I am not exactly sure of where you are going with this discussion. When I use the term RMS source voltage, I am referring to the RMS value of the source itself at its fundamental frequency which is the only drive signal present. DC voltage is not seen nor present at this point according to the written information supplied by one of the testers. The input from the socket is a sine wave during the test and was looked at visually by the same guy, so it does not make sense to consider DC voltage present.
Now, go back and recalculate whatever you have in mind with the requirement that the source is a sine wave at 50 hertz and nothing more. Then, if questions still exist, I can clear them up for you. With a pure sine wave as an input voltage source, the input power measured, delivered and calculated does not depend upon anything except the fundamental component of the current passing through it. The important current frequency will be the same as the source, in this case 50 hertz. It makes no difference how much DC or other harmonic currents are flowing through the source due to rectification by the load. Notice that this anticipated problem is due only to rectification by the load, not hidden at the source. DC supplies hidden at the source in the wall or other rooms would indicate a scam, which would not be covered by my proof. This issue is dead and can not be used to sneak power into the test system. The guy that hypothesized this problem did not understand power delivery by sources. I proved him wrong with his hypothesis and he refused to acknowledge it even though my spice program run matched his replication. This proved to me that pseudo skeptics are not willing to admit that they harbor wrong ideas. He threatened me with a copyright notice which was loony. I made the run first then he copied me. His pride must have gotten in the way of his honor. I suspect that he was under the false impression that I was wrong and it would be easy to show that fact. The shoe ended up on the other foot. If you agree to accept that the only power source available is the sine wave voltage from the wall sockets then what I have said is absolutely correct. Let's drop any reference to a value of DC voltage appearing on the input voltage source for this discussion. Begin there and you should get the correct answer. Dave -----Original Message----- From: Berke Durak <[email protected]> To: vortex-l <[email protected]> Sent: Thu, Jun 6, 2013 12:08 am Subject: [Vo]:AC/DC, power, etc. David Roberson wrote: > I have not seen an indication that that power meter senses DC directly. The DC that flows into of from the source supply does not need to be sensed in order to calculate the power being delivered from that source. I realize that this seems contrary to common sense, but there is mathematical support as well as spice model demonstration of this behavior. > I can directly measure all of the power being given to the series diode and load resistor by the AC sine wave source by multiplying the RMS source voltage times the RMS fundamental current magnitude and taking into account the phase shift between them. All other harmonics and DC make no difference to the determination. When you say RMS source voltage, that sounds like you're including DC. It's a bit confusing. I'm talking of the ability of the power meter to sense DC voltages. We agree that the current probes used don't provide DC current, however AFAIK it is perfectly possible that the power meter can read DC ***VOLTAGES***. Consider two circuits connected by a pair of wires. Assuming circuits do not accumulate charge nor radiate, whatever current goes in must eventually go out, therefore it is sufficient to specify the instantaneous current I(t) in one wire. If we take one of the wires as a voltage reference then let U(t) be the instantaneous voltage difference between the two. The instantaneous power exchanged between the two circuits circuits is then P(t) = U(t) * I(t). Assuming again that the system is stationary, each quantity has a DC component and an AC component: U(t) = U_DC + U_AC(t) and I(t) = I_DC + I_AC(t). It then follows that P(t) = (U_DC + U_AC(t))*(I_DC + I_AC(t)) = U_DC * I_DC + U_DC * I_AC(t) + U_AC(t) * I_DC + U_AC(t) * I_AC(t) We have a power meter that measures voltage and current separately to calculate instantaneous power. If it cannot measure DC currents NOR voltages, the meter will only be using U_AC(t) and I_AC(t) and the estimated power will be P_est_1(t) = U_AC(t) * I_AC(t) and the error will be P(t) - P_est_1(t) = U_DC * I_DC + U_DC * I_AC(t) + I_DC * U_AC(t). If the power meter can measure DC voltages but not currents, the estimated power will be P_est_2(t) = U_DC * I_AC(t) + U_AC(t) * I_AC(t) and the error will be P(t) - P_est_2(t) = U_DC * I_DC + I_DC * U_AC(t). If the power meter cannot see DC voltages then a DC voltage can be injected between the two wires; this will produce a DC current which won't be seen either. For example if we assume the coils are purely resistive then a power component P_DC = U_DC^2 / R will be provided to the e-Cat but invisible. If the power meter can see DC voltages, then the goal of the scammer is to make P(t) bigger than P_est_2(t) that is to maximize I_DC*(U_DC + U_AC(t)). U_AC(t) is fixed by the mains which can be assumed to be a perfect voltage source. Hence a DC current must flow, which requires a DC voltage. But in addition to power, U_DC is also observed. Any significant offset voltage would be detected. So we need a fake-Cat that has sufficiently low DC offset to not cause alarm. As the voltage amplitude is on the order of 300 volts, a few volts could be written off as measurement error or leakage. Let's say up to 10 V DC could be injected without raising suspicions. If we need to produce 2-3 kW, this means by R_DC=U^2/P that R_DC=33 mohm. In that case the current will be 300 amperes. I don't think that's feasible given that the wires would dissipate a lot of heat and be caught on the thermal cameras — unless they are room temperature superconductors (maybe that's Rossi's next scam?). So Rossi can't cheat if DC *voltages* can be measured. Now I haven't double-checked the above, so consider this preliminary, but it's very first principles and I'm sure any mistake I made will be caught by the official or unofficial EEs lurking here. But it seems that: * If the power analyzer is blind to DC *voltages*, then DC voltages can be injected but the scammer will have to watch the experimenters 24/7 to make sure that the phantom DC is switched off if someone connects something else... like a $20 multimeter. And high DC voltages are much more dangerous than AC voltages. * If the power analyzer is *not* blind to DC *voltages* then cheating seems to be impossible because very high currents would be needed, causing wire heating that would be caught by the thermal cameras, strong magnetic fields on wires and other poltergeist effects. That's why we need to know if the power analyzer senses DC voltages. -- Berke Durak
