FTL evidence patent Einstein false science .................

[John Larkin]

On Tue, 14 Oct 2003 13:27:13 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Tue, 14 Oct 2003 14:52:55 +0200, "Mathew Orman" <orman@nospam.com
wrote:


20 m long, 10MBit/sec, 6ns transient.

---
Never mind the bullshit, here's what I'd like to see:

PULSE GEN
+----------+
| OUT|-------+--[50R]----[50 OHM COAX]----------+
| GND|--+ | |
+----------+ | +--[50R]----[ORMAN BSCABLE]--+ |
| | | |
| +--+------------+ | |
| | TRIG A VERT|---------------+ |
| | | | |
| | B VERT|---------------------|
| +---------------+ | |
| [50R] [50R]
| | |
GND GND GND

The pulse generator is a voltage source with the output driving 50 ohm
resistors in series with the center conductors of two _equal_ lengths of
real 50 ohm and your BSFTL cable. You say you've got a working length
of 20 meters, so that would be fine.

The output of the pulse gen also drives the trigger input of the scope,
which is located right at the output of the pulse gen. The assumption
is that the system is entirely coaxial, the intent being to determine
the difference between the arrival time of the leading edge of pulse
propagating down the real cable and the BSFTL cable.

Send me one of your 20 meter cables and I'll test it and post the
results here, including screen shots of the scope output and test setup.

John,

He has done basicly this, and posted the waveforms. The "FTL" pulse
edge does indeed appear to precede the regular coax edge by an
impressive amount. The key to the illusion is that a carefully
selected risetime, partially differentiated by an unterminated coax,
superficially seems to rise sooner. This is, of course, only an
illusion; you can get the same effect with a simple RC network.

John

 

[Mathew Orman]

"John Larkin" <jjlarkin@highSNIPlandTHIStechPLEASEnology.com> wrote in
message news:1fioov8u1bjb5uime759dpjephe33q8bqc@4ax.com...

On Tue, 14 Oct 2003 13:27:13 -0500, John Fields
jfields@austininstruments.com> wrote:

On Tue, 14 Oct 2003 14:52:55 +0200, "Mathew Orman" <orman@nospam.com
wrote:


20 m long, 10MBit/sec, 6ns transient.

---
Never mind the bullshit, here's what I'd like to see:

PULSE GEN
+----------+
| OUT|-------+--[50R]----[50 OHM COAX]----------+
| GND|--+ | |
+----------+ | +--[50R]----[ORMAN BSCABLE]--+ |
| | | |
| +--+------------+ | |
| | TRIG A VERT|---------------+ |
| | | | |
| | B VERT|---------------------|
| +---------------+ | |
| [50R] [50R]
| | |
GND GND GND

The pulse generator is a voltage source with the output driving 50 ohm
resistors in series with the center conductors of two _equal_ lengths of
real 50 ohm and your BSFTL cable. You say you've got a working length
of 20 meters, so that would be fine.

The output of the pulse gen also drives the trigger input of the scope,
which is located right at the output of the pulse gen. The assumption
is that the system is entirely coaxial, the intent being to determine
the difference between the arrival time of the leading edge of pulse
propagating down the real cable and the BSFTL cable.

Send me one of your 20 meter cables and I'll test it and post the
results here, including screen shots of the scope output and test setup.

John,

He has done basicly this, and posted the waveforms. The "FTL" pulse
edge does indeed appear to precede the regular coax edge by an
impressive amount. The key to the illusion is that a carefully
selected risetime, partially differentiated by an unterminated coax,
superficially seems to rise sooner. This is, of course, only an
illusion; you can get the same effect with a simple RC network.

John

It is not a key to an inclusion but requirements specified in the patent
application
and it states that the input signal frequency components must be smaller
than the first resonant frequency of the coax segment.

Also if you run a sinusoidal frequency sweep you'll find out that from 0 to
the first resonance
it is the same for both coax segment and equivalent lump RLC circuit.

The key component of the invention is not disclosed.
But I can hint that it is the displacement current effect that must be
overcome in order to
enable method of constructing continuous FTL data transmission line.


Sincerely,

Mathew Orman
www.ultra-faster-than-light.com
www.radio-faster-than-light.com

 

[John Larkin]

On Tue, 14 Oct 2003 21:27:31 +0200, "Mathew Orman" <orman@nospam.com>
wrote:

It is not a key to an inclusion but requirements specified in the patent
application
and it states that the input signal frequency components must be smaller
than the first resonant frequency of the coax segment.

Also if you run a sinusoidal frequency sweep you'll find out that from 0 to
the first resonance
it is the same for both coax segment and equivalent lump RLC circuit.

The key component of the invention is not disclosed.
But I can hint that it is the displacement current effect that must be
overcome in order to
enable method of constructing continuous FTL data transmission line.


Sincerely,

Mathew Orman
www.ultra-faster-than-light.com
www.radio-faster-than-light.com

Stick to distributing porn; that's a better match to your skills.

John

 

[John Fields]

On Tue, 14 Oct 2003 12:14:32 -0700, John Larkin
<jjlarkin@highSNIPlandTHIStechPLEASEnology.com> wrote:

On Tue, 14 Oct 2003 13:27:13 -0500, John Fields
jfields@austininstruments.com> wrote:

On Tue, 14 Oct 2003 14:52:55 +0200, "Mathew Orman" <orman@nospam.com
wrote:


20 m long, 10MBit/sec, 6ns transient.

---
Never mind the bullshit, here's what I'd like to see:

PULSE GEN
+----------+
| OUT|-------+--[50R]----[50 OHM COAX]----------+
| GND|--+ | |
+----------+ | +--[50R]----[ORMAN BSCABLE]--+ |
| | | |
| +--+------------+ | |
| | TRIG A VERT|---------------+ |
| | | | |
| | B VERT|---------------------|
| +---------------+ | |
| [50R] [50R]
| | |
GND GND GND

The pulse generator is a voltage source with the output driving 50 ohm
resistors in series with the center conductors of two _equal_ lengths of
real 50 ohm and your BSFTL cable. You say you've got a working length
of 20 meters, so that would be fine.

The output of the pulse gen also drives the trigger input of the scope,
which is located right at the output of the pulse gen. The assumption
is that the system is entirely coaxial, the intent being to determine
the difference between the arrival time of the leading edge of pulse
propagating down the real cable and the BSFTL cable.

Send me one of your 20 meter cables and I'll test it and post the
results here, including screen shots of the scope output and test setup.

John,

He has done basicly this, and posted the waveforms. The "FTL" pulse
edge does indeed appear to precede the regular coax edge by an
impressive amount. The key to the illusion is that a carefully
selected risetime, partially differentiated by an unterminated coax,
superficially seems to rise sooner. This is, of course, only an
illusion; you can get the same effect with a simple RC network.

---
I remember seeing the waveforms he published, but ISTR that his
accompanying "explanation" was clearly nebulous, so since he's
claiming 10C for his 20 meter cable, I'd expect to see something like
Belden 8219 (RG58A/U "type" with a foam polyethylene core) with a
nominal velocity of propagation of about 0.75C exhibit a delay of about
120ns for a 30 meter length of cable, and about 12ns for a 30 meter
length of his in a rig like shown above. Seem reasonable?

--
John Fields

 

[Nico Coesel]

John Fields <jfields@austininstruments.com> wrote:

On Tue, 14 Oct 2003 14:52:55 +0200, "Mathew Orman" <orman@nospam.com
wrote:


20 m long, 10MBit/sec, 6ns transient.

---
Never mind the bullshit, here's what I'd like to see:

PULSE GEN
+----------+
| OUT|-------+--[50R]----[50 OHM COAX]----------+
| GND|--+ | |
+----------+ | +--[50R]----[ORMAN BSCABLE]--+ |
| | | |
| +--+------------+ | |
| | TRIG A VERT|---------------+ |
| | | | |
| | B VERT|---------------------|
| +---------------+ | |
| [50R] [50R]
| | |
GND GND GND

The pulse generator is a voltage source with the output driving 50 ohm
resistors in series with the center conductors of two _equal_ lengths of
real 50 ohm and your BSFTL cable. You say you've got a working length
of 20 meters, so that would be fine.

Now you make the same mistake as Mathew. The pulse should be measured
at the output of the generator!

--
Reply to nico@nctdevpuntnl (punt=.)
Bedrijven en winkels vindt U op www.adresboekje.nl

 

[John Fields]

On Tue, 14 Oct 2003 21:00:12 +0200, "Mathew Orman" <orman@nospam.com>
wrote:

Send me your shipping address.

--
OK, but first you have to agree to:

1. Hold me blameless for any damage the cable may incur from the time it
leaves your possession until the time it returns to your possession.

2. Pay for freight and insurance for the cable from your dock to mine
and back to yours.

3. Pay for 30 meters of Belden 8219 to be used as a reference delay
line.

4. Accept the test results without question.


If you agree I'll invoice you for the cable and return freight for the
cable, and when your check clears I'll start the tests at my convenience
and, once I've finished, I'll post the results and photos to
alt.binaries.schematics.electronic.

--
John Fields

 

[Mathew Orman]

"John Fields" <jfields@austininstruments.com> wrote in message
news:61noovcmqjh0ifttrn4otmc7fultkuat5t@4ax.com...

On Tue, 14 Oct 2003 21:00:12 +0200, "Mathew Orman" <orman@nospam.com
wrote:


Send me your shipping address.

--
OK, but first you have to agree to:

1. Hold me blameless for any damage the cable may incur from the time it
leaves your possession until the time it returns to your possession.

2. Pay for freight and insurance for the cable from your dock to mine
and back to yours.

3. Pay for 30 meters of Belden 8219 to be used as a reference delay
line.

4. Accept the test results without question.


If you agree I'll invoice you for the cable and return freight for the
cable, and when your check clears I'll start the tests at my convenience
and, once I've finished, I'll post the results and photos to
alt.binaries.schematics.electronic.

--
John Fields

The 20 m reference cable is included in the setup as well as the low-pass
filter
with T splitter in case you do not have an arbitrary waveform generator.
All costs are paid and you can damage it, no problem, it can be easily
repaired.

Sincerely,

Mathew Orman
www.ultra-faster-than-light.com
www.radio-faster-than-light.com

 

[John Fields]

On Tue, 14 Oct 2003 20:24:45 GMT, nico@puntnl.niks (Nico Coesel) wrote:

John Fields <jfields@austininstruments.com> wrote:

On Tue, 14 Oct 2003 14:52:55 +0200, "Mathew Orman" <orman@nospam.com
wrote:


20 m long, 10MBit/sec, 6ns transient.

---
Never mind the bullshit, here's what I'd like to see:

PULSE GEN
+----------+
| OUT|-------+--[50R]----[50 OHM COAX]----------+
| GND|--+ | |
+----------+ | +--[50R]----[ORMAN BSCABLE]--+ |
| | | |
| +--+------------+ | |
| | TRIG A VERT|---------------+ |
| | | | |
| | B VERT|---------------------|
| +---------------+ | |
| [50R] [50R]
| | |
GND GND GND

The pulse generator is a voltage source with the output driving 50 ohm
resistors in series with the center conductors of two _equal_ lengths of
real 50 ohm and your BSFTL cable. You say you've got a working length
of 20 meters, so that would be fine.

Now you make the same mistake as Mathew. The pulse should be measured
at the output of the generator!

---
I don't see why.

Assuming that a pulse is launched into each cable at exactly the same
time and that both cables are the same length and exhibit the same
velocity of propagation, won't the resultant pulses reach the far ends
of their cables at the same time?

--
John Fields

 

[Frank Buss]

John Fields <jfields@austininstruments.com> wrote:

Never mind the bullshit, here's what I'd like to see:

I don't think that this is enough to prove it. I'm sure you know that the
speed of a signal in copper is less than the speed of light:

http://www.jimloy.com/physics/electric.htm

Another source of error could be that the signal in the FTL cable could be
delayed for 1.5 wavelength, which appears to be 0.5 wavelength faster. So
you need a vacuum tunnel and a FTL cable parallel. First use a short tunnel
and short cable. Feed a signal in the tunnel (perhaps with laser light) and
in the cable. Calibrate the outputs to the same phase. Then use a longer
tunnel and a longer cable and test the outputs.

--
Frank Buß, fb@frank-buss.de
http://www.frank-buss.de, http://www.it4-systems.de

 

[John Fields]

On Tue, 14 Oct 2003 23:01:32 +0200, "Mathew Orman" <orman@nospam.com>
wrote:

The 20 m reference cable is included in the setup as well as the low-pass
filter
with T splitter in case you do not have an arbitrary waveform generator.
All costs are paid and you can damage it, no problem, it can be easily
repaired.

---
In order for any anomalies which may have crept into your test system
not to affect any testing done here I would prefer that everything
associated with the test, other than the DUT, be purchased and assembled
here.

I have all the test equipment necessary to perform the test, and will
not be using a filter of any type which could mask the actual results
obtained.

My test setup will be precisely as indicated in my previous post, and if
you wish for me to proceed you must still agree to the remainder of my
requirements, from my previous post, namely:


2. Pay for freight and insurance for the cable from your dock to mine
and back to yours.

3. Pay for 30 meters of Belden 8219 to be used as a reference delay
line.

4. Accept the test results without question.


If you agree I'll invoice you for the reference cable and return freight
for your reference cable, and when your check clears I'll start the
tests at my convenience and, once I've finished, I'll post the results
and photos to alt.binaries.schematics.electronic.

--
John Fields

 

[Winfield Hill]

John Fields wrote...

My test setup will be precisely as indicated in my previous post, and
if you wish for me to proceed you must still agree to the remainder
of my requirements, from my previous post, namely:

John, you shouldn't create two separate 50-ohm signal paths simply
by paralleling the two coax lines. Instead you need at least a
splitter, or better, a splitter with each output followed by a 20dB
attenuator to insure completely independent pathways. Furthermore
you need to measure the impedance of the line under test to verify
it's 50 ohms (higher impedances move closer to the speed of light).

Thanks,
- Win

whill_at_picovolt-dot-com

 

[John Larkin]

On 15 Oct 2003 07:33:19 -0700, Winfield Hill
<Winfield_member@newsguy.com> wrote:

John Fields wrote...

My test setup will be precisely as indicated in my previous post, and
if you wish for me to proceed you must still agree to the remainder
of my requirements, from my previous post, namely:

John, you shouldn't create two separate 50-ohm signal paths simply
by paralleling the two coax lines. Instead you need at least a
splitter, or better, a splitter with each output followed by a 20dB
attenuator to insure completely independent pathways. Furthermore
you need to measure the impedance of the line under test to verify
it's 50 ohms (higher impedances move closer to the speed of light).

Thanks,
- Win

whill_at_picovolt-dot-com

Why not just drive the stupid thing from a signal generator and poke
the same hi-z scope probe on the input and output, and note the
waveforms? Since the FTL cable is flexible, presumably one can locate
the termination near to the source, at least to within scope probe
range.

Oh, in coax, the prop velocity depends on the dielectric constant of
the insulator, not on the impedance.

John

 

[John Fields]

On 15 Oct 2003 07:33:19 -0700, Winfield Hill
<Winfield_member@newsguy.com> wrote:

John Fields wrote...

My test setup will be precisely as indicated in my previous post, and
if you wish for me to proceed you must still agree to the remainder
of my requirements, from my previous post, namely:

John, you shouldn't create two separate 50-ohm signal paths simply
by paralleling the two coax lines. Instead you need at least a
splitter, or better, a splitter with each output followed by a 20dB
attenuator to insure completely independent pathways. Furthermore
you need to measure the impedance of the line under test to verify
it's 50 ohms (higher impedances move closer to the speed of light).

---
OK, although he's already claimed 50 ohms along with his claimed
velocity of propagation of 10C, so I don't see much problem
differentiating between about 40ns to get through the reference cable
and 4ns to get through his, do you?

--
John Fields

 

[John Fields]

On Wed, 15 Oct 2003 08:21:14 -0700, John Larkin
<jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

Why not just drive the stupid thing from a signal generator and poke
the same hi-z scope probe on the input and output, and note the
waveforms? Since the FTL cable is flexible, presumably one can locate
the termination near to the source, at least to within scope probe
range.

---
Yeah, but I want to see the actual difference in transmission times of
the two pulses and if I trigger the scope with the original pulse then
the sweep will catch them both. I could also put channel 3 on the pulse
gen input and display trigger off of it as well as diplay it to give me
an initial visual pulse edge for a T0 reference.
---

Oh, in coax, the prop velocity depends on the dielectric constant of
the insulator, not on the impedance.

---
OK, but not enough to make the pulse travel through the pipe at 10C.


--
John Fields

 

[Mathew Orman]

"Winfield Hill" <Winfield_member@newsguy.com> wrote in message
news:bmjlrf011d3@drn.newsguy.com...

John Fields wrote...

My test setup will be precisely as indicated in my previous post, and
if you wish for me to proceed you must still agree to the remainder
of my requirements, from my previous post, namely:

John, you shouldn't create two separate 50-ohm signal paths simply
by paralleling the two coax lines. Instead you need at least a
splitter, or better, a splitter with each output followed by a 20dB

Incorrect!
Attenuator R and coax input C will produce additional RC phase shift.

In the setup the reference path will have additional RC phase shifter of one
value
and because the input to my FTL data transmission line is only 10pF the
additional RC
constant will be much smaller.
I have no problem with that but it will artificially increase the FTL speed
as calculated
using the difference in delay times and known reference cable delay.
So, to have a correct FTL speed one must precalculate the delays due to the
attenuators phase shifts.

See example of measurement method proposed by German scientist at:
http://www.ultra-faster-than-light.com/ftlspeed.htm
And the example test diagram:
http://www.ultra-faster-than-light.com/test_diagram.pdf
And the required waveform for 10MBit/sec FTL cables:
http://www.ultra-faster-than-light.com/sharp_corners.pdf
With it's FFT :
http://www.ultra-faster-than-light.com/Filtered_FFT.JPG

attenuator to insure completely independent pathways. Furthermore
you need to measure the impedance of the line under test to verify
it's 50 ohms (higher impedances move closer to the speed of light).

Correct!

Thanks,
- Win

whill_at_picovolt-dot-com

Sincerely,

Mathew Orman
www.ultra-faster-than-light.com
www.radio-faster-than-light.com

 

[Mathew Orman]

"John Larkin" <jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote in
message news:v5pqov8b9bp3ssh2mp18duovn36m72j6v5@4ax.com...

On 15 Oct 2003 07:33:19 -0700, Winfield Hill
Winfield_member@newsguy.com> wrote:

John Fields wrote...

My test setup will be precisely as indicated in my previous post, and
if you wish for me to proceed you must still agree to the remainder
of my requirements, from my previous post, namely:

John, you shouldn't create two separate 50-ohm signal paths simply
by paralleling the two coax lines. Instead you need at least a
splitter, or better, a splitter with each output followed by a 20dB
attenuator to insure completely independent pathways. Furthermore
you need to measure the impedance of the line under test to verify
it's 50 ohms (higher impedances move closer to the speed of light).

Thanks,
- Win

whill_at_picovolt-dot-com

Why not just drive the stupid thing from a signal generator and poke
the same hi-z scope probe on the input and output, and note the
waveforms? Since the FTL cable is flexible, presumably one can locate
the termination near to the source, at least to within scope probe
range.

Oh, in coax, the prop velocity depends on the dielectric constant of
the insulator, not on the impedance.

John

You are confusing EM waves propagation with transmission line waveform
propagation.
The later depends on nominal RLC only!
Also one must drive it with the signal that defines the coax segment as
"Electrically Short and Open Ended".

Sincerely,

Mathew Orman
www.ultra-faster-than-light.com
www.radio-faster-than-light.com

 

[Winfield Hill]

Mathew Orman wrote...

Winfield Hill wrote ...
John Fields wrote...

My test setup will be precisely as indicated in my previous post, and
if you wish for me to proceed you must still agree to the remainder
of my requirements, from my previous post, namely:

John, you shouldn't create two separate 50-ohm signal paths simply
by paralleling the two coax lines. Instead you need at least a
splitter, or better, a splitter with each output followed by a 20dB

Incorrect! Attenuator R and coax input C will produce additional RC
phase shift.

I agree, no additional capacitance should be added. However passive
splitters and attenuators don't have any. (They are transmission-line
components with a DC-to-3GHz bandwidth, and very fast 100ps risetime
typically. If they aren't used one will have reflections and standing
waves, etc.) They do have some very slight time delay, but identical
parts are used in both pathways. One takes two sets of measurements
by exchanging the two cables to eliminate any possible bias.

As an simple alternate one path can be used, first with one cable,
saving the traces, then with the other to compare time delays.

Thanks,
- Win

whill_at_picovolt-dot-com

 

[Keith R. Williams]

In article <l5sqovcviu5b9ti4ct8n09gil7pjj3g96b@4ax.com>,
jfields@austininstruments.com says...

On Wed, 15 Oct 2003 08:21:14 -0700, John Larkin
jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

Oh, in coax, the prop velocity depends on the dielectric constant of
the insulator, not on the impedance.

---
OK, but not enough to make the pulse travel through the pipe at 10C.

Even if the insulator has a dielectric constant of .01? (Spice is
wonderful ;-)

I bet I could sell some of that insulator! It would be better than a
look-ahead gate, or a negative delay line.

--
Keith

 

[Mathew Orman]

"Winfield Hill" <Winfield_member@newsguy.com> wrote in message
news:bmk12g028eo@drn.newsguy.com...

Mathew Orman wrote...

Winfield Hill wrote ...
John Fields wrote...

My test setup will be precisely as indicated in my previous post, and
if you wish for me to proceed you must still agree to the remainder
of my requirements, from my previous post, namely:

John, you shouldn't create two separate 50-ohm signal paths simply
by paralleling the two coax lines. Instead you need at least a
splitter, or better, a splitter with each output followed by a 20dB

Incorrect! Attenuator R and coax input C will produce additional RC
phase shift.

I agree, no additional capacitance should be added. However passive
splitters and attenuators don't have any. (They are transmission-line
components with a DC-to-3GHz bandwidth, and very fast 100ps risetime
typically. If they aren't used one will have reflections and standing
waves, etc.) They do have some very slight time delay, but identical
parts are used in both pathways. One takes two sets of measurements
by exchanging the two cables to eliminate any possible bias.

As an simple alternate one path can be used, first with one cable,
saving the traces, then with the other to compare time delays.

Thanks,
- Win

whill_at_picovolt-dot-com

I was thinking about attenuator that is constructed with two resistors.
But if one has expensive attenuator with 0 phase shift than that is perfect.

Sincerely,

Mathew Orman
www.ultra-faster-than-light.com
www.radio-faster-than-light.com

 

[John Fields]

On Wed, 15 Oct 2003 20:13:25 +0200, "Mathew Orman" <orman@nospam.com>
wrote:

I was thinking about attenuator that is constructed with two resistors.
But if one has expensive attenuator with 0 phase shift than that is perfect.

---
If you claim to be playing with transmission lines capable of
propagating signals at superluminal speeds and you don't know what a
coaxial attenuator is, then I claim that you're not playing with a full
deck.

--
John Fields