## Paul Nicholson on Alternator Resonance

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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
Message 1 of 1 , May 21, 2002
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
watts.
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
list
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.
http://groups.yahoo.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
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
http://groups.yahoo.com/group/teslafy/files/RI/Dsc00178.jpg
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
source, and instead THE SAME LEVEL OF VOLTAGE BY DSR1 INDUCTION ALONE
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
up.
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
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
as I squared R; PROVIDED THE RESONANT COMPONENT GOES TO ITS OHMIC
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.
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
secondary.
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
ratio.
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 =
59.5.
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
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}
Paul;
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
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.
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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