Hope springs eternal

Rossi-Blog Comment Discussion on LENR Forum. This starts with Alan Smith’s announcement that he will be attending the Rossi Quark-X demo in Miami (apparently), but then looks at discussion and general insanity around the Quark-X electrical measurements. Many other topics intruded into the thread, but mostly I stuck with the electrical issue.

Alan Smith, Nov 8th, 2017, 10:35 am (+1)

TRIP To DEMO – THANK YOU ALL.

My thanks to those members ( and lurkers) who have donated so generously (and unexpectedly) toward the cost of my trip to the Andrea Rossi show on the 24th. I am most grateful, and am now in a position to say ‘no more!’ The hotel and travel bill is covered – I am happy to eat at my own expense as always. The demo is (hopefully) going to be livestreamed as it happens, but whatever, I will try to file some kind of report ASAP after the show -which I guess will then segue into dinner. There is unlikely to be WiFi available to all at the venue itself so it will have to wait until I get back to my hotel. I expect to produce a more considered write-up shortly after the day.

November 24, Rossi’s Quark-X “show.” I’m glad to see someone attending, and Smith may be the best we could get. I didn’t try. I do have funding for travel, but the expected return here will be too low, particularly if there are any other reasonably objective observers, and Smith is at least close to that.

I found the announcement looking over that discussion for details and bloviation about the Quark-X. I will start with:

Alan Smith wrote:

Nov 8th 2017 3:59 pm

interested observer wrote:

@Alan: […] Please enlighten me: what past successes can you cite for Rossi? Clearly he has impressed you with his accomplishments. What are they?

A lot of people much smarter (and better physicists) than me think that Rossi has something. That is what impresses me, as well as the increasing number of commercial and other researchers looking at Ni/H as well as Pd/D.

This is also what impressed Darden and Vaughn in 2012, leading them to invest a total of about $20 million in Rossi technology. Plus legal expenses when they found it didn’t work and refused to give Rossi another $89 million, and he sued, claiming fraud. The end of that: the suit was dropped, and the parties walked, Rossi getting his reactors back, and IH surrendered the license, demonstrating clearly that, to them, the reactors were worthless. Not worth spending a few more million in legal fees that they might never see come back, though Rossi’s case was going down in flames.

Physicists are not the kind of experts one wants: as Jed Rothwell has pointed out, one would want HVAC engineers and I would add chemists, and for the Fleischmann-Pons experiment, electrochemists. I notice that Smith doesn’t name anyone. Gullstrom, perhaps? The PhD student at Uppsala who put his name on a collection of unintelligible nonsense, based on Rossi Says? Or how about Levi, Essen, et al, who also put their names on preposterous nonsense (that looked sort-of interesting when it first appeared, but which fell apart on examination), which they have not bothered to defend (the Lugano report).

Sure. It was indeed ‘impressive.’ Five years ago, that is, and there is a lot of water under the bridge since then. What the Industrial Heat (Darden et al) affair showed was (1) Per attested documents, his own emails, etc., Rossi lies. (2) A serious effort to confirm and replicate, well-funded, was unable to do so, when it was Rossi’s job to fully assist them as needed. Instead, there was obfuscation and more obfuscation, all now public record.

Jed Rothwell wrote: (November 8, 2017, 3:44 pm)

axil wrote:

The IH doral test was a battle to thwart the theft of Rossi’s IP by venture capitalist who had no intention of making good on their commitment to pay when all means were fair to use in that struggle.

This is batty on many levels.

It is Sifferkollian, one of the highest levels of batty!

First, there is no evidence that I.H. did not intend to pay. On the contrary, they did pay $11 million, which indicates they intended to pay more.

They put about $20 million into the technology, not just the $11.5 million  they gave Rossi. They worked for years attempting to confirm the technology transfer. They received $50 million from Woodford Fund, which was for other LENR technology, not Rossi (and that clearly set Rossi off), and they had received a further commitment of $150 million if needed. They were willing to pay Rossi even though Rossi’s friends (not Industrial Heat) had shot down the Second Amendment that allowed postponing the Guaranteed Performance Test, designed to trigger the $89 million payment. But they had one condition:

That Rossi show them how to make devices that work. The fact is that without that, they could not, in conscience and commitment to their investors, pay anything more. With it, $89 million would be cheap, and they knew it.

That “they never intended to pay” was the basis for Rossi’s fraud claim. It was legally and massively defective, but that story comes from him, and there is no evidence for it that appeared in Rossi v. Darden. There is plenty of evidence in the other direction.

Second, even if we assume for the sake of argument that I.H. intended to steal the product, Rossi’s antics would not have prevented that. I.H. would have known from their own tests that the machine works. Rossi’s fake tests, his fake customer, his imaginary Mezzanine heat exchanger, and his preposterous Penon report would not have prevented I.H. from stealing anything.

Unless Rossi defrauded them, the technology passed to IH in 2013, with the $10 million payment. What appears clear is that there was no commercial technology. People may argue forever about whether or not there is any Rossi Effect, but Rossi is now, apparently, acknowledging that the Doral reactor (which was the successor to his original 1 MW plant, supposedly ready for sale and use in 2011) was not ready for commercial application, and he points out that to make it work he had to be there continuously, and it was breaking down, failing, was about to die at the end of the “test.” The question remains if that reactor produced any heating other than input power, the test protocol was vulnerable to many possible artifacts and errors, and Rothwell quite correctly points out that everyone in the building would have died from the heat if it had actually functioned.

Rossi did not appear to recognize this in 2016, when he filed the suit. (At that point he gave a series of shifting answers to the heat dissipation problem. First it was that he was running an “endothermic reaction” with the heat. Totally inadequate, that was, so he then wrote that it was a combination of dissipation mechanisms, endothermy, okay, but also heat out the back doors and through the vents. Also inadequate. By 2017, he had to do something. So then we get the story of the Amazing Vanishing Heat Exchanger, that nobody saw, and that may or may not have worked as described if built. Given the host of clear lies that came out (Rossi was pushing perjury), I very much doubt that a jury would have believed this story, and the IH case was very simple and easy to understand.

But those who want to believe in Rossi invent or swallow complex Rube Goldberg schemes, a conspiracy theory.

Third, this cannot be falsified.

Ah ha ha ha! ROTFL! Jed expects other than pseudoscience from an anonymous troll, who never backs up his comments with verificable fact?

Anything that Rossi did might fit this imaginary scenario. If doing a fake test or fraudulently representing yourself as a third-party customer would somehow prevent I.H. from stealing the technology (which it would not), he might just as well have done other unethical and illegal things, such as signing up new customers in violation of agreements with I.H., or burning down the warehouse. There is nothing he could do that would not somehow fit this imaginary scenario.

Rossi appears to have retroactively justified withholding the technology from IH because of his idea that they never intended to pay. His original filings claimed that they had no means to pay, that when they entered the agreement, they didn’t have the $89 million. This was a crazy argument, this would be like claiming fraud in a mortgage agreement with a balloon payment in ten years, based on the mortgagor not having the money when signing the agreement. (And actually based on failure to pay later — which was the core Rossi allegation, and they actually did not pay. As pointed out, by IH counsel and others, failure to pay is not evidence of fraud. It is a simple contract issue. The judge decided, temporarily, to allow the fraud claim because maybe Rossi would find something in discovery showing intention to defraud. He didn’t, apparently.

But now Axil reasserts this, like Sifferkoll before him and others dwelling on Planet Rossi.

There are many interesting byways in that LENR Forum thread, but I want to focus on descriptions of the Quark-X and the basis for them.

Adrian Ashfield wrote, November 8, 2017:

LENR Calender wrote:

I am hoping the demo won’t have obvious flaws such as:

– questions about quality of the steam

– voltage being measured on a resistance and not on the reactor

What flaws?

How about the ones he mentioned, or others?

If you had been following the subject you would know the calorimetry does not involve steam.

I think I’ve seen that. This is an irritating characteristic of LENR Forum discussions. Facts are alleged without citations. “following” where? There are three basic locations where this would be discussed: Rossi’s blog, JONP, E-Catworld, and LENR Forum — massive –, plus there is the Gullstrom paper, (the newest revision?). The calorimetry I have seen asserted was unclear. It appeared to be looking at temperature rise in a very short time of a heat transfer fluid (not water). It was more or less a joke, the information was so sparse.

What is wrong with measuring the voltage across a known resistor?

It’s not “wrong,” but it provides no measurement of power input to the reactor, only current at some point in the process, not specified. It was a 1 ohm resistor, and DC voltage, measured with two meters, no less, was about 100 mV, so current was about 0.100 mA. This tells us nothing about the power input to the reactor.

Apparently the reactor has close to zero resistance. Measuring the power going into the controller would be a good idea, but that’s not what you said.

Measuring the power going into the controller would be a check, but what appears likely from the data we have is that the “controller” is an adjustable constant-current source, common with power supplies, and it was set to a current of 100 mA. The voltage will then adjust until the current setting is reached. As described in the paper, the device is two nickel rods with a “space” of 1.5 cm between them. Depending on what that space is filled with, the resistance is likely to start out high, unless the voltage reaches “striking voltage,” where the gas enclosed is ionized.

For this to work, then, the “controller” must first create the striking voltage, which could be hundreds of volts. As the plasma is formed, resistance lowers. This is called “negative resistance,” but that name can be misleading. The resistance is never negative, it remains a positive value.

If the gas and other conditions were known, resistance and voltage drop across the device could be estimated, but they are not known. It is very unlikely that a current of 100 mA would make such a strong conductor out of the plasma that resistance would be effectively zero. (Normally, with zero resistance, there would be no plasma heating and the temperature would fall. Of course, the idea is that something in the device maintains the plasma. However, a problem would be a lack of control. Very low resistance means very low control voltage, and if power from a reaction is high, the control power would be irrelevant.

I covered these measurements in this blog post, back in July.

The data in the Gullstrom and Rossi paper is utterly inadequate to allow calculation of device input power. Yet that doesn’t stop our resident pseudoscientists from bloviating.

As I pointed out before, the setup, with two meters measuring current, is very odd. As these kinds of meters are quite reliable, I have never seen such a redundant measurement. Rather, what would be normal is a measure of the voltage across the device, combined with a measurement of the current through the device. The 1 ohm resistor is, then, a current sense resistor, very normal. What I suspect is that, to conceal the actual input voltage (sensitive information!), the voltage sense lead was moved to measure the drop across the resistor instead of the device, the load. The actual text of the Gullstrom and Rossi paper:

Experiment with energy measurement done with a heat exchanger

The system is displayed in figure 5. In the figure, the yellow thermometer measures the temperature of the oil inside the heat exchanger.

This is an ordinary digital thermometer with an LCD display. It appears to be a Taylor 9847N. These are very cheap! Resolution is 0.1 C, and it updates once per second.

In the left in the figure there are two voltmeters that measure the mV of the current [sic] passing through the 1 Ohm brown resistance.

This means “the voltage drop in millivolts generated by the current passing through …” As it is a 1 ohm resistor, it could more simply have said “that display the current in MA passing through…” (But the meters are actually measuring millivolts.)

In order to not burn the equipmet [sic], the experiment was set in a state with lower output power.

This is not an actual experimental result, but some kind of explanation provided, i.e., chatter. A more thorough report might have shown values at various control levels, showing higher output power, and then stating, perhaps, that “this series was not continued because of concerns about exceeding allowed operating temperatures” or something like that. Here, this only looks like unsubstantiated hype. Not surprising, since we have an indication from Mats Lewan that Gullstrom did not actually witness a test. This might show that, but it’s a bit weak, so what I’ve read, if based on that, is not entirely clear. But this is what we have seen in Rossi tests (such as Lugano). The reports gives information that must have been supplied by Rossi, as if it were simply fact.

Calculations of the calorimetry made by the heat exchanger:

Terrible English. Heat hxchangers do not make calculations, people do. However, this would mean “using the heat exchanger.”

efficiency of the heat exchanger: 90%

I would think this means that 90% of the heat emitted by the device shows up in the oil as a temperature rise. How this was measured is not stated. This is, again, common with Rossi demonstrations: there is no control experiment or calibration.

Primary heat exchange fluid: lubricant oil ( Shell mineral oil )

Uh, then, what’s the “secondary”? This is simply poor language, I think. There is no secondary fluid. The “heat exchanger” appears to be the T-joint, containing the oil, and the device is immersed in it. The efficiency is puzzling, unless there is an insulating sleeve that inhibits heat transfer to the T joint. That would cause faster temperature rise for a given level of power dissipation. Poorer insulation would allow longer operation (before it gets too hot) and possibly more precise measurements, but that’s a bit speculative.

Characteristics of the lubricant oil: D = 0.9 Specific Heat: 0.5

No units given. Shoddy. D is presumably density, or grams/cm3, and Specific Heat would be cal/g-oC.

Calorimetric data of the fluid: 0,5 Kcal/h = 0.57 Wh/h

Why am I not surprised that the paper uses Rossi’s trope: Wh/h, which is simply W? A Watt is one joule/second. It is thus a rate of adding heat to the fluid. 0.57 W is 490 calories/h. 0.5 Kcal is close enough. This is not a characteristic of the fluid, at all, it’s simply the relationship of the units.

Flow heating: 1.58 C / 1.8″ x 11 g

As I read this, it’s puzzling. The symbol ” appears to be used for “seconds,” which is an informal shorthand not normally used for time, but for angles. The SI unit is the second, abbreviation s or sometimes sec. Then 1.8 seconds is a very short time period for a measurement. What was actually measured and how was it measured? If heat rise was measured for a longer period, which I would expect, then the total rise for the total period would be reported, but this is what this looks like. i.e., rise/period. Or to find the heating rate, one might then divide the temperature rise by the number of seconds, but in that case, what would be reported would be so many degrees per second, not per 1.8 seconds, and it is give to two decimal places, when the thermometer only reads to one — and updates once a second, so the error could be considerable.

Consider the practical difficulty of using a digital thermometer for 1.8 seconds. How would it be done? There is no electronic control or readout. Human reaction time is a substantial chunk of a second.

However, accepting the numbers, with 11 grams of fluid,

0.5 cal/g-oC x 11 g gives us 5.5 cal/oC, and a temperature rise of 1.58 C/1.8 sec  is then 4.83 cal/sec, or 17.4 Kcal/h, or 20 watts.

So the math is correct for those figures:

Resulting rating: 20 Wh/h

Or just 20 watts. It would be 20 w-s/s or 20 w-min/min or 20 w-day/day.

Energy input: V=0.1 R=1 Ohm → W=0.01

This, however, is not stated as Wh/h. The difference is?

However, it really doesn’t matter. This is not the “energy input,” and that Gullstrom signed off on this demonstrates that at best, he’s not paying attention to what is published with his name. This is the energy dissipated in the resistor, it doesn’t go into the device at all. If the resistance of the device is really very low, then the input power would also be very low, but there is a huge problem. There is no sign of any control inputs other than the current. As well, it takes substantial voltage to strike a plasma.

What is missing, radically, is an actual description of the experiment. Planet Rossi excuses this by “industrial secret.” Bottom line, though, we can tell almost nothing from this information. The calorimetry would be punk and quite inaccurate, for no apparent reason. (Why not let the temperature rise by a hundred degrees? It’s oil, it would not boil. Or at least a larger amount than “1.58 C” which is miniscule, with a 2% digitization error, plus other possible measurement problems.)

Alan Smith wrote: (November 8, 2017)

LENR Calender wrote:

The reactor having zero resistance is not “apparent” to me. It will be apparent if Rossi plugs a voltmeter across it.

How does that work then?

When someone like Smith asks such a terminally dumb question, I’m very tempted to call it “trolling.”

Adrian Ashfield wrote: (November 8,. 2017)

THHuxleynew wrote:

Rossi is up a creek with no paddle on this one. What he wrote in the TWO research papers is just wrong, the “low resistance” excuse came after much ECW worry and request for post-hoc justifications. And the resistance across the reactor is not known that is Rossi (as often he does) misleading. Very low is vague, unless it is quantified with a definite limit, in which case we would not say unknown, but known to a given precision.

Sounds like FUD to me. What creek is Rossi supposed to be without a paddle?

It’s an expression, I think it’s American. That sentence can be removed without changing the meaning much.

As far as I know, no LENR theory has yet been accepted and there are plenty of them. Your theory that it is simply not possible is popular and probably wrong too.

THH did not mention any theory, and I don’t think he believes it is not possible, and he’s certainly been willing to look at possibilities. So this is a non-sequitur response. But it does attempt to raise one actual issue:

“…with a fluorescent bulb, you will find that the ratio of applied voltage to resulting current decreases with increasing applied voltage.

That’s true, because a flourescent tube is a plasma device, and plasmas generally exhibit “negative resistance,” but this does not mean that resistance goes to zero, it doesn’t. It generally remains substantial. Consider that flourescent tube. Say it is a big one, 40 watts. The resistance before the plasma has been struck is very high. Once there is a plasma, the resistance drops. It’s not linear through the AC cycle, but as a rough idea, the wattage indicates an average current, with 115 VAC, of about 1/3 ampere, RMS. So the resistance must be about 350 ohms.

Couple that with a device that produces electricity and I suspect the answer is complicated. When I said the resistance was not known I meant not published, I assume Rossi has a pretty good idea what it is.

If a device in the circuit produces “electricity”, it all becomes unknown. What Rossi has said about the resistance does not particularly make sense, and devices that produce current still have an internal resistance. This comes up in the discussion, there is more and more nonsense. I would also assume that Rossi knows, and that he is simply lying without lying by being vague, which is what THH was writing about, not “theory.” Or to be more charitable, keeping his secrets.

Alan Smith wrote: (November 9, 2017)

LENR Calender wrote:

I’m not sure if I understand your question.

But if Rossi claims that a certain voltage is ~0 , he can certainly demonstrate it.

And I’m sure you can figure out better than me whether Rossi is providing an adequate measurement of input power.

My electrical engineering is rusty, but I don’t think the Rossi/Gullstrom paper had a proper setup. It only worked on the assumption that V=0 across the reactor. Do you disagree?

Not at all. But as you point out above, this is not a good method in the absence of other tests.

But the question Alan asked was “How does that work, then,” about measuring the voltage across the device. LENR Calender  is correct, the Gullstrom paper appears to assume (it “works”) if the voltage across the reactor is zero. We apparently know the current through the reactor (if not, then something is even more misleading than thought).

“Not a good method” understates the matter. This was a horrible way to measure and report input power to the device. It flat-out doesn’t work, and surely Smith knows that. So why the useless conversation? I don’t know why, but Smith does it frequently.

Adrian Ashfield wrote: (November 9, 2017)

THHuxleynew wrote:

As I said, in not one but two research papers with experimental write-ups added, he explains what he measures, and it is not the reactor input power. Now, maybe what he really measured is different from what those papers say. But he got criticised in the first one, and repeated it in the second.

I’ve seen several mentions of “two research papers.” I know of only one, which exists in three versions.

When your latest best device’s only experimental test write-ups do not measure input power (but measure something else and claim it is input power) you have the creek/paddle scenario.

Right. It’s a blatant error. The “input power” measured, as 10 mW, is not power going into the device, it is the power dissipated in a 1 ohm current sense resistor, a standard setup, but with the other measurement generally made with it (voltage across the device, or total power supply voltage, either one would do) missing. Normally, in that kind of setup, the power dissipated in the sense resistor is neglected. For accuracy, the voltage is measured across the device under test, but normally a sense resistor will be chosen so that the voltage drop across it is small compared to the voltage across the device.

Maybe the upcoming test will correct this? As somone with a respect for history, I’d expect not. We will see.

The history, in this case, suggests that strong optimism would be misplaced. There is no basis to suggest some major change in approach.

There was little description of the experiments in the theoretical papers.

That is correct. But what was there was incorrect.

I would guess at that time he was using a large controller that used quite a lot of power even when it was not powering the reactor.

That would be irrelevant, because an ordinary measurement of input power to a device would measure current through the device and voltage across the device. The power supply current to other equipment is only relevant to an attempt to detect fraud. If the device under test is clearly displayed with no hidden wires, etc., this is not necessary with a first-level test. However, these tests are not independent and half-measures would be useless. A serious investor, given the history, so much of which is now public record because of the lawsuit, will want to see independent tests … or risk being the next big loser in a series of them.

(Industrial Heat did not actually “lose,” because their boldness with Rossi, their willingness to risk their own $20 million, led to a larger investment by Woodford that did not go to Rossi. And they understand all this, I’m sure, as a learning experience, getting ready for an actual commercial opportunity to appear, learning how to recognize it, and how to avoid what is not real.)

What Rossi did by measuring the voltage across the 1 ohm resistor was probably good enough to measure the actual power used by the E-Cat QX.

That measurement provides no information on the power consumption of the QX. It could be anything from zero (at zero resistance) to quite high. Suppose the power supply was 200 V. At 100 mA, the measured current, the power input would be 20 W. In other words, zero excess heat. There are other issues about duty cycle; bottom line, we don’t have a reliable report, and the Gullstrom paper appears to be a slightly more sophisticated — and deceptive, because of Gullstrom’s signing off on it — Rossi Says.

A photo of a power supply allegedly used for the QX test shows a CS  3005T benchtop supply.  This would be 30 V at up to 5A, current or voltage controlled. However, that would not be enough voltage to strike a plasma, so some sort of buck or boost must be used.

Axil wrote:

Adrian Ashfield wrote:

I see the trolls are back.

For those that have suggested a light show, if the E-Cat QX shell is transparent boron nitride, it is not stable in air at those temperatures and requires a sheath. Possibly a quartz sheath filled with flowing clear oil might work.

Alan Smith has reported first hand observation of light produced by the Quark. There could also be UV light produced to account for the report of radiation damaging the unshielded eye. Boron nitride is very transparent to UV if I recall.

When we are utterly convinced of something, we don’t notice errors in statements that seem to support it. Trolls will also make statements that are wildly defective. I have not seen the statement from Smith, but he did not see the light, he would have, at best, reported that he was told by someone that they saw it “first hand.” So his report would be “second hand,” and now it is “third hand” hearsay from Axil with no source other than a name.

Someone’s eye was damaged?

Paradigmnoia wrote:

I am a bit tired today, so excuse me if this seems like a silly question:

If the total input voltage is 0.1 V, and the voltage drops 0.1 V across the 1 ohm series resistor, then does that not mean the voltage is zero on the other side of the resistor?

Yes, it does, given standard meanings. Voltage is a difference, a “drop” or a “kind of pressure). It is always measured relative to something, which is why voltmeters have two leads! If the return in the circuit is ground, and the voltage at one end of the resistor is 0.1 V, with respect to ground, and the voltage drop is 0.1 V, then, yes, the voltage at the other end is zero with respect to ground. This is the very meaning of “voltage drop.”

It really doesn’t matter what else happens in that circuit, if those are DC voltages, as imagined. Watch as some attempt to confuse this.

Axil wrote:

keep in mind that the Quark is generating electrical power from electron creation and some portions of its circuits may have negative resistance.

(1) whether or not the Quark is generating “electrical power” — from anything — is irrelevant to the question of voltage drop across the resistor, which will depend only on the current through the resistor, that’s basic electronics and physics.

(2) “negative resistance” in some element in the circuit is also irrelevant, and Axil may be playing on ignorance of the meaning of “negative resistance,” i.e., an incorrect idea that this would then generate “negative voltage.” But even if it did (imagine a battery in the circuit) it would not change the fact about the voltage drop in the resistor.

So I become very, very suspicious of Axil, more than I have ever been. He is creating obfuscation. He was apparently impersonated here, recently, by a troll with some skill. But he is verified on LENR Forum.

Paradigmnoia wrote: (November 10, 2017)

Adrian Ashfield wrote:

No.

This was in response to the question about the voltage on the other side of the resistor.

Would you please elaborate on your answer a bit? Nothing too complicated. A simple example would suffice.

One might think….

Adrian Ashfield wrote: (November 10, 2017)

The voltage drop across the resistor only tells you the current going through the reactor, not the applied voltage across it.

This was not a “simple example” or even an answer to the question, which was about the voltage on the other side of the resistor, which, given the condition of the question (it’s quoted above), is very simple. 0.0 volt. Because of various possible complications, the assumption of 0.1 volt input may be incorrect, and there might be a negative voltage, for example, on the other terminal of the reactor (it is described as a two-terminal device),  but it is absolutely correct that the voltage drop, the data we were given, does not tell us the applied voltage across the reactor, and therefore it does not tell us the input power to the reactor, thus the paper was grossly incorrect and misleading.

Ashfield is arguing from personal certainty, otherwise he would see these simple facts as such. A fact is a fact regardless of how someone might attempt to use it, but fanatics argue against what they might consider inconvenient facts. A fact can be misleading, in context. But instead of pointing this out (when it is possible), fanatics attack the facts and the reporters of fact.

Paradigmnoia wrote: (November 10, 2017)

As I thought I understood things, the total voltage in a series circuit is equal to the sum of the voltages across all the resistances.

This is true, but doesn’t really address the issue: endless argument without any rational basis.

bocijn wrote: (November 10, 2017)

[the name Kirchoff’s Law and a diagram. I provide instead a link to the Wikipedia article.)]

Axil wrote, in response to bocijn, (November 10, 2017):

Would covering that circuit with a 100 kiloVolt electrostatic field change the behavior of the circuit any?

He is again trolling. No, it would not change the behavior at all. (Axil also gives the potential of the field but not what this potential is relative to. He does not state the variation in the field, nor how this is established. However, Kirchoff’s law still holds, regardless. The sum of the voltages around the circuit will be zero (if the source is included). Paradigmnoia calls the source voltage the “total voltage,” which may lead to some confusion. What this means, in a circuit with one source, is that the the source voltage equals the voltage drops across the other elements.

bocijn wrote (November 10, 2017)

I don’t think so

He is thinking correctly as to the well-known laws.

But a varying magnetic field would affect Kirchoffs assumptions.

Not really. A varying magnetic field would induce currents and voltages in the circuit elements, which would need to be considered. For example, suppose the power source is a transformer, wherein a varying magnetic field induces currents, which then supplies power to the circuit. Kirchoff’s laws hold as to instantaneous currents. If we want to look for an exemption, we would need to look at, say, electron creation and injection (which Axil may be leading to). This would create an instantaneous voltage at the point of injection, and it would take time to propagate (lower than the speed of light, by the way, because this would not be a vacuum). The problems in the electron creation idea are enormous.

Adrian Ashfield wrote: (November 10, 2017) [replying to Paradigmnoia)

I didn’t think Rossi gave the total voltage. I would also like to see the voltage across the reactor: if it is a plasma and also generates electricity I suspect it is a little more complicated.

Again, not wrong, but utterly missing the point, and swallowing a big pile of Rossi Says. Rossi did not give the total voltage, that is true. However, he did give information that, if put together, implies it. This is the logical progression: The voltage across the sense resistor is 0.1 v. If the circuit contains three elements, being:

ground — power sourcereactorsense resistor — ground

(sense resistors are normally placed in the return line, because then one may measure all voltages with respect to ground. Normally, the voltage drop in the sense resistor is small, it has been chosen to make it so. With this circuit, two voltmeters, each measuring voltage with respect to ground, can measure all voltages in the circuit. By moving one problem, then, one can convert the two meters into what is shown in the Gullstrom paper, to conceal the input power — or, same thing, the voltage across the reactor.)

Rossi was asked about the resistance of the reactor. His first answer was “zero.” When even some of his followers exploded from the preposterousness, he then said it was very low, comparable to the resistance of silver (which is indeed very low). If that is true, then there would be no measurable voltage drop across the reactor (unless the “silver” were an extremely thin silver wire, which is surely not what he meant; that would be deceptive, eh?). The only voltage drop in the circuit, then, is that across the sense resistor, and therefore the supply voltage is 0.1 volt.

That is a result obtained by assuming that what Rossi Says — and what the paper has — is true.

Now, we don’t actually know that there are only the three elements, though that is the appearance of the Gullstrom paper. In particular, the “power supply” may be more complex, but it would still amount to a single overall source.

Ashfield is ignoring that he made a gross error in claiming that the voltage on the other side of the resistor would not be “0.0” under the stated conditions. It would be, no matter what other complications exist (such as current generation in the reactor.)

Paradigmnoia wrote: (November 10, 2017)

So maybe I worded something wrong along the way…

0.1 V total volts, plus 0.1 V drop across the resistor, = zero volts, no?

Where voltage drop is -0.1 V.

Alternately, adding the voltages measured across the series resistors, when using a positive value, results in the positive voltage value of the supply.

Actually, he didn’t word anything “wrong,” but his statements were simple and trolls can then twist them. His language is getting less clear, not more. Yes, for Kirchoff’s laws, one proceeds around the circuit, and the direction of voltage matters. Going around the circuit, from node to node, perhaps we might call a voltage drop “negative.” A voltage source, with a positive voltage, would then be “positive.

So ground – power supply output (+V)

drop in reactor (0?)

drop in resistor (-0.1 v)

So V – 0 – 0.1 = 0, by Kirchoff’s law, and V is then 0.1 V.

But it is very unlikely that the drop across the reactor is actually zero.

This is probably getting worse.

Alan Smith wrote: (November 10, 2017)

Why not? 10% duty cycle remember.

Without an explanation of what this means (source?), this is an injection of pure confusion. It is meaningless in the discussion under way. Duty cycle without repetition rate is meaningless. The Gullstrom paper implies DC input and voltages. Otherwise it is meaningless. (The meters were set for DC, if I’m correct). Duty cycle would be relevant to average power, but the power measurement (per temperature rise in the oil) was for a 1.8 second period.

Planet Rossi has been fed inintelligible bullshit by the Master, and they lap it up, an spectacular demonstration of recycling. And then they argue endlessly about details, trying desperately to find some way to explain Rossi’s bullshit to make it true. And Axil tosses in his pretend knowledge, adding to the confusion, and Alan Smith maintains just enough reserve to cover his ass, just in case, and he’s been doing it for a long time now.

These people are not seeking clarity. Some Forum participants are.

Paradigmnoia wrote: (November 10, 2017)

Rossi used the voltage drop (reported as a positive voltage) across the 1 ohm resistor to calculate the power consumed by the Quark. Of course it just calculates the power dissipated by the resistor, although it should give the correct current to the device (or for Axil’s sake, maybe the current coming out of it).

He gives away too much. Current is current, and in a loop circuit, current is the same everywhere in the circuit. It does not matter what circuit element is the “source.” Current is movement of charge, and charge is conserved. (Period!) Axil is going to suggest electron creation, as if this would work.

Yes, the data given by Gullstrom and Rossi just gives the power dissipated by the resistor, and not what was called, in the paper, the “energy input” of the reactor that was superficially measured at generating 20 W.:

Energy input: V=0.1 R=1 Ohm → W=0.01

This could hardly be a more blatant error. Yet, we see the die-hards arguing it endlessly. Very simple: error. Example from Rossi’s history: his mismeasurement of input power for the Hydro Fusion test. We have a detailed account by Mats Lewan of what he said at the time. And then we have what he wrote to Industrial heat about it.

The man lies, makes arguments that he knows are wrong (or when he figures out they are wrong, then he creates a justification, in this case, that he wanted Hydro Fusion to reject the test, that it was deliberate, a “magnificence,” a plan to avoid violating the contract by causing them to reject it. Either way, he was not going to admit a mistake. Rossi Does Not Admit Errors. Except of course, errors like “I was wrong to trust those snakes!”

Adrian Ashfield wrote: (November 10, 2017)

A plasma is not the same as a simple resistor. It seems that the resistance of the reactor is at least “low.” Did Rossi say the total voltage was exactly 0.1v? I would still like to see a direct measurement.

By refusing to understand Kirchoff’s Laws, which are terminally simple as they applies here, Ashfield can create enough confusion in his own mind to allow him to think that Rossi Truth is possible.

Nobody has mentioned “simple resistor.” Kirchoff’s Laws apply to all circuits, including negative resistances (setting aside only transient effects, irrelevant here, that can occur because of charge accumulation before current can propagate. That is, current can increase in a small segment of a circuit, and that increase will propagate at a velocity. It’s fast, delay might be nanoseconds at most. This is a DC circuit and the paper states that explicitly.

No, Rossi did not say that total voltage was exactly 0.1 V. The paper claims that the voltage drop across the resistor was 100 mV, as shown by two meters (which vary, by 0.4 % — i.e., one reads 100.2 mV, the other 99.8 mV).

What leads to 0.1 V total is the claim of zero resistance in the reactor, later modified to “very low, the resistance of silver,” implying that the reactor is a very good conductor and then this leads to a total voltage of 0.1 V. Ashfield seems to be unwilling to notice this. Instead, he is looking for outs, excuses.

What all this actually shows is two-fold: First, Rossi Says is unreliable. He’s not careful, and we also know, from other evidence, that he is deliberately deceptive, attempting to create false impressions in others. (He may rationalize these to himself as harmless exaggerations. He didn’t actually have Johnson Matthey as a customer, but perhaps intended to connect with them and get them to agree and work with him. The conversation with “customer management” was just with himself, but perhaps he put on different hats and had a conversation with himself. Or imagined it.

And perhaps the Russians did steal the fuel from that dummy reactor with a high COP. But what difference would that make? Did the Russians also manage to create the fake results? This is why I explain Rossi’s behavior as “insane.” It’s more explanatory than simple “fraud.” Mats Lewan has never confronted him, and that is very likely because Lewan knows what Rossi would do if he actually asked the difficult questions.

Adrian Ashfield wrote: (November 10, 2017) [responding to himself]

On the surface it looks like the 1 ohm resistor is simply to prevent the power pack from seeing a dead short.

Seems to me it is possible to control the current output directly by the power pack.

Ashfield is displaying ignorance of electronics and how measurements are made in electronic circuits. First of all, it appears that the power supply (“power pack”) can be set as a current source, so one would set the current for this application to 100 mA. The voltage will then vary as needed to create that current. (If the resistance is high enough, the voltage would go up to the supply voltage limit, which is for this supply, I think, 30 V. If higher voltage is needed, say to strike a plasma, that could be created by another element in the circuit that creates a boost voltage. Current would still be limited to 100 mA, the setting.

There is no problem with a dead short, then. Current simply will be 100 mA regardless.

(If the power supply is put into voltage mode, current would go to the limit, 5 A. No harm would be done to the supply. Supplies are protected against this by the current limit, just as the people working with the supply are protected by the voltage limit. Otherwise, open circuit, the voltage would go very high! 100 mA could easily kill you if it is forced through your body. 30 V is not enough to do that.)

The one ohm resistor is a standard method of measuring current, in the range involved. It is a “current sense resistor.” There is nothing mysterious about it, other than it probably being redundant to the power supply current display. What is missing is voltage sensing other than that voltage drop. That would also be displayed by the power supply. So Rossi set up obfuscation instead of verification.

Paradigmnoia wrote: (November 10, 2017) (responding to Adrian Ashfield)

There is a photo of two parallel voltmeters on the reactor in the Gullstrom-Rossi paper (v3).

There is. The text says:

In the left in the figure there are two voltmeters that measure the mV of the current passing through the 1 Ohm brown resistance.

However, this can be expected to confuse Ashfield. Those are not “on the reactor,” and Ashfield just said he’d like to see the voltage on the reactor.

Adrian Ashfield responded: (November 10, 2017)

My eyesight isn’t good enough to read the meters or even see to what they are connected.

The problem is not only his eyesight…. Nobody can see what the meters are connected to, but the text tells us, as I quoted above.

Paradigmnoia replied (November 10, 2017)

[with an explanation that the image can be blown up — I use control-mousewheel for this in my browser –, showing it blown up, and then with two images of Rossi explaining the reactor, showing the circuit.]

There are three elements in the circuit, which is the same as I showed above: the power source, shown with a standard symbol for a DC source (same symbol as a battery), it looks like the positive terminal is then connected to the “plasma,” and then the other terminal of the plasma is connected to the 1 ohm resistor, which is then connected to the negative terminal of the power source. The sense resistor is in what could be considered the ground return, and in a safe system this would normally be actually grounded. (Boost voltage would be added to the segment between the current-regulated power supply and the plasma.)

This is an extremely simple circuit, even if the plasma itself is complex. It is still a two-terminal device. Current through the resistor will be current through the plasma. With a normal plasma, though, I’d expect the voltage across the plasma to be around 100 V or more, unless plasma conductivity is being maintained some other way. A “new reaction” might possibly do that, keeping the plasma hot and thus ionized. Or it might do it for a time, depending on the heat flow, hence the “duty cycle.” Boost the plasma voltage to strike it and get it hot, then turn off the boost voltage. But this is no longer a DC supply as claimed in the paper.

Then the killer post from Paradigmnoia, November 11, 2017, showing some of Rossi’s JONP conversations on the issue of the voltage measurements.

First conversation:

Oystein Lande July 23, 2017 at 1:18 PM

Regarding the Rossi-Gullstrom paper:

The total power supply was not mentioned, but some say you used a 24 V battery?

If this is correct your total input power to the experiment setup would be 24V * 0,1 A = 2,4 Watts.

Can you confirm the battery voltage?

The discussion on ECW was pretty intense by this time, seeking to know the input voltage to the device. Lande asks the wrong question, and Rossi takes full advantage of that. Whether a battery was used or not was not the real issue. The real issue would be the voltage across the reactor, which would then allow the calculation of input power.

Andrea Rossi July 23, 2017 at 1:57 PM

Our power source can be either 120 or 220 V AC, or we can use 24V DC batteries.

Obviously your calculation is wrong, because one thing is the voltage at the power source, a totally different thing is the voltage that goes to the E-Cat through the circuitry of the control system.

The calculation was correct, given an appropriate understanding of “the experiment setup.” Which is not what Rossi here calls “our power source.” It is the output of the device Rossi has shown with a battery symbol in his whiteboard explanation. That symbol simply means a DC source, not necessarily an actual battery.

I found this common, particularly when Rossi was told (often inaccurately), something I had written. He declares it “wrong,” even when what I’ve written was standard physics and electrical engineering. In one fun example, he then cited a well-known physicis for an explanation, and that physicist actually wrote the same thing I had written. Rossi is insane, ignorant, or a fraud, or some combination, yet his followers read him claiming that he was about to win a Nobel Prize until those nasty IH snakes interfered, and they believe him, which is insane six ways till Sunday.

Then Rossi misdirects attention:

In the same Gullstroem-Rossi paper you can read the voltage measured by the 2 voltmeters.

That is a voltage, to be sure, but not a voltage that shows the “total input power,” but only the voltage across the current sense resistor, then allowing calculation of power dissipation in the resistor, but not in the “experiment set-up,” and not, specifically, in the device, the “plasma.”

Second conversation:

Silver August 15, 2017 at 6:12 AM

Your measurement system described in the Gullstrom-Rossi paper is perfect. The circuit is very simple, the plasma is a conductor, as everybody knows, therefore placing a resistance with a known ohmage and measuring the voltage across it, the current is obtained by the Ohm equation, as well as the wattage.

Fascinating. All resistors are conductors, the question is the actual conductance. Plasmas show negative resistance. A plasma device, until the plasma is “struck,” will show very high resistance. A high voltage (with a common neon bulb, about 90 volts, as I recall, but this plasma device has a much larger terminal separation, so I’d expect the striking voltage to be substantially higher) causes ionization in the plasma gas, and as ionization is established, current increases, causing more ionization; rapidly the plasma resistance goes substantially down. But it never becomes a truly good conductor. I know of no examples. So what “everybody knows” is not known by me, and I was effectively an electronics engineer.

The action described will only measure the “wattage” if the plasma resistance is zero (or certainly much lower than one ohm), when, in reality, it would be far higher than that. In other words, this “measurement system” is far from “perfect,” but rather depends on assumptions about the device that are probably counter-factual. In fact, the same voltmeters, with one probe moved, would have given the full information. That one moved probe would be placed on the a connection between the power source and the plasma, measuring the full supply voltage. The voltage across the plasma would then be the difference between the two voltages.

Why was this not done? My guess is that was the actual setup, which Rossi then changed to conceal the full voltage. Whether or not that is sane is entirely a different question. The fact is that the “wattage” that matters was not measured, it was assumed, and it was effectively assumed to be zero. It is very hard to input much power to a device with very low resistance!

Power dissipated in the current sense resistor is not dissipated in the device.

Rossi’s answer:

Andrea Rossi August 15, 2017 at 7:47 AM

Correct.

So Rossi is agreeing with total bullshit, but, hey, it’s his bullshit.

Third conversation:.

Mario Marini August 4, 2017 at 5:57 AM

Looking at the paper Gullstrom Rossi I understand you measured the wattage across the 1 Ohm resistance to determine the wattage in the circuit, based on the rule of the circuit with two resistances of which one has a known value in Ohm, the other has not. I learned this at the school of electrotecnics: when a circuit is made by a power source and 2 reasistances, to know how much is the energy in the circuit you can measure the voltage across a resistance with well known ohms and get the amps from the ohm’s equation. Multiplying V x A you know the amount of energy in the circuit, less the dissipation caused by the resistance.

Am I correct?

If he gave an answer like this on a test in that “school,” he’d have flunked. His use of language is entirely wonky. “Wattage” is not measured “across” a resistor, it is is calculated from the current, and I could go on and on. V x A in this case gives the dissipation in the resistor, and no information about the dissipation in the rest of the circuit. The measurement is of power, not “energy.” Suppose the rest of the circuit’s resistance was a megohm. At 100 mA, the voltage across the megohm would be 100 KV. The power would then be 10 KW. On the other hand, if the plasma was a dead short, the power in it would be zero.

Power measured in a circuit element tells us nothing about the rest of the circuit, unless we have more information, such as the resistance of the rest of the circuit, or the voltage across it. What is correct is that the single measurement across the current sense resistor gives us the current through the rest of the circuit. With the voltage across the rest of the circuit (or the source voltage, which will be a little higher because of the voltage drop across the sense resistor), one can then calculate the resistance of the rest, or with the resistance, one can then calculate the voltage, and the former might be what this person is remembering. If this is a real person and not just a Rossi sock.

Andrea Rossi August 4, 2017 at 2:20 PM

Exactly.

Exactly insane. “Insane” because we would think, with his experience, that Rossi would know better, much better. And then it’s insane for others, when all this is clearly explained, by people who know what they are talking about, instead of understanding, cleave to Rossi Says. THH is expert in electrical engineering. I don’t know about Paradigmnoia, but he seems competent, and so do others.

This has little to do with whether or not the Rossi Effect is real, someone could be quite ignorant and stumble across something. At this point, however, all the information we have about the Quark-X is coming through someone who does not understand basic measurement issues.

Surely if someone wants to support Rossi, they could understand that. If Rossi would listen to me, I’d attempt to explain these things to him. However, Rossi was so convinced that I was paid to attack him that he interpreted everything as hostile. In person he was polite, though.

The biggest problem I see with Rossi is that he is not willing to admit errors, and he certainly made plenty of them. Admitting error is the fastest way to move forward. Rossi, by his behavior, regardless of whether or not he had a real effect, convinced people who were strongly motivated to confirm his work that, instead, there was nothing worthwhile there. And we can go back and see so many situations where Rossi created the reaction to him; this is a major part of the training I have been through. We cannot go forward without completing the past, while we maintain denial that we are responsible.

Adrian Ashfield wrote: (November 11, 2017)

Paradigmnoia wrote:

One could use the zoom function on the PDF. The photo has fairly high resolution.)

Thanks for the enlargement but I’m sorry to say I still can’t read them without fiddling with them more. I have macular degeneration.

It happens to people our age, not uncommonly. However, then, what is functional is to ask people what is there. Presumably, Ashfield can read the comments (he hasn’t said how, but maybe he blows them way up, and we could think he could do that with the images.)

What has been visible is that Ashfield is in kind of a fog, doesn’t clearly “see” what is going on, and doesn’t trust what he’s being told. There are some commentators on LF that cannot be trusted, so crucial would be understanding who can be trusted. There are people who would rather die than mislead, and others who mislead routinely (or who are so heavily biased that everything is constructed and interpreted through the bias. These are fairly obvious, though. Be careful! Someone who is routinely deceptive might agree with us!

This is one reason why anonymous participation is problematic. With anonymity, there is little or no responsibility.

THHuxleynew wrote: (November 11, 2017)

[A good post repeating the circuit details. From it:]

Here, the whiteboard calculation for E1 (input power), the whiteboard circuit diagram, the text in the research papers, and even what Rossi says, are all consistent. The only thing that does not work is Rossi’s calculation of the Quark-X input power, which is wrong.

[…]

Rossi’s miscalculation here is quite blatant, and he does not recant. He just asserts something wrong and lets those who support him try to find some way to make this fit.

The Gullstrom paper was first published March 15, 2017. The latest version (v3) was published August 16, 2017. That was plenty of time in which to identify and correct errors.  But Rossi, from JONP, is heavily committed to at least one major error. Take the error out, the paper would look obviously incomplete. (Unless Rossi supplies the power supply voltage, say. It is hard to imagine that this would reveal some important secret. I’m guessing that there is a boost voltage, though, transiently applied to strike the plasma, and then 30 V seems inadequate for plasma maintenance.)

IH Fanboy wrote: (November 11, 2017)

@THHuxleynew

If the resistance of the QuarkX cell itself is negligible during operation (as Rossi had claimed), then there is no inconsistency, as I have pointed out to you before.

No. What Rossi wrote on JONP, and what is in the Gullstrom paper, is blatantly incorrect, no matter what IHFB has “pointed out.” As the cell resistance goes to zero, the cell dissipation goes to zero. Rossi has claimed, repeatedly, that measuring the voltage across the 1 ohm sense resistor is enough to give the cell dissipation, but that only works with an assumption about resistance, which was not stated in those cases. Rossi claimed actual measurement, not inference from an assumption about resistance, an assumption that would be, in itself,l quite remarkable.

Yes, I can imagine that somehow the device shows very low resistance, but that would still leave Rossi’s assertion about the measurements as incorrect and misleading.

Once again, the agenda of “What Rossi Says is correct, so how can we justify it?” is quite visible. The error is obvious to anyone who understands electrical engineering, this is not some esoteric idea. The error remains a blatant error, even if Quark-X is the Next Big Thing.

Adrian Ashfield wrote: (November 11, 2017)

axil wrote:

[R]ossi’s calculation of COP based on the assumption that the light that the quark produces is blackbody is wrong.

Yeah. Axil is not pure Rossi Says. He is more “Axil is Right” and “Everyone Else is Wrong.”

As Rossi has surrounded the E-Cat QX with an oil(?) jacket and is measuring the power by temperature rise in the fluid, it doesn’t matter if it is a black body or not for the demo.

It does matter because the black-body calculation is used with the heat rise calculation, as mutually confirming.

This is a much more accurate way of measuring the excess heat period.

It could be, but for “excess heat” one needs to know the input power, which is missing, and then if the black body estimate is badly off, as seems likely, this calls into question the heat measurement. It was already problematic, as I explain above. Measuring a temperature rise for 1.8 seconds? Bad Idea!

Maybe Rossi will do better, but maybe not! I get the impression nobody with any sense is advising him, since Lukacs pulled him out of that lawsuit mess.

 

 

 

 

 

Author: Abd ulRahman Lomax

See http://coldfusioncommunity.net/biography-abd-ul-rahman-lomax/

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