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Paul Nicholson on Alternator Resonance

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  • harvich
    Paul made the following replies on tesla list concerning the Q factor of the secondary being 8.7. I initially misinterpreted his responce, as apparently his
    Message 1 of 1 , May 21, 2002
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      Paul made the following replies on tesla list concerning the Q factor
      of the secondary being 8.7. I initially misinterpreted his responce,
      as apparently his method of deriving a Q seems different from mine. I
      have given further thought to this subject, and even posted a
      careless mistake, but if we go by I^2R as the real power inputs...
      However given the same method used for the primary resistance as an
      indicator of "real" power input, this comes to (.35)^2*1.1 = 0.135
      The secondary of 1000 ohms with 4 ma conduction by I^2R then only
      yeilds .016 watts.
      Thus here it has already been agreed by other reasonings that .58
      watts, is the ACTUAL real power input, so how can this be? To retrace
      things here there undoubtably is some confusion regarding reactive
      current measurements. I had initially posted the following to tesla
      Rewrite of Mutual Inductance Laws for Tesla List
      The laws of Mutual Inductance for air core coils only implies a
      several percentage points effeiciency; TRANSLATION; 100 REACTIVE
      WATTS IN- 5 WATTS REACTIVE WATTS OUT However those laws only specify
      L1L2 in air core, and with increased frequency of input for L1L2
      coils if we give each of these L quantities an associated C1 and C2
      values, we should expect the mutual induction to increase, or more
      properly the ability of the L1C1 primary to excite the L2C2 secondary
      for comparisons here.
      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
      Since the volt-amperes measurements only indicate possible deviations
      from actual real power input, when it it is shown that the components
      are actually matched to be as completely resonant as possible, the
      reactive power arguments completely fail to show how the the output
      coils appear to be greater than the input. The simple statement of
      the fact is that the unenergized field of this 7 pole face rotor @
      480 hz can produce only 1.67 volts enabling .35 A on the 14 gauge
      coil in DSR1 resonance, but over 1250 volts to light a one ended
      neon leakage current across the 1 nf capacity occurs on the secondary
      side. When the field is actually energized the output side does get
      reduced according to the increased neon leakage current.
      15 Volt Operation
      In that operation there having over 6000 volts across the
      caps,experimentation suggests that all three voltage lines of
      conduction making that high voltage can be cut from its supply
      WILL APPEAR! Pray tell us what is the K mutual inductance factor here
      in this coefficient by calculations here? The ordinary L1L2 mutual
      inducatnce equations do not account for the increased efficiency
      afforded by giving each L quantity a C to resonate with the imposed
      frequency. Indeed here is an alternator with no energized field
      exercizing resonance of L1C1 through space to L2C2 and producing one
      ended neon discharge! So to complain at least to science law makers
      of presumptions it can be said that air core mutual inductance laws
      do not preclude a non performance factor. HDN
      Paul replied back
      Unfortunately, the jpg doesn't make clear the circuit arrangement,
      so we're not sure just where the V and I measurements were picked
      Reply; 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)
      Paul; 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.
      Reply; 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...
      NOTE; Here is where a complication comes into being. Ordinarily the
      reactive input measured as VI will = the real power input measured
      VALUE OF CONDUCTION AT RESONANCE. However on the primary, and
      especially secondary sides in this example, the components come no
      where near the ideal ohmic conductions at resonance. Because however
      near equal reactances are recorded for the primary we have to assume
      that the phase angles are correct, and because the resultant amperage
      must be closely in phase with the applied voltage, that VI
      measurement might be construed as the real power input, despite the
      fact that the expected currents did not develope. Because those
      currents do not develope, the I^2 R component, is .135 watts instead
      of .58 watts, but this still leaves us wondering as to whether
      that .135 watts is the actual real power input?
      Paul; 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.
      Reply; 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.
      Paul; 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).
      NOTE; Here is where Pauls reply got me totally confused and I started
      to make a longer dissertation on how I calculate the Q by X(L)/R. The
      complication he may not be aware of is that real q's dont match the
      ideal ones made by calculation. The pi in his equation confuses me.
      It has been established that the VAR of L2C2 = 5.06
      How I would enquire on this situation is to find the energy storage
      itself by the voltage across C= 1nf ,(1265 volts) and finding the
      joules by (1/2)CV^2= 8.41 *10^-4 joules
      I would then multiply that by 480 *2 for the energy transfer rate,
      becoming only .807 watts
      Hence the voltage times existant amperage VAR of 5.06 is not
      corresponding to the actual energy transfer rate, where it was
      assumed the the VAR would be equal to that rate. Let us then note any
      deviation to be noted by 1/2LI^2, which represents the storage of
      magnetic field energy. Using 60 henry that storage would be 4.8 *10^-
      4 joules. This makes sence because the Acting L and real X(L)'s at
      480 hz of the 60 henry coil are about double what the equations will
      deliver. So realistically then we have already calculated the I^2R
      heat loss for the 1000 ohm coil as .016 watts, but there is
      about .807 watts energy transfer taking place. Now if I remember one
      of the definitions of Q, it is the reactive power input divided by
      the energy wasted in resistance to be indicative of the actual q of
      the circuit. So here we have already tuned the high inductance coil
      to have the equal reactance that the 1.05 nf cap has at 480 hz, or
      ~316,000 ohms. Since this is also the acting impedance of the 1000
      ohm coil, there must be a 316 fold voltage rise as internal q factor
      for the coil to come to Ohms law conduction values.
      Now as previously mentioned VI only equals I^2R if the resonance does
      come to ohms law conduction. Here we can see that it does not come
      close to conduction values, but that .016 watts is a result of the
      conduction that did develope. Let us multiply that wattage by the 316
      fold increase that might be assumed if that conduction did develope.
      316 * 0.016= 5.056, equivalent to the original measured VAR of the
      However this still does not answer the question of why does the .807
      watts calculated as the actual energy transfer not equal the VAR
      measurement of 5.06 ? To answer this seems more difficult, but if we
      consider that a limited resonance not coming to its ohmic law
      amperage levels, another consequence of this is that the voltage
      necessary to cause that conduction ALSO does not develope. So here a
      ratio can be formed between VAR(actual)/VAR(apparent)= .807/5.06
      = .159 What this ratio indicates is the degradation of the q factor.
      Thus only .159(316)= 50.4 acting q will develope according to that
      Now this can be compared to in actual operation, but not with the two
      wire model, because no "outside voltage" developes on that L2C2
      branch by line connection or little by induction alone, but
      internally the comparable actions take place, as if the circuit were
      actually driven by the needed three phase inputs. When these inputs
      are in place, this second stage of 60 degree phase shifted
      resonance will appear as voltage on the outside of the interphasing
      L2C2, and the interphase amperage reading itself serves as an
      indicator of the internal voltage rise. Then we can see if this
      theory cuts the mustard by ratios determined as degradations of this
      316 q factor.
      http://groups.yahoo.com/group/teslafy/files/RI/Dsc00167.jpg shows the
      parametric 5 wire reading for magnetic opposition.
      19.4 volts across the L2C2 enables 3.65 ma.
      According to our new designation this becomes VAR(input) as
      19.4*.00365 = .0708
      Now the VAR internal values,(as they were initially obtained for the
      2 wire case) can be compared for the 5 wire operation.
      First the voltage rise across the cap is determined by 3.65/3.16 *
      1000 =1155 volts
      Now the VAR(apparent) as 1155*.00365= 4.21 in this case
      The energy storage is 7*10^-4 joules, and energy transfer as 480*2
      or .672 watts for VAR(actual)
      The Q degradation factor then becomes VAR(actual)/VAR(apparent)*Q
      (actual)= (.672/4.21)*316 = 50.4
      Now let us calculate the acting Q by V(int)/V(source)= 1155/19.4 =
      This high parametric Q factor is quickly reduced in actual energized
      field condtions where at 15 volts operation referred above, 364.8
      interphasal volts enables 6424 volts for a reduced Q of 17.6. However
      the overall voltage rise is still comparable because of the fact that
      the preliminary outside voltage rise has risen non linearly in
      proportion, where now 15 volts is inputed that 365 for an outside Q
      of 365/15 = 24.33. The end acting voltage muliplication is then 24.33
      * 17.6 = 428.2 times the stator voltage.
      I would suspect that this same analysis could be made for the 15 volt
      case where VAR(apparent) =6424*.0203= 130.4 , but VAR(actual) show
      only .021 joules being transfered 960 times per second =20.8. Then
      20.8/130.4= .159, and again that limtation would only be a 50.4
      voltage rise, although then the actual internal voltage only appears
      as 17.6 times that inputed. In any case it is a step forward to note
      that VAR(apparent) and VAR (actual) can deviate so far in comparisons.
      Now here is Paul's final say on the matter and my comments made as a
      last reply;
      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). HDN NOTE{I still dont
      understand this, it should only circulate 2*480 times/sec}
      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.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
      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
      Pauls Conclusion;
      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 --
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