A Real Classic

From: If I told U I'd have 2 kill U (knews4u2chew_at_yahoo.com)
Date: 07/20/04 

Date: 20 Jul 2004 12:28:54 -0700

From:
http://groups.google.com/groups?q=%22flyback+transformer%22+group:sci.energy.*&hl=en&lr=&ie=UTF-8&group=sci.energy.*&selm=341E7EAA.6EC%40mailroom.worldnet.att.net&rnum=5

Bill Ward wrote:
>
> Michael Hannon <ohannon@mailroom.worldnet.att.net> wrote:
>
> <snip prior posts>
>
> >Uh, JW - Bill was lying his pants off about the (Meyer) circuitry >functioning.
>
> >That circuit I posted DOES NOT OUTPUT LOW VOLTS AT HIGH AMPERAGE, as
> >Bill Ward tried to bamboozle everyone into thinking, and as he
> >apparently suckered you into believing. He is using misinformation in
> >his descriptions, and is in no way acting as a "neutral party."
> >A flyback transformer doesn't output low voltage - it's used to provide
> >the high voltage to the picture tubes of TV sets, at very low amperages.
> >You might want to note as well that the Meyer cell electrical power
> >curcuitry HAS TO CONSUME SIMILAR AMPS AT LOW VOLTAGE in order to provide
> >those very low amps at high voltage. Read the circuit diagram at the
> >site - those 10 amps aren't in the output to the cell.
>
> OK, follow me through the DC path on the schematic: Start at the
> negative (-) terminal of the 25A bridge near the power transformer.
> Trace right, then down through the 20A f.s. current meter, down through
> the unlabeled diode, into the emitter of the 2N3055, out the collector
> of the 2N3055, right through diode D1, and up through L1. While you're
> there, make note of the fact that L1 is NOT A FLYBACK TRANSFORMER, but a
> simple inductor wound from #14 (I think - hard to read) wire. From L1,
> continue up THROUGH THE CELL, up, left, up again and right to the
> positive side (+) of the bridge.
>
> That's a complete circuit. The 10 amps the meter reads is the average DC
> current in that loop, which includes the cell. The only side path is the
> base current of the 3055 and power for the 555 and drivers - probably
> less than an amp total. The rest goes through the water in the cell,
> electrolysing it.

What happens when the rectifier goes "off" on its 50% duty cycle, and
C#
and L1 are left "holding the bag?" Do they just sit there, or does the
counter EMF built up in L1 fire backwards into C#, charging it to a
higher voltage (CEMF usually comes backing out of a flyback coil at
multiples of the voltage that charged it, having converted the current
used to charge it into EMF when the field collapses) (aren't most
flyback transformers, and ignition coils as well, autotransformers? -
so
does it really make much of a difference if L1 is a flyback
transformer
or a flyback coil in this case? No, it doesn't). The "current" from
the
3055 never gets to pass across the water, because THERE'S NO
ELECTROLYTE
IN IT, and at most, a fragmentary current is allowed through, but
completing the tank circuit between L1 and C#, thus allowing that back
EMF to build in C#, which in turn, wants to dump itself back into L1,
and pumped back into the circuit at en even higher voltage, whose
current is limited by the resistance of the cell. So, what do you get?
Lots of built-up CEMF volts, and low amps, which reach a final level
where the water breaks down under separation, lowering the resistence
of
the cell, and dumpingthe accumulated charge C# , for the next cycle.
An
oscillatory circuit such as this (thank you, George Moonhie for
refreshing my memory about this) does not consume what is built up in
energy as it oscillates, except for small losses, so that it will keep
oscillating until the proper breakdown voltage, at low current levels,
is reached -

voila, water separation from high voltage and low current. The 10 amps
current can't pass without the voltage, BECAUSE THERE IS NO
ELECTROLYTE
IN THE WATER, and the total input power to the circuit cannot be
exceeded by the output, so, if the CEMF back-voltage builds up in the
LC
resonant circuit of L1 and C#, the energy stored as a magnetic field
by
CURRENT INTO L1 gets output from it as voltage when the field
collapses.

> >28 V @ 10 amps is 280 watts, and at at a half-wave 50% duty cycle is 140
> >watts. At 10,000 volts, the amperage required to yield the same power is
> >14 ma, considering NO power losses in the circuitry, which never happens
> >- there are always losses to take into account. That makes the
> >experiment described well within the limits of experimental validity,
> >especially since there was NO current-limiting done, and absolutely no
> >tuning of the circuitry, as in the Meyer circuit.
> >As a matter of fact, that experiment shows how EASY it is to get a
> >reasonable facsimile of Meyer's results, even in Sri Lanka - this is
> >without question a green flag for anyone interested in replicating
> >Meyer's claims to reasonable satisfaction.
>
> The whole point is that the described experiment did NOT replicate or
> confirm Meyer's claims. Try reading it carefully before speculating
> about what it says, Michael.
>

Bull***, Mr. Ward
There's no other way to put it.
No electrolyte, no electrolysis,
especially at the 10 amps
you're trying to pass off here
as being conducted by water
with no electrolyte in it,
other than trace impurities,
which will DEFINITELY allow
the resonant voltage-pumping tank circuit
of the flyback coil, D5, and C#
to built up some nice high voltages
across the water cell,
until it starts to conduct as the water
breaks down into gases,
just as described by Stan Meyer..

Do you seriously think that this person,
who looks like he knew enough to
throw this circuit together,
discusses high voltage out of naivete'?
He obviously knows more about this than you do,
and so does Stan Meyer..

Regards,
OHannon

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