On Feb 23, 1:49 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.- Hide quoted text -
>
> - Show quoted text -