Demonstration of the E-Cat QX - 24 November - Summary thread

    • Official Post
    Quote from Eric Walker

    One detail to add: If Rossi was a normal person pursuing normal investment opportunities, he would try to conduct a demo that was not readily open to such basic critiques. The demo might not convince skeptics, but it would be good enough to elicit from normal, intelligent people the question of "what should be done as a next step to look into this device's performance?" The unavoidable conclusion is that the demo did not clear this basic hurdle, but not for lack of opportunity. (See Alan F.'s response above.)


    To the inevitable reply about protecting IP, sorry, this retort will be persuasive only to those who are already persuaded. But one can mention it anyway, and throw in the word "pathetic" as well, if it sounds good to one's ears.


    Eric,


    Well said, but (so far) our rather disappointing, and very subjective *reporter*, claims there were investors lined up waiting for a chance with the maestro.

    Just another issue regarding the miraculous PSU (which needs 60W cooling for the supply of less than 0.1W):


    Didn’t Rossi use a standard PSU previously, when the QuarkX was “tested” at his place (without a “dummy”-run), as suggested by this picture? - A PSU without any “magic” for absorbing electrons or whatever phantastic particles/wave allegedly created by the QuarkX.


    Quote from Shane D.

    So it is legal to scam a VC, if they were stupid enough to be scammed?

    During the subprime mortgage crisis, 10 trillion dollars were skimmed from the economies of the world. Such skimming was legal and nobody went to jail. Was it legal to sell mortgage-backed securities (MBS), collateralized mortgage obligations (CMO), and collateralized debt obligations (CDO). This bug in the capitalist system was not illegal. It was how the people in the know gamed the system.


    Today the regulations that were put in place to fix this bug in the system are being removed so that the system can be gamed again. What are you doing about it? Wasting your scam prevention efforts on Rossi?

    For the first dummy (conductor in place of Qx) the supply voltage V is measured directly from the 1 ohm resistor:


    V = 0.4V


    For the second dummy (800 ohms in place of Qx, in series with 1 ohm)


    V = I * (r1 + r2) = 0.02 * (800 + 1 ) = 16.02V


    For the Qx ? Who knows.

    https://animpossibleinvention.…cat-qx-demo-in-stockholm/


    The 300mv (0.3V) is measured across the 1 ohm .. ergo 0.3 amps.
    Hurley doesn't report on it's being chopped or sinusoidal or compound or anything.


    Andrea Rossi

    November 27, 2017 at 3:32 AM

    Damian:

    Yes, Mats Lewan has retrieved the water pump and the wooden board with the connectors for the 1 Ohm and 800 Ohms resistances at the end of the test and he can do with them whatever he wants, I do not need to have them back.

    Warm Regards,

    A.R.

    Hmmm ... looks like a Gamma L from the back ... is that pump off the 1MW Doral system? I can think of a use for it!

    3783-stockholm-demo-028-jpg

    Quote from Alan Fletcher

    The 300mv (0.3V) is measured across the 1 ohm .. ergo 0.3 amps.


    Mats Lewan verified that the QuarkX in operation has a negligible resistance by using a conductor in its place in one of the dummy tests:


    https://animpossibleinvention.…cat-qx-demo-in-stockholm/


    Quote from Mats Lewan

    Using a conductor as a dummy, the voltage across the 1-ohm resistance was about 0.4V, thus similar as with the reactor in the circuit. With the 800-ohm resistance, the voltage across the 1-ohm resistance was about 0.02V and the current thus about 0.02A. The power consumption of the 800-ohm resistance was then 0.02 x 0.02 x 800 ≈ 0.3W, thus much lower than the thermal power released by the reactor.


    .


    Quote from Alan Fletcher

    Hurley doesn't report on it's being chopped or sinusoidal or compound or anything.


    This was the typical waveform when the QX was operating. The scale of the voltage on the oscilloscope was set to 100 mV / subdivisions; here it's showing +/- 4 subdivisions, thus +/- 400 mV:


    3761-pasted-from-clipboard-png

    Quote

    Of course, across the 1 ohm resistor,

    Assuming the network is resistive only, the exact waveform doesn't matter. Apply whatever scaling factor turns you on ... sqrt(2), 1/sqrt(2) ... Ohms law still applies to both resistors. If it's NOT resistive then Rossi's concealing even more information.

    In Wolfgang Pauli's words : Not even wrong.


    Pump : Not a gamma-L, so no use to me.

    Putting the 800 ohm resistor in series seriously restricted the current flow, and the voltage was reported to 11.52 V across the entire system. Strangely, the 1 ohm resistor was reported to have about the same voltage drop as before. (This is what I heard in the video, but I see differently posted above). If we try and use the current from before, 0.09 A, the 800 ohm resistor would have to dissipate ~6.5 W, which is not possible for the physical size used. It would burn, and rapidly.


    If there is an AC component to the voltage, or rapidly fluctuating DC voltage, (or both) then the DC readings from a cheap voltmeter will be next to useless. The real voltage will be all over the place, and a cheap meter will display what it can, and there is zero guarantee that it will be representative. The 800 ohm resistor probably would introduce a lot of noise into the AM signal.

    Paradigmnoia

    As far as I understood:


    1 Ohm + QX = 300 mV drop (peak)

    1 Ohm + conductor = about the same as with the QX (300-400 mV drop)

    1 Ohm + 800 Ohm resistor = 20 mV drop


    Lewan and Fabiani said in the video after [2:43:30]:


    [Lewan] So.. whereas during experiment we've had a voltage about 300 mV [over the 1 ohm resistor], now we got about… let's say... 10 or 20?

    [Fabiani] 15…10… but doubled is 20-25.

    [Lewan] OK. I would have expected it to be divided by 800, but...

    [Fabiani] Depends on the feedback of the tones electronics.

    [Lewan] OK.

    [Fabiani] Not feel the charge, not full (?) no give answer.

    [Lewan] OK.

    [Fabiani] It's not "simple".

    Edited once, last by can: Corrected word that I previously got wrong. ().

    Quote from interested observer

    You asked why Rossi would do a demonstration if the e-cat doesn't work. The implication of your question is that if the e-cat didn't work, he would not do a demonstration. That is patently false IF he is a scammer or

    Your brain is addled. He couldn't operate a scam that way. It simply wouldn't work. There are enough warnings about Rossi that the first thing a potential investor would do would be the SIMPLE checks that are all that are needed to see if it works.


    I don't know where you get the idea the dumb rich would sit through a 3 hour presentation that would be very boring to a non tech. . The people there were technically knowledgeable and there for a reason.

    Quote from Roseland67

    Alan,


    Whatever the waveform was,

    Shouldn’t The meter have the capability of measuring and calculating the area under the voltage and current waves doing the integration and coming up with power used?

    Yes ... but if the network is resistive, Ohms-law ratios apply. The two resistors will see the exact same waveform at the exact same time. Muddle that with "feedback of the tones" and you've lost me.

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