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

  • Indeed the temperature of an object would be hotter being by limited to IR output in a spectral band portion of the IR band than it would be if it were a blackbody at the same temperature.


    I said that a bit sloppily.
    Let me rephrase:
    Indeed the temperature of an object would be hotter being limited to IR output in a limited spectral band portion of the IR band than it would be if it were a blackbody at the same level integrated radiant power.


    In fact I tested and reported on this effect a long time ago. I tested a frosted silica tube, then painted it black, using the same input power for both. The temperature dropped, and the emissivity went up when black.


  • @randombit0, I'm glad you have posted this detailed and clearly written argument, and don't find such posts with new material tiresome (though maybe many here are now sick and tired of this topic it remains interesting to me). I'm also aware that you and BA are very likely not the same real person. That does not matter to me, and my comment about "another alias" was related to the new approach taken without regard to the real identity of the author, and not meant to imply that I thought the two aliasses had the same author.


    Here is where your thinking on this matter diverges from mine, and I think others. I've emboldened the sentences in your argument where we partly disagree and will explain why.


    Quote

    Now consider a non gray body.Remember that for any frequency any body can't radiate more power then a black body. That is Quantum Mechanics.
    This means that if you have a body that radiates only in a window of the spectrum at a temperature T is not possible that at any frequency the radiated energy per unit area can['t] be higher then Plank curve.


    I've bracketed what I think is a grammatical typo above. Yes, I agree. As Abd correctly points out stimulated emission can break this because it destroys the stat mechs equilibrium assumption but it is generally true that spectral emissivity is never more than 1, and I don't expect weird laser effects in this experiment.

    Quote

    This means that if you have a body that radiates only in a window of the spectrum at a temperature T is not possible that at any frequency the radiated energy per unit area can['t] be higher then Plank curve.


    Again grammatical typo corrected by []. Yes, I absolutely agree. You correctly point out that (say) a selective emitter in the IR band with low spectral emissivity elsewhere will radiate less (at a given temperature T) than a grey body with the same IR band emittance.

    Quote

    But if the integral (limited to the emission window) of the curve must be 100 then we have that the temperature T of the body must be much higher then the temperature T of an ideal Black Body emitting the same total power per unit area.


    This is where we disagree. The problem I have with your statement here is that it is imprecise. According to one interpretation (I guess yours) it is correct, but it does not address the matter at hand. According to another it is incorrect.


    We are comparing two emitters. One (A) is a selective IR emitter with spectral emittance 1.0 in the 13-7um IR window and 0 elsewhere. The other (B) is an ideal black body emitter.


    I'll go with the first (correct but loose) interpretation first. I agree that if B and A have the same total output power then A must have a higher equilibrium temperature than B. That is not quite what you said, but it is true, and it is nearly what you said. There is no conclusion directly from this to the Lugano experiment results.


    However what you are (precisely) saying is that if B and A have the same (integral total 100) output power restricted to the IR band then A must have higher equilibrium temperature than B. This is false. In fact, since the emissivity within the IR band is identical for A and B, for the same IR band output we must have the same temperature.


    The take home here for the Lugano results is that if you enter the (grey-body equivalent) total emissivity < 1 for A into the Optris you get the wrong Optris temperature reading, because A must appear to the optris inbstrument the same as B, and B is a black body which obviously requires entered emissivity = 1.0.


    I hope this helps.


    Your final comment:

    Quote

    Total emissivity, that is the ratio of two integrals, express exactly the equilibrium condition.


    I absolutely agree with this: but it refers to the total power out not the IR band power out. The disagreement is in exactly what happens to the selective emitter IR band output relative to the black body IR output. I'm saying that for the same total output (and hence same total emissivity) the selective IR emitter will have higher IR output (and lower output at other wavelengths) than the black body emitter.


    Regards, THH

  • Your final comment:
    Total emissivity, that is the ratio of two integrals, express exactly the equilibrium condition.


    I absolutely agree with this: but it refers to the total power out not the IR band power out. The disagreement is in exactly what happens to the selective emitter IR band output relative to the black body IR output. I'm saying that for the same total output (and hence same total emissivity) the selective IR emitter will have higher IR output (and lower output at other wavelengths) than the black body emitter.


    It is said that great minds think alike. It should also be said that paranoid and obsessive conspiracy-theorists and fanatics may think alike as well. With the latter, though, it is remarkable to find high coherency, and when it is seen, it may be a sign that this is not more than one person, or that they are coordinated in some way.


    That is irrelevant to the scientific argument, of course. It is a social consideration. However, assertion of personal superiority is also a social consideration, not scientific.


    There are claims being made that "attacks" on Lugano are "paid FUD." Now, for any reader, there are a number of possibilities. I will start with one, which makes "paid" irrelevant, even if true -- and there is no significant sign of that it is.


    Can the reader follow the arguments? Are they understandable? My own experience is that I can follow any sane argument if I put in enough time. Unfortunately, with some arguments, so much time would be needed that I am not going to do it. With study and repeated reading, I can sometimes follow arguments that I do not understand enough to be able to make them independently. For example, Akito Takahashi's application of quantum field theory to a certain four-body problem, I have come to understand his papers sufficiently to notice certain things about them that Takahashi himself does not say: it seems complicated, for example for four bodies to come together in just the right way, but the situation is actually much simpler. He is describing two molecules, not four independent deuterons, with the two molecules in confinement, a somewhat pressurized situation, which requires some energy to create, which would be why this does not happen spontaneously in liquid deuterium. Takahashi is not actually proposing a cold fusion theory, per se. He does not calculate the expected incidence of the configuration, only that if it arises, collapse will take place within a femtosecond, and then fusion by tunnelling to 8Be will be complete within a femtosecond. He is studying something that might be related to the actual cold fusion mechanism. Or not.


    Here, the argument is actually quite simple and should be well within the reach of anyone with basic knowledge of physics. I think this could be explained easily enough to almost anyone, if they are willing to take the time to listen.


    Let me make another effort to simplify this even further.


    Molecules in a hot object interact and give off light. The character of this light depends on the temperature and the properties of the material. The hotter the material, the higher the light emission. Further, if we only look at a narrow color (frequency) of light, the intensity of the light of that color will vary with the temperature, predictably. From the intensity at a color, and with a knowledge of the correlation of intensity at a color for a given material, with the temperature of the material, the temperature can be determined. This is how IR "heat guns" work.


    The relevant property of the material is emissivity at the color involved, and at the temperature involved. The emissivity of materials may also vary with temperature, so one may need to calculate recursively. I.e., use an emissivity for a guessed temperature, then see what temperature the gun reads, then reset the emissivity for that temperature, and keep doing this until the temperature result settles, which is what Lugano did in studying their IR results with the "dummy reactor," which was only at 500 W input, not the full 900W. With the dummy reactor, they had "dots" of known emissivity. Those dots, however, could not handle the high temperature at full power input.


    Lugano obtained the emissivity of "alumina" at 1400 C, a value of 0.4. That was not spectral emissivity, for the detected color used in the IR heat camera, but total emissivity. Total emissivity will determine how much heat energy in the object will be radiated, total radiation, but not the radiation at a particular color.


    THH points to an extreme example, a material that has the maximum possible emissivity, a value of "1", in a narrow band, and no emissivity at any other color.


    "1" there indicates an emission in that range that matches the emission of a "black body." We think of "black," as "dark." Maybe explaining "black body" here will help. Imagine a hole in some material, that leads to a cavity inside. If light enters the hole from the outside, it becomes lost in many reflections inside the hole and because each reflection has a probability that it will be absorbed and not reflected, it never comes back out. When a material is "black," it means that all light is absorbed and not reflected. That cavity is a nearly-perfect "black body." It will be as black as black can be at its temperature. However, if the material is heated, that hole will begin to glow, visibly. It will generally glow more than the material! Black body radiation, this is called, is actually the maximum possible radiation of thermal energy. It is not "coherent," it has a spectrum exactly the same as any other perfect black body. It has emissivity of 1.0 at all frequencies (within some range, by the way, but for all practical purposes, we can say this).


    A "gray body" is like a black body, only not perfectly black. It has no special "color," because the emissivity is uniform for all colors. That is, as to white light illuminating it, it will appear gray, unless it is hot enough to be glowing.


    The Lugano report has a chart showing the total emissivity of alumina at various temperatures. Emissivity declines for alumina with temperature, until it flattens out at about 40% at the temperature they calculated. However, they never mention that their chart shows total emissivity. This chart can be used to predict how fast alumina will cool by radiation at a given temperature. It is not useful for determining temperature, because what is needed for that is emissivity within the color range which is detected by the camera (which is in the infrared, such that temperatures below those at which an object appears to glow can still be measured).


    It was a bonehead error to use total emissivity. However, realize that when one is working outside of one's speciality, it is terribly easy to make such errors. What I would find remarkable is that apparently they did not go over what was in their report with experts in thermometry by IR camera. (Another example of "bonehead error" outside the author field was the report of low-level neutron radiation by Pons and Fleischmann. That can be attributed to haste. And they retracted the result, once they realized the error. The Lugano authors went silent, which is tragic.)


    From their camera readings, if the true emissivity at the camera-sensitive band is known, the real temperature could be known. Using a value that is too low will produce a higher estimate of temperature. (If the emissivity were zero, the object would have to be at infinite temperature!)


    Using an IR camera to measure temperature works with many or most materials, reasonbly well. However, as pointed out, to be sure about the measurements, calibration of the camera with the material at the temperature involved is necessary. Lugano calibrated at a much lower temperature, where the band emissivity apparently was closer to the total emissivity, i.e, the material was more like a gray body. This was blatantly defective procedure and was the first thing to be noticed by scientists reviewing Lugano.


    Why didn't they calibrate, either at full temperature or at full input power, at least? Calibration at full temperature (i.e., what they thought was full temperature), i.e., 1400 C., would have been very difficult, for reasons that also should have made them suspicious of that result! However, calibration at full input power, about 900 W, instead of the 500 W they used, would have been easy. Why didn't they do it?


    I postulate a mental/emotional force field, generated by the presence and influence of Andrea Rossi. "Plausible explanations" that make no sense are accepted, because "Rossi says." I think I understand, at least a little, the psychology of it. But that is another topic.

  • In fact, since the emissivity within the IR band is identical for A and B, for the same IR band output we must have the same temperature.



    No. You and Lomax are missing the basic physics.
    Again ( and again)
    We have two system a) ideal BB b) Alumina with constant (known) input power 100 at equilibrium in vacuum ( no convection)
    The two system are geometrically identical.
    1) because of equilibrium condition input power must be equal to the output power.
    2) because of QM emissivity of b) can't (no typo is NOT POSSIBLE) higher then that of a) (ideal BB) at any lambda ( or frequency )
    3) because of 1) the integral over all lambda of the emitted power density must be equal so that integrating again over the body surface (identical for the two bodies ) we obtain the input power 100.
    4) because of 2) and 3) then we obtain that b) must have a much higher temperature then a.


    If you need a simplified explanation:
    5) because the actual shape of the Plank curve is not relevant to our reasoning we can suppose it constant up to a frequency Lth, Integrating the curve for a BB a) is in that case equal to calculate the area of a rectangle with base from 0 to Lth. The height of the rectangle is function of T and the total area give us the total emitted power density.
    6) suppose that you have a selective emitter b) that can emit only in the interval [La,Lb]
    7) so we compare an area of a rectangle with a long base [0, Lth] wit one with a much shorter base [0<La, Lb<Lth]. Because of 1) the two area must be equal.
    8) This implies that the height of b) rectangle must be higher and so its temperature.


    Next time we will make a complete derivation of Plank law with all the math.

  • Hello RandomBit0,


    Please check out the following link to a thread I started in which I share my thoughts about the Rossi Effect and asked for your opinion and thoughts. I'd really appreciate it very much if you would check out the thread and join me in a conversation. There is a LOT that is really coming together in my mind, and I'd really like to discuss it with you. Especially INTERGRANULAR HYDROGEN BUBBLES in nickel.


    An Open Conversation With RANDOMBIT0: Constructive Comments Appreciated -- Cynical Remarks Frowned Upon

  • @randombit0,


    Total radiant power is irrelevant to detecting temperature based on a selected IR window.
    The same Planck curves are plotted in both images below. The segment is equal to the IR camera spectral range view of the entire curves.
    I could find the blackbody temperature crossing point also, but it would not look very different than the 0.9ε curve, and the T would be a bit lower (pretty close to 985 K).

  • @Paradigmnoia


    There is no chance that a person somewhat educated in physics still misunderstands this issue.
    To simplify and condensate in one sentence: the camera senses power irradiated in a spectrum region where alumina has an emissivity much higher than average, and thus it computes a higher temperature than real if one inputs the average (i.e. total) emissivity.
    The amount of this blatant overestimation was proven experimentally by MFMP live on camera.
    One can accept that the original mistake by Levi and co was done in good faith. But the reiterated confutation by Randombit0 can only aim at confusing the uneducated.
    Randombit0 likely has her motives, but I am astonished by the silence of the coauthors of the Lugano paper.

  • Quote

    Next time we will make a complete derivation of Plank law with all the math.


    I think the first step in the derivation is to spell Planck's name correctly.


    https://en.wikipedia.org/wiki/Planck%27s_law



    Quote

    One can accept that the original mistake by Levi and co was done in good faith.


    No, one can not, when the simple act of calibrating properly could have corrected the errors. That also applies to previous tests including all of Lewan's lame demos with the original ecats and Levi's mass calorimetry with the original ecat, the data from which he refused to produce for Krivit and Josephson. One would have to believe that none of the celebrated scientists and technologists involved with Rossi have ever heard about calibration. Is calibration now a top secret concept?

  • It is not needed, and will rightly not be popular, but I find myself unable to stop from answering randombit0's latest misreading point by ghastly reiterated point:


    Quote from rb0

    We have two system a) ideal BB b) Alumina with constant (known) input power 100 at equilibrium in vacuum ( no convection)
    The two system are geometrically identical.
    1) because of equilibrium condition input power must be equal to the output power.


    yes

    Quote

    2) because of QM emissivity of b) can't (no typo is NOT POSSIBLE) higher then that of a) (ideal BB) at any lambda ( or frequency )

    OK, I'll agree since I'm not arguing this. (Technically, due to non-equilibrium effects, it is not quite true, but good enough).

    Quote

    3) because of 1) the integral over all lambda of the emitted power density must be equal so that integrating again over the body surface (identical for the two bodies ) we obtain the input power 100.


    Yes

    Quote

    4) because of 2) and 3) then we obtain that b) must have a much higher temperature then a.


    No-one disputes that. But so what? That is not the issue here. b) will also have a much higher IR radiance than a) (because of the restricted band) and therefore register on any IR camera as higher temperature than A whatever emissivity you put in. The question is: how much higher?


    Quote


    If you need a simplified explanation:


    (I don't, but either you are deliberately falsifying matters, or you do)

    Quote

    5) because the actual shape of the Plank curve is not relevant to our reasoning we can suppose it constant up to a frequency Lth, Integrating the curve for a BB a) is in that case equal to calculate the area of a rectangle with base from 0 to Lth. The height of the rectangle is function of T and the total area give us the total emitted power density.


    This is missing the point. If your point 4) were in any way relevant to the Lugano test issue (what are the combined effects of using total emissivity for object of type (b) above in an optris camera, and using the same total emissivity to claculate power out) I would have some sympathy. The conditions in point 4) are not those of the Lugano test data.


    Also it is misleading, the exact shape of the Planck curve response wrt T is very relevant to the matter at hand (though not to your point 4).

    Quote

    6) suppose that you have a selective emitter b) that can emit only in the interval [La,Lb]
    7) so we compare an area of a rectangle with a long base [0, Lth] wit one with a much shorter base [0<La, Lb<Lth]. Because of 1) the two area must be equal.


    What you argue is tautologous. If the power out is equal then the area under the radiant power density versus frequency graph (which itself is a product of the Planck curve and the spectral emissivity curve) must be equal. Therefore, equally tautologous, a material with lower total emissivity and the same power out and surface area must have higher temperature.


    No-one here is arguing against that. Your mistake is to think that this equality somehow makes IR camera calibration with total emissivity OK for non-grey-bodies. since we have (all of us, many times) taken you through in detail why this is not the case, suppose this time you fill in the dots of your own argument.


    Let us have your argument from point 4 (which we all agree) to what you want to claim about Optris camera temperature measurement. It will be interesting to see where exactly you go wrong.


    regards, THH

  • Quote from Andrea.s

    There is no chance that a person somewhat educated in physics still misunderstands this issue.


    The ability of some people to attend courses, and to pass exams, while remaining profoundly ignorant is very high. So I'd be more cautious than that!

  • The ability of some people to attend courses, and to pass exams, while remaining profoundly ignorant is very high. So I'd be more cautious than that!


    Yes, and then some read books while in prison. Some do well with this, some not so well.

    • Official Post

    The ability of some people to attend courses, and to pass exams, while remaining profoundly ignorant is very high. So I'd be more cautious than that!



    Well la te da TTH! :) There also exists an ability that some have to teach, or explain to others complex matters in simple terms, that *others* lack. It actually takes a special intelligence to do that. Condense it down to it's bare essentials. You spent 6 months immersed in this before making your fine report. When you started, you admittedly knew little, to nothing, about emissivity. Now you know a lot.


    Still, after all you learned, you struggle to put it simply. And as has been pointed out by AC/Wytenn...this is pretty basic physics. Not your fault for lacking teacher attributes, but after your belittling comment I just could not resist. ;)


    Remember, it is we peasants that are the ones that will show up at you bourgeois intelligentsia houses with the torches and pitchforks, so be careful what you say. :)

  • @Shane D.,
    This subject is fairly tricky to make simplified for those that are "not getting it".
    Here is one attempt:
    Imagine that you have special sunglasses on that only pass green light, blocking all other colors. You job is to look at various objects, and rate their greenness, say between 1 and 10. The objects in normal view can be any color, combination of color, and any brightness. The combination of sunglasses, your eyes, and your brain are now the rough equivalent of the Optris camera looking at IR radiation.


    An object with lots of green, but lots of other colours could look the same with the special glasses on as an object that is only green. They could even rate at the same greenness level. But the multicolour object will have more color power. But that is irrelevant to the greenness rating.


    Now, if the object was really bright, you may think that was more greenness. But is not, it is just as green as the same object with less brightness. So the greenness emissivity is a new factor we multiply your greenness rating by, to raise or lower the brightness effect on your greenness rating, so that the correct greenness value is found. This greenness emissivity factor is defined as between 0.01 to 1.0 . You cannot have more than 1 greenness emissivity, because that is maximum greenness.

  • @Shane


    The Lugano stuff is trickier than you would think because of the Levi Defense - which sounds like something from a chess manual and the modified Levi Defence. The Levi Defense says that all is good because of the Optris temperature/book emissivity iteration. The modified Levi Defense, while admitting that maybe it does not exactly work, notes that even if emissivity is 1 COP=2, which is positive. And finally the Cleverer than Levi Defense. This agrees that the temperature is calculated wrong but states that the change in power out due to lower emissivity exactly cancels the change in power due to higher temperature calculated from lower emissivity.


    None of these things are exactly simple. So if you have a simple explanation it will be a crowd pleaser but will not convince someone thinking a bit harder.


    Regards, THH

  • This subject is fairly tricky to make simplified for those that are "not getting it".
    Here is one attempt:


    Fail. Try again, I believe you can do better. Keep it simple. I think I understand the issue and could not follow the explanation. Bad sign.


    Shane, this stuff can be difficult to explain clearly; part of the problem would be variable understanding of the underlying physical principles in the readership. I'm going to agree that being able to express these things simply and clearly, so that almost anyone can understand who is willing to pay attention to the explanation, is a high skill, and not common.


    I will also assert that there is reason to think that such an ability is associated with depth of understanding as well. I always found Feynman to be crystal clear. I also know that one may have high understanding and still not be able to express it so grandmother can understand. Hah! Sexism! What if grandmother is a Nobelist in physics? Actually, so "a child" can understand is a better formulation.


    It can be more difficult to explain to adults, by the way, because of held preconceptions.

  • @Alan Smith,
    I don't disagree.


    It is a complex set of interactions that I thought the color thing made sense with, but became mired in strangeness when when I tried to squeeze a bit more detail in there.


    It's like taking your temperature with a typical glass thermometer instead of by sticking your body in a calorimeter on a scale to see how much heat you make, to order work out your temperature...(??)

  • The Lugano stuff is trickier than you would think because of the Levi Defense - which sounds like something from a chess manual and the modified Levi Defence. The Levi Defense says that all is good because of the Optris temperature/book emissivity iteration. The modified Levi Defense, while admitting that maybe it does not exactly work, notes that even if emissivity is 1 COP=2, which is positive. And finally the Cleverer than Levi Defense. This agrees that the temperature is calculated wrong but states that the change in power out due to lower emissivity exactly cancels the change in power due to higher temperature calculated from lower emissivity.


    None of these things are exactly simple. So if you have a simple explanation it will be a crowd pleaser but will not convince someone thinking a bit harder.


    My concern would be about writing what will make the topic accessible to anyone sufficiently interested to read it carefully. What you have above makes "it" sould complicated, but that is because more than one issue is being mashed together.


    When I saw the Lugano report, what I saw was a massively complicated explanation, difficult to follow. I don't trust complicated explanations. Maybe they are necessary sometimes, but they create hordes of opportunities for error. Ideally, peer review will vet them, but the Lugano report was not peer reviewed. So, unless one buckles down and tackles the massive detail, the Lugano report is more or less take it or leave it. So those who trust them praise the report, and those who don't perhaps notice the gaping errors, like lack of calibration.


    There are two basic issues.


    1. The temperature of the device. That is actually complex, because the device is not small and simple and at a uniform temperature. However, they claimed 1400 C, and that was claimed as external, and it must have been for a substantial area of the device. This is simpler than calculating power dissipation.


    2. The power dissipation, which they calculated from the temperature with a host of assumptions and procedures. Heat is dissipated in three ways: by conduction, by convection (in this case of air), and by radiation. The major conduction pathway is air, so this falls under convection.


    I found the report extremely difficult to read. What is in one place may be contradicted in another.


    Going from IR emission as measured by the camera and an emissivity value to a temperature is relatively easy to understand and follow, if one has a reliable emissivity value for the material in the camera band in the temperature range (that is what most conflict is about on the first level). From there to power dissipation is very complex.


    Radiated power varies directly as the fourth power of temperature, it is not linear. Radiated power is based on total emissivity, not band emissivity. The full relationship between emissivity, camera readings, and temperature, is a solvable problem, and relatively simple, but the calculation of COP is then far, far more complex.


    In normal calorimetry, the complexity is avoided by calibration. As long as conditions are stable, calibration will correlate input power with the IR camera readings, all the complexity is bypassed. One doesn't even need to know the temperature, but, of course, it is desirable. This is why failure to calibrate at operating input power was fatal to the clarity of Lugano.


    It might be worth quoting McKubre. This was written before other more focused critique was available. My emphasis. http://www.infinite-energy.com…ne/issue118/analysis.html


    Quote

    All well and good. Is this confidence justified by the words in the report? Is there evidence of excess heat? My impression is “Yes” (but see below). Is this evidence unambiguous? Not as presented. Is there evidence of nuclear transformation? Yes, very clearly, but questions remain to be answered (or, in some cases, asked). Do the heat and nuclear production correlate quantitatively? Yes, possibly. Is the report perfect? No, no report is perfect, but this one is imperfect in little ways and large. There is curious inattention to detail—surprising for a document as delayed, anticipated and important as this. When asked to provide a review (sight unseen) I agreed; this is important. But I also realized that unless the report was perfect in every detail, whatever I wrote would annoy somebody. Here goes.


    Following is the most important point he raised. Again my emphasis.


    Quote

    Of primary concern is the complete absence of relevant pre- or post-test calibration. The mathematical description of heat losses from a structure with multiple parts and awkward geometry is not difficult but contains many terms and several assumptions. Perhaps with more direct experience I could trust the equations and assumptions to do their job, but I do not have that experience. Neither do the authors, nor does most of the “audience.” The only way to handle such uncertainty is via relevant calibration, ideally under identical conditions at and surrounding the operating point. The calibration that was performed was at a single point well below the points of ultimate operation. This is bad and bad. We need to determine the shape of the performance manifold empirically, ideally bracketing all points of operation. This was not done.


    The full article is remarkable. McKubre generally accepts the result -- he is inclined to trust scientists --, but then reveals the serious problems that could possibly torpedo the whole thing. And they did, apparently.


    Scientifically, the Lugano report is useless. We then hear that Industrial Heat attempted to confirm the heat results -- they made that reactor -- and were unable to do so. This is devastating.


    It is a single experiment, it is unconfirmed, and it was defectively done. Add to this that they won't talk about it. Lugano is dead.

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