Research website of Vyacheslav Gorchilin
2017-06-12
The switching algorithms, electrostatic machine and a fork Avramenko
In a previous article, was described the algorithm switching for the two isolated tanks. It is only logical to replace the second a solitary container on a pole. This can give a saving in space for real designs, and in some cases to reduce the potential of high-voltage stress across the device. Interestingly, the first algorithm switching tanks does not change. It will consider next.
The figure shows the conversion process scheme with two secluded containers scheme with one solitary (C1) and one bipolar (C2) capacity. And secluded — formed external cylinder C1, and bipolar plates occurs between two cylinders: inner and outer. In fact, this coaxial capacitor, with the difference that its outer surface forms a solitary capacity. The ratios of the increment of energy will be according to the formula from the previous section (4.8): $K_{\eta2} = {C_1 \over C_2} \qquad (5.1)$ to see more real $$K_{\eta2}$$ according to the first algorithm, we recall the formula for finding solitary cylinder capacity and capacitance of the coaxial capacitor. $C_1 = {2\,\pi\,\varepsilon\,\ell \over \mathrm{arcsh}{\frac{\ell}{r_1}} + \sqrt{\frac{r_1^2}{\ell^2} + 1} + \frac{r_1}{\ell}} \qquad (5.2)$ where $$\varepsilon$$ — dielectric constant, $$\ell$$ is the length of the cylinder, $$r_1$$ is the radius of the outer cylinder. $C_2 = {2\,\pi\,\varepsilon\,\ell \over \ln{\frac{r_1}{r_2}}} \qquad (5.3)$ where: $$r_2$$ is the radius of the inner cylinder. And since the lengths of two cylinders are equal, the increment of energy all the way can be calculated according to the formula: $K_{\eta2} = {\ln{\frac{r_1}{r_2}} \over \mathrm{arcsh}{\frac{\ell}{r_1}} + \sqrt{\frac{r_1^2}{\ell^2} + 1} + \frac{r_1}{\ell}} \qquad (5.4)$ If the length of the cylinder greatly exceeds its radius, formulas (5.2) and (5.4) can be simplified: $C_1 \approx {2\,\pi\,\varepsilon\,\ell \over \ln{\frac{\ell}{r_1}}} \quad => \quad K_{\eta2} \approx {\ln{\frac{r_1}{r_2}} \over \ln{\frac{\ell}{r_1}}} \qquad (5.5)$
The first algorithm, and the electrostatic machine
This option is shown in the following figure (a), where charge is removed from brushes electrostatic machine (EM) comes to a large capacity C1. After accumulating a certain charge is partially transferred to C2 using the discharger FV1, which do not forget, in the General case is a breaker. High-voltage inductor L1 is used for time sharing between C2 and charge rent to the load, and is essentially a replacement key SW3. If it is a Tesla transformer (TT), then everything else, it will transform high voltage necessary for the load Rn.
In figures (b) and (c) show various embodiments of circuits according to the described algorithm. The bottom two C1 and C2 are coaxial capacitor: two pipes of different diameters inserted in one another.
In (c) as the spark gap acts as a button that is triggered once per revolution of the wheel EM and in L1 — Tesla transformer. Capacitance C2, obviously, should be approximately equal to the capacity of one plate of the uh multiplied by their total number. The author invites readers to create a symmetrical scheme of the first algorithm for EM by analogy with those given here.
The first algorithm and the Avramenko plug
Not obvious solution in the implementation described in the previous post, the algorithm is the inclusion of C1 and C2 with a fork Avramenko (see the following figure). Note that C1 and C2 in this scheme, in General, can consist of a steel condenser. As the key SW1 here is a high voltage diode VD1, and the SW2 — diode VD2. The key SW3 replaced dischargers FV1 and FV2, which in General case can be breakers.
High voltage HV in the first half cycle is positive with respect to ground, so the system charges the capacitors C1 and C2 through the diode VD1. The capacity of this system corresponds to the secluded capacitance C1, since the inside of the outer cylinder field in this moment. In the second half cycle through the diode VD2, the negatively charged outer cylinder. The capacity of this charging system will also be equal to a secluded capacitance C1. Educated thus the potential on the coaxial capacitor C2, through the dischargers FV1, and FV2, is discharged to the load Rn.
On the right figure shows a more advanced scheme, which involves a transformer L1. It converts high voltage to match the load Rn. The use of L1 as a Tesla transformer here is the obvious solution. If TT perform half the wavelength of the master oscillator, the gaps may not need. Only in this case, the solitary winding capacity of the TT should be much less than the capacity C2.
At first glance it seems that the small capacity and the lack of a galvanic circuit, will not give the output of this circuit a lot of power. Let's calculate. During one period of oscillation of the high-voltage HV generator to the load can be removed energy is equal to $W = {C_2\,U_a^2 \over 2}$ where: $$U_a$$ is the amplitude value of the voltage source HV. Consequently, power will be this: $P = f\,W = f {C_2\,U_a^2 \over 2} \qquad (5.6)$ where $$f$$ — frequency of oscillation. If now we take the following data: $\quad C_2= 20 (pF) \quad U_a= 10 (kV) \quad f= 1(MHz)$ and substitute in the formula (5.6), we obtain the output power at 1kW! Interesting may be getting the same output when reducing the frequency 100 times and the increase in the amplitude of the voltage only 10 times: $\quad C_2= 20 (pF) \quad U_a= 100 (kV) \quad f= 10(kHz).$ Of course, we need to take into account losses that may be 60-70%, but the capacity can not please seekers of free energy :)