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.

  • 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.

  • 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.

  • 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.

  • 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…

  • 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.

  • 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

    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.

  • 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.

  • 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.

  • 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?

  • 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.

  • 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).

  • 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

    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.

  • 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.

  • 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.

  • 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.

  • 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.

  • 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.

  • 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.

  • 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]

  • 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.

  • 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.

  • 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

    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.

  • 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.

  • 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

    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.

  • 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.

  • 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

    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.

  • 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.