Re: Bureaucratic refusal to admit mistakes re light speed


<r9ns@xxxxxxxxxxx> wrote in message
news:1163370571.124871.35020@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx


Why does NASA refuse formally to say that the assumption that light
speed extrapolates indefintely is disconfirmed by the Anderson
anomalous acceleration- ...

Because the Pioneer data confirms that the signals
to the craft take many hours to travel the distance,
the anomaly is a fraction of a second.

You know Ralph, it is really annoying to have to
correct so many trivial errors. You really should
know better given the amount of time we spent on
this last year.

Another argument that light speed is not nearly instantaneous was
given by George Dishman an electrical engineer familiar with radar
circuits etc. He spoke about radar data ...

Not radar data, we discussed the actual telemetry
carrier measurements also used by Anderson and
Markwardt. Ralph is aware of this, he has the same
files himself!

.. showing Doppler shifts of a
specific carrier frequency of signals sent to the Pioneer 10 Spacecraft
in 1988 and received many hours later supposedly at a space site
different from the earthsite where the signals originated. (The radar
data consisted of radar carrier frequencies sent to the spacecraft and
changed to a slightly different frequency characteristic of the
spacecraft in such a way that the phase remained the same and then sent
back, or in the absence of signals sent from earth just the
characteristic spacecraft frequency. Thus shifts over time showed the
relative earthsite-spacecraft motion.)
He pointed out that if it took only a few seconds for radiation to
make a round trip to and from a craft billions of miles away and back,
then at one site there would be Doppler shifts of the incoming
frequency as the earth rotated toward the receding craft but these
shifts would decrease and be followed by minimal Doppler shifts as the
earth spun perpendicular to the the craft-earthsite line and then
increasing Doppler shifts as the earth spun away from the craft.

Ralph uses "increasing" in a manner different from
everyone else. What I said was that the frequency
would decrease throughout the period of contact,
but Ralph calls that "increasing" when the diurnal
component of the Dopppler shift is negative. This
is one diagram I drew to explain it to him:

http://www.briar.demon.co.uk/Ralph/Generic.gif

.. He
then showed that the minimum for a similar pattern at another earth
site 12 hours away from the first earthsite occurred 13 or so hours
later not 12 hours later as it should have, he believed, if the round
trip time of the signal in each case was less than a minute.

The words "the minimum" refer to the first derivative,
the lower graph in the above diagram. The times were
appropriate for the two DSN sites that produced the
data used, Canberra and Madrid. Since Ralph believes
the received signal was sent from the receiving site,
at the time when the diurnal component is zero, the
craft must be transiting. This diagram summarised the
the results and shows why Ralph's theory produces an
inconsistent value for the right ascension of the craft:

http://www.briar.demon.co.uk/Ralph/Discrepancy.gif

The basis for his belief was that the 29.8km/s orbital motion of the
earth could be ignored since the direction of this motion changed only
1 degree per 24 hours. But obviously this did not take into account
the much larger magnitude of the earth's orbital motion vs its spin
motion, 29.8 vs .465 km/s. and the effect this could have during a 1
degree or .5 degree change in direction.

Again, Ralph has this wrong, the direction of motion
is of no relevance, it is change of Doppler shift
during the time between the measurements that matters.
The amplitude of the annual effect is 30km/s so in
a time of around 12 hours, the maximum change is
about 260 m/s. The diurnal change is of the order of
900 m/s. However, that is not the whole story.

When I pointed this out to
him, he refused to acknowledge his mistake and said that it was
necessary to compare second differences of Doppler shifts which was
nonsense, a case of lying with numbers.

From the very beginning, I pointed out that the method
was to find the minimum of the first derivative as can
be seen from the first diagram above

http://www.briar.demon.co.uk/Ralph/Generic.gif

The reason for this may not have been clear to Ralph
but is because the data available are usually only a
short segment of the falling slope due to the limited
time above the horizon and DSN scheduling. Taking the
first derivative gives a clear minimum regardless of
the relationship of the contact period to the diurnal
phase.

As is standard practice of course, the minimum is
found by taking the derivative and equating to zero
so the minimum of the first derivative was to be found
by taking the second and equating to zero. This is
schoolboy calculus.

Consider the generic equation:

f = A * sin(B*t) + C * sin(D*t)

The ratio of the amplitudes of the component frequencies
is A/C

The derivative is

f' = (B*A) * cos(B*t) + (D*C) * cos(D*t)

and the second derivative is

f" = -(B^2*A) * sin(B*t) -(D^2*C) * sin(D*t)

The ratio of the amplitudes of the components is

(B^2*A) / (D^2*C)

or

(B/D)^2 * (A/C)

What Ralph failed to understand is that each time
we differentiate the sum of two sine waves, it
multiplies the ratio of the amplitudes by the ratio
of the frequencies. Since the annual component has a
frequency lower than the diurnal by a factor of 365.25,
the improvement in the ratio of 260 / 900 is (365.25)^2
or 133408. The orbital motion thus introduces a maximum
error of (260 / 900) / 133408 or about 2 parts per
million. Despite several weeks of patient explanaton,
Ralph never understood this.

To help him see his obvious mistake, I described an example where at
a specific time, at a specific earthsite, the earth's orbital motion
and spin motion were both perpendicular to the line from this earthsite
to the spacecraft.

Unfortunately he got most of the numbers wrong. After
some time we mostly corrected that. See this message
and the 20 or so that follow in the thread if you are
a glutton for tedium:

http://groups.google.com/group/sci.astro/msg/3edd25c6166a0b0e?

In this example, the total earthsite motion at one site was
constant motion away from the craft from rise to set and at another
site 12 hours earlier a reversal of total motion between rise and set,
toward and away from the craft.

This merely emphasises Ralph's lack of understanding, the
method finds the time at which the craft is due south
independently for each site. In all cases therefore, the
site is moving towards the craft prior to that time and
away from it after the time.

And though the spin motion of the
earth was moving perpendicular to the earthsite to craft line at one
site 12 hours after the spin motion was moving perpendicular to the
earthsite to craft line at the other site, this did not show up in the
total motion as a nearly zero motion at each site 12 hours apart.

The earthsite motions in the example were for the first site
(445,0,-445.2) but for the 2nd site (192.5,-252,-697) in meters per
second at 6 hour intervals.

Quoting from
http://groups.google.com/group/sci.astro/msg/02fadd980ad13f84

George Dishman wrote:
... The values
I am using are mostly those we have discussed rather
than being perfectly accurate but will serve for this
purpose.

Declination 25.1 degrees
Ecliptic latitude 3.1 degrees
Equatorial speed 465.0 m/s
Site 1 latitude 40.4 degrees (DSN 63, Madrid)
Site 2 latitude -35.4 degrees (DSN 43, Canberra)
Orbital speed 29000.0 m/s

My numbers take the effect of the craft being out of
the planes into account. The set for the first site are
(+445.2105, 0.0000, -445.2105) and the second set from
the same site 12 hours later would be
(+192.5089, -252.6992, -697.9026)
...
Now both the above are for Madrid but we are comparing
one contact at Madrid with a second some hours later
at Canberra. The numbers for that contact are
(+215.0748, -252.6992, -720.4686)

Quoting from

http://groups.google.com/group/sci.astro/msg/aa8226eda9e7c067

r9ns@xxxxxxxxxxx wrote:

George Dishman wrote:
<r9ns@xxxxxxxxxxx> wrote in message
news:1129748326.018826.149590@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
....
Lets use your numbers(+445.2105, 0.0000, -445.2105)
(+192.5089,-252.6992 -697.9026)

Actually those are both Madrid,

Then you misrepresented what they were. ...


The numbers Ralph just quoted are for the _same_ site,
something I pointed out at the time and which he even
noted in his response at the time (while accusing me of
'misrepresenting' the numbers). Here we are a year later
and he still can't get it right.

Since Ralph seems to want to reopen this topic, I will
repeat the relevant tables and the conclusion here which
proves that the method does indeed reduce the error
resulting from the orbital motion to a negligible level
as the schoolboy maths predicted. Bear in mind that
Ralph's theory fails this test by about 100 minutes so an
error of less than 0.1 seconds is completely irrelevant:

George


On the 23rd October 2005, George Dishman wrote in message
http://groups.google.com/group/sci.astro/msg/3edd25c6166a0b0e

....
Here are the first site totals again for convenience with
some letters on the lines for ease of reference:

Time Total
a -360 449.199057
b -240 363.826463
c -120 203.539221
d 0 0.000000
e 120 -203.539221
f 240 -363.826463
g 360 -449.199057

To form the first difference column we approximate the rate four
hours before the transit as the difference between the six hour
and two hour values divided by 2, that is the difference column
for line b is half of line c minus line a:

(203.539221 - 449.199057) / 2 = -122.829918

That means the speed is falling at 122.8m/s per hour at that time.

Time Total First Diff
a -360 449.199057
b -240 363.826463 -122.829918
c -120 203.539221 -181.913232
d 0 0.000000 -203.539221
e 120 -203.539221 -181.913232
f 240 -363.826463 -122.829918
g 360 -449.199057


Now repeat that to get the second difference:

Time Total First Diff Second Diff
a -360 449.199057
b -240 363.826463 -122.829918
c -120 203.539221 -181.913232 -40.354652
d 0 0.000000 -203.539221 0.000000
e 120 -203.539221 -181.913232 40.354652
f 240 -363.826463 -122.829918
g 360 -449.199057


The time I identify is when the second difference is zero so in
this case the determination is perfect. That's actually unusual
and occurs only because Ralph happened to choose the time when
the orbital speed is zero for the example. However, it isn't a
single value that matters, we are looking for the difference
between the sites so what happens at the second site? Applying
the same method gives the following:

Time Total First Diff Second Diff
360 218.855246
480 130.484395 -128.424709
600 -37.994171 -191.603550 -43.151780
720 -252.722706 -214.728268 0.000535
840 -467.450707 -191.602481 43.152849
960 -635.927669 -128.422570
1080 -724.295848

Now we can see there is an error of 0.000535 in the final
column. The value changes from -43.151780 one hour before
transit to +43.152849 one hour after transit so that is a rate
of -83.304629 in two hours or -0.359603 per minute. If we use
a linear approximation (least squares) then the time when the
second difference is zero will be offset from the true transit
time by:

error = 0.000535 / -0.359603

= 0.001487 minutes

= 0.0892 seconds

So there you have it Ralph, your example shows the orbital
speed introduces an error of less than a tenth of a second
in the determination of the transit time.



.

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