On Feb 23, 8:00 pm, "\\frac{1}{R}\\sqrt{\\frac{L}{C}}" <b...@bla.org>
wrote:
> 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.
>
> The last statement is not true, a relativistic effect must be considered
> in GPS because you're dealing with a space based clock that travels at
> several km/s relative to an observer on Earth. Furthermore there is a
> difference in the gravitational potential between Earth and the GPS
> satellite so that you need the GR correction as well.
>
> In reality both SR and GR corrections are deliberately applied to the
> GPS satellite clock so that the thing appears to run at the proper
> frequency for an Earth fixed observer. A nice homework exercise for you
> is to compute the frequency shift required at the 10,something Mhz
> reference oscillator in the satellite.
>
> Sometimes these military guys are a bit smarter than you think.
>
> Q
>
> --
> CO2 at 390 ppm and counting, put a tiger in your tank -- ESSO commercial



You are dealing with the resident antirelativity, antiEinstein nutter,
you will not get anywhere.
Nice posts, though, you know your stuff.