| [Subsidiary
discussion.]
Make up the 40nsec piece of 50 ohm coaxial cable out of two 20nsec
lengths. Charge it to 8 volts.
The output will be a 4v pulse 80nsec long.
Now introduce a 100 ohm series resistor in the inner conductor
at the central join. The output will still be 80nsec long. [No.
Longer. IC 18.00 28july] The first 40nsec will be 4v, and the next
40nsec will be 2v. (There will be a trailing lower voltage part.)
A clue to why is Figure 17 in http://www.ivorcatt.co.uk/2_1.htm
and then Figure 14.
Ivor Catt. 1am. Tuesday 28 July 2009. [Corrected 1.30am.]
@@@@@@@@@@@@@@@
An addition;
If however the cable is made up of three 20nsec lengths, and the
100 ohm resistor put in 20nsec from the reed relay, the cable charged
up to 8v, then the pulse will be 120nsec long. The first 40nsec
will be 4v, and the rest (80nsec) 2 volts.
The first 20nsec of output had already passed the difficult 100
ohm resistor before the reed relay closed. The next 20nsec was helped
across the 100 ohm by the end portion of westward travelling energy
current. At that point, there was no more westward travelling energy
current to help the eastwards travelling energy current across the
100 ohms, which therefore halved in voltage amplitude as per Figure
14.
Note that the 100 ohms has the same effect as the combined northwards
and southwards transmission lines in Figure 14.
[Added at 18.00 on 28july09. At t20, there begins a reflection
from the 100 ohm resistor point towards the west. It is 2v high,
and lasts for 40nsec. It reflects, and returns from the west immediately
after the end of the 4v. When passing the 100 ohms, it is again
halved to 1v forward and 1v reflected.
It seems that the final output will be 20nsec at 4v, 80nsec at
2v, 80nsec at 1v, and halving every 80nsec but going on for ever.]
. Ivor Catt. 1.30am 28 July 2009.
@@@@@@@@@@@@@@@
29 July 2009
Inquest on the appearance of an idea. I thought of
the first step, which was to probe the 40nsec cable along its length.
The problem was that the scope probe would load down the line and
interfere with the initial charging of the line, before the reed
relay closes. I wold therefore probe at only one point, and then
only at one other point, etc. I saw the probes as deleterous. A
few hours later Forrest began to think positively about the probes,
that they would modify the passing contrapuntal signal. This was
a key development of the initial idea of mine. However, it introduced
the idea of a positive, useful meddling with the 40nsec line. I
then took the next step, which was that instead of seeing the effect
on the contrapuntal signal of a probe, we see the effect of a load.
Then I realised that the ideal load was a series load, not a shunt
load to earth. - IC
@@@@@@@@@@@@@@@
What follows is only for the record, to record where the ideas
came from and when. - IC.
----- Original Message -----
From: "ivor catt" <ivorcatt@electromagnetisn.demon.co.uk>
To: "Forrest Bishop" <forrestb@ix.netcom.com>
Cc: "John Raymond Dore" <johnrdore@googlemail.com>
Sent: Monday, July 27, 2009 7:01 PM
Subject: 109 and josephson
> This (below) is completely new.
> Congratulations.
> Because of the loading of a probe, I thought in terms of only
probing at one
> point at one time. Thus, I was blocked off from the ideas below
by you.
>
> If the energy current travelling in the wrong direction attenuates
in its
> travels, perhaps this will give evidence of its direction of
travel. The
> last half of the resulting 80 nsec pulse will attenuate twice,
because it
> passes the probe twice. The first half - the rightwards travelling
energy
> current will attenuate only once. My immediate reaction is
to say that the
> final 80nsec pulse should have small steps down, the first
few steps down
> being only half of the later steps down. However, the last
few sentences
> ignored the problem of initially charging to a steady unvarying
voltage
> along its length. However, I think we can still do that, before
the relay
> closes. Perhaps in this context we would consider only two
probes, one half
> way along and another at the start of the infinitely long coax
cable.
> I need to think more about this brilliant idea.
> I think the charging source is via 1 megohm, but I have not
checked to
> brochure on the www, it is only my memory telling me that.
> I have not bothered to calculate it, but found on two websites
that
> capacitance per foot is about 20pF. Thus, if the source is
through 1 Mohm,
> then the charging time constant is 20 microseconds. Now the
time avilable
> for each cycle, at 720 Hertz, is around 1 msec, sop there is
a margin of a
> factor of 50.
>
> However, that does not deal with the loading of the inner conductor
by
> probes, potentially upsetting the charging through 1 Mohm.
I need to know
> what is the resistance (?impedance?) of the highest impedance
scope probe.
> John Dore mentioned FET probe, but they may not be fast.
>
> Incidentally, I forgot that today's oscilloscopes record a
waveform, which
> it did not do when I was active. John Dore didn't understand
my talk about
> sampling and normal 'scopes, but did not realise I had forgotten
that a
> sampling scope will record a very slow waveform, so it will
be fine. We
> definitely use a sampling scope.
>
> @@@@@@@@@@
>
> It is extraordinary that I have just discovered
> http://www.ivorcatt.co.uk/981.htm among my many papers. I did
not find the
> last bit, but what is now on http://www.ivorcatt.co.uk/981.htm
is the
> crucial part.
>
> This means that I drew Josephson into cattq in 1997, long before
Nigel Cook
> pulled him in around 2004
>
> Ivor
>
>
> ----- Original Message -----
> From: "Forrest Bishop" <forrestb@ix.netcom.com>
> To: "ivor catt" <ivorcatt@electromagnetism.demon.co.uk>
> Sent: Monday, July 27, 2009 3:17 PM
> Subject: Re: 109
>
>
>> Dear Ivor
>>
>> As a refinement, or complication, of this experiment each
probe along the
>> line removes a portion of the energy from each (west and
east) TEM wave,
>> according to the relative impedances (cf Catt on how a
voltmeter works).
>> The initial experiment, analysis, and animations should
ignore this of
>> course. A later experiment could introduce a lower impedance
probe at one
>> or more of the test stations. The resulting attenuation
of each of the TEM
>> waves would then be seen at the next adjacent stations,
to the west and to
>> the east. In this way the lower impedance probe is setting
a definite
>> identification marker for both of the waves. Each of the
probes could do
>> the same, by making their input impedances all different
from each other.
>>
>> Forrest
>>
>
----- Original Message -----
From: "ivor catt" <ivorcatt@electromagnetisn.demon.co.uk>
To: "Forrest Bishop" <forrestb@ix.netcom.com>
Cc: "John Raymond Dore" <johnrdore@googlemail.com>
Sent: Monday, July 27, 2009 7:01 PM
Subject: 109 and josephson
> This (below) is completely new.
> Congratulations.
> Because of the loading of a probe, I thought in terms of only
probing at one
> point at one time. Thus, I was blocked off from the ideas below
by you.
>
> If the energy current travelling in the wrong direction attenuates
in its
> travels, perhaps this will give evidence of its direction of
travel. The
> last half of the resulting 80 nsec pulse will attenuate twice,
because it
> passes the probe twice. The first half - the rightwards travelling
energy
> current will attenuate only once. My immediate reaction is
to say that the
> final 80nsec pulse should have small steps down, the first
few steps down
> being only half of the later steps down. However, the last
few sentences
> ignored the problem of initially charging to a steady unvarying
voltage
> along its length. However, I think we can still do that, before
the relay
> closes. Perhaps in this context we would consider only two
probes, one half
> way along and another at the start of the infinitely long coax
cable.
> I need to think more about this brilliant idea.
> I think the charging source is via 1 megohm, but I have not
checked to
> brochure on the www, it is only my memory telling me that.
> I have not bothered to calculate it, but found on two websites
that
> capacitance per foot is about 20pF. Thus, if the source is
through 1 Mohm,
> then the charging time constant is 20 microseconds. Now the
time avilable
> for each cycle, at 720 Hertz, is around 1 msec, sop there is
a margin of a
> factor of 50.
>
> However, that does not deal with the loading of the inner conductor
by
> probes, potentially upsetting the charging through 1 Mohm.
I need to know
> what is the resistance (?impedance?) of the highest impedance
scope probe.
> John Dore mentioned FET probe, but they may not be fast.
>
> Incidentally, I forgot that today's oscilloscopes record a
waveform, which
> it did not do when I was active. John Dore didn't understand
my talk about
> sampling and normal 'scopes, but did not realise I had forgotten
that a
> sampling scope will record a very slow waveform, so it will
be fine. We
> definitely use a sampling scope.
>
> @@@@@@@@@@
>
> It is extraordinary that I have just discovered
> http://www.ivorcatt.co.uk/981.htm among my many papers. I did
not find the
> last bit, but what is now on http://www.ivorcatt.co.uk/981.htm
is the
> crucial part.
>
> This means that I drew Josephson into cattq in 1997, long before
Nigel Cook
> pulled him in around 2004
>
> Ivor
>
>
> ----- Original Message -----
> From: "Forrest Bishop" <forrestb@ix.netcom.com>
> To: "ivor catt" <ivorcatt@electromagnetism.demon.co.uk>
> Sent: Monday, July 27, 2009 3:17 PM
> Subject: Re: 109
>
>
>> Dear Ivor
>>
>> As a refinement, or complication, of this experiment each
probe along the
>> line removes a portion of the energy from each (west and
east) TEM wave,
>> according to the relative impedances (cf Catt on how a
voltmeter works).
>> The initial experiment, analysis, and animations should
ignore this of
>> course. A later experiment could introduce a lower impedance
probe at one
>> or more of the test stations. The resulting attenuation
of each of the TEM
>> waves would then be seen at the next adjacent stations,
to the west and to
>> the east. In this way the lower impedance probe is setting
a definite
>> identification marker for both of the waves. Each of the
probes could do
>> the same, by making their input impedances all different
from each other.
>>
>> Forrest
>>
>
----- Original Message -----
From: "ivor catt" <ivorcatt@electromagnetisn.demon.co.uk>
To: "Forrest Bishop" <forrestb@ix.netcom.com>
Cc: "John Raymond Dore" <johnrdore@googlemail.com>
Sent: Monday, July 27, 2009 7:01 PM
Subject: 109 and josephson
> This (below) is completely new.
> Congratulations.
> Because of the loading of a probe, I thought in terms of only
probing at one
> point at one time. Thus, I was blocked off from the ideas below
by you.
>
> If the energy current travelling in the wrong direction attenuates
in its
> travels, perhaps this will give evidence of its direction of
travel. The
> last half of the resulting 80 nsec pulse will attenuate twice,
because it
> passes the probe twice. The first half - the rightwards travelling
energy
> current will attenuate only once. My immediate reaction is
to say that the
> final 80nsec pulse should have small steps down, the first
few steps down
> being only half of the later steps down. However, the last
few sentences
> ignored the problem of initially charging to a steady unvarying
voltage
> along its length. However, I think we can still do that, before
the relay
> closes. Perhaps in this context we would consider only two
probes, one half
> way along and another at the start of the infinitely long coax
cable.
> I need to think more about this brilliant idea.
> I think the charging source is via 1 megohm, but I have not
checked to
> brochure on the www, it is only my memory telling me that.
> I have not bothered to calculate it, but found on two websites
that
> capacitance per foot is about 20pF. Thus, if the source is
through 1 Mohm,
> then the charging time constant is 20 microseconds. Now the
time avilable
> for each cycle, at 720 Hertz, is around 1 msec, sop there is
a margin of a
> factor of 50.
>
> However, that does not deal with the loading of the inner conductor
by
> probes, potentially upsetting the charging through 1 Mohm.
I need to know
> what is the resistance (?impedance?) of the highest impedance
scope probe.
> John Dore mentioned FET probe, but they may not be fast.
>
> Incidentally, I forgot that today's oscilloscopes record a
waveform, which
> it did not do when I was active. John Dore didn't understand
my talk about
> sampling and normal 'scopes, but did not realise I had forgotten
that a
> sampling scope will record a very slow waveform, so it will
be fine. We
> definitely use a sampling scope.
>
> @@@@@@@@@@
>
> It is extraordinary that I have just discovered
> http://www.ivorcatt.co.uk/981.htm among my many papers. I did
not find the
> last bit, but what is now on http://www.ivorcatt.co.uk/981.htm
is the
> crucial part.
>
> This means that I drew Josephson into cattq in 1997, long before
Nigel Cook
> pulled him in around 2004
>
> Ivor
>
>
> ----- Original Message -----
> From: "Forrest Bishop" <forrestb@ix.netcom.com>
> To: "ivor catt" <ivorcatt@electromagnetism.demon.co.uk>
> Sent: Monday, July 27, 2009 3:17 PM
> Subject: Re: 109
>
>
>> Dear Ivor
>>
>> As a refinement, or complication, of this experiment each
probe along the
>> line removes a portion of the energy from each (west and
east) TEM wave,
>> according to the relative impedances (cf Catt on how a
voltmeter works).
>> The initial experiment, analysis, and animations should
ignore this of
>> course. A later experiment could introduce a lower impedance
probe at one
>> or more of the test stations. The resulting attenuation
of each of the TEM
>> waves would then be seen at the next adjacent stations,
to the west and to
>> the east. In this way the lower impedance probe is setting
a definite
>> identification marker for both of the waves. Each of the
probes could do
>> the same, by making their input impedances all different
from each other.
>>
>> Forrest
>>
>
----- Original Message -----
From: "Forrest Bishop" <forrestb@ix.netcom.com>
To: "ivor catt" <ivorcatt@electromagnetism.demon.co.uk>
Sent: Monday, July 27, 2009 3:17 PM
Subject: Re: 109
> Dear Ivor
>
> As a refinement, or complication, of this experiment each probe
along the line removes a portion of the energy from each (west and
east) TEM wave, according to the relative impedances (cf Catt on
how a voltmeter works). The initial experiment, analysis, and animations
should ignore this of course. A later experiment could introduce
a lower impedance probe at one or more of the test stations. The
resulting attenuation of each of the TEM waves would then be seen
at the next adjacent stations, to the west and to the east. In this
way the lower impedance probe is setting a definite identification
marker for both of the waves. Each of the probes could do the same,
by making their input impedances all different from each other.
>
> Forrest
>
> -----Original Message-----
>>From: ivor catt <ivorcatt@electromagnetisn.demon.co.uk>
>>Sent: Jul 26, 2009 6:01 PM
>>To: Forrest Bishop <forrestb@ix.netcom.com>
>>Subject: Re: 109
>>
>>Do animations in the way you see best.
>>We are here all September, and after the second half of
October.
>>John is keeping the 109 at his house. I don't see any point
in shipping it
>>before we know where the experiment will be done.
>>I have just added more to http://www.ivorcatt.co.uk/965.htm
. See
>>http://www.ivorcatt.co.uk/9659.jpg
>>I have tried to block in in colour the oscilloscope waveforms
>>http://www.ivorcatt.co.uk/9653.jpg to show where Catt says
each segment
>>goes, and find the task terribly confusing. I will try some
more.
>>"the Cube project (this looks to be coming to fruition
at last)." - FB
>>I am very glad to hear this because you have been so heavily
invested in it.
>>Ivor
>>
>>
>>----- Original Message -----
>>From: "Forrest Bishop" <forrestb@ix.netcom.com>
>>To: "ivor catt" <ivorcatt@electromagnetism.demon.co.uk>
>>Sent: Sunday, July 26, 2009 4:26 PM
>>Subject: Re: 109
>>
>>
>>> Dear Ivor,
>>>
>>> I also agree the experiment is significant. It is one
of the first I
>>> would like to do if I ever finish the Cube project
(this looks to be
>>> coming to fruition at last).
>>>
>>> The animations I already made do show this, though
in a different
>>> proportion than the 1:4 asked for. I might be able
to use that same
>>> "digital set" (called a "scene")
that I created at your house, if that is
>>> ok with you. It may be easier to reuse this scene than
to create a new
>>> one, though it took a lot of time to animate it. Or
maybe a new scene with
>>> a better animation technique would take less time.
>>>
>>> Did John Dore send you a Tektronics 109? I see from
you itinerary that
>>> you are not around much the rest of the summer. I'll
fly over at some
>>> point, depending partly on how these Cubes tun out.
>>>
>>>
>>> Forrest
>>>
>>> -----Original Message-----
>>>>From: ivor catt <ivorcatt@electromagnetisn.demon.co.uk>
>>>>Sent: Jul 26, 2009 5:35 AM
>>>>To: Forrest Bishop <forrestb@ix.netcom.com>
>>>>Subject: 109
>>>>
>>>>Three parties, Walton, Dore and Foggitt, think the
proposed experiment is
>>>>very significant.
>>>>
>>>>I point you to http://www.ivorcatt.co.uk/965.htm
>>>>It hyperlinks to http://www.ivorcatt.co.uk/9653.jpg
and
>>>>http://www.ivorcatt.co.uk/9656.jpg
>>>>It would be very valuable if you created an animation
based on these
>>>>pictures.
>>>>
>>>>Assume 40nsec cable charged up to 10. I plan to
make and upload further
>>>>pictures illustrating the contrapuntal model, giving
each of the eight 5v
>>>>(half size) 10 nsec long portions a different colour
and showing their
>>>>progress. The first graph will be at t-20, then
t-10, then on to t0. This
>>>>will indicate graphically that before the switch
closed, there was already
>>>>movement.
>>>>
>>>>None of Walton, Dore and Foggitt can see how any
other model than the
>>>>contrapuntal can explain the waveforms. I think
it's a winner.
>>>>
>>>>Ivor
.
.
|