The science of getting free energy


     Papalashvili Dimitri         Georgia, Tbilisi,   2459226        E-mail: d170347@gmail.com

Getting free energy scientifically.

The great Nikola Tesla, despite his six-foot height, a talented engineer and scientist.
He had an extraordinary memory, and all or most of the calculations carried out in his remarkable mind.
So he did not leave records with the calculations.
For example, as he did Airconditioning, using hot and cold current.
It is known only that he did this by changing the parameters of impulses or the time of their actions.

Most of the seekers of free energy are searching for it by method of trial and error.
We - the participants of the forum will try to analyze the method to obtain the free energy from the point of view of the theoretical foundations of electrical engineering .
The participants of the forum who are not familiar with the higher mathematics and theoretical electro-technique , can contribute, through simple experiments.

If it is a 100% frequency of occurrence of the results of the experiment, it means we are on the right path.
It is important, as well, to experiments were carried out by a few people.
To do this you need a tester, oscilloscope and a soldering iron.
Need the same book of problems on the electro-technique with the decisions. A very useful book.
At the beginning of each Chapter provides a summary of the theory.
The Author Шебес М.Р. You can download it from here.
We consider the problem of 8.58 out of a book of Shebes.


Here, during the action of a rectangular pulse, voltage and current have a positive sign.
At the end of the pulse is a power surge down the voltage and current have opposite signs.
If the voltage divided by current, then we obtain a negative resistance.
Of course, in the nature of the negative resistance does not happen.

A minus sign indicates that the energy of a resistor is returned to the environment.
Thus, we get a cold and a warm current.
Hot voltage, more cold in this example.
So, where is the free energy?
All the talk about resonance, although no one really knows how it arises.
We don't know, and, therefore, on this question the answer specialists on the airwaves.

Parallel resonance.



The formula of the energy in the inductor


The current in the inductor


Find energy in the form of voltage


The received power


For capacitor


Here the number of pi =3.14

Resonance frequency


Inductance in the case of resonance, if С=const , then


Capacity in the case of resonance, if



Check the received formulas
Take inductance L=0.25*10-3 Hn and the frequency of F=10000 Hz
Resistors: R1=10 Ohm и R2=10000 Ohm.VoltageUbc=1000 v
According to the formula (8) we define the required capacity of the capacitor.C=0.0253/0.25*10-3*100002=10-6 farads, or 1mF.
Resistance inductance and capacitance are equal XL=XС=2*3.14*10000*0.25*10-3=15.7 Ohm

The current in the inductor by the formula (2)
i2= Ubc/ XL=1000 / 15.7=63.7 A

According to the formula (4) we find the power in the circuit
P=0.0126*10002/ 0.25*10-3*10000=5000 w, или 5kW.

Strange, but the late don Smith argued that power is directly proportional to the square of the frequency.
Then in the example presented here, the power will be equal to 50 MWt
And to obtain a 5 kW, it will be enough voltage 10 volts. Very strange!
We will check it out.

Analyze the formula (4) и (5)
Power inductor


For capacitor


For both formulas power is directly proportional to the square of the voltage.
For capacitor - power is directly proportional to the frequency and capacity.

The increase in the frequency - resonance inductance is reduced in proportion to the square of the frequency.

That according to the formula (4) is good.

What he had in mind Tesla, when he said that a large capacity and a little inductance - is that bad?
Perhaps he had in mind serial resonance.


Our working scheme.


Make a generator for car Tesla - it is very simple!


Consider the transitional processes taking place in parallel resonant circuit.

Operational impedance of the circuit is of the form:



Resistance of the circuit bc

:

The roots of the characteristic equation, we obtain, by equating to zero resistance of the circuit:




It is known that the roots of this equation can be: valid and different, valid and equal, and also complex.
Find a condition in which the roots will take the specified values.
The radicand equate to zero, we obtain


Here, if R > Rcr , we obtain the complex roots, and in terms of currents and voltages appear harmonic components (in the General case - sine wave).First, find the in operator form. Find expressions of currents and voltage Ubc









For Ubc

F1(p)=U/RC

F2(p)=p2+p/RC+1/LC



For a rectangular pulse U=const, we obtain the solution in the General form:



Here we notice that the decrease of the b , voltage Ubc increases.

Near the values of critical resistance, voltage increases to thousands of times.

This task with the decision is in the book of Shebes. You can download it here.

The task of 8.26 solved by the classical method, and 9.4 - operator.

It considers three options roots.

We are interested in the third variant.

In this variant of the R = 100 Ohm, L = 40 mH и C = 5 mF.

Resonance frequency f = 331 Hz

Critical resistance Rcr= 44.721 Ohm

In the task at the specified parameters - the amplitude of the voltage Ubc equal to 125 volts.

If instead of R = 100 Ом take R = 50 Ohm, get the amplitude Ubc equal to 500 volts.

Lovers of strong sensations can take R = 44.73 Ohms, and get 12600 volts.

That in 100 times more input voltage and reactive energy circuit is increased to 10000 times.


Thus the phenomenon can be called energy resonance, or "resonance in resonance" by Tariel Kapanadze.

Energy resonance, or "resonance in resonance" by Tariel Kapanadze.

How to get energy out of the parallel-resonant circuit about this in the continuation.

Find i2(t)









Consider the case of real parameters.
f = 30000 Hz, L=5010-6 Henry, C = 0.5610-6 microfarads.
Rcr= 4.72 ohm, take R= 5 ohm.
XL=XC=9.4 ohm.

ω=2πf=2π×30000=188500

The roots


Voltage Ubc



Current I2



In the initial phase of the voltageUbc is zero.
In the initial phase of the current I2 is small, which means that at the initial time the voltage and current are zero.
Despite their large amplitude, the energy of a coil and a capacitor can not pull out.

Unfortunately the desired critical mode has no effect.
In the sequel we consider the ordinary resonance circuits, and how to use it.


I wish you success in the experiments!


To be continued


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