"Androcles" wrote in message
k...

"George Dishman" wrote in message
...

"Androcles" wrote in message
. uk...

"George Dishman" wrote in message
...

"The Ghost In The Machine" wrote in
message ...
In sci.physics.relativity, N:dlzc D:aol T:com (dlzc)

snip your answer to Ghost, he can reply to that

Isn't the answer somewhat simpler:

"Androcles" wrote in message
. uk...


Why did Einstein say
eta = y,
zeta = z?
Did he not know how to move sideways or up?

Whether the motion is sideways or up, in his
derivation, he chose to define the x axis as
parallel to the direction of motion. That
distinguishes the x axis from the others and
hence

Ever travelled in a circle? ...

His derivation of the transforms uses the example
of linear motion. The example is sufficient for the
purpose. Later your claim that "if one is wrong,
all are wrong" rests on an unstated assumption of
symmetry but the situation is not symmetrical since
the x axis is defined as the direction of motion,
(whether that is sideways or up or any direction
you like) so your argument fails.


Proof you are an idiot (I'm being nice, it's Xmas,otherwise I'd use
stronger language):

"If we assume that the result proved for a polygonal line ...
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

As I said, but you know that, you're just trying
to change the subject because ...

... is also valid for a continuously curved line, we arrive at this
result: If one of two synchronous clocks at A is moved in a closed curve
with constant velocity until it returns to A, the journey lasting t
seconds, then by the clock which has remained at rest the travelled clock
on its arrival at A will be 1/2t.v^/c^2 second slow. Thence we conclude
that a balance-clock at the equator must go more slowly, by a very small
amount, than a precisely similar clock situated at one of the poles under
otherwise identical conditions."

Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/

Note the 'assume', note the change in direction, further note that a
sundial at the equator must go at exactly the same speed, by a very zero
amount, as a precisely similar sundial situated at one of the poles under
otherwise identical conditions. Note that you are a moron who wants to
kiss Einstein's arse to seem as if you are as stupid as he was in the hope
of looking good in the eyes of the other sheep all doing the same thing.
So why don't you just shut up talking to me, I KNOW you are an ineducable
moron. Ignorance is educable, stupidity is forever.-- Uncle Al.


But you know all this, it's just a smokescreen to
hide the fact that you now know that ballistic light
cannot explain the phase shift measured by the
photodiode in an iFOG because it is fused to the
light source.

It certainly can, ...

Go ahead then, show the derivation and you will win
the argument.

That fact is it doesn't, it predicts a null shift
and you know that, that's why you are deperately
trying to change the subject.

... because it must.

Exactly, if it is valid, it must correctly predict
the phase shift, but it doesn't so Ritz is falsified
by theSagnac effect as I said and you now understand.

Bye bye.

As I said, just a smokescreen to avoid admitting
you cannot explain why the Sagnac effect occurs in
a fibre gyro after all you said to Henri. At least
he is trying even if he can't succeed.

Have a Merry Christmas anyway.

George


Androcles

Your argument was lost 92 years ago
when Sagnac published his results and no amount of
debate here will change that.

George


... I do it once a day at latitude
51 degrees North, and my sundial agrees perfectly with an
African sundial at the same longitude. If I move my sundial
1/2 an inch to the East it gains 38,000 nanoseconds over
the African sundial. I know because I checked it using GPS.
I've even travelled south or north while travelling east or west.



tau = (t-vx/c2)/sqrt(1-v2/c2)

does not imply

tau = (t-vy/c2)/sqrt(1-u2/c2)
tau = (t-vz/c2)/sqrt(1-w2/c2)

Correct. It implies
tau = (t-uy/c²)/sqrt(1-u²/c²)
tau = (t-wz/c²)/sqrt(1-w²/c²)
Thank you for pointing out that typo.

so the statement

If one is right they all are, if one is wrong they all are.

is a baseless and naive assertion.

Not at all. One is wrong, so they all are.

What IS a baseless and extremely naive ASSUMPTION is
"In agreement with experience we further assume the quantity
2AB/(t'A-tA) = c,
to be a universal constant--the velocity of light in empty space." --
Einstein.

In agreement with experience and without any assumption,
BA = -AB,
2AB = AC,

[AB +BA]/(t'A-tA) = 0
Hence c = 0 in Einstein's math.

That makes YOU baseless and naive, not me.


I'm not going to get into this sidetrack as A.
only raised it as a diversion to avoid admitting
he had been proved wrong on several points and
owed Henri a bottle, but I can't imagine this
hasn't been pointed out to him many times over
the years.

Merry Christmas all.

Bah! Humbug!
Androcles.

"George Dishman" wrote in message
...

"Androcles" wrote in message
k...

"George Dishman" wrote in message
...

"Androcles" wrote in message
. uk...

"George Dishman" wrote in message
...

"The Ghost In The Machine" wrote in
message ...
In sci.physics.relativity, N:dlzc D:aol T:com (dlzc)

snip your answer to Ghost, he can reply to that

Isn't the answer somewhat simpler:

"Androcles" wrote in message
. uk...


Why did Einstein say
eta = y,
zeta = z?
Did he not know how to move sideways or up?

Whether the motion is sideways or up, in his
derivation, he chose to define the x axis as
parallel to the direction of motion. That
distinguishes the x axis from the others and
hence

Ever travelled in a circle? ...

His derivation of the transforms uses the example
of linear motion. The example is sufficient for the
purpose. Later your claim that "if one is wrong,
all are wrong" rests on an unstated assumption of
symmetry but the situation is not symmetrical since
the x axis is defined as the direction of motion,
(whether that is sideways or up or any direction
you like) so your argument fails.


Proof you are an idiot (I'm being nice, it's Xmas,otherwise I'd use
stronger language):

"If we assume that the result proved for a polygonal line ...
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

As I said, but you know that,

Tha'rs right. Those are hos words. Einstein doesn't know that. Point made.
And it makes you an idiot for not recognizing it as a quotation.

you're just trying
to change the subject because ...

... is also valid for a continuously curved line, we arrive at this
result: If one of two synchronous clocks at A is moved in a closed curve
with constant velocity until it returns to A, the journey lasting t
seconds, then by the clock which has remained at rest the travelled clock
on its arrival at A will be 1/2t.v^/c^2 second slow. Thence we conclude
that a balance-clock at the equator must go more slowly, by a very small
amount, than a precisely similar clock situated at one of the poles under
otherwise identical conditions."

Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/

Note the 'assume', note the change in direction, further note that a
sundial at the equator must go at exactly the same speed, by a very zero
amount, as a precisely similar sundial situated at one of the poles under
otherwise identical conditions. Note that you are a moron who wants to
kiss Einstein's arse to seem as if you are as stupid as he was in the
hope
of looking good in the eyes of the other sheep all doing the same thing.
So why don't you just shut up talking to me, I KNOW you are an ineducable
moron. Ignorance is educable, stupidity is forever.-- Uncle Al.


But you know all this, it's just a smokescreen to
hide the fact that you now know that ballistic light
cannot explain the phase shift measured by the
photodiode in an iFOG because it is fused to the
light source.

It certainly can, ...

Go ahead then, show the derivation and you will win
the argument.

I already did. You have such a pathetically short attention span,
being an absolute idiot, that you can't read or understand plain English or
mathematics.
Androcles.



That fact is it doesn't, it predicts a null shift
and you know that, that's why you are deperately
trying to change the subject.

... because it must.

Exactly, if it is valid, it must correctly predict
the phase shift, but it doesn't so Ritz is falsified
by theSagnac effect as I said and you now understand.

Bye bye.

As I said, just a smokescreen to avoid admitting
you cannot explain why the Sagnac effect occurs in
a fibre gyro after all you said to Henri. At least
he is trying even if he can't succeed.

Have a Merry Christmas anyway.

George


Androcles

Your argument was lost 92 years ago
when Sagnac published his results and no amount of
debate here will change that.

George


... I do it once a day at latitude
51 degrees North, and my sundial agrees perfectly with an
African sundial at the same longitude. If I move my sundial
1/2 an inch to the East it gains 38,000 nanoseconds over
the African sundial. I know because I checked it using GPS.
I've even travelled south or north while travelling east or west.



tau = (t-vx/c2)/sqrt(1-v2/c2)

does not imply

tau = (t-vy/c2)/sqrt(1-u2/c2)
tau = (t-vz/c2)/sqrt(1-w2/c2)

Correct. It implies
tau = (t-uy/c²)/sqrt(1-u²/c²)
tau = (t-wz/c²)/sqrt(1-w²/c²)
Thank you for pointing out that typo.

so the statement

If one is right they all are, if one is wrong they all are.

is a baseless and naive assertion.

Not at all. One is wrong, so they all are.

What IS a baseless and extremely naive ASSUMPTION is
"In agreement with experience we further assume the quantity
2AB/(t'A-tA) = c,
to be a universal constant--the velocity of light in empty space." --
Einstein.

In agreement with experience and without any assumption,
BA = -AB,
2AB = AC,

[AB +BA]/(t'A-tA) = 0
Hence c = 0 in Einstein's math.

That makes YOU baseless and naive, not me.


I'm not going to get into this sidetrack as A.
only raised it as a diversion to avoid admitting
he had been proved wrong on several points and
owed Henri a bottle, but I can't imagine this
hasn't been pointed out to him many times over
the years.

Merry Christmas all.

Bah! Humbug!
Androcles.

"Androcles" wrote in message
k...

"George Dishman" wrote in message
...

"Androcles" wrote in message
k...

"George Dishman" wrote in message
...
....
But you know all this, it's just a smokescreen to
hide the fact that you now know that ballistic light
cannot explain the phase shift measured by the
photodiode in an iFOG because it is fused to the
light source.

It certainly can, ...

Go ahead then, show the derivation and you will win
the argument.

I already did. You have such a pathetically short attention span,
being an absolute idiot, that you can't read or understand plain English
or mathematics.

All your posts assumed the detector was
not on the turntable. This is typical and
from just four days ago and you said "I see
now what the problem is, you both think the
detector is moving." so you certainly hadn't
posted a valid analysis before that.

"Androcles" wrote in message
k...

"George Dishman" wrote in message
...
....
http://en.wikipedia.org/wiki/Image:S...rferometer.png

No indication of what is turning. That's MMX, not Sagnac.
MMX, no shift.


Now assuming that the source, mirrors and screen
are all on the turntable, should there be a beat
frequency on the screen?

Nope.
....
I see now what the problem is, you both think the detector
is moving. Just think of it as the light strobing on and off.
....
No but the photodetector on the end of a fibre ring
gyro is, and it is "on the turntable". That's where
this conversation started.

I don't know where you get that idea from. The photodetector
that is on the turntable is MMX, not Sagnac, and no fringe shift
is observed.
Grandpa stands beside a carousel, the two kids climb aboard,
start beside him, and walk around it in opposite directions.
They cross on the opposite side and meet again with a small
offset from grandpa, caused by the rotation of the carousel.
It doesn't matter whether they walk in 4 straight lines along chords
or a curve around the perimeter.

Henri insists Grandpa rides the carousel, and I am NOT
getting on it. I rather suspect he got that idea from listening
to you, judging from your insistence above.
He hasn't a hope in hell of programming a Sagnac model
with Grandpa riding along. The kids meet where they started
on the carousel, they do not meet at Grandpa.

We both knew the detector is moving. As you
said, "The kids meet where they started on
the carousel, ..." but Grandpa is riding on
it too, so they reach him at the same time.
Ritz predicts no time difference in the
Sagnac configuration. You haven't posted
any derivation since then in sci.astro that
says anything different and I agree with
everything above, Ritz predicts a null
result and is consequently falsified. That's
also what Henri's program shows.

George

"George Dishman" wrote in message
...

"Androcles" wrote in message
k...

"George Dishman" wrote in message
...

"Androcles" wrote in message
k...

"George Dishman" wrote in message
...
...
But you know all this, it's just a smokescreen to
hide the fact that you now know that ballistic light
cannot explain the phase shift measured by the
photodiode in an iFOG because it is fused to the
light source.

It certainly can, ...

Go ahead then, show the derivation and you will win
the argument.

I already did. You have such a pathetically short attention span,
being an absolute idiot, that you can't read or understand plain English
or mathematics.


All your posts assumed the detector was
not on the turntable. This is typical and
from just four days ago and you said "I see
now what the problem is, you both think the
detector is moving." so you certainly hadn't
posted a valid analysis before that.

Two frames of reference, you need two detectors to compare
them.

All your posts assumed there is only one detector.
This is typical and from just right now ago you said
"All your posts assumed THE detector was not on the turntable"
so you certainly hadn't posted ANY analysis before that.
Albert Michelson is one observer, Georges Sagnac is
the other. The phase is not the same from one to the other,
or between any pair of mirrors.
Wilson owes me a case of Glenlivet that you are ethically bound
to pay half of on his behalf.


"Androcles" wrote in message
k...

"George Dishman" wrote in message
...
...
http://en.wikipedia.org/wiki/Image:S...rferometer.png

No indication of what is turning. That's MMX, not Sagnac.
MMX, no shift.


Now assuming that the source, mirrors and screen
are all on the turntable, should there be a beat
frequency on the screen?

Nope.
...
I see now what the problem is, you both think the detector
is moving. Just think of it as the light strobing on and off.
...
No but the photodetector on the end of a fibre ring
gyro is, and it is "on the turntable". That's where
this conversation started.

I don't know where you get that idea from. The photodetector
that is on the turntable is MMX, not Sagnac, and no fringe shift
is observed.
Grandpa stands beside a carousel, the two kids climb aboard,
start beside him, and walk around it in opposite directions.
They cross on the opposite side and meet again with a small
offset from grandpa, caused by the rotation of the carousel.
It doesn't matter whether they walk in 4 straight lines along chords
or a curve around the perimeter.

Henri insists Grandpa rides the carousel, and I am NOT
getting on it. I rather suspect he got that idea from listening
to you, judging from your insistence above.
He hasn't a hope in hell of programming a Sagnac model
with Grandpa riding along. The kids meet where they started
on the carousel, they do not meet at Grandpa.

We both knew the detector is moving.

That depends on which detector you mean, Michelson or Sagnac.


As you
said, "The kids meet where they started on
the carousel, ..." but Grandpa is riding on
it too, so they reach him at the same time.

You are a ****.

Ritz predicts no time difference in the
Sagnac configuration.

You are a ****.

You haven't posted
any derivation since then in sci.astro that
says anything different and I agree with
everything above, Ritz predicts a null
result and is consequently falsified. That's
also what Henri's program shows.

You are still a snipping, arrogant, illiterate, innumerate, illogical,
**** without a scrap of logic in you, you whining little toad.
You don't have an inkling about mathematics or physics
and live in the vain hope some moron will think you are clever,
Mr SmartArse who pretends he understands relativity and
doesn't have a clue how to synchronize his watch to Cassini
time.

Hey dumb****! Do you know how to move sideways or up?

tau = (t-vx/c²)/sqrt(1-v²/c²)
tau = (t-vy/c²)/sqrt(1-u²/c²)
tau = (t-vz/c²)/sqrt(1-w²/c²)
xi = (x-vt)/sqrt(1-v²/c²)
eta = (y-ut)/sqrt(1-u²/c²)
zeta= (z-wt)/sqrt(1-w²/c²)

Right or wrong, dumb****?
If one is right they all are, if one is wrong they all are,
pathetic ****head.

Einstein said
eta = y,
zeta = z
because he didn not know how to move sideways or up,
anencephalous cretin.


[quote]
we establish by definition that the "time" required by a crab to travel
from A to B equals the "time" it requires to travel from B to A.
[end quote]
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/

Einstein can prove nothing can go faster than a crab.

Oops!... Did I say 'a crab'? Sorry...'light'.


"In agreement with experience we further assume the quantity
2AB/(t'A-tA) = c,
to be a universal constant--the velocity of light in empty space." --
Einstein.

In agreement with experience and without any assumption,
BA = -AB,
2AB = AC,

[AB +BA]/(t'A-tA) = 0
Hence c = 0 in Einstein's math.

Observation:
http://www.britastro.org/vss/gifc/00918-ck.gif
Explanation:
http://www.ebicom.net/~rsf1/sekerin.htm (fig 3)

(Or stars explode twice in three months, which is stupid).

In agreement with experience and without any assumption,
you remain a snipping, arrogant, illiterate, innumerate, illogical,
incompetent **** without a scrap of logic in you,
you whining little toad.
You don't have an inkling about mathematics or physics
and live in the vain hope some moron will think you are clever,
Mr SmartArse who pretends he understands physics and
doesn't have a clue how to synchronize his watch to Cassini
time.

**** off, useless tord!
Androcles.

"Androcles" wrote in message
k...

"George Dishman" wrote in message
...

"Androcles" wrote in message
k...

"George Dishman" wrote in message
...

"Androcles" wrote in message
k...

"George Dishman" wrote in message
...
...
But you know all this, it's just a smokescreen to
hide the fact that you now know that ballistic light
cannot explain the phase shift measured by the
photodiode in an iFOG because it is fused to the
light source.

It certainly can, ...

Go ahead then, show the derivation and you will win
the argument.

I already did. You have such a pathetically short attention span,
being an absolute idiot, that you can't read or understand plain English
or mathematics.


All your posts assumed the detector was
not on the turntable. This is typical and
from just four days ago and you said "I see
now what the problem is, you both think the
detector is moving." so you certainly hadn't
posted a valid analysis before that.

Two frames of reference, you need two detectors to compare
them.

All your posts assumed there is only one detector.

That is correct, there is only one detector. It
is mounted on the back of the laser diode.

This is typical and from just right now ago you said
"All your posts assumed THE detector was not on the turntable"
so you certainly hadn't posted ANY analysis before that.
Albert Michelson is one observer, Georges Sagnac is
the other. The phase is not the same from one to the other,
or between any pair of mirrors.

Commercial iFOG devices have just one detector and
give an output which is proportional to ratation
rate, there is no MMX structure in them.

http://www.kvh.com/pdf/aiaa98.pdf

Gives a lot more details.

Wilson owes me a case of Glenlivet that you are ethically bound
to pay half of on his behalf.

If you are referring to your recent argument about
"in phase in one frame, not in phase in another"
then you lost. I can't comment on any previous bets
you may have made, and I am not part of your bet,
but if you feel I should be, you owe me a bottle
too.


"Androcles" wrote in message
k...

"George Dishman" wrote in message
...
...
http://en.wikipedia.org/wiki/Image:S...rferometer.png

No indication of what is turning. That's MMX, not Sagnac.
MMX, no shift.


Now assuming that the source, mirrors and screen
are all on the turntable, should there be a beat
frequency on the screen?

Nope.
...
I see now what the problem is, you both think the detector
is moving. Just think of it as the light strobing on and off.
...
No but the photodetector on the end of a fibre ring
gyro is, and it is "on the turntable". That's where
this conversation started.

I don't know where you get that idea from. The photodetector
that is on the turntable is MMX, not Sagnac, and no fringe shift
is observed.
Grandpa stands beside a carousel, the two kids climb aboard,
start beside him, and walk around it in opposite directions.
They cross on the opposite side and meet again with a small
offset from grandpa, caused by the rotation of the carousel.
It doesn't matter whether they walk in 4 straight lines along chords
or a curve around the perimeter.

Henri insists Grandpa rides the carousel, and I am NOT
getting on it. I rather suspect he got that idea from listening
to you, judging from your insistence above.
He hasn't a hope in hell of programming a Sagnac model
with Grandpa riding along. The kids meet where they started
on the carousel, they do not meet at Grandpa.

We both knew the detector is moving.

That depends on which detector you mean, Michelson or Sagnac.

Henri and I are discussing Sagnac only, it
has only one detector.

As you
said, "The kids meet where they started on
the carousel, ..." but Grandpa is riding on
it too, so they reach him at the same time.

You are a ****.

I see you cannot refute the point.

Ritz predicts no time difference in the
Sagnac configuration.

You are a ****.


I see you cannot refute the point.

You haven't posted
any derivation since then in sci.astro that
says anything different and I agree with
everything above, Ritz predicts a null
result and is consequently falsified. That's
also what Henri's program shows.

You are still a snipping, arrogant, illiterate, innumerate, illogical,
**** without a scrap of logic in you, you whining little toad.
You don't have an inkling about mathematics or physics ...

You were unaware the detector was on the turntable
in the Sagnac experiment. Now you imagine it has two
detectors instead of one. You got the beam splitter
at 90 degrees to the correct orientation. You didn't
understand the implication for coherence length. You
thought signals that reached a point on a mirror in
phase as seen in one frame would be out of phase as
seen from another. Henri or I had to correct you on
all those points so the evidence speaks for itself.

snip diversion

George

"George Dishman" wrote in message
...

"Androcles" wrote in message
k...

"George Dishman" wrote in message
...

"Androcles" wrote in message
k...

"George Dishman" wrote in message
...

"Androcles" wrote in message
k...

"George Dishman" wrote in message
...
...
But you know all this, it's just a smokescreen to
hide the fact that you now know that ballistic light
cannot explain the phase shift measured by the
photodiode in an iFOG because it is fused to the
light source.

It certainly can, ...

Go ahead then, show the derivation and you will win
the argument.

I already did. You have such a pathetically short attention span,
being an absolute idiot, that you can't read or understand plain
English or mathematics.


All your posts assumed the detector was
not on the turntable. This is typical and
from just four days ago and you said "I see
now what the problem is, you both think the
detector is moving." so you certainly hadn't
posted a valid analysis before that.

Two frames of reference, you need two detectors to compare
them.

All your posts assumed there is only one detector.

That is correct, there is only one detector. It
is mounted on the back of the laser diode.

Not my fault you are too stupid to realise two are needed.
Even the phuckwit Einstein used tau and t, troll.



This is typical and from just right now ago you said
"All your posts assumed THE detector was not on the turntable"
so you certainly hadn't posted ANY analysis before that.
Albert Michelson is one observer, Georges Sagnac is
the other. The phase is not the same from one to the other,
or between any pair of mirrors.

Commercial iFOG devices have just one detector and
give an output which is proportional to ratation
rate, there is no MMX structure in them.

http://www.kvh.com/pdf/aiaa98.pdf

Gives a lot more details.

Wilson owes me a case of Glenlivet that you are ethically bound
to pay half of on his behalf.

If you are referring to your recent argument about
"in phase in one frame, not in phase in another"
then you lost.

****ing lying troll. You are a ****, Dishman.

I can't comment on any previous bets
you may have made, and I am not part of your bet,
but if you feel I should be, you owe me a bottle
too.

Prove it. Oh wait, you already did.


"Androcles" wrote in message
k...

"George Dishman" wrote in message
...
...
http://en.wikipedia.org/wiki/Image:S...rferometer.png

No indication of what is turning. That's MMX, not Sagnac.
MMX, no shift.


Now assuming that the source, mirrors and screen
are all on the turntable, should there be a beat
frequency on the screen?

Nope.
...
I see now what the problem is, you both think the detector
is moving. Just think of it as the light strobing on and off.
...
No but the photodetector on the end of a fibre ring
gyro is, and it is "on the turntable". That's where
this conversation started.

I don't know where you get that idea from. The photodetector
that is on the turntable is MMX, not Sagnac, and no fringe shift
is observed.
Grandpa stands beside a carousel, the two kids climb aboard,
start beside him, and walk around it in opposite directions.
They cross on the opposite side and meet again with a small
offset from grandpa, caused by the rotation of the carousel.
It doesn't matter whether they walk in 4 straight lines along chords
or a curve around the perimeter.

Henri insists Grandpa rides the carousel, and I am NOT
getting on it. I rather suspect he got that idea from listening
to you, judging from your insistence above.
He hasn't a hope in hell of programming a Sagnac model
with Grandpa riding along. The kids meet where they started
on the carousel, they do not meet at Grandpa.

We both knew the detector is moving.

That depends on which detector you mean, Michelson or Sagnac.

Henri and I are discussing Sagnac only, it
has only one detector.

The result from the other detector was already proven by Michelson,
Sagnac didn't need two. You are a ****, Dishman.

As you
said, "The kids meet where they started on
the carousel, ..." but Grandpa is riding on
it too, so they reach him at the same time.

You are a ****.

I see you cannot refute the point.
What point, Dishman? I see no point to refute.
You are a ****, Dishman, I see you cannot refute the point.




Ritz predicts no time difference in the
Sagnac configuration.

You are a ****.


I see you cannot refute the point.
You are a ****, Dishman, I see you cannot refute the point.


You haven't posted
any derivation since then in sci.astro that
says anything different and I agree with
everything above, Ritz predicts a null
result and is consequently falsified. That's
also what Henri's program shows.

You are still a snipping, arrogant, illiterate, innumerate, illogical,
**** without a scrap of logic in you, you whining little toad.
You don't have an inkling about mathematics or physics ...

You were unaware the detector was on the turntable
in the Sagnac experiment. Now you imagine it has two
detectors instead of one. You got the beam splitter
at 90 degrees to the correct orientation. You didn't
understand the implication for coherence length. You
thought signals that reached a point on a mirror in
phase as seen in one frame would be out of phase as
seen from another. Henri or I had to correct you on
all those points so the evidence speaks for itself.


I don't play poker with my cards face up on the table, either.
You are so ****ing stupid it's incredible. I'm now up to 30 bottles.

snip diversion
snip lying **** Dishman
*plonk*
Androcles.

"Androcles" wrote in message k...

"George Dishman" wrote in message
...

[snip]

snip diversion

snip lying **** Dishman
*plonk*
Androcles.

George, that means that he pretends he can't read your
messages anymore. He still reads them, so you can now
safely talk to him without being insulted. It usually takes
a few weeks or months... then he forgets that he allegedly
killfiled you, and everything starts all over...

Dirk Vdm

"Dirk Van de moortel" wrote
in message ...

"Androcles" wrote in message
k...

"George Dishman" wrote in message
...

[snip]

snip diversion

snip lying **** Dishman
*plonk*
Androcles.

George, that means that he pretends he can't read your
messages anymore. He still reads them, so you can now
safely talk to him without being insulted. It usually takes
a few weeks or months... then he forgets that he allegedly
killfiled you, and everything starts all over...

I suspected as much but he has nothing to say
really anyway. I am certain he knows that there
cannot be a phase shift with a single detector on
the turntable and the technical papers from KVH
make that structure completely clear. If anyone
killfiles me, I do the same, there's no point in
replying to someone who isn't listening, but I
will occasionally see his posts if I use Google.

In the meantime I'll just go back to the discussion
with Henri, he understands how the apparatus is
configured and with a little advice his program is
close to illustrating the consequences.

Thanks for the hint though.

Have a great Christmas Dirk.
George

"George Dishman" wrote in message ...

"Dirk Van de moortel" wrote
in message ...

"Androcles" wrote in message
k...

"George Dishman" wrote in message
...

[snip]

snip diversion

snip lying **** Dishman
*plonk*
Androcles.

George, that means that he pretends he can't read your
messages anymore. He still reads them, so you can now
safely talk to him without being insulted. It usually takes
a few weeks or months... then he forgets that he allegedly
killfiled you, and everything starts all over...

I suspected as much but he has nothing to say
really anyway. I am certain he knows that there
cannot be a phase shift with a single detector on
the turntable and the technical papers from KVH
make that structure completely clear. If anyone
killfiles me, I do the same, there's no point in
replying to someone who isn't listening, but I
will occasionally see his posts if I use Google.

But he *is* listening - that's the entire point ;-)


In the meantime I'll just go back to the discussion
with Henri, he understands how the apparatus is
configured and with a little advice his program is
close to illustrating the consequences.

Okay, enjoy Ralph then.
He is clearly far less insane than Androcles, so he
definitely is better conversation...


Thanks for the hint though.

Have a great Christmas Dirk.
George

Likewise, cheers,
Dirk Vdm

"Androcles" wrote in message
k...

[hanson]
ahahaha... AHAHAHA... ahahaha... Let me interlude, lewd, but
by no means nude, and not trying to created a feud, as I will
only make an account of "****-a-count" for your benefit, Andro:

"George Dishman" wrote in message
...
"Androcles" wrote in message
k...

****ing lying troll. You are a **** [1], Dishman.

You are a **** [2], Dishman.

You are a ****. [3]

You are a **** [4], Dishman, I see you cannot refute the point.

You are a **** [5].

You are a **** [6], Dishman, I see you cannot refute the point.

snip lying **** [7] Dishman

[hanson]
After the magic seventh ****-try, Andro, doesn't it dawn on you
that your landsman, Dishman, does NOT know what a **** is?
Dishman may be ****less, or worse, ****ignorant indeed....
So, why **** with him?
ahahaha... ahahaha... Happy 2006 to you guys... and thanks
for the laughs... ahahaha... ahahanson

PS: I'll respond to your GPS and REL issues when I can get
off the floor from the ROTFL-ing which you guys have caused.
Maybe Dishman and Stocky came out of the same **** that
made them so unaccountably ****able, but Stocklbauer is at
least a hands-on ****or... Gash, finally, I've caught up: 7 to 7
ahahaha... ahahaha...

BTW: I wonder every now and then why certain things are
so engrained, determined and limited by language like
"a man rules in his mansion" .... but women go whole hog
and demand for "a **** to rule the entire country" .....
So much for the knight in shining armor.... ahahaha....