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
