Discussion:
A Lesson In Sagnac and Rotating frames, For Paul, Tom and Jerry.
(too old to reply)
Henry Wilson DSc
2010-02-11 23:33:09 UTC
Permalink
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.

//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)

(the surface is drawn flat for convenience. The fibre goes right around the
planet)

Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre. The pulses are labelled according to their emission times and move at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.

Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p. At any instant, 40 million pulses are in transit
around the planet in each direction.

Tom observes that the pulses arriving simultaneously from both fibres carry
identical labels.

Using powerful rockets, the planet is sent into rotation with a period of 1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.

In this new situation, the pulses that arrive at C when it is at point p, were
not emitted at that point but rather from an imaginary point q, which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.

Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.

Since the distance between pulses is still 1 metre in the C's frame, there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.

"That's odd", says Tom, "now, for some reason, the pulses coming from the left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"

"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".



Henry Wilson...

.......provider of free physics lessons
Henry Wilson DSc
2010-02-12 21:27:47 UTC
Permalink
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre. The pulses are labelled according to their emission times and move at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p. At any instant, 40 million pulses are in transit
around the planet in each direction.
Tom observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period of 1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
In this new situation, the pulses that arrive at C when it is at point p, were
not emitted at that point but rather from an imaginary point q, which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
Since the distance between pulses is still 1 metre in the C's frame, there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
"That's odd", says Tom, "now, for some reason, the pulses coming from the left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
Henry Wilson...
.......provider of free physics lessons
Not ONE ANSWER!!!!!!!!

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-12 22:35:45 UTC
Permalink
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
What is q and v?
Post by Henry Wilson DSc
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre. The pulses are labelled according to their emission times and move at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p.
Doesn't matter if they are detected at p or at C
Post by Henry Wilson DSc
At any instant, 40 million pulses are in transit
around the planet in each direction.
Tom
Tom? I thought it was Paul and Jerry?
Post by Henry Wilson DSc
observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period of 1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
OK .. assuming you got the math right
Post by Henry Wilson DSc
In this new situation, the pulses that arrive at C when it is at point p, were
not emitted at that point but rather from an imaginary point q,
Not an imaginary point at all. It is as real and physical point as p is.
All you are saying is that C has moved (and continues to move)
Post by Henry Wilson DSc
which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Angular speed I think you would mean, it is moving around the ring according
to C's rotating frame.
Post by Henry Wilson DSc
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
So you're using an emission theory analysus, which does say that the time
taken for the two paths is the same.
Post by Henry Wilson DSc
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
I'll take your word for it.
Post by Henry Wilson DSc
Since the distance between pulses is still 1 metre in the C's frame,
That's what emission theory would say
Post by Henry Wilson DSc
there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
WRONG. Emmision theory does not say that. It says that the distance in C's
frame between pulses is one meter. The distance the pulses travel in C's
frame is the same. So the number of pulses in the fibre at any given time
is the same.

You can also tell it is wrong because the same number of pulses has been
emitted in each direction and you just said earlier that they take the same
time to travel around the ring. So there must be the same number of them in
there. Or does your explanation have pulse fairies that eat up the pulses?
Post by Henry Wilson DSc
"That's odd", says Tom, "now, for some reason, the pulses coming from the left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
Emmision theory doesn't say that will happen at all. It says what you said
earlier .. the time take for them to travel is the same.

What you are now suddenly describing is what SR or an aether theory would
predict. You've changed theories in mid analysis !!!
Post by Henry Wilson DSc
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
No .. because your analysis above used SR at the end. You switched theories
mid way thru.

You started saying the time for the pulses to travel in both directions
(with the rotating tube) was the same. That is what emission theory says.
And you said the distance between the pulses (in the rotating frame) is the
same. That is what emission theory says. The consequence of that is that
there MUST be the same number of pulses in each direction in the tube at any
given time. And they MUST arrive at the detector with the same label. And
so emission theory says that a sagnac device will NOT detect rotation.

You've just admitted that your theory is refuted.

The *real* Sagnac devices have proven that THAT theory is NOT correct, and
it is NOT the same time required for the pulses (which is what SR predicts).

BAHAHAHHA
BURT
2010-02-12 22:47:36 UTC
Permalink
Post by Inertial
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
What is q and v?
Post by Henry Wilson DSc
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre.  The pulses are labelled according to their emission times and move
at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p.
Doesn't matter if they are detected at p or at C
Post by Henry Wilson DSc
At any instant, 40 million pulses are in transit
around the planet in each direction.
Tom
Tom?  I thought it was Paul and Jerry?
Post by Henry Wilson DSc
observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period of 1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
OK .. assuming you got the math right
Post by Henry Wilson DSc
In this new situation, the pulses that arrive at C when it is at point p, were
not emitted at that point but rather from an imaginary point q,
Not an imaginary point at all.  It is as real and physical point as p is.
All you are saying is that C has moved (and continues to move)
Post by Henry Wilson DSc
which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Angular speed I think you would mean, it is moving around the ring according
to C's rotating frame.
Post by Henry Wilson DSc
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
So you're using an emission theory analysus, which does say that the time
taken for the two paths is the same.
Post by Henry Wilson DSc
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
I'll take your word for it.
Post by Henry Wilson DSc
Since the distance between pulses is still 1 metre in the C's frame,
That's what emission theory would say
Post by Henry Wilson DSc
there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
WRONG. Emmision theory does not say that.  It says that the distance in C's
frame between pulses is one meter.  The distance the pulses travel in C's
frame is the same.  So the number of pulses in the fibre at any given time
is the same.
You can also tell it is wrong because the same number of pulses has been
emitted in each direction and you just said earlier that they take the same
time to travel around the ring.  So there must be the same number of them in
there.  Or does your explanation have pulse fairies that eat up the pulses?
Post by Henry Wilson DSc
"That's odd", says Tom, "now, for some reason, the pulses coming from the left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
Emmision theory doesn't say that will happen at all.  It says what you said
earlier .. the time take for them to travel is the same.
What you are now suddenly describing is what SR or an aether theory would
predict.  You've changed theories in mid analysis !!!
Post by Henry Wilson DSc
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
No .. because your analysis above used SR at the end.  You switched theories
mid way thru.
You started saying the time for the pulses to travel in both directions
(with the rotating tube) was the same.  That is what emission theory says.
And you said the distance between the pulses (in the rotating frame) is the
same.  That is what emission theory says.  The consequence of that is that
there MUST be the same number of pulses in each direction in the tube at any
given time.  And they MUST arrive at the detector with the same label.  And
so emission theory says that a sagnac device will NOT detect rotation.
You've just admitted that your theory is refuted.
The *real* Sagnac devices have proven that THAT theory is NOT correct, and
it is NOT the same time required for the pulses (which is what SR predicts).
BAHAHAHHA- Hide quoted text -
- Show quoted text -
Einstein could never reconcile the photon with the light wave. I ask
which wave is the photon in?
The magnetic or electric? Light is electric energy oscillating with
magnetism in a Dual Unfied force.

Mitch Raemsch
Henry Wilson DSc
2010-02-13 05:18:54 UTC
Permalink
Post by Inertial
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
What is q and v?
Post by Henry Wilson DSc
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre. The pulses are labelled according to their emission times and move at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p.
Doesn't matter if they are detected at p or at C
C is at point p.
Post by Inertial
Post by Henry Wilson DSc
At any instant, 40 million pulses are in transit
around the planet in each direction.
Tom
Tom? I thought it was Paul and Jerry?
Tom reckon the rotating frame analysis of Sagnac proves BaTh wrong.
Post by Inertial
Post by Henry Wilson DSc
observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period of 1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
OK .. assuming you got the math right
Post by Henry Wilson DSc
In this new situation, the pulses that arrive at C when it is at point p, were
not emitted at that point but rather from an imaginary point q,
Not an imaginary point at all. It is as real and physical point as p is.
All you are saying is that C has moved (and continues to move)
Hey dopey, can you see a point marked where you were 10 seconds ago?
Post by Inertial
Post by Henry Wilson DSc
which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Angular speed I think you would mean, it is moving around the ring according
to C's rotating frame.
v is its peripheral speed wrt the nonR frame.
Post by Inertial
Post by Henry Wilson DSc
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
So you're using an emission theory analysus, which does say that the time
taken for the two paths is the same.
It certainly is.
Post by Inertial
Post by Henry Wilson DSc
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
I'll take your word for it.
Post by Henry Wilson DSc
Since the distance between pulses is still 1 metre in the C's frame,
That's what emission theory would say
that's what is happening.
Post by Inertial
Post by Henry Wilson DSc
there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
WRONG. Emmision theory does not say that. It says that the distance in C's
frame between pulses is one meter. The distance the pulses travel in C's
frame is the same.
No dopey, the travel times are the same. The path lengths are different by +/-
62m
Post by Inertial
So the number of pulses in the fibre at any given time
is the same.
that's correct. that is why the fringes on a sagnac interferometer do not move
during constant rotation.
Post by Inertial
You can also tell it is wrong because the same number of pulses has been
emitted in each direction and you just said earlier that they take the same
time to travel around the ring. So there must be the same number of them in
there.
No it doesn't mean that at all. I have explained perfectly well what happens.
It is only during a CHANGE of speed that pulses move in and out of the two
paths.
Post by Inertial
Or does your explanation have pulse fairies that eat up the pulses?
No the explanation is that you are plain dopey.
Post by Inertial
Post by Henry Wilson DSc
"That's odd", says Tom, "now, for some reason, the pulses coming from the left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
Emmision theory doesn't say that will happen at all. It says what you said
earlier .. the time take for them to travel is the same.
Correct.
Post by Inertial
What you are now suddenly describing is what SR or an aether theory would
predict. You've changed theories in mid analysis !!!
No dopey. You are incapable of understanding what i said.
Post by Inertial
Post by Henry Wilson DSc
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
No .. because your analysis above used SR at the end. You switched theories
mid way thru.
You started saying the time for the pulses to travel in both directions
(with the rotating tube) was the same. That is what emission theory says.
And you said the distance between the pulses (in the rotating frame) is the
same. That is what emission theory says. The consequence of that is that
there MUST be the same number of pulses in each direction in the tube at any
given time.
Only when it is not rotating.
Post by Inertial
And they MUST arrive at the detector with the same label. And
so emission theory says that a sagnac device will NOT detect rotation.
No dopey , it says the sagnac effect is too hard for you to understand.
Post by Inertial
You've just admitted that your theory is refuted.
The *real* Sagnac devices have proven that THAT theory is NOT correct, and
it is NOT the same time required for the pulses (which is what SR predicts).
BAHAHAHHA
come back when you have learnt to use your brain

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-14 11:25:20 UTC
Permalink
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT
ROTATING.
Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
What is q and v?
Post by Henry Wilson DSc
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre. The pulses are labelled according to their emission times and
move
at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p.
Doesn't matter if they are detected at p or at C
C is at point p.
Yeup
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
At any instant, 40 million pulses are in transit
around the planet in each direction.
Tom
Tom? I thought it was Paul and Jerry?
Tom reckon the rotating frame analysis of Sagnac proves BaTh wrong.
You just muddled up your example by changing Jerry's name :) Guess you're
not able to admit something like that though.
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period
of
1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
OK .. assuming you got the math right
Post by Henry Wilson DSc
In this new situation, the pulses that arrive at C when it is at point
p,
were
not emitted at that point but rather from an imaginary point q,
Not an imaginary point at all. It is as real and physical point as p is.
All you are saying is that C has moved (and continues to move)
Hey dopey, can you see a point marked where you were 10 seconds ago?
Why do you think points need to be marked in order not to be imaginary?
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Angular speed I think you would mean, it is moving around the ring according
to C's rotating frame.
v is its peripheral speed wrt the nonR frame.
Fine .. I'm happy with that. Though a constant speed, it is not a constant
velocity.
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
So you're using an emission theory analysus, which does say that the time
taken for the two paths is the same.
It certainly is.
Good
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
I'll take your word for it.
Post by Henry Wilson DSc
Since the distance between pulses is still 1 metre in the C's frame,
That's what emission theory would say
that's what is happening.
Good
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
WRONG. Emmision theory does not say that. It says that the distance in C's
frame between pulses is one meter. The distance the pulses travel in C's
frame is the same.
No dopey, the travel times are the same. The path lengths are different by +/-
62m
In which frame?
Post by Henry Wilson DSc
Post by Inertial
So the number of pulses in the fibre at any given time
is the same.
that's correct. that is why the fringes on a sagnac interferometer do not move
during constant rotation.
I meant in each direction.
Post by Henry Wilson DSc
Post by Inertial
You can also tell it is wrong because the same number of pulses has been
emitted in each direction and you just said earlier that they take the same
time to travel around the ring. So there must be the same number of them in
there.
No it doesn't mean that at all.
Yes it does.
Post by Henry Wilson DSc
I have explained perfectly well what happens.
And it is self-contradictory
Post by Henry Wilson DSc
It is only during a CHANGE of speed that pulses move in and out of the two
paths.
Irrelevant
Post by Henry Wilson DSc
Post by Inertial
Or does your explanation have pulse fairies that eat up the pulses?
No the explanation is that you are plain dopey.
No .. it is that you can't put forward a logically consistent argument
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
"That's odd", says Tom, "now, for some reason, the pulses coming from
the
left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
Emmision theory doesn't say that will happen at all. It says what you said
earlier .. the time take for them to travel is the same.
Correct.
So you cannot measure the absolute rotation according to emission theory.
Real sagnac devices operate as SR predicts, and you CAN measure it
Post by Henry Wilson DSc
Post by Inertial
What you are now suddenly describing is what SR or an aether theory would
predict. You've changed theories in mid analysis !!!
No dopey. You are incapable of understanding what i said.
I understand just fine. Which is why i know you are wrong .. yet again
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
No .. because your analysis above used SR at the end. You switched theories
mid way thru.
You started saying the time for the pulses to travel in both directions
(with the rotating tube) was the same. That is what emission theory says.
And you said the distance between the pulses (in the rotating frame) is the
same. That is what emission theory says. The consequence of that is that
there MUST be the same number of pulses in each direction in the tube at any
given time.
Only when it is not rotating.
Not according to emission theory. if the pulses take the same time to get
back to point C, and they travel at the same speed in C's frame, then they
must have the same time stamp on them and must have travelled the same
distance (in that frame)
Post by Henry Wilson DSc
Post by Inertial
And they MUST arrive at the detector with the same label. And
so emission theory says that a sagnac device will NOT detect rotation.
No dopey , it says the sagnac effect is too hard for you to understand.
No .. yhou've just been getting it wrong for years .. and continue to post
your nonsese
Post by Henry Wilson DSc
Post by Inertial
You've just admitted that your theory is refuted.
The *real* Sagnac devices have proven that THAT theory is NOT correct, and
it is NOT the same time required for the pulses (which is what SR predicts).
BAHAHAHHA
come back when you have learnt to use your brain
At least I have one that works
Henry Wilson DSc
2010-02-14 21:49:15 UTC
Permalink
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT
ROTATING.
Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
What is q and v?
Post by Henry Wilson DSc
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre. The pulses are labelled according to their emission times and
move
at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p.
Doesn't matter if they are detected at p or at C
C is at point p.
Yeup
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
At any instant, 40 million pulses are in transit
around the planet in each direction.
Tom
Tom? I thought it was Paul and Jerry?
Tom reckon the rotating frame analysis of Sagnac proves BaTh wrong.
You just muddled up your example by changing Jerry's name :) Guess you're
not able to admit something like that though.
Tom has always believed that is true.
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period
of
1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
OK .. assuming you got the math right
Post by Henry Wilson DSc
In this new situation, the pulses that arrive at C when it is at point
p,
were
not emitted at that point but rather from an imaginary point q,
Not an imaginary point at all. It is as real and physical point as p is.
All you are saying is that C has moved (and continues to move)
Hey dopey, can you see a point marked where you were 10 seconds ago?
Why do you think points need to be marked in order not to be imaginary?
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Angular speed I think you would mean, it is moving around the ring according
to C's rotating frame.
v is its peripheral speed wrt the nonR frame.
Fine .. I'm happy with that. Though a constant speed, it is not a constant
velocity.
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
So you're using an emission theory analysus, which does say that the time
taken for the two paths is the same.
It certainly is.
Good
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
I'll take your word for it.
Post by Henry Wilson DSc
Since the distance between pulses is still 1 metre in the C's frame,
That's what emission theory would say
that's what is happening.
Good
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
WRONG. Emmision theory does not say that. It says that the distance in C's
frame between pulses is one meter. The distance the pulses travel in C's
frame is the same.
No dopey, the travel times are the same. The path lengths are different by +/-
62m
In which frame?
Post by Henry Wilson DSc
Post by Inertial
So the number of pulses in the fibre at any given time
is the same.
that's correct. that is why the fringes on a sagnac interferometer do not move
during constant rotation.
I meant in each direction.
Post by Henry Wilson DSc
Post by Inertial
You can also tell it is wrong because the same number of pulses has been
emitted in each direction and you just said earlier that they take the same
time to travel around the ring. So there must be the same number of them in
there.
No it doesn't mean that at all.
Yes it does.
Post by Henry Wilson DSc
I have explained perfectly well what happens.
And it is self-contradictory
Post by Henry Wilson DSc
It is only during a CHANGE of speed that pulses move in and out of the two
paths.
Irrelevant
Post by Henry Wilson DSc
Post by Inertial
Or does your explanation have pulse fairies that eat up the pulses?
No the explanation is that you are plain dopey.
No .. it is that you can't put forward a logically consistent argument
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
"That's odd", says Tom, "now, for some reason, the pulses coming from
the
left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
Emmision theory doesn't say that will happen at all. It says what you said
earlier .. the time take for them to travel is the same.
Correct.
So you cannot measure the absolute rotation according to emission theory.
Real sagnac devices operate as SR predicts, and you CAN measure it
Post by Henry Wilson DSc
Post by Inertial
What you are now suddenly describing is what SR or an aether theory would
predict. You've changed theories in mid analysis !!!
No dopey. You are incapable of understanding what i said.
I understand just fine. Which is why i know you are wrong .. yet again
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
No .. because your analysis above used SR at the end. You switched theories
mid way thru.
You started saying the time for the pulses to travel in both directions
(with the rotating tube) was the same. That is what emission theory says.
And you said the distance between the pulses (in the rotating frame) is the
same. That is what emission theory says. The consequence of that is that
there MUST be the same number of pulses in each direction in the tube at any
given time.
Only when it is not rotating.
Not according to emission theory. if the pulses take the same time to get
back to point C, and they travel at the same speed in C's frame, then they
must have the same time stamp on them and must have travelled the same
distance (in that frame)
Post by Henry Wilson DSc
Post by Inertial
And they MUST arrive at the detector with the same label. And
so emission theory says that a sagnac device will NOT detect rotation.
No dopey , it says the sagnac effect is too hard for you to understand.
No .. yhou've just been getting it wrong for years .. and continue to post
your nonsese
Post by Henry Wilson DSc
Post by Inertial
You've just admitted that your theory is refuted.
The *real* Sagnac devices have proven that THAT theory is NOT correct, and
it is NOT the same time required for the pulses (which is what SR predicts).
BAHAHAHHA
come back when you have learnt to use your brain
At least I have one that works
I admitted my mistakes and corrected them in the next post.

the labels are the same at all speeds.

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-14 22:44:32 UTC
Permalink
"Henry Wilson DSc" <***@..> wrote in message news:***@4ax.com...
[snip]
Post by Henry Wilson DSc
I admitted my mistakes and corrected them in the next post.
If so, that is a pleasant change for you .. Where is the next post?
Henry Wilson DSc
2010-02-14 23:07:39 UTC
Permalink
Post by Inertial
[snip]
Post by Henry Wilson DSc
I admitted my mistakes and corrected them in the next post.
If so, that is a pleasant change for you .. Where is the next post?
Just below the first one. posted on 14th

here's a copy.
***************
Post by Inertial
Post by Henry Wilson DSc
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
What is q and v?
Post by Henry Wilson DSc
Post by Henry Wilson DSc
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre. The pulses are labelled according to their emission times and move at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p.
Doesn't matter if they are detected at p or at C
I'll withdraw my previous reply because it was wrong. So was my initial message
because I stuffed up what I wanted to say.

p is the point where the pulses emitted at q are received.
Post by Inertial
Post by Henry Wilson DSc
Post by Henry Wilson DSc
observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period of 1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
OK .. assuming you got the math right
Post by Henry Wilson DSc
Post by Henry Wilson DSc
In this new situation, the pulses that arrive at C when it is at point p, were
not emitted at that point but rather from an imaginary point q,
Not an imaginary point at all. It is as real and physical point as p is.
All you are saying is that C has moved (and continues to move)
It is imaginary in the R frame...REAL in the nonR frame.
Post by Inertial
Post by Henry Wilson DSc
Post by Henry Wilson DSc
which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Angular speed I think you would mean, it is moving around the ring according
to C's rotating frame.
yes
Post by Inertial
Post by Henry Wilson DSc
Post by Henry Wilson DSc
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
So you're using an emission theory analysus, which does say that the time
taken for the two paths is the same.
Yes. The path lengths differ by 124 metres but one pulse moves at c+v and the
other at c-v, so they both arrive at p together.
Post by Inertial
Post by Henry Wilson DSc
Post by Henry Wilson DSc
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
I'll take your word for it.
Post by Henry Wilson DSc
Post by Henry Wilson DSc
Since the distance between pulses is still 1 metre in the C's frame,
That's what emission theory would say
Post by Henry Wilson DSc
Post by Henry Wilson DSc
there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
WRONG. Emmision theory does not say that. It says that the distance in C's
frame between pulses is one meter. The distance the pulses travel in C's
frame is the same. So the number of pulses in the fibre at any given time
is the same.
But I'm talking about the total number of pulses in transit in each path at any
instant. there are 124 more in one path than the other.
Post by Inertial
You can also tell it is wrong because the same number of pulses has been
emitted in each direction and you just said earlier that they take the same
time to travel around the ring. So there must be the same number of them in
there. Or does your explanation have pulse fairies that eat up the pulses?
Now you are being silly.

I made a mistake in my original post. This is regrettable but doesn't present a
problem for Bath.
As you rightly said, the pulses that leave together DO always arrive together
at the detector.
Post by Inertial
Post by Henry Wilson DSc
Post by Henry Wilson DSc
"That's odd", says Tom, "now, for some reason, the pulses coming from the left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
This is wrong....and is not what I wanted to point out.
Post by Inertial
Emmision theory doesn't say that will happen at all. It says what you said
earlier .. the time take for them to travel is the same.
It is.....and the label on each arriving pulse is the same.
Post by Inertial
What you are now suddenly describing is what SR or an aether theory would
predict. You've changed theories in mid analysis !!!
Yes, basically I did. I apologise for any inconvenience.
Post by Inertial
Post by Henry Wilson DSc
Post by Henry Wilson DSc
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
No .. because your analysis above used SR at the end. You switched theories
mid way thru.
Yes I did.
Post by Inertial
You started saying the time for the pulses to travel in both directions
(with the rotating tube) was the same. That is what emission theory says.
And you said the distance between the pulses (in the rotating frame) is the
same. That is what emission theory says.
It is.
Post by Inertial
The consequence of that is that
there MUST be the same number of pulses in each direction in the tube at any
given time.
No that is YOUR mistake. The path lengths are different so there are more
pulses in one path than the other.
Post by Inertial
And they MUST arrive at the detector with the same label.
They do. I was wrong, before.
Post by Inertial
And
so emission theory says that a sagnac device will NOT detect rotation.
You've just admitted that your theory is refuted.
Now I'll tell you why that is wrong.
Post by Inertial
The *real* Sagnac devices have proven that THAT theory is NOT correct, and
it is NOT the same time required for the pulses (which is what SR predicts).
BAHAHAHHA
The true fact is that because the number of pulses is different in each path,
THE RATE at which they arrive at the detector is DIFFERENT for each path.
Between the instant of emission from point q and the detection at point p,
40000062 pulses arrive from one direction but only 39999938 from the other.
Those numbers will change only during a rotational speed change.

It is that difference in number that is analogous to the classical 'phase
difference' of the two light rays in a ring gyro.

Two different arrival rates produces a beat frequency at the detector.
Exactly how that gives rise to interference fringes I'm not sure at this stage
but the fact that BaTh predicts such a difference and produces the correct
equation for fringe displacement is sufficient to demonstrate that Sagnac
definitely does NOT refute BaTh.



Henry Wilson...

.......provider of free physics lessons

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-14 23:21:23 UTC
Permalink
Post by Henry Wilson DSc
Post by Inertial
[snip]
Post by Henry Wilson DSc
I admitted my mistakes and corrected them in the next post.
If so, that is a pleasant change for you .. Where is the next post?
Just below the first one. posted on 14th
here's a copy.
***************
Post by Inertial
Post by Henry Wilson DSc
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT
ROTATING.
Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
What is q and v?
Post by Henry Wilson DSc
Post by Henry Wilson DSc
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre. The pulses are labelled according to their emission times and
move
at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p.
Doesn't matter if they are detected at p or at C
I'll withdraw my previous reply because it was wrong. So was my initial message
because I stuffed up what I wanted to say.
That's fine. Everyone makes mistakes at some time or another. I am always
willing and happy to take that into account when it is admitted in that way.
Inertial
2010-02-14 23:28:50 UTC
Permalink
Post by Henry Wilson DSc
Two different arrival rates produces a beat frequency at the detector.
There is no beat frequency in a Sagnac device. The two beams have the same
frequency at the detector. This is the case in both emission theory and SR.
SR, however, predicts differing arrival TIMES for pulses emitted at the same
time .. hence the shift. Emission theory says they arrive at the same time
(as you agree) .. hence there is no shift predicted (this refuting the
theory).

That your theory is even worse and somehow predicts differing arrival rates
and a beat frequency that does not occur is enough proof to refute it.
Henry Wilson DSc
2010-02-15 00:32:32 UTC
Permalink
Post by Inertial
Post by Henry Wilson DSc
Two different arrival rates produces a beat frequency at the detector.
There is no beat frequency in a Sagnac device. The two beams have the same
frequency at the detector. This is the case in both emission theory and SR.
SR, however, predicts differing arrival TIMES for pulses emitted at the same
time .. hence the shift. Emission theory says they arrive at the same time
(as you agree) .. hence there is no shift predicted (this refuting the
theory).
That your theory is even worse and somehow predicts differing arrival rates
and a beat frequency that does not occur is enough proof to refute it.
You still don't get it.

The two beams DO NOT have the same frequency at the detector.
In the clock example I gave, 40000062 pulses arrive at the detector from one
direction but only 3999938 from the other DURING THE COMMON TRAVEL TIME of any
particular split pulse. Those numbers change during a speed CHANGE.

Convert that to 'wavecrests', in the case of light, and you get a beat
frequency that is somehow related to fringe displacement. Don't ask me how at
the moment but the reason will reveal quite a lot about the nature of a photon.

The fact that the frequencies in each direction are different annihilates the
Roberts, Jerry, Andersen 'rotating frame' argument as well as your own.



Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-15 00:50:45 UTC
Permalink
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Two different arrival rates produces a beat frequency at the detector.
There is no beat frequency in a Sagnac device. The two beams have the same
frequency at the detector. This is the case in both emission theory and SR.
SR, however, predicts differing arrival TIMES for pulses emitted at the same
time .. hence the shift. Emission theory says they arrive at the same time
(as you agree) .. hence there is no shift predicted (this refuting the
theory).
That your theory is even worse and somehow predicts differing arrival rates
and a beat frequency that does not occur is enough proof to refute it.
You still don't get it.
I get it just fine
Post by Henry Wilson DSc
The two beams DO NOT have the same frequency at the detector.
Yes, they do in reality. That your theory predicts otherwise shows it to be
wrong.
Post by Henry Wilson DSc
In the clock example I gave, 40000062 pulses arrive at the detector from one
direction but only 3999938 from the other DURING THE COMMON TRAVEL TIME of any
particular split pulse. Those numbers change during a speed CHANGE.
And that is nonsense .. as the time taken for each pulse to make the trip is
the same (in both directions), and the time between pulses being emitted is
the same (in both directions), then in a given time, the same number of
pulses arrive (in both directions).
Post by Henry Wilson DSc
Convert that to 'wavecrests', in the case of light, and you get a beat
frequency that is somehow related to fringe displacement. Don't ask me how at
the moment but the reason will reveal quite a lot about the nature of a photon.
Because it is nonnsese. There is no beat frequency.
Post by Henry Wilson DSc
The fact
Lie
Post by Henry Wilson DSc
that the frequencies in each direction are different
They are not.
Post by Henry Wilson DSc
annihilates the
Roberts, Jerry, Andersen 'rotating frame' argument as well as your own.
Nope .. you've just proven your own argument wrong. Well done.
Henry Wilson DSc
2010-02-15 02:21:39 UTC
Permalink
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Two different arrival rates produces a beat frequency at the detector.
There is no beat frequency in a Sagnac device. The two beams have the same
frequency at the detector. This is the case in both emission theory and SR.
SR, however, predicts differing arrival TIMES for pulses emitted at the same
time .. hence the shift. Emission theory says they arrive at the same time
(as you agree) .. hence there is no shift predicted (this refuting the
theory).
That your theory is even worse and somehow predicts differing arrival rates
and a beat frequency that does not occur is enough proof to refute it.
You still don't get it.
I get it just fine
Post by Henry Wilson DSc
The two beams DO NOT have the same frequency at the detector.
Yes, they do in reality. That your theory predicts otherwise shows it to be
wrong.
Post by Henry Wilson DSc
In the clock example I gave, 40000062 pulses arrive at the detector from one
direction but only 3999938 from the other DURING THE COMMON TRAVEL TIME of any
particular split pulse. Those numbers change during a speed CHANGE.
And that is nonsense .. as the time taken for each pulse to make the trip is
the same (in both directions), and the time between pulses being emitted is
the same (in both directions), then in a given time, the same number of
pulses arrive (in both directions).
Try thinking for once. I know this is difficult...but you should be able to
understand.
The path lengths are different. The distance between pulses is 1 metre in all
frames. The number of pulses in each path differs by 124. When a pulse is split
into two and sent down both fibres, it takes the same time to get to the
detector in both directions. Both halves arrive together.
So 124 more pulses have arrived from one direction than the other.
IN OTHER WORDS, the pulse arrival rate is different from both paths.
Post by Inertial
Post by Henry Wilson DSc
Convert that to 'wavecrests', in the case of light, and you get a beat
frequency that is somehow related to fringe displacement. Don't ask me how at
the moment but the reason will reveal quite a lot about the nature of a photon.
Because it is nonnsese. There is no beat frequency.
Post by Henry Wilson DSc
The fact
Lie
Post by Henry Wilson DSc
that the frequencies in each direction are different
They are not.
Post by Henry Wilson DSc
annihilates the
Roberts, Jerry, Andersen 'rotating frame' argument as well as your own.
Nope .. you've just proven your own argument wrong. Well done.
Use your brain...

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-15 02:32:23 UTC
Permalink
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Two different arrival rates produces a beat frequency at the detector.
There is no beat frequency in a Sagnac device. The two beams have the same
frequency at the detector. This is the case in both emission theory and SR.
SR, however, predicts differing arrival TIMES for pulses emitted at the same
time .. hence the shift. Emission theory says they arrive at the same time
(as you agree) .. hence there is no shift predicted (this refuting the
theory).
That your theory is even worse and somehow predicts differing arrival rates
and a beat frequency that does not occur is enough proof to refute it.
You still don't get it.
I get it just fine
Post by Henry Wilson DSc
The two beams DO NOT have the same frequency at the detector.
Yes, they do in reality. That your theory predicts otherwise shows it to be
wrong.
Post by Henry Wilson DSc
In the clock example I gave, 40000062 pulses arrive at the detector from one
direction but only 3999938 from the other DURING THE COMMON TRAVEL TIME
of
any
particular split pulse. Those numbers change during a speed CHANGE.
And that is nonsense .. as the time taken for each pulse to make the trip is
the same (in both directions), and the time between pulses being emitted is
the same (in both directions), then in a given time, the same number of
pulses arrive (in both directions).
Try thinking for once. I know this is difficult...but you should be able to
understand.
The path lengths are different. The distance between pulses is 1 metre in all
frames. The number of pulses in each path differs by 124. When a pulse is split
into two and sent down both fibres, it takes the same time to get to the
detector in both directions. Both halves arrive together.
So 124 more pulses have arrived from one direction than the other.
IN OTHER WORDS, the pulse arrival rate is different from both paths.
Post by Inertial
Post by Henry Wilson DSc
Convert that to 'wavecrests', in the case of light, and you get a beat
frequency that is somehow related to fringe displacement. Don't ask me
how
at
the moment but the reason will reveal quite a lot about the nature of a photon.
Because it is nonnsese. There is no beat frequency.
Post by Henry Wilson DSc
The fact
Lie
Post by Henry Wilson DSc
that the frequencies in each direction are different
They are not.
Post by Henry Wilson DSc
annihilates the
Roberts, Jerry, Andersen 'rotating frame' argument as well as your own.
Nope .. you've just proven your own argument wrong. Well done.
Use your brain...
I did .. and you just refuted your own theory. Saying 'it predicts
something that doesn't happen and somehow instead what is observed does
happen anyway and my theory cannot explain why' means your theory is
refuted.
Henry Wilson DSc
2010-02-15 02:45:18 UTC
Permalink
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
annihilates the
Roberts, Jerry, Andersen 'rotating frame' argument as well as your own.
Nope .. you've just proven your own argument wrong. Well done.
Use your brain...
I did .. and you just refuted your own theory. Saying 'it predicts
something that doesn't happen and somehow instead what is observed does
happen anyway and my theory cannot explain why' means your theory is
refuted.
Don't kid yourself.

By counting the pulses arriving from the two directions, I can calculate the
earth's rotation speed. I don't need an interferometer. A clock will do just as
well.

Sagnac fully supports BaTh.

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-15 02:55:08 UTC
Permalink
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
annihilates the
Roberts, Jerry, Andersen 'rotating frame' argument as well as your own.
Nope .. you've just proven your own argument wrong. Well done.
Use your brain...
I did .. and you just refuted your own theory. Saying 'it predicts
something that doesn't happen and somehow instead what is observed does
happen anyway and my theory cannot explain why' means your theory is
refuted.
Don't kid yourself.
I'm not
Post by Henry Wilson DSc
By counting the pulses arriving from the two directions, I can calculate the
earth's rotation speed.
Nope. as the tube has a fixed length, the same number must arrive per
second as are sent per second. Otherwise either pulses are disappearing
along the way, or being created from nowhere. You theory relies on pulse
fairies.
Post by Henry Wilson DSc
I don't need an interferometer. A clock will do just as
well.
Sagnac fully supports BaTh.
Nope .. your supposed 'BaTh' is pure impossible nonsense.
Inertial
2010-02-15 02:52:59 UTC
Permalink
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Two different arrival rates produces a beat frequency at the detector.
There is no beat frequency in a Sagnac device. The two beams have the same
frequency at the detector. This is the case in both emission theory and SR.
SR, however, predicts differing arrival TIMES for pulses emitted at the same
time .. hence the shift. Emission theory says they arrive at the same time
(as you agree) .. hence there is no shift predicted (this refuting the
theory).
That your theory is even worse and somehow predicts differing arrival rates
and a beat frequency that does not occur is enough proof to refute it.
You still don't get it.
I get it just fine
Post by Henry Wilson DSc
The two beams DO NOT have the same frequency at the detector.
Yes, they do in reality. That your theory predicts otherwise shows it to be
wrong.
Post by Henry Wilson DSc
In the clock example I gave, 40000062 pulses arrive at the detector from one
direction but only 3999938 from the other DURING THE COMMON TRAVEL TIME
of
any
particular split pulse. Those numbers change during a speed CHANGE.
And that is nonsense .. as the time taken for each pulse to make the trip is
the same (in both directions), and the time between pulses being emitted is
the same (in both directions), then in a given time, the same number of
pulses arrive (in both directions).
Try thinking for once. I know this is difficult...
It is for you
Post by Henry Wilson DSc
but you should be able to
understand.
I do
Post by Henry Wilson DSc
The path lengths are different.
The distance between pulses is 1 metre in all
frames. The number of pulses in each path differs by 124. When a pulse is split
into two and sent down both fibres, it takes the same time to get to the
detector in both directions. Both halves arrive together.
So 124 more pulses have arrived from one direction than the other.
IN OTHER WORDS, the pulse arrival rate is different from both paths.
So the following is equivalent to what you have here:

You have two paths from (say) A to B.

You have n objects leaving A over a period of 1 second down one path and
taking time T to get to B. The first object arrives at time T, the last at
time T+1.

You have another n objects leaving A over a period of 1 second down the
other path and taking the same time T to get to B. The first object arrives
at time T, the last at time T+1.

Yet by your logic, you would claim that more would arrive per second at B
from one path than the other.

That is just pure impossible nonsense.

You also claim that even though in any given time you emit the same number
of pulses in both directions, and that they are all in the tube for the same
time, that there are more pulses in the tube going in one direction than
their are in the other. Where do the extra pulses come from?

That is also just pure impossible nonsense.

At any given time the tube has the same length in both directions .. the
length is the circumference of the tube, it does not change. And as you say
that the pulses are the same distance apart, there must be the same number
of them in the tube at a time. But you claim that at any given time there
are more pulses in the tube going in one direction than that other.

That is also just pure impossible nonsense.

You really have just totally refuted your 'thoery'. It is all based on pure
impossible nonsense.
Henry Wilson DSc
2010-02-15 05:44:11 UTC
Permalink
Post by Inertial
Post by Henry Wilson DSc
The path lengths are different.
The distance between pulses is 1 metre in all
frames. The number of pulses in each path differs by 124. When a pulse is split
into two and sent down both fibres, it takes the same time to get to the
detector in both directions. Both halves arrive together.
So 124 more pulses have arrived from one direction than the other.
IN OTHER WORDS, the pulse arrival rate is different from both paths.
You have two paths from (say) A to B.
You have n objects leaving A over a period of 1 second down one path and
taking time T to get to B. The first object arrives at time T, the last at
time T+1.
You have another n objects leaving A over a period of 1 second down the
other path and taking the same time T to get to B. The first object arrives
at time T, the last at time T+1.
THIS IS WRONG. HOW LONG DOES IT TAKE FOR SOMETHING TO SINK INTO YOUR BRAIN.

You have n+m objects in one path and n-m objects in the other.

One path moves faster than the other so the nth object in each path reunites
with its counterpart simultaneously at the detector. While it is traveling, n+m
objects reach the detector one way and n-m the other way.
Post by Inertial
Yet by your logic, you would claim that more would arrive per second at B
from one path than the other.
Do you know what a bicycle chain looks like? The faster you move it the more
links pass a particular point.
Post by Inertial
That is just pure impossible nonsense.
What you are saying is pure nonsense , yes.
Post by Inertial
You also claim that even though in any given time you emit the same number
of pulses in both directions, and that they are all in the tube for the same
time, that there are more pulses in the tube going in one direction than
their are in the other. Where do the extra pulses come from?
They enter or leave each path during a speed change...That is when the fringes
are seen to MOVE.
Post by Inertial
That is also just pure impossible nonsense.
What you are saying is pure nonsense, yes.
Post by Inertial
At any given time the tube has the same length in both directions .. the
length is the circumference of the tube, it does not change. And as you say
that the pulses are the same distance apart, there must be the same number
of them in the tube at a time.
Only when it is not rotating. and p = q.

You have to consider the distance C moves while a pulse is in transit. That is
the reason for the path length difference.

I might remind you that your own SR says exactly that too.
Post by Inertial
But you claim that at any given time there
are more pulses in the tube going in one direction than that other.
That is also just pure impossible nonsense.
You really have just totally refuted your 'thoery'. It is all based on pure
impossible nonsense.
You will probably feel very embarrased when and if the penny drops but I doubt
if you have the intelligence to make that happen.


Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-15 06:12:31 UTC
Permalink
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
The path lengths are different.
The distance between pulses is 1 metre in all
frames. The number of pulses in each path differs by 124. When a pulse
is
split
into two and sent down both fibres, it takes the same time to get to the
detector in both directions. Both halves arrive together.
So 124 more pulses have arrived from one direction than the other.
IN OTHER WORDS, the pulse arrival rate is different from both paths.
You have two paths from (say) A to B.
You have n objects leaving A over a period of 1 second down one path and
taking time T to get to B. The first object arrives at time T, the last at
time T+1.
You have another n objects leaving A over a period of 1 second down the
other path and taking the same time T to get to B. The first object arrives
at time T, the last at time T+1.
THIS IS WRONG. HOW LONG DOES IT TAKE FOR SOMETHING TO SINK INTO YOUR BRAIN.
No .. its not wrong.
Post by Henry Wilson DSc
You have n+m objects in one path and n-m objects in the other.
Where did the m extra objects come from? The same number go down each path
and the same number arrive at the other end. So each pair that leave, they
arrive together. Where are these extra/missing pulses?
Post by Henry Wilson DSc
One path moves faster than the other so the nth object in each path reunites
with its counterpart simultaneously at the detector.
So that means there must be the same number of objects in the path.
Post by Henry Wilson DSc
While it is traveling, n+m
objects reach the detector one way and n-m the other way.
Nope. Impossible. For each pulse entering path 1, a pulse enters path 2.
For each pulse exiting path 1, a pulse exits path 2. When do the m extra
pulses enter one path, and when do m fewer enter the other?
Post by Henry Wilson DSc
Post by Inertial
Yet by your logic, you would claim that more would arrive per second at B
from one path than the other.
Do you know what a bicycle chain looks like? The faster you move it the more
links pass a particular point.
Irrelevant
Post by Henry Wilson DSc
Post by Inertial
That is just pure impossible nonsense.
What you are saying is pure nonsense , yes.
I'm only repeating what you are claiming
Post by Henry Wilson DSc
Post by Inertial
You also claim that even though in any given time you emit the same number
of pulses in both directions, and that they are all in the tube for the same
time, that there are more pulses in the tube going in one direction than
their are in the other. Where do the extra pulses come from?
They enter or leave each path during a speed change...
There is no speed change. The rotation is a constant rate.
Post by Henry Wilson DSc
That is when the fringes
are seen to MOVE.
Post by Inertial
That is also just pure impossible nonsense.
What you are saying is pure nonsense, yes.
I'm only repeating what you are claiming
Post by Henry Wilson DSc
Post by Inertial
At any given time the tube has the same length in both directions .. the
length is the circumference of the tube, it does not change. And as you say
that the pulses are the same distance apart, there must be the same number
of them in the tube at a time.
Only when it is not rotating. and p = q.
Wrong
Post by Henry Wilson DSc
You have to consider the distance C moves while a pulse is in transit. That is
the reason for the path length difference.
Path length is different .. BUT length of tube is unchanged. The distance
between pulses is the same .. so number of pulses in each direction in the
tube must be the same.
Post by Henry Wilson DSc
I might remind you that your own SR says exactly that too.
SR's correctness is not relevant in a discussion of your nonsense.
Post by Henry Wilson DSc
Post by Inertial
But you claim that at any given time there
are more pulses in the tube going in one direction than that other.
That is also just pure impossible nonsense.
You really have just totally refuted your 'thoery'. It is all based on pure
impossible nonsense.
You will probably feel very embarrased when and if the penny drops but I doubt
if you have the intelligence to make that happen.
You won't ever admit your mistakes .. you're too stupid a moron to realise
and too dishonest to admit it if you did.
Henry Wilson DSc
2010-02-15 10:33:10 UTC
Permalink
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
You have two paths from (say) A to B.
You have n objects leaving A over a period of 1 second down one path and
taking time T to get to B. The first object arrives at time T, the last at
time T+1.
You have another n objects leaving A over a period of 1 second down the
other path and taking the same time T to get to B. The first object arrives
at time T, the last at time T+1.
THIS IS WRONG. HOW LONG DOES IT TAKE FOR SOMETHING TO SINK INTO YOUR BRAIN.
No .. its not wrong.
Post by Henry Wilson DSc
You have n+m objects in one path and n-m objects in the other.
Where did the m extra objects come from? The same number go down each path
and the same number arrive at the other end. So each pair that leave, they
arrive together. Where are these extra/missing pulses?
Look, that is the case when it is not rotating.

What you and many others cannot get into your heads is that when it IS
rotating, the source/detector moves a distance vt during the travel time of a
pulse.

This is what makes the path lengths unequal...and results in more pulses being
present in one path than the other.

I know this is difficult but you should at least try to understand it.
It is the basis of the SR sagnac explanation as well as BaTh's.
Post by Inertial
Post by Henry Wilson DSc
One path moves faster than the other so the nth object in each path reunites
with its counterpart simultaneously at the detector.
So that means there must be the same number of objects in the path.
You are ignoring the distance vt. In both SR and BaTh, the path lengths are
2piR + vt and 2piR - vt.
Post by Inertial
Post by Henry Wilson DSc
While it is traveling, n+m
objects reach the detector one way and n-m the other way.
Nope. Impossible. For each pulse entering path 1, a pulse enters path 2.
For each pulse exiting path 1, a pulse exits path 2. When do the m extra
pulses enter one path, and when do m fewer enter the other?
I have told you already...during an acceleration.
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Yet by your logic, you would claim that more would arrive per second at B
from one path than the other.
Do you know what a bicycle chain looks like? The faster you move it the more
links pass a particular point.
Irrelevant
no it isn't . You can obviously only understand very simple language,
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
time, that there are more pulses in the tube going in one direction than
their are in the other. Where do the extra pulses come from?
They enter or leave each path during a speed change...
There is no speed change. The rotation is a constant rate.
At constant rate, vt is constant and there is fringe displacement but no fringe
movement.
Post by Inertial
Post by Henry Wilson DSc
That is when the fringes
are seen to MOVE.
Post by Inertial
That is also just pure impossible nonsense.
What you are saying is pure nonsense, yes.
I'm only repeating what you are claiming
Post by Henry Wilson DSc
Post by Inertial
At any given time the tube has the same length in both directions .. the
length is the circumference of the tube, it does not change. And as you say
that the pulses are the same distance apart, there must be the same number
of them in the tube at a time.
Only when it is not rotating. and p = q.
Wrong
Post by Henry Wilson DSc
You have to consider the distance C moves while a pulse is in transit. That is
the reason for the path length difference.
Path length is different .. BUT length of tube is unchanged. The distance
between pulses is the same .. so number of pulses in each direction in the
tube must be the same.
The length of the fibre is NOT the path length. You are ignoring the distance
vt.
You are also ignoring the fact that every rotating frame has an associated
intertial frame.
Post by Inertial
Post by Henry Wilson DSc
I might remind you that your own SR says exactly that too.
SR's correctness is not relevant in a discussion of your nonsense.
I was merely emphasising your inability to understand plain physics.
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
But you claim that at any given time there
are more pulses in the tube going in one direction than that other.
That is also just pure impossible nonsense.
You really have just totally refuted your 'thoery'. It is all based on pure
impossible nonsense.
You will probably feel very embarrased when and if the penny drops but I doubt
if you have the intelligence to make that happen.
You won't ever admit your mistakes .. you're too stupid a moron to realise
and too dishonest to admit it if you did.
You are as useless as Einstein...zero spatial and logical ability.


Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-15 10:59:11 UTC
Permalink
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
You have two paths from (say) A to B.
You have n objects leaving A over a period of 1 second down one path and
taking time T to get to B. The first object arrives at time T, the last at
time T+1.
You have another n objects leaving A over a period of 1 second down the
other path and taking the same time T to get to B. The first object arrives
at time T, the last at time T+1.
THIS IS WRONG. HOW LONG DOES IT TAKE FOR SOMETHING TO SINK INTO YOUR BRAIN.
No .. its not wrong.
Post by Henry Wilson DSc
You have n+m objects in one path and n-m objects in the other.
Where did the m extra objects come from? The same number go down each path
and the same number arrive at the other end. So each pair that leave, they
arrive together. Where are these extra/missing pulses?
Look, that is the case when it is not rotating.
Rotation doesn't change it.
Post by Henry Wilson DSc
What you and many others cannot get into your heads is that when it IS
rotating, the source/detector moves a distance vt during the travel time of a
pulse.
Of course they do
Post by Henry Wilson DSc
This is what makes the path lengths unequal...
Of course they are
Post by Henry Wilson DSc
and results in more pulses being
present in one path than the other.
The number of pulses in the tube at any given time going one way is the same
as the number going the other.
Post by Henry Wilson DSc
I know this is difficult but you should at least try to understand it.
It is the basis of the SR sagnac explanation as well as BaTh's.
Nope. Your lies are not a basis for anything
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
One path moves faster than the other so the nth object in each path reunites
with its counterpart simultaneously at the detector.
So that means there must be the same number of objects in the path.
You are ignoring the distance vt.
Not ignoring it. It is compensated by the pulses moving faster along the
longer path.
Post by Henry Wilson DSc
In both SR and BaTh, the path lengths are
2piR + vt and 2piR - vt.
Nope. SR and Emission theory have different pairs of values for the two
path lengths for the two path lengths.
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
While it is traveling, n+m
objects reach the detector one way and n-m the other way.
Nope. Impossible. For each pulse entering path 1, a pulse enters path 2.
For each pulse exiting path 1, a pulse exits path 2. When do the m extra
pulses enter one path, and when do m fewer enter the other?
I have told you already...during an acceleration.
There is no acceleration of the tube rotation rate. The tube rotates at a
constant speed
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Yet by your logic, you would claim that more would arrive per second at B
from one path than the other.
Do you know what a bicycle chain looks like? The faster you move it the more
links pass a particular point.
Irrelevant
no it isn't .
It isn't relevant.
Post by Henry Wilson DSc
You can obviously only understand very simple language,
I can understand things that you have no hope of.
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
time, that there are more pulses in the tube going in one direction than
their are in the other. Where do the extra pulses come from?
They enter or leave each path during a speed change...
There is no speed change. The rotation is a constant rate.
At constant rate, vt is constant and there is fringe displacement but no fringe
movement.
Your theory does not predict that. You say that there is a frequency beat
which means a constantly changing fringe displacement.
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
That is when the fringes
are seen to MOVE.
Post by Inertial
That is also just pure impossible nonsense.
What you are saying is pure nonsense, yes.
I'm only repeating what you are claiming
Post by Henry Wilson DSc
Post by Inertial
At any given time the tube has the same length in both directions .. the
length is the circumference of the tube, it does not change. And as you say
that the pulses are the same distance apart, there must be the same number
of them in the tube at a time.
Only when it is not rotating. and p = q.
Wrong
Post by Henry Wilson DSc
You have to consider the distance C moves while a pulse is in transit. That is
the reason for the path length difference.
Path length is different .. BUT length of tube is unchanged. The distance
between pulses is the same .. so number of pulses in each direction in the
tube must be the same.
The length of the fibre is NOT the path length.
I didn't say it was
Post by Henry Wilson DSc
You are ignoring the distance
vt.
No .. it is simply not relevant to the number of pulses in the tube at a
given time. Only the tube length is relevant.
Post by Henry Wilson DSc
You are also ignoring the fact that every rotating frame has an associated
intertial frame.
Irrelevant.
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
I might remind you that your own SR says exactly that too.
SR's correctness is not relevant in a discussion of your nonsense.
I was merely emphasising your inability to understand plain physics.
I understand it just fine. And I can see your nonsense very clearly. Its
hilarious
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
But you claim that at any given time there
are more pulses in the tube going in one direction than that other.
That is also just pure impossible nonsense.
You really have just totally refuted your 'thoery'. It is all based on pure
impossible nonsense.
You will probably feel very embarrased when and if the penny drops but I doubt
if you have the intelligence to make that happen.
You won't ever admit your mistakes .. you're too stupid a moron to realise
and too dishonest to admit it if you did.
You are as useless as Einstein...zero spatial and logical ability.
You're the one who has pulse fairies creating pulses and fringe fairies
stoping the beat frequency changing the fringe displacement. Your theory is
just a work of fiction and unrelated to reality.
Henry Wilson DSc
2010-02-15 22:17:32 UTC
Permalink
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
You have two paths from (say) A to B.
You have n objects leaving A over a period of 1 second down one path and
taking time T to get to B. The first object arrives at time T, the last at
time T+1.
You have another n objects leaving A over a period of 1 second down the
other path and taking the same time T to get to B. The first object arrives
at time T, the last at time T+1.
THIS IS WRONG. HOW LONG DOES IT TAKE FOR SOMETHING TO SINK INTO YOUR BRAIN.
No .. its not wrong.
Post by Henry Wilson DSc
You have n+m objects in one path and n-m objects in the other.
Where did the m extra objects come from? The same number go down each path
and the same number arrive at the other end. So each pair that leave, they
arrive together. Where are these extra/missing pulses?
Look, that is the case when it is not rotating.
Rotation doesn't change it.
It changes the path lengths. Are you not aware that light takes time to travel
and during its passage around the Earth, the Earth moves a distance vt.
Gawd, just look up any SR analysis of sagnac and you will see this.
Post by Inertial
Post by Henry Wilson DSc
What you and many others cannot get into your heads is that when it IS
rotating, the source/detector moves a distance vt during the travel time of a
pulse.
Of course they do
Post by Henry Wilson DSc
This is what makes the path lengths unequal...
Of course they are
Post by Henry Wilson DSc
and results in more pulses being
present in one path than the other.
The number of pulses in the tube at any given time going one way is the same
as the number going the other.
It fucking is NOT.
The numbers arriving from each direction between the emission of pulse X and
its arrival at the detector differ by 124 in my example.
If you cannot understand this I will waste no further time on you.
Post by Inertial
Post by Henry Wilson DSc
I know this is difficult but you should at least try to understand it.
It is the basis of the SR sagnac explanation as well as BaTh's.
Nope. Your lies are not a basis for anything
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
One path moves faster than the other so the nth object in each path reunites
with its counterpart simultaneously at the detector.
So that means there must be the same number of objects in the path.
You are ignoring the distance vt.
Not ignoring it. It is compensated by the pulses moving faster along the
longer path.
They DO move faster. That's why more arrive at the detector in the same time.
Post by Inertial
Post by Henry Wilson DSc
In both SR and BaTh, the path lengths are
2piR + vt and 2piR - vt.
Nope. SR and Emission theory have different pairs of values for the two
path lengths for the two path lengths.
They do not. Both theories asy the path lengths are 2piR+/vt

I think it's time you did some reading and used your brain...if you have one.
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
While it is traveling, n+m
objects reach the detector one way and n-m the other way.
Nope. Impossible. For each pulse entering path 1, a pulse enters path 2.
For each pulse exiting path 1, a pulse exits path 2. When do the m extra
pulses enter one path, and when do m fewer enter the other?
I have told you already...during an acceleration.
There is no acceleration of the tube rotation rate. The tube rotates at a
constant speed
gawd! what an idiot.

go away you are not capable of discussing physics...


Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-15 22:25:09 UTC
Permalink
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
You have two paths from (say) A to B.
You have n objects leaving A over a period of 1 second down one path and
taking time T to get to B. The first object arrives at time T, the
last
at
time T+1.
You have another n objects leaving A over a period of 1 second down the
other path and taking the same time T to get to B. The first object arrives
at time T, the last at time T+1.
THIS IS WRONG. HOW LONG DOES IT TAKE FOR SOMETHING TO SINK INTO YOUR BRAIN.
No .. its not wrong.
Post by Henry Wilson DSc
You have n+m objects in one path and n-m objects in the other.
Where did the m extra objects come from? The same number go down each path
and the same number arrive at the other end. So each pair that leave, they
arrive together. Where are these extra/missing pulses?
Look, that is the case when it is not rotating.
Rotation doesn't change it.
It changes the path lengths.
It doesn't change the length of the tube, or the number of pulses in transit
at any time nor the arrival rate.
Post by Henry Wilson DSc
Are you not aware that light takes time to travel
and during its passage around the Earth, the Earth moves a distance vt.
Gawd, just look up any SR analysis of sagnac and you will see this.
You don't understand the SR analysis .. nor SR itself.
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
What you and many others cannot get into your heads is that when it IS
rotating, the source/detector moves a distance vt during the travel time of a
pulse.
Of course they do
Post by Henry Wilson DSc
This is what makes the path lengths unequal...
Of course they are
Post by Henry Wilson DSc
and results in more pulses being
present in one path than the other.
The number of pulses in the tube at any given time going one way is the same
as the number going the other.
It fucking is NOT.
Of course it is .. if you put the same number of pulses into a fixed length
tube and they are in the tube for the same time, its the same number of
pulses
Post by Henry Wilson DSc
The numbers arriving from each direction between the emission of pulse X and
its arrival at the detector differ by 124 in my example.
Nopwe .. you're a blatant liar or a fool. My bet is both.
Post by Henry Wilson DSc
If you cannot understand this I will waste no further time on you.
So you run away when you are proven wrong. Typical cowardly lying crackpot
Henry.
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
I know this is difficult but you should at least try to understand it.
It is the basis of the SR sagnac explanation as well as BaTh's.
Nope. Your lies are not a basis for anything
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
One path moves faster than the other so the nth object in each path reunites
with its counterpart simultaneously at the detector.
So that means there must be the same number of objects in the path.
You are ignoring the distance vt.
Not ignoring it. It is compensated by the pulses moving faster along the
longer path.
They DO move faster. That's why more arrive at the detector in the same time.
No .. they don't. The same number of pulses are emitted per second, so the
same number must be detected per second
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
In both SR and BaTh, the path lengths are
2piR + vt and 2piR - vt.
Nope. SR and Emission theory have different pairs of values for the two
path lengths for the two path lengths.
They do not.
Yes .. the ydo have different values
Post by Henry Wilson DSc
Both theories asy the path lengths are 2piR+/vt
The t values are different for the two pulses in SR, not in emission theory

You just have no idea.
Post by Henry Wilson DSc
I think it's time you did some reading and used your brain...if you have one.
I use mine .. your only use of your brain is to lie and cheat. What a
waste.
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
While it is traveling, n+m
objects reach the detector one way and n-m the other way.
Nope. Impossible. For each pulse entering path 1, a pulse enters path 2.
For each pulse exiting path 1, a pulse exits path 2. When do the m extra
pulses enter one path, and when do m fewer enter the other?
I have told you already...during an acceleration.
There is no acceleration of the tube rotation rate. The tube rotates at a
constant speed
gawd! what an idiot.
Indeed you are
Post by Henry Wilson DSc
go away you are not capable of discussing physics...
A dsicssion of physics with a moron like you is pretty pointless. But it is
worth having your lies exposed for the benefit of others.

And they have been .. your assertions are soundly refuted.
Inertial
2010-02-15 22:31:25 UTC
Permalink
Post by Henry Wilson DSc
The numbers arriving from each direction between the emission of pulse X and
its arrival at the detector differ by 124 in my example.
So, according to you in some fixed period of time, there are 62 more pulses
arriving in one direction than were emitted in that same period of time. So
if you leave it running for a while, you end up with more pulses coming out
of the device than have gone into it. You've created free energy.
Congratulations.
Henry Wilson DSc
2010-02-15 23:04:25 UTC
Permalink
Post by Inertial
Post by Henry Wilson DSc
The numbers arriving from each direction between the emission of pulse X and
its arrival at the detector differ by 124 in my example.
So, according to you in some fixed period of time, there are 62 more pulses
arriving in one direction than were emitted in that same period of time.
Yes. It's called doppler shift.
Post by Inertial
So
if you leave it running for a while, you end up with more pulses coming out
of the device than have gone into it. You've created free energy.
Congratulations.
moron. The increased number in one path is matched by a decrease in the other.
Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-15 23:11:37 UTC
Permalink
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
The numbers arriving from each direction between the emission of pulse X and
its arrival at the detector differ by 124 in my example.
So, according to you in some fixed period of time, there are 62 more pulses
arriving in one direction than were emitted in that same period of time.
Yes. It's called doppler shift.
There is no Doppler shift, you moron.
Post by Henry Wilson DSc
Post by Inertial
So
if you leave it running for a while, you end up with more pulses coming out
of the device than have gone into it. You've created free energy.
Congratulations.
moron. The increased number in one path is matched by a decrease in the other.
So .. you only emit down the path with more pulses coming out than going in
and you create free energy.

You really are an idiot if you continue this pretense that what you said was
right.

If two pulses go into a fixed length circular tube (in opposite directions)
and travel for the same time, then they MUST arrive at the detector at the
same time and so the arrival rate of successive pulses MUST be the same.
Your claim to the contrary is ridiculous.
Henry Wilson DSc
2010-02-16 02:58:03 UTC
Permalink
Post by Inertial
Post by Henry Wilson DSc
moron. The increased number in one path is matched by a decrease in the other.
So .. you only emit down the path with more pulses coming out than going in
and you create free energy.
The extra pulses are already in there.
Post by Inertial
You really are an idiot if you continue this pretense that what you said was
right.
If two pulses go into a fixed length circular tube (in opposite directions)
and travel for the same time, then they MUST arrive at the detector at the
same time and so the arrival rate of successive pulses MUST be the same.
Your claim to the contrary is ridiculous.
This is just far too hard for you I'm afraid.

You cannot get it into your head that, when it is rotating, the path length
DOES NOT EQUAL the tube length.

Here it is, SR style: http://en.wikipedia.org/wiki/Sagnac_effect

Go down to 'theory'. SEE THE DIAGRAM WITH THE OMEGA IN THE MIDDLE?


Henry Wilson...

.......provider of free physics lessons
eric gisse
2010-02-16 03:34:13 UTC
Permalink
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
moron. The increased number in one path is matched by a decrease in the other.
So .. you only emit down the path with more pulses coming out than going
in and you create free energy.
The extra pulses are already in there.
...and where do they come from?

[...]
Henry Wilson DSc
2010-02-16 06:04:32 UTC
Permalink
Post by eric gisse
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
moron. The increased number in one path is matched by a decrease in the other.
So .. you only emit down the path with more pulses coming out than going
in and you create free energy.
The extra pulses are already in there.
...and where do they come from?
From the other path of course. Aren't you capable of thinking at all?
Post by eric gisse
[...]
Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-16 06:34:00 UTC
Permalink
Post by Henry Wilson DSc
Post by eric gisse
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
moron. The increased number in one path is matched by a decrease in
the
other.
So .. you only emit down the path with more pulses coming out than going
in and you create free energy.
The extra pulses are already in there.
...and where do they come from?
From the other path of course.
BAHAHA.

So if you don't shine any light pulses down the other path .. what happens?
Post by Henry Wilson DSc
Aren't you capable of thinking at all?
No .. we ARE capable .. and do. If you had thought before posting you
wouldn't have made so many incredibly stupid bungles, some of which you are
STILL defending. You're a joke.
eric gisse
2010-02-16 07:15:23 UTC
Permalink
Post by Henry Wilson DSc
Post by eric gisse
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
moron. The increased number in one path is matched by a decrease in the other.
So .. you only emit down the path with more pulses coming out than going
in and you create free energy.
The extra pulses are already in there.
...and where do they come from?
From the other path of course. Aren't you capable of thinking at all?
I don't know Henri, are you capable of being not pathologically dishonest?
Post by Henry Wilson DSc
Post by eric gisse
[...]
Henry Wilson...
.......provider of free physics lessons
Inertial
2010-02-16 03:36:22 UTC
Permalink
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
moron. The increased number in one path is matched by a decrease in the other.
So .. you only emit down the path with more pulses coming out than going in
and you create free energy.
The extra pulses are already in there.
Post by Inertial
You really are an idiot if you continue this pretense that what you said was
right.
If two pulses go into a fixed length circular tube (in opposite directions)
and travel for the same time, then they MUST arrive at the detector at the
same time and so the arrival rate of successive pulses MUST be the same.
Your claim to the contrary is ridiculous.
This is just far too hard for you I'm afraid.
No .. you're just wrong. And I (and everyone else here but you) sees that
clearly.
Post by Henry Wilson DSc
You cannot get it into your head that, when it is rotating, the path length
DOES NOT EQUAL the tube length.
Of course it is different .. I didn't say it was the same. That doesn't
make any difference to the complete fuckup you made here. You claim of
different arrival rates for the pulses is contradicted by your own claim
that pairs of pulses that are emitted at the same time arrive at the same
time.
Post by Henry Wilson DSc
Here it is, SR style: http://en.wikipedia.org/wiki/Sagnac_effect
Go down to 'theory'. SEE THE DIAGRAM WITH THE OMEGA IN THE MIDDLE?
No need. I'm familiar with Sagnac, and unlike you, I understand it (and SR,
which you never could understand because when you tried to it was just
incoherent to you. Fortunately, those of us with a bit more brain power and
understanding of physics find it quite easy to follow).
Henry Wilson DSc
2010-02-16 06:03:39 UTC
Permalink
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
moron. The increased number in one path is matched by a decrease in the other.
So .. you only emit down the path with more pulses coming out than going in
and you create free energy.
The extra pulses are already in there.
Post by Inertial
You really are an idiot if you continue this pretense that what you said was
right.
If two pulses go into a fixed length circular tube (in opposite directions)
and travel for the same time, then they MUST arrive at the detector at the
same time and so the arrival rate of successive pulses MUST be the same.
Your claim to the contrary is ridiculous.
This is just far too hard for you I'm afraid.
No .. you're just wrong. And I (and everyone else here but you) sees that
clearly.
Post by Henry Wilson DSc
You cannot get it into your head that, when it is rotating, the path length
DOES NOT EQUAL the tube length.
Of course it is different .. I didn't say it was the same. That doesn't
make any difference to the complete fuckup you made here. You claim of
different arrival rates for the pulses is contradicted by your own claim
that pairs of pulses that are emitted at the same time arrive at the same
time.
Post by Henry Wilson DSc
Here it is, SR style: http://en.wikipedia.org/wiki/Sagnac_effect
Go down to 'theory'. SEE THE DIAGRAM WITH THE OMEGA IN THE MIDDLE?
No need. I'm familiar with Sagnac, and unlike you, I understand it (and SR,
which you never could understand because when you tried to it was just
incoherent to you. Fortunately, those of us with a bit more brain power and
understanding of physics find it quite easy to follow).
you don't have a clue about any branch of physics...
you are a moron.

<plonk>

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-16 06:32:53 UTC
Permalink
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
moron. The increased number in one path is matched by a decrease in
the
other.
So .. you only emit down the path with more pulses coming out than going in
and you create free energy.
The extra pulses are already in there.
Post by Inertial
You really are an idiot if you continue this pretense that what you said was
right.
If two pulses go into a fixed length circular tube (in opposite directions)
and travel for the same time, then they MUST arrive at the detector at the
same time and so the arrival rate of successive pulses MUST be the same.
Your claim to the contrary is ridiculous.
This is just far too hard for you I'm afraid.
No .. you're just wrong. And I (and everyone else here but you) sees that
clearly.
Post by Henry Wilson DSc
You cannot get it into your head that, when it is rotating, the path length
DOES NOT EQUAL the tube length.
Of course it is different .. I didn't say it was the same. That doesn't
make any difference to the complete fuckup you made here. You claim of
different arrival rates for the pulses is contradicted by your own claim
that pairs of pulses that are emitted at the same time arrive at the same
time.
Post by Henry Wilson DSc
Here it is, SR style: http://en.wikipedia.org/wiki/Sagnac_effect
Go down to 'theory'. SEE THE DIAGRAM WITH THE OMEGA IN THE MIDDLE?
No need. I'm familiar with Sagnac, and unlike you, I understand it (and SR,
which you never could understand because when you tried to it was just
incoherent to you. Fortunately, those of us with a bit more brain power and
understanding of physics find it quite easy to follow).
you don't have a clue about any branch of physics...
you are a moron.
<plonk>
So .. you cannot refute any of my claims and proofs that you are WRONG.

If the pulses leave in pairs (as you imply), and take the same time to
travel all the way around their path and arrive back again (as emission
theory says), then they MUST arrive back in the same pairs they were
emitted, and so the SAME NUMBER OF PULSES arrive back from each fibre path
in any given time interval.

You cannot determine the rotation of the device (according to emission
theory) by looking at the arrival times and rates of the pulses. And the
fact that your CAN in reality (with a real sagnac device) is resounding and
refutation of emission theory.

Why do you keep lying, Henry?
eric gisse
2010-02-16 07:01:08 UTC
Permalink
***@..(Henry Wilson DSc) wrote:

[...]
Post by Henry Wilson DSc
you don't have a clue about any branch of physics...
you are a moron.
<plonk>
Henry Wilson...
.......provider of free physics lessons
Looks like the 'provider of free physics lessons' couldn't find a way out.
Henry Wilson DSc
2010-02-16 09:58:36 UTC
Permalink
Post by eric gisse
[...]
Post by Henry Wilson DSc
you don't have a clue about any branch of physics...
you are a moron.
<plonk>
Henry Wilson...
.......provider of free physics lessons
Looks like the 'provider of free physics lessons' couldn't find a way out.
poor little eric,.... still hasn't learnt how to make an intelligent
statement...

Henry Wilson...

.......provider of free physics lessons
eric gisse
2010-02-15 00:57:14 UTC
Permalink
***@..(Henry Wilson DSc) wrote:

[...]
Post by Henry Wilson DSc
The fact that the frequencies in each direction are different annihilates
the Roberts, Jerry, Andersen 'rotating frame' argument as well as your
own.
...and your evidence that the frequencies are different is based on what?
Post by Henry Wilson DSc
Henry Wilson...
.......provider of free physics lessons
Henry Wilson DSc
2010-02-13 21:53:21 UTC
Permalink
Post by Inertial
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
What is q and v?
Post by Henry Wilson DSc
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre. The pulses are labelled according to their emission times and move at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p.
Doesn't matter if they are detected at p or at C
I'll withdraw my previous reply because it was wrong. So was my initial message
because I stuffed up what I wanted to say.

p is the point where the pulses emitted at q are received.
Post by Inertial
Post by Henry Wilson DSc
observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period of 1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
OK .. assuming you got the math right
Post by Henry Wilson DSc
In this new situation, the pulses that arrive at C when it is at point p, were
not emitted at that point but rather from an imaginary point q,
Not an imaginary point at all. It is as real and physical point as p is.
All you are saying is that C has moved (and continues to move)
It is imaginary in the R frame...REAL in the nonR frame.
Post by Inertial
Post by Henry Wilson DSc
which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Angular speed I think you would mean, it is moving around the ring according
to C's rotating frame.
yes
Post by Inertial
Post by Henry Wilson DSc
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
So you're using an emission theory analysus, which does say that the time
taken for the two paths is the same.
Yes. The path lengths differ by 124 metres but one pulse moves at c+v and the
other at c-v, so they both arrive at p together.
Post by Inertial
Post by Henry Wilson DSc
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
I'll take your word for it.
Post by Henry Wilson DSc
Since the distance between pulses is still 1 metre in the C's frame,
That's what emission theory would say
Post by Henry Wilson DSc
there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
WRONG. Emmision theory does not say that. It says that the distance in C's
frame between pulses is one meter. The distance the pulses travel in C's
frame is the same. So the number of pulses in the fibre at any given time
is the same.
But I'm talking about the total number of pulses in transit in each path at any
instant. there are 124 more in one path than the other.
Post by Inertial
You can also tell it is wrong because the same number of pulses has been
emitted in each direction and you just said earlier that they take the same
time to travel around the ring. So there must be the same number of them in
there. Or does your explanation have pulse fairies that eat up the pulses?
Now you are being silly.

I made a mistake in my original post. This is regrettable but doesn't present a
problem for Bath.
As you rightly said, the pulses that leave together DO always arrive together
at the detector.
Post by Inertial
Post by Henry Wilson DSc
"That's odd", says Tom, "now, for some reason, the pulses coming from the left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
This is wrong....and is not what I wanted to point out.
Post by Inertial
Emmision theory doesn't say that will happen at all. It says what you said
earlier .. the time take for them to travel is the same.
It is.....and the label on each arriving pulse is the same.
Post by Inertial
What you are now suddenly describing is what SR or an aether theory would
predict. You've changed theories in mid analysis !!!
Yes, basically I did. I apologise for any inconvenience.
Post by Inertial
Post by Henry Wilson DSc
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
No .. because your analysis above used SR at the end. You switched theories
mid way thru.
Yes I did.
Post by Inertial
You started saying the time for the pulses to travel in both directions
(with the rotating tube) was the same. That is what emission theory says.
And you said the distance between the pulses (in the rotating frame) is the
same. That is what emission theory says.
It is.
Post by Inertial
The consequence of that is that
there MUST be the same number of pulses in each direction in the tube at any
given time.
No that is YOUR mistake. The path lengths are different so there are more
pulses in one path than the other.
Post by Inertial
And they MUST arrive at the detector with the same label.
They do. I was wrong, before.
Post by Inertial
And
so emission theory says that a sagnac device will NOT detect rotation.
You've just admitted that your theory is refuted.
Now I'll tell you why that is wrong.
Post by Inertial
The *real* Sagnac devices have proven that THAT theory is NOT correct, and
it is NOT the same time required for the pulses (which is what SR predicts).
BAHAHAHHA
The true fact is that because the number of pulses is different in each path,
THE RATE at which they arrive at the detector is DIFFERENT for each path.
Between the instant of emission from point q and the detection at point p,
40000062 pulses arrive from one direction but only 39999938 from the other.
Those numbers will change only during a rotational speed change.

It is that difference in number that is analogous to the classical 'phase
difference' of the two light rays in a ring gyro.

Two different arrival rates produces a beat frequency at the detector.
Exactly how that gives rise to interference fringes I'm not sure at this stage
but the fact that BaTh predicts such a difference and produces the correct
equation for fringe displacement is sufficient to demonstrate that Sagnac
definitely does NOT refute BaTh.



Henry Wilson...

.......provider of free physics lessons
Androcles
2010-02-13 22:37:33 UTC
Permalink
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT
ROTATING.
Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
What is q and v?
Post by Henry Wilson DSc
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre. The pulses are labelled according to their emission times and
move
at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p.
Doesn't matter if they are detected at p or at C
I'll withdraw my previous reply because it was wrong. So was my initial message
because I stuffed up what I wanted to say.
p is the point where the pulses emitted at q are received.
Post by Inertial
Post by Henry Wilson DSc
observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period
of
1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
OK .. assuming you got the math right
Post by Henry Wilson DSc
In this new situation, the pulses that arrive at C when it is at point
p,
were
not emitted at that point but rather from an imaginary point q,
Not an imaginary point at all. It is as real and physical point as p is.
All you are saying is that C has moved (and continues to move)
It is imaginary in the R frame...REAL in the nonR frame.
Post by Inertial
Post by Henry Wilson DSc
which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Angular speed I think you would mean, it is moving around the ring according
to C's rotating frame.
yes
Post by Inertial
Post by Henry Wilson DSc
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
So you're using an emission theory analysus, which does say that the time
taken for the two paths is the same.
Yes. The path lengths differ by 124 metres but one pulse moves at c+v and the
other at c-v, so they both arrive at p together.
Post by Inertial
Post by Henry Wilson DSc
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
I'll take your word for it.
Post by Henry Wilson DSc
Since the distance between pulses is still 1 metre in the C's frame,
That's what emission theory would say
Post by Henry Wilson DSc
there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
WRONG. Emmision theory does not say that. It says that the distance in C's
frame between pulses is one meter. The distance the pulses travel in C's
frame is the same. So the number of pulses in the fibre at any given time
is the same.
But I'm talking about the total number of pulses in transit in each path at any
instant. there are 124 more in one path than the other.
His "Emmision" theory has nothing to do with emission theory.
Post by Henry Wilson DSc
Post by Inertial
You can also tell it is wrong because the same number of pulses has been
emitted in each direction and you just said earlier that they take the same
time to travel around the ring. So there must be the same number of them in
there. Or does your explanation have pulse fairies that eat up the pulses?
Now you are being silly.
I made a mistake in my original post. This is regrettable but doesn't present a
problem for Bath.
As you rightly said, the pulses that leave together DO always arrive together
at the detector.
Post by Inertial
Post by Henry Wilson DSc
"That's odd", says Tom, "now, for some reason, the pulses coming from
the
left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
This is wrong....and is not what I wanted to point out.
Post by Inertial
Emmision theory doesn't say that will happen at all. It says what you said
earlier .. the time take for them to travel is the same.
It is.....and the label on each arriving pulse is the same.
Post by Inertial
What you are now suddenly describing is what SR or an aether theory would
predict. You've changed theories in mid analysis !!!
Yes, basically I did. I apologise for any inconvenience.
Post by Inertial
Post by Henry Wilson DSc
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
No .. because your analysis above used SR at the end. You switched theories
mid way thru.
Yes I did.
Post by Inertial
You started saying the time for the pulses to travel in both directions
(with the rotating tube) was the same. That is what emission theory says.
And you said the distance between the pulses (in the rotating frame) is the
same. That is what emission theory says.
It is.
Post by Inertial
The consequence of that is that
there MUST be the same number of pulses in each direction in the tube at any
given time.
No that is YOUR mistake. The path lengths are different so there are more
pulses in one path than the other.
Post by Inertial
And they MUST arrive at the detector with the same label.
They do. I was wrong, before.
Post by Inertial
And
so emission theory says that a sagnac device will NOT detect rotation.
You've just admitted that your theory is refuted.
Now I'll tell you why that is wrong.
Post by Inertial
The *real* Sagnac devices have proven that THAT theory is NOT correct, and
it is NOT the same time required for the pulses (which is what SR predicts).
BAHAHAHHA
The true fact is that because the number of pulses is different in each path,
THE RATE at which they arrive at the detector is DIFFERENT for each path.
Between the instant of emission from point q and the detection at point p,
40000062 pulses arrive from one direction but only 39999938 from the other.
Those numbers will change only during a rotational speed change.
It is that difference in number that is analogous to the classical 'phase
difference' of the two light rays in a ring gyro.
Two different arrival rates produces a beat frequency at the detector.
Exactly how that gives rise to interference fringes I'm not sure at this stage
It's simple enough.
There are two beat frequencies, the sum and the difference.
http://en.wikipedia.org/wiki/Circle_of_fifths
So if you have 4 Hz and 5 Hz you'll also get 1 Hz and 9 Hz, but it won't
be very musical.
Loading Image...
Post by Henry Wilson DSc
but the fact that BaTh predicts such a difference and produces the correct
equation for fringe displacement is sufficient to demonstrate that Sagnac
definitely does NOT refute BaTh.
Henry Wilson...
.......provider of free physics lessons
Henry Wilson DSc
2010-02-13 23:43:15 UTC
Permalink
Post by Androcles
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT
ROTATING.
Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
What is q and v?
Post by Henry Wilson DSc
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre. The pulses are labelled according to their emission times and
move
at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p.
Doesn't matter if they are detected at p or at C
I'll withdraw my previous reply because it was wrong. So was my initial message
because I stuffed up what I wanted to say.
p is the point where the pulses emitted at q are received.
Post by Inertial
Post by Henry Wilson DSc
observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period
of
1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
OK .. assuming you got the math right
Post by Henry Wilson DSc
In this new situation, the pulses that arrive at C when it is at point
p,
were
not emitted at that point but rather from an imaginary point q,
Not an imaginary point at all. It is as real and physical point as p is.
All you are saying is that C has moved (and continues to move)
It is imaginary in the R frame...REAL in the nonR frame.
Post by Inertial
Post by Henry Wilson DSc
which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Angular speed I think you would mean, it is moving around the ring according
to C's rotating frame.
yes
Post by Inertial
Post by Henry Wilson DSc
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
So you're using an emission theory analysus, which does say that the time
taken for the two paths is the same.
Yes. The path lengths differ by 124 metres but one pulse moves at c+v and the
other at c-v, so they both arrive at p together.
Post by Inertial
Post by Henry Wilson DSc
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
I'll take your word for it.
Post by Henry Wilson DSc
Since the distance between pulses is still 1 metre in the C's frame,
That's what emission theory would say
Post by Henry Wilson DSc
there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
WRONG. Emmision theory does not say that. It says that the distance in C's
frame between pulses is one meter. The distance the pulses travel in C's
frame is the same. So the number of pulses in the fibre at any given time
is the same.
But I'm talking about the total number of pulses in transit in each path at any
instant. there are 124 more in one path than the other.
His "Emmision" theory has nothing to do with emission theory.
....do you know what 'typophobia' is? You should ask your doctor about it...
Post by Androcles
Post by Henry Wilson DSc
Now you are being silly.
I made a mistake in my original post. This is regrettable but doesn't present a
problem for Bath.
As you rightly said, the pulses that leave together DO always arrive together
at the detector.
Post by Inertial
Post by Henry Wilson DSc
"That's odd", says Tom, "now, for some reason, the pulses coming from
the
left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
This is wrong....and is not what I wanted to point out.
Post by Inertial
Emmision theory doesn't say that will happen at all. It says what you said
earlier .. the time take for them to travel is the same.
It is.....and the label on each arriving pulse is the same.
Post by Inertial
What you are now suddenly describing is what SR or an aether theory would
predict. You've changed theories in mid analysis !!!
Yes, basically I did. I apologise for any inconvenience.
Post by Inertial
Post by Henry Wilson DSc
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
No .. because your analysis above used SR at the end. You switched theories
mid way thru.
Yes I did.
Post by Inertial
You started saying the time for the pulses to travel in both directions
(with the rotating tube) was the same. That is what emission theory says.
And you said the distance between the pulses (in the rotating frame) is the
same. That is what emission theory says.
It is.
Post by Inertial
The consequence of that is that
there MUST be the same number of pulses in each direction in the tube at any
given time.
No that is YOUR mistake. The path lengths are different so there are more
pulses in one path than the other.
Post by Inertial
And they MUST arrive at the detector with the same label.
They do. I was wrong, before.
Post by Inertial
And
so emission theory says that a sagnac device will NOT detect rotation.
You've just admitted that your theory is refuted.
Now I'll tell you why that is wrong.
Post by Inertial
The *real* Sagnac devices have proven that THAT theory is NOT correct, and
it is NOT the same time required for the pulses (which is what SR predicts).
BAHAHAHHA
The true fact is that because the number of pulses is different in each path,
THE RATE at which they arrive at the detector is DIFFERENT for each path.
Between the instant of emission from point q and the detection at point p,
40000062 pulses arrive from one direction but only 39999938 from the other.
Those numbers will change only during a rotational speed change.
It is that difference in number that is analogous to the classical 'phase
difference' of the two light rays in a ring gyro.
Two different arrival rates produces a beat frequency at the detector.
Exactly how that gives rise to interference fringes I'm not sure at this stage
It's simple enough.
There are two beat frequencies, the sum and the difference.
http://en.wikipedia.org/wiki/Circle_of_fifths
So if you have 4 Hz and 5 Hz you'll also get 1 Hz and 9 Hz, but it won't
be very musical.
http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/MechSagnac.gif
In the case of light it will be the difference that creates the pattern.
....and fringe DISPLACEMENT will be proportional to beat frequency.



Henry Wilson...

.......provider of free physics lessons
Androcles
2010-02-14 01:20:44 UTC
Permalink
On Sat, 13 Feb 2010 22:37:33 -0000, "Androcles"
Post by Androcles
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT
ROTATING.
Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
What is q and v?
Post by Henry Wilson DSc
(the surface is drawn flat for convenience. The fibre goes right
around
the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both
ends
of the
fibre. The pulses are labelled according to their emission times and
move
at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were
emitted
and
detected by Jerry at point p.
Doesn't matter if they are detected at p or at C
I'll withdraw my previous reply because it was wrong. So was my initial message
because I stuffed up what I wanted to say.
p is the point where the pulses emitted at q are received.
Post by Inertial
Post by Henry Wilson DSc
observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period
of
1
day, (86400 seconds). Its surface rotation speed 'v' wrt the
nonrotating
frame
is 465 m/s.
OK .. assuming you got the math right
Post by Henry Wilson DSc
In this new situation, the pulses that arrive at C when it is at point
p,
were
not emitted at that point but rather from an imaginary point q,
Not an imaginary point at all. It is as real and physical point as p is.
All you are saying is that C has moved (and continues to move)
It is imaginary in the R frame...REAL in the nonR frame.
Post by Inertial
Post by Henry Wilson DSc
which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Angular speed I think you would mean, it is moving around the ring according
to C's rotating frame.
yes
Post by Inertial
Post by Henry Wilson DSc
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
So you're using an emission theory analysus, which does say that the time
taken for the two paths is the same.
Yes. The path lengths differ by 124 metres but one pulse moves at c+v
and
the
other at c-v, so they both arrive at p together.
Post by Inertial
Post by Henry Wilson DSc
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
I'll take your word for it.
Post by Henry Wilson DSc
Since the distance between pulses is still 1 metre in the C's frame,
That's what emission theory would say
Post by Henry Wilson DSc
there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
WRONG. Emmision theory does not say that. It says that the distance in C's
frame between pulses is one meter. The distance the pulses travel in C's
frame is the same. So the number of pulses in the fibre at any given time
is the same.
But I'm talking about the total number of pulses in transit in each path at any
instant. there are 124 more in one path than the other.
His "Emmision" theory has nothing to do with emission theory.
....do you know what 'typophobia' is? You should ask your doctor about it...
Post by Androcles
Post by Henry Wilson DSc
Now you are being silly.
I made a mistake in my original post. This is regrettable but doesn't present a
problem for Bath.
As you rightly said, the pulses that leave together DO always arrive together
at the detector.
Post by Inertial
Post by Henry Wilson DSc
"That's odd", says Tom, "now, for some reason, the pulses coming from
the
left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
This is wrong....and is not what I wanted to point out.
Post by Inertial
Emmision theory doesn't say that will happen at all. It says what you said
earlier .. the time take for them to travel is the same.
It is.....and the label on each arriving pulse is the same.
Post by Inertial
What you are now suddenly describing is what SR or an aether theory would
predict. You've changed theories in mid analysis !!!
Yes, basically I did. I apologise for any inconvenience.
Post by Inertial
Post by Henry Wilson DSc
"This is great", says Jerry after some deliberation, "now we can build
a
ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
No .. because your analysis above used SR at the end. You switched theories
mid way thru.
Yes I did.
Post by Inertial
You started saying the time for the pulses to travel in both directions
(with the rotating tube) was the same. That is what emission theory says.
And you said the distance between the pulses (in the rotating frame) is the
same. That is what emission theory says.
It is.
Post by Inertial
The consequence of that is that
there MUST be the same number of pulses in each direction in the tube at any
given time.
No that is YOUR mistake. The path lengths are different so there are more
pulses in one path than the other.
Post by Inertial
And they MUST arrive at the detector with the same label.
They do. I was wrong, before.
Post by Inertial
And
so emission theory says that a sagnac device will NOT detect rotation.
You've just admitted that your theory is refuted.
Now I'll tell you why that is wrong.
Post by Inertial
The *real* Sagnac devices have proven that THAT theory is NOT correct, and
it is NOT the same time required for the pulses (which is what SR predicts).
BAHAHAHHA
The true fact is that because the number of pulses is different in each path,
THE RATE at which they arrive at the detector is DIFFERENT for each path.
Between the instant of emission from point q and the detection at point p,
40000062 pulses arrive from one direction but only 39999938 from the other.
Those numbers will change only during a rotational speed change.
It is that difference in number that is analogous to the classical 'phase
difference' of the two light rays in a ring gyro.
Two different arrival rates produces a beat frequency at the detector.
Exactly how that gives rise to interference fringes I'm not sure at this stage
It's simple enough.
There are two beat frequencies, the sum and the difference.
http://en.wikipedia.org/wiki/Circle_of_fifths
So if you have 4 Hz and 5 Hz you'll also get 1 Hz and 9 Hz, but it won't
be very musical.
http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/MechSagnac.gif
In the case of light it will be the difference that creates the pattern.
Yeah,
Loading Image...
....and fringe DISPLACEMENT will be proportional to beat frequency.
Bullshit. You've never worked with an oscilloscope.
How come you never produce a web page to back up your stupid claims?
Henry Wilson DSc
2010-02-14 06:27:48 UTC
Permalink
Post by Androcles
On Sat, 13 Feb 2010 22:37:33 -0000, "Androcles"
Post by Androcles
It's simple enough.
There are two beat frequencies, the sum and the difference.
http://en.wikipedia.org/wiki/Circle_of_fifths
So if you have 4 Hz and 5 Hz you'll also get 1 Hz and 9 Hz, but it won't
be very musical.
http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/MechSagnac.gif
In the case of light it will be the difference that creates the pattern.
Yeah,
http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/Sagnac2.JPG
Yes. That shows the difference.
Post by Androcles
....and fringe DISPLACEMENT will be proportional to beat frequency.
Bullshit. You've never worked with an oscilloscope.
How come you never produce a web page to back up your stupid claims?
...you're starting to sound more like little eric every day.....

Henry Wilson...

.......provider of free physics lessons
eric gisse
2010-02-14 07:19:17 UTC
Permalink
***@..(Henry Wilson DSc) wrote:
[...]
Post by Henry Wilson DSc
Post by Androcles
Post by Henry Wilson DSc
....and fringe DISPLACEMENT will be proportional to beat frequency.
Bullshit. You've never worked with an oscilloscope.
How come you never produce a web page to back up your stupid claims?
...you're starting to sound more like little eric every day.....
Which makes me wonder why you post here at all. Nobody here likes you, you
don't like anyone here, and you have your own webspace. Why not just post to
it and be smug in the knowledge that you are right and the world is wrong?
Post by Henry Wilson DSc
Henry Wilson...
.......provider of free physics lessons
Henry Wilson DSc
2010-02-14 07:28:53 UTC
Permalink
Post by eric gisse
[...]
Post by Henry Wilson DSc
Post by Androcles
Post by Henry Wilson DSc
....and fringe DISPLACEMENT will be proportional to beat frequency.
Bullshit. You've never worked with an oscilloscope.
How come you never produce a web page to back up your stupid claims?
...you're starting to sound more like little eric every day.....
Which makes me wonder why you post here at all. Nobody here likes you, you
don't like anyone here, and you have your own webspace. Why not just post to
it and be smug in the knowledge that you are right and the world is wrong?
when are you going to say something intelligent?
Post by eric gisse
Post by Henry Wilson DSc
Henry Wilson...
.......provider of free physics lessons
Henry Wilson...

.......provider of free physics lessons
eric gisse
2010-02-14 09:13:33 UTC
Permalink
Post by Henry Wilson DSc
Post by eric gisse
[...]
Post by Henry Wilson DSc
Post by Androcles
Post by Henry Wilson DSc
....and fringe DISPLACEMENT will be proportional to beat frequency.
Bullshit. You've never worked with an oscilloscope.
How come you never produce a web page to back up your stupid claims?
...you're starting to sound more like little eric every day.....
Which makes me wonder why you post here at all. Nobody here likes you, you
don't like anyone here, and you have your own webspace. Why not just post
to it and be smug in the knowledge that you are right and the world is
wrong?
when are you going to say something intelligent?
When are you going to apologize for posting forged degrees?
Post by Henry Wilson DSc
Post by eric gisse
Post by Henry Wilson DSc
Henry Wilson...
.......provider of free physics lessons
Henry Wilson...
.......provider of free physics lessons
Jerry
2010-02-15 04:42:10 UTC
Permalink
Post by eric gisse
When are you going to apologize for posting forged degrees?
That's not going to work, Eric.

You see, Ralph really believes that his degree is legitimate. An
advanced degree, such as a PhD or DSc, is conferred upon a person
by a committee of the degree candidate's peers in the subject of
entry, who agree that the candidate's accomplishments, [1] as
manifest in a tangible work of art (such as a dissertation,
composition, work of craft etc. as appropriate for the subject)
[2] and, as may or may not be necessary depending on the subject,
defended in an appropriate venue (such as a dissertation defense),
[3] are such that he/she is worthy of welcome to the community of
peers.

Ralph has a body of work (i.e. his online writings and computer
programs) which in his mind, he has successfully defended against
all criticism. Since there is nobody in the world whom Ralph
respects as a peer, he has no choice but to set himself as head
of the degree-conferring organization.

Deluded? You bet.

Jerry
Inertial
2010-02-15 05:11:17 UTC
Permalink
Post by Jerry
Post by eric gisse
When are you going to apologize for posting forged degrees?
That's not going to work, Eric.
You see, Ralph really believes that his degree is legitimate. An
advanced degree, such as a PhD or DSc, is conferred upon a person
by a committee of the degree candidate's peers in the subject of
entry, who agree that the candidate's accomplishments, [1] as
manifest in a tangible work of art (such as a dissertation,
composition, work of craft etc. as appropriate for the subject)
[2] and, as may or may not be necessary depending on the subject,
defended in an appropriate venue (such as a dissertation defense),
[3] are such that he/she is worthy of welcome to the community of
peers.
Ralph has a body of work (i.e. his online writings and computer
programs) which in his mind, he has successfully defended against
all criticism. Since there is nobody in the world whom Ralph
respects as a peer, he has no choice but to set himself as head
of the degree-conferring organization.
Deluded? You bet.
BAHAHAHA. Clever :) (you, that is, not Henry)
Henry Wilson DSc
2010-02-15 05:46:44 UTC
Permalink
On Sun, 14 Feb 2010 20:42:10 -0800 (PST), Jerry
Post by Jerry
Post by eric gisse
When are you going to apologize for posting forged degrees?
That's not going to work, Eric.
You see, Henry really believes that his degree is legitimate. An
advanced degree, such as a PhD or DSc, is conferred upon a person
by a committee of the degree candidate's peers in the subject of
entry, who agree that the candidate's accomplishments, [1] as
manifest in a tangible work of art (such as a dissertation,
composition, work of craft etc. as appropriate for the subject)
[2] and, as may or may not be necessary depending on the subject,
defended in an appropriate venue (such as a dissertation defense),
[3] are such that he/she is worthy of welcome to the community of
peers.
Henry has a body of work (i.e. his online writings and computer
programs) which in his mind, he has successfully defended against
all criticism. Since there is nobody in the world whom Henry
respects as a peer, he has no choice but to set himself as head
of the degree-conferring organization.
Deluded? You bet.
Note, I made a glaring error in my original message.
I have admitted the same and corrected it.

Sagnac fully supports BaTh.
Post by Jerry
Jerry
Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-15 06:13:26 UTC
Permalink
Post by Henry Wilson DSc
On Sun, 14 Feb 2010 20:42:10 -0800 (PST), Jerry
Post by Jerry
Post by eric gisse
When are you going to apologize for posting forged degrees?
That's not going to work, Eric.
You see, Henry really believes that his degree is legitimate. An
advanced degree, such as a PhD or DSc, is conferred upon a person
by a committee of the degree candidate's peers in the subject of
entry, who agree that the candidate's accomplishments, [1] as
manifest in a tangible work of art (such as a dissertation,
composition, work of craft etc. as appropriate for the subject)
[2] and, as may or may not be necessary depending on the subject,
defended in an appropriate venue (such as a dissertation defense),
[3] are such that he/she is worthy of welcome to the community of
peers.
Henry has a body of work (i.e. his online writings and computer
programs) which in his mind, he has successfully defended against
all criticism. Since there is nobody in the world whom Henry
respects as a peer, he has no choice but to set himself as head
of the degree-conferring organization.
Deluded? You bet.
Note, I made a glaring error in my original message.
I have admitted the same and corrected it.
Sagnac fully supports BaTh.
That's still total crap. Correcting one error in a post doesn't fix up the
HUGE error in your so-called theory. Namely that it just DOES NOT WORK.
eric gisse
2010-02-15 08:30:46 UTC
Permalink
Post by Henry Wilson DSc
On Sun, 14 Feb 2010 20:42:10 -0800 (PST), Jerry
Post by Jerry
Post by eric gisse
When are you going to apologize for posting forged degrees?
That's not going to work, Eric.
You see, Henry really believes that his degree is legitimate. An
advanced degree, such as a PhD or DSc, is conferred upon a person
by a committee of the degree candidate's peers in the subject of
entry, who agree that the candidate's accomplishments, [1] as
manifest in a tangible work of art (such as a dissertation,
composition, work of craft etc. as appropriate for the subject)
[2] and, as may or may not be necessary depending on the subject,
defended in an appropriate venue (such as a dissertation defense),
[3] are such that he/she is worthy of welcome to the community of
peers.
Henry has a body of work (i.e. his online writings and computer
programs) which in his mind, he has successfully defended against
all criticism. Since there is nobody in the world whom Henry
respects as a peer, he has no choice but to set himself as head
of the degree-conferring organization.
Deluded? You bet.
Note, I made a glaring error in my original message.
I have admitted the same and corrected it.
Sagnac fully supports BaTh.
Post by Jerry
Jerry
Henry Wilson...
.......provider of free physics lessons
Imagine that - Henri ignores the discussion of his faked degrees to discuss
something nobody really cares about.
Jerry
2010-02-15 10:49:13 UTC
Permalink
Post by eric gisse
Post by Henry Wilson DSc
On Sun, 14 Feb 2010 20:42:10 -0800 (PST), Jerry
Post by Jerry
Post by eric gisse
When are you going to apologize for posting forged degrees?
That's not going to work, Eric.
You see, Henry really believes that his degree is legitimate. An
advanced degree, such as a PhD or DSc, is conferred upon a person
by a committee of the degree candidate's peers in the subject of
entry, who agree that the candidate's accomplishments, [1] as
manifest in a tangible work of art (such as a dissertation,
composition, work of craft etc. as appropriate for the subject)
[2] and, as may or may not be necessary depending on the subject,
defended in an appropriate venue (such as a dissertation defense),
[3] are such that he/she is worthy of welcome to the community of
peers.
Henry has a body of work (i.e. his online writings and computer
programs) which in his mind, he has successfully defended against
all criticism. Since there is nobody in the world whom Henry
respects as a peer, he has no choice but to set himself as head
of the degree-conferring organization.
Deluded? You bet.
Note, I made a glaring error in my original message.
I have admitted the same and corrected it.
Sagnac fully supports BaTh.
Post by Jerry
Jerry
Henry Wilson...
.......provider of free physics lessons
Imagine that - Henri ignores the discussion of his faked degrees to discuss
something nobody really cares about.- Hide quoted text -
Note that Henri has not denied any of my analysis.
He -believes- it...everything except the "Deluded? You bet" part.

Jerry
Henry Wilson DSc
2010-02-15 22:20:15 UTC
Permalink
On Mon, 15 Feb 2010 02:49:13 -0800 (PST), Jerry
Post by Jerry
Post by eric gisse
Imagine that - Henri ignores the discussion of his faked degrees to discuss
something nobody really cares about.- Hide quoted text -
Note that Henri has not denied any of my analysis.
He -believes- it...everything except the "Deluded? You bet" part.
Moron Jerry, the bedpan expert, is upset by the fact that I can calculate the
earth's rotation rate using a clock, an optical fibre and the BaTh version of
Sagnac.
Post by Jerry
Jerry
Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-15 22:32:04 UTC
Permalink
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 02:49:13 -0800 (PST), Jerry
Post by Jerry
Post by eric gisse
Imagine that - Henri ignores the discussion of his faked degrees to discuss
something nobody really cares about.- Hide quoted text -
Note that Henri has not denied any of my analysis.
He -believes- it...everything except the "Deluded? You bet" part.
Moron Jerry, the bedpan expert, is upset by the fact that I can calculate the
earth's rotation rate using a clock, an optical fibre and the BaTh version of
Sagnac.
Nope .. I've refuted your assertion already. You continue to lie about it
though.
Jerry
2010-02-16 02:00:42 UTC
Permalink
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 02:49:13 -0800 (PST), Jerry
Post by Jerry
Post by eric gisse
Imagine that - Henri ignores the discussion of his faked degrees to discuss
something nobody really cares about.
Note that Henri has not denied any of my analysis.
He -believes- it...everything except the "Deluded? You bet" part.
Moron Jerry, the bedpan expert, is upset by the fact that I can calculate the
earth's rotation rate using a clock, an optical fibre and the BaTh version of
Sagnac.
I note that although you throw insults, you do not deny any of
my analysis of why you feels justified in awarding yourself a
doctorate.

Your "rayphases" model has been refuted many times over, by me
and by many others employing different arguments and methods of
analysis. It is self-contradictory and predicts effects that have
never been seen. In this thread alone, I've seen you squirm and
deny assertions that you made mere hours previously. Your model
is HOPELESS.
http://mysite.verizon.net/cephalobus_alienus/henri/HWFantasy.htm

Jerry
Androcles
2010-02-16 02:11:32 UTC
Permalink
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 02:49:13 -0800 (PST), Jerry
Post by Jerry
Post by eric gisse
Imagine that - Henri ignores the discussion of his faked degrees to discuss
something nobody really cares about.
Note that Henri has not denied any of my analysis.
He -believes- it...everything except the "Deluded? You bet" part.
Moron Jerry, the bedpan expert, is upset by the fact that I can calculate the
earth's rotation rate using a clock, an optical fibre and the BaTh version of
Sagnac.
I note that although you throw insults, you do not deny any of
my analysis of why you feels justified in awarding yourself a
doctorate.

Your "rayphases" model has been refuted many times over, by me
and by many others employing different arguments and methods of
analysis. It is self-contradictory and predicts effects that have
never been seen. In this thread alone, I've seen you squirm and
deny assertions that you made mere hours previously. Your model
is HOPELESS.
http://mysite.verizon.net/cephalobus_alienus/henri/HWFantasy.htm

Jerry
============================================
If you had any integrity, you ugly fuck, you'd model it correctly
instead of sneering, you fuckin' poofter.
Jerry
2010-02-16 02:47:03 UTC
Permalink
Post by Androcles
Post by Jerry
I note that although you throw insults, you do not deny any of
my analysis of why you feels justified in awarding yourself a
doctorate.
Your "rayphases" model has been refuted many times over, by me
and by many others employing different arguments and methods of
analysis. It is self-contradictory and predicts effects that have
never been seen. In this thread alone, I've seen you squirm and
deny assertions that you made mere hours previously. Your model
is HOPELESS.http://mysite.verizon.net/cephalobus_alienus/henri/HWFantasy.htm
Jerry
============================================
If you had any integrity, you ugly fuck, you'd model it correctly
instead of sneering, you fuckin' poofter.
I've already done a simple computer animation:
http://mysite.verizon.net/cephalobus_alienus/sagnac/BallisticSagnac.htm

Others more skilled than me in math have performed a detailed
analysis INCLUDING SECOND-ORDER EFFECTS, thus demonstrating that
the "Coriolis" effects that you repeatedly cite are negligible
EXCEPT AT RING SPEEDS THAT ARE A SIGNIFICANT FRACTION OF THE
SPEED OF LIGHT:
http://home.c2i.net/pb_andersen/pdf/four_mirror_sagnac.pdf

Paul Andersen has done a really nice animation of a four-mirror
Sagnac ring that I doubt that I'm capable of significantly
improving on, so I haven't tried:
http://home.c2i.net/pb_andersen/FourMirrorSagnac.html

Then, of course, I've done a fair amount of debunking of other
fanciful ballistic Sagnac schemes that don't work. For example,
one of the seven "plonks" that you've awarded me came when I
proved as utter nonsense your notion that "when the final beam-
splitter is reached and the two beams combined, still in phase,
one departs with velocity c+v and the other with velocity c-v.
Upon reaching the detector the two beams are necessarily out of
phase."
http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/Sagnac.htm

Why don't your plonks persist, by the way? It would be much
easier on you. Just "plonk" and dream on forever that you've
won the argument, rather than persist in getting your face
mashed into the dirt...

Jerry
Androcles
2010-02-16 03:22:03 UTC
Permalink
Post by Androcles
Post by Jerry
I note that although you throw insults, you do not deny any of
my analysis of why you feels justified in awarding yourself a
doctorate.
Your "rayphases" model has been refuted many times over, by me
and by many others employing different arguments and methods of
analysis. It is self-contradictory and predicts effects that have
never been seen. In this thread alone, I've seen you squirm and
deny assertions that you made mere hours previously. Your model
is
HOPELESS.http://mysite.verizon.net/cephalobus_alienus/henri/HWFantasy.htm
Jerry
============================================
If you had any integrity, you ugly fuck, you'd model it correctly
instead of sneering, you fuckin' poofter.
I've already done a simple computer animation:
http://mysite.verizon.net/cephalobus_alienus/sagnac/BallisticSagnac.htm

Others more skilled than me in math have performed a detailed
analysis INCLUDING SECOND-ORDER EFFECTS, thus demonstrating that
the "Coriolis" effects that you repeatedly cite are negligible

===================================================
If you had any integrity, you ugly STOOOPID fuck, you'd know I said the
Coriolis effect WAS the Sagnac effect years ago.
Loading Image...



EXCEPT AT RING SPEEDS THAT ARE A SIGNIFICANT FRACTION OF THE
SPEED OF LIGHT:
http://home.c2i.net/pb_andersen/pdf/four_mirror_sagnac.pdf
===============================================
There are no exceptions, shithead.




Paul Andersen has done a really nice animation of a four-mirror
Sagnac ring that I doubt that I'm capable of significantly
improving on, so I haven't tried:
http://home.c2i.net/pb_andersen/FourMirrorSagnac.html
================================================
Andersen is as stooopid as you, dumbfuck.



Then, of course, I've done a fair amount of debunking of other
fanciful ballistic Sagnac schemes that don't work. For example,
one of the seven "plonks" that you've awarded me came when I
proved as utter nonsense your notion that "when the final beam-
splitter is reached and the two beams combined, still in phase,
one departs with velocity c+v and the other with velocity c-v.
Upon reaching the detector the two beams are necessarily out of
phase."
http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/Sagnac.htm

================================================
Your assertions are not a proof and debunk nothing, you fuckin' cretin.


Why don't your plonks persist, by the way?
=================================================
The kill-file will be cleared annually with spring cleaning or whenever I
purchase a new computer or hard drive.
Update: the last clearance was 25/12/09.


It would be much
easier on you. Just "plonk" and dream on forever that you've
won the argument, rather than persist in getting your face
mashed into the dirt...

Jerry
==================================================
An ignorant, lying, snipping faggot with no integrity like you
couldn't come close, deranged bedpan-changing "little sister"
Tommy General Crank.
-- Androcles
'By denying scientific principles, one may maintain any paradox.' - Galileo
Galilei
Androcles
2010-02-16 03:56:57 UTC
Permalink
"Jerry" <***@comcast.net> wrote in message news:897a3ae0-12e2-413a-85d2-***@t1g2000vbq.googlegroups.com...


Paul Andersen has done a really nice animation of a four-mirror
Sagnac ring that I doubt that I'm capable of significantly
improving on, so I haven't tried:
http://home.c2i.net/pb_andersen/FourMirrorSagnac.html

=================================================
The sudden jumps at the mirrors prove he's a fraud, you dumb bastard.
Jerry
2010-02-16 05:00:37 UTC
Permalink
Post by Androcles
Post by Jerry
Paul Andersen has done a really nice animation of a four-mirror
Sagnac ring that I doubt that I'm capable of significantly
improving on, so I haven't tried:http://home.c2i.net/pb_andersen/FourMirrorSagnac.html
=================================================
The sudden jumps at the mirrors prove he's a fraud, you dumb bastard.
Nope. I have an early version of Paul's source code because I was
helping him out on a trivial programming issue...two eyes are
better than one...and like you, I was bothered a bit by the
discontinuities at the mirrors and tried to see if there was any
way to improve on the appearance.

The jumps are a VISUAL ARTIFACT of the graphical representation
of the waves as sinusoidal squiggles. In reality, the path
lengths are accurate to the practical limits of floating point
arithmetic and finite pixel size.

I experimented a bit with alternate means of representing wave
shape via color coding along straight rays. The results were not
especially satisfactory because of aliasing effects. Straight
slanted lines look "jaggy" and throw the eyes off. So I tried
out some shareware graphical toolkits and although the results
were better, they still weren't perfect because my dots changed
shape as they moved. If I had discovered any really decent
solution to the problem, I would have let Paul know, but I never
found a perfect solution so I didn't.

Whatever. Paul's math is accurate, even if the graphs look a
bit funky at the mirror surfaces, and that's the important thing.

Jerry
Androcles
2010-02-16 08:23:22 UTC
Permalink
Post by Androcles
Post by Jerry
Paul Andersen has done a really nice animation of a four-mirror
Sagnac ring that I doubt that I'm capable of significantly
http://home.c2i.net/pb_andersen/FourMirrorSagnac.html
Post by Androcles
=================================================
The sudden jumps at the mirrors prove he's a fraud, you dumb bastard.
Nope. I have an early version of Paul's source code because I was
helping him out on a trivial programming issue...two eyes are
better than one...and like you, I was bothered a bit by the
discontinuities at the mirrors and tried to see if there was any
way to improve on the appearance.

The jumps are a VISUAL ARTIFACT of the graphical representation
of the waves as sinusoidal squiggles. In reality, the path
lengths are accurate to the practical limits of floating point
arithmetic and finite pixel size.
===================================================
In a simulation "practical" means rotate faster and slow the simulated
light down so the viewer can easily see what is going on, you ignorant
arse.
This is Andersen:
http://home.c2i.net/pb_andersen/pdf/four_mirror_sagnac.pdf
Look at figure 3:
Read what Andersen says:
"The beams are slightly curved because..."
But he doesn't animate it.

I've animated it.
In reality, the path lengths are accurate to the practical limits of finite
pixel size,
but the velocities are not - and neither are Tusseladd's.
Mine works because I chose an exact integer wavelength in fixed frame.
LIKE THIS, you worthless bigot:
http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/CoriSag.gif
You and Tusseladd are frauds!

AND you ARE capable of significantly improving on it, you haven't tried
because
1) You are too fuckin' LAZY,
2) You've already made up your pathetic mind, you fuckin' BIGOT!

I'll give Tusseladd a tad more credit than you, at least he's not as
lazy as you are even if he's just as much a bigot.

Your mission (should you decide to accept it) is to animate Tusseladd's
Fig 3. using Java, allowing the user to set the speed to see what is going
on in the rotating frame, the lab frame and the SR (=aether) frame.
This tape will self-destruct in five months, that should be enough
time.

You may outdo Wilson, you lazy fraudulent bigot, and you can outdo
my animation, but you will not defeat me.
"Nope" is not an acceptable reply, nor are phase-shift jumps at
the corners acceptable.
The gauntlet is down. Shut the fuck up and get on with it.
Jerry
2010-02-16 12:43:39 UTC
Permalink
Post by Androcles
Post by Jerry
Post by Androcles
The sudden jumps at the mirrors prove he's a fraud, you dumb bastard.
Nope. I have an early version of Paul's source code because I was
helping him out on a trivial programming issue...two eyes are
better than one...and like you, I was bothered a bit by the
discontinuities at the mirrors and tried to see if there was any
way to improve on the appearance.
The jumps are a VISUAL ARTIFACT of the graphical representation
of the waves as sinusoidal squiggles. In reality, the path
lengths are accurate to the practical limits of floating point
arithmetic and finite pixel size.
===================================================
In a simulation  "practical" means rotate faster and slow the simulated
light down so the viewer can easily see what is going on, you ignorant
arse.
   http://home.c2i.net/pb_andersen/pdf/four_mirror_sagnac.pdf
 "The beams are slightly curved because..."
But he doesn't animate it.
I've animated it.
In reality, the path lengths are accurate to the practical limits of finite
pixel size,
but the velocities are not - and neither are Tusseladd's.
Mine works because I chose an exact integer wavelength in fixed frame.
   http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/CoriSag.gif
You and Tusseladd are frauds!
The only reason Coriolis curvature SEEMS to add significantly to
path length in your animation is that you spin the wheel at some
huge fraction of the speed of light.

Coriolis curvature is a second-order effect. That means that at
REALISTIC speeds, the relative contribution of the SECOND-ORDER
Coriolis curvature becomes negligible relative to the FIRST-ORDER
Sagnac effect.

Work it out. If you spin a four-mirror Sagnac ring at 1 m/s, then
Coriolis curvature adds approximately 1e-18 meters to the path
length. That's a tiny, tiny, tiny, TINY fraction of the typical
distance between atoms in a solid.

Modern-day ring gyros are sensitive enough to detect the rotation
of the Earth. How many radians per second is that? What would be
the contribution of second-order Coriolis curvature relative to
the first order Sagnac effect in this scenario?

Who is the fraud, Androcles?

I'm ignoring your childish challenge, by the way:

[SNIP stupid ignorant childish frothing-at-the-mouth challenge]

Jerry
Jerry
2010-02-16 12:48:14 UTC
Permalink
Post by Jerry
Work it out. If you spin a four-mirror Sagnac ring
with mirrors spaced 1 meter apart
Post by Jerry
at 1 m/s, then
Coriolis curvature adds approximately 1e-18 meters to the path
length. That's a tiny, tiny, tiny, TINY fraction of the typical
distance between atoms in a solid.
Jerry
Androcles
2010-02-16 13:43:05 UTC
Permalink
"Jerry" <***@comcast.net> wrote in message news:74fc8891-d6ce-4b60-a908-***@b7g2000yqd.googlegroups.com...
On Feb 16, 6:43 am, Jerry <***@comcast.net> wrote:

Left out a parameter:

Yes, you did. Coriolis is ably and undeniably demonstrated here, bigot.
http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/gifs/coriolis.mov
and here, arsehole:


You are a third order ignoramus and first order troll.
Androcles
2010-02-16 13:36:11 UTC
Permalink
Post by Androcles
Post by Jerry
Post by Androcles
The sudden jumps at the mirrors prove he's a fraud, you dumb bastard.
Nope. I have an early version of Paul's source code because I was
helping him out on a trivial programming issue...two eyes are
better than one...and like you, I was bothered a bit by the
discontinuities at the mirrors and tried to see if there was any
way to improve on the appearance.
The jumps are a VISUAL ARTIFACT of the graphical representation
of the waves as sinusoidal squiggles. In reality, the path
lengths are accurate to the practical limits of floating point
arithmetic and finite pixel size.
===================================================
In a simulation "practical" means rotate faster and slow the simulated
light down so the viewer can easily see what is going on, you ignorant
arse.
http://home.c2i.net/pb_andersen/pdf/four_mirror_sagnac.pdf
"The beams are slightly curved because..."
But he doesn't animate it.
I've animated it.
In reality, the path lengths are accurate to the practical limits of finite
pixel size,
but the velocities are not - and neither are Tusseladd's.
Mine works because I chose an exact integer wavelength in fixed frame.
http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/CoriSag.gif
You and Tusseladd are frauds!
The only reason Coriolis curvature SEEMS to add significantly to
path length in your animation is that you spin the wheel at some
huge fraction of the speed of light.

Coriolis curvature is a second-order effect.
=======================================
Liar.
<Irrelevant rant snipped>


Who is the fraud, Androcles?
====================================
You are.
Coriolis is ably demonstrated here, bigot.
http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/gifs/coriolis.mov
Jerry
2010-02-16 20:22:26 UTC
Permalink
Post by Androcles
Post by Jerry
The only reason Coriolis curvature SEEMS to add significantly to
path length in your animation is that you spin the wheel at some
huge fraction of the speed of light.
Coriolis curvature is a second-order effect.
=======================================
Liar.
<Irrelevant rant snipped>
Consider the path of light between two mirrors in a high speed
Sagnac apparatus spinning at 72,000 rpm. This extremely high rate
of rotation results in the beam path following an arc with an
apparent radius of 250000 meters as observed from the rotating
frame.

If we assume that the two mirrors are 1 meter apart, the distance
from the center of the chord representing the straight line
distance between the two mirrors to the center of the arc
representing the apparent path of the light between the two
mirrors would be 1 micron.

Q: What is the total distance traversed by the beam of light
between the two mirrors as measured in the rotating frame?

A: The half-angle between the two endpoints of the arc has a sine
of 0.5/250000. Setting this equal to the first two terms of the
Taylor series expansion for sine
x - x^3/6 = 0.000002
we find that the half-angle between the two endpoints of the
arc is (0.000002 + 1.33e-18) radians. Multiplying this by
twice the radius, we find that the apparent total distance
traversed by the beam of light between the two mirrors as
measured in the rotating frame is (1 + 6.7e-13) meters, which
is immeasurably different from 1 meter even though we are
spinning the turntable at such high speed that any real device
would probably fly apart.

Jerry
Androcles
2010-02-16 21:10:38 UTC
Permalink
Post by Jerry
Post by Androcles
Post by Jerry
The only reason Coriolis curvature SEEMS to add significantly to
path length in your animation is that you spin the wheel at some
huge fraction of the speed of light.
Coriolis curvature is a second-order effect.
=======================================
Liar.
<Irrelevant rant snipped>
Consider the path of light between two mirrors in a high speed
Sagnac apparatus spinning at 72,000 rpm.
No, I will not. The only reason you get any fringe shift is that the
frequency of the light is in the order of 10^14 Hz.
Coriolis is a first-order effect, you lying incompetent tord,
and I can snip better than you, fuckin' useless bigot.

"I doubt that I'm capable of significantly
improving on, so I haven't tried" - Lazy fuckin' arsehole Tom&Jeery,
the cartoon pussy.
Jerry
2010-02-16 21:23:17 UTC
Permalink
Post by Androcles
Post by Jerry
Post by Androcles
Post by Jerry
The only reason Coriolis curvature SEEMS to add significantly to
path length in your animation is that you spin the wheel at some
huge fraction of the speed of light.
Coriolis curvature is a second-order effect.
=======================================
Liar.
<Irrelevant rant snipped>
Consider the path of light between two mirrors in a high speed
Sagnac apparatus spinning at 72,000 rpm.
No, I will not.
Coward.
Post by Androcles
The only reason you get any fringe shift is that the
frequency of the light is in the order of 10^14 Hz.
Coriolis is a first-order effect, you lying incompetent tord,
The horizontal displacement of the beam would be first order.
The increase in path length would be second order.

What is relevant here is the increase in path length.

Consider the path of light between two mirrors in a high speed
Sagnac apparatus spinning at 72,000 rpm. This extremely high rate
of rotation results in the beam path following an arc with an
apparent radius of 250000 meters as observed from the rotating
frame.

If we assume that the two mirrors are 1 meter apart, the distance
from the center of the chord representing the straight line
distance between the two mirrors to the center of the arc
representing the apparent path of the light between the two
mirrors would be 1 micron.

Q: What is the total distance traversed by the beam of light
between the two mirrors as measured in the rotating frame?

A: The half-angle between the two endpoints of the arc has a sine
of 0.5/250000. Setting this equal to the first two terms of the
Taylor series expansion for sine
x - x^3/6 = 0.000002
we find that the half-angle between the two endpoints of the
arc is (0.000002 + 1.33e-18) radians. Multiplying this by
twice the radius, we find that the apparent total distance
traversed by the beam of light between the two mirrors as
measured in the rotating frame is (1 + 6.7e-13) meters, which
is immeasurably different from 1 meter even though we are
spinning the turntable at such high speed that any real device
would probably fly apart.

Jerry
Androcles
2010-02-16 21:24:17 UTC
Permalink
Post by Androcles
Post by Jerry
Post by Androcles
Post by Jerry
The only reason Coriolis curvature SEEMS to add significantly to
path length in your animation is that you spin the wheel at some
huge fraction of the speed of light.
Coriolis curvature is a second-order effect.
=======================================
Liar.
<Irrelevant rant snipped>
Consider the path of light between two mirrors in a high speed
Sagnac apparatus spinning at 72,000 rpm.
No, I will not.
Coward.

=====================
Stupid cunt.
Jerry
2010-02-16 21:25:38 UTC
Permalink
Post by Jerry
Post by Androcles
Post by Jerry
Post by Androcles
Post by Jerry
The only reason Coriolis curvature SEEMS to add significantly to
path length in your animation is that you spin the wheel at some
huge fraction of the speed of light.
Coriolis curvature is a second-order effect.
=======================================
Liar.
<Irrelevant rant snipped>
Consider the path of light between two mirrors in a high speed
Sagnac apparatus spinning at 72,000 rpm.
No, I will not.
Coward.
=====================
Stupid cunt.
Consider the path of light between two mirrors in a high speed
Sagnac apparatus spinning at 72,000 rpm. This extremely high rate
of rotation results in the beam path following an arc with an
apparent radius of 250000 meters as observed from the rotating
frame.

If we assume that the two mirrors are 1 meter apart, the distance
from the center of the chord representing the straight line
distance between the two mirrors to the center of the arc
representing the apparent path of the light between the two
mirrors would be 1 micron.

Q: What is the total distance traversed by the beam of light
between the two mirrors as measured in the rotating frame?

A: The half-angle between the two endpoints of the arc has a sine
of 0.5/250000. Setting this equal to the first two terms of the
Taylor series expansion for sine
x - x^3/6 = 0.000002
we find that the half-angle between the two endpoints of the
arc is (0.000002 + 1.33e-18) radians. Multiplying this by
twice the radius, we find that the apparent total distance
traversed by the beam of light between the two mirrors as
measured in the rotating frame is (1 + 6.7e-13) meters, which
is immeasurably different from 1 meter even though we are
spinning the turntable at such high speed that any real device
would probably fly apart.

Jerry
Androcles
2010-02-16 21:29:54 UTC
Permalink
Post by Jerry
Post by Androcles
Post by Jerry
Post by Androcles
Post by Jerry
The only reason Coriolis curvature SEEMS to add significantly to
path length in your animation is that you spin the wheel at some
huge fraction of the speed of light.
Coriolis curvature is a second-order effect.
=======================================
Liar.
<Irrelevant rant snipped>
Consider the path of light between two mirrors in a high speed
Sagnac apparatus spinning at 72,000 rpm.
No, I will not.
Coward.
=====================
Stupid cunt.
Consider the path ...
===================================================
The gauntlet is down. Shut the fuck up and get on with it, YOU LYING COWARD.

In a simulation "practical" means rotate faster and slow the simulated
light down so the viewer can easily see what is going on, you ignorant
arse.
This is Andersen:
http://home.c2i.net/pb_andersen/pdf/four_mirror_sagnac.pdf
Look at figure 3:
Read what Andersen says:
"The beams are slightly curved because..."
But he doesn't animate it.

I've animated it.
In reality, the path lengths are accurate to the practical limits of finite
pixel size, but the velocities are not - and neither are Tusseladd's.
Mine works because I chose an exact integer wavelength in fixed frame.
LIKE THIS, you worthless bigot:
http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/CoriSag.gif
You and Tusseladd are frauds!

AND you ARE capable of significantly improving on it, you haven't tried
because
1) You are too fuckin' LAZY,
2) You've already made up your pathetic mind, you fuckin' BIGOT!

I'll give Tusseladd a tad more credit than you, at least he's not as
lazy as you are even if he's just as much a bigot.

Your mission (should you decide to accept it) is to animate Tusseladd's
Fig 3. using Java, allowing the user to set the speed to see what is going
on in the rotating frame, the lab frame and the SR (=aether) frame.
This tape will self-destruct in five months, that should be enough
time.

You may outdo Wilson, you lazy fraudulent bigot, and you can outdo
my animation, but you will not defeat me.
"Nope" is not an acceptable reply, nor are phase-shift jumps at
the corners acceptable.
The gauntlet is down. Shut the fuck up and get on with it YOU COWARD.
Jerry
2010-02-16 21:36:53 UTC
Permalink
Post by Androcles
You may outdo Wilson,
Of course.
Post by Androcles
you lazy fraudulent bigot, and you can outdo
my animation,
Already have.
Post by Androcles
but you will not defeat me.
"Nope" is not an acceptable reply, nor are phase-shift jumps at
the corners acceptable.
The gauntlet is down. Shut the fuck up and get on with it YOU COWARD.
Then defeat my simple math.

Consider the path of light between two mirrors in a high speed
Sagnac apparatus spinning at 72,000 rpm. This extremely high rate
of rotation results in the beam path following an arc with an
apparent radius of 250000 meters as observed from the rotating
frame.

If we assume that the two mirrors are 1 meter apart, the distance
from the center of the chord representing the straight line
distance between the two mirrors to the center of the arc
representing the apparent path of the light between the two
mirrors would be 1 micron.

Q: What is the total distance traversed by the beam of light
between the two mirrors as measured in the rotating frame?

A: The half-angle between the two endpoints of the arc has a sine
of 0.5/250000. Setting this equal to the first two terms of the
Taylor series expansion for sine
x - x^3/6 = 0.000002
we find that the half-angle between the two endpoints of the
arc is (0.000002 + 1.33e-18) radians. Multiplying this by
twice the radius, we find that the apparent total distance
traversed by the beam of light between the two mirrors as
measured in the rotating frame is (1 + 6.7e-13) meters, which
is immeasurably different from 1 meter even though we are
spinning the turntable at such high speed that any real device
would probably fly apart.

Jerry
Androcles
2010-02-16 21:39:36 UTC
Permalink
Post by Androcles
You may outdo Wilson,
Of course.
Post by Androcles
you lazy fraudulent bigot, and you can outdo
my animation,
Already have.
====================================
Lying cowardly snipping troll.

In a simulation "practical" means rotate faster and slow the simulated
light down so the viewer can easily see what is going on, you ignorant
arse.
This is Andersen:
http://home.c2i.net/pb_andersen/pdf/four_mirror_sagnac.pdf
Look at figure 3:
Read what Andersen says:
"The beams are slightly curved because..."
But he doesn't animate it.

I've animated it.
In reality, the path lengths are accurate to the practical limits of finite
pixel size, but the velocities are not - and neither are Tusseladd's.
Mine works because I chose an exact integer wavelength in fixed frame.
LIKE THIS, you worthless bigot:
http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/CoriSag.gif
You and Tusseladd are frauds!

AND you ARE capable of significantly improving on it, you haven't tried
because
1) You are too fuckin' LAZY,
2) You've already made up your pathetic mind, you fuckin' BIGOT!

I'll give Tusseladd a tad more credit than you, at least he's not as
lazy as you are even if he's just as much a bigot.

Your mission (should you decide to accept it) is to animate Tusseladd's
Fig 3. using Java, allowing the user to set the speed to see what is going
on in the rotating frame, the lab frame and the SR (=aether) frame.
This tape will self-destruct in five months, that should be enough
time.

You may outdo Wilson, you lazy fraudulent bigot, and you can outdo
my animation, but you will not defeat me.
"Nope" is not an acceptable reply, nor are phase-shift jumps at
the corners acceptable.
The gauntlet is down. Shut the fuck up and get on with it, COWARD!
Jerry
2010-02-16 21:45:18 UTC
Permalink
Post by Androcles
You may outdo Wilson, you lazy fraudulent bigot, and you can outdo
my animation, but you will not defeat me.
"Nope" is not an acceptable reply, nor are phase-shift jumps at
the corners acceptable.
The gauntlet is down. Shut the fuck up and get on with it, COWARD!
Why are you so scared of a few paragraphs of simple math?
The most complex concept is a simple Taylor series.
Surely you can spend ten minutes perusing what I have to say.
If you can defeat my arguments, do so.

Consider the path of light between two mirrors in a high speed
Sagnac apparatus spinning at 72,000 rpm. This extremely high rate
of rotation results in the beam path following an arc with an
apparent radius of 250000 meters as observed from the rotating
frame.

If we assume that the two mirrors are 1 meter apart, the distance
from the center of the chord representing the straight line
distance between the two mirrors to the center of the arc
representing the apparent path of the light between the two
mirrors would be 1 micron.

Q: What is the total distance traversed by the beam of light
between the two mirrors as measured in the rotating frame?

A: The half-angle between the two endpoints of the arc has a sine
of 0.5/250000. Setting this equal to the first two terms of the
Taylor series expansion for sine
x - x^3/6 = 0.000002
we find that the half-angle between the two endpoints of the
arc is (0.000002 + 1.33e-18) radians. Multiplying this by
twice the radius, we find that the apparent total distance
traversed by the beam of light between the two mirrors as
measured in the rotating frame is (1 + 6.7e-13) meters, which
is immeasurably different from 1 meter even though we are
spinning the turntable at such high speed that any real device
would probably fly apart.

Jerry
Androcles
2010-02-16 21:55:35 UTC
Permalink
"Jerry" <***@comcast.net> wrote in message news:25ff7824-d721-4b46-9ce4-***@c10g2000vbr.googlegroups.com...
On Feb 16, 3:39 pm, "Androcles" <***@Hogwarts.physics_u> wrote:

Lying cowardly snipping troll.

In a simulation "practical" means rotate faster and slow the simulated
light down so the viewer can easily see what is going on, you ignorant
arse.
This is Andersen:
http://home.c2i.net/pb_andersen/pdf/four_mirror_sagnac.pdf
Look at figure 3:
Read what Andersen says:
"The beams are slightly curved because..."
But he doesn't animate it.

I've animated it.
In reality, the path lengths are accurate to the practical limits of finite
pixel size, but the velocities are not - and neither are Tusseladd's.
Mine works because I chose an exact integer wavelength in fixed frame.
LIKE THIS, you worthless bigot:
http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/CoriSag.gif
You and Tusseladd are frauds!

AND you ARE capable of significantly improving on it, you haven't tried
because
1) You are too fuckin' LAZY,
2) You've already made up your pathetic mind, you fuckin' BIGOT!

I'll give Tusseladd a tad more credit than you, at least he's not as
lazy as you are even if he's just as much a bigot.

Your mission (should you decide to accept it) is to animate Tusseladd's
Fig 3. using Java, allowing the user to set the speed to see what is going
on in the rotating frame, the lab frame and the SR (=aether) frame.
This tape will self-destruct in five months, that should be enough
time.

You may outdo Wilson, you lazy fraudulent bigot, and you can outdo
my animation, but you will not defeat me.
"Nope" is not an acceptable reply, nor are phase-shift jumps at
the corners acceptable.
The gauntlet is down. Shut the fuck up and get on with it, COWARD!
Jerry
2010-02-16 22:00:18 UTC
Permalink
Post by Androcles
You may outdo Wilson, you lazy fraudulent bigot, and you can outdo
my animation, but you will not defeat me.
"Nope" is not an acceptable reply, nor are phase-shift jumps at
the corners acceptable.
The gauntlet is down. Shut the fuck up and get on with it, COWARD!
This is getting tiresome.
You've lost and you know that you've lost. Your silly behavior
in changing the followup proves it.

Go home and sulk.

Consider the path of light between two mirrors in a high speed
Sagnac apparatus spinning at 72,000 rpm. This extremely high rate
of rotation results in the beam path following an arc with an
apparent radius of 250000 meters as observed from the rotating
frame.

If we assume that the two mirrors are 1 meter apart, the distance
from the center of the chord representing the straight line
distance between the two mirrors to the center of the arc
representing the apparent path of the light between the two
mirrors would be 1 micron.

Q: What is the total distance traversed by the beam of light
between the two mirrors as measured in the rotating frame?

A: The half-angle between the two endpoints of the arc has a sine
of 0.5/250000. Setting this equal to the first two terms of the
Taylor series expansion for sine
x - x^3/6 = 0.000002
we find that the half-angle between the two endpoints of the
arc is (0.000002 + 1.33e-18) radians. Multiplying this by
twice the radius, we find that the apparent total distance
traversed by the beam of light between the two mirrors as
measured in the rotating frame is (1 + 6.7e-13) meters, which
is immeasurably different from 1 meter even though we are
spinning the turntable at such high speed that any real device
would probably fly apart.

Jerry

Henry Wilson DSc
2010-02-16 06:00:09 UTC
Permalink
On Mon, 15 Feb 2010 18:00:42 -0800 (PST), Jerry
Post by Jerry
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 02:49:13 -0800 (PST), Jerry
Your "rayphases" model has been refuted many times over, by me
and by many others employing different arguments and methods of
analysis. It is self-contradictory and predicts effects that have
never been seen. In this thread alone, I've seen you squirm and
deny assertions that you made mere hours previously. Your model
is HOPELESS.
http://mysite.verizon.net/cephalobus_alienus/henri/HWFantasy.htm
Well, this experiment requires no light rays or interferometers.
It is simple. Just wrap two optical fibres around the equator and send time
labelled light pulses every 3.33 nanoseconds down each one in opposite
directions. The pulses will be 1 metre apart in both fibres.

According to BaTh, the Earth rotates by 62 metres during the time it takes for
a pulse to travel around the ring. So one light path will be 4000062 metres
long and the other 3999938.

Therefore, between the time pulse N is emitted and received, 4000062 pulses
arrive from one fibre and only 3999938 from the other. All you need is a
counter.
Post by Jerry
Jerry
Henry Wilson...

.......provider of free physics lessons
Jerry
2010-02-16 06:25:28 UTC
Permalink
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 18:00:42 -0800 (PST), Jerry
Post by Jerry
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 02:49:13 -0800 (PST), Jerry
Your "rayphases" model has been refuted many times over, by me
and by many others employing different arguments and methods of
analysis. It is self-contradictory and predicts effects that have
never been seen. In this thread alone, I've seen you squirm and
deny assertions that you made mere hours previously. Your model
is HOPELESS.
http://mysite.verizon.net/cephalobus_alienus/henri/HWFantasy.htm
Well, this experiment requires no light rays or interferometers.
It is simple. Just wrap two optical fibres around the equator and send time
labelled light pulses every 3.33 nanoseconds down each one in opposite
directions. The pulses will be 1 metre apart in both fibres.
According to BaTh, the Earth rotates by 62 metres during the time it takes for
a pulse to travel around the ring. So one light path will be 4000062 metres
long and the other 3999938.
Therefore, between the time pulse N is emitted and received, 4000062 pulses
arrive from one fibre and only 3999938 from the other. All you need is a
counter.
Good grief-and-a-half!
When your stupidity finally becomes apparent to you, I predict
that you will claim to have been joking, that you were merely
testing whether any of us "relativists" were capable of reasoning
out a simple trick question. You will laugh about how you put one
over on us, and you'll call everybody except yourself a moron for
being so gullible...

Jerry
Inertial
2010-02-16 06:31:51 UTC
Permalink
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 18:00:42 -0800 (PST), Jerry
Post by Jerry
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 02:49:13 -0800 (PST), Jerry
Your "rayphases" model has been refuted many times over, by me
and by many others employing different arguments and methods of
analysis. It is self-contradictory and predicts effects that have
never been seen. In this thread alone, I've seen you squirm and
deny assertions that you made mere hours previously. Your model
is HOPELESS.
http://mysite.verizon.net/cephalobus_alienus/henri/HWFantasy.htm
Well, this experiment requires no light rays or interferometers.
So a light pulse isn't just a short light ray?
Post by Henry Wilson DSc
It is simple. Just wrap two optical fibres around the equator and send time
labelled light pulses every 3.33 nanoseconds down each one in opposite
directions. The pulses will be 1 metre apart in both fibres.
OK
Post by Henry Wilson DSc
According to BaTh, the Earth rotates by 62 metres during the time it takes for
a pulse to travel around the ring. So one light path will be 4000062 metres
long and the other 3999938.
But it will travel the longer path quicker, so they arrive at the same time
with the same arrival rate
Post by Henry Wilson DSc
Therefore, between the time pulse N is emitted and received, 4000062 pulses
arrive from one fibre and only 3999938 from the other.
WRONG.

If the pulses leave in pairs (as you imply), and take the same time to
travel all the way around their path and arrive back again (as emission
theory says), then they MUST arrive back in the same pairs they were
emitted, and so the SAME NUMBER OF PULSES arrive back from each fibre path
in any given time interval.

You cannot determine the rotation of the device (according to emission
theory) by looking at the arrival times and rates of the pulses. And the
fact that your CAN in reality (with a real sagnac device) is resounding and
refutation of emission theory.

Why do you keep lying, Henry?
Post by Henry Wilson DSc
All you need is a
counter.
All you need is SR .. which DOES let you determine the absolute rotation.
Henry Wilson DSc
2010-02-16 10:02:32 UTC
Permalink
Post by Inertial
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 18:00:42 -0800 (PST), Jerry
Post by Jerry
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 02:49:13 -0800 (PST), Jerry
Your "rayphases" model has been refuted many times over, by me
and by many others employing different arguments and methods of
analysis. It is self-contradictory and predicts effects that have
never been seen. In this thread alone, I've seen you squirm and
deny assertions that you made mere hours previously. Your model
is HOPELESS.
http://mysite.verizon.net/cephalobus_alienus/henri/HWFantasy.htm
Well, this experiment requires no light rays or interferometers.
So a light pulse isn't just a short light ray?
Post by Henry Wilson DSc
It is simple. Just wrap two optical fibres around the equator and send time
labelled light pulses every 3.33 nanoseconds down each one in opposite
directions. The pulses will be 1 metre apart in both fibres.
OK
Post by Henry Wilson DSc
According to BaTh, the Earth rotates by 62 metres during the time it takes for
a pulse to travel around the ring. So one light path will be 4000062 metres
long and the other 3999938.
But it will travel the longer path quicker, so they arrive at the same time
with the same arrival rate
Post by Henry Wilson DSc
Therefore, between the time pulse N is emitted and received, 4000062 pulses
arrive from one fibre and only 3999938 from the other.
WRONG.
If the pulses leave in pairs (as you imply), and take the same time to
travel all the way around their path and arrive back again (as emission
theory says), then they MUST arrive back in the same pairs they were
emitted, and so the SAME NUMBER OF PULSES arrive back from each fibre path
in any given time interval.
You cannot determine the rotation of the device (according to emission
theory) by looking at the arrival times and rates of the pulses. And the
fact that your CAN in reality (with a real sagnac device) is resounding and
refutation of emission theory.
Why do you keep lying, Henry?
Post by Henry Wilson DSc
All you need is a
counter.
All you need is SR .. which DOES let you determine the absolute rotation.
You will remain plonked until your brain decides to function

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-16 11:33:36 UTC
Permalink
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 18:00:42 -0800 (PST), Jerry
Post by Jerry
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 02:49:13 -0800 (PST), Jerry
Your "rayphases" model has been refuted many times over, by me
and by many others employing different arguments and methods of
analysis. It is self-contradictory and predicts effects that have
never been seen. In this thread alone, I've seen you squirm and
deny assertions that you made mere hours previously. Your model
is HOPELESS.
http://mysite.verizon.net/cephalobus_alienus/henri/HWFantasy.htm
Well, this experiment requires no light rays or interferometers.
So a light pulse isn't just a short light ray?
Post by Henry Wilson DSc
It is simple. Just wrap two optical fibres around the equator and send time
labelled light pulses every 3.33 nanoseconds down each one in opposite
directions. The pulses will be 1 metre apart in both fibres.
OK
Post by Henry Wilson DSc
According to BaTh, the Earth rotates by 62 metres during the time it
takes
for
a pulse to travel around the ring. So one light path will be 4000062 metres
long and the other 3999938.
But it will travel the longer path quicker, so they arrive at the same time
with the same arrival rate
Post by Henry Wilson DSc
Therefore, between the time pulse N is emitted and received, 4000062 pulses
arrive from one fibre and only 3999938 from the other.
WRONG.
If the pulses leave in pairs (as you imply), and take the same time to
travel all the way around their path and arrive back again (as emission
theory says), then they MUST arrive back in the same pairs they were
emitted, and so the SAME NUMBER OF PULSES arrive back from each fibre path
in any given time interval.
You cannot determine the rotation of the device (according to emission
theory) by looking at the arrival times and rates of the pulses. And the
fact that your CAN in reality (with a real sagnac device) is resounding and
refutation of emission theory.
Why do you keep lying, Henry?
Post by Henry Wilson DSc
All you need is a
counter.
All you need is SR .. which DOES let you determine the absolute rotation.
You will remain plonked until your brain decides to function
When will yours .. it hasn't shown any sign of rational though in the last
couple of days (in particular). I've refuted your theory .. get a new
hobby.
Henry Wilson DSc
2010-02-16 10:26:54 UTC
Permalink
On Mon, 15 Feb 2010 18:00:42 -0800 (PST), Jerry
Post by Jerry
Your "rayphases" model has been refuted many times over, by me
and by many others employing different arguments and methods of
analysis. It is self-contradictory and predicts effects that have
never been seen. In this thread alone, I've seen you squirm and
deny assertions that you made mere hours previously. Your model
is HOPELESS.
http://mysite.verizon.net/cephalobus_alienus/henri/HWFantasy.htm
It is true that I made a late addition to that program which is rather
misleading...but that matters not.

The reason you are having trouble understanding the Bath explanation is that
three separate oscillation periods are involved.
There is the the rotation of the Earth, the rotation of the pulses around the
ring and the oscillator that causes pulses to be emitted at regular intervals.
The fact that movement is circular rather than linear further complicates the
issue.

Only those with superior intelligence should even try to fathom what is
actually happening here. Certainly, neither a raw graduate nor anyone with
early dementia would, have a hope.
Post by Jerry
Jerry
Henry Wilson...

.......provider of free physics lessons
eric gisse
2010-02-16 02:28:58 UTC
Permalink
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 02:49:13 -0800 (PST), Jerry
Post by Jerry
Post by eric gisse
Imagine that - Henri ignores the discussion of his faked degrees to
discuss something nobody really cares about.- Hide quoted text -
Note that Henri has not denied any of my analysis.
He -believes- it...everything except the "Deluded? You bet" part.
Moron Jerry, the bedpan expert [...]
You have one of these for everyone. Perhaps you'd be happier posting
somewhere else?
eric gisse
2010-02-15 05:54:40 UTC
Permalink
Post by Jerry
Post by eric gisse
When are you going to apologize for posting forged degrees?
That's not going to work, Eric.
Its' been working great. It shuts him down immediately. He stops with his
little one-liner 'give me the last word!!!' idiocies rather quick once I
started doing that.

In regular discussions he snips all references to his forged degrees and
refuses to discuss it. He won't engage in any discussion when that's the
only topic.

[..]
Post by Jerry
Deluded? You bet.
Jerry
Oh yeah. But he still has enough sense of shame to be embarrassed about it
when confronted.
Inertial
2010-02-15 06:14:33 UTC
Permalink
Post by eric gisse
Post by Jerry
Post by eric gisse
When are you going to apologize for posting forged degrees?
That's not going to work, Eric.
Its' been working great. It shuts him down immediately. He stops with his
little one-liner 'give me the last word!!!' idiocies rather quick once I
started doing that.
In regular discussions he snips all references to his forged degrees and
refuses to discuss it. He won't engage in any discussion when that's the
only topic.
[..]
Post by Jerry
Deluded? You bet.
Jerry
Oh yeah. But he still has enough sense of shame to be embarrassed about it
when confronted.
Nope .. like any facts he doesn't like .. he ignores it. If he ignores it,
then it doesn't exist and there is no need for him to be embarrassed (which
would require him to have some integrity .. he clearly has none).
Henry Wilson DSc
2010-02-15 10:34:28 UTC
Permalink
Post by Inertial
Post by eric gisse
Post by Jerry
Post by eric gisse
When are you going to apologize for posting forged degrees?
That's not going to work, Eric.
Its' been working great. It shuts him down immediately. He stops with his
little one-liner 'give me the last word!!!' idiocies rather quick once I
started doing that.
In regular discussions he snips all references to his forged degrees and
refuses to discuss it. He won't engage in any discussion when that's the
only topic.
[..]
Post by Jerry
Deluded? You bet.
Jerry
Oh yeah. But he still has enough sense of shame to be embarrassed about it
when confronted.
Nope .. like any facts he doesn't like .. he ignores it. If he ignores it,
then it doesn't exist and there is no need for him to be embarrassed (which
would require him to have some integrity .. he clearly has none).
When are you going to say something intelligent?

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-15 11:34:52 UTC
Permalink
Post by Henry Wilson DSc
Post by Inertial
Post by eric gisse
Post by Jerry
Post by eric gisse
When are you going to apologize for posting forged degrees?
That's not going to work, Eric.
Its' been working great. It shuts him down immediately. He stops with his
little one-liner 'give me the last word!!!' idiocies rather quick once I
started doing that.
In regular discussions he snips all references to his forged degrees and
refuses to discuss it. He won't engage in any discussion when that's the
only topic.
[..]
Post by Jerry
Deluded? You bet.
Jerry
Oh yeah. But he still has enough sense of shame to be embarrassed about it
when confronted.
Nope .. like any facts he doesn't like .. he ignores it. If he ignores it,
then it doesn't exist and there is no need for him to be embarrassed (which
would require him to have some integrity .. he clearly has none).
When are you going to say something intelligent?
Case proven.
Henry Wilson DSc
2010-02-15 10:34:10 UTC
Permalink
Post by eric gisse
Post by Jerry
Post by eric gisse
When are you going to apologize for posting forged degrees?
That's not going to work, Eric.
Its' been working great. It shuts him down immediately. He stops with his
little one-liner 'give me the last word!!!' idiocies rather quick once I
started doing that.
In regular discussions he snips all references to his forged degrees and
refuses to discuss it. He won't engage in any discussion when that's the
only topic.
[..]
Post by Jerry
Deluded? You bet.
Jerry
Oh yeah. But he still has enough sense of shame to be embarrassed about it
when confronted.
When are you going to say something intelligent?


Henry Wilson...

.......provider of free physics lessons
eric gisse
2010-02-15 16:04:17 UTC
Permalink
Post by Henry Wilson DSc
Post by eric gisse
Post by Jerry
Post by eric gisse
When are you going to apologize for posting forged degrees?
That's not going to work, Eric.
Its' been working great. It shuts him down immediately. He stops with his
little one-liner 'give me the last word!!!' idiocies rather quick once I
started doing that.
In regular discussions he snips all references to his forged degrees and
refuses to discuss it. He won't engage in any discussion when that's the
only topic.
[..]
Post by Jerry
Deluded? You bet.
Jerry
Oh yeah. But he still has enough sense of shame to be embarrassed about it
when confronted.
When are you going to say something intelligent?
When are you going to apologize for posting forged degrees?
Post by Henry Wilson DSc
Henry Wilson...
.......provider of free physics lessons
Androcles
2010-02-14 07:31:01 UTC
Permalink
On Sun, 14 Feb 2010 01:20:44 -0000, "Androcles"
Post by Androcles
On Sat, 13 Feb 2010 22:37:33 -0000, "Androcles"
Post by Androcles
It's simple enough.
There are two beat frequencies, the sum and the difference.
http://en.wikipedia.org/wiki/Circle_of_fifths
So if you have 4 Hz and 5 Hz you'll also get 1 Hz and 9 Hz, but it won't
be very musical.
http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/MechSagnac.gif
In the case of light it will be the difference that creates the pattern.
Yeah,
http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/Sagnac2.JPG
Yes. That shows the difference.
Post by Androcles
....and fringe DISPLACEMENT will be proportional to beat frequency.
Bullshit. You've never worked with an oscilloscope.
How come you never produce a web page to back up your stupid claims?
...you're starting to sound more like little eric every day.....
...you're starting to sound more like little Einstein every day.....
--
Androcles
.......provider of expensive physics lessons Awilson can't afford.
bert
2010-02-13 18:48:44 UTC
Permalink
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre.  The pulses are labelled according to their emission times and move at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p. At any instant, 40 million pulses are in transit
around the planet in each direction.
Tom observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period of 1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
In this new situation, the pulses that arrive at C when it is at point p, were
not emitted at that point but rather from an imaginary point q, which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
Since the distance between pulses is still 1 metre in the C's frame, there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
"That's odd", says Tom, "now, for some reason, the pulses coming from the left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
Henry Wilson...
.......provider of free physics lessons
My "Spin is in Theory" Tells it all. The universe is #1 energy,and
rotational spin is in reality what all there is is all about TreBert
BURT
2010-02-13 19:11:32 UTC
Permalink
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre.  The pulses are labelled according to their emission times and move at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p. At any instant, 40 million pulses are in transit
around the planet in each direction.
Tom observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period of 1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
In this new situation, the pulses that arrive at C when it is at point p, were
not emitted at that point but rather from an imaginary point q, which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
Since the distance between pulses is still 1 metre in the C's frame, there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
"That's odd", says Tom, "now, for some reason, the pulses coming from the left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
Henry Wilson...
.......provider of free physics lessons
My "Spin is in Theory"  Tells it all.  The universe is #1 energy,and
rotational spin is in reality what all there is is all about  TreBert- Hide quoted text -
- Show quoted text -
Spin is misused for Rotation. A rotation speed flow that changes sizes
is what spin is.

Mitch Raemsch
bert
2010-02-13 19:44:33 UTC
Permalink
Post by BURT
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre.  The pulses are labelled according to their emission times and move at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p. At any instant, 40 million pulses are in transit
around the planet in each direction.
Tom observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period of 1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
In this new situation, the pulses that arrive at C when it is at point p, were
not emitted at that point but rather from an imaginary point q, which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
Since the distance between pulses is still 1 metre in the C's frame, there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
"That's odd", says Tom, "now, for some reason, the pulses coming from the left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
Henry Wilson...
.......provider of free physics lessons
My "Spin is in Theory"  Tells it all.  The universe is #1 energy,and
rotational spin is in reality what all there is is all about  TreBert- Hide quoted text -
- Show quoted text -
Spin is misused for Rotation. A rotation speed flow that changes sizes
is what spin is.
Mitch Raemsch- Hide quoted text -
- Show quoted text -
Burt lectron Speed rotation is the heart of my c speed of c TreBert
Paul B. Andersen
2010-02-15 13:13:02 UTC
Permalink
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
(the surface is drawn flat for convenience. The fibre goes right around the
planet)
Paul's clock emits light pulses every 3.33 nanoseconds through both ends of the
fibre. The pulses are labelled according to their emission times and move at c
wrt the fibre and the clock. The linear distance between each pulse is
therefore 1 metre. Each pulse takes ~4E7/3E8 (0.133) seconds to pass through
the fibre in either direction.
Since the planet is not rotating, the pulses returning to C were emitted and
detected by Jerry at point p. At any instant, 40 million pulses are in transit
around the planet in each direction.
Tom observes that the pulses arriving simultaneously from both fibres carry
identical labels.
Using powerful rockets, the planet is sent into rotation with a period of 1
day, (86400 seconds). Its surface rotation speed 'v' wrt the nonrotating frame
is 465 m/s.
In this new situation, the pulses that arrive at C when it is at point p, were
not emitted at that point but rather from an imaginary point q, which is at
rest in the nonR frame and moving away from C at speed -v in C's R frame.
Each pulse takes still takes (0.133) seconds to pass through the fibre in
either direction.
Therefore the distance p-q = 465.(4/30) metres.....or 62 metres.
Since the distance between pulses is still 1 metre in the C's frame, there are
now 40000062 pulses moving through the fibre in one direction but only 39999938
moving in the other.
"That's odd", says Tom, "now, for some reason, the pulses coming from the left
fiber left Paul's clock 0.266 seconds ahead of those arriving from the right.
Which one should I use to synch my clock?"
"This is great", says Jerry after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Henry Wilson's
BaTh".
Good grief, Ralph. :-)

Yet again you have made a giant fool of yourself.
I wonder how long it will take this time before you
realize that you have made a giant, unbelievable stupid blunder.

I bet it will take a very long time, and if it eventually should
dawn to you, you will never admit it. What will you do then?
Start a new thread with a modified 'experiment'? :-)

I won't bother to tell you what your blunder is.

I will tell you a story, though.
I have a circular tube. It is not rotating.
At one point in that tube, there is a source/target device (STD)
which is emitting marbles at a constant speed 1 m/s relative
to the STD. When a marble has moved around the circuit
and gets back to the STD, it is re-emitted by the STD.
The circumference of the circle is 10m. So there are at any time
10 marbles in the tube, and it takes 10 seconds for a marble to
go around the circle.

Now we let the circle rotate with one rotation in 10 seconds,
in the same direction as the marbles are moving, which means that
the STD is moving at 1 m/s in the non rotating frame.

In this new situation, the marble that arrive at the STD when it is at point p,
were not emitted at that point but rather from an imaginary point q, which is at
rest in the nonR frame and moving away from the STD at speed -1 m/s in the STD R frame.

Each marble still takes 10 seconds to pass through the tube.
Therefore the distance p-q = (1 m/s * 1 second) metres.....or 1 metre.

Since the distance between marbles is still 1 metre in the STD's frame,
there are now 11 marbles moving through the tube.

"This is great", says Ralph after some deliberation, "now we can build a ring
gyro that will detect absolute rotation, based entirely on Ralph Rabbidge's
marble creating fairies.
--
Paul

http://home.c2i.net/pb_andersen/
Henry Wilson DSc
2010-02-15 22:24:16 UTC
Permalink
On Mon, 15 Feb 2010 14:13:02 +0100, "Paul B. Andersen"
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
I recognized my mistake and immediately corrected it.

The pulses that leave together arrive together.

If you want to comment, use my later version in which the pulse arrival rates
are different from both directions. The theory should be obvious to anyone with
a higher IQ than inertial.


Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-15 22:35:15 UTC
Permalink
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 14:13:02 +0100, "Paul B. Andersen"
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
I recognized my mistake and immediately corrected it.
The pulses that leave together arrive together.
So the SAME NUMBER of pulses arrive in both directions per second. So your
original claim was wrong and you cannot measure absolute rotation with the
device.
Post by Henry Wilson DSc
If you want to comment, use my later version in which the pulse arrival rates
are different from both directions.
But you just said "The pulses that leave together arrive together."

So that means they arrive with the same arrival rate. Otherwise they can't
arrive in pairs. Unless you have pulse fairies creating extra pulses from
nowhere.
Post by Henry Wilson DSc
The theory should be obvious to anyone with
a higher IQ than inertial.
Your blatant error should be obvious to even you. Yet either you're a
bigger moron than I thought, or as big a liar.
Henry Wilson DSc
2010-02-15 23:05:43 UTC
Permalink
Post by Inertial
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 14:13:02 +0100, "Paul B. Andersen"
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
I recognized my mistake and immediately corrected it.
The pulses that leave together arrive together.
So the SAME NUMBER of pulses arrive in both directions per second. So your
original claim was wrong and you cannot measure absolute rotation with the
device.
Post by Henry Wilson DSc
If you want to comment, use my later version in which the pulse arrival rates
are different from both directions.
But you just said "The pulses that leave together arrive together."
So that means they arrive with the same arrival rate. Otherwise they can't
arrive in pairs. Unless you have pulse fairies creating extra pulses from
nowhere.
Post by Henry Wilson DSc
The theory should be obvious to anyone with
a higher IQ than inertial.
Your blatant error should be obvious to even you. Yet either you're a
bigger moron than I thought, or as big a liar.
When are you going to say something intelligent?

Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-15 23:22:09 UTC
Permalink
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 14:13:02 +0100, "Paul B. Andersen"
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
I recognized my mistake and immediately corrected it.
The pulses that leave together arrive together.
So the SAME NUMBER of pulses arrive in both directions per second. So your
original claim was wrong and you cannot measure absolute rotation with the
device.
Post by Henry Wilson DSc
If you want to comment, use my later version in which the pulse arrival rates
are different from both directions.
But you just said "The pulses that leave together arrive together."
So that means they arrive with the same arrival rate. Otherwise they can't
arrive in pairs. Unless you have pulse fairies creating extra pulses from
nowhere.
Post by Henry Wilson DSc
The theory should be obvious to anyone with
a higher IQ than inertial.
Your blatant error should be obvious to even you. Yet either you're a
bigger moron than I thought, or as big a liar.
When are you going to say something intelligent?
This is Henry's typical results when he knows that he is beaten.

Of course, everything I say is intelligent. But you are just too big of a
moron to understand it.

And here's something else intelligent for you:

If two pulses go into a fixed length circular tube (in opposite directions)
at the same time, and travel for the same duration, then they MUST arrive at
the detector at the same time and so the arrival rate of successive pairs of
pulses MUST be the same. Your claim to the contrary is ridiculous.

At least you've retracted your claim that there are differing numbers of
pulses in the tube going one way to the other. Are you ready to retract
your ridiculous claim that the arrival rates are different when you admit
that every pair of pulses arrives at the same time?

Note that even with what you have admitted now, there can be no phase shift
because the pairs of pulses emitted together arrive together at the same
time. Phase shift occurs when pulses arrive at different times (as, for
example, SR predicts). Emission theory is once again soundly refuted, using
your own argument.
Henry Wilson DSc
2010-02-16 10:34:02 UTC
Permalink
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 14:13:02 +0100, "Paul B. Andersen"
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
I recognized my mistake and immediately corrected it.
The pulses that leave together arrive together.
So the SAME NUMBER of pulses arrive in both directions per second. So your
original claim was wrong and you cannot measure absolute rotation with the
device.
Post by Henry Wilson DSc
If you want to comment, use my later version in which the pulse arrival rates
are different from both directions.
But you just said "The pulses that leave together arrive together."
So that means they arrive with the same arrival rate. Otherwise they can't
arrive in pairs. Unless you have pulse fairies creating extra pulses from
nowhere.
Post by Henry Wilson DSc
The theory should be obvious to anyone with
a higher IQ than inertial.
Your blatant error should be obvious to even you. Yet either you're a
bigger moron than I thought, or as big a liar.
When are you going to say something intelligent?
This is Henry's typical results when he knows that he is beaten.
Of course, everything I say is intelligent. But you are just too big of a
moron to understand it.
If two pulses go into a fixed length circular tube (in opposite directions)
at the same time, and travel for the same duration, then they MUST arrive at
the detector at the same time and so the arrival rate of successive pairs of
pulses MUST be the same. Your claim to the contrary is ridiculous.
At least you've retracted your claim that there are differing numbers of
pulses in the tube going one way to the other. Are you ready to retract
your ridiculous claim that the arrival rates are different when you admit
that every pair of pulses arrives at the same time?
Note that even with what you have admitted now, there can be no phase shift
because the pairs of pulses emitted together arrive together at the same
time. Phase shift occurs when pulses arrive at different times (as, for
example, SR predicts). Emission theory is once again soundly refuted, using
your own argument.
I see I'll have to unplonk you for your own sake. You need help.

1st question: Do you understand how and why the path lengths are different in
the nonR frame?


Henry Wilson...

.......provider of free physics lessons
Inertial
2010-02-16 11:39:14 UTC
Permalink
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
Post by Inertial
Post by Henry Wilson DSc
On Mon, 15 Feb 2010 14:13:02 +0100, "Paul B. Andersen"
Post by Henry Wilson DSc
Paul and Jerry are standing together on the equator of an Earthlike planet,
next to an optical fibre that encircles the planet, WHICH IS NOT ROTATING. Its
circumference is 40 million metres.
//===================C======================// optical fibre around equator
_____________v<-_____p___q_______________________surface (actually curved)
I recognized my mistake and immediately corrected it.
The pulses that leave together arrive together.
So the SAME NUMBER of pulses arrive in both directions per second. So your
original claim was wrong and you cannot measure absolute rotation with the
device.
Post by Henry Wilson DSc
If you want to comment, use my later version in which the pulse
arrival
rates
are different from both directions.
But you just said "The pulses that leave together arrive together."
So that means they arrive with the same arrival rate. Otherwise they can't
arrive in pairs. Unless you have pulse fairies creating extra pulses from
nowhere.
Post by Henry Wilson DSc
The theory should be obvious to anyone with
a higher IQ than inertial.
Your blatant error should be obvious to even you. Yet either you're a
bigger moron than I thought, or as big a liar.
When are you going to say something intelligent?
This is Henry's typical results when he knows that he is beaten.
Of course, everything I say is intelligent. But you are just too big of a
moron to understand it.
If two pulses go into a fixed length circular tube (in opposite directions)
at the same time, and travel for the same duration, then they MUST arrive at
the detector at the same time and so the arrival rate of successive pairs of
pulses MUST be the same. Your claim to the contrary is ridiculous.
At least you've retracted your claim that there are differing numbers of
pulses in the tube going one way to the other. Are you ready to retract
your ridiculous claim that the arrival rates are different when you admit
that every pair of pulses arrives at the same time?
Note that even with what you have admitted now, there can be no phase shift
because the pairs of pulses emitted together arrive together at the same
time. Phase shift occurs when pulses arrive at different times (as, for
example, SR predicts). Emission theory is once again soundly refuted, using
your own argument.
I see I'll have to unplonk you for your own sake. You need help.
1st question: Do you understand how and why the path lengths are different in
the nonR frame?
Of course .. I understand all this far better than you.

Do YOU understand that the number of particles (at any given time) in the
tube/fibre going in one direction is the same number as going in the other?

Do YOU understand that every pulse has its own path?

Do YOU understand that the pulses with the longer paths travel faster than
the ones with the slower paths so that the time to travel the whole path is
the same in either direction?

Do YOU understand that a pair of pulses emitted at the same time in opposite
directions will therefore take the same time to travel along their
respective paths and will arrive at the same time?

Do YOU understand the pairs of pulses arriving at the same time means the
arrival rates are the same for both paths?

Do YOU understand that there is only one pulse on each path, because every
pulse has its own path?
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