On Feb 23, 2:09 pm, PD <TheDraperFam...@gmail.com> wrote:
> On Feb 23, 12:49 pm, koobee.wub...@gmail.com wrote:
>
>
>
>
>
> > On Feb 20, 4:33 am, "Jon" <jon8...@peoplepc.com> wrote:
>
> > > Supposing three satellites transmit signals with time stamps that are picked
> > > up by a receiver.  If the clocks on the satellites are closely synchronized,
> > > then
>
> > > |AD|=|AB|+t_AB
> > > |BD|=|BC|+t_BC
> > > |CD|=|CA|+t_CA
>
> > > Where
> > > A,B,C=position vectors of three satellites
> > > AB,BC,CA=distances between satellites
> > > t_AB,t_BC,t_CA=time differences between signals at D.
> > > D=position vector of receiver.
>
> > > Once |AD|,|BD|,|CD| are found, the coordinates of D can be derived.
>
> > > Is this how it is done?
>
> > > summarized at
>
> > >http://mypeoplepc.com/members/jon8338/math/id8.html
>
> > You need four satellites where each satellite should know its time and
> > position.  Time can just be a counter with all the satellites
> > synchronized to within a count of each other.  Each satellite then
> > just broadcast its time and position information as almanac signal to
> > whoever wants to receive it.
>
> > Since the distance is traveled by light with a known speed, you can
> > then easily form a set of four equations with four unknowns.  The four
> > unknowns are your time (relative to the counter in each satellite) and
> > position.
>
> > **  c^2 (t1 – t)^2 = (x1 – x)^2 + (y1 – y)^2 + (z1 – z)^2
> > **  c^2 (t2 – t)^2 = (x2 – x)^2 + (y2 – y)^2 + (z2 – z)^2
> > **  c^2 (t3 – t)^2 = (x3 – x)^2 + (y3 – y)^2 + (z3 – z)^2
> > **  c^2 (t4 – t)^2 = (x4 – x)^2 + (y4 – y)^2 + (z4 – z)^2
>
> > Where
>
> > **  (t1, x1, y1, z1) = Time and position of satellite 1
> > **  (t2, x2, y2, z2) = Time and position of satellite 2
> > **  (t3, x3, y3, z3) = Time and position of satellite 3
> > **  (t4, x4, y4, z4) = Time and position of satellite 4
> > **  (t, x, y, z) = Time and position of the receiver
>
> > Notice that relativistic effect is never needed.
>
> Except that t1, t2, t3, and t4 are Earth-referenced times, not the
> native time of the clocks on the satellites, which are t1', t2', t3',
> t4'. To get from t1', t2', t3', t4' to t1, t2, t3, t4, the satellites
> have programmed in a correction that comes from understanding
> relativistic effects.
>
> PD- Hide quoted text -
>
> - Show quoted text -

xxein:  You can call them relativistic effects all you want but that
only satisfies how the theory is understood with it's conjugate math.
I can use a 50% different math and get the same (perhaps more
accurate) answer because of fewer "unnecessary" variables introduced.
This is an ideal formulation that still must be corrected with a
feedback because of the un-ideal reality of the orbits, but with a
regularity, it will produce a higher degree of precision.

There is still one issue that seemingly cancels out for either
method.  The Shapiro delay and the bending of light in gravity are
factors that don't cancel out.  What this means is that light does not
go from the Earth's surface in the same time it takes light to return
from the satellite.  This comes from the Pound-Rebka experiments.  It
is still not recognized as a logical flaw.  But Carroll Alley pointed
out a flaw in the original development of GPS that might have had
something to do with that line of logical thought.  It is sort of a
throw back to doubling light delay time of light from 1921 by
Eddington.

Somehow physicists don't see the conection.  If light bends in gravity
with a delay, what is the "+bend" coming from a satellite directly
toward us?  What is the "-bend" from us to the satellite?  Now
consider that the satellites are never at zenith to GPS receivers on
Earth.  Is that delay acceptable or can we work it out to more
accuracy?

If I had more calculus prowess,...  But I don't.  I can "see" it
clearly though with logic.