Re: GPS vs. Source Dependency
jgreenfield_at_seol.net.au
Date: 02/27/05
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Date: 27 Feb 2005 15:57:38 -0800
Paul B. Andersen wrote:
> Henri Wilson wrote:
> > On 26 Feb 2005 21:22:35 -0800, "Sbharris[atsign]ix.netcom.com"
> > <sbharris@ix.netcom.com> wrote:
> >
> >
> >>>You don't know what you're talking about. GPS may correct for
phase
> >>>shift of frequency, but it uses time-of-flight of what is
essentially
> >>
> >>a
> >>
> >>>time-stamp signal. And it calculates distance from time of flight
and
> >>>speed of signal, which it presumes to be c. Not c+v or c-v. Just
c.
> >>
> >>
> >>It makes little difference which one you use. The satellite moves
only
> >>about
> >>1cm in that kind of time difference. That's the order of the error
you
> >>will get
> >>if you use c+v.
> >>
> >>
> >>COMMENT:
> >>
> >>
> >>What? Try again. If your value of c is wrong by 1.5 ppm over a
> >>distance of 26 million meters, then the flight time you measure
will
> >>give you an incorrect distance by 40 meters. And that's your error
due
> >>to using a value of c-v or c+v when v is 470 m/sec, the maximal
radial
> >>GPS velocity. This has nothing to do with how far the SATELLITE
moves.
> >>The error is due to the difference in how far the SIGNAL moves in
the
> >>given time, if you have the wrong velocity for it.
> >>
> >>d1= ct
> >>d2 = (c+v)t
> >>
> >>t is not a variable, it is a known. So you do the math and
subtract d1
> >
> >>from d2 to get your error in d from using the wrong signal
velocity.
> >
> > The orbit is wrongly calculated using c instead of c+v. This
cancels out the
> > 40m error.
> > and if all the orbiting clocks are in close synch, the positioning
error due to
> > the omission of the 'free fall correction' is only 7 cms.
>
> Do you really not understand how ridiculous this answer is, Henri?
:-)
> There are five monitoring stations spread all around the Earth.
> Don't you understand how inconsistent the measurements from
> these stations would be if the speed of light were c+v, and this
> was not accounted for?
But it IS accounted for! They ALL observe a Doppler signal alteration,
right?
And that alteration is due to c = f x u ,where c alters, u remains
the same, causing f to Doppler. It is seen continuously- but not
RECOGNIZED/INTERPRETED as to WHY!
Jim G
c'=c+v
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