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Measurements using field probe
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From: D.C. Cox [SMTP:DR.RESONANCE-at-next-wave-dot-net]
Sent: Monday, June 29, 1998 8:08 AM
To: Tesla List
Subject: Re: Measurements using field probe
to: Bert
It would not be reasonable to assume the 144 Ohm resistance for the 420 BPS
of operation. This value would only be true for the first few firings of
the cap as the "hot" resistance of the bulb is considerably less than its
initial "cold" measured DC resistance. After a few seconds this value
would be lower.
DR.RESONANCE-at-next-wave-dot-net
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> From: Tesla List <tesla-at-pupman-dot-com>
> To: 'Tesla List' <tesla-at-pupman-dot-com>
> Subject: Measurements using field probe
> Date: Sunday, June 28, 1998 10:20 PM
>
>
> ----------
> From: Bert Hickman [SMTP:bert.hickman-at-aquila-dot-com]
> Sent: Sunday, June 28, 1998 1:36 PM
> To: Tesla List
> Subject: Re: Measurements using field probe
>
> Tesla List wrote:
> >
> > ----------
> > From: terryf-at-verinet-dot-com [SMTP:terryf-at-verinet-dot-com]
> > Sent: Saturday, June 27, 1998 12:34 PM
> > To: Tesla List
> > Subject: Re: Measurements using field probe
> >
> > Hi Bert and All,
> >
> > Since this has become a public post I will answer it here.
There is a lot
> > of good science in this many will be interested in. My comments are at
the
> > end regarding the light bulb test.
> >
> <SNIP>
>
> > I believe the 1000 amp spike level is far too high.
> >
> > A 144 ohm filament (100 Watts 120 Volt) with 1000 amps going through it
for
> > 1.5 uS will dissipate I^2Rt joules. That works out to 216 joules per
arc.
> > At 420 BPS we would dissipate 97720 watts in the bulb!!
> >
>
> Excellent points! This morning I remeasured the quartz halogen lamp
> under the 400 mA average current condition and found that it developed
> about 28 volts across the filament. So the resistance at the filament's
> operating temperature in the experiment is around 70 ohms. However, the
> tungsten filament's R versus T function is very non-linear, with a very
> high positive temperature coefficient - the "cold" resistance is only
> about 13 ohms or less than 1/10 of its fully-lit resistance. 28 Volts at
> 0.4A implies that an "average" of about 11.2 watts was being dissipated
> in the lamp, and the remainder in the power arc itself.
>
> > Also, the top toroid is a capacitor. The charge on a capacitor is
equal to
> > the capacitance multiplied by the voltage (Q = C x V). The charge is
also
> > equal to the current multiplied by the time. (Q = I x t). For 1000 amp
and
> > 1.5 uS the charge would be 1500 uCoul. For a toroid of say 50pF the
voltage
> > works out to 30 million volts!! That's too high! But a good try :-))
>
> Clearly something was out of whack! My coil runs pretty well... but not
> THAT well! 1000 amps for the entire 1.5 uS period would clearly be too
> high. :^) But could we have large current peaks for a much shorter
> period? Revised calculations to follow...
>
> >
> > The problem was introduced when you assumed the following:
> >
> > "I then connected the same bulb to a variac
> > and an AC ammeter to estimate the average current necessary to light
the
> > bulb to an equivalent degree. This level was reached at about 400 MA,
> > implying that the lamp was seeing an "average" current level of about
> > 400 MA"
> >
> > Actually the power was equivalent not the current. The power is
> > proportional but current is a squared function. This threw everything
off
> > by a squared function.
> >
>
> Aha!! Excellent point! I was attempting to duplicate the function of a
> hot-wire ammeter in the experiment, but didn't look at the power
> relationship! :^(
>
> > If we go back and redo the previous calculations with the square root
of
> > current ( SQRT(1000) = 31.67 ) we get the following.
> >
> > A 144 ohm filament with 31.62 amps going through it for 1.5 uS will
> > dissipate I^2Rt joules. That works out to 0.216 joules per arc. At
420 BPS
> > we would dissipate 90.72 watts in the bulb - Much more believable!
>
> With Rfil of 70 ohms, an measured average power of 11.2 watts in lamp,
> and a breakrate of 420 BPS, this is about 0.0266 Joules/bang being
> dissipated in the lamp. Solving for I if this is done in 1.5 uSec, we
> get a current of about 16 Amps. However, it's more likely that most of
> this current flows for only 1-2 time constants, or about 600 nSec.
> Solving for the current, we get an average current of about 28 Amps.
> Much more reasonable that 1000 Amps (in hindsight)...
>
> >
> > Also, the top toroid is a capacitor. The charge on a capacitor is
equal to
> > the capacitance multiplied by the voltage (Q = C x V). The charge is
also
> > equal to the current multiplied by the time. (Q = I x t). For 31.62
amp and
> > 1.5 uS the charge would be 47.43 uCoul. For a toroid of say 50pF the
> > voltage works out to 950 kilovolts - A much better number.
>
> This is probably too high an output voltage for my 10" coil. I estimate
> that my tank circuit provides about 3.7 joules/bang (0.0205 uF at 19 KV
> Vgap). At 420 BPS, the system is delivering about 1550 total watts of
> average primary power. Most of this energy, about 80%, or about 3
> Joules/bang, appears to be making it to the secondary:toroid based upon
> independent measurements of the amplitudes of successive pri-sec energy
> interchanges. The combined topload and secondary self-C of this system
> is about 43 pF. This implies a peak voltage of about 370 kV at the
> toroid - which seems to be in the ballpark for the 60-65" streamers that
> this system produces.
>
> Now all of the secondary energy is observed to be dissipated within 1.5
> uSec (about 5 time constants) during a power arc. In fact, if the
> discharge has a decaying exponential envelope [or mostly linear?], more
> than 98% of this energy will have been dissipated in about 2 time
> constants, or around 600 nSec. At 420 BPS, this implies that the RF
> power that is used to "incinerate the air" amounts to about 1260 watts.
> During power arcs we know that the lamp will be dissipating an average
> of about 11.2 watts across 70 ohms, or slightly less than 1% of the
> total power. The remaining 99% of the power must be dissipated in the
> arc. Assuming that the current is the same for both (whatever it is...),
> the implied average Rarc must be about 100X that of the filament, or
> about 7000 Ohms. Also, doing a "ballpark" sanity check, 370 kV/7000 = 52
> Amps - not unreasonable.
>
> >
> > Also the top terminal capacitance and the arc form a RC network that
should
> > discharge in 5 RC time constants. This would imply a resistance of:
> >
> > 1.5 uS = 5 x R x C
> >
> > With say a 50 pF top terminal the resistance implied is 6000 ohms!
This is
> > comparable to the values I see of ~2000 ohms for arc current! I think
we
> > are getting some interesting correlation here!!
>
> I agree! Based upon the observed time constant of about 300nSec and 43
> pF of total C, I also get an "average" Rarc of about 7000 Ohms. Note
> that this matches up with the power estimates and implied Rarc
> calculated above.
>
> Terry, it certainly sounds like we're in the same ballpark! I also seem
> to remember Greg Leyh doing some measurements of high-current discharges
> which implied arc resistance in the kilo-ohm range...
>
> >
> > Note that 950 kV / 6000 ohms gives a current of 158 amps. Higher
than
> > 31.67 and lower than 1000. The linear decrement and the guesses I have
made
> > have introduced some error. However, I believe the principles are
sound.
>
> I agree! At 950 kV, the terminal voltage is probably too high, but with
> the adjusted numbers the results for Rarc are still amazingly close!
>
> > We are at least in the hundreds of percent. Usually this stuff has
been
> > many orders of magnitude off in the past!!! We are getting much
better!!
> >
> > To make a long story short... just remember that the light bulb
illuminates
> > in proportion to the current squared and all should work out reasonably
well.
> >
> > The fact that we independently are getting resistance values for arcs
to
> > ground in the 1 to 10 kiloohm range is very significant! I should
point out
> > that without the antenna being properly terminated, the antenna and
cable
> > will ring from the impulse of an arc. This false ring may make the
> > decrement look about 250% longer than it really is. If this affected
your
> > measurements, the real numbers would work out very accurately to what I
> > would think is really going on. My plane wave antennas had this
problem in
> > the beginning. Thus, all the coax-matching resistors were added to it.
> >
> > Thanks for sharing this neat experiment. Hopefully it can now be well
> > understood what it was showing. It is wonderful that we are know
seeing the
> > first bits of good data coming in regarding output arcs. Once such
things
> > are understood, we can make design changes to really push the arcs. I
am
> > intrigued by a scope graph I have taped to my computer that shows the
burst
> > during an arc producing 3.5 times the original secondary voltage. If
this
> > effect can be understood and optimized...........
> >
> > All the best
> >
> > Terry Fritz
> >
>
> Thanks for pointing out the problems in the earlier calculations.
> Interesting comments about the ringdown durations versus antenna
> termination and the waveform showing 3.5X Vsec... Something to look at
> in a future experiment when I get some time. It's really too bad we have
> to work for a living - doing this kind of research & theory is really
> fun!
>
> Safe coilin' to you!
>
> -- Bert --
>