Some Points Regarding a Recent Presentation at ICCF20 on the ‘Lugano Report’ (Rainer Rander)

  • Because there is substantial agreement here, in spite of some sputtering, I'm responding directly to Wyttenbach.


    With the latest findings - Lugano = Levi (COP 1.4,.. 1.5) - no provable progress since ten years, for me the storybook is closed.


    The way I put it, Lugano is dead. It is possible to argue until the cows come home, and all night long, but the substance is as Wyttenbach agrees.


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    To reopen the discussion somebody had to provide a new reactor to an independent and knowledgeable lab to evaluate it.


    We are free to discuss whatever we choose, here. However, if the condition described in not fulfilled, none of this discussion could lead to general acceptance of Lugano.


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    The worst experience about Lugano was the invention of new facts, about all kind of errors, proofs (irrefutable ones)...


    By whom? I will not expect an answer. If it was someone else, Wyttenbach seems willing to let it go, and if it was him, I do not like to rub peoples faces in errors, beyond necessity. I am not going to follow someone down the street, shouting "you were wrong." In a discussion, I will correct what I see as errors. It is part of how I learn, and once in a while someone shows me my own errors. I consider that a favor, regardless of their motives.


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    Finally the Optris issue (wrong emissivity) could be resolved with a support request by Optris.


    I am not going to ask Optris about something where the matter is clear, and it is clear on the issues involved. They are not experts on alumina, per se, and what Wyttenbach seems to be talking about is what might be called "off-label" use of the camera and general information they provided. If someone wants to duplicate what MFMP did with the Lugano reactor, their simulation, yes, they might well contact Optris, even to just cover all the bases. That's doing thorough work. What THH wrote about a report of a conversation was cogent. What we saw here was a willingness to "translate" what was actually said in the manual, to make it appear to be what was being said by the author. That, in an academic environment, would be a major offense. Quote, and provide full context. Then interpret, separately. Link to sources used and be specific, this all allows others to easily check.


    Many times, I was going to write something that I'd read, but my habit of providing sources caused me to reread the source. Oops! It didn't say what I remembered! Learning to mistrust oneself, in this productive way, is part of developing scientific objectivity. Making it easier for others to check what is being said leads to more rapid correction, when that is needed. Sometimes I go back and read what I wrote years ago. On occasion, I notice an error. And nobody pointed it out. Damn! WTF?


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    One thing everybody should have learned: Without proper calibration, exact documentation and independent reviews no further results will be accepted.


    Bingo! The duck comes down with $50 in his mouth.


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    Lugano was a first IH/Rossi clash story: May be, in ten years we will know the true intentions behind this farce...


    This could be a little misleading. We have seen no sign of IH clash with Rossi over Lugano, at the time. Yes. We may suspect that IH, attempting to confirm Lugano, had questions for Rossi. We may suspect some sort of clash arising, but IH, at that point, was not about to push Rossi too hard. Rossi, I suspect, started to think that he could get away with anything with IH, they were so accommodating. The cads! They tricked him by being nice!


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    To be honest: Lugano was just some kind of distraction - not more: And one last thing you can believe me: LENR needs "no" carrier (Ni,Pd,Zr, Ti...) for its ignition. Just watch the BLP/Mills video of the self sustain mode. Or, if you reverse the logic: All the mentioned carriers will work and any COP is possible!


    Sigh. I'll leave it at that. Just "Sigh!"

  • What do you think about the idea of high pressure intragranular hydrogen bubbles being the trigger location of the Rossi Effect?


    In fact the main place where Hydrogen is absorbed are the vitreous inter-granular regions. (few nm compared to the tens of nm of the granules )
    Many groups try to let Nickel absorb the Hydrogen at low temperature before heating, but all the process must be done in a controlled atmosphere.
    A layer of Nickel oxide would make Hydrogen absorption quite more difficult.
    Is nice to write about real science.

  • reductio ad absurdem


    Dear Abdul, the right spelling is Reductio ad Absurdum or Argomentum ad absurdum.
    If you want to disprove an hypothesis you first suppose it true then using logic reasoning yo must come to a conclusion that is clearly not true.
    You have not used this way of reasoning in your writing.


    They did not appear to be aware of the band vs. total emissivity problem, and appear to have used the same value for both, leading to major error.


    Let us now demonstrate that your affirmation is false using the Reductio ad Absurdum, with simple words.


    First we must suppose that the argument is true so that we have to use the BB emissivity for Alumina in the range of the Camera.
    Then we repeat the reasoning we have done already:
    Let us have two bodies a) and b) with same geometrical shape and with the same input power:
    a) being a perfect Black body and
    b) made of alumina
    We let the two bodies reach the equilibrium and suppose that they are in vacuum ( at least 10^-4 to stop convection ) so that the only way they can emit energy is radiation. Because of this equilibrium condition the power output from the two bodies will be the same.
    This lead to the conclusion that b) must have a much higher temperature then a) but because of our first hypothesis the Camera would measure a temperature almost equal. This is absurd because is contrary to energy conservation so we conclude that the first hypothesis is not correct.


    Quite sure that I will have to repeat all that to you again.


  • It would be absurd, yes. But it relies on your faulty idea that radiation in the IR band depends only on input power and band emissivity - so that a and b (with the same input power) would have the same band radiation.


    Whereas the rest of the world knows that radiation in the IR a band depends on temperature and band emissivity. In your case the radiation from b (at a higher equilibrium temperature) would therefore be higher. And the camera would correctly read the higher temperature.


    Your repeated inability to correct this erroneous argument raises the question: is this some cognitive lack: or are you a troll pure and simple?




    Quite sure that I will have to repeat all that to you again.

    • Official Post

    It would be absurd, yes. But it relies on your faulty idea that radiation in the IR band depends only on input power and band emissivity - so that a and b (with the same input power) would have the same band radiation.


    Whereas the rest of the world knows that radiation in the IR a band depends on temperature and band emissivity.



    Worth mentioning, that using electrrical resistance heating, there is a close ( though not linear) correlation between temperature and power input. I think this detail is where the devil lives. :evil:

  • Worth mentioning, that using electrrical resistance heating, there is a close ( though not linear) correlation between temperature and power input. I think this detail is where the devil lives.


    @Alan: For breakfast I prefer red-current over the Ampere one, but on the other side, the Alumina I swallow after two pots of coffee, must have a negative emissivity, as it reduces the heat in the stomach significantly.


    Conclusion: Not all currents/aluminas are equal...

  • Quote

    there is a close ( though not linear) correlation between temperature and power input.


    Surely not?


    In this case at even 800C radiation is the dominant dissipation route and that scales as T^4 (for a black body) and close to this for alumina since T^4 is much stronger than the total emissivity change with T. The ambient correction is pretty small.


    So to 1st approx we have T ~ P^(1/4).

  • Abd Ul-Rahman Lomax wrote:


    Dear Abdul, the right spelling is Reductio ad Absurdum or Argomentum ad absurdum.


    Zero gives the correct spelling. The error appears to be fairly common, though. The substance here was not quoted, only the spelling correction.


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    If you want to disprove an hypothesis you first suppose it true then using logic reasoning yo must come to a conclusion that is clearly not true.
    You have not used this way of reasoning in your writing.


    What I provided was an example that simplified the problem, that could then be used for a reductio argument. Because what I wrote was not quoted, I'm not going over it, but will instead look at what Zero provided.


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    Abd Ul-Rahman Lomax wrote:
    They did not appear to be aware of the band vs. total emissivity problem, and appear to have used the same value for both, leading to major error.


    Now, this is very clear as an error. Zero is trying to negate it. Zero is strongly attempting to make the question be about power, but what has been emphasized is about temperature. Yes, that's related to power, but one step at a time! By mixing these up, Zero is able to present an argument that might convince the unwary.


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    Let us now demonstrate that your affirmation is false using the Reductio ad Absurdum, with simple words.


    Yum!


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    First we must suppose that the argument is true so that we have to use the BB emissivity for Alumina in the range of the Camera.


    That is, we are assuming that band emissivity, in the camera band, is 1.0


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    Then we repeat the reasoning we have done already:
    Let us have two bodies a) and b) with same geometrical shape and with the same input power:
    a) being a perfect Black body and
    b) made of alumina


    Okay. Clear.


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    We let the two bodies reach the equilibrium and suppose that they are in vacuum ( at least 10^-4 to stop convection ) so that the only way they can emit energy is radiation. Because of this equilibrium condition the power output from the two bodies will be the same.


    Yes.


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    This lead to the conclusion that b) must have a much higher temperature then a) but because of our first hypothesis the Camera would measure a temperature almost equal. This is absurd because is contrary to energy conservation so we conclude that the first hypothesis is not correct.


    By the conditions of the problem, the alumina has lower total emissivity. With equal input power, with lower total emissivity, the temperature of the alumina would be higher, a fact which Zero has often emphasized. It would show the band irradiance of a hotter body, not equal as claimed by Zero, because it would actually be hotter. The camera would correctly measure the temperature if the band emissivity is properly set to match the band emissivity of the material (in this case, that of a black body in the camera band).


    To reiterate, Zero has stated contradictory conditions:
    Black body vs non-gray body with total emissivity much less than 1.0, but band emissivity of 1.0. (That is the reductio situation I set up, simply taking band emissivity to a very narrow band that would not represent much power at all.)
    Same power input to both bodies.
    Therefore the non-gray body will be hotter, because for this purpose, it is total emissivity that matters, not band emissivity.
    But Zero has the camera show the same reading, i.e., the same temperature. He made that up, it's impossible, for the reasons he has amply shown.


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    Quite sure that I will have to repeat all that to you again.


    "Have to" indicates compulsion, so this will reveal that something is driving Zero to keep him making the same mistake over and over. And over.

  • Worth mentioning, that using electrrical resistance heating, there is a close ( though not linear) correlation between temperature and power input. I think this detail is where the devil lives.


    I rather doubt it. However, It may look more or less that way within a certain range. I studied the first Parkhomov device in detail. See https://en.wikiversity.org/wik…hydrogen_system/Parkhomov


    The chart on this page shows the Parkhomov temperatures plotted against power input. It does look more or less linear at the lower power inputs. And, in fact, the down-curve appears to hit at about the point where the boiler temperature reached boiling, thus nailing it to a specific temperature, which would more effectively cool the device.


    That chart, by the way, showed no room for LENR heat, which contradicted the evaporative calorimetry. This was the first sign I saw that there might be a problem with Parkhomov. supposedly the XP increased drastically above a certain temperature, but the temperature of the device did not go up any more than might be expected from electrical heating. Parkhomov had not calibrated, and never really did calibrate this design.


    Before I created that plot, I'd been quite enthusiastic about Parkhomov! Then Ed pointed out something fishy about the COP vs input power, and so I checked in detail.


    In any case, on the primary question, the power dissipation by radiation should go as the fourth power of the temperature. But ... there is also conduction and convection to consider in dissipating the input power -- or any generated power (the issue is simply heating power, not the source.)

  • There is also the issue of changing heat data quantization coinciding with data glitches and places where the power in/out behavior changes. How can the quantization change?? Its almost as if Parkhomov created the data by scaling sections and then finally cut and pasted them together. Perhaps there's a simple explanation, but I couldn't see it.

  • There is also the issue of changing heat data quantization coinciding with data glitches and places where the power in/out behavior changes. How can the quantization change?? Its almost as if Parkhomov created the data by scaling sections and using cut and paste. Perhaps there's a simple explanation, but I couldn't see it.


    https://www.lenr-forum.com/for…ParkhomovPowerTempV3-zip/


    Please, when posting data here, provide a source. The data in the spreadsheet linked isn't sourced. Examining it, it is not from the same experiment, it is probably from a later one, where the spreadsheet data was provided. Parkhomov did copy and paste data for a segment of his report, one time, a major blunder, though not of major import, in itself. His notebook ran out of battery power and he didn't want to leave the period blank. Why not? That's a whole interesting question, and he never explained that part, and I think I have an idea why. (It would have opened a can of worms. Why was he running the notebook on battert power? I think it was because he had so much noise on the thermocouples from his heater, a nice little transformer, so he floated that circuit. At least that is what occurred to me.)


    Parkhomov acted like a bumbling amateur, unfortunately, on occasion. Falsifying data just to make it look better? Some of his later work seems better.


    His data in some later work was extremely noisy. He started using thermostatic temperature control, very bad idea. So his input power was constantly flipping, creating a huge mess.

  • The spreadsheet posted is my analysis of the original data. I'll post the original tomorrow morning as a reference. if you look near the end of the sheet you will see the very large gap in data where the doctored graphical output caused so much concern.

  • There is also the issue of changing heat data quantization coinciding with data glitches and places where the power in/out behavior changes. How can the quantization change?? Its almost as if Parkhomov created the data by scaling sections and then finally cut and pasted them together. Perhaps there's a simple explanation, but I couldn't see it.


    I'm not entirely certain what you did, but in talking about temperature quantization, there could be an easy explanation. He was using a limited resolution analog to digital converter. The one-bit increment will correspond to an increment of a certain temperature. You are probably seeing quantization error. It doesn't particularly mean anything other than indicating the resolution of his temperature measurements. This could be checked, going over the data.


    In his first experiment, there was, at a certain point toward the end of the run, massive rapid change. His thermocouple appears to be failing, as the heater was about to blow out. So the "thermal arrest" that meant so much to so many may simply have been thermocouple failure.

  • A first note about Zero.
    Zero is the most important digit in any numbering system. His history is quite fascinating https://en.wikipedia.org/wiki/0_(number) and was imported in Europe by the Italian mathematician Fibonacci who invented the name Zero because that number was "as light as a Zephyr wind".
    Zero symbol also has a terrific importance in mathematics from calculus ( Zero of a function ) to set theory ( element Zero, Zero (empty) set ).


    But also remember that Zero was a long range fighter aircraft of the WWII https://en.wikipedia.org/wiki/Mitsubishi_A6M_Zero.
    It was made ( what a coincidence! ) by the Mitsubishi Heavy Industries.


    It would be absurd, yes. But it relies on your faulty idea that radiation in the IR band depends only on input power and band emissivity - so that a and b (with the same input power) would have the same band radiation.


    Oh ho ! You are making a lot of confusion here.
    What you must have because of equilibrium condition and energy conservation is that two bodies ( even not identical ) with the same input power should emit the same power. Power emission has three terms: Radiation, Conduction and Convection.
    We have introduced the hypothesis that they are in vacuum in order to ignore the convection and conduction terms so that the two bodies must radiate the same total amount of power.


    Only in this point we introduced the hypothesis that the bodies are identical in shape and that their material has equal emissivity in the camera range (as it happens for Alumina and an ideal BB ).


    By those hypothesis we can arrive to the conclusion that using the BB emissivity (band) would give the wrong temperature reading. So one must use total emissivity.


    "Have to" indicates compulsion,


    No compulsion, dear Abdul, I write much less and much short comments then yours so, if any, I'm not the one who has a compulsive disorder among you and me.
    But:
    "There is none so deaf as those who will not hear" and you don't want to hear.
    May be you are a "selective listener"...... or don't want to hear because you must not.

  • Hi Abd,
    Yes, of course there will be quantization due to analog to digital conversion, but this would remain fixed with values always being multiples of the quantization resolution. For example if we had an analog to digital converter with 1.2V quantization, when converting a voltage ramp from 0 to 10V we would get:
    0, 1.2, 2.4, 3.6, 4.8, 6, 7.2, 8.4, and 9.6. Any complex wave ranging from 0 to 10 volts fed in to this converter would always produce these values. We could determine the quantization of the converter by post analysis of the data by looking at the deltas between samples which in this case would be multiples of 1.2V. This is fixed and can only change if the analog to digital convertor is physically replaced or the data is post processed and scaled mathematically.
    What we find with Parkhomov's data are such changes with 10-30 minute chunks showing different quantization values when looking at the delta's of the temperature data. Also I find it curious that most of the quantization changes occur at the points of minor or major data loss.
    I still cannot see how this is possible.

  • @randombit0,


    If the two objects have the same input power, and same effective spectral emissivity in the same band that is measured by a camera, in a vacuum, but one object is a different temperature due to less emissivity over some other un-measured area, then the spectral blackbody temperature for the measured band will be higher for this object.


    The entire object becomes hotter.


    So even the bands that have the same emissivity will report a higher radiant output from the hotter object. They will have spectral black body equivalents, but the two black body equivalents will not be equivalent to each other. They will be black body equivalents of two different temperatures.


    Even though the two objects have the same radiant power in total, each IR band expresses only a portion of the total power. But each band power portion is proportional to the object temperature. And the power emitted in each band is a portion of, or is the same as, what a blackbody could radiant from that band.


    The blackbody power and temperature for any given band does not have to be equivalent to that of a black body for the entire object. The use of a black body simply inserts a standard gauge into the calculations. It can be used for bands of IR, one wavelength, or for total emissivity.


    In fact, it may be constructive to consider single wavelength IR temperature measurements using this conversational model of two objectswith same input power in a vacuum.


    Obviously we we do not use the total emissive power and total emissivity to calculate temperature when using only one IR wavelength. This does not mean all power must be measured in one wavelength. We are not talking about an IR laser, just a hot object.

  • THHuxley wrote:


    Oh ho ! You are making a lot of confusion here.
    What you must have because of equilibrium condition and energy conservation is that two bodies ( even not identical ) with the same input power should emit the same power. Power emission has three terms: Radiation, Conduction and Convection.


    Zero is repeating this over and over. It's completely correct, with the specification given:


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    We have introduced the hypothesis that they are in vacuum in order to ignore the convection and conduction terms so that the two bodies must radiate the same total amount of power.


    We would also need to eliminate any energy storage or production in the object. But that is understood for this discussion. That is, "input power" includes any locally generated energy, and equilibrium rules out any energy storage, which could not be indefinitely maintained.


    Yes. Zero has said this over and over as if it is not being accepted, but it is. Completely. It's understood and we have repeated it, many times.


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    Only in this point we introduced the hypothesis that the bodies are identical in shape and that their material has equal emissivity in the camera range (as it happens for Alumina and an ideal BB ).


    Yes. Again, this is totally understood and accepted as assumptions for the purpose of discussion. Notice that Zero accepts that the material emissivity matches that of a black body in the camera range. That is crucial.


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    By those hypothesis we can arrive to the conclusion that using the BB emissivity (band) would give the wrong temperature reading. So one must use total emissivity.


    Which is utterly false and a blatant non sequitur, that totally denies how these cameras are used. (I interpret Zero's statement about "use" to mean how the camera is set for the determination of temperature.)


    If an object has the same total emissivity as another, and the same power is input, the bodies must be, at equilibrium, at the same temperature. Yes.


    If, however, an object has lower total emissivity than that of a black body, as all non-black bodies do, with the same input power, it must be, at equilibrium, at a higher temperature. Zero has stated this over and over, again, as if we did not understand this. Conversely, if the object with lower total emissivity is at the same temperature as the black body, the input power must be lower.


    What the camera will see is the result of band emissivity in the camera band which the camera bolometer sees. By the conditions of this problem, this is 1.0, i.e., the emissivity of a black body *in that band*. It is the same if there is a black body "dot" placed on the body, or a black body created with a drilled hole, or some high-emissivity paint. That dot will radiate according to black body radiation. But the temperature of the body, with the same input power, because of lower total emissivity of the entire object, will be hotter. And then the black body dot will show black body radiation in the band corresponding to the higher temperature.


    Zero has set up contradictory conditions: differing total emissivity (black body and alumina), same band emissivity (black body and alumina), same input power, and same temperature as shown by an IR camera set for emissivity 1.0 (black body).


    And then:


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    Abd Ul-Rahman Lomax wrote:


    No compulsion, dear Abdul, I write much less and much short comments then yours so, if any, I'm not the one who has a compulsive disorder among you and me.


    I did not mention a compulsive disorder, only compulsion which is, indeed, indicated by the language Zero used, "have to." That has nothing to do with the relative quantity of text. I am retired, and currently must restrict physical activity for health reasons, so I have a lot of time to research and write, and I have always written as others might climb Mount Everest. Because it's there.


    I do notice compulsion from time to time, and usually stop writing and do something else until it passes. My best writing is not done under compulsion; in fact, it can be my worst.


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    But:
    "There is none so deaf as those who will not hear" and you don't want to hear.
    May be you are a "selective listener"...... or don't want to hear because you must not.


    What I am has no relationship to what Zero is, except that I assume we are both human. What I see here is an argument maintained beyond all reason and sanity, where the topic is one that I understand with high coherence, where others agree, and where there is nobody apparently neutral correcting my understanding, only someone with an obvious axe to grind, who at the same time drips with condescending contempt, as if dealing with complete idiots. Does Zero understand how obvious all this is?


    Would Zero tell us if he did? Some people lie so intensively and for so long that they lose the ability to distinguish truth. The end of these people is rarely good. If ever.


    As usual, I invite correction.

  • OK.

    Here is the old thread.

    The IR T transformations between different emissivities can be discussed here.


    So why does Optris say that n=3 in the 7.5 to 13 um ( Rossi vs. Darden developments - Part 2 ) = 3, and also why are the n=17 and n=21 where they are?

    ..............................

    It is T^4 for a blackbody, integrating all wavelengths. So T^21 seems pretty extreme.

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