123571297_1.pdf

    Basically I convinced myself (a couple weeks ago) that the peak polarity in figure 5 is not necessarily wrong, one could rather argue that it shows that the I3 clamp was ok.

    But the current pulses width on the time axis of figure 5 matches a 3kW consumption, i.e. more than three times the consumption declared in the TPR2, consistently with a constant resistance of the (inconel) coils. If one believes the story of the doped semiconductor heaters whose resistance drops by a factor 3, the current pulses would be a lot narrower than shown in figure 5.

    And a mistake by a factor 3 could still be explained by a reversed clamp elsewhere (I1, I2 if we exclude I3), which would give the same rms current but one third of the power.

    • Obvious

      Since you are pretty good as this, what do you think the would the meter read like if all V clamps were removed and both I1 and I2 also?
      (just for the photo)

      • andrea.s

        I don’t own a PCE-830 and the manual doesn’t say, but I suppose that one can plot a single current waveform asynchronous to all others, and most likely the display will trigger on the first upwards transition. In this case one simply does not have a phase information, but the pulsewidth remains and reveals the power consumption.

        • Obvious

          I was wondering if OL would show up on the empty line info spaces. My though experiment seemed to suggest that I# would still report a current value, though. Without a PCE-830 to work with, or some handy three phase resistive load, its just theory I suppose.

    • ivanc

      Good job, I actually suspect something more serious, my speculation, repeat speculation is that the controller permutes the phase of two lines (remote control?). and the testers did set up the circuit correctly.

    • Dr. Mike

      andrea s.
      Thanks for taking the time to put together this really excellent analysis. I found one typo: On page 4 in the paragraph above Figure 4, I believe “dissipating 3W” should be “dissipating 3KW”. I’m sure that all of the EE’s will understand your analysis, but I’m not sure about those with less electrical engineering background.
      I am in agreement with your analysis and probably more important am in agreement with what you say on page 10 in the section on “General remarks”.
      Dr. Mike

      • andrea.s

        thanks Dr. Mike, yes it is 3KW !

    • Thomas Clarke

      andrea.s

      This is a very authoritative analysis – thanks for doing it – a lot of work.

      I agree with all you say, some small additions:

      The error in the Joule powers must be explained, regardless of whether a reversed clamp or some other possible mechanism has been ruled out.

      If the error comes from an unusual heater element then the active run warm-up current and power in figures will show this by giving a variable resistance. The authors could resolve the matter quickly by showing this.

      The X3.3 error could be reversed clamp, or single line power recorded mistakenly for total power, during the active test. I guess it could be many other things too. The 10% above X3 is most likely a slight phase imbalance in the resistances. This could be confirmed or denied by looking at the three active run line powers, which would therefore differ by just under 10%.

      I have confirmed (though it is also standard theory) that the instantaneous power in this circuit when the triacs are on is the same for any phase. So the SCR duty cycle is proportional to power out for given resistance and it together with the known information about currents, powers does give useful information.

      The X3.3 anomaly has been validated by many people here and is not in doubt.

      Best wishes, Tom

    • Thomas Clarke

      Just another small typo:

      bottom page 3:

      “undergoes a three-fold drop above 1250°C and stays constant up to 1400°C”

      Should be:

      “undergoes a three-fold drop between 500C and 1250C, and then stays constant up to 1400C”

      • andrea.s

        Hello Thomas,

        thanks for your remarks. I sent the note to the authors and to Rossi. I got a reply by Rossi, not in the best of moods, but I wasn’t expecting to be congratulated. As usual, silence from the Authors.

  • Dr. Mike

    ivanc,
    Everything looks good in your Tables to me. About the only thing you didn’t calculate is the real Cu wire Joule heating numbers when the correct 1/SQRT(3) factor is used for the C2 currents. I’m sure the Lugano authors will make this correction in their revised report. This correction doesn’t affect any of the results where ratios of Joule heating are used to calculate currents. It will slightly change the COP numbers.
    Dr. Mike

  • Obvious

    I will scroll down for another look at your script. I have another bigger version also.
    I need a break from this to start fresh, right from square one with all assumptions examined one by one. Methodically.
    I have pages of crap scribbled out, each one with probably one important piece of the puzzle on it.
    The phase amps fix some things and complicates others.
    The problem, I think, begins when we assume that 19.7 A is a meaningful number that represents all the power in the circuit at the same time for the dummy, and can be simply extrapolated from. It can be useful, but how useful is something of an enigma.
    I am also going to attempt to de-integrate the reported input Watts and see what terrible path that leads down. If extrapolated resistance climbs by 2sqrt(3), then I might get somewhere.

  • Obvious

    The difference is probably that the professors used the recorded V and I to measure power. And we are trying to make a voltmeter from a spreadsheet.

  • Obvious

    Since you are pretty good as this, what do you think the would the meter read like if all V clamps were removed and both I1 and I2 also?
    (just for the photo)

  • Obvious

    I have some funny formatting issues, mostly with comments, but copy-paste worked overall.

  • Obvious

    Rossi was right. The C2 cables are 1/2 of C1, on average.
    The whole phase thing just adds needless complexity. At least for the cables.
    I have gone through my old assumption, where I had exactly half the Jh, and I need to double it to match the total power using the method I used.
    The logic is so simple:
    If each C1 reads 19.7 A on average (RMS if you prefer) then so do all the others. So the phase calculations, or worse mixed phase and line calculations MUST be identical for the whole set of all cables when summed, if phase current is applied correctly.
    If they are not, then the total for phase math part is wrong, or the division between phase and line current is wrong.
    If the C1 cables see 19.7 A RMS, then sliding the clamp further down until reading the attached twinned C2 cables, the reading will be same. Try it. It is a fact.
    If each C1, and also each pair of C2 cables see 19.7 A when combined (twinned), then so do all the others, on average.
    This means the RMS of each set is identical. This requires that 1/2 of the current measured in one C1 MUST be in the respective attached C2 cables, not at the same instant, but on average nevertheless.

    • Dr. Mike

      Obvious,
      I don’t believe Rossi was involved in the calculation of the C2 currents. This is an error made just by the professors due to their lack of familiarity of 3-phase circuit calculations.
      Although one of your statements is correct, the RMS current in the C2 cables is still equal to the RMS C1 current divided by SQRT(3). Your true statement is: “If the C1 cables see 19.7 A RMS, then sliding the clamp further down
      until reading the attached twinned C2 cables, the reading will be same”. A RMS ammeter clamped around both C2 cables will read the same RMS current as the C1 cable feeding it. However, if the RMS ammeter is clamped around just one of the C2 lines the RMS current will be measured as the C1 current divided by SQRT(3). The instantaneous current in the C1 line will equal the sum of the instantaneous currents in the two C2 lines, two currents that are 120 deg out of phase. If you calculate the RMS (root-mean-squared) of this C1 current, you will find it to be equal to SQRT(3) times the RMS value of either of the C2 current, which will be equal for a balanced load.

      Dr. Mike

      • Obvious

        With all due respect, NO. (sorry, Pointless)
        One C2 cable might vary all over the place relative to the C1 it is attached to.
        But two of them side by side cancel some of the back-and forth “noise”.
        This is because some of that current is destined for another phase, and is just “en-route”. This “en-route” current is entirely made up for, on average at one of the other cables. And they all average exactly the same over time, and certainly over hours.
        RMS is what you guys drilled, over and over.
        The RMS of any one C2 cable is identical to the RMS of every other one. They MUST be, since it is a balanced circuit.
        Instantaneous is what scraps the entire calculations using 19.7.
        If you want phase current, then use the vectors. As is required. Then the C1 cables MUST be left as line current-fed. Or else figure out how to split the load at the C1’s relative to the C2’s.
        Where I messed up earlier was imagining dragging the clamp across the entire circuit. This works until you get to the center. Then things get complicated.
        So just stop at the center. We have a measurement that is good right to the center of the delta.
        A minor error may be encountered where the two C2 cables may be in parallel, rather than see 19.7A each. However, the professors have used 1/2 line current, This is the same, exactly, as two parallel C2 cables at 19.7 A.
        Edit: If you do the correct math for phase, the results will be identical to the 1/2 IL values for C2 in the report. If you do not have these values, STOP. Something is wrong. Find out what is wrong.

        • Dr. Mike

          Obvious,
          If you do the correct math accounting for the for the phase difference, that is 2*pi/3 or 120 deg, you will find that the RMS C1 current is SQRT(3) times the RMS C2 current. If you can prove otherwise, you need to re-write the electrical engineering textbooks on 3-phase delta power circuits.
          Dr. Mike

          • Obvious

            That is true. But pointless.
            I am not re-writing textbooks.
            The Iphase C2 story is totally a pointless waste of calculation time for the Joule heat.
            Two C2’s in series at 19.7 A is the same as two in parallel at 9.5 A, which is the same as one cable with the same total resistance even if it was made of cheese strings.
            These cables, from the viewpoint of one 19.7 measurement, all see the same thing from that perspective. What happens here, on average, is exactly what happens to the other two corners, but rotated out 120° in time. But on average, RMS, these cables have no 120°, or 10°, or anything like it. They are the same.
            Time averaging has wiped out each and all of the 120° phase separations into an average angle of 180°. Exactly one half of 360°.

          • Dr. Mike

            Obvious,
            I think you need to look up the definition of RMS current. Caluulate the RMS value of the C2 current, assuming the C2 current is the instantaneous sum of two sine wave currents that are 120 deg. out of phase. The math is not trivial. You will need to be able to do an integration of the square of the absolute value of the superposition of the two C1 waveforms over one complete cycle.
            Dr. Mike

          • Obvious

            The math is not trivial.
            Then why the heck would anyone want to do it?
            Why invoke it?

          • Dr. Mike

            Obvious,
            You have to do the math to determine the relationship between the RMS current in the C1 line compared to the RMS current in the C2 line. What I was saying about the math not being trivial was that if you are not familiar with integral calculus and trig function substitutions, you would not understand how the SQRT(3) factor is calculated, even if a detailed derivation is presented. Do you understand the basic calculation of RMS current for a sine wave current, that is, if i = a sin(wt), then the RMS current is I(RMS) = a/SQRT(2)? You need to do this same calculation for the C1 line where the current is i =a sin(wt) +a sin(wt+2*pi/3). The RMS current for this superposition of two currents can be calculated to be a*SQRT(3/2). The ratio of the C1:C2 RMS currents is a*SQRT(3/2) / a/SQRT(2) = SQRT(3).
            Dr. Mike

          • Mark Szl

            Crystal clear Mike and anyone can Google the same for this balanced 3 phase delta circuit.

            I am really getting a bad feeling about this latest report and Ecat. What a blunder!!!

          • Dr. Mike

            Mark,
            It wouldn’t be a total blunder. I will still wait to see what the professors have to say about the power vs current discrepancy. Even if the COP is determined not to be what was originally claimed, the “ash” analysis results alone are enough to show that nuclear reactions were taking place. Did you read Andrea S.’s report given in the link in 5 or so comments below? If you haven’t, please read it. I was really surprised there might also be an error in the COP calculation of the report from last year.
            Dr. Mike

          • US_Citizen71

            “Figure 4 details the electrical connections of all elements of the experimental setup.” – Denial isn’t just a river in Egypt.

            ” but as I say even if it was checked it could suffer the same error (maybe naturally would) as the other one.” – Without the how that is simply speculation, deflection and denial, not supported by any data.

          • Dr. Mike

            US_Citizen71,
            Do you really believe “Figure 4 details the electrical connections of all elements of the experimental setup”? What about the 4th sentence in the Introduction on page 1: “In addition, the resistor coils are fed with specific electromagnetic pulses”? The gray box with the potentiometer on top in the Figure 3 set-up appears to be the controller. What is the brown box? Where is it in the Figure 4 electrical diagram? If the brown box is a pulse generator, is it properly connected to the rest of the circuit? I believe Rossi has every right to keep proprietary the nature of those “specific electromagnetic pulses” fed to the resistor coils; however, the wiring diagram is incomplete without including the brown box.
            There are at least two things missing from the report that could have a significant bearing on the results. First the authors fail to state if the “specific electromagnetic pulses” were also added in the dummy run. Second, if additional pulses were fed to the coil, we know that some additional energy was added in those pulses. If the nature of those pulses are proprietary, the authors should have both stated this fact and included an upper limit for the amount of power they added to the input power.
            Dr. Mike

          • US_Citizen71

            Yes I believe the figure shows all connections the testers made. The boxes you refer to are part of the controls(control box). If it was all put in a shell it would be one box and still have interconnections inside. An IC has all kinds of wires to interconnect the parts onboard do we view it as millions of parts or as one?

          • Obvious

            OK. …
            On your Joule heat calculations, divide “Power In” by “I in”. Now you have LV.
            Now look at your Vph number.
            Explain the 10 V difference for the dummy. I did all it the way down the group. 1.4041304 difference LV to Vph, top to bottom.
            Vph = VL in a delta

          • Obvious

            Dr. Mike,
            When you asked yourself, as I suggested, “Why cannot I add the three
            19.7 A RMS values from each corner of the delta together?”, what was the answer?
            Why then does dividing 19.7 by anything other than 3, then adding them up, not make any sense?
            Everything else gives the wrong answer. That’s why.
            No matter how you slice up the 19.7 A for IL, if you cannot arrive back to the same IL by summing the results of the pieces, the math is wrong. Period.
            It cannot be more, it cannot be less. You cannot add a factor or quotient to only one side of the equation and get a balanced equation. Period. You must perform the equivalent operation to both sides of an equation in order for it to balance.
            This should be obvious.
            The RMS current value for IL is the sum of both positive and negative currents. It is a magnitude twice as high as the average negative and positive sides. It does not care whether one side of a delta is doing something more than another. It just chops the sign off and adds it to the total, then divides it by the number of parts. Even if all 19.7 A came from one opposite corner, and none came from anywhere else, the RMS answer is 19.7. If one side was -19.7 A, and the other was +19.7 A, the RMS answer is still 19.7 A. The total of the parts MUST add back up to 19.7. The RMS of all three 19.7 A corners is still 19.7 A.
            All the corners MUST add up to 19.7 A at any one time, RMS, if the circuit is balanced. No matter what they are doing at any one point in time, because RMS averages over time. They are the same. They are equal.
            Adding them is like adding all the V’s measured across a bunch of parallel resistors. The sum of 1000 Volt measurements across 1000 parallel resistors is not 1000 times the voltage found at one, or any one, or any combination of V measurements. It doesn’t even matter what the 1000 resistors’ resistances are. They could be anything. V is V.

            Just because sqrt(3) is a fancy number does not make the RMS IL current bigger. EVER.
            Dividing IL by sqrt(3) and adding up three of these pieces, (each greater than 1/3) makes IL bigger, because the parts add up to more than the original value. This is wrong. Period. It is an unbalanced equation.
            Non sequitur. It does not follow. One side of the operation MUST equal the other. Or it is just nonsense with numbers attached.
            3(19.7)/sqrt(3) = 3(19.7)/sqrt(3)———–YES
            .
            19.7 = 3(19.7/sqrt(3))———————-NO
            19.7 =/= 34.121
            Square the right side of this lopsided equation, and bingo-bango
            the power is three times higher!!!!, and then because R gets unequal treatment, it looks 3 times smaller!!!!

            Work out where it is going wrong.
            I have explained the reason several times.

    • Obvious

      A piece of the upcoming opus, so it is not so vastus all at once.

      Using phase current for the delta correctly:
      JhC1 = 3*(IL^2)(RC1) = 3(19.7^2(0.002811)) = 3.27464 W
      JhC2= 6*(Ip^2)(RC2) = 6(11.37~^2(0.004375)) = 3.39774 W
      Jhtotal = JhC1 + JhC2
      Jhtotal = 6.67 W

      Using the report method, exactly as in the report.
      WC1 = 3(R1I1²) = 3(4.375 ∙ 10–3 ∙ (19.7)²) = 5.1 [W]
      WC2 = 6(R2I2²) = 6(2.811 ∙ 10–3 ∙ (9.85)²) = 1.6 [W]
      Wtot.dummy = 5.1 + 1.6 = 6.7 ≈ 7[W]

  • Obvious

    Rossi was right. The C2 cables are 1/2 of C1, on average.
    The whole phase thing just adds needless complexity. At least for the cables.
    I have gone through my old assumption, where I had exactly half the Jh, and I need to double it to match the total power using the method I used.
    The logic is so simple:
    If each C1 reads 19.7 A on average (RMS if you prefer) then so do all the others. So the phase calculations, (or worse) mixed phase and line calculations MUST be identical for the whole set of all cables when summed, if phase current is applied correctly.
    If they are not, then the total for phase math part is wrong, or the division between phase and line current is wrong.
    If the C1 cables see 19.7 A RMS, then sliding the clamp further down until reading the attached twinned C2 cables, the reading will be same. Try it. It is a fact.
    If each C1, and also each pair of C2 cables see 19.7 A when combined (twinned), then so do all the others, on average.
    This means the RMS of each set is identical. This requires that 1/2 of the current measured in one C1 MUST be in the respective attached C2 cables, not at the same instant, but on average nevertheless.

    • Dr. Mike

      Obvious,
      I don’t believe Rossi was involved in the calculation of the C2 currents. This is an error made just by the professors due to their lack of familiarity of 3-phase circuit calculations.
      Although one of your statements is correct, the RMS current in the C2 cables is still equal to the RMS C1 current divided by SQRT(3). Your true statement is: “If the C1 cables see 19.7 A RMS, then sliding the clamp further down
      until reading the attached twinned C2 cables, the reading will be same”. A RMS ammeter clamped around both C2 cables will read the same RMS current as the C1 cable feeding it. However, if the RMS ammeter is clamped around just one of the C2 lines the RMS current will be measured as the C1 current divided by SQRT(3). The instantaneous current in the C1 line will equal the sum of the instantaneous currents in the two C2 lines, two currents that are 120 deg out of phase. If you calculate the RMS (root-mean-squared) of this C1 current, you will find it to be equal to SQRT(3) times the RMS value of either of the C2 current, which will be equal for a balanced load.

      Dr. Mike

      • Obvious

        With all due respect, NO. (sorry, Pointless)
        One C2 cable might vary all over the place relative to the C1 it is attached to.
        But two of them side by side cancel some of the back-and forth “noise”.
        This is because some of that current is destined for another phase, and is just “en-route”. This “en-route” current is entirely made up for, on average at one of the other cables. And they all average exactly the same over time, and certainly over hours.
        RMS is what you guys drilled, over and over.
        The RMS of any one C2 cable is identical to the RMS of every other one. They MUST be, since it is a balanced circuit.
        Instantaneous is what scraps the entire calculations using 19.7.
        If you want phase current, then use the vectors. As is required. Then the C1 cables MUST be left as line current-fed. Or else figure out how to split the load at the C1’s relative to the C2’s.
        Where I messed up earlier was imagining dragging the clamp across the entire circuit. This works until you get to the center. Then things get complicated.
        So just stop at the center. We have a measurement that is good right to the center of the delta.
        A minor error may be encountered where the two C2 cables may be in parallel, rather than see 19.7A each. However, the professors have used 1/2 line current, This is the same, exactly, as two parallel C2 cables at 19.7 A.
        Edit: If you do the correct math for phase, the results will be identical to the 1/2 IL values for C2 in the report (at least in sum). If you do not have these values, STOP. Something is wrong with the new phase math. Find out what is wrong.
        Edit2: If you want to split the C1 cable mathematically so that the phases are entirely separated, then a suggestion is to mathematically chop each C1 cable in half (lengthwise). One half for each phase end.
        Edit3: Ask yourself “Why cannot I add the three 19.7 A together?”. You already know the answer. It is important.

        • Dr. Mike

          Obvious,
          If you do the correct math accounting for the for the phase difference, that is 2*pi/3 or 120 deg, you will find that the RMS C1 current is SQRT(3) times the RMS C2 current. If you can prove otherwise, you need to re-write the electrical engineering textbooks on 3-phase delta power circuits.
          Dr. Mike

          • Obvious

            That is true. But pointless.
            I am not re-writing textbooks.
            The Iphase C2 story is totally a pointless waste of calculation time for the Joule heat.
            Two C2’s in series at 19.7 A is the same as two in parallel at 9.5 A, which is the same as one cable with the same total resistance even if it was made of cheese strings.
            These cables, from the viewpoint of one 19.7 measurement, all see the same thing from that perspective. What happens here, on average, is exactly what happens to the other two corners, but rotated out 120° in time. But on average, RMS, these cables have no 120°, or 10°, or anything like it. They are the same.
            Time averaging has wiped out each and all of the 120° phase separations into an average angle of 180°. Exactly one half of 360°.

            Edit: Re-introduce angles at your peril. Sqrt(3) is a vector sum in a set of two very specific triangles. It has sign if there are more than one. It requires four points and at least one named side length in order to be described specifically.
            And that’s just one of them.

          • ivanc

            Obvious you giving me head aches!!!!
            You seem a great guy with a lot of enthusiasm but there is something fundamental you missing. “you can not mix RMS with vectors”
            In reality the vectors are the one working in an electrical alternate system, but the RMS is a method to simplify calculations. they are like water and oil. they do not mix.You have absolutely no data to work with vectors.

          • Obvious

            That’s right, Ivan.

            We have no data for vectors. But sqrt(3) describes a vector. You must understand why Ip is sqrt(3) of IL.

            It is not just a pretty handy number to make phase current.

            It is an important number.

            RMS wipes out vector values, because squaring numbers removes sign, and turns them into magnitudes.

            If you cannot add together all three 19.7 A from the three corners, what in the world makes you think it is OK to divide them by 1.73 and then it is OK to add 3 of them up?

            Using sqrt(3) puts the sign back in, and turns the magnitude back into a vector.
            You can use RMS with vectors all you want. But once you put the vector in a number you have a real mess getting the vectors back out and still have a useable number, unless you do vector math which includes:

            (as simple as possible, not the real formula)

            ILa = Σ Ipa + Ipb + Ipc
            Ipa = IL/sqrt(3) < -120°
            Ipb = IL/sqrt(3) < 0
            Ipc = IL/sqrt(3) < +120°

            To make not zero: (and fix RMS to back linear signs)

            Ipa = IL/sqrt(3) 180°/ <-120°
            Ipb = IL/sqrt(3) 180°/ origin < 0
            Ipc = IL/sqrt(3) 180°/ < +120°
            I think.
            Now you have one corner done.
            Still ILb and ILc remaining to deal with(?). Using same origin, they should end up in the same place if balanced.
            If you correctly repeat with the Ipb and Ipc origins at 120° CCW or CW for each delta corner, then the sum of three complex manipulations of the three delta corner IL measurements resulting in 3 sets of Ip vector calculations should go back to zero again.

          • ivanc

            Yes, that factor had been derived using vector analysis in a delta 3phase system. It is not a vector, is a scalar number, has no direction or angle.

          • Obvious

            Nonsense.
            Sqrt(3) is a vector. 1/3 of at least three of them, each with sign, to order to complete a delta equation correctly.
            Sqrt(3) needs an origin.
            19.7 A RMS is a magnitude. It has no direction.
            A magnitude times a vector is still a vector.
            Repeat it until you believe it, or better, understand it.
            Or please explain how
            ….IL = 3(IL/sqrt(3)
            19.7 = 34.12..!!!!…………where does this re-balance?
            ( I posted a more frustrated answer to Dr. Mike. I hope he doesn’t take it personally.)
            Edit: scroll down to figure 5 here:
            http://ece.k-state.edu/~starret/581/3phase.html
            And this is better: (delta at bottom)
            http://elearning.vtu.ac.in/e-con/EEE/html/0054.htm

          • Dr. Mike

            Obvious,
            Please read the very last equation in your second reference!
            Dr. Mike

          • Dr. Mike

            Obvious,
            I think you need to look up the definition of RMS current. Caluulate the RMS value of the C2 current, assuming the C2 current is the instantaneous sum of two sine wave currents that are 120 deg. out of phase. The math is not trivial. You will need to be able to do an integration of the square of the absolute value of the superposition of the two C1 waveforms over one complete cycle.
            Dr. Mike

          • Obvious

            The math is not trivial.
            Then why the heck would anyone want to do it?
            Why invoke it?

          • Dr. Mike

            Obvious,
            You have to do the math to determine the relationship between the RMS current in the C1 line compared to the RMS current in the C2 line. What I was saying about the math not being trivial was that if you are not familiar with integral calculus and trig function substitutions, you would not understand how the SQRT(3) factor is calculated, even if a detailed derivation is presented. Do you understand the basic calculation of RMS current for a sine wave current, that is, if i = a sin(wt), then the RMS current is I(RMS) = a/SQRT(2)? You need to do this same calculation for the C1 line where the current is i =a sin(wt) +a sin(wt+2*pi/3). The RMS current for this superposition of two currents can be calculated to be a*SQRT(3/2). The ratio of the C1:C2 RMS currents is a*SQRT(3/2) / a/SQRT(2) = SQRT(3).
            Dr. Mike

          • Mark Szl

            Crystal clear Mike and anyone can Google the same for this balanced 3 phase delta circuit.

            I am really getting a bad feeling about this latest report and Ecat. What a blunder!!!

          • Dr. Mike

            Mark,
            It wouldn’t be a total blunder. I will still wait to see what the professors have to say about the power vs current discrepancy. Even if the COP is determined not to be what was originally claimed, the “ash” analysis results alone are enough to show that nuclear reactions were taking place. Did you read Andrea S.’s report given in the link in 5 or so comments below? If you haven’t, please read it. I was really surprised there might also be an error in the COP calculation of the report from last year.
            Dr. Mike

          • Mark Szl

            No i am rather new to reading these reports and never seen the last one or even fully read this one. Just waiting to see what people say before looking deeper. My EE training is to many decades ago and i am to busy to brush up on it again. Just being lazy.

            In any case, it sounds like you believe some LENR is happening but this device had a significantly lower COP than 3.

            Interesting to find out the source of the error because it is being repeated unknowning … we hope.

          • Obvious

            Dr. Mike,
            When you asked yourself, as I suggested, “Why cannot I add the three
            19.7 A RMS values from each corner of the delta together?”, what was the answer?
            Why then does dividing 19.7 by anything other than 3, then adding them up, not make any sense?
            Everything else gives the wrong answer. That’s why.
            No matter how you slice up the 19.7 A for IL, if you cannot arrive back to the same IL by summing the results of the pieces, the math is wrong. Period.
            It cannot be more, it cannot be less. You cannot add a factor or quotient to only one side of the equation and get a balanced equation. Period. You must perform the equivalent operation to both sides of an equation in order for it to balance.
            This should be obvious.
            The RMS current value for IL is the sum of both positive and negative currents. It is a magnitude twice as high as the average negative and positive sides. It does not care whether one side of a delta is doing something more than another. It just chops the sign off and adds it to the total, then divides it by the number of parts. Even if all 19.7 A came from one opposite corner, and none came from anywhere else, the RMS answer is 19.7. If one side was -19.7 A, and the other was +19.7 A, the RMS answer is still 19.7 A. The total of the parts MUST add back up to 19.7. The RMS of all three 19.7 A corners is still 19.7 A.
            All the corners MUST add up to 19.7 A at any one time, RMS, if the circuit is balanced. No matter what they are doing at any one point in time, because RMS averages over time. They are the same. They are equal.
            Adding them is like adding all the V’s measured across a bunch of parallel resistors. The sum of 1000 Volt measurements across 1000 parallel resistors is not 1000 times the voltage found at one, or any one, or any combination of V measurements. It doesn’t even matter what the 1000 resistors’ resistances are. They could be anything. V is V (in this case).

            Just because sqrt(3) is a fancy number does not make the RMS IL current bigger. EVER.
            Dividing IL by sqrt(3) and adding up three of these pieces, (each greater than 1/3) makes IL bigger, because the parts add up to more than the original value. This is wrong. Period. It is an unbalanced equation.
            Non sequitur. It does not follow. One side of the operation MUST equal the other. Or it is just nonsense with numbers attached.
            3(19.7)/sqrt(3) = 3(19.7)/sqrt(3)———–YES
            .
            19.7 = 3(19.7/sqrt(3))————————-NO
            19.7 =/= 34.121
            Square the right side of this lopsided equation, and bingo-bango
            the power is three times higher!!!!, and then because R gets unequal treatment, it looks 3 times smaller!!!! (relative to power)

            Work out where it is going wrong.
            I have explained the reason several times.

          • Mark Szl

            Static Analysis Of Power Systems:

            https://www.kth.se/polopoly_fs/1.353002!/Menu/general/column-content/attachment/comp_stat.pdf

            Looks like chapter 3 is the relevant chapter. How that relates to Joule heating I really don’t know.

          • Dr. Mike

            Mark,
            This book doesn’t specifically discuss 3-phase delta power systems, but you could use the equations in the book for calculating RMS current to calculate the RMS current ratio of the line current to the phase current and find it to be SQRT(3). This relates to Joule heating because the Joule heating in the wire is simply the wire resistance times the RMS current squared. If you will look at the Table Obvious has prepared, you will see that his calculated values for the Cu wire Joule heating (column H) based on the reported input powers do not match the Joule heating numbers in Table 7 of the Lugano report.

            Dr. Mike

          • Mark Szl

            Sure and if you just Google 3-phase delta power systems then you will find stuff everywhere. Including SQRT(3). Thanks Mike.

          • Obvious

            Please examine the following: (scroll down to the delta).

            http://elearning.vtu.ac.in/e-con/EEE/html/0054.htm

          • Dr. Mike

            Obvious,
            Thank you for directing me to the article. Now i would like for you to go back to this article and read the very last equation given:

            I(phase) = I(line) / SQRT(3)

            Your own referenced article not only gives you the result I have been trying to explain to you, it gives a fairly simple derivation that doesn’t involve calculating the RMS C1 currents from first principles.
            Dr. Mike

          • Obvious

            Yes. I understand fully about Phase current being IL/sqrt(3).
            I never doubted it.
            But they must be used “just so” or everything they modify becomes junk.
            It is 11.37 A in phase, but what is that relative to the circuit?
            We already have a measurement. 19.7.
            That Phase current is primarily for sizing wires so they don’t burn off, due to higher currents in windings, etc.
            Why would anyone want to apply that crazy stuff to an existing measurement? Where instead things are squared and rooted, and cosine, and tan-ed all over, messing with significant figures left and right and blowing precision away?
            Congratulations, you now have not one current, but no less than 10 to deal with, and must re-orient them all back to one location.
            Note that magnitude Ip = Ib = Iy = Ir, but all of the last three are 120° apart from each other.

    • Thomas Clarke

      @disqus_ZGXR2keJAu:disqus

      I’m afraid figures trump words here. Go to my post below which shows the results of a python program (together with the program source) that I wrote.

      http://www.e-catworld.com/2014/11/15/mats-lewan-testers-rule-out-inverted-clamp-hypothesis-rossi-comments-mark-e-kitiman/#comment-1698979189

      I work out, for every possible phase value of the thee line voltages:

      (1) the currents through each delta resistor (from ohms law)
      (2) the sum of each pair of these currents – which is the currents at each line.

      I then take the sum of squares of the 6 C2 delta currents, divide by the number of C2 wires (6) and sqrt.

      I then take the sum of squares of the line (C1) currents, divide by the number of C1 wires (3) and sqrt.

      I work out the ratio, which is alweays sqrt(3).

      Please go to this and tell me what is wrong, or use it to correct your own calculations. The bit that actually calculates the ratio is only about three lines long. You can download python 2.7 and run it yourself.

      BTW – the error in your argument above is that the instantaneous currents in the delta resistors are never all equal. So when you sqaure them, and then square pairwise sums of them, you get an answer different from what you expect.

      Tom

      • Obvious

        (as simple as possible, not the real formula)

        IL = Σ Ipa + Ipb + Ipc
        Ipa = IL/sqrt(3) < -120°
        Ipb = IL/sqrt(3) < 0
        Ipc = IL/sqrt(3) < +120°

        Edit: see below response to Ivan for more detail.

      • Obvious

        The real resistor value problem is a minefield of wrong turns if not considered carefully.

    • andrea.s

      Obvious, you played yourself a cheese video trick.
      Suppose the thyristor switch on the V1 line is open, whereas current flows out of V2 and into V3, as in one of the six pulses.
      There is no current on the “C1” cable on the V1.
      The current from V2 is split , 2/3 on the resistor connecting V2 and V3 , and 1/3 on the series of the other two resistors and connecting “C2” wires, going in and out of the clamp which reads zero (cheese video trick).
      This current you must root-square-sum to the ones you read during the other pulses within a period.

      • Obvious

        All of the signs are eliminated by RMS. The RMS for one line, over time, is identical to the others.
        Square, then mean, then root is the proper order.
        The diametrically opposed version of the cheese trick is 3 times current (or infinity).

        Edit: And therefore, to avoid the infinity, 3 times, or zero power cheese trick, something is wrong when the sum of three opposed sqrt(3) vectors
        DO NOT = zero.

    • Obvious

      That was full of errors….deleted.

  • Dr. Mike

    andrea s.
    Thanks for taking the time to put together this really excellent analysis. I found one typo: On page 4 in the paragraph above Figure 4, I believe “dissipating 3W” should be “dissipating 3KW”. I’m sure that all of the EE’s will understand your analysis, but I’m not sure about those with less electrical engineering background.
    I am in agreement with your analysis and probably more important am in agreement with what you say on page 10 in the section on “General remarks”.
    Dr. Mike

  • US_Citizen71

    I updated my sheet with the the c1 current, the joule heating formula and then back solved the joule heating formula for the c1 current. Doing so showed me why ivanc’s equation can’t work. In order to solve for the c1 current with just the joule heat you would need to have the c1 in the formula itself.

    https://docs.google.com/spreadsheets/d/1b_bXWnPXdsvUQT0ng1F2vATFwfQK_mXQw4nw4y4Safo/edit?usp=sharing

    • Obvious

      That crashed my Excel and also had your name on it.
      Maybe you were editing it.
      I would delete and re-upload.

  • US_Citizen71

    I updated my sheet with the the c1 current, the joule heating formula and then back solved the joule heating formula for the c1 current. Doing so showed me why ivanc’s equation can’t work. In order to solve for the c1 current(Ic1) with just the joule heat you would need to have the Ic1 in the formula itself.

    https://docs.google.com/spreadsheets/d/1b_bXWnPXdsvUQT0ng1F2vATFwfQK_mXQw4nw4y4Safo/edit?usp=sharing

    • ivanc

      Which formula can not work? could you put the formula in your post?… Please.
      You have calculate the IL and is almost half of what it should be. ( they say 40 to 50)

    • Obvious

      That crashed my Excel and also had your name on it.
      Maybe you were editing it.
      I would delete and re-upload.

    • Thomas Clarke

      us_citizen.

      The spreadsheet is now correct, assuming the heater resistance stays constant.

      Your calculated Joule power is 3.3 X lower than that given in the report for the active run.

      However ivanc’s spreadsheet is also correct. He does not assume constant resistance, but calculates resistance from the other data, which is maybe why it looks different.

      • US_Citizen71

        The large difference between actual joule heating and the reported joule heating makes me wonder if there is a variable that we are not considering and using that is the culprit. Impedance from the heating coils acting as inductors comes to mind. But, they do not use it in the sample joule heating equation but maybe they did for the active run equation? Most of my practical electrical circuit experience is DC so I would need to do a little research to answer that myself. The last time a I dealt with AC in a significant way was academically 20+ years ago.

        • ivanc

          Inductance and capacitance will add to R as they also oppose the flux of current, so no point to go that way. so will increase the impedance not reduce it.
          Good to know you finally see there is a problem with the data!

          • US_Citizen71

            I agree that there is a problem with the joule heating data and according to the power systems engineer I consulted we really shouldn’t even have that data to play with since the power is less than one kilowatt. They appear to have used 6 gauge (awg) and that gauge caused a drop of 1.5% of the power over the course of about 5 meters total from my calculations. Relating it to home use the same rate of drop over the wires leading from the pole to the average outlet in a home would be a 30% drop more or less. They should have gone with 4 or 2 gauge wires to make the drop completely insignificant. I have yet to find a reason to not believe the total power number being close other than the difference in the joule heating being subtracted, since there was two True-RMS power meters one at the wall and one after the control box. The report claims that the two were insync and the difference between the two were within 360W at all times. The 360W is the drop of the control box so at most the real life COP measure would be heat observed divided by the power coming from the wall. Which still is way over unity.

          • ivanc

            You missing the point, I agree the joule doest not have significant value.
            but let us know the other measurement they did.
            and is quiet important because is the IL that was feeding the system.
            so there is not really only one measurement (power like you think)
            is two (power and current), both must be in harmony.
            And remember that power has been contested. but luckily or unluckily depends of your point of view we have the IL to double check.

          • US_Citizen71

            And you still haven’t answered my original question. What is wrong with the measurements made by the PCE 830 connected from the wall? If you do not have an answer for that then I have to conclude this whole exchange by you is nothing but FUD. I’m sure you will ignore this again like you have the last 4 times I have asked but it is worth a fifth attempt.

          • ivanc

            where in the report is that data?
            my formers aswers have not been publised.

          • US_Citizen71
    • ivanc

      OK show the equation and state why does not work!… Please!.

      Why you make an open statement without being specific on were is the error?

      • US_Citizen71

        Ivanc it is the last column on my sheet. It won’t work because the current for c2 is part of the equation and to find Ic2 you need Ic1 unless you derive it from the provided power which is what you said you want to use the derived Ic1 to verify.

        Ic1= √((JH/3-2*Rc2*Ic2²)/Rc1))

        • ivanc

          This equation is my equation (a) , but you missing one detail,
          Ic2=Ic1/2 (This is because we reverse engineering the report)
          it should have been Ic2=Ic1/sqrt(3), but we must do the same as them to find the current they read.

          Then you have to reorder your expression, factorize Ic1 and the equation is solved. you do have two variables and two equations, so what is the problem?
          is from this equation and after factorizing Ic1 that for me is IL, that I derive my joule factor, replacing Rc1 and Rc2 values so is easy to compute 😉

          reordering:
          3 Ic1^2 Rc1 + 6 Ic2^2 Rc2 = JH
          replacing:
          3 Ic1^2 Rc1 + 6 (Ic1/2)^2 Rc2 = JH
          simplifying:

          3 Ic1^2 Rc1 + 6 Ic1^2/4 Rc2 = JH

          factorizing Ic1^2:
          Ic1^2(3 Rc1 + 6/4 Rc2) = JH
          solving:
          Ic1=sqrt( JH / (3 Rc1 + 6/4 Rc2) )
          Done.

          so the equation turns into Ic1=sqrt(JH*some_factor_I_called_Joule_factor)

  • Obvious

    Some big problem here.

    I/sqrt(3) = 3(I/sqrt(3))
    Looks ugly?, but then:

    ip = IL/sqrt(3)
    ipwr = 3ip^2*ecat_R ……………this is OK?
    ipwr = 3(IL/sqrt(3))^2*R….substitute the ip equation back in…..

    divide by 1.73, then multiply 3 times…= 1.73 times IL, squared…times R..

  • US_Citizen71

    Ivanc it is the last column on my sheet. It won’t work because the current for c2 is part of the equation and to find Ic2 you need the current for Ic1 unless you derive it from the provided power which is what you said you want to use the derived Ic1 for.

    Ic1= √((JH/3-2*Rc2*Ic2²)/Rc1))

  • Dr. Mike

    Mark,
    This book doesn’t specifically discuss 3-phase delta power systems, but you could use the equations in the book for calculating RMS current to calculate the RMS current ratio of the line current to the phase current and find it to be SQRT(3). This relates to Joule heating because the Joule heating in the wire is simply the wire resistance times the RMS current squared. If you will look at the Table Obvious has prepared, you will see that his calculated values for the Cu wire Joule heating (column H) based on the reported input powers do not match the Joule heating numbers in Table 7 of the Lugano report.

    Dr. Mike

    • Obvious

      Please examine the following: (scroll down to the delta).

      http://elearning.vtu.ac.in/e-con/EEE/html/0054.htm

      • Dr. Mike

        Obvious,
        Thank you for directing me to the article. Now i would like for you to go back to this article and read the very last equation given:

        I(phase) = I(line) / SQRT(3)

        Your own referenced article not only gives you the result I have been trying to explain to you, it gives a fairly simple derivation that doesn’t involve calculating the RMS C1 currents from first principles.
        Dr. Mike

        • Obvious

          Yes. I understand fully about Phase current being IL/sqrt(3).
          I never doubted it.
          But they must be used “just so” or everything they modify becomes junk.
          It is 11.37 A in phase, but what is that relative to the circuit?
          We already have a measurement. 19.7.
          That Phase current is primarily for sizing wires so they don’t burn off, due to higher currents in windings, etc.
          Why would anyone want to apply that crazy stuff to an existing measurement? Where instead things are squared and rooted, and cosine, and tan-ed all over, messing with significant figures left and right and blowing precision away?
          Congratulations, you now have not one current, but no less than 10 to deal with, and must re-orient them all back to one location.

  • Obvious

    I was wondering if OL would show up on the empty line info spaces. My though experiment seemed to suggest that I# would still report a current value, though. Without a PCE-830 to work with, or some handy three phase resistive load, its just theory I suppose.

  • Obvious

    That’s right, Ivan.

    We have no data for vectors. But sqrt(3) describes a vector. You must understand why Ip is sqrt(3) if IL.

    It is not just a pretty handy number to make phase current.

    It is an important number.

    RMS wipes out vector values, because squaring numbers removes sign, and turns them into magnitudes.

    If you cannot add together all three 19.7 A from the three corners, what in the world makes you think it is OK to divide them by 1.73 and then it is OK to add 3 of them up?

    Using sqrt(3) puts the sign back in, and turns the magnitude back into a vector.
    You use RMS with vectors all you want. But once you put the vector in a number you have a real mess getting the vectors back out and still have a useable number, unless you do vector math which includes:

    (as simple as possible)

    IL = Σ Ipa + Ipb + Ipc
    Ipa = IL/sqrt(3) < -120°
    Ipb = IL/sqrt(3) < 0
    Ipc = IL/sqrt(3) < +120°

  • Obvious

    (as simple as possible)

    IL = Σ Ipa + Ipb + Ipc
    Ipa = IL/sqrt(3) < -120°
    Ipb = IL/sqrt(3) < 0
    Ipc = IL/sqrt(3) < +120°

  • US_Citizen71

    The difference makes me wonder if there is a variable that we are not considering and using that is the culprit. Impedance from the heating coils acting as inductors comes to mind. But, they do not use it in the sample joule heating equation but maybe they did for the active run equation? Most of my practical electrical circuit experience is DC so I would need to do a little research to answer that myself. The last time a I dealt with AC in a significant way was academically 20+ years ago.

  • ivanc

    The equation of Us_citizenxx

    Ic1= √((JH/3-2*Rc2*Ic2²)/Rc1))

    Is my equation (a) , but you missing one detail,
    Ic2=Ic1/2 (This is because we reverse engineering the report)
    it should have been Ic2=Ic1/sqrt(3), but we must do the same as them to find the current they read.

    Then
    you have to reorder your expression, factorize Ic1 and the equation is
    solved. you do have two variables and two equations, so what is the
    problem?
    is from this equation and after factorizing Ic1 that for me
    is IL, that I derive my joule factor, replacing Rc1 and Rc2 values so is
    easy to compute 😉

    reordering:
    3 Ic1^2 Rc1 + 6 Ic2^2 Rc2 = JH
    replacing:
    3 Ic1^2 Rc1 + 6 (Ic1/2)^2 Rc2 = JH
    simplifying:

    3 Ic1^2 Rc1 + 6 Ic1^2/4 Rc2 = JH

    factorizing Ic1^2:
    Ic1^2(3 Rc1 + 6/4 Rc2) = JH
    solving:
    Ic1=sqrt( JH / (3 Rc1 + 6/4 Rc2) )
    Done.

    so the equation turns into Ic1=sqrt(JH*some_factor_I_called_Joule_factor)

  • Obvious

    Nonsense.
    Sqrt(3) is a vector. 1/3 of at least three of them, each with sign, to order to complete a delta equation correctly.
    Sqrt(3) needs an origin.
    19.7 A RMS is a magnitude. It has no direction.
    Repeat it until you believe it, or better, understand it.
    Or please explain how
    ….IL = 3(IL/sqrt(3)
    19.7 = 34.12..!!!!…………where does this re-balance?
    ( I posted a more frustrated answer to Dr. Mike. I hope he doesn’t take it personally.)

    • Dr. Mike

      Obvious,
      Please read the very last equation in your second reference!
      Dr. Mike

  • Obvious

    LOL
    Check this:
    Use the Lugano Theorem to derive the current for the active run….(fight over this later).
    P = IL^2 = (1/2IL)^2
    therefore: IL = sqrt(Lugano Input W/3*2)…….this returns the current value required by the Theorem
    I Lugano T = sqrt(LWi/3*2)
    This gives a decent range of values, but leaves the dummy run a tiny bit short. The Joule heat is involved here I’m sure…..anyways
    Now, test the Lugano Theorem values against P = I^2R…..This will be effective R
    1.1234 not so hot……0.712…not so hot…..0.411…nope….1.854……hmmmm….1.5….whoa.
    Not so shabby at all.
    The values are the natural distributions for the P formula in a delta, unforced, except a sliding R, which is identical for all readings.
    Yes it is derived from the Watt values given.
    These are measurements which trump all second-hand derivations.
    This means, if nothing else, a linear approximation is achievable with the values given.
    The dummy heat low A value derived does deserve some attention.
    This does possibly indicate something different here is happening.
    The difference, it turns out, is that the R is 1.234 in the dummy.***This is the total inductance***
    And the Active Run it is 1.5. INCREASING.
    Run……Report W………………Lugano Theorem Amps…………P = ILT^2R…………..R
    1………………815.16………………………….23.32……………………………….815.16…………..1.5
    5………………785.79………………………….22.89……………………………….785.89…………..1.5
    6………………923.71………………………….24.82……………………………….923.71…………..1.5
    7………………906.31………………………….24.58……………………………….906.31…………..1.5
    dummy……..486………………………………19.7 (report)……………………….486………………1.23 calc
    dummy……..486………………………………18.00 Lugano Theorem ……..486……………….1.5

    The spreadsheet gang here can whip the whole works out in a few minutes I’m sure.

    • Dr. Mike

      Obvious,
      Since there is no such thing as “Lugano Theorem Amps” your calculations are meaningless. Ivanc has correctly calculated the RMS phase currents (C2 currents) from both the reported input powers and from the reported Cu wire Joule heating values.
      Dr. Mike

      • Obvious

        Then why do I now have a consistent simulation (newer than posted) that uses only the report data, other than a slightly high R that now predicts all active run values within less than 0.5W, and a dummy run within 22.4 Watts (high), which is within the error range?
        And everyone else has 3X the reported values and a floating R?

        • Dr. Mike

          Obvious,
          I didn’t say you opinion is meaningless, I said your calculations are meaningless. You can’t invent equations that are irrelevant to science to fit data and claim you have a good equation. Thomas Clarke has explained this to you and I have explained this to you.
          Dr. Mike

          • Obvious

            The framework of my Theorem is entirely based on a consistent description of current flow and distribution, which defies no science whatsoever. It is not a random correlation. It is directly tied to the inevitable and perfect geometry of the delta configuration, balanced three phase power, and resistance. All of which are in this case guaranteed by the integrated average of current, and the fact that resistance does not change much for most materials.
            Importantly, it respects that both P=IV and P=I^2R must be true at the same time.

          • Obvious

            Sorry, had an = instead of + above. That does make it look strange.
            Tell you what. I’ll work on the idea some more, prove it out better, and write a paper if the idea doesn’t totally fall apart after some more work. Then you can review it and see if you can make head from tails of it.

          • Dr. Mike

            Obvious,
            As I explained to you in a previous lengthy reply, Your equation:

            P = IL^2 + (1/2IL)^2 can not be correct because the units om each side of the equation do not match, that is, Watts do not equal Amps squared.
            Dr. Mike

          • Obvious

            Nobody seems to think it is strange to convert to phase current, then immediately turn it back into Line Current again.
            Or invent phase voltage, when we know it is already Line voltage, but for some reason it is OK to make it 10 Volts less when it is convenient.
            Or add vectors as if they were magnitudes in the same direction as line current.
            I’d be more worried about that.

          • Obvious

            I think I have the real answer this time.
            I had an error in my equations that originally appeared to show a perfect fit to the data. (This was the version I was promissing a day or two ago.)
            Then I fixed that during proofreading, and then all the nonsense started up again.
            Then just on a hunch, I put the error back in, and everything seems to work again. (I’m still beta testing the error version again. I wish I kept the original instead of “fixing” it and saving over it….)
            I think this idea duplicates the professors’ error that makes a mess of Joule heat slope and makes the fake negative R happen.
            You still need to deconstruct the bad data to get the active run current. I ran a back-check on that, and was able to extract almost exactly the combined resistance of a set of three wires, namely a C1 and two parallel C2’s (correct to 5 decimal places).

  • Obvious

    LOL
    Check this:
    Use the Lugano Theorem to derive the current for the active run….(fight over this later).
    P = IL^2 + (1/2IL)^2
    therefore: IL = sqrt(Lugano Input W/3*2)…….this returns the current value required by the Theorem
    I Lugano T = sqrt(LWi/3*2)
    This gives a decent range of values, but leaves the dummy run a tiny bit short. The Joule heat is involved here I’m sure…..anyways
    Now, test the Lugano Theorem values against P = I^2R…..This will be effective R
    1.1234 not so hot……0.712…not so hot…..0.411…nope….1.854……hmmmm….1.5….whoa.
    Not so shabby at all.
    The values are the natural distributions for the P formula in a delta, unforced, except a sliding R, which is identical for all readings.
    Yes it is derived from the Watt values given.
    These are measurements which trump all second-hand derivations.
    This means, if nothing else, a linear approximation is achievable with the values given.
    The dummy heat low A value derived does deserve some attention.
    This does possibly indicate something different here is happening.
    One difference, it turns out, is that the R is 1.234 in the dummy.***This is the total impedance***
    And the Active Run it is 1.5. INCREASING. (Some adjustments still not included. No warranty.)
    The high dummy W test is interesting.
    Run……Report W………………Lugano Theorem Amps………………..P = ILT^2R…………..R
    1……………….815.16………………………….23.32……………………………….815.16…………..1.5
    5……………….785.79………………………….22.89……………………………….785.89…………..1.5
    6……………….923.71………………………….24.82……………………………….923.71…………..1.5
    7……………….906.31………………………….24.58……………………………….906.31…………..1.5
    dummy…R….486………………………………19.7 (report)………………………486………………1.23 calc
    dummy……….486………………………………18.00 Lugano Theorem ……..486………………1.5
    dummy.diff1..582..forced value to getA..19.70 ……………………………….582………………1.5
    The spreadsheet gang here can whip the whole works out in a few minutes I’m sure.

    • Dr. Mike

      Obvious,
      Since there is no such thing as “Lugano Theorem Amps” your calculations are meaningless. Ivanc has correctly calculated the RMS phase currents (C2 currents) from both the reported input powers and from the reported Cu wire Joule heating values.
      Dr. Mike

      • Obvious

        Then why do I now have a consistent simulation (newer than posted) that uses only the report data, other than a slightly high R that now predicts all active run values within less than 0.5W, and a dummy run within 22.4 Watts (high), which is within the error range?
        And everyone else has 3X the reported values and a floating R?

        Edit: As a good scientist, I have posted the method. Feel free to try it.
        Your opinion is that it is meaningless. That has nothing to do with whether it does or does not actually work.

        The Lugano Theorem can extract the Joule heat amps also, just as easily. These fix the larger dummy error (above) quite nicely, without strange values.
        If it is nonsense, why does it work?

        • Dr. Mike

          Obvious,
          I didn’t say you opinion is meaningless, I said your calculations are meaningless. You can’t invent equations that are irrelevant to science to fit data and claim you have a good equation. Thomas Clarke has explained this to you and I have explained this to you.
          Dr. Mike

          • Obvious

            The framework of my Theorem is entirely based on a consistent description of current flow and distribution, which defies no science whatsoever. It is not a random correlation. It is directly tied to the inevitable and perfect geometry of the delta configuration, balanced three phase power, and resistance. All of which are in this case guaranteed by the integrated average of current, and the fact that resistance does not change much for most materials.
            Importantly, it respects that both P=IV and P=I^2R must be true at the same time.

          • Obvious

            Sorry, had an = instead of + above. That does make it look strange.
            Tell you what. I’ll work on the idea some more, prove it out better, and write a paper if the idea doesn’t totally fall apart after some more work. Then you can review it and see if you can make head from tails of it.

            Edit: Also had “inductance” where I meant “impedance”.

          • Dr. Mike

            Obvious,
            As I explained to you in a previous lengthy reply, Your equation:

            P = IL^2 + (1/2IL)^2 can not be correct because the units om each side of the equation do not match, that is, Watts do not equal Amps squared.
            Dr. Mike

          • Obvious

            Nobody seems to think it is strange to convert to phase current, then immediately turn it back into Line Current again.
            Or invent phase voltage, when we know it is already Line voltage, but for some reason it is OK to make it 10 Volts less when it is convenient.
            Or add vectors as if they were magnitudes in the same direction as line current.
            I’d be more worried about that.

  • Obvious

    If I am not totally messed up, Ivan, you have a negative COP of 1.4 in you Joule heat calculations using phase [~ rows 92-110 in my version of yours] due to low phase voltage, which should be the same as line voltage. (I used report P/IL you calculated by IL = Ip*sqrt(3)) which should be fine).
    And possibly a Voltage COP of 3.26 in your main sheet of estimates.
    You know your spreadsheet better.
    Divide your estimated W by known IL and projected Line and Phase currents to get a line and Phase V for each P in the sheet.
    Divide the report W and projected W by known IL and/or any projected IL.
    Recalculate P by VI. Compare. Double-check the Phase Volts. They should be the same as Line Volts .Find any factor of V to V and P to P where there are differences.
    Just poking around with the range of minimum power to maximum power in the active run I found huge W increases developed from very little incremental IV in your sheet.
    Between runs 5 and 6 I had 4.732 V (extrapolated from your figures, I could be wrong) increase and 3.82 A increase which is only 18 W, while there is 449 W increase in estimated power.

  • Obvious

    If I am not totally messed up, Ivan, you have a negative COP of 1.4 in your Joule heat calculations using phase [~ rows 92-110 in my version of yours] due to low phase voltage, which should be the same as line voltage. (I used report P/IL you calculated by IL = Ip*sqrt(3)) which should be fine).
    And possibly a Voltage COP of 3.26 in your main sheet of estimates.
    You know your spreadsheet better.
    Divide your estimated W by known IL and projected Line and Phase currents to get a line and Phase V for each P in the sheet.
    Divide the report W and projected W by known IL and/or any projected IL.
    Recalculate P by VI. Compare. Double-check the Phase Volts. They should be the same as Line Volts .Find any factor of V to V and P to P where there are differences.
    Just poking around with the range of minimum power to maximum power in the active run I found huge W increases developed from very little incremental IV in your sheet.
    Between runs 5 and 6 I had 4.732 V (extrapolated from your figures, I could be wrong) increase and 3.82 A increase which is only 18 W, while there is 449 W increase in estimated power.

    • ivanc

      ja , ja , ja, ja,,,,,:) , yes you are!!!!!!

      • Obvious

        OK. …
        On your Joule heat calculations, divide “Power In” by “I in”. Now you have LV.
        Now look at your Vph number.
        Explain the 10 V difference for the dummy. I did all it the way down the group. 1.4041304 (factor) difference LV to Vph, top to bottom.
        Vph = VL in a real delta

        • ivanc

          I do not understand your question, you assuming Vph constant during the tests?

          • Obvious

            That’s not what I meant. Obviously V changes from run to run.

            What I mean is that your ratio of VL to Vp is not 1:1, which it should be. Whatever the error is there, it makes Vp lower than VL, (ie: VL/Vp =~1.4) which it cannot be in a delta. The error is the same in all the runs and the dummy, on your sheet.

          • ivanc

            vl=2 vL1+ 2 vcu_ph + vresistence_ecat, there will be small difference because the wires are having a small drop in voltage.

          • Obvious

            Nearly a 50% drop?

          • ivanc

            No way, your calculation is wrong the difference is really small, because the resistance in the cables is much lower than the resistance of the ecat resistor
            Your methods are so far from reality than I can not even give you a feedback as I said before it gave me a headache.
            Get a 3phase circuit and do the measurements yourself, your theories and drawings have not theoretical support, in electricity all theory is supported by lab experiments.

          • Obvious

            I divided 485 by 19.7 to get volts, and compared to your table.
            P = IV
            V = P/I
            Supported by millions of experiments and billions of devices.

          • ivanc

            This is why is wrong: you using the total power.
            but each phase is only 1/3 of the total power.
            and the feeding cables see IL , and the ones in the delta itself see IL/sqrt(3)
            🙂 but good for keep trying.!!!!

          • Obvious

            OK.
            Then we are on the same track now, sort of, regarding V.
            I ended up with the lower voltage as a requirement late yesterday after I caught an error in my stuff. The math wouldn’t hold together with the higher voltage. I get the IL/sqrt(3) for phase idea.

          • ivanc

            Good on you.!!!

  • US_Citizen71

    Until a reasonable scientific answer is given by joule heating crowd as to why the PCE 830 attached to line running from the did not register the 3X+ power they claim but a value within 360 watts of the one attached between the control box and the reactor their comments and theories should be viewed as nothing but disinformation. The unit attached to the line in from the wall is the double check on the system. Everything else is part of an experimental brand new technology and there is no published rules and laws for the workings of cold fusion reactor setup. The values we have been playing with give us some insight into the workings of the reactor setup but they only begin to scratch the surface of the truth of how it operates.

    • Dr. Mike

      Us_Citizen71,
      Where in the report is the data for the PCE-830 that is measuring the power before the control box, that is, PCE 830 A in Figure 4 on page 5? I have read over the report several times and have apparently missed the Table showing this data.
      Dr. Mike

      • US_Citizen71

        If you reread the begining of my comment from 5 days ago just after we stared down this rabbit hole I cover the 360 watts of the control box and the PCE 830 units.

        http://www.e-catworld.com/2014/11/15/mats-lewan-testers-rule-out-inverted-clamp-hypothesis-rossi-comments-mark-e-kitiman/#comment-1696526975

        Since there is no table of data the conservative assumption is that the control box pulls 360 watts at all times.

        • Obvious

          The Compact Fusion Operator’s Manual suggests 1.3 W of dissipation per amp of load current.
          The Compact Fusion Maintenance Manual has a range, depending on options, but peak “typical” suggested is 67.2 W at 50A load. Minimum is 23.4 W.
          I don’t know if this test qualifies as typical….

    • Obvious

      Since we have one data point for the dummy, and 16 for the active run, shouldn’t then the active run W be the reference point? Put in your favorite R and make a line. See what it takes to make it to the dummy.

  • US_Citizen71

    Until a reasonable scientific answer is given by joule heating crowd as to why the PCE 830 attached to line running from the wall did not register the 3X+ power they claim but a value within 360 watts of the one attached between the control box and the reactor their comments and theories should be viewed as nothing but disinformation. The unit attached to the line in from the wall is the double check on the system. Everything else is part of an experimental brand new technology and there is no published rules and laws for the workings of cold fusion reactor setup. The values we have been playing with give us some insight into the workings of the reactor setup but they only begin to scratch the surface of the truth of how it operates.

    • Dr. Mike

      Us_Citizen71,
      Where in the report is the data for the PCE-830 that is measuring the power before the control box, that is, PCE 830 A in Figure 4 on page 5? I have read over the report several times and have apparently missed the Table showing this data.
      Dr. Mike

      • US_Citizen71

        If you reread the begining of my comment from 5 days ago just after we stared down this rabbit hole I cover the 360 watts of the control box and the PCE 830 units.

        http://www.e-catworld.com/2014/11/15/mats-lewan-testers-rule-out-inverted-clamp-hypothesis-rossi-comments-mark-e-kitiman/#comment-1696526975

        Since there is no table of data the conservative assumption is that the control box pulls 360 watts at all times.

        • Obvious

          The Compact Fusion Operator’s Manual suggests 1.3 W of dissipation per amp of load current.
          The Compact Fusion Maintenance Manual has a range, depending on options, but peak “typical” suggested is 67.2 W at 50A load. Minimum is 23.4 W.
          I don’t know if this test qualifies as typical….

    • Obvious

      Since we have one data point for the dummy, and 16 for the active run, shouldn’t then the active run W be the reference point? Put in your favorite R and make a line. See what it takes to make it to the dummy.

      • Thomas Clarke

        That is an interesting speculation. However we then have a problem – the dummy run would have a COP of 3!!

        the authors carefully validated power for the dummy run and found that it matched.

        • Obvious

          Worse yet, I tried that last night.
          The intercept is so bad I would normally throw the data point out.
          There would have to a serious curve between the dummy and the active run to fit that.
          One attempt that seemed to fix it pretty well is if the dummy W is times two.

          Edit:LOL.
          Or Active run W/2 for a COP of ~6.5?
          I’ll fiddle with the intercepts for real some time in the next couple of days once I clean up data and headings.

  • Thomas Clarke

    US_Citizen,

    Without further info from the authors of the report we only know what they have given us. They state that the two PCE-830s were used to check the control box but not that this was done during the active test. Their general approach has been to test all the issues on the dummy test and assume all remains the same for the active test so it would be entirely consistent (though not what I’d like) for no checking to be done for the active test.

    Even if they did check the PCE-830 then whatever mistake altered the power from one could be repeated to alter the power from the other.

    There are two obvious options:
    reversed clamp
    line power taken as total power

    I agree that the authors could resolve this matter by checking their stored data and adding to the report. We must hope they will do this. Until they do you cannot make assumptions. Or rather you can, but you would be logically wrong to do so!

    • US_Citizen71

      We know it was attached during the active test. Figure 4 of the report is the wiring diagram of the setup and it shows both PCE 830 units. The description below the figure and the quote below stated this.

      ‘Figure 4 details the electrical connections of all elements of the experimental setup. The two PCEs were inserted one upstream and one downstream of the control unit: the first allowed us to measure the current, voltage and power supplied to the system by the power mains; the second measured these same quantities as input to the reactor. Readings were consistent, showing the same current waveform; furthermore, they enabled us to measure the power consumption of the control system, which, at full capacity, was seen to be the same as the nominal value declared by the manufacturer.” – Page 5 Lugano Report

      edit:
      “There are two obvious options:
      reversed clamp
      line power taken as total power”

      reversed clamp – This requires both PCE 830 units to have a reversed clamp at the very least, there may be more requirements as I haven’t done a circuit analysis of the setup that way. Both being reversed is very unlikely, this theory is more innuendo than science. As I stated in my comment 5+ days ago they showed they could attach the units correctly to the measure the mains power twice. There is no reason to assume they couldn’t do it X number of times again, but they likely didn’t need to setup to measure the mains power more than twice. Once the 2nd unit was setup and measured the wall power and they determined that it was Industrial 3 phase power and the values agreed with the first unit, there wouldn’t be any reason to disconnect it until disassembly of the setup at the end of the test.

      line power taken as total power – This is nothing but innuendo and FUD.

      • Thomas Clarke

        Yes, but we do not know it was checked, or even present, in the active test. The “full capacity” test was not the active test – which was certainly not at full capacity, but something different done (I guess) during the dummy testing for a very short amount of time to avoid destruction of the device. Always good not to make assumptions, but as I say even if it was checked it could suffer the same error (maybe naturally would) as the other one.

        • US_Citizen71

          “Figure 4 details the electrical connections of all elements of the experimental setup.” – Denial isn’t just a river in Egypt.

          ” but as I say even if it was checked it could suffer the same error (maybe naturally would) as the other one.” – Without the how that is simply speculation, deflection and denial, not supported by any data.

          You can’t will the PCE 830 on the line in from the mains out of existence no matter how hard you try and want it to go away. If it doesn’t fit into your theory then the theory is wrong.

          • Dr. Mike

            US_Citizen71,
            Do you really believe “Figure 4 details the electrical connections of all elements of the experimental setup”? What about the 4th sentence in the Introduction on page 1: “In addition, the resistor coils are fed with specific electromagnetic pulses”? The gray box with the potentiometer on top in the Figure 3 set-up appears to be the controller. What is the brown box? Where is it in the Figure 4 electrical diagram? If the brown box is a pulse generator, is it properly connected to the rest of the circuit? I believe Rossi has every right to keep proprietary the nature of those “specific electromagnetic pulses” fed to the resistor coils; however, the wiring diagram is incomplete without including the brown box.
            There are at least two things missing from the report that could have a significant bearing on the results. First the authors fail to state if the “specific electromagnetic pulses” were also added in the dummy run. Second, if additional pulses were fed to the coil, we know that some additional energy was added in those pulses. If the nature of those pulses are proprietary, the authors should have both stated this fact and included an upper limit for the amount of power they added to the input power.
            Dr. Mike

          • US_Citizen71

            Yes I believe the figure shows all connections the testers made. The boxes you refer to are part of the controls(control box). If it was all put in a shell it would be one box and still have interconnections inside. An IC has all kinds of wires to interconnect the parts onboard do we view it as millions of parts or as one?

          • Thomas Clarke

            I’m not willing it out of existence. And if you read what I have written you will see that my hypothesis here does not depend on the matter, so the criticism is unjustified.

            The issue is which of the data that was available they actually looked at during the active test.

            I’m unwilling to speculate – we know the anomalies indicate that they checked things less than would, in retrospect, be advisable. We do not know what they checked, except where this is explicitly mentioned in the report.

      • Thomas Clarke

        No – we cannot know that the second PCE-830 was checked on the active run. Why would they? I, for example, wouyld have required a control at the same temperature as the active run, and would have wanted all teh power measurements and checked for consistency.

        But I am not them.

        • US_Citizen71

          Why would they check the unit they installed to double check the figures of the test? Do you really need an answer to that?

          They likely did collect more data than we have seen, including measurements of every last thing that could be measured. The test wasn’t done for publication, it was done for Elforsk you know the people that paid them to do the test and published the results on their own web server. “Readings were consistent, showing the same current waveform; furthermore, they enabled us to measure the power consumption of the control system, which, at full capacity, was seen to be the same as the nominal value declared by the manufacturer” The Dummy run was the lowest power run so how could it be at full capacity?

          • Dr. Mike

            US_Citizen71,
            In your statement: “They likely did collect more data than we have seen, including measurements of every last thing that could be measured”, I would have to agree with you that the professors probably collected a lot more data than we’ve seen. In fact, if they would reveal some of that data the electrical engineers that have reviewed the report could understand the discrepancies in the data that was presented in the report. I certainly disagree with you that they measured every last thing. For one they did not measure the room temperature resistance of the heater coils. This should have been done to verify that measured powers and currents were in agreement. The second thing the professors failed to measure is the C2 currents. If they had measured a single C2 current, they would have found that the RMS current in the C2 lines was not 1/2 of the RMS current in the C1 lines.
            Dr. Mike

          • US_Citizen71

            I believe that there was a partial NDA that covered the inner working of the reactor and controls. Values that would help reverse engineer the device were banned from publication. Values that show the input power and heat output were free to use. This might explain the joule heating as well. The value for joule heating was the only value rounded in a way that didn’t move the COP lower. They rounded 6.7 to 7 the real number was 7.27. This allows them to end up with the same number they would have rounded to if they were being conservative but not reveal the true value per my proposed NDA.

          • Thomas Clarke

            Neither run was at full capacity – the testers tell us they did not want to do that. However they could turn the potentiometer up full while testing before or after the dummy run – probably given their concerns for only 1s or so.

          • US_Citizen71

            The same could have been done on an active run. The full capacity may also to refer to the highest setting they used. It may in truth always pull 360 watts. We do not know the values for it but it doesn’t matter both PCE 830 units were attached during all runs. They comment that doing so allowed them to evaluate the power of the control box so they would have seen a difference of 3X plus. They observed this setup for 32 days and you don’t think they looked at every screen and instrument many times. It is heater it is like watching grass grow.

  • Obvious

    Worse yet, I tried that last night.
    The intercept is so bad I would normally throw the data point out.
    There would have to a serious curve between the dummy and the active run to fit that.
    One attempt that seemed to fix it pretty well is if the dummy W is times two.

  • US_Citizen71

    We know it was attached during the active test. Figure 4 of the report is the wiring diagram of the setup and it shows both PCE 830 units. The description below the figure and the quote below stated this.

    ‘Figure 4 details the electrical connections of all elements of the experimental setup. The two PCEs were inserted one upstream and one downstream of the control unit: the first allowed us to measure the current, voltage and power supplied to the system by the power mains; the second measured these same quantities as input to the reactor. Readings were consistent, showing the same current waveform; furthermore, they enabled us to measure the power consumption of the control system, which, at full capacity, was seen to be the same as the nominal value declared by the manufacturer.” – Page 5 Lugano Report

  • US_Citizen71

    They likely did collect more data than we have seen, including measurements of every last thing that could be measured. The test wasn’t done for publication, it was done for Elforsk you know the people that paid them to do the test and published the results on their own web server. “Readings were consistent, showing the same current waveform; furthermore, they enabled us to measure the power consumption of the control system, which, at full capacity, was seen to be the same as the nominal value declared by the manufacturer” The Dummy run was the lowest power run so how could it be at full capacity?

    • Dr. Mike

      US_Citizen71,
      In your statement: “They likely did collect more data than we have seen, including measurements of every last thing that could be measured”, I would have to agree with you that the professors probably collected a lot more data than we’ve seen. In fact, if they would reveal some of that data the electrical engineers that have reviewed the report could understand the discrepancies in the data that was presented in the report. I certainly disagree with you that they measured every last thing. For one they did not measure the room temperature resistance of the heater coils. This should have been done to verify that measured powers and currents were in agreement. The second thing the professors failed to measure is the C2 currents. If they had measured a single C2 current, they would have found that the RMS current in the C2 lines was not 1/2 of the RMS current in the C1 lines.
      Dr. Mike

      • US_Citizen71

        I believe that there was a partial NDA that covered the inner working of the reactor and controls. Values that would help reverse engineer the device were banned from publication. Values that show the input power and heat output were free to use. This might explain the joule heating as well. The value for joule heating was the only value rounded in a way that didn’t move the COP lower. They rounded 6.7 to 7 the real number was 7.27. This allows them to end up with the same number they would have rounded to if they were being conservative but not reveal the true value per my proposed NDA.

  • Andreas Moraitis

    I am not sure if the electrical setup is really a standard three-phase system. The tree coils might have different functions. For example, two of them could be used primarily for heating, the third one (which might be the “mouse”) could mainly provide the “electromagnetic pulses”. This would require different voltages, currents, and waveforms. Perhaps some of the coils are using pulsed DC instead of AC. It is even possible that the function of the coils changes in the course of the operation. We have neither enough data nor do we know anything about the inner workings of the gray (but epistemologically ‘black’) control box. Therefore, all conjectures which are based on the assumption that the three “phases” are equally ranked could be wrong.

    • Andreas Moraitis

      After looking again at the wiring diagram I see that it is more complicated than I have initially assumed, since each C1 line is connected to two other C1 lines via two coils. But anyway there might be a chance that the control box feeds the three output lines in a different way.

  • Andreas Moraitis

    I am not sure if the electrical setup is really a standard three-phase system. The tree coils might have different functions. For example, two of them could be used primarily for heating, the third one (which might be the “mouse”) could mainly provide the “electromagnetic pulses”. This would require different voltages, currents, and waveforms. Perhaps some of the coils are using pulsed DC instead of AC. It is even possible that the function of the coils changes in the course of the operation. We have neither enough data nor do we know anything about the inner workings of the gray (but epistemologically ‘black’) control box. Therefore, all conjectures which are based on the assumption that the three “phases” are equally ranked could be wrong.

    • Andreas Moraitis

      After looking again at the wiring diagram I see that it is more complicated than I have initially assumed, since each C1 line is connected to two other C1 lines via two coils. But anyway there might be a chance that the control box feeds the three output lines in a different way.

  • Thomas Clarke

    Andreas,

    The setup is a bog standard delta connection, in which each of the three lines connects to two of the load resistors.

    Giancarlo – who knows the Fusion controller that makes the standard, no weird electromagnetic pulses, waveforms – says from teh shown waveform that they show a two-on one-off switching pattern. In every cycle each of the three load resistors are connected via two on triacs to two lines at different times. because of the delta connection there is some current (R is double, so I is 1/2) in the other two resistive loads.

    This switching method is not quite normal three phase, but commonly used.

    At each switch on time C1 currents are 1,1,0 (0 for the switched off line).
    The corresponding C2 currents are:

    1 -> 2/3, 1/3
    1 -> 2/3,1/3
    0 -> 1/3,1/3

    RMS comparison is thus:

    C1: sqrt[(1+1)/3]
    C2: sqrt[(4+4+1+1+1+1)/6] = sqrt(2).

    What a surprise! Still 1/sqrt(3).

    Rossi does not use non-standard SCR control, quite sensibly he is using an industry SCR controller, and driving it from a microprocessor to implement a constant temperature as measured by the thermocouple.

    • Andreas Moraitis

      You are right with regard to the configuration, I should have looked at the diagram before I posted my comment. It is not possible to control each of the coils independently from the others. Nevertheless, a change of the voltage in one of the phases would affect two coils. These coils could even be completely deactivated by switching the phase off. So it might still be possible that not all coils have been working in the same way during the test.

    • Dr. Mike

      Thomas,
      I don’t follow your RMS ratio calculation. If C1 = SQRT(2/3) and C2 = SQRT(2), then C1= C2/SQRT(3), rather than C2 x SQRT(3). I calculated the relative RMS currents as :

      C1 = k * I * SQRT(1^2 + 1^2 + 0^2) = k * I * SQRT(2)

      C2 = k * I * SQRT[ (2/3)^2 + (1/3)^2 + (1/3)^2] = k * I * SQRT(2/3)

      For this calculation C2 = C1 / SQRT(3)
      Dr. Mike

      • Thomas Clarke

        Mike,
        That is what I calculated (it may be was not clear, but it still looks OK to me going back to it). I was saying the ratio was still 1/sqrt(3) – as expected.

        Thomas

    • Obvious

      I have just finished re-doing the Jh calcs and extrapolation from scratch.
      I have total agreement with the report for Input W.
      My average extrapolated variance is:New Jh ratio Active run W – Reported W = 0.17 W
      Variation due to rounding, etc. is: dummy report – new calcs dummy = 0.09075 W

      A lot of folks are just plain doing the Phase-Line calculations wrong, including the professors. They grossly underestimated the Dummy Joule heat, and not-so-grossly overestimate the Active Run Joule heat, by not doing the 3 phase math right.
      The Active Run Input W was measured, as was the Dummy Input W.
      But the three phase resistance and power calculations for Joule heat are messed up in the report, since the Professors applied an incorrect formula for Joule heat by not dividing up the power contributions from each Phase and Line correctly.

      Ptotal = P1 + P2
      P1 = sqrt(3)Ptotal
      P2 = sqrt(3)Ptotal
      P3 = P2 = P1 = PL = Pp

      And so averagePL = 1/2(sqrt(3)Ptotal)….not sure I’m explaining this well. This was a complex derivation from the delta to Wye transformation math, using the Two Wattmeter Method I suggested two weeks ago, but I was using it wrong at the time.

      P1 = Vsub12(Isub1) cos (30° + θ) = VL + IL cos (30° + θ)
      P2 = Vsub32(Isub3) cos (30° + θ) = VL + IL cos (30° – θ)
      Ptotal = P1 + P2 = sqrt(3)(VL)(IL)
      So:
      P1 + P2 = sqrt(3)(Ptotal)
      Then
      P3 = P2 – P1 = (VL)(IL)sinθ
      P3 = Ptotal (sqrt(3)) cosθ

      P3 = (1/sqrt(3))Ptotal
      And
      Ptotal = sqrt(3)(VL)(IL)cosθ

      Ptotal = sqrt(3)(PL)

      The general idea is that 479 = P1+P2 so reverse this to get P1, then Pp
      PL= Pp = Ptotal/sqrt(3)
      P3 = P2 = P1 = PL = Pp

      (My apologies if some wye calcs got pasted in by mistake earlier, I think I got them all out now. Let me know if you see something funny, or ask if you have a problem with a part of this.)

      Ip = IL/sqrt(3)
      Vp = Pp/Ip
      Vp = VL
      Therefore Pp = 479/sqrt(3) (for the dummy)
      And P1 = 276.55 W (for the dummy)
      Now see if you can work it out.
      Be sure to derive R phase using Vp and Ip. It is not R apparent (whole device) nor One single reactor resistor.

      The Joule heat has been done wrong in the report. The correct Joule heat for the Dummy should be 20.017 W.
      The correct Joule heat for Run 6 is 36.828 W. I have done them all, these are examples.
      The C1 cables are 3*IL^2(Rc1) W.
      The C2 cables are 6*Ip^2(Rc2) W.
      Edit: note how now the dummy Jh is ~1/2 of the maximum run Jh, and the dummy input W is ~1/2 the maximum active run (Run 6) W input.
      Edit: when using Phase calculations, use Phase resistance for the reactor.
      When doing total power calculations, use apparent resistance for the entire reactor .
      Each resistor is actually 3/2 Phase resistance.
      This is because the 3 resistors are series-parallel.

      • Thomas Clarke

        However you do the math, the Joule heating power is proportional to the square of the line wire current, which they measured directly. Therefore unless the profs had amnesia between calculating dummy and active powers, AND got the equation wrong for one or the other, it does not change the anomaly.

        The real power does not depend on what the load is – they measured this directly with the PCE-830. However you count it the Joule heating correction is a relatively small one.

        I don’t myself believe they could possibly be so inconsistent as to use different formulae for the Joule power dummy and active.. Especially because they explicitly say all the active calculations are done like the dummy ones.

        If they really made that mistake then nothing they write is safe.

        • Obvious

          My correction above assumes only that the professors used only Line Current and Average Power to determine Joule heat, exactly as written in the report. They used the same method each time. They used real measurements for power and Line amps (presumably LV also), then applied the wrong formula to make the Joule heat calculation (every time, exactly the same way).

          Once the fact that phase current is higher than one line (since it is the sum of two lines) is used to correct Joule heat for the dummy, Joule heat rises by three times total for the dummy because there are more active cables.
          However, phase voltage is the same as Line voltage, so the higher line current does not translate to the same rate of increase for the C2 cables when Line amps increase.

          In this way, the phase power portion of Joule heating does not rise as fast as the rise in current might suggest, when more current is applied due to the higher Line voltage. This effectively reduces the Active run Joule heat compared to the incorrect method described in the report.
          Extrapolating the line current again from the newly corrected Joule heat values, squaring and multiplying by R gives exactly the same values used in the report for Input power. Only derived from the improved dummy run Joule heat ratio.

          No tricks. No new science. No nonsense.

          You can derive exactly the same values as I did, easily, once the smoke clears from what was said about measurements vs calculations described in the report.
          Look up the Two Wattmeter method calculations and you will see that this can be used to double-check the professors’ assumption.
          If you do not get VL = Vp in your calculations, there is an error.
          That does not occur in my calculations. V is a critical component of deriving the correct answer.

          • Thomas Clarke

            “No tricks. No new science. No nonense”

            I beg to differ. You claim the profs used line current and average power to estimate Joule heating. They did not, you can easily check this. They used line (C1) current and C2 current (a constant multiple of line current) and calculated the wire resistance which they assume correctly does not change. Power does not come into it.

            You claim that the load is not linear – that is, that the C2 currents are not proportional to the line current but increase slower than linear.

            That is impossible from a system containing wires and resistors. The profs, applying the same equation for Joule heating, get:
            power = IL^2R1

            The heater has power:
            power = IL^2R2

            The exact values of R1 and R2 depend on a more complex calculation but they cannot change as V changes. There is no mechanism for that to happen, and electricity 101 says it does not.

            If you disagree try writing out the equations you are using for V & I, and then either they will be linear, or you will be breaking one of:
            Ohms Law
            Kirchoff’s current Law
            Kirchojff’s Voltage Law

            The profs, whatever mistakes they made, at least knew better than that!

          • Obvious

            I have a fully consistent version. Consistent with Ohms Law, KVL, KCL, Joule, etc.
            Using textbook equations.
            No BS phase voltage differing from line voltage.
            No magic resistors.
            No strange assumptions.
            No control box losses required.
            No strange AC waveforms are required (they are averaged away).
            Fully linear W to W to Jh to Jh to Jh to W.
            Only one minor error in the report Joule heat calculations method claimed.
            No change to COP (maybe + 0.1 more COP than reported).
            No change to reported Input W, dummy or active run.
            No change to reported Line I.

            Just because you cannot do the same does not mean I cannot.

            I will further postulate that one resistor, disconnected, at 20° C will be almost exactly 3.0 Ω, with a very slight increase at ca. 450° C.

            Edit: (I am double-checking my last postulate now, but I am 95% certain it is correct.)

          • Thomas Clarke

            We know the Joule heat calculation method. They provided it for the dummy.

            (1) measure C1 current (I1)
            (2) calculate C2 current (I2 = I1*0.5)
            (3) calculate R1, R2 from geometry and resistivity of copper
            (4) Pjoule = 3*I1^2*R1 + 6*I2^2*R2

            This leads to the anomaly since Pjoule scales as I1^2, the same as total power which is
            Ptot = I1^2(3R1 + 6R2 + 3Rheater’)

            [Rheater’ is equivalent heater resistance taking into account the voltages across the three heater elements]

            Note that the exact constants (3, 6, 1/2) don’t alter the result. 1/2 is actually wrong, it should be 1/sqrt(3).

            Perhaps you would like to describe your idea for how they calculated Pjoule in similar detail so that we can compare? Remember they do say in the report how the calculate it for the dummy, and we agree no difference for active.

          • Obvious

            I’m making a proper answer. I have to do it separate, then paste it here.
            I have lost it twice already with errant backspaces wiping out my post for some reason. I caught one minor error I might have made earlier, which is good, though. Not sure where it leads but I’ll see. It might not be a real error after all.
            Let you know in while. Got some things to do offline…

  • Andreas Moraitis

    You are right with regard to the configuration, I should have looked at the diagram before I posted my comment. It is not possible to control each of the coils independently from the others. Nevertheless, a change of the voltage in one of the phases would affect two coils. These coils could even be completely deactivated by switching the phase off. So it might still be possible that not all coils have been working in the same way during the test.

  • Dr. Mike

    Thomas,
    I don’t follow your RMS ratio calculation. If C1 = SQRT(2/3) and C2 = SQRT(2), then C1= C2/SQRT(3), rather than C2 x SQRT(3). I calculated the relative RMS currents as :

    C1 = k * I * SQRT(1^2 + 1^2 + 0^2) = k * I * SQRT(2)

    C2 = k * I * SQRT[ (2/3)^2 + (1/3)^2 + (1/3)^2] = k * I * SQRT(2/3)

    For this calculation C2 = C1 / SQRT(3)
    Dr. Mike

  • Obvious

    I have just finished re-doing the Jh calcs and extrapolation from scratch.
    I have total agreement with the report for Input W.
    A lot of folks are just plain doing the Phase-Line calculations wrong, including the professors. They grossly underestimated the Dummy Joule heat, and grossly overestimate the Active Run Joule heat, by not doing the 3 phase math right.
    The Active Run Input was measured, as was the Dummy.
    But the three phase resistance and power calculations are messed up in the report, since the Professors derived an incorrect formula for Joule heat by not dividing up the power contributions form each phase correctly.
    Ptotal = P1 + P2
    P1 = sqrt(3)Ptotal
    P2 = sqrt(3)Ptotal
    P3 = P2 = P1 = P
    And P = Ptotal/sqrt(3)
    Ip = IL/sqrt(3)
    Vp = Pp/Ip
    Vp = VL
    Therefore Pp = 486/sqrt(3)
    And P1 = 276.55 W
    Now see if you can work it out.
    Be sure to derive R phase using Vp and Ip. It is not R apparent (whole device) nor One single reactor resistor.
    The Joule heat has been done wrong in the report. The correct Joule heat for the Dummy should be 20.017 W
    The correct Joule heat for Run 6 is 36.828 W
    The C1 cables are 3*IL^2(R C1) W.
    The C2 cables are 6*Ip^2(Rc2) W

  • Obvious

    My correction above assumes only that the professors used only Line Current and Average Power to determine Joule heat, exactly as written in the report. They used the same method each time. They used real measurements for power, then applied the wrong formula to make the Joule heat calculation.

    Once the fact that phase current is higher than one line (since it is the sum of two lines) is used to correct Joule heat for the dummy, Joule heat rises by three times total for the dummy because there are more cables.
    However, phase voltage is the same as Line voltage, so the higher line current does not translate to the same rate of increase for the C2 cables when Line amps increase.

    In this way, the phase power portion of Joule heating does not rise as fast as the rise in current might suggest, when more current is applied due to the higher Line voltage. This effectively reduces the Active run Joule heat compared to the incorrect method described in the report.
    Extrapolating the line current again from the newly corrected Joule heat values, squaring and multiplying by R gives exactly the same values used in the report for Input power.

    No tricks. No new science. No nonsense.

    You can derive exactly the same values as I did, easily, once the smoke clears from what was said about measurements vs calculations described in the report.
    Look up the Two Wattmeter method calculations and you will see that this can be used to double-check the professors’ assumption.
    If you do not get VL = Vp in your calculations, there is an error.
    That does not occur in my calculations. V is a critical component of deriving the correct answer.

    • Thomas Clarke

      “No tricks. No new science. No nonense”

      I beg to differ. You claim the profs used line current and average power to estimate Joule heating. They did not, you can easily check this. They used line (C1) current and C2 current (a constant multiple of line current) and calculated the wire resistance which they assume correctly does not change. Power does not come into it.

      You claim that the load is not linear – that is, that the C2 currents are not proportional to the line current but increase slower than linear.

      That is impossible from a system containing wires and resistors. The profs, applying the same equation for Joule heating, get:
      power = IL^2R1

      The heater has power:
      power = IL^2R2

      The exact values of R1 and R2 depend on a more complex calculation but they cannot change as V changes. There is no mechanism for that to happen, and electricity 101 says it does not.

      If you disagree try writing out the equations you are using for V & I, and then either they will be linear, or you will be breaking one of:
      Ohms Law
      Kirchoff’s current Law
      Kirchojff’s Voltage Law

      The profs, whatever mistakes they made, at least knew better than that!

      • Obvious

        I have a fully consistent version. Consistent with Ohms Law, KVL, KCL, Joule, etc.
        Textbook equations.
        No BS phase voltage differing from line voltage.
        No magic resistors.
        No strange assumptions.
        Fully linear W to W to Jh to Jh to Jh to W.
        Only one minor error in the report calculations method.
        No change to COP (maybe 0.1).
        No change to Input W, dummy or active run,
        Just because you cannot do the same does not mean I cannot.

  • US_Citizen71

    The same could have been done on an active run. The full capacity may also to refer to the highest setting they used. It may in truth always pull 360 watts. We do not know the values for it but it doesn’t matter both PCE 830 units were attached during all runs. They comment that doing so allowed them to evaluate the power of the control box so they would have seen a difference of 3X plus. They observed this setup for 32 days and you don’t think they looked at every screen and instrument many times. It is heater it is like watching grass grow.

  • ivanc

    US_Citizen71

    ivanc

    2 days ago

    And you still haven’t answered my original question. What is wrong
    with the measurements made by the PCE 830 connected from the wall?
    If you do not have an answer for that then I have to conclude this whole
    exchange by you is nothing but FUD. I’m sure you will ignore this again
    like you have the last 4 times I have asked but it is worth a fifth
    attempt.

    I did, but my post have to go to the sanitation committee first!!!!!!!!

    • Andreas Moraitis

      Do you mean the PCE before the control box? From where do you take the data? The calculations of the COP are most likely based on the readings of the second meter.

      • ivanc

        this was us_citizen71 remarks, I agree with you, the data cames from the second meter, there i nothing about the 1st., I answered this before but frank is stopping my post.
        but you could see my spreadsheet, I will copy my comments there.

      • US_Citizen71

        Ivan’s response was due to me calling him out for not explaining why he thinks the first PCE 830 did not measure 3X power that he claims must be there and as you notice he still didn’t answer the question but deflected. I laid out pieces and parts of the report to show the first meter showed a reading that was within 360 watts of the measurement of the meter after the control box at the link below.

        http://www.e-catworld.com/2014/11/15/mats-lewan-testers-rule-out-inverted-clamp-hypothesis-rossi-comments-mark-e-kitiman/#comment-1696526975

        • ivanc

          I did, basically said, there is no data for that meter.

          but you did not answer why you think the expresion below is wrong:

          The equation of Us_citizenxx

          Ic1= √((JH/3-2*Rc2*Ic2²)/Rc1))

          “This equation is my equation (a) , but you missing one detail,
          Ic2=Ic1/2 (This is because we reverse engineering the report)
          it should have been Ic2=Ic1/sqrt(3), but we must do the same as them to find the current they read.

          Then
          you have to reorder your expression, factorize Ic1 and the equation is
          solved. you do have two variables and two equations, so what is the
          problem?
          is from this equation and after factorizing Ic1 that for me
          is IL, that I derive my joule factor, replacing Rc1 and Rc2 values so is
          easy to compute 😉

          reordering:
          3 Ic1^2 Rc1 + 6 Ic2^2 Rc2 = JH
          replacing:
          3 Ic1^2 Rc1 + 6 (Ic1/2)^2 Rc2 = JH
          simplifying:

          3 Ic1^2 Rc1 + 6 Ic1^2/4 Rc2 = JH

          factorizing Ic1^2:
          Ic1^2(3 Rc1 + 6/4 Rc2) = JH
          solving:
          Ic1=sqrt( JH / (3 Rc1 + 6/4 Rc2) )
          Done.

          so the equation turns into Ic1=sqrt(JH*some_factor_I_called_Joule_factor)”

          • US_Citizen71

            No data listed in the report is correct but is stated that the reading on the first meter never exceed a difference of 360 watts of the 2nd meter. The readings of the 2nd meter are power levels in figure 7. So the power in from the wall never exceeded 360 watts more than the power levels listed in the report. Proof that the power was not 3X plus as your equation and theory suggest. The world has moved on beyond this as well as it has been shown by many posters that high temperature heating elements that drop in resistance at temperature are actually quite available and used in lots of different products.

          • ivanc

            still, speculation on your part, if I have to speculate I would say, the phases were changed in the controller or prior to it in the power supply. but who knows is pure speculation, when I see the reading of that data I will agree or dis_agree with you in this.

          • Andreas Moraitis

            Yes!

            „The two PCEs were inserted one upstream and one downstream of the control unit: the first allowed us to measure the current, voltage and power supplied to the system by the power mains; the second measured these same quantities as input to the reactor. Readings were consistent, showing the same current waveform; furthermore, they enabled us to measure the power consumption of the control system, which, at full capacity, was seen to be the same as the nominal value declared by the manufacturer.“ (p. 5)

            The term “at full capacity” indicates that they must have controlled this during the active run. There is no way to explain the accordance of the measured difference with the nominal power consumption under the condition that there have been inverted clamps.

  • US_Citizen71
  • Obvious

    I’m making a proper answer. I have to do it separate, then paste it here.
    I have lost it twice already with errant backspaces wiping out my post for some reason. I caught one minor error I might have made earlier, which is good, though. Not sure where it leads but I’ll see. It might not be a real error after all.
    Let you know in while. Got some things to do offline…

  • US_Citizen71

    No data listed in the report is correct but is stated that the reading on the first meter never exceed a difference of 360 watts of the 2nd meter. The readings of the 2nd meter are power levels in figure 7. So the power in from the wall never exceeded 360 watts more than the power levels listed in the report. Proof that the power was not 3X plus as your equation and theory suggest. The world has moved on beyond this as well as it has been shown by many posters that high temperature heating elements that drop in resistance at temperature are actually quite available and used in lots of different products.

    • Andreas Moraitis

      Yes!

      „The two PCEs were inserted one upstream and one downstream of the control unit: the first allowed us to measure the current, voltage and power supplied to the system by the power mains; the second measured these same quantities as input to the reactor. Readings were consistent, showing the same current waveform; furthermore, they enabled us to measure the power consumption of the control system, which, at full capacity, was seen to be the same as the nominal value declared by the manufacturer.“ (p. 5)

      The term “at full capacity” indicates that they must have controlled this during the active run. There is no way to explain the accordance of the measured difference with the nominal power consumption under the condition that there have been inverted clamps.

  • Obvious

    That’s not what I meant. Obviously V changes from run to run.

    What I mean is that your ratio of VL to Vp is not 1:1, which it should be. Whatever the error is there, it makes Vp lower than VL, (ie: VL/Vp =~1.4) which it cannot be in a delta. The error is the same in all the runs and the dummy, on your sheet.

  • Obvious

    Nearly a 50% drop?

  • Obvious

    I divided 485 by 19.7 to get volts, and compared to your table.

  • Andreas Moraitis

    Thanks to cobraf, AlainCo & others:

    http://www.cobraf.com/forum/immagini/R_123571969_1.pdf

    This SiC heating element seems to match perfectly the data from the report.

    • Andreas Moraitis

      I’m not sure about the absolute values per wire, but the relationships are apparently correct. The resistance drops about 1/3.5 from 450 to 1200 deg C and remains then constant until 1400 deg C.

    • Obvious

      That is really neat.
      Once I fix all my calculations, which I just had to repair, I’ll weigh in on whether I think it is required or not.
      Maybe you guys just helped Rossi along. I bet that heats up mighty quick.

  • Andreas Moraitis

    Thanks to cobraf, AlainCo & others:

    http://www.cobraf.com/forum/immagini/R_123571969_1.pdf

    This SiC heating element seems to match perfectly the data from the report.

    • Andreas Moraitis

      I’m not sure about the absolute values per wire, but the relationships are apparently correct. The resistance drops about 1/3.5 from 450 to 1200 deg C and remains then constant until 1400 deg C.

    • Mark Szl

      Wow! Great find!!!

      Thanks AlainCo & others.

      That now shifts the case back towards the non-fraud/non-screw-up side of the debate.

      • Mark Szl

        Guess that SiC element post is a hoax.

        It’s shifted back to the fraud / screw-up side of the fence. LOL

    • Obvious

      That is really neat.
      Once I fix all my calculations, which I just had to repair, I’ll weigh in on whether I think it is required or not.
      Maybe you guys just helped Rossi along. I bet that heats up mighty quick.

    • ivanc

      except that you need to tell Rossi, that SiC is not an alloy!!!!!! 🙂
      maybe the difference between a ceramic and metal is not obvious!!!!!
      Maybe he pay for an metal and somebody shipped a ceramic element…..
      wonderful word of speculation

    • Thomas Clarke

      And it is a practical joke! When something looks to good to be true that is often the case!

  • Obvious

    OK.
    Then we are on the same track now, sort of, regarding V.
    I ended up with the lower voltage as a requirement late yesterday after I caught an error in my stuff. The math wouldn’t hold together with the higher voltage. I get the IL/sqrt(3) for phase idea.

    • ivanc

      Good on you.!!!

  • Obvious

    I think I have the real answer this time.
    I had an error in my equations that originally appeared to show a perfect fit to the data. (This was the version I was promising a day or two ago.)
    Then I fixed that during proofreading, and then all the nonsense started up again.
    Then just on a hunch, I put the error back in, and everything seems to work again. (I’m still beta testing the error version again. I wish I kept the original instead of “fixing” it and saving over it….)
    I think this idea duplicates the professors’ error that makes a mess of Joule heat slope and makes the fake negative R happen.
    You still need to deconstruct the bad data to get the active run current. I ran a back-check on that, and was able to extract almost exactly the combined resistance of a set of three wires, namely a C1 and two parallel C2’s (correct to 5 decimal places).

    • Thomas Clarke

      When I see it I will go through the figures and assumptions, and tell you how it differs from reality!

      • Obvious

        You bet.
        I just rebuilt the spreadsheet from scratch, (again), which looks good, and now I am working on re-creating the reported Jh values for the active run in order to duplicate the error to verify that the entire sheet works.

      • Obvious

        Reply 2:
        I see what the professors did.
        Now I’m just just trying to rationalize why they did it.

        Check this out (using Run1):
        Extrapolated I for Run1 = 46.669 A ; therefore Ip = 46.669/sqrt(3) = 26.994 A
        Jhc1 = 3*((Ip^2)*(0.004375)) = 9.528804 W
        Jhc2 = 6*((Ip/2)^2)*(0.002811) = 3.061196 W
        Jhtotal = Jhc1 + Jhc2 =12.59 W
        Then Jhtotal * 3 = 37.77 W

        Using this same method, Jhdummy = 6.73 W
        But the line current is of course what they used (or seemed to describe) in the report for the dummy.
        The line current calculation for the dummy does give the same answer, if not multiplied by three.
        Now you have to decide what is the right one for the report…..

        Cheers! Enjoy your weekend.
        Edit: It looks like the Jhtotal for the dummy must be multiplied by 3 to fix this up…

        • Thomas Clarke

          Let me see if I follow this.

          You are saying that a X3 difference in Joule heating powers between dummy and active tests could be explained by the fact that they need to multiply power X3 to get from one line data to total data. If they did this in the active test but not the dummy test that would explain the anomaly.

          I agree this would be a plausible explanation. I don’t agree it can be a plausible explanation because they explicitly show the X3 calculation in the dummy test working (equations 9,10,11 in the report)..

          The error they get by NOT doing X3 on the Joule heat for the active test is the wrong way round making the anomaly worse by X3. Otherwise we have no error from this mechanism.

          • Obvious

            It is certainly strange, no matter how you slice it.

            I look forward to the official explanation.

            But all the math works this way, including extraction of cable resistance from the Joule heat. The same amount for the entire data set, using three times Joule heat for the dummy, or instead 1/3 Joule heat for the active run and the same dummy Joule heat. Resistance for the reactor stays stable both ways, with ~0.1 to 0.2 Ω drop over the whole data set.

          • Thomas Clarke

            I think you are now agreeing with ivanc, andreas.s, Dr Mike and me that the X3.3 anomaly is a real thing that requires and explanation from the testers – for example showing that the heater is not Inconel but some unusual material with a X3.3 resistance change from 500C to 1250C?

          • Obvious

            I don’t believe in the 3x resistance drop at all.
            The professors have done something strange, somehow. Almost certainly accidentally (IMO).
            Probably they multiplied the Jh current by three, twice somehow in their spreadsheet, or somehow used phase current in one set of calculations, and line current in another.
            It is a very specific error. Maybe a cut-pasted formula grabbing data from a wrong column, something like that. Only the professors know, or have the data to sort it out. They must patch this up before a final, shorter press version gets published.
            Since I can extract the cable set resistances from their published data, consistently, exactly, I am certain that they believed the values they were using were the correct ones. Even a real negative resistance cannot be that linear (totally straight-line) for the entire active run data set, and still do a 1/3 drop (or even better a 2(sqrt(3)) drop) cleanly from one range to another, and not have that show up in the devolution of the Jh data somehow. Especially with the ITP picking the input heat without (as far as we know) knowing exactly where the resistance change will occur, and getting it accidentally exactly right.
            And my corrections still incorporate a COP of 0.93 or 0.94 for the dummy run.
            When I get a bit more time, I will go over my two versions (one with 3x dummy Jh, the other with 1/3 active run Jh) and see if one must be more physical than the other. There is probably a hint that one is either impossible or far more likely compared to the other. But maybe not….

            Edit: This means yes. I mostly agree.
            In my opinion: The anomaly on paper is real. I think the heater input is just heater input. It should make sense. The “missing” 3x power increase (that would make COP=1) and/or the amazing 1/3 inverse resistor are both artifacts of a paper error. The source of the error from which all the others are extrapolated is now almost indisputably within the Joule heat values reported.

          • Thomas Clarke

            Good.

            So we agree there is an error in one of the figures presented. And we cannot know which one.

            However the matter can easily be clarified by the authors. They can check and correct, whether X3 change in resistance or error in figures.

            Till then we cannot know which figure is calculated differently in the report between dummy and active runs.

          • Obvious

            I think we can tell which one, if we consider the evidence carefully.

            There are other sqrt(3) anomalies in the data, related to the main error.
            For example, I found another one when using the corrected Joule heat (Jhc1 using IL, and Jhc2 using Ip) for the dummy, and then comparing to the reported active run. It will be interesting to see wye they got the numbers mixed up.

        • ivanc

          obvious, jhc1 equation is wrong,
          it should be:
          jhc1=3*46.669 ^2 * .004375 = ?
          this is because the joule depends on the current circulating, and the feeding cables see IL not Ip

          • Obvious

            I tend to think so also. But there is something still weird about how the Joule heating was handled regardless, besides that.
            If 19.7 A RMS was measured in all three lines, where does it go out? The answer to that is important.

  • Obvious

    You bet.
    I just rebuilt the spreadsheet from scratch, (again), which looks good, and now I am working on re-creating the reported Jh values for the active run from in order to duplicate the error to verify the entire sheet works.

  • Obvious

    Reply 2:
    I see what they did, now I’m just just trying to rationalize why they did it;
    Check this out (using Run1):
    Extrapolated I for Run1 = 46.669 therefore Ip = 46.669/sqrt(3) = 26.994 A
    Jhc1 = 3*((Ip^2)*(0.004375)) = 9.528804 W
    Jhc2 = 6*((Ip/2)^2)*(0.002811) = 3.061196 W
    Jhtotal = Jhc1 + Jhc2 =12.59
    Then Jhtotal * 3 = 37.77
    Using this same method, Jhdummy =6.73 W
    But the line current is of course what they used in the report for the dummy.
    The line current calculation for the dummy does give the same answer, if not multiplied by three.
    Now you have to decide what is the right one for the report…..

    Cheers!, Enjoy your weekend.

    • Thomas Clarke

      Let me see if I follow this.

      You are saying that a X3 difference in Joule heating powers between dummy and active tests could be explained by the fact that they need to multiply power X3 to get from one line data to total data. If they did this in the active test but not the dummy test that would explain the anomaly.

      I agree this would be a plausible explanation. I don’t agree it can be a plausible explanation because they explicitly show the X3 calculation in the dummy test working (equations 9,10,11 in the report)..

      The error they get by NOT doing X3 on the Joule heat for the active test is the wrong way round making the anomaly worse by X3. Otherwise we have no error from this mechanism.

      • Obvious

        It is certainly strange, no matter how you slice it.

        I look forward to the official explanation.
        But all the math works this way, including extraction of cable resistance from the Joule heat. The same amount for the entire data set, using three times Joule heat for the dummy, or instead 1/3 Joule heat for the active run and the same dummy Joule heat. Resistance for the reactor stays stable both ways, with ~0.1 to 0.2 drop over the whole data set.

  • Andreas Moraitis

    There is another possible reason for the resistance drop. The triac cuts the sinusoidal signal into sections of different length, depending on the desired output. This adds harmonics to the signal, its spectrum changes dramatically. The spectrum will also be altered whenever the output is modified: The shorter the ‘remaining’ sections of the original, sinusoidal signal, the higher the relative strength of the harmonics, and vice versa. That is, if the output of the triac is increased, the influence of the harmonics will be reduced in favour of the fundamental frequency. Therefore, the impedance of the coils should drop. Since we do not know exactly the resistance and inductance of the coils and which waveforms have been applied in the different test runs, the significance of this effect is difficult to estimate. But MFMP could determine it by means of a comparative test with DC. At the same wire temperature the resistance should be lower than in a run with chopped AC, but it should not drop by the same amount if the temperature is increased.

    • Obvious

      Due to averaging the input signal over 20+ hours, any strange waveform, harmonics, etc., should be expected to be averaged into irrelevance in a final reported value.

      The resistance could (for example) fluctuate from triple to 1/3 of its off-the-shelf value hundreds of times in the total measurement time period, and all we can calculate with is the puréed final average.

      • Andreas Moraitis

        Even if the chopping pattern is not synchronized with the original signal (so that the waveform would change periodically) I think one should see differences in the average values when runs with lower and higher output are compared. But there are other open questions. For example, we do not know if and how the readings of the thermocouple influence the behaviour of the control box. There might be a regulation mechanism that is more complex than a simple emergency cut-out. We should also not forget the “electromagnetic pulses” which are mentioned in the report. Maybe the have not been used in the dummy run.

  • Andreas Moraitis

    There is another possible reason for the resistance drop. The triac cuts the sinusoidal signal into sections of different length, depending on the desired output. This adds harmonics to the signal, its spectrum changes dramatically. The spectrum will also be altered whenever the output is modified: The shorter the ‘remaining’ sections of the original, sinusoidal signal, the higher the relative strength of the harmonics, and vice versa. That is, if the output of the triac is increased, the influence of the harmonics will be reduced in favour of the fundamental frequency. Therefore, the impedance of the coils should drop. Since we do not know exactly the resistance and inductance of the coils and which waveforms have been applied in the different test runs, the significance of this effect is difficult to estimate. But MFMP could determine it by means of a comparative test with DC. At the same wire temperature the resistance should be lower than in a run with chopped AC, but it should not drop by the same amount if the temperature is increased.

    • Obvious

      Due to averaging the input signal over 20+ hours, any strange waveform, harmonics, etc., should be expected to be averaged into irrelevance in a final reported value.

      The resistance could (for example) fluctuate from triple to 1/3 of its off-the-shelf value hundreds of times in the total measurement time period, and all we can calculate with is the puréed final average.

      • Andreas Moraitis

        Even if the chopping pattern is not synchronized with the original signal (so that the waveform would change periodically) I think one should see differences in the average values when runs with lower and higher output are compared. But there are other open questions. For example, we do not know if and how the readings of the thermocouple influence the behaviour of the control box. There might be a regulation mechanism that is more complex than a simple emergency cut-out. We should also not forget the “electromagnetic pulses” which are mentioned in the report. Maybe the have not been used in the dummy run.

    • ivanc

      If any of you s ay make any sennse then t eproblem would not be a drop in resistence, but a problem measuring voltage and current. then they would be using the wrong meter.
      RMS is sthe equivalent to DC ideally if you use the proper RMS meter, the harmonics and pulses aare expresed in the measurment. if what you say is correct then the report have qbsolutely no validity.

  • Obvious

    I don’t believe in the 3x resistance drop at all.
    The professors have done something strange, somehow. Almost certainly accidentally (IMO).
    Probably they multiplied the Jh current by three, twice somehow in their spreadsheet, or somehow used phase current in one set of calculations, and line current in another.
    It is a very specific error. Maybe a cut-pasted formula grabbing data from a wrong column, something like that. Only the professors know, or have the data to sort it out. They must patch this up before a final, shorter press version gets published.
    Since I can extract the cable set resistances from their published data, consistently, exactly, I am certain that they believed the values they were using were the correct ones. Even a real negative resistance cannot be that linear (totally straight-line) for the entire active run data set, and still do a 1/3 drop (or even better a 2(sqrt(3)) drop) cleanly from one range to another, and not have that show up in the devolution of the Jh data somehow. Especially with the ITP picking the input heat without (as far as we know) knowing exactly where the change will occur, and getting it accidentally exactly right.

  • Obvious

    I think we can tell which one, if we consider the evidence carefully.

    There are other sqrt(3) anomalies in the data, related to the main error.
    For example, I found another one when using the corrected Joule heat (Jhc1 using IL, and Jhc2 using Ip) for the dummy, and then comparing to the reported active run.

  • Obvious

    I tend to think so also. But there is something still weird about how the Joule heating was handled regardless, besides that.
    If 19.7 A RMS was measured in all three lines, where does it go out? The answer to that is important.