Research website of Vyacheslav Gorchilin
2015-10-06
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The free energy of the electron

This method can be called in different ways: dopolnuvanje energy of the electron, its release, and even the opening free energy of the electron. The point is that humanity still has not learned to use its full potential. But all we need to ensure energy independence is to look at old things and concepts in new ways. Next, we show this with examples and simple mathematical calculations. As always, try not to go far beyond high-school physics :)

The method consists in the redistribution of charges along a long line (DL) and eat them in the load at certain points in time. But if pereraspredelenie charge is due to the interference of standing waves is by its nature reactive, you eat is already in the active load. When a certain combination of standing waves increases an efficiency of the second kind $$\eta_{2}$$, which leads to the energy gain $$K_{\eta2}$$ of the entire device.

Redistributed charges

For understanding the method will start with simple puzzles. There are five identical capacitors of $$C1..C5$$. Will give them a charge $$Q$$ value of 5 units (for simplicity, while we use relative units). Since the same container, the charge is evenly distributed between them, and the voltage across each of them becomes equal to 1. Accordingly, the potential energy $$W$$ of each capacitor will be equal to 0.5. This is depicted in the figure on the left.

Recall that the charge associated with the voltage so: $Q = C\,U,$ and the potential energy of the capacitor is according to the following formulas: $W = \frac {Q\,U} {2} = \frac {Q^{2}} {2\,C} = \frac {C\,U^{2}} {2}.$

Redistribute the charge as follows: the upper capacitor $$C1$$ will receive 3 units, $$C2$$ — 2 units, and at the bottom it will be zero. Note that the number of electrons in the system remains the same, only changed their location as shown at the right. The total potential energy of the capacitors of $$W_{gen}$$ in the first case — 2.5 units in the second — already 6.5 units. Due to the redistribution of the charge we received an increase of energy 2.6 times.

We immediately note that for such a redistribution of charge using a conventional circuit design may require energy exactly equal to the raise. Below will show you how to do it relatively inexpensively. In fact, this is the whole "trick" of this method.

The above model of the charge distribution is also interesting because we can use it similarly to a system of resonant circuits, or a long line (DL). While we consider FOR lossless, and thus can enter an equivalent circuit consisting only of capacitances and inductances. But we need the whole process fluctuations of DL, but only moments in which the entire charge is concentrated in containers, so we can further oprostiti its equivalent circuit. As is known [1], the total energy of the oscillatory circuit is equal to $W = \frac {Q_m^{2}} {2\,C}$ where: $$Q_m$$ is the maximum value of the charge on the capacitor oscillating circuit.

Since we are interested in a snapshot of the processes in the system, then it is legitimate to apply an equivalent circuit of a long line, which is shown in Fig. For our tasks ahead this will be enough.

Now we can consider a long line of fairly simple mathematical tools. A special case is the Tesla Transformer (TT) [2], the mathematical model which we will later build than scattering all sorts of rumors about him "magic capabilities", and at the same time confirm some real guesswork.

A little math
Consider a system of identical capacitors: $C1 = C2 = C_{i} = C \qquad (1.1)$ where: $$i$$ is the number of the capacitor in the interval $$1..N$$, and $$N$$ is total number, capacity $$C$$ is the known value. Also, we are aware of the charge $$Q_g$$ that we transfer this system of capacitors, and the distribution function of voltage (charge) across this system is $$f(x)$$. To simplify reasoning, we assume that $$x$$ is a relative value showing the measuring point of the charge, voltage or energy along the capacitors, varying from zero to unity.
To summarize, we need the discrete values at each point (hereafter we will show that for continuous $$x$$, formulae are derived similarly). Inject it: $x_i = \frac {i} {N}$ where: $$i$$ is the number of the condenser. Then the voltage on each capacitor of $$C_i$$ will be: $U_i = U_m \, f(x_i) \qquad (1.2)$ Express $$Q_g$$ using the known $$C_g$$ and $$f(x)$$, and find an unknown $$U_m$$ is the amplitude of the voltage. To do this, first determine the charge on either of capacitors. It will be equal to: $Q_i = C \, U_i = C \, U_m \, f(x_i),$ then the total charge will be expressed as follows: $Q_g = \sum^N_{i=1} |Q_i| = C\,U_m\sum^N_{i=1} \big|f(x_i)\big|, \qquad (1.3)$ $$Q_i$$, which stands under the sign of the amount recorded for the General case — module, since we need to sum up all the charges: positive and negative. Now we can find the unknown value of: $U_m = \frac {Q_g} {C} {1 \over \sum_{i=1}^{N} \big|f(x_i)\big|} \qquad (1.4)$ We've after got to the potential energy in each capacitor, which, based on the previous formulas are equal: $W_i = \frac {Q_i \, U_i} {2} = \frac {C \, U_m^{2} \, f(x_i)^{2}} {2},$ and can now calculate the total potential energy of the entire system of capacitors: $W_g = \frac {C \, U_m^{2}} {2} \sum_{i=1}^{N} f(x_i)^{2}$ Substituting here the result from (1.1) and (1.4), we obtain the important formula: $W_g = \frac {Q_g^{2} \, N} {2 \, C_g} {\sum_{i=1}^{N} f(x_i)^{2} \over \left[ \sum_{i=1}^{N} \big|f(x_i)\big| \right]^{2}} \qquad (1.5)$ This formula is the initial energy of the system (left side) multiplied by the ratio of the sum of squares to the square of the sum (right part). To bring this model to the actual DL can be even more, if you increase $$N$$ (number of capacitors) up to extremely large, and $$x_i$$ to an extremely low value. Turning to the limits and integrating we get the result in the General integral form: $W_g = \frac {Q_g^{2}} {2 \, C_g} {\int^1_0 f(x)^2 \, dx \over \left[\int^1_0 \big|f(x)\big| \, dx \right]^2} \qquad (1.6)$ this formula we later will come back, and yet we find the energy increment relative to the system with uniform distribution of voltage (charge). Because of the following formula, in fact, all this "fuss"=> $K_{\eta2} = {\int^1_0 f(x)^2 \, dx \over \left[\int^1_0 \big|f(x)\big| \, dx \right]^2} \qquad (1.7)$ where: $$K_{\eta2}$$ is the magnification factor $$\eta_2$$, or energy increment in the system. About $$\eta_{2}$$ can be read here.
The following table shows the values of $$K_{\eta2}$$ depending on the distribution function $$f(x)$$. Well seen pattern: the faster changes the function $$f(x)$$, the greater the value of $$K_{\eta2}$$.
It's fun! Note the function $$\sin(x)$$, value $$K_{\eta2}$$ which consists of numbers of the natural numbers.
 $$f(x)$$ 1 $$\sin(x)$$ $$\sin(x)^2$$ $$x$$ $$x^2$$ $$x^3$$ $$K_{\eta2}$$ 1 1.2345 1.668 1.3333 1.8 2.286
The energy gain

In this work we do not consider possible additional increment of energy in the system due to the capture of electrons (and other particles) from the atmosphere or from the earth.

Look closely at the numerator and denominator of the formula (1.7). The numerator — not that other, as the specific output energy (specific energy of removal), and the denominator is the specific input energy (specific energy of excitation DL). Then the formula can be further simplified: $K_{\eta2} = \frac {W_{out}} {W_{in}} \qquad (1.8)$ This ratio can be modeled in this project. It is possible to obtain different combinations of standing waves and consequently a different distribution of charges along the DL. In this simulation, real-time counts $$W_{in}$$, $$W_{out}$$ and $$K_{\eta2}$$ in the above formulas.
The output power of the device
The power output is quite simple: multiply the energy removal from DL $$W_g$$ and the fundamental frequency of the excitation $$F$$. Remembering the formula (1.6), we find the power: $P_g = W_g \, F = \frac {Q_g^{2}} {2C_g} \, K_{\eta2} \, F \qquad (1.9)$ In fact, $$\frac {Q_g^{2}} {2C_g} \, F$$ is the input power, the excitation power DL. If it is denoted as $$P_{in}$$, we can obtain a very simple formula of relations of input and output: $P_g = P_{in} \, K_{\eta2} \qquad (1.10)$
It should be noted that in reality there is still a loss to really do DL, it supplies power to devices in the removable systems, etc., which can reach 30..50%. Considering this efficiency, it is possible to deduce the most General formula for the operation of our device: $P_g = P_{in} \, K_{\eta2} \, \eta_1 \qquad (1.11)$ where: $$\eta_1$$ — efficiency of the first kind in relative units.
A little about the Tesla transformer and the excitation
The Tesla transformer (TT) is a special case of long lines and all recommendations for initiation apply equally to any DL. All the nuances of this transformer is impossible to describe even within a few articles, but some of them will try to illuminate. We start from our Mat. model, and more precisely, from the formula (1.9). Pay attention to her left side: $P_g \sim \frac {Q_g^{2}} {C_g}$ Recall that $$P_g$$ is removed from the TT power which is proportional to the square of the charge $$Q_g$$ and inversely proportional to the capacitance of the coil TT. $$Q_g$$ is nothing like the energy of the impulse from the inductor, which is equal to the product of $$C_I \, U_I$$, where $$C_I$$ is the capacitance in the circuit inductor, and $$U_I$$ is the voltage at the time of opening the key. Therefore, we need to use the highest possible voltage to the inductor and to reduce inter-turn capacity on the TT, and the increase in $$U_I$$ is much more important than the decrease in $$C_g$$. The capacity of the circuit inductor, $$C_I$$ will depend on the power key and selected under its options.

It goes without saying that when you increase the resonance frequency of the coil of the TT we proportionally increase and the power output. It will be limited only by circuitry of our device.

As shown by the formula (1.9) — we need maximum transfer charge from the inductor to the coil TT, not to transform one voltage to another, as happens in an ordinary transformer. Therefore, PVC winding connection between the inductor and the secondary coil of TT should be possible small.

Another tip is not associated with our calculations, but greatly affects the efficiency of the TT is the quality factor of the secondary coil. It is clear that it is necessary to wind a coil to a coil, and the Council — shakes with Litz wire — Litz wire, each vein which are isolated. The fact that at high frequencies the currents mostly on the surface of the conductor, therefore, the more area of its surface [4], the better. Litz wire increases the area several times.

Quality, in the broadest sense, is also associated and a combination of two resonances: LC and wave. In their intersection can be as close as possible to the coincidence of the above-described calculations with real data.

A little about renting

Methods of energy extraction from a long line — a separate topic for research. Therefore, we mention only some well-known approaches to this problem. The easiest way of removal is capacitive coupling between the DL and the metal mesh (foil, coil removal) which, through the key at certain points in time and removed the energy to the load. This method is shown in the layout. We must not forget that the mesh should be continuous, for example, in this simulation you need two grids — two half DL, and the load should turn on between them.

As the key can act as a discharger [5], and electronic circuit. The discharger has plus in its simplicity and operation in range of a sufficiently high voltage. Disadvantage — difficult adjustment and low stability. Electronic circuit runs with less voltage, but more stable and can carry the current in the load not only with a cutoff at some threshold voltage, but in strictly defined points in time. By the way, if the circuit operates with a voltage cut off, it must have a small hysteresis.

Another way is fairly well known is to eat 6-7 coils wound as well as the main TT; it is located in the center and coils around the circumference. Each coil contributes to the overall increase. The disadvantage of this method is the involvement of a fairly large space and the electric field throughout, so — quite a big loss.

All calculations given above were obtained for capacitances and voltages. But they can be transferred to inductances and currents — the result is the same. From this follows directly the second method of removal is the open circuit at the antinode of current in certain points in time, and passing it through the load. The moments of the circuit exactly the same as for tension [6].

Apparently, the ideal method of removal will be the synthesis of these two methods.

Vyacheslav Gorchilin, 2015
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