New E-Cat QX Picture and New Rossi-Gullstrom Paper (Very high COP reported with Calorimetry)
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Seems like you both see what you want to see. A bit like when the disussion was about frames or no frames in the window. In my opinon it is not possible to see in the picture if the voltage is measured across the resistor or reactor. The green connector seems to be connected to both the resistor and the reactor. And the red connector disapear behind the box. Can any of you explain how you come to your conclusion?
It is a bit hard to see in the photo, but it is more or less visible. The brown resistor is connected to the dual female wire. In the photo, it is not connected to either voltmeter. Zoom in on the voltmeter connections, and you can see that both volt meters are measuring the voltage drop across the reactor.
QuoteTo me it really doesn't matter what can be seen in the picture or not. It is clearly described in the report Page 18:
"In the left in the figure there is two voltmeters that measure the mV of the current passing through the 1 Ohm brown resistance."This quote from the paper is consistent with what I am trying to explain to you. When the quote says "that measure" it doesn't mean presently, i.e., as in this very moment. The two voltmeters are shown in the photo that measure [not presently] the mV of the current passing through the 1 Ohm brown resistance [because then you know the current through the reactor].
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That is consistent with what I wrote above. Please read carefully.
It is both. The photo shows both voltmeters measuring the voltage across the resistor. At some point in time (not shown in photo), at least one of the voltmeters can be used to measure the voltage across the brown resistor to determine the current through the reactor, in order to sanity check the power calculation of the reactor. It is that simple.
IHFB - do you note the line in the paper V=0.1, R = 1 P=0.01?
(1) Do you agree that from this Rossi concludes the power in to the plasma generator is 10mW (absurd BTW)
(2) Do you agree that the only way you can get P = 0.01 from these figures is V*V/R where V = 0.1?
(3) You may argue that there are two voltages: V1 = 100mV across the brown resistor giving I = V/R = 100mA, and V2 = 100mV across the plasma generator. What do you think are the chances of that happening - as a matter of electrical circuits the same current through the resistor and the plasma generator means no relationship between the voltages - you would need the plasma generator to have an impedance when on of exactly 1 ohm? Also - if it really were that surely Rossi would make it clear e.g. V1=100mV, V2=100mV.
I'm quite sory about this - you have a position that is totally untenable - which you stand on as far as I can see through lack of technical knowledge and/or a surfeit of determination to see this matter only one way (most people other than you call this bias).
I think it is overreaching for you to criticise others on this matter till you understand it yourself.
THH
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(3) You may argue that there are two voltages: V1 = 100mV across the brown resistor giving I = V/R = 100mA, and V2 = 100mV across the plasma generator. What do you think are the chances of that happening - as a matter of electrical circuits the same current through the resistor and the plasma generator means no relationship between the voltages - you would need the plasma generator to have an impedance when on of exactly 1 ohm? Also - if it really were that surely Rossi would make it clear e.g. V1=100mV, V2=100mV.
Yes, 3 is correct. What you call overreaching, I call plain. And Rossi has confirmed it. He matched the resistance (what he called impedance) of the reactor to that of the brown resistor.
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For clarity here is the ceramic resistor sharing a common connection with the negative terminal of both meters. As DNI pointed out the positive (red) wire disappears behind the box somewhere, but is consistent with the papers diagram and description of measurment. The idea of 100mV being across the reactor is just not possible if the reactor is relying on an arc.
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Yes, 3 is correct. What you call overreaching, I call plain. And Rossi has confirmed it. He matched the resistance (what he called impedance) of the reactor to that of the brown resistor.
IHFB - that is absurd. He is not measuring the plasma generator impedance - and remember it is undergoing never before seen LENR activity - so there is no stable theory. How can he match an unknown and probably changing impedance? And, with respect, if you read the paper he does not say that - he says the opposite.
Quite apart from the fact that I don't know if you realise but you do not get plasma in that size of tube from 100mV. But you maybe do get it from 200V...
You are trolling here. It is beneath you. I hope.
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IHFB - that is absurd. He is not measuring the plasma generator impedance - and remember it is undergoing never before seen LENR activity - so there is no stable theory. How can he match an unknown and probably changing impedance? And, with respect, if you read the paper he does not say that - he says the opposite.
Quite apart from the fact that I don't know if you realise but you do not get plasma in that size of tube from 100mV. But you maybe do get it from 200V...
You are trolling here. It is beneath you. I hope.
What I stated all makes complete sense. You are offended by it because of its plainness, and because it threatens your world view.
As for the plasma discussion (and I doubt this is your specialty), I don't know how the reaction is triggered. But I suppose the plasma could be maintained with a fairly low voltage.
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What I stated all makes complete sense. You are offended by it because of its plainness, and because it threatens your world view.
As for the plasma discussion (and I doubt this is your specialty), I don't know how the reaction is triggered. But I suppose the plasma could be maintained with a fairly low voltage.
OK - I'll give this one up. But not for the reason you think! Does anyone else here understand IHFB's thought processes on this one?
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Sorry, IHFB seems to have lost it here. I've just read the paper again and just cannot believe what I'm seeing, calculated input power is just plain wrong. Perhaps Gullstrom is smart in his field but hopeless on a practical level and hasn't studied basic electrical theory.
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No, no way to align this with what I have learned in my career about solid state physics and mathematics ...
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Frank Acland ECW Admin
Andrea Rossi explained this to me in an interview I had with him yesterday. I am still transcribing it, but here is an excerpt, since it is relevant to the discussion here:
"We have measured only that resistance [the 1 Ohm resistor] because that is the only resistance we have in the circuit. If the E-Cat has a resistance, that makes our calculations more conservative, because, as you well know, the resistance goes in the denominator when you make the calculus of the amps. You have volts as the numerator, and the resistance as the denominator. So the bigger the resistance, the smaller is the amount of amps.
"To be conservative, since the datum of the resistance of the E-Cat QX is confidential, we just do not consider the resistance. Because correctly we should have to make the sum of the resistance of the resistor that has been put in the circuit, and the resistance of the E-Cat. So we should have amps = volts/R1 (the resistor)+R2 (the resistance of the E-Cat). But we do not consider the resistance of the E-Cat, we consider it as if it is a perfect conductor, and we only consider the one 1 Ohm to make the calculation of the amps."
Frank Acland ECW Admin •
AR also told me that "we use two voltmeters to make a double check. The difference of the measurement is the margin of error of two different voltmeters ( several mV )" -
Sorry, IHFB seems to have lost it here. I've just read the paper again and just cannot believe what I'm seeing, calculated input power is just plain wrong. Perhaps Gullstrom is smart in his field but hopeless on a practical level and hasn't studied basic electrical theory.
Please be more specific.
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Frank Acland ECW Admin
... and we only consider the one 1 Ohm to make the calculation of the amps."As I predicted.
QuoteAR also told me that "we use two voltmeters to make a double check. The difference of the measurement is the margin of error of two different voltmeters ( several mV )"
And as I predicted.
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It seems that the power going to the reactor is very small indeed. Hence the high COP
Perhaps the critics can now move on to explain why the calorimetry must be wrong, before what is happening is actually explained enough to understand it.
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I'd like to revise my understanding of the circuit. While I still am of the opinion that the ceramic resistor is being used to determine the current through the reactor, and that the dual voltmeters are being used in a redundant configuration, I now think that this explanation better represents the configuration (from ECW):
Thomas Kaminski GiveADogABone • an hour ago
Actually, the way I interpret Frank's comment from Rossi is that the resistance of the QuarkX is in series with the 1-ohm resistor, but they are measuring the voltage across the combined load. That would drop the power estimate, because any additional resistance will lead to a smaller current. The voltage is the voltage across both.
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"If the E-Cat has a resistance, that makes our calculations more conservative
Wrong wrong wrong!!!
Its just the opposite. Here he is admitting they don't know the resistance, worst apparently it doesn't matter.
Rossi calculates input power using his 1R resistor and 0.1V a current of 0.1A. This gives 0.01W
However suppose the reactor had an impedance of 1K. Here we have given 0.1A a voltage of 100V and reactor power of 10W
Wow a COP of 1000 !!!!
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Quote
maryyugo wrote:
I have no idea what a decentralized Augur prediction market platform is.QuoteDisplay MoreThe Augur API is a set of JavaScript bindings for the methods encoded in Augur’s smart contracts. The API method name, as well as its parameters, are generally identical to those of the underlying smart contract method.
Augur’s core contracts exist on Ethereum’s decentralized network. The various serialization, networking, and formatting tasks required to communicate with the Augur contracts from a web application are carried out by Augur’s middleware.
augur.js is the Augur JavaScript SDK, and is the user-facing component of the middleware. If you want to interact with the Augur contracts from a custom application, augur.js is the recommended way to do so. The easiest way to install augur.js is using npm:
$ npm install augur.jsTo use the Augur API, augur.js must connect to an Ethereum node, which can be either remote (hosted) or local. To specify the connection endpoint, pass your RPC connection info to
augur.connect.augur.connectalso accepts a second argument specifying the path to geth’sOh, thanks! THAT explains it.
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He matched the resistance (what he called impedance) of the reactor to that of the brown resistor.
Wow! Another feat of genius for the Maestro of Bodena. He created nickel62 cheaply in his lab, he sold many megawatt plants, he charmed the military, he built robotic factories, and now he conquers plasmas with a brown resistor and a pair of $5 Chinese DVM's! Nobel material.
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@MY,
It will be easy to use. You don't need to learn the developer side of it.
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I note that Rossi does not believe in running a dummy to check that his setup is working properly.
Sorry but:
NO CONTROL = NO RESULT
