carmack je napisao/la:
Kolika bi bila realna korisnost (Pizlazno/Pulazno) tog sistema?
Dobro pitanje. Prvo ću citirat Teslu nekoliko puta:
„It will be of interest to compare my system as first described in a Belgian patent of 1897 with the Hertz-wave system of that period. The significant differences between them will be observed at a glance. The first enables us to transmit economically energy to any distance and is of inestimable value; the latter is capable of a radius of only a few miles and is worthless. In the first there are no spark-gaps and the actions are enormously magnified by resonance. In both transmitter and receiver the currents are transformed and rendered more effective and suitable for the operation of any desired device. Properly constructed, my system is safe against static and other interference and the amount of energy which may be transmitted is billions of times greater than with the Hertzian...“
„In 1900, however, after I had evolved a wireless transmitter which enabled me to obtain electro-magnetic activities of many millions of horse-power....“
„In my first efforts, of course I simply contemplated to disturb effectively the earth, sufficiently to operate instruments. Well, you know you must first learn how to walk before you can fly. As I perfected my apparatus, I saw clearly that I can recover, of that energy which goes in all directions, a large amount, for the simple reason that in the system I have devised, once that current got into the earth it had no chance of escaping, because my frequency was low; hence, the electro-magnetic radiation was low. The potential, the electric potential, is like temperature. We might as well call potential electric temperature. The earth is a vast body. The potential differences in the earth are small, radiation is very small. Therefore, if I pass my current into the earth, the energy of the current is stored there as electromagnetic momentum of the vibrations and is not consumed until I put a receiver at a distance, when it will begin to draw the energy and it will go to that point and nowhere else“
„Suppose that the earth were an elastic bag filled with water. My transmitter is equivalent to a pump. I put it on a point of the globe, and work my little piston so as to create a disturbance of that water. If the piston moves slowly, so that the time is long enough for the disturbance to spread over the globe, then what will be the result of my working this pump? The result will be that the bag will expand and contract rhythmically with the motions of the piston, you see. So that, at any point of that bag, there will be a rhythmical movement due to the pulsations of the pump.
That is only, however, when the period is long. If I were to work this pump very rapidly, then I would create impulses, and the ripples would spread in circles over the surface of the globe. The globe will no longer expand and contract in its entirety, but it will be subject to these outgoing, rippling waves.
Remember, now, that the water is incompressible, that the bag is perfectly elastic, that there are no hysteretic losses in the bag due to these expansions and contractions; and remember also, that there is a vacuum, in infinite space, so that the energy cannot be lost in waves of sound. Then, if I put at a distant point another little pump, and tune it to the rhythmical pulses of the pump at the central plant, I will excite strong vibrations and will recover power from them, sufficient to operate a receiver. But, if I have no pump there to receive these oscillations, if there is nowhere a place where this elastic energy is transferred into frictional energy (we always use in our devices frictional energy -- everything is lost through friction), then there is no loss, and if I have a plant of 1,000 horsepower and I operate it to full capacity, that plant does not take power, it runs idle, exactly as the plant at Niagara. If I do not put any motors or any lamps on the circuit, the plant runs idle. There is a 5,000 horsepower turbine going, but no power is supplied to the turbine except such power as is necessary to overcome the frictional losses.
Now the vast difference between the scheme of radio engineers and my scheme is this. If you generate electromagnetic waves with a plant of 1,000 horsepower, you are using 1,000 horsepower right along -- whether there is any receiving being done or not. You have to supply this 1,000 horsepower, exactly as you have to supply coal to keep your stove going, or else no heat goes out. That is the vast difference. In my case, I conserve the energy; in the other case, the energy is all lost.“
Ako pogledaš prethodne šeme predajnik1 i prijemnik3 treba samo sabrati sve gubitke i ograničiti se na sigurno.
Npr.
-ako je transformator snage npr. 2kW obilježit ćemo P1=2kW
-snaga motora koji pogoni rotacioni prekidač P2=1kW
Sumarno ulažem Pul=P1+P2=3kW a kolika će snaga se razviti u primarnom krugu i koliko će se prenijeti na sekundar to prvo zavisi od izbora kapaciteta i induktiviteta primara i sekundara pa zatim od podešenosti rezonancije odnosno podešenja vrijednosti Q faktora.
Neka je vrijednost Q faktora sekundara samo Q=500 puta
Sada neka je primarni napon dobiven preko ulaznog transformatora sa 220V podignut na npr. 5kV i njime se pune primarni kondenzatori i prazne preko primarnog induktiviteta.
Uzet ćemo u obzir i gubitke na rotacionom prekidaču i samoj sprezi primara sa sekundarom i neka se od onih 5kV sa primara prenese samo 2kV na sekundar, odnosno samo na par namotaja sekundara (labava ukupna sprega a jaka parcijalna sprega)!
Sada imamo situaciju da je sekundar ustvari otvoreno RLC kolo i da imamo generator 2kV spojen a R-L-C komponente mi sami biramo ali tako da uspostavimo bar približnu rezonanciju da imamo koliko-toliko velik Q faktor, kao u ovom slučaju Q = 500.
Sada dobivamo da će max napon na sekundaru biti oko Uc=Us=2kV*500=1MV
(Uc-napon na kapacitetu odnosno TOP-LOAD, Us-napon na sekundaru)
Npr. ako uzmem da je Cs=150pF, fs=50kHz
onda je snaga u sekundaru: Ps=Cs*fs*(Us)^2=7,5 MW ili 7500kW

Ovo je reaktivna oscilirajuća energija koju je moguće primiti sa Teslinim prijemnikom kao na šemi prijemnik3 ili neki drugi.
Neka sada jedan prijemnik bude podešen da prima samo 1% od te ukupne energije odnosno samo 75kW i ako oduzmemo gubitke u primaru neka su kao i u predajniku zbog pogona rotacinog prekidača 1kW imamo na raspolaganju 74kW

što daje traženi odnos
Pizlazno/Pulazno = 24 puta
Ovdje treba naglasiti da TeslaCoil predajnik je ustvari predajnik reaktivne energije a prijemnik je takođe prijemnik reaktivne energije koju on efikasno pretvara u aktivnu.
Ali prijemnik radi tako da ne remeti rad predajnika, odnosno neće se desiti kao kod standardnog RLC kola kojeg smo učili u školi, da čim nešto spojiš paralelno na RLC kolo pobrkaš rezonanciju i nemožeš pokupit onaj očigledni višak. Jer npr. postignuta je rezonancija u standardnom školskom RLC kolu i spojiš voltmetar na kondenzator i mjeriš na njemu i 200 puta veći napon od napona izvora, struja je maksimalna, i proračun ti kaže da je reaktivna snaga veća 200 puta ali džaba, nemožeš je pokupiti u tom zatvorenom krugu!!! Prvi korak je da bude RLC kolo otvorenog tipa odnosno da se ide ka jednožičnom prenosu pa onda sve ostalo!