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## Re: [jlnlabs] Don Smith devices www.altenergy-pro.com

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• ... This is what needs to be investigated further. From what I fathom I completely agree, the true power input is only the I squared R heating loss of the
Message 1 of 1 , Jul 7, 2005
--- "vcrepair@..." <vcrepair@...> wrote:

> Are you familiar with basic electrical theory and
> Faraday's law, Lenz law eddy currents, etc?
>
> VOLTS * AMPS = WATTS
>
> Volts or Amperes of current by themselves is NOT
> power. Battery can have VOLTAGE, but no power output
> if nothing is connected to it. With super conductor
> you have current with no power loss. BUT even with
> regular conductors, if well designed, there is
> little
> power loss from I^2R or Current * Current *
> Resistance
> = Power. A good wire doesn't "loose" or dissipate
> much
> power. Most of the current and voltage goes through
> the wire.
>
> In brief, if you get current from a generator it
> creates an opposing force to the changing magnetic
> field. That requires more mechanical input power to
> the generator.
This is what needs to be investigated further. From
what I fathom I completely agree, the "true" power
input is only the I squared R heating loss of the
wire. However if the coiled wire is resonant to the
frequency of the generator, and has a coresponding C
value in series with the L value, an energy transfer
between magnetic and electric fields can exist. This
can be computed into joules/sec and compared to the
heat loss wattage. In the case for the single L value
as a reactive state; whereby the current lags behind
the voltage according to the phase angle of the
inductor, the amount of energy storage is .5LI^2. This
energy storage is said to exist as borrowed and
returned energy to the generator. Thus "How Much"
mechanical drag will the generator have when part of
this energy transfer from the generator consists of
that "borrowed and returned" energy transfer? The net
effect of then removing this effect of obtaining the
energy transfer of magnetic/electric field oscillation
itself from the generator is that remarkably the
amount of energy transfer occuring in its reactive
state is enhanced q times, where the theoretical q
factor is simply X(L)/R: but more importantly it is
now an oscillation of energy "between" mangnetic AND
electric fields, rather then an oscillation of
borrowed and returned energy with the generator. As
such the load demand as that removal of power source
would dictate should have gone down at the mechanical
input end. In construction of a maximum power transfer
resonance at 480 hz, it seemed remarkable also that
the alternator could output ten amps per phase, a
total of 30 amps, without any appreciable stator core
heating. What seemes bizaare is that these circuits,
having a q of 5, or 8.5 between phases; is that one
can take a 3/8 inch sample of ferrite between these
phases of 5 fold voltage rise, and it will quickly
glow hot within minutes, so much so that a 900 degree
F temperature is achieved. In this situation the
acting resistance of the ferrite has been changed from
20-30,000 ohms to 7 ohms where the 21 volts difference
between phases enables 3 amp of current to be drawn
through the ferrite. The ferrite glow is accomplished
with 63 watts of real observable heat power: BUT
according to phase angle theory the outside supply AC
circuits that can enable these ferrite rectifications
between them must have had its phase angle changed
since it now consumes much less current then it did in
the series resonant case; and essentially then the
load between these phases has driven the circuit into
a power factor correction direction. If we then take
this into consideration, and use the phase angle
method to compute the true power input for the ferrite
heating process, we find that the output exceeds the
"true" power input when those laws are used. I then
think that for this particular example those laws do
not apply, and the apparent power input as VI must
also be considered the true power input. Otherwise we
have a true paradox. The only way all these things
could be known for sure is to measure the mechanical
input power for all these cases.

These circuits I used were of 7 ohms impedance @ 480
hz, which means the load resistance was matched to the
supply lines impedance. The circuits having a q of 5,
have a 5 fold increase of both current and internal
voltage rise within the circuit to accomplish this.
Those outside circuits only behave that way if their
is no load betweeen the phases, and If a trisectional
wye is employed between the phases as a short between
the delta phases voltage rise; this topologically
changes the entire circuit from three phases of series
resonances procurred in Delta to three tank circuits
procurred in WYE, with each of these tank circuits
having shared internal pathways beween phases. In this
situation then 5 times more current exists in the
loops as what the stator lines input.(The resonant
rise of amperage in a tank circuit as governed by q
factor) By using a load that is impedance matched to
the supply lines it also seems practical to note that
if the ferrite heating process is shorted, it only
draws a current slightly higher that what the heating
effect will deliver. Another paradox exists with
regard to the true power input to the tank circuit.
Evidence suggests that the current is 180 out of phase
to the impressed voltage in this case. The phase angle
laws of considering the "instanteous" voltage and
amperage as the true power input seems inadequate for
this situation.
HDN

Tesla Research Group; Pioneering the Applications of Interphasal Resonances http://groups.yahoo.com/group/teslafy/
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