On Sat, 18 Aug 2007 11:34:42 +0100, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
.. .
On Sun, 12 Aug 2007 10:58:09 +0100, "George Dishman"
wrote:
"Henri Wilson" HW@.... wrote in message
...

George, please calculate the fractional speed change as light falls
from
26000kms to Earth. Then compare it with the GR correction.
Can't you do it?

Apples and pears Henry, compare the ballistic frequency
change to the GR frequency change. They are different.

Identical George.

No Henry, zero is not identical to non-zero,
you are wrong.

You can't do it can you George. You know nothing at all about basic
physics.

I can Henry, the answer is zero and it is you who
is demonstrating that you know nothing because you
are trying to equate a speed change to a frequency
change when in fact it causes a wavelength change.

You don't even know how to do the calculation so how would you know....

There is no calculation to do, ballistic theory
does not change frequencies, it changes speed
and hence wavelength.

It changes individual photon frequencies. I have told you why many times.

No, you have introduced the "K" factor to prevent
it changing frequencies, but that id for Doppler
and we are talking about the gravitational potential
effect neglecting relative motion so even that
doesn't apply.

I think we are talking about different things here.

My theory states that a photon has an intrinsic frequency based on some kind of
standing wave or spinning effect. This gives the appearance of a moving spatial
pattern... like a moving saw blade.
It also says that observed frequency is really the 'wavecrest arrival
frequency' at the observer. The equation is f = h(c+v)/Lambda, where Lambda is
ABSOLUTE wavelength.

That wavelength can differ from the norm in the source frame if: 1) the source
is accelerating (ADoppler x K) or 2) if light speed varies during travel
(similar to VDoppler)



Correct. Funny how you know what the postulate is
when it suits you. From that and a few other basics,
it follows that it has the same speed in _all_
inertial frames _simultaneously_.

Which is what my illustration shows.
Isn't the postulate ridiculous when you see it in action?

Nothing in your illustration conflicts with how
reality behaves, it is only your insistance that
the world cannot be the way it is because you
can't cope with it that is ridiculous.

see George, you apparently cannot drag yourself away from the belief that space
has absolute properties which detrmine light speed.

That means properties of space determine light speed.

Yes, the Riemann geonetry of spacetime determines that
behaviour.

I don't need Reimann or any other weird geometry to show the physical
meaning
of the second postulate.

Yes you do, your alternative is merely LET.

alias SR.

George, we live in a world of 3 space and 1 time.

Yes, and those four dimensions are described locally
by that Reimann geometry.

That's how we do our experiments. Any physical experiment must relate to
the
universe we know.

Right, that is why all physical laws should be locally
Lorentz Invariant.


postulating again..

YOU say
Maxwell's equation explains this.

No, It say they tell you the same thing, they don't
provide an explanation. The geometry gives that.

Crap.

It is beyond your level of maths.

It is nothing more than a complicated way of expressing the unproven 2nd
postulate.

What you say is irrelevant until you provide experimental
data to refute the conventional view.

I have been doing that George. I have shown that most variable star curves
are
due to c+v effects and not intrinsic variations.

No, you have modelled a temperature variation with
a curve generating program that as easily modelled
the theme from Close Encounters :-) so you have
proved nothing.

I have explained the correct physical approach to
you several times - model the velocity curve first
on the assumption that it is due to velocity and
second assuming it is due to acceleration. For each
predict the radius curve and compare against the
known measurements. Only one will work if your
program is sufficiently restricted and that will
tell you which assumption is valid. Only then do
you predict the luminosity curve and try to
correlate it with the observed curve, the residual
being the intrinsic thermal effect.

George, with due respects, you haven't a clue.


My tilted arrow theory explains Sagnac.

Rubbish, it does not even _attempt_ to address the
problem of time of arrival.

The concept is established.

Garbage, the photons arrive at the same time
regardless of what you imagine happens to
their intrinsic properties.

No George, it becomes quite complicated, he end result being that the travel
times of the two rays are different..

George




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On Wed, 15 Aug 2007 02:05:25 -0700, George Dishman
wrote:


Henri Wilson wrote:

On Sun, 12 Aug 2007 11:01:10 +0100, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in message ...

You are falling into a common trap, GR is not a
collection of separate effects like your bodged
theory, it is a single set of equations which
on their own give the accurate answer.

..not that I accept the actual error
has ever been measured accurately.

One part in 10,000 for GPS as I said.

never tested.

Tested continuously every day, the clocks are
kept within nanoseconds of the right time and
no drift of 38us per day is seen as should be
the case in ballistic theory.

Nobody really cares...

Every user of a sat nav cares, you just want to
bury your head as usual.

Nobody cares.

You may not, real users care.

The clocks are all mutually synched continuously.

That's right, and they only do that because real
users care about the accuracy. That process
involves adjustments at a level about four orders
of magnitude below the GR correction so confirms
it to around 0.01%. Burry your head all you like,
GR is tested that way and its prediction is
confirmed.

Sorry, GR is not involved in any of this.

George



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On Wed, 15 Aug 2007 02:30:18 -0700, George Dishman
wrote:


Henri Wilson wrote:
On Sun, 12 Aug 2007 11:04:13 +0100, "George Dishman" wrote:
"Henri Wilson" HW@.... wrote in message ...
On Sat, 11 Aug 2007 11:39:14 +0100, "George Dishman" wrote:


Does that mean the maths is correct for the right or
wrong reason?

It means the maths is right, period.

The maths might be right.

Yes Henry, the maths IS right, not "at least 20% out" as
you claimed above.

After all its predictions are exactly the same as the
BaTh in most cases.

Nope, most are completely different and ballistic
theory gets them wrong. Sagnac, Shapiro, the GPS
accuracy, pulsars, contact binaries and so on all
get incorrect predictions using ballistic theory and
your excuses that 'something else is not being taken
into account' doesn't solve that problem. Using SR or
GR gives correct answers which is all anyone cares
about.

But why make a simple process appear incredibly difficult?

Simple answer: it's the only maths that gives the right
answers. It isn't that difficult in theory, the ultimate
equation is just one line:

http://math.ucr.edu/home/baez/einstein/node1.html

How simple is that? It is the solutions of that equation
that are the hard part, but only because humans haven't
evolved with maths co-processors in our brains.

The reason it is used is because it covers a topic that
ballistic theory doesn't address, GR is primarily a model
for gravity and it is far better than Newton's.

Oh crap. It doesn't tell us anything about gravity.
It merely describes how space would have to be distorted IF light speed was
constant in a gravity well.

The only
other mathematical alternatives not already falsified are
limited in scope to regions where they are indistinguishable
from GR, and in any conditions where a testable difference
occurs, GR has been shown to be the one that is correct.
Look up MOND if you are interested. Both SR and
Newtonian gravity can be derived as asymptotic limits
of GR and SR merged with QM gives QED which predicts
everything about light and its interaction with matter with
incredible accuracy.

Bottom line? GR and QED work, the alternatives don't.

The true alternatives haven't been correctly applied.

George



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On Wed, 15 Aug 2007 21:06:06 +0200, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Tue, 14 Aug 2007 15:35:59 +0200, "Paul B. Andersen"

IF YOU CLAIM THE CALCULATION TO BE INCONSISTENT WITH THE POSTULATES
OF SR - POINT OUT _EXACTLY_ WHERE THE INCONSISTENCY IS.

Try again, please.

Below is a description of the standard Sagnac diagram,
and a calculation of what SR predicts for the Sagnac experiment.

- Given an inertial frame which is the reference
for all speeds mentioned below.
That is, all speeds are relative to this frame.
- Given a stationary circle with radius r.
- Given a light source moving at the speed v around the circle.
- Assume the light is moving around the circle (infinite number of mirrors).
- According to SR, the speed of the light is c.
- Let tf be the time the light emittet in the forward direction
uses to catch up with the source.
- Let tb be the time the light emittet in the backward direction
uses to meet the source.

So we have:
2*pi*r + tf*v = tf*c
tf = 2*pi*r/(c-v)

2*pi*r - tb*v = tb*c
tb = 2*pi*r/(c+v)

You have just accepted and used the velocity terms c+v and c-v that I have been
talking about all along.

Don't they teach you to read in Norway?

Are you seriously claiming that any of these equations
are inconsistent with the postulates of SR just because
the terms (c-v) and (c+v) appear in them?

They obviously don't teach people to read OR THINK in Norway.
WHERE DO THE C+V AND C-V TERMS COME FROM PAUL?

You have to do better than this, Henri.

delta_t = tf - tb = 4*pi*r*v/(c^2 - v^2)

Setting w = v/r, A = pi*r^2, g = (1 - v^2/c^2)^-0.5
we get:

delta_t = (4Aw/c^2)* g^2

The g^2 will obviously be unmeasureable different from 1
for any practical Sagnac experiment.

So SR predicts delta_t = 4Aw/c^2 which is in accordance
with enumerable practical experiments.

You are effectively applying the circularly derived SR speed addition formula.

Assume w always equals c, where w is photon speed in all frames. (second
postulate)

So:
w = c
= c(c+v)/(c+v)
= (c+v)/(1 + cv/c^2)......Einstein's formula for u = c.
= c
= w
Fantastic!!! I have just proved the second postulate using nothing but circular
maths....!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!

Hahahahaohohohoh!

Quite. Hilarious.

Please stop the mindless babble, and address the issue:

IF YOU CLAIM THE CALCULATION TO BE INCONSISTENT WITH THE POSTULATES
OF SR - POINT OUT _EXACTLY_ WHERE THE INCONSISTENCY IS.

Circularity implies consistency, Paul.

Stupidities like:
"(c+/-v) appear in the equations, that is inconsistent with SR"
won't do.

So try again, please.

Below is a description of the standard Sagnac diagram,
and a calculation of what SR predicts for the Sagnac experiment.

- Given an inertial frame which is the reference
for all speeds mentioned below.
That is, all speeds are relative to this frame.
- Given a stationary circle with radius r.
- Given a light source moving at the speed v around the circle.
- Assume the light is moving around the circle (infinite number of mirrors).
- According to SR, the speed of the light is c.
- Let tf be the time the light emittet in the forward direction
uses to catch up with the source.
- Let tb be the time the light emittet in the backward direction
uses to meet the source.

None of that has any physical significancce.

So we have:
2*pi*r + tf*v = tf*c
tf = 2*pi*r/(c-v)

2*pi*r - tb*v = tb*c
tb = 2*pi*r/(c+v)

delta_t = tf - tb = 4*pi*r*v/(c^2 - v^2)

Setting w = v/r, A = pi*r^2, g = (1 - v^2/c^2)^-0.5
we get:

delta_t = (4Aw/c^2)* g^2

The g^2 will obviously be unmeasurable different from 1
for any practical Sagnac experiment.

So SR predicts delta_t = 4Aw/c^2 which is in accordance
with enumerable practical experiments.

It isn't SR. It uses a theory based on rotation being absolute...which it is.
My 'photon arrow' theory explains the Sagnac effect..

Paul



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On Wed, 15 Aug 2007 05:55:29 -0700, George Dishman
wrote:


Henri Wilson wrote:

On Tue, 14 Aug 2007 16:28:28 +0200, "Paul B. Andersen" wrote:
Henri Wilson wrote:
On Mon, 13 Aug 2007 20:24:19 +0200, "Paul B. Andersen" wrote:

So we have:
2*pi*r + tf*v = tf*c
tf = 2*pi*r/(c-v)

2*pi*r - tb*v = tb*c
tb = 2*pi*r/(c+v)

I repeat, what do (c+v and (c-v) represent Paul?

They are a sum and a difference between two speeds, obviously!

CORRECT!!!!!!!

Yes, Paul is correct.

...the speeds of the rays relative to the source....

No, you are wrong. The values (c+v) and (c-v) are the sum
and differnce in the lab frame, not the speed in the source
frame or "relative to the source" as you put it.

Good, you finally understand how light speed can be different from c.

This is called an equation, Henri:
2*pi*r + tf*v = tf*c
the v is the speed of the source, and the c is
the speed of the light. Nothing is moving at
any other speed than c and v.

Then why the hell do you use c+v and c-v instead of c?

Algebra. The distance between two object changes
at a rate that depends on the sum or difference of
their speeds, not just the speed of either.

George, you are going in circles...like the light...

Repeat:
THE LIGHT IS MOVING WITH THE SPEED c IN THE INERTIAL FRAME!

Repeat, Then why the hell do you use c+v and c-v instead of c?

Because both motins change the distance between them.

Good you finally seem to be learning something.
Now can you see why the BaTh predicts variable star curves by using exactly the
principle you just mentioned?

The _solution_ of the equation is:
tf = 2*pi*r/(c-v)

How the hell can you claim that the speed of light
in the inertial frame is different from c just because
the difference (c-v) appear in the solution of the equation
WHICH IS BASED ON THE ASSUMPTION THAT THE SPEED OF LIGHT IS
c IN THE INERTIAL FRAME?

How stupid can you get?

Yes you are.
You claim to be using c when you plainly use c+/-v in your equations.

He didn't claim to be using c in the equation, he
correctly pointed out that the speed of the light is c.
Only you made the mistake of confusing the algebraic
sum and difference in the lab frame with the speed in
the source frame.

Whether or not the OW light speed of both rays is c IN THE SOURCE FRAME, as
Einstein's silly second postulate says, is of no importance.

If the speed of light had been (c+/-v), as predicted
by the BaTh, the equations would have been:

2*pi*r + tf*v = tf*(c+v)
tf = 2*pi*r/c

2*pi*r - tb*v = tb*(c-v)
tb = 2*pi*r/c

So the BaTh predicts: delta_t = tf - tb = 0

Enumerable practical experiment show that delta_t = 4Aw/c^2

The BaTh is falsified.

You don't stand a hope in hell of proving the BaTh wrong.

He just did, the theory predicts delta_t = 0 when it is
measured to be 4Aw/c^2 as predicted by SR.

That analysis has nought to do with SR.

If the equation is correct then my theory will naturally arrive at the same
answer.

George



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On Wed, 15 Aug 2007 21:45:38 +0200, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Tue, 14 Aug 2007 16:28:28 +0200, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Mon, 13 Aug 2007 20:24:19 +0200, "Paul B. Andersen"
wrote:


So we have:
2*pi*r + tf*v = tf*c
tf = 2*pi*r/(c-v)

2*pi*r - tb*v = tb*c
tb = 2*pi*r/(c+v)
I repeat, what do (c+v and (c-v) represent Paul?
They are a sum and a difference between two speeds, obviously!

CORRECT!!!!!!!
...the speeds of the rays relative to the source....

This is called an equation, Henri:
2*pi*r + tf*v = tf*c
the v is the speed of the source, and the c is
the speed of the light. Nothing is moving at
any other speed than c and v.

Then why the hell do you use c+v and c-v instead of c?

Repeat:
THE LIGHT IS MOVING WITH THE SPEED c IN THE INERTIAL FRAME!

Repeat, Then why the hell do you use c+v and c-v instead of c?

The _solution_ of the equation is:
tf = 2*pi*r/(c-v)

How the hell can you claim that the speed of light
in the inertial frame is different from c just because
the difference (c-v) appear in the solution of the equation
WHICH IS BASED ON THE ASSUMPTION THAT THE SPEED OF LIGHT IS
c IN THE INERTIAL FRAME?

How stupid can you get?

Yes you are.
You claim to be using c when you plainly use c+/-v in your equations.

I think George has answered this well enough.

But I must say that I find your ranting incredible stupid.
The equations where the speeds appear a
2*pi*r + tf*v = tf*c
2*pi*r - tb*v = tb*c
I used c for the speed of light and v for the speed
of the source. Nowhere did I "use c+v and c-v instead of c".

It is the BaTh that "use c+v and c-v instead of c".

correctly...

And you know what that lead to:
If the speed of light had been (c+/-v), as predicted
by the BaTh, the equations would have been:

2*pi*r + tf*v = tf*(c+v)
tf = 2*pi*r/c

2*pi*r - tb*v = tb*(c-v)
tb = 2*pi*r/c

THERE THEY ARE AGAIN...C+V AND C-V


So the BaTh predicts: delta_t = tf - tb = 0

Enumerable practical experiment show that delta_t = 4Aw/c^2

The BaTh is falsified.

You don't stand a hope in hell of proving the BaTh wrong.

You sure are right about that, but I don't have to.
All I did was to calculate what the BaTh predicts for the Sagnac.
Enumerable practical experiments already done, falsify the BaTh.
The very fact that the inertial navigation system in thousands
of aeroplanes work as I write this, falsifies the BaTh right now.

Oh rubbish. The BaTh is fully supported by all known experiments...and guess
what, none of your claims is related to relativity...

And now.

And now.

....

Paul



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On Fri, 17 Aug 2007 14:39:48 +0100, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
.. .
...
The number of cycles counted divided by the total number in a chip is
actually
a measure of phase.

Right, but since the C/A code repeats rapidly,
the phase alone is useless for position, it
tells you where you are within a small square
like a chessboard, but which square. The time
difference from the clocks is the primary
measure hence the accuracy of the location
confirms the time it took the signal to cover
a known distance, i.e. it is a direct measure
of the speed.

....and so, as long as the clocks are all mutually in synch, the system will
work properly, the ground clock is irrelevant.
GR has nothing to do with the operation of GPS.

George




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On Fri, 17 Aug 2007 14:36:57 +0100, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
.. .


Would you please explain the logic behind that conclusion, Henri? :-)


From the fact that the receiver determines precisely "the actual clock
reading"
when the signal was sent, you conclude the 'GR correction' is totally
unnecessary,
and the explanation for that conclusion is:

Hahaha! That's not how it works.
The phasing of signals from three or four clocks is compared. The actual
times
are only important for the purpose of synching them all with each other.

Wrong Henry, the time stamps in the signals are subtracted
from a current reference time on a local clock to get the
apparent distance travelled. Those times are used to solve
four equations for four unknowns, the actual time and the
'pseudo-ranges' which is why four satellites are needed.
The carrier cycles merely improve the resolution of the
time measurement.

The local clock time cancels out. GR is totally unnecessary.



Still - after all these years - amazing.

It is of no real consequence anyway. A small 'free fall' correction is
built
into the clocks before launch. After that the clock are regularly synched
in
orbit.

Still repeating that same crap George?




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On Fri, 17 Aug 2007 23:07:42 +0200, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
On Wed, 15 Aug 2007 22:25:49 +0200, "Paul B. Andersen"

Would you please explain the logic behind that conclusion, Henri? :-)

From the fact that the receiver determines precisely "the actual clock reading"
when the signal was sent, you conclude the 'GR correction' is totally unnecessary,
and the explanation for that conclusion is:

Hahaha! That's not how it works.
The phasing of signals from three or four clocks is compared. The actual times
are only important for the purpose of synching them all with each other.

So you are confirming my suspicion that you didn't understand a single
word of what George and I told you. You _still_ believe that the phase of
the signal from different satellites are compared!
Amazing! Unbelievable!

'time difference', phase difference'...is there a difference in practice?

I can only repeat:
I find it almost unbelievable that it is possible to discuss the GPS
for years, and still be completely ignorant of how it works.

How the hell do you manage to stay _this_ ignorant, Henri?

Plain and simple stupidity?

Why the hell do I waste my time like this?

Plain and simple stupidity?

Paul, simply stupid

GR is totally unnecessary for the operation of the GPS system.


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On Fri, 17 Aug 2007 14:30:13 +0100, "George Dishman"
wrote:


"Henri Wilson" HW@.... wrote in message
.. .
On Mon, 13 Aug 2007 22:08:18 +0200, "Paul B. Andersen"

George agreed with me.

I said the carrier is used to improve the
resolution of the time difference measurements
which provide the pseudo-ranges. It's like using
the second hand to improve a measurement of time
on a clock beyond that given by the hour and
minute hands. Paul's description is fine, and in
a way your is too, the number of whole cycles of
the carrier is a measure of the time as a fraction
of the chip cycle time.

So to determine the exact time when the signal
received at an instant was sent, the receiver is
counting chips in the PRN-signal.

....or part of a chip, ie., phase difference.

Right, phase difference is a finer-grained measure
of time.

For the C/A channel, this gives a resolution of 0.98 us.

Assuming the orbit is known precisely.

Which it is by continuous monitoring.

Since all frequencies in the signal are derived from
the same frequency standard, it will be a specific number of carrier
cycles in each chip (1540 cycles of the L1 carrier in C/A PRN chip).
So to increase the time resolution, it is possible to count
the carrier cycles within a chip.
Since a carrier cycle is less than 0.1 ns, I think it
is more "counting the cycles" than it is actually
measuring the phase within a cycle.
(But counting cycles is sort of measuring the phase.)

Which all goes to prove that the 'GR correction' is totally unnecessary...

Nonsense. If all the clocks were synchronised but
they ran without the GR correction, the location
of a receiver would be determined just as
accurately but the use of GPS as a time reference
would be hopeless, it would drift by 38us a day.

George, there is already a huge discrepancy between GPS and UTC times. As long
as all the clock differences are known accurately they can easily be software
corrected out. That is what hapens.

George




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