Re: Pioneer 10 test of light speed delay
From: George Dishman (george_at_briar.demon.co.uk)
Date: 11/30/04
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Date: Tue, 30 Nov 2004 22:45:35 -0000
"r9ns" <r9ns@verizon.net> wrote in message
news:30e10dee.0411281253.395abd1d@posting.google.com...
> "George Dishman" <george@briar.demon.co.uk> wrote in message
> news:<coacum$lco$1@news.freedom2surf.net>...
>> "r9ns" <r9ns@verizon.net> wrote in message
>> news:30e10dee.0411220713.4ff400f6@posting.google.com...
>
> George, let me address your last comment first and maybe it will help
> to show my claim that the directional change in the frequency
> differences occurring 26 degrees more than the time between the
> Canberra and Madrid meridians does not indicate that the received
> frequencies could not have been transmitted a few seconds before from
> the same site.
Your comments don't address my method at all so don't
shed any light on why you think it incorrect. However
I'll reply to yours as you have cleared up a problem.
I'll take your last point first again as it is relevant
later.
>> >> > The maximal angle between lines from Pluto at 40AU to the earth
>> >> > and
>> >> > to the sun is 1.4 deg not 11 deg.
>> >>
>> >
>> > But that was not the question
>>
>> There was no question, I just commented on what you
>> said above: "the craft is moving away the sun at about
>> 13.059km/sec and the projection of this on the earthsite
>> to craft line is, through a nearly zero angle, 13.059."
>>
>> The angle between the motion of the craft and the site-
>> craft line is around 11.4 degrees (plus or minus 1.3)
>
> I think you may be misquoting me.
I trimmed out some of your very long sentence to make
it clearer but really it's academic. The point is
that you must project the velocity of the craft onto
the site-craft line, not the Sun-craft line, to find
the contribution to the Doppler.
> In any case I was referring in
> the context to the difference in angle between the line from the
> craft or its velocity at any one point and a line to the sun or to the
> earth which are at the maximum about 1 and half degrees apart.
Yes that's correct but it is not what you need in the
equation below, you need the projection onto the
site-craft line. Write that down somewhere, I've said
it a thousand times ;-)
>> Both the angle and its cosine are dimensionless. Your equation
>> still has different units on the two sides and is wrong unless
>> I have misunderstood what you mean by (T), c or R1. ....
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
That was the problem, I said some posts ago I assumed T
was time and R1 was range but you have cleared that up:
> .. T and R1 are the transmission and reception
> frequencies at this time. ...
In that case the units are fine and I follow what you are
doing.
> To see that the Anderson et al claim of anomalous acceleration could
> also be due to the fact that light speed
There's no chance of doing that, the speed change is only
about 10mm/s per year so you need another three decimals
on the speed to even see it, and there are far too many
approximations that give much bigger errors. However,
let's go through your equation, it's ok in principle:
> (T)(1+2(K1*V1-13.059)/c)=R1, so ((R1-T)c+2T(13.059))/2V1T=K1*
To make typing easier, define the one-way fractional
frequency shift z as
z = (R1-T)/2T
then your first equation is
z = (K1*V1-13.059)/c
or
cz = K1*V1-13.059
V1 is actually the sum of the orbital motion of the Earth
which I will call V2 and the velocity of the site relative
to the earth's centre, call that V3. K1 is the cos of an
angle so I'll call the angle between V2 and the craft-site
line A2, and similarly A3 for the angle to V3. Your final
term is the speed of the craft multiplied by the cos of the
angle between the craft velocity and the site-craft line,
call that angle A4 and call the speed V4. The general form
is then:
cz = V2*cos(A2) + V3*cos(A3) + V4*cos(A4)
The signs are positive if velocity is defined in the
same direction for all. This is ignoring elevation of
course but I'm keeping it simple to illustrate the key
points. Do you follow so far? All I have done is
rearrange a bit to make it more readable and break out
the components of the site velocity. All the angles are
relative to the site-craft line.
> The arccos of K1 is the angle between, V1, the velocity of
> the earth site wrt the sun and the line to the craft from the receiver
> site at this time. 13.059 is an initial estimate of craft velocity wrt
> the sun and the earth, T and R1 are the transmission and reception
> frequencies at this time. Note a decrease of 1.5km per sec would
> produce a change of K1 from eg .848 to .818 etc.
We know V2 is ~29.67km/s and V3 is ~353m/s for Madrid,
and at any moment there is a simple relationship between
A2 and A3 though the actual values depend on the location
of the craft. You don't know that location, or V4, or A4.
A4 is the angle we discussed above and the JPL trajectory
would mean A4 was about 12.7 degrees in March 1988 but
about 10.1 degrees six months later so you should have used
13.059*cos(10.1) = 12.8566
and
13.059*cos(12.7) = 12.7395
Notice even this change is more than 10,000 times larger
than the anomaly.
However, those values are obtained from the conventional
theory, not yours so you can't use them anyway, you have
to solve for them. I've snipped the rest as it is based
on these numbers and I'll look at solving the problem
next if you have followed this.
>> Your comments suggest perhaps I wasn't clear. I am not
>> saying the values vary with time in the way that azimuth
>> would, the RA and Dec given by the Madrid site
>
> It has nothing to do with the site.
You can obtain the values from a single contact and they
should of course be the same regardless of site. That is
the test I am applying to your theory. If your theory is
right, the values from all sites will be the same. If
they differ, your theory is wrong.
George
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