Re: P10 Acceleration: Light Speed Doesn't Extrapolate

From: Ralph Sansbury (r9ns_at_bestweb.net)
Date: 08/10/04 

Date: 10 Aug 2004 06:08:14 -0700

Jonathan Silverlight <jsilverlight@spam.merseia.fsnet.co.uk.invalid> wrote in message news:<AwIwfZbNIAGBFwNj@merseia.fsnet.co.uk>...
> In message <e988316b.0408090642.18da4440@posting.google.com>, Ralph
> Sansbury <r9ns@bestweb.net> writes
> >
> > What are the formulae for the errors in the lidar and radar
> >reflections? The references and discussion on my web page show they
> >are large. That is some of the moon reflections are from secondary
> >scattering etc and the radar reflections from Venus etc supposedly
> >showing different surface heights for different pixels of the image
> >have larger error bars so that they are indistinguishable from noise.
>
> How about some references to back up that assertion? Last time this came
> up you said Craig Markwardt told you, but we know that was mistaken, to
> be charitable.

   Ask him again. He did say so. And many others say so. Dont get hung
up on exact quotes. The meaning of what he said is exactly equivalent
to what I said. The original articles by Pettingill et al indicate
that this is so.
   The point is that the variations of the radar indicated surface
heights at various pixes are such that you cant say that there is a
real difference. All agree on this. I go further and say that the
standard method has some problems as discussed on the web page.

> Is the noise figure in one of the original 1960s papers, or the more
> recent Magellan or Pioneer results?

   I believe it is indicated. Also I have no quarrel with such radar
reflections from much smaller distances which are certainly less than
a second in light speed delay.

But the point of this thread is as follows:
    
(3)The trajectory of the P10 craft is based on assumptions about its
 velocity and position at earlier times and assumptions about the
 earthsite motions at uplink and downlink. That is the craft is placed
 at successive positions in such a way as to confirm the standard
light
 speed delay model more than any other. The approx. 16Hz average
 difference compared to an approx 2500Hz average difference between
the
 conventional model prediction and the nearly instantaneous model is
 due to placing the craft at positions that make the conventional
 prediction match the observed radar. But this is just a circular
argument obscured by the numerical complexity. If A is true than B is
true and If B is true then A is true. So A is true.
  The problem with the conventional model is that the radar data
does not, based on the archived records, remain consistent with the
model and these assumptions and the lack of fit increases even more
than the anomalous accelerationcan account for. This is shown on the
spread*** at http://www.bestweb.net/~sansbury.
That is, the discrepancy of 16Hz increases in this hour and a half
period to 25Hz etc when the discrepancy should be initially zero and
should remain so except for the 10^-7 fraction of 1Hz associated with
the claimed anomalous acceleration.

.
    How to fix this?
 
    We can change the craft trajectory starting at any point in time
assuming tentatively that the craft is at the same distance,R*, from
the earth and R from the sun(R(21:27)= 2684810625, but at a possibly
different angle to the craft-earthsite line etc so that the predicted
frequencies match the observed frequencies according to the nearly
instantaneous model transmitter earthsite motions.
   And if our initial assumption of R needs to be adjusted to produce
the proper motions of the craft toward the earth(less than the
constant motion of the craft away from the earth) so as to produce a
match between the radar and predicted frequencies, then we can do so
later as shown below.
   To see this, we note first that the observed Doppler count
frequency subtracted from 1MHz, a so called bias, decreases minute by
minute starting at 21:27(row14) by 12Hz per minute but that if we do
not do this subtraction, the observed Doppler count frequency
increases minute by minute by, initially, 12Hz. increasing to
16Hz/minute, 78minutes later which correlates with the successive
increases, 1m/s to= .6m/s increase in the earth motion toward the
craft minute by minute. How to make the correlation perfect?

   Note, (1+.33(10^-8))T = T+7.66Hz corresponds to 1m/s when the
transmitter frequency is at is here T=2.291944138GHz . If the
direction to the craft was the same as the direction of the 1m/s
velocity of the earth, then an uplink and same downlink velocity total
of 2m/s would produce a shift of twice 7.66=15.32 Hz in the radar
   The observed increase in frequency of 12Hz from 21:27 to 21:28 is
associated with a velocity difference of 12/7.66=1.5666 m/s. This
could be caused by a motion of the earth toward the craft of half this
on the uplink,.7833m/s and .7833 m/s again on the downlink during the
same minute according to the nearly instantaneous model.
  But we must allow for the increase of the total uplink and downlink
velocity caused by the slowing down of the craft due to the
gravitational effect of the sun,( a = kM/R^2, (a)(t2-t1)/2 =.1m/s for
the assumed R. That is, .1m/s must be subtracted from the uplink and
from the downlink to find out how much of the actual difference in
velocity must be projected onto the earthsite-craft line. Thus .6833
m/s is the difference we require and the actual ephemeris difference
is 1m/s so we have to multiply .6833 times the ephemeris velocity
value for 21:27 and the value for 10:28. This means projecting the
earthsite velocity line through an angle, arccos(.6833)=46.91deg thus
determining the new craft position.
    (If the craft was at only .46R of the assumed distance from the
sun, the velocity to the earth due to this implied acceleration would
be 2 times .2285m/s=.457m/s. so .7833-.2285= .5548 is the the
difference we require and so we have to multiply .5548 times the
ephemeris value of 1m/s. This means projecting the earthsite velocity
lines through and angle arccos(.5548)=56.30deg )

   An hour and 15 minutes later, from 22:43 to 22:44 the earthsite
velocity increases 30.09135 to 30.09194 or .59m/s or 4.52Hz to which
we must add 3.48Hz to get half of 16Hz radar observed. If the craft
acceleration and so velocity to the earth was increased to 3.498Hz or
.457m/s by reducing the assumed distance R to .46R and the
craft-earthsite angle to the earthsite velocity was reduced to zero
then we would have a perfect match.
   Of course we should look at later received radar frequencies to see
if the pattern of increasing frequency stops and a pattern of
decreasing frequency begins.
   Also we now have to increase the initial angle as was shown above
to be 57.55deg and over the 78 minutes from 21:27 to 22:45, the angle
decreases continuously by .738 degrees per minute.

Thus the instantaneous model leads to a more accurate trajectory than
the conventional model without the need for an anomalous acceleration-
at least for this 78 minutes.

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