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

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

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.

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

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

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}

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

Reply;

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 --