>From what I gather, there seems to be a host of
methods for making colloidal silver water, including
even using the higher voltages of NST's, which are
current limited transformers that can take a dead
short. I now have several large induction coils set up
for allowing the passage of a "dead short" between
these two coil systems that acts like a backwards
electrical system. Following is a post made to another
list concerning how these induction coil systems could
be put into a Binary Resonant System, (BRS),posted
using 6 500 ft coils of 14 gauge wire. I am using 30
of these coils in my system, which means it can be
safely driven at 120 volt AC.
from post;
However what seems to perplex most people
who dont understand what I am talking about is that
the BRS is a backwards electrical system. It is the
same thing from the inside out, as the outside in,
because it has point symmetry. In this condition of
making silver water, as the solution becomes more
conductive, instead of the BRS input comsuming more
amperage, it will consume less. In fact it seems every
reason to believe that as soon as the solution reaches
a certain level of conductivity, the power input would
become so low as to always make a certain ppm quality,
irregardless of the time the water was exposed. These
are future issues to explore, but first the
replication issue needs to be addressed.
To replicate the BRS I will first describe the safe
and then the catastrophic version, which may be easier
for others to build. The problem with making a BRS is
the tremendous cost of the coils and capacitors. Since
we are dealing with 60 hz resonance in the US these L
and C quantities become fairly huge. The first
sensible thing that should be done is to aquire a 4:1
step down transformer so that the input AC voltage
will be about 30 volts. Now an example is made for 3
14 gauge 500 ft coils in series, for each side of the
circuits series resonance. Each side must then have a
minimum of 33 mH for the BRS effect to easily be seen
at 60 hz. This example then uses a total of 6 coils,
but any other large inductances can be used. Three of
the 14 gauge coils in series comes to about 33 mH. The
ohmic resistance of these 3 coils in series are about
4 ohms, allowing for 30/4=7.5 Amps at series
resonance. This occurs on both sides for a total of 15
amps input. This then is a borderline safe scenario,
as we dont want more than 10 amps in those 14 gauge
wires. Now let us find the real impedance and capacity
to use to resonate each coil system;
First the Inductive reactance is found by the equation
X(L)=2 (pi)(freq)L=(6.28)(60)(.033)=12.43 ohms
Now oftentimes the impedance can be estimated as the
inductive reactance if the resistance is small in
comparison but here we see that this is not the case
here because of the small quantity of inductance we
are using in this case example. Here the resistance is
one quarter the X(L) quantity, so the true impedance
is the square root of both of these quantities squared
or
Sq. Root{12.4^2+4^2}=Sq rt{153.76+16}~13 ohms
This 13 ohms represents the amount of current each
coil side will allow from the ohmic estimation made by
impedance calculation. Thus the coils themselves will
conduct 30/13= 2.3 Amps in a non resonant condition.
Now the capacity to resonate must be found, to do this
first the capacitive reactance must equal the 12.4 ohm
X(L) value.
x(C)=1/2(pi)(freq)(C) where C is to be found;
12.4=1/(6.28)(60)C ; 376.8*12.4=1/C
C=.000214 F= 214 UF(microfarad) [a fairly large
quantity to be used on each side using 33 mH as L]
Each coil branch gets this capacity to series
resonate, each placed on opposite ends of the parallel
connections to the source AC input. This makes each
series resonance 180 out of phase, consuming 15 amps
in that condition.
Now the voltage rise for the system can be found by
the Q, the ratio of X(L)/R or 12.4/4=3.1 Thus 3.1*30
volts input is risen within the circuit in both
opposite polarities to yeild 2*93= 186 volts unobvious
potential beween the midpoints. These midpoints go to
the silver pieces.
Now if we attach an ampmeter between the silver pieces
we should find a maximum of 2.3 amps, this is the
maximum current that will be allowed between these
opposite branches by parallel resonant current
limiting. Connecting the midpoints of the circuit
changes the entire circuit into a figure 8 tank
circuit with twice the internal resistance. Thus only
1.15 amps will be in the actual tank circuit. And the
actual input will be Q or 3 times less than this, or
only around .38 Amps. Now if we immerse these
electrodes in the distilled water, the 186 some volts
should immediately drop to a much lower level, and if
we plot this voltage over time we should see a
significant drop in voltage and amperage consumption
during this time period. I suspect that method should
make batches fairly quickly compared to present day
methods, but I have barely scraped the obtuse
descriptions found on the silver water list.
Now to understand in this poor example of what can be
a catastrophic binary resonant circuit we can operate
the above circuit just fine without stepping the
voltage down! Just make sure the switch is closed!
Then the consumption will be .38Amps*4=1.53 Amps
There will be 4.6 amps in the wire. But if the switch
comes open 30 amps will immediately start melting your
wire!, a very dangerous circuit! But the 60 amps
demanded from the circuit is supposed to throw your
circuit breaker anyways.
Sincerely HDN
=====
Binary Resonant Systemhttp://www.insidetheweb.com/mbs.cgi/mb124201
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