Re: GPS question
On Feb 23, 8:00 pm, <b...@bla.org> wrote:
> koobee.wub...@gmail.com wrote:
> > 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.
Hmmm... You never understood what I said and what the Wikipedia
article said below. <shrug>
http://en.wikipedia.org/wiki/Gps
> [Rest of nonsense snipped]
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