Pioneering the Applications of Interphasal Resonances
http://tech.groups.yahoo.com/group/teslafy/
On Saturday, April 2, 2016 9:02 PM, "[email protected] [teslafy]"
<[email protected]> wrote:
Let me reassert my earlier prognosis discovered in the work with
resonating a 500 ft length spool of 14 gauge wire @ 60 hz: that once a q factor
of pi is exceeded by the amount of inductance employed: the energy transfer
between L and C in joules/sec will exceed the energy transfer as joules/sec to
the heat loss as I^2R of the coil. The spec.s for 14 gauge wire indicate 2.6
ohms/1000 ft: so a 500 ft spool of 14 gauge wire can be estimated at 1.3 ohms.
The inductance can be measured around 11 mh or 0.011 H. Now let us find the
theoretical impedance Z given these values.
Z= Sq rt { X(L)^2 + R^2} X(L) = 2pi*F*(L) X(C) = 1/{2pi* F* (C) and X(L) = X(C)
as the conditions for resonance. First we find a value for Z given these
parameters. To do that we first need to find X(L). X(L) =6.28*60(hz)*.011 =
4.14 Q = X(L)/R = 4.14/1.3 = 3.18. So we can see that resonating a hardware
store bought coil spool of 500 ft of 14 gauge wire @ 60 hz already produces a
theoretical Q value very near the value of Pi. Next we need to look at the
formulas for the L and C energy storage in joules. For brevity since we know
that for the conditions for resonance we will be choosing a C value that has a
reactance equalling that of the inductor X(L) value which we have just obtained
as 4.14 ohms, we can skip the C side of things,or perhaps return to that side
of things later for verification of results: and simply find the energy stored
in the magnetic field from L. That formula is given in joules by E =
(1/2)(LI^2) and to find I the current we need to find Z the impedance. We can
treat Z in alternating currents just like we treat R in direct currents by Ohms
Law where V =IR. Again for simplicity we will assume we have a 10 volt rms AC
input. The further complication here is that because this is stated as an rms
value, the actual peak of the voltage will be the sq rt of 2, or 1.41 times
higher then the rms value. This also applies equally with the rms amperage
measurements. When we are using the formula for the energy in a magnetic field
where E=(.5)LI^2 the AMOUNT of magnetic field energy in oscillation with it's
electric field counterpart dictated by E=(.5)CV^2 is dependent on the
instantaneous time when we measure it. Since the energy expression is going
from the kinetic magnetic value to a potential electric field value , to find
the maximum amount of energy itself in oscillation we must measure that as the
peak value which is 41% higher then what the meter will record as the rms
value. So this must be factored into the equation. To employ the amperage as
the I figure in the equation E=(.5)LI^2 we need to find the amperage given a 10
volt(rms) input. To find this we need to calculate the impedance Z= sq
rt{R^2+X(L)^2}. We have a inductive reactance of 4.14 ohms and a resistance of
1.3 ohms. This becomes sq rt{4.14^2 +1.3^2} = 4.34 ohms
The measure of current may predict the voltage across the component when the
reactance is known. This fact becomes critical for the measurements of higher
voltages in series resonance. Essentially for a given amount of alternating
current the pressure to make that current across the reactance should be
proportionally linear in resultant actions. Here a 10 volt (rms) source will
allow a 2.3 Amp (rms) current with a peak value of (2.3 *1.4 = 3.22A) given the
4.34 ohm impedance. The energy in the magnetic field as E=(.5)LI^2. We have
derived I(peak) and next need L for the equation already specified as 11 mh.
Thus given the equation Joules=(.5)LI(peak)^2 we have (.5)*.011*(3.22^2) = .057
joules. This amount of energy translates back and forth as (kinetic) magnetic
expression and (potential) electric storage 120 times a second or .057*120=
6.84 joules /sec of energy movement in time and space. Now to compare this to
the actual I^2R heat release to the environment we use the rms (averaged) I
value of 2.3 A. (2.3)^2* 1.3 = 6.88 joules /sec. Now also the apparent power
equals the true one, and an "entitlement" of equal energy transfer in time andf
space is available for further secondary applications whereby the secondary
output may appear greater then the input on a true power basis.Like · Reply · 2
hrsHarvey Norris 2nd comment; The obvious conclusion for these comparisons is
similar to social security entitlements. People protest that the payments
should not be called "entitlements" because it was their money being borrowed
and returned. But people forget that their input money was also matched by
their employer, which then becomes an "entitlement" as part of the equation.
Now the analogy here differs because to start out with "Only a potion or % of
ALL of the energy was borrowed and returned from the source and just like a
banker interest was applied to the loan.'' This was called the "apparent power"
Now because of the loans and hypothetically attached,interest portion; the
apparent power always cost more then the true power at delivery. Now the
bankers sudenly realized that their extra costs of delivery line charge on
every electric bill if eliminated would allow a larger amount of cash flow to
be achieved in the transaction. Exactly PI times more amperage now exists AND
NOW the apparent power equals the true power at delivery. WE CALL THIS THE PI
BREAK-EVEN state whereby the series reonance has produced UNOBVIOUS internal
pressure enabled by cancelling reactances to produce PI times more amperage
then its reactive state. A sort of "entitlement" has been added by NATURE
whereby now because the amount of energy transfer is the same for both sides,
and the apparent power equals the tue power transfer, an "entitlement" is
achieved on the delivery side. By judicious applications of air core mutual
inductions betwen systems it is well within logical definitions to show and
demonstrate true power inputs where the secondary vibrational loss as heat is
greater then the sending primary coil's vibrational loss as true power input
shown as heat. This would then appear to show a circumstance of reactive energy
magnification as regards to what is actually inputed by the source.Left as 2nd
and 3rd comments past first
one.http://www.veteranstoday.com/2015/10/06/pi2/?fb_action_ids=930360377001232&fb_action_types=og.comments
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