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*To*: tesla-at-pupman-dot-com*Subject*: Re: Rewrite of Mutual Inductance Laws for Tesla List*From*: "Tesla list" <tesla-at-pupman-dot-com>*Date*: Fri, 17 May 2002 18:33:17 -0600*Resent-Date*: Fri, 17 May 2002 18:32:29 -0600*Resent-From*: tesla-at-pupman-dot-com*Resent-Message-ID*: <oDi5i.A.rwG.aEa58-at-poodle>*Resent-Sender*: tesla-request-at-pupman-dot-com

Original poster: "harvey norris by way of Terry Fritz <twftesla-at-qwest-dot-net>" <harvich-at-yahoo-dot-com> --- Tesla list <tesla-at-pupman-dot-com> wrote: > Original poster: "Paul Nicholson by way of Terry > Fritz <twftesla-at-qwest-dot-net>" > <paul-at-abelian.demon.co.uk> > > Harvey, > > > > http://groups.yahoo-dot-com/group/teslafy/files/RI/Dsc00184.jpg > > > L1 =10.8 mh, C1 =14 uf > > L2= 60 henry, C2= 1 nf > > > Input = 1.67 volts *.35 A = .58 VAR > > Output=(4/3.16*1000)*.004A = 5.06 VAR > > Unfortunately, the jpg doesn't make clear the > circuit arrangement, so > we're not sure just where the V and I measurements > were picked up. Only one element is engaged for the primary, an amperage meter in line measures that, and also a voltage meter across that element (primary LC input) > But it's clear you've got a pair of coupled LC's > both resonant near > your alternator frequency. > > I suspect that your input measurements are made > between the alternator > and L1C1, in which case if L1C1 is tuned to the > alternator frequency > these values will represent *real* power, not VAR. Exactly as intended, this only shows what is inputed,but because of resonance that input is equivalent to the real power input. This is only made to counter the first argument that would be made that reactive power input does not equal the real one, but for resonance they are equal. So far so good... > > Whereas with the L2C2 measurements, the current can > only be the reactive > circulating current (since you've not mentioned a > load), and therefore > the 'output' reading is really the VAR of the stored > energy in L2C2 > resonator. Unfortunately there is no load, only the L2C2 quantity as that load, where L2 contains 1000 ohms resistance to be factored in as a load. If C2 has an arc gap across it, then we can deal with that load of arcing resistance itself circumventing the resonance, in which case the delivery is reduced, and everyone becomes happy with the reduced delivery. > > On this basis, we can estimate the Q factor of your > combined L1C1/L2C2 > system. Your 5.06 VAR represents a stored energy of > 1680uJ (since this > energy circulates back and forth between L2 and C2 > some 2 * pi * 480 > times per second). > > If 0.58 Watts of real power are necessary to > replenish the leakage from > your store of 1680uJ, then the Q is given by > > Q = 2 * pi * frequency * stored energy / input > power > = 2 * 3.141 * 480 * 1680e-6/0.58 > = 8.7 Exactly as noted formerly, on the internal voltage side of L2C2, that voltage will rise ~ 8.7 times, (or lower) then that inputed: but the point being made here.(with 3 phase inputs) is that the q can go higher than that circumstance. Let us again look at the actual operation at 15 volts stator input; http://groups.yahoo-dot-com/group/teslafy/files/RI/Dsc00178.jpg Here 15.03 volts stator enables a 364.8=~365 volts across the interphasing. That q is 365/15= 24.3 which sounds good, but is not that available if the connections were measured with no load at all. In that case if the voltage were measured without the interphasing across it, the voltage reading would be higher. THE CONTENTION THAT NO LOAD IS ACROSS THE MIDPOINTS DOES NOT CONSIDER THE FACT THAT THE OHMIC RESISTANCE OF THE ELEMENT BEING RESONATED ITSELF IS THE ACTUAL LOAD. Again if we simply placed the ohmic resistance, (impedance) of the voltage meter itself as the load, a higher voltage would be delivered. We can determine this quickly for the case of book values approximated as a near "no load" situation. For X(L)~= Z, .15 henry with 12.5 ohms resistance will be by X(L)= 2 pi F L= 6.28* 480* .15 = 452 ohms. The q factor determining internal voltage rise for each DSR2 side to accomplish conduction then equals 452/12.5 = 36.17. The voltage across the interphasing should be 1.7 times this value at 61.5 times the input voltage. The reason that only 24.3 times the input voltage occurs is that the R(int) value of the load to be resonated IS ALREADY TAKEN INTO CONSIDERATION. Therefore to say that no load exists is quite redundant, any component to be resonated has its internal resistance, which as you have formerly aptly pointed out, that R (int) value must also be taken into consideration in the final resonance, here shown to be reduced at ~ 1/3 to the expected voltage value if that component itself had no resistance. Now let us look at the final q value in real operation at 15 volts. The first stage of resonance has increased the voltage input from 15 volts to 365. The second voltage rise increases this to that made with 20.3 ma across the 1 nf capacity, equivalent to 6400 volts. The inner Q normally only exibits a 8 to 9 q factor. 9 * 365 = 3285 volts. but 6400 volts is appearing. This means a q factor 1.9 times that, almost double to your figurings... So then can you revise your estimations...? As I have said the method increases the available q factor that would normally be available, and it is most logical that when we make a magnetic field in opposition to resonance, this method countermands the devilish limitations of resonance inherent with internal capacity. If a better method is available of cancelling the internal capacity of a coil delimiting a q factor, to make a better Q factor overall, I also would like to know about that one. HDN, AKA Dr Wrongway. > So I'm afraid I don't see anything in your post > which indicates a > problem with mutual inductance. At times like this > you always have to > ask yourself which is the most likely of two > possibilities: Either > you've misunderstood the operation of the circuit, > or some very well > tried and tested laws of physics are wrong. > -- > Paul Nicholson > -- ===== Tesla Research Group; Pioneering the Applications of Interphasal Resonances http://groups.yahoo-dot-com/group/teslafy/

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