Practical Guige to Free Energy Devices

joule of energy at any desired time rate because the permanent magnet stores that energy unlimited and so if we
want a power output of 1 KW as the power P we calculate:
P = dU/dt
For P = 1 KW , we need pulse 1 joule of energy for only 1 millisecond.
In the same way, if we can get power of the same levels from Earth's magnetic field, we must calculate the
volume of the air core coil. By using the same equation, we see that
(0.5) 2 x (5 x 10 -2 ) x (2 x 10 -2 ) x (10 -2 ) = (10 -4 ) 2 x V
V is the volume of the coil we need for get the same magnetic energy levels, and in this case, V = 250 m 3
That is to say, a coil of 6.3 m diameter and 6.3 m length, placed parallel to the Earth's magnetic field, can store
the same energy as that little 5,000 gauss permanent magnet which we considered for a MEG device.
But it is not necessary build a huge coil, we can use a smaller coil. The enclosed magnetic energy will be lower,
but as P = dU/dt we must raise the frequency of the pulses to obtain the same power level coming from a bigger
coil. For example, an air core coil of 1 meter diameter and 1 meter length according to equation (1), stores an
energy of:
U = 1 / (8 x pi x 10 -7 ) x ( 10 -4 ) 2 x pi x 1 / 4 x 1 = 0.003 Joules
If we pulse that energy level at 330 kHz, then we will get 1 kW, and at 660 kHz, 2 kW, etc., thus a higher
frequency yields more power.
Then the question becomes, how can we pulse the constant magnetic field inside the coil? The answer is simple:
by using an external source, we can cancel the Earth's magnetic field inside the coil. There must be power and
energy amplification with respect to the external input source. To realise that power amplification, we must do the
following:
Let the magnetic field variation inside the air coil be given by:
B(t) = Bo + Bf x Sin( w x t )
Where
Bo is the constant of Earth's magnetic field,
Bf is the magnetic field in the coil created by the external power source, and
w is the angular frequency of the external source.
Replacing B(t) from equation (1) we get the energy variation with time, U(t), and then we can calculate the power
as P = dU/dt resulting in:
P(t) = Bf x w x V x (Bo + Bf x Sin(w x t) x Cos( w x t ) ) / muo ........ (2)
Remember that V is the volume inside the coil.
We see here that the output power depends on Bo, the Earth's magnetic field, just as in the case of Bearden's
MEG it depends on the magnetic field intensity of the permanent magnet in the circuit.
So we can now calculate a COP value with Bo and without Bo, or Bo = 0
Calculating the RMS power for both cases (not reproduced here because it corresponds to a case of basic
differential calculus) and using the ratio, the result for the COP is:
COP = ( 1 + ( 2 x Bo / Bf ) 2 ) 0.5
We see then power amplification, and of course if Bo=0 and not a permanent magnetic field, the maximum COP is
1, input and output powers are equal. In the case of Bearden's MEG, the condition is Bo = Bf for not degaussing
the permanent magnet and in that case we have a COP = square root of (5), which is a value between 2 and 3
which corresponds to the practical results for this classic calculation.
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