statements, but I figured I better mull things over first... I may

have a conceptial difference of opinion with Paul, but perhaps I

dont, which is the reason for mulling it over.

The first problem I have is with the way he views energy transfer. I

view it as the amount of energy times the number of transfers divided

by that time period. Paul however uses pi times that amount, and

finally does give a good explanation as to why this is so. Even

though he has explained why pi times more energy transfer takes place

with a capacitive reanctance, I am unsatisfied with that explanation,

for the simple fact that the actual energy transfer can be any

quantity with respect to the energy input, and this is only

negotiated by the q of the coil. In my thinking the VAR input then

does not have to equal the actual energy transfer, and for the case

of a capacitive reactance this VAR quantity actually is pi times

more than the noted energy transfer, but to make the VAR equation

equal, the pi multiplier is made in place by equation.

Here is Paul's Reply from tesla list;

Date:Thu, 30 May 2002 17:27:49 -0600

From:"Tesla list" <tesla@...>

To:tesla@...

Subject:Re: VA and stored energy in capacitors

Jolyon wrote:

Since relationship between apparent power (VA), frequency (F) and

stored energy (E) in a capacitor in an AC circuit is VA=4pFE, where

it is assumed that F is the line frequency (50 to 60 Hz) during the

period when the capacitor is being charged by the transformer prior

to the firing of the spark-gap, ...

I had wrote back asking why this was 4 pi *f*E where E is the energy

storage,(.5CV^2), where in my thinking it should be 2*f*E. Pauls

reply shows that I did not factor in Vpeak, and instead used Vrms,

and he also shows where this pi comes from...

In general for a capacitor carrying a sinusoidal current, the stored

energy is 0.5 * C * Vp^2 where Vp is the peak voltage. This peak

voltage is sqrt(2) * Vrms, so in terms of RMS voltage, E = C *

Vrms^2. The associated RMS current is 2 * pi * F * C * Vrms, so we

have

VA(R)

= Vrms * Irms

= V * 2 * pi * F * C * Vrms

= 2 * pi * F * C * Vrms^2

= 2 * pi * F * E

I put the (R) in VA(R) to remind that we're talking reactive VA, not

real power - the energy delivered to the C in one quarter-cycle is

returned to the source during the next so the net energy transfer is

zero. For example, a resonating LC behaves as if the stored energy E

shuffles back and forth 2 * pi * F times per second. The above

doesn't quite apply to an idealised TC charging circuit, because the

discharge of the C is via another path. The line-side RMS ammeter is

only registering the charging current, not the two-way charge-

discharge current. To a first approximation the RMS ammeter reading

is halved to pi * F * C * Vrms. So now VA = Vrms * Irms = V * pi * F

* C * Vrms = pi * F * C * Vrms^2 = pi * F * E where I've dropped the

(R) because now some portion of the VA is real and some reactive.

This wouldn't be expected to apply to a real charging circuit as it

stands because the currents and voltages are no longer sinusoidal -

form factors other than sqrt(2) relate RMS readings to peak values,

readings depend on just where in the circuit you put the metering,

and the effect of chokes and ballasts all come into play. Best thing

to do is to model your particular circuit to estimate how your actual

meter readings relate to the peak voltages and stored energy.> doesn't the same equation apply when the gap fires when F becomes

After the gap fires the low resistance of the arc effectively

> the resonant frequency of the Tesla coil primary?

decouples the charging circuit from the primary LC. Things get a lot

simpler, the LC currents and voltages are sinusoidal, and RMS meter

readings in the LC circuit would obey VA(R) = 2 * pi * F * E during

each cycle of the ringdown. Of course, real meters report an RMS

value averaged over the response time of the meter's pointer which is

typically much longer than the whole firing event, so again some

circuit modeling would have to be applied to make sense of real

measurements

. > Does this not go some way into accounting for high peak powers> observed in TC discharges?

It certainly describes the 'power compression' taking place in the

charging/firing/primary circuit. Roughly speaking, energy stored in C

over a quarter-cycle of line frequency is released over a quarter-

cycle of the resonant frequency, so you're getting a 'compression

ratio' in the order of F_res/F_line. However, the discharges from the

secondary are compressed further still, because once the bulk of your

stored energy has made its way to the topload, it can all discharge

to ground in nano-seconds. Again, the formula VA = 2 * pi * F * E

could be said to apply during the arc discharge, but this time the F

is the frequency of oscillation of the discharge itself, which

involves the topload C and the self-inductance of the arc and its

ground return path - a very high frequency. Therefore ultimately, the

peak power in the discharge could be compressed by up to

F_discharge/F_line. -- Paul Nicholson --

Heres a reply I never sent because maybe I dont know what I"m talking

about!

Thank You Paul for this great explanation of energy tranfer! It

solves a paradox that has bothered me for some time. We know that for

a LC series resonance as the C quantity approaches resonance I^2R as

real power input approaches VI as apparent power input, and that at

resonance they become equal for ideal resonators. The

misunderstanding I had was that VI exceeded I^2R in reactive

measurements, because this accounts for the "extra" energy taken by

the fields in oscillation, and that apparent power measurements

exceed the true power as heat loss because of this. If we think like

that, we end up wondering how more energy can be going into the the

resonant fields exchanging energy, when no portion of energy is

allocated for that situation because VI=I^2R in ideal resonance. You

have answered this question very well with the realization that the

net higher amounts of energy transfer found in resonance actually

consists of that made by that "borrowed and returned" net zero

energy transfer of the source.

What this then implies to me is that while we have always considered

a reactance a case where energy is borrowed and returned from the

source, leading to the observation that VI > I^2R, the implication

that this ceases to happen in resonance is wrong!

Essentially in this correct understanding, in resonance we are only

enabling the reactive components to borrow and return energy Q times

the former levels. VI can = I^2R at resonance, but this does not

cancel the energy going into the magnetic and electric fields,

instead it maximizes it, where for very large L values generally the

amount of energy transfer at resonance can exceed the actual perfect

resonance energy transfer as heat.

Paul has noted that for a capacitive reactance the energy transfer in

VARS is given as 2 pi * E(rms)* f

But logically I see the actual energy transfer as 2*f*(CV^2), where

(CV^2) is the correction of the normal Vrms energy storage as .5CV^2,

for the peak voltage value which will contain twice the energy,

although only 40% more voltage, but because of the exponential term

V^2, this is so. So my estimate is pi times less (ACTUAL) energy

transfer than what the VAR input indicates, and again this is easily

possible since the actual energy transfer can be ANY quantity, and

that this is negotiated to be within practical parameters by how the

acting Q will differ from the real predicted Q actions, and it is

this predicted Q that determines the ratio between the apparent power

and that of the real. So by X(L)/R, since concievably this can be

made to any ratio, then any ratio of energy transfer vs power inputed

can then be obtained,(within practical limits).

To see whether this is actually so let us find the condition where

the VAR input actually equals the energy transfer,(according to my

hypothesis). Since we know that by my definition the capacitive

reactance is only allowing a current that shows pi times less energy

transfer than the VAR readings indicate, again to show what Paul has

done is the following;

First he has taken the formula for capacitive reactance X(C)= 1/(2

pi*f*C),

Where Z= X(C), and solved for I(rms)

Since V=IZ, I = V/Z = V(rms)/X(C) = V(rms)*2 pi*f*C

or,

I(rms) =V(rms)*2 pi*f*C

which is a nice formula for determining the current in a capacity

without having to resort to finding the capacitive reactance of the

capacity to determine the conduction at the impressed voltage.

Since apparent power = V(rms)*I(rms), by replacing I(rms) by the

above substitution we have;

Apparent Power = V(rms)* V(rms)*2 pi*f*C, or V(rms)^2*2 pi*f*C, or

2 pi*f*(CV(rms)^2)

Now then this is again showing that VAR is pi times the actual

energy transfer for the capacitive reactance. Let us first find out

if this is true for an inductive reactance, where now we can specify

the inductor to have a q of pi.

Since 3.14 =X(L)/R, for a 1 ohm inductor to simplify calculations of

I^2R, we now need to find the inductance for 60 hz. X(L)= 2pi*F*L

3.14 = 6.28*60*L where L = 8.3 mH

Now we need to specify an input voltage to make comparisons of

reactive power vs real, and again convenience will allow us to use 1

volt(rms) We also know that for capacity the true energy storage is

not .5CV^2 but CV^2 to account for Vpeak, so the same routine should

be employed to account for the energy storage of .5LI^2 becoming

double at I(peak). Thus the energy to be moved 120 times per second

becomes LI^2. Now we need to find Z for the conduction at 1 volt.

Z = sq rt{ 3.14^2*1^2} = 3.295, where the inverse of this becomes the

conduction of .303 A at 1 volt

We then have a VAR of .303

LI^2 =.0083*(.303)^2 = 7.64 *10^-4, where moving this 2 f times

is .0917 joules/sec, and moving it 2 pi f times is .288 joules/sec,

clearly under the VAR input of .303

Now let us find the real power input by I^2R as .0918 watts, so that

3.3 times the real power is shown by the VI of the apparent.

Apparently,(no pun intended!), for a inductor with a q of pi, this

will cause the the I^2R heating losses to be approximately

equivalent to the energy transfer, since R=1 and also (L)* 2f = .0083

* 120 = .996 If we then selected the inductor to have a higher q

than pi, say 2 pi for example, in one example this would cause R to

be reduced to .5 ohms, so for the VAR input, the contribution of

resistance to the impedance Z would be minimalized, and Z becomes

closer to the sq rt of Pi squared, or pi itself, or the value of X(L)

itself. We are not changing the inductive value to gain an increased

q, only the resistance of that inductor is being changed, so while R

is varied to lower resistances to gain a higher Q for the coil, the

inductance L or inductive reactance X(L) remains unchanged. Since the

conduction of the reactance determines the I in the VI apparent

power, as the resistance is minimized, the current is not maximized

and only approaches the highest possible value of 1 volt across pi

ohms of impedance, or it approaches a current of the inverse of pi

or pi^-1 = .318 A Thus as the resistance is lowered the apparent

power input approaches .318 VAR, but the I^2R quantity as real power

does get lowered proportionally to the decrease in resistance. Thus

the ratio of energy transfer vs heat losses was equal for a inductor

with a q of pi, so for an inductor with a q of 2 pi, the ratio of

energy transfer to heat loss must also ~ double. Thus we now have a

reliable mechanism to estimate the ratio of energy transfer to that

of heat losses, by making a ratio of the coils Q factor to that of Pi.

Now let us return to the case of X(C), where the energy transfer has

according to my hypothesis shown itself to be pi times less then the

VAR input. It seems easy to speculate that by giving X(C) a

cancelling reactance X(L): then the VAR input will change on account

of the rising I as a consequence of ohms law conduction value being

made, but V input remains identical so the left side of the equation

VI =2f(CV^2) will have its value increased Q or pi fold. However the

right term (CV^2) will have its value increased exponentially pi

squared, leading to the observation that the energy transfer should

now equal the VAR input, since originally it was thought to be pi

times less then that inputed, and the situation has been corrected by

now allowing pi times more multiplication for the right term in the

expression.

Thus for the original inductor of 1 ohm with 1 volt impressed

resulting in 1 amp conduction at resonance we have a simple VAR

input of 1, also equalling the heat losses of I^2R, another

tripartide unity of triple ones, and if the inductor has a Q of Pi,

the energy transfer is then 2f*C*pi^2. To find out if this also

equals one we must assign a value to C given these circumstances of

using a 8.3 mh inductor of 1 ohm @ 60 hz resonance.

Since both the conduction and impressed voltage are known, we can try

Paul's derived formula to find the unknown C, where I(rms) =V(rms)*2

pi*f*C

1= 1*6.28*60*C, C =(6.28*60)^-1 = 2654 uf, or .00265 F, thus in

actuality we CANNOT use that formula to determine an unknown capacity

in resonance, we can only use it to find an unknown capacity in

actual reactance current tests.

Finding X(L) for 8.3 mh; this has already been noted to be Pi, thus

we must also make an X(C)= Pi

Pi =1/2 Pi*f*C , Pi squared=1/2fC

2fC= 1/pi^2, C=1/2f pi^2= 8.44 *10^-4 Farad

By using Thompson Resonance formula

R(F)= 1/(2 pi*sq rt {LC}) we find that the needed C value is 848 uf

where the energy storage is then .0083 joules which transfering 120

times /sec yeilding 1 watt of energy transfer, thus at resonance with

an inductor of pi q factor the apparent power, the real power. and

the energy transfer all equal one watt.

To find the situation where the energy transfer is twice as great as

the ohmic losses, and also the apparent power, we need only make the

inductor have a q of 2 pi. The amperage term in VI will be doubled by

using half the resistance, for a new q of 2 pi. Formerly the energy

transfer was equal to the apparent power with an inductor with a q of

pi, so with twice the q factor, and twice the resonant rise of

voltage, we have 4 times the energy storage by V^2, and also I^2 in

each of the terms CV^2 and LI^2 as those field energies in

transition. Since the left term became doubled, but the right term

quadrupled this then indicates twice the energy in transition than is

being inputed. Thus it seems to make better sense to assume that the

VAR input can enable any energy transfer, according to the q of the

inductor being used.

Sincerely HDN