Re: Sync Gap Timing (phase angle considerations)

["Tesla list" <tesla-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:

 > >This has been asked before, but it keeps coming up,
 > and always bugs me:
 > >Do you want the gap to fire at the AC peaks and
 > troughs, or at the zero
 > >crossings?
Look at the components of a tesla primary. It consists
of a relatively low inductive reactance L, where X(L)=
2pi*F*L and a very high capacitive reactance in C.
Capacitive reactance is made by the formula
X(C)=1/(2pi*F*C) Reactance is measured in ohms. These
equations use henries for L and Farads for C. Since C
is typically a very low number as expressed in Farads,
it being in the denominator means that X(C) will be a
very high number value in ohms of reactance, and since
X(C)>> X(L) -at- 60 hz, the fact that L and C are in
series implies that the total series arrangement is
predominantly capacitive reactance at 60 hz, so we can
treat it simply as a capacity.

With an inductive reactance it is commonly stated that
the resultant amperage lags the impressed voltage by
90 degrees on the AC cycle. This may not be entirely
true as it should lag the impressed voltage by the
phase angle that inductor makes by plotting out its
resistance and its inductive reactance with y and x
axis', but here we are only concerned with how the
amperage in time from the displacement current on the
capacity will act with respect to its voltage input.
It does the opposite effect as the inductor, so it is
said that the voltage across the cap will LEAD the
impressed voltage by ninety degrees. We can ask the
question, when in  the time of the AC current through
a capacity  does the capacity contain the greatest
voltage? Obviously in this case the greatest voltage
across the cap concurrently occurs with the point when
greatest amperage is occuring across the cap. But IT
IS THAT AMPERAGE WITH RESPECT TO THE INPUT VOLTAGE
THAT IS ITSELF 90 degrees out of phase with the input.

Jackson in "Introduction to ELECTRIC CIRCUITS" notes
the following; pg 334..

the alternating current "through" a pure capacitance
leads the voltage across it by 90 degrees.

Now for this case a special further statement can be
made. Since we are equating  in time the voltage
stored on the capacitor with the amperage "through"
the capacitor, and the amperage through that capacitor
is also 90 degrees out of phase with the input
voltage, we can also state the obvious; The voltage
across the capacitor is 90 degrees out of phase with
the source AC voltage.

What this means is that when the arc gaps are brought
together, the voltage on the capacitor plates will
naturally wish to fire when they contain the most
voltage, and with respect to the source AC voltage
cycle that will be near its zero crossing point, since
these relationships are already 90 degrees out of
phase.

All of this can also easily be realized by simple
argument, whereby it should be realized that in the
first half of the source frequency AC cycle, (when it
is now making its zero crossing point), the cap has
been charged in one polarity , and if a gap is then
available for disharge, it would happen in that time
period. If no gap were available it is in the second
half of the source frequency AC cycle where the stored
energy is returned to the source. Jackson also notes;

inductance and capacitance alternately store energy
during the charging quarter cycle, only to return this
energy to the circuit during the next (discharging)
quarter cycle.

The potential confusion that may develope with that
statement is that the quarter cycles being refered to
are NOT the quarter cycles of the source voltage AC,
but the ACTUAL AC on the component itself acting in 90
degree phased time with respect to the source. So it
is very obvious that (without the complication of
having a gap and consequent quenching considerations)
after the zero crossing point of the input source AC,
the capacitor starts returning energy as the input
polarity has changed, but with respect to the AC
voltage itself on that capacitor, this is the portion
of time between the first and second quarter cycle.

Let us hope the textbook has cleared up any potential
confusion on this matter...

However there is one more important point to be dealt
with where Jackson ALSO states...
the current through a PURE inductor lags the voltage
across it by 90 degrees.
The very evident problem with this statement is that
"pure" inductances in the real world dont exist,
unless we are dealing with a superconductor of no
resistance. We have to actually form a phase angle
using the resistance of the inductor to find out how
far the amperage will lag the voltage source. What
this actually means is that in the real world with
most values of inductance at 60 hz, if the inductance
is not a very high value, the phase angle lag would
actually be MUCH smaller than 90 degrees. This is
quoted from;
HW Jackson on AC Power/Reactive vs Real Phase Angles.
http://groups.yahoo-dot-com/group/teslafy/message/4

Jackson now describes the condition of power in a
circuit containing (equal)resistance and
(inductive)reactance.
"If a circuit consists of equal resistance and
inductive reactance in series, the current lags the
the applied emf by 45 degrees."

"Now in this circumstance the Q of the coils, or the
ratio of X(L)/R seems to be the parameter for
determining the true phase angle."

This article also shows the procedure for taking the
arctan of Q to find the phase angle involved with an
inductor.

"Thus here in this example we can see the larger the Q
the greater the phase angle of the reactive currents.
Speaking of reactive currents being 90 degrees out of
phase with the voltage is then strictly a misnomer,
Imagined as the condition of a described IDEAL
component of zero resistance and not a REAL component
having resistance, as these calculations have shown at
60 hz it takes a VERY LARGE inductance to even become
close to a REAL 45 degree phase angle in its reactive
condition."

Actually that statement only applies for inductors of
smaller gauge wire. A quick example here can be made
here for a 500 ft spool of 14 gauge wire having 11 mh
and 1.2 ohms.
X(L) = 2 pi*F*L = 6.28*60*.011= 4.1448 ohms
Q = X(L)/R =4.1448/1.2 = 3.454

Now in trigometry the Y coordinate [X(L)] and the X
coordinate R gives a tangent as Y/X

What we need to know to find that phase angle is the
degrees necessary to have a tangent of 3.454

To easily find this on a calculator we can instead
employ the "inverse tangent" trigonometric function.

This is commonly shown as arctan, or tan^-1 on
calculators where tan^-1 (3.454) = 1.289 radians

Trigonometric functions on calculators always usually
deal in radians, where there are 2 pi radians/ 360
degrees.  Since the answer is specified in radians we
can divide by the conversion factor, which is the same
thing as multiplying by 180 degrees/pi radians* 1.289
radians = (180*1.289)/3.14 = 73.89 degree phase angle.

If the same inductor had more resistance, this phase
angle would not be so high.

Sincerely HDN