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.

  • Obvious

    I tend to think so also. But there is something still weird about how the Joule heating was handled regardless, besides that.
    If 19.7 A RMS was measured in all three lines, where does it go out? The answer to that is important.

  • Obvious

    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 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.

  • 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

    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.

  • 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

    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

    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.!!!

  • 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

    I divided 485 by 19.7 to get volts, and compared to your table.

  • Obvious

    Nearly a 50% drop?

  • 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.

  • 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

    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…

  • 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

  • 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
  • US_Citizen71

    The same could have been done on an active run. The full capacity may also to refer to the highest setting they used. It may in truth always pull 360 watts. We do not know the values for it but it doesn’t matter both PCE 830 units were attached during all runs. They comment that doing so allowed them to evaluate the power of the control box so they would have seen a difference of 3X plus. They observed this setup for 32 days and you don’t think they looked at every screen and instrument many times. It is heater it is like watching grass grow.

  • Obvious

    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.

  • 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

  • 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

  • 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.

  • 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.