[Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
Actually it is not beyond the bounds of possibility to set up such a demonstration. What exactly do you have in mind, and who would be interested in seeing such a demo? Do you have any contacts on the Rossi team? I don't think Rossi would travel to the USA to see such a demo. Electrical Engineers already know that a diode will convert AC to DC. Pretty much all scientists know that an AC current clamp will not measure DC. (Of course, DC rated Hall effect clamps are available but were not used in the demo, partially because Rossi appears to believe that an AC outlet will only deliver AC current - this is far from being the case). So who would your intended audience be for such a demonstration? Duncan On 5/26/2013 7:26 PM, David Roberson wrote: Not my position. You need to show how it was done. Dave -Original Message- From: Duncan Cumming spacedr...@cumming.info To: vortex-l vortex-l@eskimo.com Sent: Sun, May 26, 2013 9:47 pm Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments So is it your position that a current clamp without a Hall effect unit can measure DC? Mine is that it cannot. Duncan On 5/26/2013 5:34 PM, David Roberson wrote: How do we know that your diode trick will actually do what you think? You need to prove that this is possible, otherwise anyone can make the assumption that it might not work just as with the ECAT tests. If you do not prove that this will work, then why should we accept it as a possibility? A lot of time and energy is being wasted trying to see if bull frogs can fly. Some might actually be born with wings. Have we proven that none of them can fly? Rossi and the testers have done a lot to prove that the ECAT works. No one has proven that it does not. The only offers from the other side of the table assume fraud. Is this a valid position for them to take? Dave -Original Message- From: Duncan Cumming spacedr...@cumming.info To: vortex-l vortex-l@eskimo.com Sent: Sun, May 26, 2013 8:18 pm Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments I am not trying to assert anything as fact. I am merely pointing out that a simple diode inside the controller box (to which access was forbidden by Rossi) COULD HAVE given the observed results. I am NOT saying that it, in fact, did, merely speculating that it could have. For any scientific experiment, the onus is on the experimenters to produce the result. The best way to do this is to provide sufficient information for others to replicate the experiment. Duncan On 5/26/2013 5:07 PM, David Roberson wrote: Perhaps you should build one of these scam machines and prove that it will work without being detected. That would be the best way to show that it is possible. Why should we accept this assertion as fact any more than believing that the testers missed finding the scam? We can spend an equal amount of time knocking down any theory that is put forth as others can spend assuming they are real. Dave -Original Message- From: Duncan Cumming spacedr...@cumming.info To: vortex-l vortex-l@eskimo.com Sent: Sun, May 26, 2013 7:59 pm Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments The only possibility to fool the power-meter then is to raise the DC voltage on all the four lines This turns out not to be the case. You could also draw DC current through any of the lines, which current would not register on the clamps. The simplest way to do this would be just to use a diode in series with the heating element. Since power = current x voltage x pf, it is NOT necessary to change the voltage in order to change the power. Duncan On 5/26/2013 2:21 PM, Jed Rothwell wrote: A Swedish correspondent sent me this link: http://www.energikatalysatorn.se/forum/viewtopic.php?f=2t=560sid=5450c28dab532569dee72f88a43a56f0start=330 This is a discussion in Swedish, which Google does a good job translating. Before you translate it, you will see that in the middle of it is a message from one of the authors, Torbjörn Hartman, in English. Here it is, with a few typos corrected. QUOTE: Remember that there were not only three clamps to measure the current on three phases but also four connectors to measure the voltage on the three phases and the zero/ground line. The protective ground line was not used and laid curled up on the bench. The only possibility to fool the power-meter then is to raise the DC voltage on all the four lines but that also means that the current must have an other way to leave the system and I tried to find such hidden connections when we were there. The control box had no connections through the wood on the table. All cables in and out were accounted for. The E-cat was just lying on the metal frame that was only free-standing on the floor with no cables going to it. The little socket, where the mains cables from the wall connector where connected with the cables to the box and where we had
Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
What I am proposing is a lot simpler than that. No bridge rectifier, no capacitor, just a simple diode. I am saying that given a diode in series with a resistor, it is not possible to measure the power using a clamp on ammeter. I am not suggesting that anybody has performed a scam. I am suggesting that the equipment used would not have measured the power consumed by the resistor if rectification were present in the controller box. Is there anybody reading this that can do SPICE simulations? Might it be possible to simulate a resistor in series with a diode and determine the actual and apparent power if an AC coupled current meter is used? Duncan P.S. I never mentioned either bridge rectifiers or capacitors. In the case of a bridge rectifier type power supply, then a clamp on ammeter will work OK. I do not suspect such a thing in the demo. On 5/26/2013 7:35 PM, David Roberson wrote: Assume that you have a bridge rectifier in the blue box. This is followed by a filtering capacitor. The DC is then used by the electronics connected to the capacitor. Are you saying that it is not possible to determine the power input to this type of network by measuring the input AC voltage and current? Or are you saying that someone has performed a scam and put a DC supply in series with the normal AC voltage? You do know that this could easily be measured by a simple DC voltmeter, right? Dave -Original Message- From: Duncan Cumming spacedr...@cumming.info To: vortex-l vortex-l@eskimo.com Sent: Sun, May 26, 2013 10:01 pm Subject: Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments Almost. The power being fed to the heater exceeds that measured at the wall, because the sensor used (an AC current clamp) cannot sense the direct current being drawn from the wall socket. Some people find the difference between current and voltage confusing. What I am saying here is that if you connect a resistor in series with a diode to a wall socket, then the CURRENT drawn is direct even though the VOLTAGE at the socket is alternating. (Rossi does not seem to understand this concept judging by his message that got posted today). So unless you use a DC rated current meter (such as a shunt) you will not sense all of the current, and hence power, drawn from the wall socket. The electrical power meter in your house certainloy IS rated for DC, so you will certainly be BILLED for the power even though you didn't measure it yourself! V = IR Power = Voltage * Current * Power Factor Duncan On 5/26/2013 5:57 PM, Eric Walker wrote: I wrote: On Sun, May 26, 2013 at 5:18 PM, Duncan Cumming spacedr...@cumming.info mailto:spacedr...@cumming.info wrote: I am not trying to assert anything as fact. I am merely pointing out that a simple diode inside the controller box (to which access was forbidden by Rossi) COULD HAVE given the observed results. I am NOT saying that it, in fact, did, merely speculating that it could have. Am I right in understanding that this line of reasoning requires tampering with the mains itself, where the electrical measurements were made, in addition to any sly customizations that might have been made at the controller? I think I'm starting to understand. This is a separate line of reasoning to the one about the possibility of hidden DC and RF passing undetected through the clamp meters at the mains. In this line of reasoning, the duty cycle (35 percent ON) is misunderstood, and there is a hidden DC component from the controller delivering power to the E-Cat, but not above what was read from the wall -- am I describing this right? Eric
Re: [Vo]:RE: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
My source was Hanno Essen, one of the authors. He answered a question asked by email by one Sterling D. Allan http://sterlingdallan.com/ / of Pure Energy Systems News/, reported earlier in this list. /4. Have you tried to test the output of the power supply to exclude that/ /also a DC current is supplied to the device, which clamp amperometers/ /could not detect?/ No, we did not think of that. The power came from a normal wall socket and there did not seem to be any reason to suspect that it was manipulated in some special way. Now that the point is raised we can check this in future tests. The PCE clamp to which you link is, indeed, DC rated. But Essen does seem to believe, erroneously, that it is not possible to draw direct current from an ordinary AC outlet. In fact, a simple diode does enable one to take fluctuating DC from an AC outlet, which outlet has not been manipulated in any special way. Maybe Essen does not have an EE background? On 5/26/2013 7:53 PM, Jones Beene wrote: *From:*Duncan Cumming So is it your position that a current clamp without a Hall effect unit can measure DC? Mine is that it cannot. Did you actually check the PCE site? It looks to me like all the current clamps on the PCE power analyzer site measure both AC and DC http://www.industrial-needs.com/technical-data/current-detector-PCE-DC-3.htm
Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
Yes, Robin is correct. Duncan On 5/26/2013 8:08 PM, mix...@bigpond.com wrote: In reply to David Roberson's message of Sun, 26 May 2013 22:35:09 -0400 (EDT): Hi, This is a little different. A full bridge rectifier will allow for both halves of the AC current to pass, and so it should be measured as little different to a purely resistive load. However a single diode will only allow one half to pass, which *may* mess up magnetic field based current measurements. (I guess whether if does or not depends on the sophistication of the device.) Assume that you have a bridge rectifier in the blue box. This is followed by a filtering capacitor. The DC is then used by the electronics connected to the capacitor. Are you saying that it is not possible to determine the power input to this type of network by measuring the input AC voltage and current? Or are you saying that someone has performed a scam and put a DC supply in series with the normal AC voltage? You do know that this could easily be measured by a simple DC voltmeter, right? Dave [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
[Vo]:Re: [Vo]:Torbjörn Hartman describes
With chopped DC, a clamp on ammeter will show the AC component. So if you had 0 to 1 amp chopped, the ammeter would show 0.5 amps peak AC. So you get a partial reading, substantially less than the true current that is actually flowing. IMHO, this could have happened at the demo. I am not saying that it did (I was not there), merely that it could have. Duncan On 5/26/2013 8:09 PM, a.ashfield wrote: Duncan Cumming No, it does not. What happens is that the diode rectifies the mains to DC, and the DC is not sensed by the clamp-type current meter. What would the clamp on meter show with chopped DC?
[Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
I am not sure if I count as a skeptic, because I am not saying that any kind of scam was perpetrated. I am certainly not suggesting that there was a DC power supply hidden in the wall! My doubts are related to the electrical engineering skills evident in the published paper, attempting the notoriously difficult task of measuring three phase non sinusoidal power. Not only is the waveform non sinusoidal, it is a trade secret! I am merely saying that rectification will cause a misleadingly low value of current to be registered using a clamp on ammeter. Since the DC is not smooth, there will, indeed, be a small reading from the ammeter but substantially lower than the actual current. This will, in turn, lead to a misleadingly low power measurement. Duncan On 5/26/2013 8:46 PM, David Roberson wrote: Robin, The problem at hand is that the skeptic claims that power due to the DC current can be very large and not detected. There has been no discussion of the AC current reading being affected by the DC so far. That is a different issue entirely. I would like for them to answer the questions because then they might realize that their position is invalid. I can explain this if required. No one is suggesting that Rossi actually has a DC power supply hidden within the wall I hope. This would be beyond reality since it would be so easy to measure with a voltmeter or any monitor that looks at the voltage. The testers did a visual look at the voltage from what I have determined. So, skeptics, what say you? Dave -Original Message- From: mixent mix...@bigpond.com To: vortex-l vortex-l@eskimo.com Sent: Sun, May 26, 2013 11:08 pm Subject: Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments In reply to David Roberson's message of Sun, 26 May 2013 22:35:09 -0400 (EDT): Hi, This is a little different. A full bridge rectifier will allow for both halves of the AC current to pass, and so it should be measured as little different to a purely resistive load. However a single diode will only allow one half to pass, which *may* mess up magnetic field based current measurements. (I guess whether if does or not depends on the sophistication of the device.) Assume that you have a bridge rectifier in the blue box. This is followed by a filtering capacitor. The DC is then used by the electronics connected to the capacitor. Are you saying that it is not possible to determine the power input to this type of network by measuring the input AC voltage and current? Or are you saying that someone has performed a scam and put a DC supply in series with the normal AC voltage? You do know that this could easily be measured by a simple DC voltmeter, right? Dave [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
[Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
I will give it my best shot. Consider a diode in series with a resistor, and connected to an AC outlet. For the first half of the cycle the diode conducts, and a positive current flows. For the second half, the diode does not conduct and NO NEGATIVE CURRENT FLOWS, even though a negative voltage is present. This is the function of a diode. So what you have is an intermittent flow of positive current, which delivers power to the load resistors. The magnitude of this power is given by I^2*R. If you were to measure the current using a clamp on ammeter which is DC rated, then the current would be determined accurately and the RMS value determined by the digital voltmeter (which must be a true RMS type, of course). Multiplying this by the voltage gives the power dissipated in the resistor. If you were to measure the current using a clamp on ammeter which is NOT DC rated, then only the fluctuations in current would be measured, which fluctuations would be a lot less than the true value of current. So a misleadingly low value of current would be measured, leading to a substantial under estimate of the power dissipated in the resistor. This is electrical engineering 101, but it shows some of the problems involved in measuring AC power with a non-sinusoidal waveform. There are those who assume that a commercially available power meter measures just that, but in fact this is a difficult task that should be undertaken by a qualified engineer with knowledge of the waveform that he is measuring. The specs of such an instrument clearly indicate the limitations to which it is subject, and one must be careful not to exceed these limitations. An absence of DC sensing capability is one such limitation. In the diode example above, the diode itself does not provide any power whatever. It merely confuses some types of power meters. On 5/26/2013 9:51 PM, David Roberson wrote: It does not make any difference whether or not the instrument measures DC current through the input power cables. That issue is dead unless someone wants to insist that Rossi or one of his partners hid a DC supply inside the wall, or in some other place which allows the DC to appear at the power input terminals. This would have been obvious to anyone looking at the voltage. Andrew or Duncan please explain how the DC current through the input power cable is able to deliver a large power to the load resistors? It can not be done with any type of diode hidden within the blue box. Are you ready to concede the point? Dave -Original Message- From: Alan Goldwater a...@magicsound.us To: vortex-l vortex-l@eskimo.com Sent: Mon, May 27, 2013 12:19 am Subject: Re: [Vo]:RE: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments That is a different instrument. The one used in the tests (PCE-830 http://www.industrial-needs.com/technical-data/power-anlayser-PCE-830.htm) does not measure DC. On 5/26/2013 7:53 PM, Jones Beene wrote: Did you actually check the PCE site? It looks to me like all the current clamps on the PCE power analyzer site measure both AC and DC http://www.industrial-needs.com/technical-data/current-detector-PCE-DC-3.htm
[Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
This is only true for sinusoidal waveforms. As soon as you introduce non-sinusoidal waveforms, such as by using a diode, then different calculations must be used. http://en.wikipedia.org/wiki/Power_factor On 5/27/2013 7:49 AM, David Roberson wrote: All of the power being delivered into the resistor from the wall socket can be determined by taking the AC voltage which is a sine wave and multiplying it by the fundamental frequency of the AC current(also a sine wave). This must be adjusted by multiplication by the cosine of the phase angle between the supply voltage and fundamental current.
Re: [Vo]:The inanity of the hidden input power hypothesis
I am not suggesting that there was any modification of the laboratory wiring, such a thing would be ridiculous as you correctly point out. What I AM suggesting is that an oscilloscope be used to measure the CURRENT waveform at the electrical outlet, not the voltage. The voltage is obviously a sine wave as with any electrical outlet socket. As you also correctly point out - Rossi did not want an oscilloscope present - period. I would expand upon this by saying that he will also probably not permit an osciloscope in the future, particularly measuring current (using a DC rated current probe,of course). We seem to mostly be in agreement about the facts here, only the motivation for Rossi and his critics seems to be in dispute. Duncan On 5/27/2013 8:01 AM, Jones Beene wrote: Whoa. Someone is building a mountain out of a molehill here - and for what purpose? To show that a that cheating could have been accomplished - as an exercise in remote possibilities or magic tricks? ... or is it to express frustration that the poster does not understand the experiment? Rossi did not want an oscilloscope present - period. This has nothing to do with its placement. Of course there would be little apparent harm to connect a scope to the same wall plug to which the power input for the E-cat is connected, but if a scope is present anywhere, then it can be used to inadvertently expose a trade secret. Thus - no scope permitted, only power analyzers. To go further than what an o'scope could tell us that a power analyzer could not exposes bias. Not that bias needs exposing, since this entire thread is surely the pinnacle of lame bickering over nothing of importance. Never did Rossi say that DC capable clamps would not be allowed. In fact he would have expected that DC capable clamps could have been used - had he taken the time to reflect on the issue. To think that any scammer risks exposure by rewiring the lab is absurd - since the independent testers were permitted to have a DC capable clamp or power analyzers that could have measured DC, even if this one did not. This whole collection of dozens of needless postings is itself the pathetic invention of frustrated skeptics who think that Rossi must be cheating - but cannot prove it ... so they are grasping at straws. If Rossi had altered the wiring with DC or RF, it could have been discovered with a permitted instrument, over which AR had no control. Moreover, if Rossi cheated in this way, it could have physically injured the participants (given that skeptics are looking for an extra kilowatt or more of input). Does he risk that? No way! To say that he does risk it - exposes the silliness of this stance, since there is no real motive. If there is a mistake in measurement, it is most likely on the output side, not the input. In short: Get over it! There is NO MODIFICATION OF THE LAB WIRING. Move on to something has a minimum level of credulity! Jones -Original Message- From: Rob Dingemans Hi, Duncan Cumming wrote: Now for the argument that Rossi runs the risk that somebody will try a type B meter (DC capable), or, for that matter, a simple oscilloscope. He simply does not permit such things. He claims not to allow an oscilloscope because it would reveal a proprietary waveform. By keeping tight control over the test conditions, he is able to ensure that his questionable power measurements are not exposed. By not allowing inspection of the heater controller, he keeps the diode (or asymmetrical firing of the Triacs) from public view. Rossi behaves as if a mundane heater control is super-secret technology - does nobody else find this strange? I can hardly believe that when you connect a scope to the same wall plug as to which the input for the E-cat is connected that Andrea will not allow this. If my assumption is right that: a: the proprietary waveform is of a much higher frequency/waveform then the AC from the wall plug, b: Andrea might be afraid for feedback signals coming from the E-cat control box back into the grid, then a low-pass filter (up to ~ 50 Hz) between the wall plug and the E-cat control box should be sufficient for: a: the scope not being able to detect the proprietary waveform generated in the control box and fed back to the grid, b: at the same time still be able to detect any possible strange waveforms trying to being inserted through the wall plug into the control box of the E-cat, c: and also preventing any strange waveforms to be passing through the low-pass filter into the control box of the E-cat :-) . B.t.w. if Andrea is afraid of the proprietary waveform generated in the control box and fed back to the grid from happening he should redesign his control box and include the low-pass filter as a part of the internal circuitry. Kind regards, Rob
Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
Yes, Hall effect clamps are readily available, I am not disputing that. They used to suffer from drift problems, but these problems have pretty much been solved. The one that you show has a 3% accuracy and 8 digits of drift - not bad. The only info I have about Rossi came from a single post on this list, his opinion may well be different than it appears - for one thing, there were some translation problems. So we are in agreement on these points. Here are a few more points - can we agree on these as well? 1) I am not saying that Rossi is now, or has ever previously done anything fraudulent. 2) I am not saying that anything except regular AC voltage was present at the wall socket during the demo. 3) I am not claiming that the lab was in any way re-wired in an attempt to perpetrate a fraud. 4) I am not speculating that either bridge rectifiers or smoothing capacitors may have been used. Duncan On 5/27/2013 11:36 AM, Jed Rothwell wrote: Duncan Cumming wrote: (Of course, DC rated Hall effect clamps are available but were not used in the demo, partially because Rossi appears to believe that an AC outlet will only deliver AC current - this is far from being the case). 1. People have been measuring DC amperage by measuring a magnetic field since 1820. 2. These are Hall effect clamps. See the specifications They are rated for very low DC power: http://www.industrial-needs.com/manual/manual-clamp-meter-pce-cd3.pdf 3. Rossi played no role in this. His beliefs about AC are probably not as you describe them, but in any case he had no say in the matter. - Jed
Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
OK, I will tackle this problem head-on using the Socratic method in stages. First, consider a wire carrying 100 amps of direct current, plus one amp of pure sinusoidal AC current at 60Hz. What is the AC component of the current? Duncan P.S. Don't worry, we will get to the diode later. On 5/27/2013 11:57 AM, David Roberson wrote: Duncan, I hate to keep repeating myself that the power can be measured by analyzing the AC components only. When will you guys show why this is not true? I suggest that you start with the simple system you proposed of a diode in series with a resistor driven by an AC wall socket. Explain how it works as you say and I promise to show you the error of your calculations. Dave -Original Message- From: Duncan Cumming spacedr...@cumming.info To: vortex-l vortex-l@eskimo.com Sent: Mon, May 27, 2013 2:38 pm Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments I am not sure if I count as a skeptic, because I am not saying that any kind of scam was perpetrated. I am certainly not suggesting that there was a DC power supply hidden in the wall! My doubts are related to the electrical engineering skills evident in the published paper, attempting the notoriously difficult task of measuring three phase non sinusoidal power. Not only is the waveform non sinusoidal, it is a trade secret! I am merely saying that rectification will cause a misleadingly low value of current to be registered using a clamp on ammeter. Since the DC is not smooth, there will, indeed, be a small reading from the ammeter but substantially lower than the actual current. This will, in turn, lead to a misleadingly low power measurement. Duncan On 5/26/2013 8:46 PM, David Roberson wrote: Robin, The problem at hand is that the skeptic claims that power due to the DC current can be very large and not detected. There has been no discussion of the AC current reading being affected by the DC so far. That is a different issue entirely. I would like for them to answer the questions because then they might realize that their position is invalid. I can explain this if required. No one is suggesting that Rossi actually has a DC power supply hidden within the wall I hope. This would be beyond reality since it would be so easy to measure with a voltmeter or any monitor that looks at the voltage. The testers did a visual look at the voltage from what I have determined. So, skeptics, what say you? Dave -Original Message- From: mixent mix...@bigpond.com To: vortex-l vortex-l@eskimo.com Sent: Sun, May 26, 2013 11:08 pm Subject: Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments In reply to David Roberson's message of Sun, 26 May 2013 22:35:09 -0400 (EDT): Hi, This is a little different. A full bridge rectifier will allow for both halves of the AC current to pass, and so it should be measured as little different to a purely resistive load. However a single diode will only allow one half to pass, which *may* mess up magnetic field based current measurements. (I guess whether if does or not depends on the sophistication of the device.) Assume that you have a bridge rectifier in the blue box. This is followed by a filtering capacitor. The DC is then used by the electronics connected to the capacitor. Are you saying that it is not possible to determine the power input to this type of network by measuring the input AC voltage and current? Or are you saying that someone has performed a scam and put a DC supply in series with the normal AC voltage? You do know that this could easily be measured by a simple DC voltmeter, right? Dave [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
[Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
Although it is true that the DC component of voltage is 0, the DC component of current is not. Since the current is non-sinusoidal, it is not possible to analyze it using only the fundamental frequency. This is the whole issue of power supply design. Consider an old-school power supply using a half wave rectifier (diode) a smoothing capacitor, and a load. This supply produces output power, even though the DC component of voltage at the input is zero. Where does this power come from, if not from the AC outlet? The fundamental component of a non-sinusoidal waveform represents only a fraction of that waveform. The rest of it is represented by the harmonics. If you are interested in power supply design, the following text book is excellent: http://www.amazon.com/Switching-Power-Supply-Design-ebook/dp/B001AO0GDG/ref=dp_kinw_strp_1 Duncan On 5/27/2013 12:17 PM, David Roberson wrote: That is a good try. I agree with all that you say except for one key item. 1). No negative current flows due to the diode. 2). The instantaneous power being delivered to the resistor is I^2*R as you suggest. 3). The DC rated clamp on meter should measure the total RMS current provided it can handle distorted AC waveforms. Now, here is where you have a problem with the measurement. You say to multiply the true RMS current by the voltage and that is where the problem arises. I am confident that you realized that as soon as you said it! All of the power that is applied to the resistor comes from the wall socket. The voltage at this location is a sinewave at the frequency supplied by the electrical service. There is no DC voltage component, so the DC power being supplied is 0. The AC component of the input frequency is the only one that can have power supplied and that can only be given to current at its fundamental frequency. So, to determine how much power the resistor absorbs you must take the fundamental current component and multiply it by the fundamental voltage supplied by the wall socket. This needs to be corrected for phase shift if any exists with the product by the cosine of the difference in phase of the two components. Therefore, the DC flowing through the rectifier does not contribute to the measurement. Dave -Original Message- From: Duncan Cumming spacedr...@cumming.info To: vortex-l vortex-l@eskimo.com Sent: Mon, May 27, 2013 2:56 pm Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments I will give it my best shot. Consider a diode in series with a resistor, and connected to an AC outlet. For the first half of the cycle the diode conducts, and a positive current flows. For the second half, the diode does not conduct and NO NEGATIVE CURRENT FLOWS, even though a negative voltage is present. This is the function of a diode. So what you have is an intermittent flow of positive current, which delivers power to the load resistors. The magnitude of this power is given by I^2*R. If you were to measure the current using a clamp on ammeter which is DC rated, then the current would be determined accurately and the RMS value determined by the digital voltmeter (which must be a true RMS type, of course). Multiplying this by the voltage gives the power dissipated in the resistor. If you were to measure the current using a clamp on ammeter which is NOT DC rated, then only the fluctuations in current would be measured, which fluctuations would be a lot less than the true value of current. So a misleadingly low value of current would be measured, leading to a substantial under estimate of the power dissipated in the resistor. This is electrical engineering 101, but it shows some of the problems involved in measuring AC power with a non-sinusoidal waveform. There are those who assume that a commercially available power meter measures just that, but in fact this is a difficult task that should be undertaken by a qualified engineer with knowledge of the waveform that he is measuring. The specs of such an instrument clearly indicate the limitations to which it is subject, and one must be careful not to exceed these limitations. An absence of DC sensing capability is one such limitation. In the diode example above, the diode itself does not provide any power whatever. It merely confuses some types of power meters. On 5/26/2013 9:51 PM, David Roberson wrote: It does not make any difference whether or not the instrument measures DC current through the input power cables. That issue is dead unless someone wants to insist that Rossi or one of his partners hid a DC supply inside the wall, or in some other place which allows the DC to appear at the power input terminals. This would have been obvious to anyone looking at the voltage. Andrew or Duncan please explain how the DC current through the input power cable is able to deliver a large power to the load resistors? It can not be done with any type of diode hidden within the blue
Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
Why this hangup about fundamental components? I can extract current from an AC waveform any way I want. Switched mode power supplies usually do this at 20kHz or so, even though the fundamental component is 60Hz. But you are right about one thing - we may as well end this discussion. It is like trying to explain the purpose and function of a Dewar vessel to an ant! Duncan On 5/27/2013 1:55 PM, David Roberson wrote: If you do not understand what I have already written then it is not going to help to go over it again. I leave this discussion by asking you one pertinent question. Where do you think the power comes from that ends up in the resistor? There is only one source and it is the AC mains. Power from an AC source can only be extracted by the fundamental component of that source, period. All others, including DC balance out over the long run and can not make a long term contribution. Once you realize that this is true, which is common theory, it will become clear to you that a measurement of these two waveforms is all that is required. Forget the nonsense about diodes faking out good AC true RMS instruments. It don't happen. Dave -Original Message- From: Duncan Cumming spacedr...@cumming.info To: vortex-l vortex-l@eskimo.com Sent: Mon, May 27, 2013 4:32 pm Subject: Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments OK, I will tackle this problem head-on using the Socratic method in stages. First, consider a wire carrying 100 amps of direct current, plus one amp of pure sinusoidal AC current at 60Hz. What is the AC component of the current? Duncan P.S. Don't worry, we will get to the diode later. On 5/27/2013 11:57 AM, David Roberson wrote: Duncan, I hate to keep repeating myself that the power can be measured by analyzing the AC components only. When will you guys show why this is not true? I suggest that you start with the simple system you proposed of a diode in series with a resistor driven by an AC wall socket. Explain how it works as you say and I promise to show you the error of your calculations. Dave -Original Message- From: Duncan Cumming spacedr...@cumming.info To: vortex-l vortex-l@eskimo.com Sent: Mon, May 27, 2013 2:38 pm Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments I am not sure if I count as a skeptic, because I am not saying that any kind of scam was perpetrated. I am certainly not suggesting that there was a DC power supply hidden in the wall! My doubts are related to the electrical engineering skills evident in the published paper, attempting the notoriously difficult task of measuring three phase non sinusoidal power. Not only is the waveform non sinusoidal, it is a trade secret! I am merely saying that rectification will cause a misleadingly low value of current to be registered using a clamp on ammeter. Since the DC is not smooth, there will, indeed, be a small reading from the ammeter but substantially lower than the actual current. This will, in turn, lead to a misleadingly low power measurement. Duncan On 5/26/2013 8:46 PM, David Roberson wrote: Robin, The problem at hand is that the skeptic claims that power due to the DC current can be very large and not detected. There has been no discussion of the AC current reading being affected by the DC so far. That is a different issue entirely. I would like for them to answer the questions because then they might realize that their position is invalid. I can explain this if required. No one is suggesting that Rossi actually has a DC power supply hidden within the wall I hope. This would be beyond reality since it would be so easy to measure with a voltmeter or any monitor that looks at the voltage. The testers did a visual look at the voltage from what I have determined. So, skeptics, what say you? Dave -Original Message- From: mixent mix...@bigpond.com To: vortex-l vortex-l@eskimo.com Sent: Sun, May 26, 2013 11:08 pm Subject: Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments In reply to David Roberson's message of Sun, 26 May 2013 22:35:09 -0400 (EDT): Hi, This is a little different. A full bridge rectifier will allow for both halves of the AC current to pass, and so it should be measured as little different to a purely resistive load. However a single diode will only allow one half to pass, which *may* mess up magnetic field based current measurements. (I guess whether if does or not depends on the sophistication of the device.) Assume that you have a bridge rectifier in the blue box. This is followed by a filtering capacitor. The DC is then used by the electronics connected to the capacitor. Are you saying that it is not possible to determine the power input to this type of network by measuring the input AC voltage and current? Or are you saying that someone has performed a scam and put a DC supply in series with the normal AC voltage? You do know
Re: [Vo]:The inanity of the hidden input power hypothesis
What WOULD go a long way towards convincing engineers that the effect is real is the simple use of DC power on the heaters. That completely eliminates all of this meter A / meter B nonsense. Plus some transparency from Rossi, allowing the engineers to measure what they want to measure, such as waveforms and so forth. He can use an NDA if he is worried about trade secrets. For a strong proof, use an insulated chamber (tube furnace) with cooling coils, as opposed to an uninsulated chamber just sitting in the lab. A few days of self sustaining running (cooling only without electrical input) would be pretty convincing. Use electrical heaters only for starting the reaction. http://www.sentrotech.com/high-temperature-tube-furnace To an engineer, calorimetric measurements made on a hot tube that is completely uninsulated appear to lack accuracy. Not to mention the safety issues of a red hot tube just hanging out there in the lab. The slightest draft will change the cooling rate. For good accuracy, cooling should be done in a controlled manner with proper cooling coils. A longer test, just more of the same, will be no more convincing than a shorter test. I, for one, WANT to be convinced! Duncan On 5/26/2013 12:02 AM, Eric Walker wrote: On Sat, May 25, 2013 at 11:54 PM, Andrew andrew...@att.net mailto:andrew...@att.net wrote: The *only* way to convince the scientific community is via evidence. They will be carrying out a much longer experiment in the future. If they were to have an electrical engineer take a close look at the input power across the entire range of interest and rule out input fake, after which they were to report results similar to the ones that were reported this time around, would this be considered adequate evidence for a prima facie conclusion that Rossi's device is producing excess heat? Eric
Re: [Vo]:The inanity of the hidden input power hypothesis
Alan: Perhaps you can explain something to me. With the device in the ambient laboratory atmosphere, it loses a certain number of watts of heat to the air, but this is subject to variation from drafts and so forth. But even so, it is still possible to control the e-cat by modulating the electrical input. Now consider an insulated furnace equipped with cooling tubes. The device still loses the exact same number of watts to the cooling coils that it previously lost to the ambient air, only now the heat losses are controlled and measurable using ordinary calorimetry. Why is the control problem any better or worse? You can still modulate the heating power exactly as before, and the losses are exactly the same as they were before. The only difference is that you no longer need to make assumptions about emissivity and convection currents. Duncan On 5/26/2013 1:14 PM, Alan Fletcher wrote: From: Duncan Cumming spacedr...@cumming.info Sent: Sunday, May 26, 2013 12:53:45 PM A few days of self sustaining running (cooling only without electrical input) would be pretty convincing. Use electrical heaters only for starting the reaction. I think it's been pretty conclusively demonstrated that the eCat operates in a narrow band of stability. (See the November melt-down, for example). Lewan's self-running test showed a peak and then a decay -- and certainly wouldn't have lasted a few days. Make the peak too high, and you've lost control. Rossi has apparently found a way of controlling the reaction, in a range up to COP=6, by stimulating the reaction in a series of rise-and-decay bursts. Siegel on http://scienceblogs.com/startswithabang/2013/05/21/the-e-cat-is-back-and-people-are-still-falling-for-it/ demands a closed-loop test. But with a COP of 6, how are you going to get the feedback? (Turbines are being tested, but not yet formally announced). So ... you're left with thermoelectric, at what .. 10% efficiency? So to prove the impossible eCat you have to violate the laws of thermodynamics. The gas-fired eCat is still a mystery. In addition to the thermal input, does it still need pulses or waveforms? In my opinion, we're left with either wiring fakes (eg coax, or double-running wires to hide the current), or DC. DC can, and should have been checked, on the input lines. Preferably with in-line resistors, rather than current loops. Both of these (wiring or other power fakes) are easily eliminated by using a motor generator. Heck, even COAL plants take their control energy from the Grid.
Re: [Vo]:The inanity of the hidden input power hypothesis
No extra weight whatever. The NDA would not be for the purposes of convincing anybody about the demo. It would be for the purposes of allowing engineering measurements (waveforms, DC components and so forth) to be made. If Levi measured, for example, that there was no DC offset, chances are Rossi would either allow him to say that or choose not to release any test data. We would avoid the ridiculous position that we are in now, when some data is released (e.g. power output) but other data is withheld (e.g. power input), for who knows what reason? On 5/26/2013 1:16 PM, Alan Fletcher wrote: From: Duncan Cumming spacedr...@cumming.info Sent: Sunday, May 26, 2013 12:53:45 PM He can use an NDA if he is worried about trade secrets. With whom? Levi et al? Since they're being called co-conspirators, what extra weight would derive from their saying I saw the sekrit data under an NDA.
[Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
The only possibility to fool the power-meter then is to raise the DC voltage on all the four lines This turns out not to be the case. You could also draw DC current through any of the lines, which current would not register on the clamps. The simplest way to do this would be just to use a diode in series with the heating element. Since power = current x voltage x pf, it is NOT necessary to change the voltage in order to change the power. Duncan On 5/26/2013 2:21 PM, Jed Rothwell wrote: A Swedish correspondent sent me this link: http://www.energikatalysatorn.se/forum/viewtopic.php?f=2t=560sid=5450c28dab532569dee72f88a43a56f0start=330 This is a discussion in Swedish, which Google does a good job translating. Before you translate it, you will see that in the middle of it is a message from one of the authors, Torbjörn Hartman, in English. Here it is, with a few typos corrected. QUOTE: Remember that there were not only three clamps to measure the current on three phases but also four connectors to measure the voltage on the three phases and the zero/ground line. The protective ground line was not used and laid curled up on the bench. The only possibility to fool the power-meter then is to raise the DC voltage on all the four lines but that also means that the current must have an other way to leave the system and I tried to find such hidden connections when we were there. The control box had no connections through the wood on the table. All cables in and out were accounted for. The E-cat was just lying on the metal frame that was only free-standing on the floor with no cables going to it. The little socket, where the mains cables from the wall connector where connected with the cables to the box and where we had the clamps, was screwed to the wood of the bench but there was no screws going through the metal sheet under the bench. The sheet showed no marks on it under the interesting parts (or elsewhere as I remember it). Of course, if the white little socket was rigged inside and the metal screws was long enough to go just through the wood, touching the metal sheet underneath, then the bench itself could lead current. I do not remember if I actually checked the bench frame for cables connected to it but I probably did. However, I have a close-up picture of the socket and it looks normal and the screws appear to be of normal size. I also have pictures of all the connectors going to the powermeter and of the frame on the floor. I took a picture every day of the connectors and cables to the powermeter in case anyone would tamper with them when we were out. I lifted the control box to check what was under it and when doing so I tried to measure the weight and it is muck lighter than a car battery. The box itself has a weight, of course, and what is in it can not be much. All these observations take away a number of ways to tamper with our measurements but there can still be things that we didn't think of and that is the reason why we only can claim indications of and not proof of anomalous heat production. We must have more control over the whole situation before we can talk about proof. Best regards, Torbjörn END QUOTE - Jed
Re: [Vo]:The inanity of the hidden input power hypothesis
They could make any measurements with any equipment they chose Jed, am I mistaken here? I thought it said in the paper that they were NOT permitted to measure the heater waveforms, because they were proprietary. Am I misinformed about this? Also my scenario is a simple diode. Hardly outlandish, you can buy them at Radio Shack... Duncan On 5/26/2013 4:59 PM, Jed Rothwell wrote: Duncan Cumming spacedr...@cumming.info mailto:spacedr...@cumming.info wrote: We would avoid the ridiculous position that we are in now, when some data is released (e.g. power output) but other data is withheld (e.g. power input), for who knows what reason? Nothing is being withheld. Levi et al. were given unimpeded access to the input power. They could make any measurements with any equipment they chose. If it can be shown that they made inadequate measurements with the wrong instrument, that is their fault, and their's alone. Rossi withheld nothing. For who knows what reason has no meaning. I repeat, there was no reason and nothing was withheld. Some skeptics have proposed highly unlikely scenarios that they imagine might be possible. This is the only reason for this discussion, and it has nothing to do with Rossi or Levi et al. Skeptics will continue to propose ever-more outlandish reasons to reject this until the day the scientific establishment admits it is real. At that moment, the skeptics will claim they believed it all along, and they will modestly take credit for it. - Jed
Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
I am not trying to assert anything as fact. I am merely pointing out that a simple diode inside the controller box (to which access was forbidden by Rossi) COULD HAVE given the observed results. I am NOT saying that it, in fact, did, merely speculating that it could have. For any scientific experiment, the onus is on the experimenters to produce the result. The best way to do this is to provide sufficient information for others to replicate the experiment. Duncan On 5/26/2013 5:07 PM, David Roberson wrote: Perhaps you should build one of these scam machines and prove that it will work without being detected. That would be the best way to show that it is possible. Why should we accept this assertion as fact any more than believing that the testers missed finding the scam? We can spend an equal amount of time knocking down any theory that is put forth as others can spend assuming they are real. Dave -Original Message- From: Duncan Cumming spacedr...@cumming.info To: vortex-l vortex-l@eskimo.com Sent: Sun, May 26, 2013 7:59 pm Subject: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments The only possibility to fool the power-meter then is to raise the DC voltage on all the four lines This turns out not to be the case. You could also draw DC current through any of the lines, which current would not register on the clamps. The simplest way to do this would be just to use a diode in series with the heating element. Since power = current x voltage x pf, it is NOT necessary to change the voltage in order to change the power. Duncan On 5/26/2013 2:21 PM, Jed Rothwell wrote: A Swedish correspondent sent me this link: http://www.energikatalysatorn.se/forum/viewtopic.php?f=2t=560sid=5450c28dab532569dee72f88a43a56f0start=330 This is a discussion in Swedish, which Google does a good job translating. Before you translate it, you will see that in the middle of it is a message from one of the authors, Torbjörn Hartman, in English. Here it is, with a few typos corrected. QUOTE: Remember that there were not only three clamps to measure the current on three phases but also four connectors to measure the voltage on the three phases and the zero/ground line. The protective ground line was not used and laid curled up on the bench. The only possibility to fool the power-meter then is to raise the DC voltage on all the four lines but that also means that the current must have an other way to leave the system and I tried to find such hidden connections when we were there. The control box had no connections through the wood on the table. All cables in and out were accounted for. The E-cat was just lying on the metal frame that was only free-standing on the floor with no cables going to it. The little socket, where the mains cables from the wall connector where connected with the cables to the box and where we had the clamps, was screwed to the wood of the bench but there was no screws going through the metal sheet under the bench. The sheet showed no marks on it under the interesting parts (or elsewhere as I remember it). Of course, if the white little socket was rigged inside and the metal screws was long enough to go just through the wood, touching the metal sheet underneath, then the bench itself could lead current. I do not remember if I actually checked the bench frame for cables connected to it but I probably did. However, I have a close-up picture of the socket and it looks normal and the screws appear to be of normal size. I also have pictures of all the connectors going to the powermeter and of the frame on the floor. I took a picture every day of the connectors and cables to the powermeter in case anyone would tamper with them when we were out. I lifted the control box to check what was under it and when doing so I tried to measure the weight and it is muck lighter than a car battery. The box itself has a weight, of course, and what is in it can not be much. All these observations take away a number of ways to tamper with our measurements but there can still be things that we didn't think of and that is the reason why we only can claim indications of and not proof of anomalous heat production. We must have more control over the whole situation before we can talk about proof. Best regards, Torbjörn END QUOTE - Jed
Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
No, it does not. What happens is that the diode rectifies the mains to DC, and the DC is not sensed by the clamp-type current meter. http://en.wikipedia.org/wiki/Current_clamp No tampering with the mains itself would be necessary if one were to use a diode. Duncan On 5/26/2013 5:34 PM, Eric Walker wrote: On Sun, May 26, 2013 at 5:18 PM, Duncan Cumming spacedr...@cumming.info mailto:spacedr...@cumming.info wrote: I am not trying to assert anything as fact. I am merely pointing out that a simple diode inside the controller box (to which access was forbidden by Rossi) COULD HAVE given the observed results. I am NOT saying that it, in fact, did, merely speculating that it could have. Am I right in understanding that this line of reasoning requires tampering with the mains itself, where the electrical measurements were made, in addition to any sly customizations that might have been made at the controller? Eric
Re: [Vo]:Re: [Vo]:Re: [Vo]:Re: [Vo]:Torbjörn Hartman describes power measurments
Almost. The power being fed to the heater exceeds that measured at the wall, because the sensor used (an AC current clamp) cannot sense the direct current being drawn from the wall socket. Some people find the difference between current and voltage confusing. What I am saying here is that if you connect a resistor in series with a diode to a wall socket, then the CURRENT drawn is direct even though the VOLTAGE at the socket is alternating. (Rossi does not seem to understand this concept judging by his message that got posted today). So unless you use a DC rated current meter (such as a shunt) you will not sense all of the current, and hence power, drawn from the wall socket. The electrical power meter in your house certainloy IS rated for DC, so you will certainly be BILLED for the power even though you didn't measure it yourself! V = IR Power = Voltage * Current * Power Factor Duncan On 5/26/2013 5:57 PM, Eric Walker wrote: I wrote: On Sun, May 26, 2013 at 5:18 PM, Duncan Cumming spacedr...@cumming.info mailto:spacedr...@cumming.info wrote: I am not trying to assert anything as fact. I am merely pointing out that a simple diode inside the controller box (to which access was forbidden by Rossi) COULD HAVE given the observed results. I am NOT saying that it, in fact, did, merely speculating that it could have. Am I right in understanding that this line of reasoning requires tampering with the mains itself, where the electrical measurements were made, in addition to any sly customizations that might have been made at the controller? I think I'm starting to understand. This is a separate line of reasoning to the one about the possibility of hidden DC and RF passing undetected through the clamp meters at the mains. In this line of reasoning, the duty cycle (35 percent ON) is misunderstood, and there is a hidden DC component from the controller delivering power to the E-Cat, but not above what was read from the wall -- am I describing this right? Eric
Re: [Vo]:The inanity of the hidden input power hypothesis
I myself am somewhat doubtful about the power measurements, and would like to consider the meter A / meter B issue. There is nothing at all mysterious about this. Meter A is a current clamp, incapable of detecting DC. Meter B is a current shunt or hall effect clamp, capable of detecting DC. The way to bamboozle meter A is a simple diode in series with the load, costing under a dollar. Hardly rocket science. There is, of course, a simple way to uncover such a fraud - just use an oscilloscope to measure the current waveform. It is much cheaper and easier to procure meter A than meter B, and also much easier to use. It is a pain to break the cables and insert current shunts, plus some power is wasted in the shunts. Also, you need a floating power supply and true differential amplifier to power the amplifiers after the shunts. All of this is possible, but a lot more difficult than a simple clamp ammeter. So Rossi would make a good guess that meter A (not DC capable) would be used for the test. Now for the argument that Rossi runs the risk that somebody will try a type B meter (DC capable), or, for that matter, a simple oscilloscope. He simply does not permit such things. He claims not to allow an oscilloscope because it would reveal a proprietary waveform. By keeping tight control over the test conditions, he is able to ensure that his questionable power measurements are not exposed. By not allowing inspection of the heater controller, he keeps the diode (or asymmetrical firing of the Triacs) from public view. Rossi behaves as if a mundane heater control is super-secret technology - does nobody else find this strange? As to the hypothesis that only a fool would give money to an inventor without independent testing, I can only agree. Duncan On 5/24/2013 6:27 PM, Jed Rothwell wrote: Several people have proposed that Rossi has secretly installed equipment in the wall circuit to deliver more electricity than the power meter shows. Common sense considerations show that this is so unlikely we can dismiss it. People should do a reality check. First, let us define the hypothesis, in general terms. You say there is a method of arranging electricity with hidden DC or something else that will fool a certain kind of power meter. Let us call it meter Type A. There must also be a meter of Type B that will detect this trick. You do not assert that it impossible to detect this power with any instrument on the market. That would be absurd. You are saying that Levi et al. brought the wrong kind of meter. Here are some problems with this hypothesis: Rossi did not know what kind of meter they intended to bring. He might have gone to a lot of trouble to fool Type A only to see them show up with Type B. His scheme would fall apart. Rossi does not know what kind of meter they will bring to the next test. They might show up with Type B, putting an end to his scheme a few weeks from now. Sooner or later, someone is bound to try Type B. Or they will try plugging it into another circuit. Despite all the blather to the contrary, it is a fact that Rossi has allowed several completely independent tests of his machines, in Italy and the U.S. He was not present. He wasn't even on the same continent. They plugged the machines into their own wall sockets. There is not the slightest chance anyone will give him a large sum or money without independent testing. I know some of the people who might give him money, and who have given him money. They are not fools. Perhaps you assert that Levi may have brought Type A because he is in cahoots with Rossi. The same set of conditions apply. Sooner or later someone will try power meter Type B and the scam will collapse instantly. Levi knows that. If he knows how to conspire to select the wrong kind of meter, he will also know the right kind, and he will know there is no chance of keeping this under wraps indefinitely, and no chance of cashing in on it. He knows that he will be caught sooner or later. This applies to all of the other far fetched notions about IR lasers and so on. I would also point out that despite all the noise from Krivit, neither he nor anyone else has caught Rossi cheating so far. They have caught him making stupid mistakes, with a plugged up reactor. Suppose Rossi had allowed me to come with my instruments. Or suppose that I had gone with Krivit and used Rossi's instruments. I would measured a few things, sparged the water, and I would have said, Andrea, this thing is not working. It is plugged up. That is exactly what happened to the people at NASA. It took them little time to figure this out. It would not have taken me much longer. I have spent several months making similar measurements. I may not know much, but I can tell when X liters per minute are going in but only a fraction of X is coming out, and I darn well would check for that. Anyone who has ever done flow calorimetry would. The
[Vo]:Why did Rossi prevent detailed measurement of the power input? : Magic Tricks!
I propose a simple design for a Rossi controller: A diode in series with the resistor. This will draw a small AC component at 60Hz, combined with a large DC component. So any kind of power meter using clamp on ammeters will register some power, but nowhere near the full power consumed. Simply short out the diode for the calibration run, and voila. Rossi does not need to know the exact type of power meter to be used in advance. I should point out that this design is not original. At Cambridge University Engineering Department, as undergraduates we were shown a baffling demo of two light bulbs in series. If you unscrewed one bulb, the other one still worked! It was built using thick copper wire on a plexiglass base, very baffling. It was done using concealed diodes in the light fixtures and also in the bulbs themselves. A 'scope would have given the game away immediately, but of course one didn't happen to be available. They liked to train us for real-world scenarios such as this one! Here is a light bulb illusion on YouTube, not quite as good (no plexiglass, unfortunately) but better showmanship and nice music: http://www.youtube.com/watch?v=KkfdN3QfQIY I should also point out that the guy on You Tube calls it a Light Bulb Trick, qualifying himself as an illusionist. Anybody claiming that the effect is real would, of course, be a charlatan. Challenge: Can anybody perform this trick using a plexiglass base, solid copper wire connections, and a glass table? Duncan On 5/24/2013 1:00 AM, Robert Lynn wrote: This has only just occurred to me, but in my mind is a bit of a red flag: The reactor vessel is a sealed metal container, no electrical or magnetic signal of any frequency will penetrate it (It is a faraday cage). And all of the resistive heating elements are positioned around it, so they do nothing but deliver heat to the reactor contents - no special magnetic or electrical excitation can pass through the reactor vessel. All of these configurational details were revealed to the testers by Rossi. So why did Rossi feel the need to prevent detailed analysis of the input power to these resistors that are no more than resistive heaters? We know he ran it in at least a partially pulsed 35% on 65% off mode with period of about 6 minutes from the thermography. So what possible harm could have come from allowing continuous measurement of voltage drop and current flow through the resistors? As such preventing that measurement serves no sensible purpose that I, or any other engineer/scientist could see, it is a pointless obfuscation. All it achieves is raising suspicion about just what electrical power is really flowing through those resistors.
Re: EXTERNAL: [Vo]:My evaluation of the Rossi test
Would you believe that it is possible to melt tungsten (MP 3422 deg C) using only 10 watts? Try putting 10 watts into a 1 watt flashlight bulb. It will burn out immediately as the tungsten filament melts. No LENR reaction, just straight resistive heating. Power and temperature are not directly related, the temperature also depends on a host of other variables. A 110V plug-in arc welder can produce a plasma temperature of well over 6,000 deg C for only 2kW of power. Duncan On 5/24/2013 11:38 AM, Axil Axil wrote: The temperature difference between the melting point of stainless steel and ceramic is 600 degrees C. To produce this temperature difference beyond the melting point of nickel powder and stainless steel requires a continuing LENR reaction, IMHO.
Re: [Vo]:Levi Hot Cat paper is a gem
Original Message Subject:Fwd: Re: [Vo]:Levi Hot Cat paper is a gem Date: Thu, 23 May 2013 10:20:27 -0700 From: Duncan Cumming spacedr...@cumming.info To: vortex-l-requ...@eskimo.com I am acting as devils advocate here for a minute. Had the demo been intentionally faked, there are a lot of much easier ways to do it than re-wiring the building! Power measurement was done using a wide band 3 phase power meter, a notoriously difficult instrument to use. A slight slackening of one of the current sensing clamps, a particle of grit (or Scotch tape) on the clamp face, or mis-threading of the cables through the clamps would give lower than actual power readings. A controller could easily be designed to bamboozle such a power meter, by exceeding either the shape factor or the bandwidth spec of the power meter. No measurements were made of the current waveform, which measurements would have immediately exposed such chicanery. In short, the power measurement could have been fiddled very easily. Now I am not saying that it was, merely that it would have been easy to do so. The way to avoid such problems in the future would be simply to use DC to power the heaters. Or have the reactor tube tested at somebody else's facility, with a manufacturer's rep present to ensure that nobody saws the tube in half. Or to use an ordinary tube furnace with cooling coils for a self sustaining test. In other words, if the manufacturer really wanted to test the reactor properly, they could - easily. Duncan Original Message Subject:Re: [Vo]:Levi Hot Cat paper is a gem Resent-Date:Thu, 23 May 2013 09:01:42 -0700 Resent-From:vortex-l@eskimo.com Date: Thu, 23 May 2013 08:59:25 -0700 (PDT) From: Alan Fletcher a...@well.com Reply-To: vortex-l@eskimo.com To: vortex-l@eskimo.com From: Eric Walkereric.wal...@gmail.com Sent: Wednesday, May 22, 2013 11:00:43 PM Alan (or someone) made the point that everything, laptop and all, were plugged into the same power supply. Would hidden DC or AC above or below the range of the meter hurt the laptop? That was me -- and only a couple of things were plugged into the same socket -- the meter and a camera. The laptops were further over on a separate plug. And of course, since the whole building was wired for the power-input fake, just that ONE socket for the controller would have been rigged, set up before the test team arrived. (Certainly for the December test -- they said it was already running.)
[Vo]:Levi Hot Cat paper is a gem:cooling coils
Hi, guys! This is my first post, Andrew invited me in to the list. I read the Levi et al paper, and I noticed that they measured an input power of around 400 watts and an output power of around 2000 watts. But (for control simplicity reasons) they had no insulation on their tube furnace, and had to keep input power connected so as to keep the reaction going. Why not use an ordinary, properly insulated electric tube furnace fitted with cooling coils on the inside? That way, the input power could be disconnected entirely once the reactor starts to run, and the temperature controlled by the coolant flow. A run of 96 hours producing 2000 watts with no input power would be a lot more convincing to many people. The reason is that, even if the power measurement is inaccurate, the fact that the machine is self sustaining means that some power must be being produced. Probably compressed air in stainless steel coils would be a suitable coolant for this application. I think that a self sustaining reactor running for a few days at a few kW would go a long way towards improving the credibility of this particular device. There is no need to generate electricity, just thermal output without electrical input, only using electrical energy to start the reactor. Duncan Cumming Original Message Subject:Re: [Vo]:Levi Hot Cat paper is a gem : power conditioner needed Resent-Date:Wed, 22 May 2013 13:32:49 -0700 (PDT) Resent-From:vortex-l@eskimo.com Date: Wed, 22 May 2013 13:33:57 -0700 From: Andrew andrew...@att.net Reply-To: vortex-l@eskimo.com To: vortex-l@eskimo.com I doubt that Rossi would allow a power conditioner, because he himself states that there is some snip Andrew
