Re: Sagnac Fully Supports the BaTh.

On Sep 30, 12:57 am, HW@....(Dr. Henri Wilson) wrote:
On Sat, 29 Sep 2007 21:24:59 -0700, Jerry <Cephalobus_alie...@xxxxxxxxxxx>
wrote:





On Sep 29, 10:46 pm, HW@....(Dr. Henri Wilson) wrote:
On Sat, 29 Sep 2007 19:59:25 -0700, Jerry <Cephalobus_alie...@xxxxxxxxxxx>
wrote:

On Sep 29, 5:19 pm, HW@....(Dr. Henri Wilson) wrote:
In case you missed it:http://www.users.bigpond.com/hewn/sagnac.jpg

hahahahaohohohohooho!
Henri Wilson. ASTC,BSc,DSc(T)

Do you realize how many entirely different and totally
incompatible BaTh "explanations" of Sagnac you've come up
with? As soon as you realize that one explanation is
unworkable, you dream up another equally stupid explanation.

Ah but the amazing feature of this one is that it produces
the right answer.

Only because you pulled 8vR/c for the path length difference
out of "thin air". You have NO DERIVATION for this expression.
You copied this from the conventional SR analysis and assert
without proof that your analysis reaches the same conclusions.

Can't you read ?
The full derivation is given in the illustration.
I knew it would be too hard for someone like you.

* Your attempted explanation that the fringe shifts were
due to the beams being laterally displaced was complete
nonsense.
* Do you remember your explanation that the Sagnac effect
was the result of centrifugal force acting on the beams?
* Whatever happened to your twisting photon axes?
* How about pressures above the WDT and "fields" forcing
the photons to travel at c relative to the observer in
the stationary frame?

...all part of the investigative process.
Relating Sagnac with the BaTh has always been a problem.
I have found the answer.

No more an answer than any of your previous attempts...

It is a simple, concise and complete answer. I know this comes as a shock to
you. You have been making a complete fool of yourself for years. Admit it
Crank..

Your latest explanation is just as absurd as your previous
ones.

So absurd it models the physics correctly and produces the
right results. :)

SR produces the right results, and the mathematical
demonstration that it does so is trivial.

You assert by fiat that your explanation produces the same
formula as SR, and the mathematical demonstration that it
does so is non-existent.

When you learnt to read and think you will be able to see for yourself why I'm
obviously right.
It's terribly simple really. All you have to do is work out the different path
lengths in both rays and divide by the length of a 'wavelength'.





You state that "during acceleration, the 'wavecrests'
that are entering the detector are traveling at a speed that
is not the current one."

that's correct.

During a speed CHANGE the photons arriving at the detector
are traveling at their orignal c+v.root2 but the apparatus
is now moving at a different 'v'.
...so a few wavecrests move in or out of each ray. The
number of wavelengths in each overall beam changes. This
is surely not too hard for you to understand.

Dear Quacker Henri,

The light has NO MEMORY. Turn the light source off.
Turn the light source on again.

In this scenario, the light will effectively have ALWAYS
been traveling at c+/-v, and the apparatus will effectively
have ALWAYS been turning at v. There will have been no
acceleration, yet the fringes will still be displaced.

Has the sudden realisation that sagnac fully supports BaTh
sapped your brain of whatever energy it had?
The light only moves at c towards the first mirror when
the apparatus is not rotating. Otherwise it moves at
c+v.root2 in the nonrotating frame (plus a very
small second order term). It retains that speed throughout
the whole trip. The length of all legs of either ray is
the same...but as you can see from my exaggerated diagram,
the blue path is longer than the green one. Path length is
proportional to rotation speed.

If acceleration is occuring, the light leaving the source
will have a different speed from that which arrives after
traveling around the loop....small but
exactly what is expected and measured.

....and if no acceleration is occurring?

What happens when the turntable is no longer accelerating
and reaches steady state, Quacker Wilson? When the turntable
reaches steady state constant rotation, all the wavecrests
will be traveling at c +/- v, and the fringe shifts will
be zero again!

That's what one would expect, you dope. That's what happens.
Fringes don't move during a period of CONSTANT angular
velocity. Both path lengths change during - and only during -
an acceleration and that's when the number of wavelengths
in each beam changes. In other words, fringe movement occurs.
A simple gyro counts the number moved and converts that into
angle.

How about Michelson-Gale? The Earth was ALWAYS rotating
at constant angular velocity throughout the course of the
experiment.

Naturally it returned a null result...pure ballistic stuff
also..

MICHELSON-GALE RETURNED A POSITIVE RESULT!!!!
MICHELSON-GALE RETURNED A POSITIVE RESULT!!!!
MICHELSON-GALE RETURNED A POSITIVE RESULT!!!!
MICHELSON-GALE RETURNED A POSITIVE RESULT!!!!
MICHELSON-GALE RETURNED A POSITIVE RESULT!!!!
MICHELSON-GALE RETURNED A POSITIVE RESULT!!!!

http://adsabs.harvard.edu/abs/1925ApJ....61..140M
http://www.mathpages.com/rr/s2-07/2-07.htm

Jerry
Henri Wilson's Lies
(1)Fakes Diploma (2)Uses Deceptive Language (3)Fakes Program
(4)Intentionally Misquotes (5)Snips (6)Accuses Others of Lying
1 http://mysite.verizon.net/cephalobus_alienus/henri/diploma.htm
2 http://mysite.verizon.net/cephalobus_alienus/henri/deception.htm
3 http://mysite.verizon.net/cephalobus_alienus/henri/rt_aurigae.htm
4 http://mysite.verizon.net/cephalobus_alienus/henri/history.htm
5 http://mysite.verizon.net/cephalobus_alienus/henri/snips.htm
6 http://mysite.verizon.net/cephalobus_alienus/henri/accuses.htm

.

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