Re: c = constant is still under discussion in this group

Dear Rudolf Drabek:

"Rudolf Drabek" <newsrudy@xxxxxx> wrote in message
news:1151593222.112796.211450@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

N:dlzc D:aol T:com (dlzc) schrieb:
....
But you don't seem to want to retain this,
so let's see what kind of apparatus you come
up with...

I thought the description is clear.
1. Two antennas at a cerrtain distance

Distance = Remote synchronized clock.

If I understand you right you say : one clock
generator connected to both antennas is the
same as 2 synchronized clocks at the antennas.

With a synchronization procedure that places "both antennas
equidistant" and "separated by about 30 km", yes.

O.k. but with my setup I have synch. with no
drift between the clocks (temperature
differences of the fibres excluded and I have not
calculated the influence).

You cannot compensate for the LET aether, just so you realize...

Originally I started the setup with 2 clocks
but later for simplicity I changed to one clock
in the middle.

You could just as easily place the clock at one antenna, and
synchronize the remote one. This is very similar to the
operation of the Very Long Baseline Array, where antenna
"recordings" are matched based on recorded clock pulses. And
images of very remote objects are evoked from these recordings.

great enough to have a resolution a
magnitude greater than the wanted
detectable v, chosen to be +3 km/sec.
So if the distance is 9 km light needs 30 us.
This distance need not to be very accurate,
because it's calibrated out, using a non
moving signal source on Earth, with a
resolution of 1ppm is 9mm according.
If c+v would be possible, then it would
need 9/300003 is 30 us - 0.3ns = 9cm
deviation

Finding an anisotropy is one thing. Going from
the anisotropy to "c+/-v" requires knowing the
distance. Distance is TWLS.

I think here I did a trick: At the starting post there
are the formulas.
Both signals are using the same distance. It is
only necessary that for both signals (spacecraft
and calibration) the distance is the same and
not an exact defined distance.
For clarification again:

spacecraft | measuring distance |
or cal. signal antenna1 +9km
antenna2

*------->--------------------------|---------------------------|
if cal. signal te = s/c
=
30us
if spacecraft with some v tn = s/(c+-v)

if c+-v would be possible then tn = te(1 +-v/c) for
v<<c
What we wait for is that tn is always te.

And it was when you synched the two antenna arrays, and in both
SR and LET, they stay synched.

Even with an unaccurate clock I can reach
sufficient accuracy. The clock signal is always
the same at both locations.

Is there still something I have overseen?

You are expecting logic to influence those that do not respect
it, because they don't choose to believe the consequences of the
data.

How does light know how to travel at a certain
speed in a medium... like the air envelope
around the surface of the Earth? Do you think
that will affect what you will measure?

To cancel this effect I can use the spacecraft
itself for calibration at certain time of the year,
when Doppler is Zero. Then the
athmossphere should have no effect.

Density (and hence c_medium) is a function of temperature and
(barometric) pressure. The ballistic and dragged-aether guys
have already used this excuse. You'll have to repeat on the Moon
or some other airless body to satisfy this dodge on their part.

David A. Smith


.

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