Power from the Earth Magnetic Field

[Tesla_2006]

Many people think is not possible get power from the Earth magnetic field because that field intensity
is too low and the energy levels don?t work for useful applications, that is a bad claim

  I build many coils get many kilowatts for useful purposes from the Earth magnetic field and here show the basic concept and some formulas fundament all this

Earth magnetic field influence is not for ignore, in times of hard sun activity the Earth magnetic field oscillates and in a long transmission power line there is surges
of voltage , overvoltages cause many trouble, breakdown and stoping the electrical sources, for the Faraday law that induction are gived for the following

V = 2 * pi * f * B * A

B will be the Earth magnetic field , f the frequency of the fluctuations and A the surface across the field go


For calculation purposes we can approach the Earth magnetic field to 1 Gauss ( 10 ^ -4 Teslas)

If we think in a long common transmission line of about 10 Km and the lines are spaced to 1 mt we have

V = 6,28 * 10^-4 * 10 ^ 4 * f = 6,28 * f

if the flutuations should be  f = 10 Hz, we have a overvoltage of 63 V !, for 100 Hz then 630 V, etc,....

And so if the Earth magnetic field is lower in intensity is considerable his effect in a great surface rande and great volume range

For energy and power considerations, we can see that field is lower than common magnets in our houses, but the volume of space is too large, the energy of a magnetic field is not in the field intensity only, the volume across that field expands is important
  The energy stored in a magnetic permanent field B across a space volume V is

U = 1/(2*muo) * B^2 * V,  (1)

muo is the magnetic permitivity of the vacuum

Common permanent magnets store energy we can use for get unlimited power like the MEG of Bearden , then the Earth magnetic field across of a air core coil can do the same result

Now we can do a comparation between a permanent magnet and a coil oriented to the Earth magnetic field for get the same energy levels

Let's take in consideration a powerful permanent magnet as the used in a MEG of 5000 gauss and dimentions of 50 mm x 20 mm x 10 mm , according of that values the energy stored in the permanent
magnet will be using (1) :

U = 1/(8*pi*10^-7) * (0,5)^2 * (5*10^-2)*(2*10^-2)*(10^-2)

U = 0,995 Joule

  That is say 1 Joule aprox of energy, however it may be a low energy level, but devices like the MEG with that permanent magnet get too many kilowatts, the reason is because that magnetic energy is constant.
If we close that magnetic field in a core or magnetic circuit and we pulse that field we get 1 joule of energy at any desired time rate because the permanent magnet store that energy unlimitied 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 milisecond

Of the same way if we can get power levels of the same levels from the Earth magnetic field we must calculate the adecuate volume of the air coil, by using the same equation we see


  (0,5)^2 * (5*10^-2)*(2*10^-2)*(10^-2) = ( 10^-4)^2 * V

V is the volume of the coil we need for get the same magnetic energy levels


resulting...... V = 250 m3


that is say a cubic coil of 6,3 mts of diameter and 6,3 mts of long oriented paralell to the Earth magnetic field can store the same energy of that little 5000 gauss permanent magnetc we consider for a MEG device

But is not necessary build a too large coil , we can build a more little coil, the enclosed magnetic energy will be lower but as P = dU/dt we must raise the frequency of the pulses for get the same power level for a more bigger coil, for example an air coil of 1 meter of diameter and 1 meter of long according to (1) store an energy


U = 1/(8*pi*10^-7) * ( 10^-4)^2 * pi * 1 / 4 * 1 = 0,003 Joule

If we pulse that energy level to 330 KHz we get 1 KW, 2 KW to 660 KHz, etc,.....more frequency more power


Then the question is how we pulse the constant magnetic field inside of the coil and the answer is single, with a external source can cancel the Earth magnetic field inside of the coil
The external source need a power level and we need the output power can be more great than the input external source power, that is say, there must be power and energy amplification respect
to the input external source, for see that power amplification we must do the following

Let the magnetic field variation inside of the air coil gived for:


B(t) = Bo + Bf * Sin( w * t )


  Where Bo is the constant Earth magnetic field, Bf the magnetic field in the coil caused for the external power source for cancel the inside Earth magnetic field, and w the angular frequency of the external source

Replacing B(t) in the equation (1) we get  energy variation in time, U(t), then we can calculate the power as P = dU/dt resulting



P(t) = Bf * w * V * ( Bo + Bf * Sin( w * t) * Cos( w * t ) ) / mu0,  (2)



  Remember V is the volumen inside of the coil

We see here the output power go in dependence of Bo , the Earth magnetic field, such as in the Bearden Meg go in dependence of the magnetic field intensity of the magnet in the magnetic circuit

Then we can calculate a COP rate with Bo and without Bo, or Bo = 0

Calculating the rms power for both cases ( not reproduced here beacause it correspond to a couse of basical diferential calculus )  and making the ratio, the result for the COP is



COP = SQRT( 1 + ( 2 * Bo / Bf )^2 ) , SQRT denotes square root


We see then power amplification ,and of corse if Bo=0, not permanent magnetic field,  the maximal COP is 1, input and output powers are equal, in the Bearden MEG the condition is Bo=Bf for not degaussing the permanent magnet and in that case we have  COP = SQRT(5) a value between 2 and 3 corresponding to the real and practical results for this classic calculations


But for the Earth magnetic field we can get values more elevated because never we can cause the desmagnetization of the Earth magnetic field

How many turns of the coil, frequency, diameter, longs, etc, we need?


The input power for cancel the Earth magnetic inside the coil, or the condition  Bf = Bo , we calculate using the rms component of (2) doing Bo = 0, we have


P = 0,05 * f * A * L ,  P in watts


A is the section of the coil in mts2, f the frequency in Hertz, L the long in mts


For cancel the Earth magnetic field inside of the coil the condition based in the ampere law is

 
N * i / L = 100


N the number of turns of the coil and i the current in amperes



Finally the induced voltage in the coil will be


V = 2 * pi * f * N * Bo * A



Filling with values we have for example


N = 1000 turns, f = 100 KHz , diameter coil = 1 mt , L = 1 mt , i = 100 mA = 0,1 A, Bo = Earth magnetic field

V = aprox 4000 Volts


Input power then for this example is about 4 KW, output power for the COP factor will be a maximal of 12 KW, then in closed loop operation we give the needed 4 KW to the input for get an autopowered device resulting in a static generator delivers 8 KW of unlimitied power output, my practical results go according this calculations

How is the practical building?


By using a large plastic pipe we build the primary input coil for use with the external source, for get the output power we use another pipe inside of the first pipe aprox of the same diameter and long with the adecuate turns and wire for step down the voltage, for example 110 VAC
The Hight voltage oscillator of the input source I use a resonant LC from ferrite power transformers

Is very interesting see work this devices, and in days of geomagnetic storms power levels go too many hight and I must put protections for don't damage the system and charges I've connected


Now there is another performance in this, if we use not an air coil and we use a core, steel, iron, or another the field inside increases dramatically and we get a more little coil and more power, but we don't can use high frequencies

But not air coils is other theme for other time



Any question to the following email deleting WITHOUTTHIS


enertec2200WITHOUTTHIS@yahoo.es



Good bye

[hartiberlin]

Hi,
can you show a picture of your device in a working condition
powering a load, e.g. a lamp bulb ?

Normaly when you build a transformer like this,
if you draw current from the output,
the input will need also more current and thus
more input power.
How do you avoid this ?

Many thanks.

Regards, Stefan.

[Y :-)]

Hi Stefan and all

Don't forget about this !!!
http://www.overunity.com/index.php/topic,1645.msg16747.html#msg16747

[FreeEnergy]

hmmm 

[roggy32]

Wouldn't the inter-electrode capacitance of the wire eat all of the "energy" from the spikes. Unless the voltage was above 1kv and had sufficient potential.