Merged Cold Fusion Claims

[Cuddles]

Still, just *how in the world* can one think themselves a physicist if they do not ****** know how clamp meters work.

Very easily. I'm a physicist and I have no idea how they work. Mainly because I've never even heard of them before. Being a physicist does not magically grant you knowledge of how all scientific tools work. The vast majority of physicists have probably never had any reason to hear of, let alone actually use, one of these. Now, that doesn't excuse people who are actually using one from not understanding what they're doing. But that's a very different matter from complaining that they should know simply by virtue of being physicists.

But even in that case, who says they don't know how they work? If someone isn't expecting deliberate fraud, they may well not look for it. If they're working on the assumption that this is an unadulterated mains supply, they'd have no reason to worry about their equipment only measuring AC, because that's all they're expecting to be there. More to the point, as noted earlier at least some of them have been previously associated with Rossi. They could well know exactly what the fraud is and have done this deliberately in the hopes that no-one, or at least no-one likely to give them money, notices.

 

[Yevgen Barsukov]

There's a lesson here, folks. There is one way and only one way to do a fraud-proof power measurement: you intercept all conductors between the device and ground with shunt resistors. You read out both ends of each resistor with an oscilloscope. If there's a connection (a table, a frame, a chassis, etc.) that the inventor insists is "grounded", attach a grounding strap to it and leave it there during the entire experiment. There is, as far as I know, no way to hide electrical power input to such a setup.

Exactly right. And ideally you would need to actually record all the waveforms with high sampling rate (e.g with osciloscope) and only than you can derive exactly how much power is delivered (given that they can change the waveform at any time with their black-box controller).

Btw this article shows a nice schematic how to get zero current indicated
even though current is flowing, when clamp ammeters are used:
http://scienceblogs.com/startswitha...-is-back-and-people-are-still-falling-for-it/

Regards,
Yevgen

 

[Dmytry]

Very easily. I'm a physicist and I have no idea how they work. Mainly because I've never even heard of them before. Being a physicist does not magically grant you knowledge of how all scientific tools work. The vast majority of physicists have probably never had any reason to hear of, let alone actually use, one of these.

It's not about tiny details of how it works, i.e. whenever it is using a transformer, a hall effect device, or something else... clearly, if it clamps around the wire making no actual electrical contact of any kind, all it can do is measure magnetic field somehow, with the corresponding caveats (i.e. measures total current through it's hole).

I'm not even an actual physicist (I am a video game developer) and I immediately knew the general principle how clamp meter worked the first time I seen one - I didn't know if a particular brand was using transformer or some sort of magnetic field sensor, and what sort of magnetic field sensor it'd use, of course, but the general principle (that it relies on magnetic field in a loop of ferromagnetic material around the wire) was clear.

Now, that doesn't excuse people who are actually using one from not understanding what they're doing. But that's a very different matter from complaining that they should know simply by virtue of being physicists.

But even in that case, who says they don't know how they work? If someone isn't expecting deliberate fraud, they may well not look for it. If they're working on the assumption that this is an unadulterated mains supply, they'd have no reason to worry about their equipment only measuring AC, because that's all they're expecting to be there.

No, even on an unadulterated supply (240v RMS sine wave voltage) you can connect a diode in series with your load, and then the current will have big DC component.

More to the point, as noted earlier at least some of them have been previously associated with Rossi. They could well know exactly what the fraud is and have done this deliberately in the hopes that no-one, or at least no-one likely to give them money, notices.

If they were in on the fraud, they would have told us they used in-line measurement device complete with oscilloscopes and everything as you say, and they would skip on actually doing anything because they would know it doesn't work.

 

[WhatRoughBeast]

I can think of at least 2 reasons why the report is massively suspect, but I'll concentrate on the one which, for reasons I don't comprehend, nobody seems to have noticed.

If you want to measure the temperature of an object, why not just put a thermocouple on it? During the last couple of publicized tests that's what Rossi et al did, so it's not as if the idea is new to them. For heaven's sake, you can get thermocouples for a buck apiece on eBay. Then, knowing what the temperature is, you can check (with the IR camera) the emissivity, and from that determine the radiated power.

Refusal to take obvious action casts doubt on one's motives, or competence, or both.

 

[TjW]

Duuude. Thermocouples don't give you cool false-color images to put on the website.

 

[Christian Klippel]

So, if I were trying to smuggle in energy, how would I do it. Lasers? RF transmitters? Possible, but dangerous and expensive.

[...]

Here's how I'd do it: A normal 3 phase power lead has 5 cores, and 5 pins on the plug. Neutral, Phase (hot/Live) 1, Phase 2, Phase 3 and Earth/Ground. Keep P1, P2 and P3 connected as usual, and label/colour code them appropriately. Perhaps, I'll conveniently separate the wires out, ready to have probes connected. Meanwhile, I'll disconnect the earth wire, and rewire it as a "secret" phase wire.

I'll use my "trade secret" power supply, to suck power from the monitored phases as and when required, to make things look good. But it will also suck power from my "stealth phase" wire, cunningly disguised as a safety earth.

Of course, the current flowing in my stealth wire, will appear in the neutral wire - but that's alright, you don't need to measure neutral current to measure power - in fact most power meters *don't* measure the neutral current, because it is only in rare conditions that measuring it is of any use (and energy metering is not one).

With regards to RF powering, that is not hard at all. Take a look at an induction cooktop. It doesn't even need to be that much of RF for them, a few tens of kHz is enough. Regarding "real" RF, heck, i even designed and built my own RF power supply unit for those Metcal soldering systems (13.56MHz) and can get the cartridges glowing dark-red after a while with a main supply to the RF stage of only 21 volts. (Of course that's not how it should be, obviously, and happens only if i mess up the feedback loop/idle supply voltage)

With regards to three-phase power and the neutral: If you have three equal (resistive) loads, each load from one of each phase to neutral, then you will have basically no current flowing at the neutral. Meassuring only one phase, of course, will not give you the total current either.

In an unknown setup, each and every wire and connection that goes to a device should be monitored, otherwise it is far too easy to play silly shenanigans.

-------

Oh, and regarding thermal imaging, here's something funny that any of you can try if you can get hold of a thermal imager. Take a stainless steel pot, fill it with ice water. Take second (preferably black) pot, fill with water and get that water cooking. Place them so that the radiant heat of the black one hits the surface of the ice-water filled one that you are imaging. Look at the results and contemplate the wonderful world of test equipment that can't differentiate between radiant and reflected heat.

Greetings,

Chris

 

[Craig B]

Can anyone who knows about these things comment on this? It's an extract from a notice on the Rossi "independent" report at http://www.extremetech.com/extreme/...ied-has-10000-times-the-energy-density-of-gas

As far as we can tell, the main barrier to cold fusion — as with normal fusion — is producing more energy than you put in. In NASA’s tests, it takes a lot more energy to fuse the nickel and hydrogen than is produced by the reaction. Rossi, it would seem, has discovered a secret sauce that significantly reduces the amount of energy required to start the reaction. As for what the secret sauce is, no one knows — in the research paper, the independent scientists simply refer to it as “unknown additives.” All told, the E-Cat seems to have a power density of 4.4×105 W/kg, and an energy density of 5.1×107 Wh/kg.

I understood that even exploding stars can't obtain energy by transmuting nickel. http://en.wikipedia.org/wiki/Silicon-burning_process

The entire silicon-burning sequence lasts about one day and stops when nickel-56 has been produced. The star can no longer release energy via nuclear fusion because a nucleus with 56 nucleons has the lowest mass per nucleon (any proton or neutron) of all the elements in the alpha process sequence.

Is it even imaginable that some simple concoction - the "sauce", however "secret" it might be - could work the trick of fusing nickel to release nuclear energy, by reducing the temperature required to initiate the reaction? By the way, I hope it can, or I won't by buying one of Rossi's heaters for my house. I don't want anything in a box under the stairs that replicates the conditions in the core of a star a few moments before it goes supernova.

 

[ben m]

Is it even imaginable that some simple concoction - the "sauce", however "secret" it might be - could work the trick of fusing nickel to release nuclear energy, by reducing the temperature required to initiate the reaction?

No, not even according to Rossi's "data". If you were somehow lowering the barrier to nuclear fusion ... well, you'd be putting in this small amount of energy and getting nuclear fusion. And we know what nuclear fusion looks like: it's a nuclear reaction, i.e. something that produces gamma rays and radioactive daughter nuclei. Go inside the Sun and you find a hydrogen plasma, which is being continually heated by high-energy gamma rays, neutrons, and beta rays from the fusion reactions. Go inside a bomb, or a tokamak, and you find the same thing. The gamma rays are not an incidental side effect, but rather are fundamental to the behavior of energetic nuclei.

So as soon as Rossi says "no gamma rays were detected", and "no radioactive products", you can discard this "lowered the fusion barrier" interpretation.

 

[Dr.Sid]

Yep .. no gamma, no fusion ..

 

[Craig B]

Yep .. no gamma, no fusion ..

A contributor to a cold fusion site Nickelpower has disagreed with me in these terms. Do you have any comments?

Also worth pointing out is that nuclear reactions can proceed without gammas being emitted. What you’ve quoted effectively says that if there aren’t any gammas then it can’t be a nuclear process. Epic failure of understanding there.

 

[ChumpusRex]

It might be better to say: "no known or theorised fusion".

Some physicists have suggested that a nuclear decay process via a large number of energy levels, each of only a few keV would not produce any detectable gamma.

That such a decay scheme has never been seen, nor is compatible with quantum mechanics, is an important caveat. However, to state "no gamma, no fusion" is, perhaps, opening oneself to criticism.

 

[nathan]

if fusion doesn't produce gamma rays, what is the mechanism for the excess energy to be released? IIUC fusion produces a few MeV's per event. If that's emitted as a whole slew of infra-red photons, we'd need a few million of them. As ChumpusRex has just said, that'd require transitioning through a few million quantum states.

 

[Oystein]

if fusion doesn't produce gamma rays, what is the mechanism for the excess energy to be released? IIUC fusion produces a few MeV's per event. If that's emitted as a whole slew of infra-red photons, we'd need a few million of them. As ChumpusRex has just said, that'd require transitioning through a few million quantum states.

Hi everyone! I am new to this thread, and pretty much ignorant of most of what's being said so far. My (modest) interest stems from a debate with a 9/11 truther who believes that Rossi-style cold fusion is the best explanation for everything.

This said, I'll re-lurk after replying to this:

Gamma rays (photons) aren't the only carrier of energy from nuclear reactions. You also have kinetic energy of things like alpha particles, electrons, positrons; also neutrinos, and what not.
Not saying that physics exists that allows for any nuclear transition to release their energy benignly without gammas or other hazardous stuff. Just saying: Gamma rays aren't strictly needed to release excess energy.

 

[WhatRoughBeast]

It might be better to say: "no known or theorised fusion".

Some physicists have suggested that a nuclear decay process via a large number of energy levels, each of only a few keV would not produce any detectable gamma.

That such a decay scheme has never been seen, nor is compatible with quantum mechanics, is an important caveat. However, to state "no gamma, no fusion" is, perhaps, opening oneself to criticism.

This very conveniently overlooks the postulated Rossi mechanism: Ni + H = Ni.
In the case of, for instance, 62Ni, the fusion produces 63Cu with some energy release, which is mysteriously not associated with the gamma radiation which is the hallmark of lesser, "conventional" fusion processes. Furthermore, the process can only produce a small amount of energy due to binding energy concerns, and may even be endothermic. The 63Cu is stable, so the process ends there and ends well. Of course, the two stable copper isotopes are produced from two nickel isotopes which have a abundance ratio of 3:1, Natural copper has the same two isotopes at a 7:3 ratio, And guess which ratio Rossi's "spent fuel" sample contained? (Hint: it wasn't 7:3)

So far, so good, and your statement stands. However, nickel has 5 stable isotopes, not 2, and the process does not end cleanly for the other 3. All 3 isotopes of copper (59Cu, 61Cu and 62Cu) undergo beta+ decay, producing a positron which immediately annihilates an electron and produces a pair of 511 kev gamma rays. Speculation that decay can occur in small steps has only one drawback: it's not observed in nature.

Another problem with this process is obvious to anyone who bothers to do a little reading. The most common isotope of nickel is 58Ni, which makes up 68% of all naturally occurring nickel. 58Ni + H = 59Cu. 59Cu then decays to 59Ni with the usual gamma release. 59Ni, however, is not stable, and decays with a half-life of 7500 years to 59Co .

Rossi's patent application remarks that the nickel powder must contain only 62Ni, as this will produce stable copper. While this is praiseworthy, it misses the fact that 64Ni will also do. What is not mentioned is that these two isotopes make up 5% of naturally occurring nickel. A low-gamma process is indeed conceivable, but it can't possibly be a mystery to Rossi. He need "only" perform isotope separation on his nickel to eliminate the 95% of the material which causes problems. And I suggest that you look around to find out what that costs. I suggest you look real hard. Given the low energy release per atom of nickel, and the high cost of processing the fuel, even if the Rossi process isn't an outright scam I can't see it being economical - let alone revolutionary.

 

[ben m]

Gamma rays (photons) aren't the only carrier of energy from nuclear reactions. You also have kinetic energy of things like alpha particles, electrons, positrons; also neutrinos, and what not.
Not saying that physics exists that allows for any nuclear transition to release their energy benignly without gammas or other hazardous stuff. Just saying: Gamma rays aren't strictly needed to release excess energy.

Almost, but not quite. Rossi is claiming nickel-proton fusion to produce copper and heat. Quite generally, by saying "heat" he's ruled out any sort of weird all-neutrino process; even k-capture produces hard x-rays. By saying "copper" he's ruled out any charged particle emission (electron, positron, alpha) which must produce something other than copper. (Electrons and positrons would produce gammas anyway). Neutron emission is energetically forbidden---in this part of the Table of Nuclides it *costs* energy to insert a proton and eject a neutron.

so that's just using conservation laws.

Using nuclear physics, we know what a (Ni + p) system looks like. If you get a nickel nucleus anywhere within a few femtometers of a proton or neutron, the result is a copper nucleus in a highly excited state. You can pretty much write down the wavefunction for this---it's called a "halo" state. The halo states are unstable and rapidly emit---you guessed it---easily-detectable gamma rays.

That's what Rossi has to fudge away. Even if he had a chemical catalyst that magically shepherded protons past the fusion barrier, he'd have to have another catalyst that modified the nuclear physics of what happens next.

How inconvenient, right? All of the laws of physics that *would* make Rossi's device verifiable---radioactivity, gamma emission, ability to heat up using non-trade-secret power supplies, or ability to maintain heat after the resistors are unplugged---have *all* gone wrong, all at the same time. What a coincidence.

 

[Craig B]

Almost, but not quite. Rossi is claiming nickel-proton fusion to produce copper and heat. Quite generally, by saying "heat" he's ruled out any sort of weird all-neutrino process; even k-capture produces hard x-rays. By saying "copper" he's ruled out any charged particle emission (electron, positron, alpha) which must produce something other than copper.

In fact his fanboys have abandoned anything as specific as that, and are now implying that Rossi invented the copper thing to put his many enemies - Big Oil, the Islamicists, the Mafia and others - off the scent, and to protect the commercial secrecy of the process. One contributor to a pro-Rossi blog is quite content to state:

I think that we the fans of “cold fusion” or LENR already knew that it was not nuclear fusion in the conventional sense of the word. It is anomalous heat, and for now that is all that we need to know. Fire had no explanation for at least a million years and yet we managed to use it to our benefit. But when we finally did get an explanation, then we were able to improve it greatly. I assume that it won’t take a million years to find an explanation for this anomalous heat.

My bold. http://nickelpower.org/2013/05/20/the-independent-report-is-here/#comment-22774

 

[Horatius]

Fire had no explanation for at least a million years and yet we managed to use it to our benefit.

How sad, when he's forced to admit that even cavemen had more engineering success with their cutting edge technologies than the Cold Fusioneers have had....

 

[ben m]

In fact his fanboys have abandoned anything as specific as that, and are now implying that Rossi invented the copper thing to put his many enemies - Big Oil, the Islamicists, the Mafia and others - off the scent, and to protect the commercial secrecy of the process.

Oy.

We've been through this before. If one want to "protect the commercial secrecy of a product", a good place to start would be not:

• setting up fakey-looking public demos showing your device puffing steam
• setting up fakey-looking "scientific" demos in which your handpicked engineer buddies videotape your already-running "demo" and publish photos on the arXiv
• writing a blog titled like an academic journal about your R&D process
• filing a fraudulent patent application
• "misdirecting your enemies". Seriously, who does this?

If you have a real product that you're trying to both sell and protect, you:

• Get a lawyer to write up a really good NDA
• Conduct ironclad private demos for customers who've signed the NDA
• Shut up in public. No blog, no demos, no preprints. Just circulate around those smoke-filled rooms and cash their checks.

If you're afraid of being assassinated by agents of Big Oil, well, Rossi is doing the opposite of the smart thing. (Isn't that how the free-energy-suppression theories start? "Oh well," some bought-and-paid-for Italian coroner would say, "It looks like this fraudster offed himself after realizing he was about to be arrested for fraud. He apparently wrapped himself in duct tape, encased his own feet in concrete, and jumped off a pier into the Thames near the BP headquarters there. Our investigators have seized his lab for evidence but found nothing, not a thing, no E-cats at all, just lots of fraud. Certainly nothing that's been disclosed or documented in a way we can't explain! Keep buying oil, folks, nothing to see.") Well, if you want to make that impossible, the correct procedure is:

• Give up on becoming a trillionaire. Settle for, I dunno, billionaire.
• File an ordinary, full-disclosure patent. As soon as you do this, you're no longer a useful target for shady conspiracies.
• Send the device for independent tests sufficient to get past the patent office's auto-rejection-of-perpetual-motion clause. Publish the results and invite replication.
• Publicly announce the terms, conditions, and fees for *licensing* any commercial E-cat installation.

 

[Small Bones]

I've found couple of simple demonstrations which show how easily power measurements can be faked and they also may hint at Rossi's secret ingredient!

 

[Oystein]

Almost, but not quite. Rossi is claiming nickel-proton fusion to produce copper and heat. Quite generally, by saying "heat" he's ruled out any sort of weird all-neutrino process; even k-capture produces hard x-rays. By saying "copper" he's ruled out any charged particle emission (electron, positron, alpha) which must produce something other than copper. (Electrons and positrons would produce gammas anyway). Neutron emission is energetically forbidden---in this part of the Table of Nuclides it *costs* energy to insert a proton and eject a neutron.

so that's just using conservation laws.

Using nuclear physics, we know what a (Ni + p) system looks like. If you get a nickel nucleus anywhere within a few femtometers of a proton or neutron, the result is a copper nucleus in a highly excited state. You can pretty much write down the wavefunction for this---it's called a "halo" state. The halo states are unstable and rapidly emit---you guessed it---easily-detectable gamma rays.

That's what Rossi has to fudge away. Even if he had a chemical catalyst that magically shepherded protons past the fusion barrier, he'd have to have another catalyst that modified the nuclear physics of what happens next.

How inconvenient, right? All of the laws of physics that *would* make Rossi's device verifiable---radioactivity, gamma emission, ability to heat up using non-trade-secret power supplies, or ability to maintain heat after the resistors are unplugged---have *all* gone wrong, all at the same time. What a coincidence.

Thanks for the summary

(But I never, not for a second, contemplated that Rossi might have found a gamma-free, radiocativity-free process with usable heat release)