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.