A Slew of Noise
The Cold Fusion cells used by researchers like SRI's Michael McKubre operate with a constant-current DC power supply. McKubre is one of the leading researchers in the Cold Fusion field. His work is funded by the Electric Power Research Institute (EPRI).
The advantage of using a constant current DC power supply is that the experimenter can control the DC drive current, which is the faradaic current that is either charging the electrodes or dissociating the electrolyte.
In all the technical models for input electrical power that I've looked at, the electrical input power is modeled as pure DC power, with no AC component. That is, the constant-current DC power supply is treated as an ideal constant current source, with no AC perturbations around the constant DC current.
However, all real constant current power supplies have a characteristic slew rate which specifies how fast they respond when the load impedance changes abruptly. The Kepco BOP 20-20M 400-watt power supply used by McKubre has a slew rate of 1.25 A/μsec in the constant-current mode and 1.0 V/μsec when operated in the constant-voltage mode.
The slew can be modeled as a linear ramp whenever the resistive load changes abruptly from one value to another. Instead of an ideal square wave (instantaneous adjustment), the rise and fall of the square wave is really a ramp, with slope given by the slew rate.
One can model AC power by using sawtooth or triangle waveforms, which are easy enough to integrate with simple calculus. Depending on the fluctuations in the resistive load, there will be corresponding fluctuations in the voltage which ramp up and down at the slew rate to maintain constant current. One can get an idea of how much AC power is going into the resistive load by computing the AC power of a simple triangle wave with a peak-to-peak voltage.
I did this for two examples, corresponding to a pair of experiments in McKubre's EPRI paper. McKubre doesn't say what the peak-to-peak voltage excursions are when his cells are bubbling, so I assumed a 2 V peak-to-peak swing around his reported DC values, to make the math easy to do. For the case of 2 A going into a nominal 2.5 Ω resistive load at 5 V for a DC power of 10 W, a 2-V peak-to-peak triangle wave riding on top of the DC comes out as 0.5 W of AC, for a total of 10.5 W electrical power. For the case of 7 A going into 6/7 Ω resistive load at 6 V for a DC power of 42 W, a 2-V peak-to-peak triangle wave riding on top of the DC comes out as 1.8 W of AC power, for a total of 43.8 W.
Thus a voltage excursion of the order of magnitude of 2 V peak-to-peak works out to about 4-5% of the electrical power being AC noise power at a frequency related to the slew rate of the power supply.
The CBS News film crew that accompanied 60 Minutes correspondent Scott Pelley to McKubre's lab at SRI may have missed an opportunity to measure the peak-to-peak AC (audio) power going into McKubre's cell. All they had to do was slap one of their audio VU meters across the terminals of McKubre's cells to see if there was any AC (audio) power that McKubre was leaving out of his calculations.
Title: Hot Vibrations
Artist: Michael McKubre
Composer: Brian Wilson, Mike Love, and Barsoom Tork Associates
YouTube: Good Vibrations — The Beach Boys
I love the effervescent fizz and spritz
And the way the sunlight plays upon the mist
I hear the sound of a gentle hiss
On the wind that lifts warm vapors with a twist
I'm picking up hot vibrations
It's giving me excitations
Click, pop pop, hot vibrations
Pop, pop, excitations
Click, pop, pop, hot vibrations
Close my eyes, I can listen closer now
Softly now, I calculate the mean
When I reckon swirling steam
It chatters me with a static whirl
I'm picking up hot vibrations
It's giving me excitations
Click, pop, pop, hot vibrations
Pop, pop, excitations
Click, pop, pop, hot vibrations
I don't know why but it sounds like fries
Oh my my my, what a sensation
Oh my my, noise generation
Got to keep those frying hot vibrations
Happening forever
CopyClef 2011 Brian Wilson, Mike Love, and Barsoom Tork Associates.
Resurrection Hackware. All songs abused.
The advantage of using a constant current DC power supply is that the experimenter can control the DC drive current, which is the faradaic current that is either charging the electrodes or dissociating the electrolyte.
In all the technical models for input electrical power that I've looked at, the electrical input power is modeled as pure DC power, with no AC component. That is, the constant-current DC power supply is treated as an ideal constant current source, with no AC perturbations around the constant DC current.
However, all real constant current power supplies have a characteristic slew rate which specifies how fast they respond when the load impedance changes abruptly. The Kepco BOP 20-20M 400-watt power supply used by McKubre has a slew rate of 1.25 A/μsec in the constant-current mode and 1.0 V/μsec when operated in the constant-voltage mode.
The slew can be modeled as a linear ramp whenever the resistive load changes abruptly from one value to another. Instead of an ideal square wave (instantaneous adjustment), the rise and fall of the square wave is really a ramp, with slope given by the slew rate.
One can model AC power by using sawtooth or triangle waveforms, which are easy enough to integrate with simple calculus. Depending on the fluctuations in the resistive load, there will be corresponding fluctuations in the voltage which ramp up and down at the slew rate to maintain constant current. One can get an idea of how much AC power is going into the resistive load by computing the AC power of a simple triangle wave with a peak-to-peak voltage.
I did this for two examples, corresponding to a pair of experiments in McKubre's EPRI paper. McKubre doesn't say what the peak-to-peak voltage excursions are when his cells are bubbling, so I assumed a 2 V peak-to-peak swing around his reported DC values, to make the math easy to do. For the case of 2 A going into a nominal 2.5 Ω resistive load at 5 V for a DC power of 10 W, a 2-V peak-to-peak triangle wave riding on top of the DC comes out as 0.5 W of AC, for a total of 10.5 W electrical power. For the case of 7 A going into 6/7 Ω resistive load at 6 V for a DC power of 42 W, a 2-V peak-to-peak triangle wave riding on top of the DC comes out as 1.8 W of AC power, for a total of 43.8 W.
Thus a voltage excursion of the order of magnitude of 2 V peak-to-peak works out to about 4-5% of the electrical power being AC noise power at a frequency related to the slew rate of the power supply.
The CBS News film crew that accompanied 60 Minutes correspondent Scott Pelley to McKubre's lab at SRI may have missed an opportunity to measure the peak-to-peak AC (audio) power going into McKubre's cell. All they had to do was slap one of their audio VU meters across the terminals of McKubre's cells to see if there was any AC (audio) power that McKubre was leaving out of his calculations.
Title: Hot Vibrations
Artist: Michael McKubre
Composer: Brian Wilson, Mike Love, and Barsoom Tork Associates
YouTube: Good Vibrations — The Beach Boys
I love the effervescent fizz and spritz
And the way the sunlight plays upon the mist
I hear the sound of a gentle hiss
On the wind that lifts warm vapors with a twist
I'm picking up hot vibrations
It's giving me excitations
Click, pop pop, hot vibrations
Pop, pop, excitations
Click, pop, pop, hot vibrations
Close my eyes, I can listen closer now
Softly now, I calculate the mean
When I reckon swirling steam
It chatters me with a static whirl
I'm picking up hot vibrations
It's giving me excitations
Click, pop, pop, hot vibrations
Pop, pop, excitations
Click, pop, pop, hot vibrations
I don't know why but it sounds like fries
Oh my my my, what a sensation
Oh my my, noise generation
Got to keep those frying hot vibrations
Happening forever
CopyClef 2011 Brian Wilson, Mike Love, and Barsoom Tork Associates.
Resurrection Hackware. All songs abused.
Good Vibrations — The Beach Boys
posted by Moulton | 4:33 PM
43 Comments:
Moulton assumes very high noise levels. I couldn't find any reports of how significant bubble noise would be, but I'd assume that any researcher who saw 2V P-P noise on a 5 volt signal would know that he had a problem, and that would only, Moulton's calculations, produce a 5% error; 5% is at the low end of what's considered significant in CF work lately.
As to the fact, however, control experiments with hydrogen and dead cathodes (inactive palladium) show that this supposed noise problem doesn't exist. They would equally be affected by bubble noise.
Add to that the confirmation of calorimetry of excess heat by the measurement of helium from the same cells, and it's iced. While individual experiments may be subject to various errors, the overall work isn't.
So how about calling some of the scientists listed on http://3.ly/ARLcf? You could ask them about the gas loading experiments which don't involve the complexities of electrochemical calorimetry. Maybe ask Mel Miles about how the alloys in his NRL boron alloy patent have been doing in the lab?
Abd, if you are not satisfied with simplifying assumption that make the calculus tractable (and why should you), then you can always go back to first principles and solve for the power by Maxwell's Equations. You can also eliminate the problem of the slew rate of the Kepco power supply by using a Van de Graaf generator as your constant current source. Now, I'm quite sure the mathematical complexities of doing that are well beyond both of our pay grades, and we can both return to preposterously dimwitted flights of fancy to escape the difficulties of solving Maxwell's Equations for a Faradaic RLC circuit operating with steady-state DC current plus gas discharge riding on top of ohmic white noise up to 1 GHz.
James, Mel Miles still hasn't responded to my previous message to him.
Miles is, what, over eighty years old? Not sure.
I can ask questions on the CMNS list, and will, when the questions are mature. That means well-discussed.
You have not shown your "calculus." You just gave results, based on unclear assumptions.
And you are simply ignoring questions that would uncover your errors.
In this case, sure, certain kinds of noise could create missed power, but it's only speculation that such noise exists, and a glance at the power supply voltage and current using tools that McKubre definitely possessed and would use, would rule it out. Indeed, if his method is applied with possibly unstated details, there would be no error.
I.e., if voltage and current are measured with fast samples, and not averaged before multiplying, there would be no error. The "average power" would then be an average of instantaneous powers, and would not miss noise power. I'll need to look at the specs for his multimeter.
Do the math, Abd. It's not that hard. You studied Physics at Cal Tech under Feynman. Surely you learned how to integrate some simple trig functions.
You do the homework problem and see if you come up with the same formula I did.
If voltage and current are measured with fast samples, and not averaged before multiplying, there would be no error.
So let's see McKubre's numbers when he goes back and changes the method of calculation of input power, to measure both the AC and the DC input power, rather than assuming the AC power is zero.
I'll need to look at the specs for his multimeter.
It's a Keithley Model 195A 5-½ digit digital multimeter with 0.01% DC volt accuracy. He averages 32 successive DC values and multiplies that by his constant current to get DC power.
1. Yes, I could do it. But I'm not going to waste the time, it's not necessary to solve every silly problem, already known, by experimental evidence, to be moot.
2. I find it more rewarding to look at problems from a more intuitive perspective, and it's pedagogically more effective as well. If there is true conflict between the math and intuition from general principles, sure, that's a problem. I do trust math, itself, but math is often manipulated and used to create an impression of certainty, when the underlying assumptions are bogus, and math in the hands of someone with a bias can also suffer from confirmation bias. I.e., make an error that favors your bias, you then report the result. If it doesn't favor your bias, you study your math to find the error, or you shut up about it. It's important to distinguish between mathematical relationships, highly useful for predictive power, and our reporting of them, the latter is quite human and quite fallible.
3. Yeah, I looked at McKubre's voltmeter already. It's possible he could have used it to more accurately measure power, but it looks like he didn't operate it and analyze the data that way. Again, what he did was simpler, experimentally and practically, and he wouldn't do the more complex thing unless he was concerned, like, for example, he looked at his power waveform with an oscilloscope and saw a 2V peak-to-peak noise signal riding on 5 VDC.
Don't you think he'd be concerned? Do you think he's an idiot? Don't you think that concerns about his power measurements haven't already been raised, ad nauseum? This one could be checked in a couple of minutes, most of it going over and pulling a scope off of the equipment shelf. I'd be utterly astonished if he never looked.
I could use my Rigol 1052E oscilloscope to capture power supply power data, but the accuracy is only 8-bit. What would be more efficient would be to use a DC offset with a known value. The sample is indeed at 1GS/sec, but it looks like it's alternating, because the two-channel specification is at 500 MS/sec. But this would capture power noise below roughly a few hundred MHz, which should be quite adequate. If used in parallel with a technique like McKubre used, this would allow high DC accuracy, and limit the amount of AC power.
But I'm not doing calorimetry, beyond the most crude, and my temperature probes aren't all that accurate. Cheap. I just want to get some rough idea, see if there is a difference visible between hydrogen and deuterium. It's not a crucial part of the experiment, the protocol didn't call for calorimetry at all.
It's not a waste of time if you are doing an educational project.
Working out the math is part of the challenge in getting the correct model to explain the observations.
Nor do you have to do calorimetry. You might have the equipment to directly measure AC power when a cell is bubbling. Can you run your scope in a mode that calculates the instantaneous product of directly measured voltage and current?
Alternatively, can you tap the cell terminals with an isolation capacitor and feed the audio signal into a high-impedance HiFi pre-amp? If there is popping and crackling, you should be able to hear it on your HiFi.
You are right, partially, about "educational project," but I'm at the point where I can't just go for a shotgun approach to education. I don't need to develop those math skills at this time, there is no other application for them. And lots of time is involved. Someone else can do it, if it needs to be done, and, in fact, if I felt this were necessary, that's exactly what I'd ask for, I'd ask a mailing list where there are numerous people with the skills.
Yes, I believe, about measuring, with my scope, simultaneous power by multiplying voltage and current measurements, limited by the fact that, apparently, the samples alternate with 2 nsec between them. More expensive scopes would have two DACs. The Keithley meter also has a trigger input, so that two of them could be used for simultaneous measurement, but I don't know the acquisition time.
If this were important, it could be done. Now, time is money and equipment is money. There is a far simpler approach that involves looking with a scope -- easy! -- to see if there is significant noise voltage and at what frequencies, and, as well, checking for the possible influence of bubble noise by measuring power in control cells with lots of bubble noise.
The control experiments establish an upper limit on such noise, and it's certainly not above other sources of error, ordinary calorimetry noise. You can see the noise in the hydrogen control that was in series with the deuterium cell. The noise increased with higher current. The source? Not clear to me, but also not showing signs of systematic increase in heat. That measurement was simultaneous with the deuterium cell, same current, that showed the excess heat. Why would bubble noise, translated to power noise in the power supply, with the same current through each cell, not produce excess heat in both cells?
There are lots of ways to demonstrate that there is no significant power supply noise power, we don't actually need more. By the way, I do intend to "listen" to my cells, literally, with transducers that go up to a gigaherz of audio bandwidth, in theory. I won't actually have them hooked up to a speaker, but directly will record the voltages with the DSO, that was why I bought it, in fact.
I should mention that I'll be recording cell voltages and power using a Labjack. The data will be recorded raw, but the bandwidth on this will not be very high, and the resolution is 12 bits. Good enough for folk music, Barry. Looks like the maximum sample rate is 50 kHz. Looks like it's got independent ADCs (??). I was going to use a single current regulator for both deuterium and hydrogen cells, but could run into power supply headroom problems, so I'm actually running two, they will be set to the same current. So I'm vulnerable to some difference there. This, though, simplifies my electronic connections, everything can be referenced to ground. Otherwise I'd be looking at the possibility of exceeding the input limitations of the LabJack. The hydrogen control is not really important in my work, it's just nailing down conditions a little. The more important controls will be radiation detectors all over the place. Not just one near the cathode!
(this is a problem with lots of work, lack of data on control detectors. So someone sees radiation. Fine. Interesting. But, what else was in this environment? Radon? A cosmic ray shower? Accumulated background on detectors? Etc. I'm not just using lots of detectors, I'm placing them in pairs, and looking for coincident tracks. I'm quite proud of the idea, though, in fact, I'm not the first person to think of it.... just, maybe, the first person to do this with a CF experiment. If I actually do it, instead of just blabbing.)
I wrote cell "power" that I'd be recording. I mean current, of course. Power will be calculated from pairs of simultaneous measurements. I'll catch low-frequency noise, but not anything above maybe 20 or 30 kHz, my guess. I can check for that with the Rigol scope, if I care.
My thermocouples have pretty lousy resolution, so.... I'm not banking on being able to measure any excess heat at all from such a dinky little experiment. But we'll see. I think I'd be silly not to at least look a little, I've got plenty of spare inputs on the LabJack. And those displays of current and voltage look nice on the computer, eh? It will create the illusion that I'm doing Real Science. Fun. When the news media shows up -- I think they will, you know -- they'll have something impressive to photograph. And by that time, I'll have images to give them of radiation tracks from prior experiments, assuming I'm lucky.
I'm hoping that someone will start raving about "cold fusion fanatics" at that time, it will help. When one is as obscure as I am, every bit of publicity will help.
Since ordinary water is free, you can run some electrolysis of it at your highest drive power and see if you can measure the AC power contribution from the bubbling. That will give you a baseline model for resistance fluctuations from bubbling.
And you might even be able to hear the crackling if you pipe the audio frequency signal into a high-impedance input to your HiFi.
Yeah, interesting idea. Easy to do, once I'm set up. Bubble noise will vary greatly with exact experimental configuration. I'll have a 250 micron (0.010 inch) gold wire with some kind of fluffy mess of palladium plating on it. These puppies can look really ugly. People have some image of some nice neat palladium plating, but the plating is a fractal, I think, or certainly not clean and neat, that may be part of how codeposition works. Very high surface area. There may be good adherence of bubbles, which should increase bubble noise, I'd expect. Bigger bubbles.
On the other hand, the large surface area may inhibit the bubbles from causing major resistance variation....
Part of my project is to set up an easy, cheap test bed to try out various ideas. At some point, I expect, I'll dump a little beryllium chloride in the cell, maybe even before the palladium chloride. I presume you can guess why.
I wouldn't waste precious metals getting some baseline measurements of AC noise power from bubbling. You can probably do that with copper or nickel electrodes and ordinary water, in an open Pyrex cup.
The most precious material, in my cells, is heavy water, maybe $15 for 25 ml if I don't have to add more. Fingers crossed.
Gold wire is, what, about $750 for 25 feet. The cell has about two inches in it, for the Galileo protocol. I'd like to reduce that, but don't dare until I have replication first. So that's about $5 for the gold. There is more for the platinum anode, maybe double that.
I have stainless steel yarn that I could use, to be sure. 316L steel, might be great. For this purpose, I don't care about contamination from various elements in the steel. I'll put some rad detectors on it, also. Cheap. (A stack of two costs me about fifty cents.) Never can tell, why not look? Shades of Oriani!
What if I find something? Oh Noes!
I'd report it, but damn! Politically inconvenient. Still, we have to deal with Nature as we find her, not as we'd like her to be. Good advice with women, also, eh?
Ah,yes, palladium chloride. And I'd want to reproduce the cathode structure. So, simple answer: stainless steel anode, fine. Silver cathode, easy, cheap. Yes, palladium chloride, I forget what it costs, but it might be $5 or less. Light water. The plastic cell is something like fifty cents. Cheap Fun. Thanks, Barry.
I have no idea what you are expecting to find with the beryllium chloride.
Do you have an hypothesis to test?
Well, read the Wikipedia article on Be, and do you remember what I'm looking for? And are you aware of the probable primary ash? Two clues. Two reasons for Be.
Um, you're looking to sell some kits to amateur hobbyists?
Oh, Beryllium is a neutron absorber that splits into a coupla Heliums plus two more neutrons.
So if you salt your CF cell with some Beryllium and catch a stray neutron, you can gin up some Helium and some more neutrons.
Terrific.
1. Yes, I'm looking to sell kits. Fun is the primary goal. Selling the kits is how I justify spending the money. I'm not selling cold fusion. I'm selling science, and a piece of scientific investigation. Exact replication of a published experiment, sort of cutting edge.
2. Market is threefold. Original idea, science fair kits for brave kids with supportive parents. Then I found that there were grad students itching to do this. Then I realized that there was a need for standard experiments that could be tried by many people, cutting out a lot of the variation, except for what's deliberate.
3. Now as to what I'd find. You are right, the Be might amplify stray neutrons. But, of course, I'd run a hydrogen control. And, of course, presumably, I've already found neutrons without the Be. Be, though, doesn't merely amplify neutrons. It also converts alphas to C12 and a neutron. Are there alphas buzzing around inside the lattice? There is some possibility that I'll be able to characterize neutron energies. Might I be able to tell the difference between multiplied neutrons and alpha neutrons? Don't know.
What can be done for cheap? Maximum fun at minimum cost? All for what I'll already be set up to do, to check out Be, I'll need very little of the stuff. But this isn't first. First I need to replicate, with my materials. Then I can make a hundred kits, assuming I find something worthwhile. If I don't, given the importance of the SPAWAR publication, the hunt will be on to find what's different. I'll need help at this point, the D2O will need to be analyzed, etc.
Yeah, results will be published no matter what.
You might have some luck capturing stray neutrons. That article doesn't quite give enough information to suggest how often that happens.
The other reaction — capturing an alpha particle — isn't gonna happen. When they say "high-energy" alpha particles, they ain't kidding. You have bigtime coulomb forces to overcome there. And then all you get is C-12, which is ubiquitous in the atmosphere as CO2, big deal.
But you might be able to find some data on how often Be captures a stray neutron and splits into He, liberating two more neutrons. It's obviously not a chain reaction, or Be wouldn't be hanging around for long.
The SPAWAR work suggests that there are energetic neutrons, at about 13 MeV, but the rate is very low. The detectors may show, off the top of my head, perhaps two tracks per hour, but the key is the spatial location of the tracks. The detectors are very close to the cathode, and the tracks are spatially associated with the cathode.
Adding a little beryllium to the electrolyte, and plating it onto the cathode, could enhance the number of neutrons. Beryllium itself won't create neutrons, and so few cosmic ray neutrons would hit a small cathode that this isn't an issue. Beryllium is, however, an efficient absorber of neutrons, capture rate might be quite high.
When the beryllium is added, by controlling where in the cathode structure the beryllium is located, could be significant. I'd expect different effects if the beryllium is on the gold, say, or if it's on the surface or in the middle. Surface might poison the reaction, though... It's also possible to consider using a beryllium wire instead of gold, as the substrate.
Now, as to alphas. If the reaction is 4D -> Be-8, one of the proposals that has some quantum field theory behind it, there may be some hot alphas in there. The Hagelstein limit -- have you seen this yet? -- suggests a limitation, from various evidence, to 20 KeV, but Hagelstein could be off a bit. Be-8 ground state decay is something like 90 KeV. This is a big problem for the Be-8 theory!
However, overcoming the Coulomb barrier for fusion takes something like 10 KeV for significant cross-section. 'Taint that hard.
What I read was that thick slabs of beryllium scatter and slow down neutrons, but few hit the bulls eye.
There is no way you are going to get four deuteriums at room temperature to rendezvous like Bob and Carol and Ted and Alice and have an orgy of simultaneous copulation.
"No way." Great, Barry. Isn't this what experiment is about? Testing predictions? Checking out possibilities?
You really don't pay attention, you focus on what you think is impossible, instead of what might be possible. This approach to science is deadly, and it really worries me that you might be communicating this to kids.
You dismiss a Be experiment because you imagine that I'm trying to "get four deuteriums to rendezvous at room temperature," as if that is what anyone is thinking. You are referring to TSC theory, which isn't four independent deuterons meeting in some plasma, where interactions are random, it's about two molecules meeting each other in a particular way, ending up in a particular physical relationship, tetrahedrally symmetric.
But that's mechanism, Barry, and mechanism isn't relevant to my work, except I'd be testing to see if there are energetic alphas inside the lattice, where they are very difficult to observe. TSC theory predicts, probably, energetic alphas. But so do most CF theories, in some way or other.
And you don't seem to get it: neutrons are already observed, clearly. I'm just attempting to replicate, and the Be idea is, then, a possible variation.
The main reason is the alpha, n reaction, not the neutron multiplication.
Your thinking about 4D theory is quite typical for skeptics, the assumptions are those of plasma, where confinement doesn't exist, and where molecules don't exist.
We already know from experiment that fusion cross-section for 3D in palladium deuteride under bombardment exceeds your naive "rendezvous" expectation by 10^26. That's hot fusion, it produces the neutrons.
Could *two molecules* meet? The issue, really, is how they meet. If they meet in the gas, and the collision velocities were such as to cause them to head toward the TS configuration, they would dissociate before they reached it. Lucky, indeed, because otherwise, if Takahashi is correct, any deuterium would fuse. It takes confinement, but in confinement, it's hard to get two molecules in the same lattice cell. Very hard. Storms thinks "too hard." But that's not based on actual calculation, as far as I know.
This is not going to be resolved by debate, it's going to be resolved by the ordinary process of forming hypotheses and testing them. And the work is very difficult, the "unknown reaction" that we want to explain takes place in the closet, so to speak, inside a solid, where it's really difficult to observe.
But people are working on it. Work to "prove cold fusion" has mostly ceased. It's over.
Your own responses show exactly why. The researchers could waste years coming up with better calorimetry, and you'd just assert more possible artifacts, an endless cycle.
Experimental work now is focusing, and has been for years, on improving reliability and scaling it up, on finding new approaches that might be less chaotic and more productive.
It's not possible to cheat Nature, Abd.
It's not possible to violate the Laws of Nature.
But hey, don't take my word for it.
Go ahead and spend the rest of your life trying to cheat Nature, trying to find phenomena that violate the Laws of Nature.
Try to drive a cell in which no AC noise power goes in, so that you can safely leave it out of the books.
Try to drive a cell where no moisture sneaks by as mist when you assumed you had fully evaporated it.
Go ahead. Be my guest.
Barry, I've certainly tried to explain this to you. You don't hear, you don't understand what I'm saying, you are unable to repeat it back clearly, you misrepresent it over and over and over. That is, the problem is manifest right here, without even needing to know the experimental evidence.
There is input noise power, the question is how much, not whether it's there or not. The experimental evidence, which you haven't addressed, is crystal clear: the noise is below a need to consider it. Input power, calculated from voltage and current measurements, is confirmed by the calorimetry, even in the presence of experimental levels of bubble noise.
There is some level, in some experiments, of mist, possible. At high gas evolution rates, it could be a considerable effect. But it's not relevant to closed cells, and not to most open cell designs; the level of possible escape is, again, too low. It could explain some results under some conditions.
You originally asserted it with respect to Zhang, where it was obviously irrelevant, misting would not have caused calorimetric error, because of the calorimeter design. All mist would evaporate within the calorimetric perimeter.
Again and again you have come up with defective criticisms, cogent only when considered in isolation and only when positive evidence is considered, i.e., the reasons you asserted the ad hoc criticism in the first place, but not when contrary evidence is considered.
This isn't scientific thinking at all. It's religion, or political fanaticism.
I'm not selling cold fusion. I'm selling science.
How much AC noise power is going into an electrolytic cell as a function of fluctuations in the ohmic resistance, as it merrily bubbles along?
This is sophomore level AC circuit analysis.
Okay, easy answer. An amount that is below significance. It is below about 1% as confirmed by the calorimetry of control cells and control periods that would have the same input noise problem.
Now, suppose the input voltage is about 5 volts. I look at it with an oscilloscope, which has high bandwidth, way beyond the possible noise from slew response to resistance variations, so I'd see the noise voltage. I see that it is below 0.1 volt. If I apply that to your unverified power calculation, that would limit the noise power to about 0.25% of DC power. But the calorimetry is only accurate to about 1%. And the power excursions are enormous compared to the 1% calorimetry accuracy. Therefore there is no significant noise power.
Your theory is based on an assumption that nobody at SRI would have thought to look at the signal, and that they would have neglected very large noise signals. And it completely neglects the controls, which you have yet to address, in spite of constant mention.
Barry, you are like a bulldog who has failed to notice that he's grabbed firmly onto his own foot.
What are the actual fluctuations in those handful of runs where McKubre reported more than 2% unexplained excess heat, over and above his reported DC input power?
In which experimental series? From a graph that we just looked at on Wikiversity, it's apparent that noise power isn't more than 10 milliwatts, compared to an excess heat signal of 500 milliwatts. But even that is an overestimation.
Frankly, I don't care what the actual noise power is, unless I see some explanation that could cover why this noise power doesn't also appear with dead cells under identical bubbling conditions, nor in a series-connected deuterium cell which would certainly show a third of the supposedly bogus excess power, but, instead showed no more than about 2%, under generous assumptions at peak input power.
And, for that matter, how and why bogus excess power, from any artifact, would end up being well-correlated with helium measurements. So far, you've sloughed this one off entirely.
Neglecting controls and correlations is not science, it's pseudoskepticism, simply trying hard to find something wrong, based on a conviction that something must be wrong, an attitude that then leads to selective consideration of the evidence, thus explaining your original mystery, Barry, why the "skeptics" and the "believers" end up in such deep disagreement.
Somebody isn't listening to all the evidence!
A few days ago, I asked you about the solubility of helium in water. Do you recall me asking you about that?
Sure. I don't know the solubility of helium in water. Why should I care? Helium would be present in controls, if contamination were the issue. It's not. Period. (Well, in some work, there is no effort to exclude atmospheric helium, they only report increase over ambient, but in most work, the materials were all handled to exclude helium contamination.)
I do heartily suggest that if you are going to critique helium work, that you look at actual experimental evidence, instead of following your usual tack of making up stuff out of whole cloth.
I don't know the solubility of helium in water.
Do you know if you read my comments here in "Tiny Bubbles" or my comments here in "Classical Gas"?
Why should I care?
Because you cared if Helium showed up in the vented gases.
Helium would be present in controls, if contamination were the issue.
Why would the same amount of Helium be found in solution in samples of heavy water as in ordinary water? Why would the same amount of Helium be released from solution in one experiment, where some amount of water is dissociated, evaporated, or carried off as mist, compared to another experiment where a different amount of (a different kind of) water is dissociated, evaporated, or carried off as mist?
Barry, you consistently assume that "controls" means "light water controls." Perhaps you should become familiar with the research. Yes, there are light water controls, but in the heat/helium work, helium measured in an experiment that showed no excess heat is also a control. It was heavy water. The same batch of heavy water, generally.
And helium introduced steadily from heavy water would behave differently from that generated depending on whatever is causing the excess heat. Excess heat is not uniformly generated during the experiment, and is not uniformly correlated with current (there is an overall correlation mostly having to do with peak current, but the same current will sometimes produce excess heat, and sometimes, not, depending on other conditions, including loading ratio, but, as well, the exact surface characteristics of the palladium.
Miles selected his palladium, he'd developed skill at that, which is why he found excess heat in 21 out of 33 cells. That was *very good* for the time.
Storms summarizes Miles work this way:
33 cells, 21 showed excess heat, so 12 did not. None of the 12 cells showed helium. (I think the 12 cells were not hydrogen controls. In McKubre's work, no hydrogen cells showed either excess heat or helium.)
Of the 21 cells with observed excess heat, 18 showed helium, at levels consistent with deuterium fusion being the source of the excess heat. 3 did not.
Storms notes that with one of the 3 exceptions, calorimetry error was suspected. The other two were a different alloy used for a cathode. Leaving us with, still, a mystery to be explained.
But not impeaching the correlation. Correlations don't have to be perfect to be conclusive. The 3 cells were not excluded from the correlation calculations.
Physicists are accustomed to setting up "perfect experiments." When Storms wrote you that it was different in chemistry and materials science, you thought he was blowing smoke. He wasn't. This issue has been discussed, again, by the sociologists of science.
No heat, no helium
No heat means no AC noise power (because AC noise power would have injected heat, joule for joule).
No AC noise power means no bubbling.
No bubbling means no dissociation of electrolyte.
No dissociation of electrolyte means no release of dissolved Helium.
(Am I boring you yet?)
No excess heat. Heat. Bubbling. If there is going to be AC power, it will be there, because the source of it is bubbling. There is dissociation of electrolyte. If the helium were in the electrolyte, helium would show up in all the cells, every one of them using the contaminated heavy water. These are all electrolytic cells, and if they are showing excess heat, they have very high loading, probably above 90%.
Yes, you are boring me, because you are responding only to your own imagination. The above was a Really Stupid Argument. Really Stupid is Boring, like shooting fish in a bucket is Boring.
If the calorimetry error is of the order of 1%, then you won't see heat from AC noise power unless the fluctuations in ohmic resistance is of the order of 10%.
Remember that everything in the bubbling model is stoichiometric, and a direct function of the drive current. As the drive current goes up, you dissociate more water, get more bubbles, get more fluctuations, more AC noise power, and the release of more dissolved helium.
So if stoichiometry bores you, and if AC circuit analysis bores you, why are you still paying attention to me?
Obviously, you are burning out, Barry, you've been reduced to repetitious stupidity.
Drive current is the same in these experiments, generally, and they are all reasonably close to maximum loading, but loading, indeed, would be another observable difference between the excess heat and other cells. However, the other cells don't fully load because the palladium won't accept it, so they are all at maximum bubbling. If the source of the helium is the heavy water, all the cells would show helium. But only those which show excess heat show helium.
Indeed, why would power supply AC noise only show up in the same set of cells as show helium? These are identical cells, as to bubbling, they are heavy water, they are palladium, and the only difference is probably the exact surface conditions, as to cracks in the palladium. Heavy cracking means heavy leakage of deuterium. I.e., it bubbles out.
Now, this could explain something, come to think of it. But it wouldn't eliminate the helium, it would reduce the quantity. Non-excess heat cells are effectively filtering the deuterium, it's going in but it's coming out somewhere else. Helium doesn't move so easily through the lattice, though it does move. Thus helium would be filtered, to some extent. With the highly loaded cells, the bubbling is more likely to be raw deuterium, thus it would carry more of the dissolved helium.
In fact, though, this would merely weaken the correlation. What happens, in fact, is that about half the helium is retained in the palladium, and analysis of the palladium shows that it's concentrated near the surface. mmmm... that supports the dissolved helium hypothesis, as far as it goes. Perhaps the helium was carried into the palladium by the deuterium, some of it.
Now, how would you tell the difference? Look, Barry, you can do much better than you have. If you are bored, perhaps it is because you *aren't interested* in the topic.
Ridiculing others over a topic that bores you is a Bad Idea. You won't understand it, or them.
The last time I checked, helium was not identical to deuterium. The last I checked, deuterium still had twice the mass and twice the density as hydrogen.
What I'm interested in (in case you forgot) was identifying the affective emotional states of those engaged in the learning journey associated with investigations into what's going on in CF cells.
Are you running on hope?
Because I hope to discover what you are running on.
I'm running on a love of truth, and a fascination with the process of how we approach truth.
In that case, I reckon you will be fascinated with "The Phantom Tollbooth" — especially when you come to the climactic battle for Truth.
How would you characterize Milo's quest for Truth?
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