Theory of LTR Discussion from the Tesla List
Updated: Sept. 4, 2006
Date: Fri, 11 Nov 2005 10:42:43 -0700
From: Tesla list <tesla@pupman.com>
To: tesla@pupman.com
Subject: Theory of LTR
Original poster: "Bob (R.A.)
Jones" <a1accounting@bellsouth.net>
Hi all,
I recently tried a different direction on the theoretical the optimum primary C
(Cp) for a given inductive ballast (L) The maths is not finished but the
direction appears productive. Here is the short word version of it, minus many
of the assumptions, for those into the theory stuff. First in a sync gap
operating at the same break rate as twice the supply frequency. The Cp and its
repeated discharge is equivalent to a square wave signal in series with Cp but
without the SG. The amplitude of the square wave is equal to the voltage on Cp
at the point of discharge and has the same phase as the discharge but opposite
polarity. Considering only the fundamental of the square wave, the square wave
lags the voltage on Cp but with the opposite polarity so equivalently it leads
the voltage. Hence the combination of Cp and squarewave generator has an
impedance (at the supply frequency) equal to a smaller C (Cequ) in parallel with
a R. (Requ)
The energy dissipated in Requ is equal to the bang energy. Therefore maximum
dissipation in Requ will be when the impedance of Cequ is equal to the impedance
of the ballast inductor. i.e. resonant. As Cequ is smaller than Cp, Cp must be
increased until its equivalent Cequ is resonant with L to obtain the maximum
power in Requ which is the maximum bang size.
I am guessing but it will be similar with a static gap. The repeated discharge
of Cp advances Cp voltage relative to its charge current so again the equivalent
impedance (at the supply frequency) is equal to a smaller capacitor and so the
actual C must be made large to obtain the maximum bang size. The above may also
explain why the maximum bang size is obtained when the current in L is not zero.
ie when the real current is a maximum there must still be some reactive current.
Robert (R. A.) Jones
Date: Mon, 14 Nov 2005 11:01:01 -0700
From: Tesla list <tesla@pupman.com>
To: tesla@pupman.com
Subject: Re: Theory of LTR
Original poster: "Dmitry (father
dest)" <dest@himplast.ru>
> Original poster: "Bob (R.A.) Jones"
<a1accounting@bellsouth.net>
> Hi all,
> I recently tried a different direction on the
theoretical the optimum
> primary C (Cp) for a given inductive ballast (L)
is it only for nst using case? coz i for example choose the ballast for the Cp
and the current i need, not vice-versa.
-----
The solution to no primary hits lay in getting rid of the primary! This is no
joke either.
20-06-96 (c) Richard Hull, TCBOR
Date: Mon, 14 Nov 2005 11:05:44 -0700
From: Tesla list <tesla@pupman.com>
To: tesla@pupman.com
Subject: Re: Theory of LTR
Original poster: "Bob (R.A.)
Jones" <a1accounting@bellsouth.net>
Hi all,
I am reasonable confident I got it right. Here is the latest. It much simpler
to consider the equivalent circuit of the SG and Cp as series Requ and Cp, where
Cp is unchanged and Requ is the product of the impedance of Cp at the supply
frequency, the sin of the firing angle relative to the charging current and the
ratio of fundamental to peak of a square wave. The power dissipated in R per
cycle is the bang energy. Note that the harmonics of the square wave are ignored
but this produces only an approximately 5% error in firing voltage because of
the amplitudes of the harmonics and the high impedance (relative to the supply
frequency) of the L at their frequency.
It relatively simply (using AC circuit theory) predicts the optimum firing
angle, C value for max bang size, bang size and PF in the synchronous SG case.
It theoretically confirms the experimental and circuit simulation results that
an LTR C has the biggest bang energy.
Robert (R. A.) Jones
----- Original Message -----
From: "Tesla list" <tesla@pupman.com>
To: <tesla@pupman.com>
Sent: Friday, November 11, 2005 9:42 AM
Subject: Theory of LTR
> Original poster: "Bob (R.A.) Jones"
<a1accounting@bellsouth.net>
>
> Hi all,
>
> I recently tried a different direction on the
theoretical the optimum primary C (Cp) for a given inductive ballast >(L) >
The maths is not finished but the direction appears productive. Here is
the
> short word version of it, minus many of the
assumptions, for those into the
> theory stuff.
> First in a sync gap operating at the same break
rate as twice the supply frequency.
> The Cp and its repeated discharge is equivalent
to a square wave signal in
> series with Cp but without the SG.
> The amplitude of the square wave is equal to the
voltage on Cp at the point
> of discharge and has the same phase as the
discharge but opposite polarity.
> Considering only the fundamental of the square
wave, the square wave lags
> the voltage on Cp but with the opposite
polarity so equivalently it leads the voltage.
> Hence the combination of Cp and squarewave
generator has an impedance (at
> the supply frequency) equal to a smaller C (Cequ)
in parallel with a R. (Requ)
> The energy dissipated in Requ is equal to the
bang energy.
> Therefore maximum dissipation in Requ will be
when the impedance of Cequ is
> equal to the impedance of the ballast inductor.
i.e. resonant.
> As Cequ is smaller than Cp, Cp must be increased
until its equivalent Cequ
> is resonant with L to obtain the maximum power
in Requ which is the maximum
> bang size.
>
> I am guessing but it will be similar with a
static gap. The repeated
> discharge of Cp advances Cp voltage relative to
its charge current so again
> the equivalent impedance (at the supply
frequency) is equal to a smaller
> capacitor and so the actual C must be made large
to obtain the maximum bang
size.
> The above may also explain why the maximum bang
size is obtained when the
> current in L is not zero. ie when the real
current is a maximum there must
> still be some reactive current.
> Robert (R. A.) Jones
Date: Mon, 14 Nov 2005 12:33:59 -0700
From: Tesla list <tesla@pupman.com>
To: tesla@pupman.com
Subject: Re: Theory of LTR
Original
poster: "Bob (R.A.) Jones" <a1accounting@bellsouth.net>
Hi Richard,
The equivalent circuit is just for the SG and C. I used it with an NST to
determine the C for the biggest bang.
But its equally applicable to any ballast including inductive and
inductive/resistive combination or primary ballast. It just requires that you
can define the ballast impedance (constant impedance) so you can put that
impedance in series with the equivalent circuit and do the ac analysis on it.
The theory is applicable to a sync gap. But as a static gap has similar
parameters in a LTR, STR sense you can use it for that as well.
Robert (R. A.) Jones
----- Original Message -----
From: "Tesla list" <tesla@pupman.com>
To: <tesla@pupman.com>
Sent: Monday, November 14, 2005 10:01 AM
Subject: Re: Theory of LTR
> Original poster: "Dmitry (father dest)" <dest@himplast.ru>
>
>
> > Original poster: "Bob (R.A.) Jones"
<a1accounting@bellsouth.net>
>
> > Hi all,
>
> > I recently tried a different direction on the
theoretical the optimum
> > primary C (Cp) for a given inductive ballast
(L)
>
> is it only for nst using case? coz i for example
choose the ballast
> for the Cp and the current i need, not
vice-versa.
>
> -----
> The solution to no primary hits lay in getting
rid of the primary!
> This is no joke either.
> 20-06-96 (c) Richard Hull, TCBOR
Date: Mon, 14 Nov 2005 21:37:26 -0700
From: Tesla list <tesla@pupman.com>
To: tesla@pupman.com
Subject: Re: Theory of LTR
Original poster: "Gerry
Reynolds" <gerryreynolds@earthlink.net>
Hi Bob,
How did your results compare to the "standard" 2.8*Cres recommendation for LTR
value when using SRSG (at 120pps).
Gerry R.
>Original poster: "Bob (R.A.) Jones"
<a1accounting@bellsouth.net>
>
>Hi Richard,
>
>The equivalent circuit is just for the SG and C.
I used it with an NST to
>determine the C for the biggest bang.
>
>But its equally applicable to any ballast
including inductive and
>inductive/resistive combination or primary
ballast.
>It just requires that you can define the ballast
impedance (constant
>impedance) so you can put that impedance in
series with the equivalent
>circuit and do the ac analysis on it.
>The theory is applicable to a sync gap. But as a
static gap has similar
>parameters in a LTR, STR sense you can use it for
that as well.
>
>Robert (R. A.) Jones
>A1 Accounting, Inc., Fl
>407 649 6400
>----- Original Message -----
>From: "Tesla list" <tesla@pupman.com>
>To: <tesla@pupman.com>
>Sent: Monday, November 14, 2005 10:01 AM
>Subject: Re: Theory of LTR
>
>
> > Original poster: "Dmitry (father dest)" <dest@himplast.ru>
> >
> >
> > > Original poster: "Bob (R.A.) Jones"
<a1accounting@bellsouth.net>
> >
> > > Hi all,
> >
> > > I recently tried a different direction on
the theoretical the optimum
> > > primary C (Cp) for a given inductive
ballast (L)
> >
> > is it only for nst using case? coz i for
example choose the ballast
> > for the Cp and the current i need, not
vice-versa.
> >
> > -----
> > The solution to no primary hits lay in getting
rid of the primary!
> > This is no joke either.
> > 20-06-96 (c) Richard Hull, TCBOR
Date: Mon, 14 Nov 2005 21:48:41 -0700
From: Tesla list <tesla@pupman.com>
To: tesla@pupman.com
Subject: Re: Theory of LTR
Original poster: Harvey Norris <harvich@yahoo.com>
--- Tesla list <tesla@pupman.com> wrote:
> It just requires that you can define the ballast
> impedance (constant
> impedance) so you can put that impedance in
series
> with the equivalent
> circuit and do the ac analysis on it.
Of quite great interest also is the ability of a field regulated alternator to
supply power. In a certain way the field regulated alternator, ( as a control
factor measure) also represents a "ballasted" supply when hooked to the pole pig
transformer. In fact at lower regulation values it may be safe to short out the
stator outputs to determine how much availability of amperage that source can
supply. Suppose the short reveals a 10 AMP supply at the specified open circuit
voltage. Now suppose that we then use a capacitive reactance that should deliver
the same 10 AMPS at the noted open circuit voltage. If fact what happens is that
excess of 10 amps is drawn under those conditions, because the chosen capacitive
reactance begins series resonating with the internal Z(int) value of the stator
windings. Thus in actuality the method of using a short to the supply to
determine maximum available amperage from the same supply is invalid; when using
matched opposite reactance values.
The objection that when NST's are used in this same kind of reactance
matching; where that objection
to overvolting the NST transformer is countered by using Larger The Resonant:
LTR cap values as a load,
would seem to be ruled unnecessary in the case of a well insulated pole pig to
achieve the same objective.
It may well be that "identical to resonant cap loads" or ITR values could then
be successively employed with
pole pigs themselves powered by the (properly) field regulated alternator. If we
look at the Z(int) value of the pole pig itself being a perfect reflector that
would cause its capacitive secondary load to achieve its same primary
consumptive value as the secondary predictions would make; this is a serious
error.
Predictions mean nothing when practical real world demonstrations show different
results. The inductive reactance of an open secondary pole pig primary amperage
measurement is SUPPOSED to be a linear relationship with regard to frequency
input. The quantity of inductive reactance present in the pole pig primary
measured by open secondary measurement at 60 hz is Supposed to be linear/
whereby 8 times the frequency input that should result in 8 times the measured
inductive reactance is entirely untrue. Likewise the same linear supposition for
large inductive reactive loads being linear is likewise entirely untrue. The
relationship is quite more unlinear then people would assume by book learning.
But assumptions themselves can be deceiving. If a 30 ma rated short NST is
driven at 8 times the pole
pig imparted frequency by a smaller voltage inputed alternator, the shorted NST
output THEN does show the
same linear relationship, in that formerely at 60 hz I shows 30 ma on short,
with 120 volt input/ but now it
shows 30 ma/8 currents at short.( with the same 120 volt input @ 480 hz) In
contrast however the same
(resonant)capacity chosen as the pole pig load may provide a significant voltage
rise.
These things may sound confusing
however certain non- linearities are present when assuming X(L) to be
a constant linear increased value with increase of frequency input. There is far
more here to the eye then would initially appear.
Sincerely HDN