On Feb 23, 12:01 pm, Eric Gisse wrote:
> On Feb 23, 9:49 am, 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.
>
> ...even tnough t1-4 are wrong by roughly 50,000ns/day?

So, the college drop-out sleeps until 11AM Alaska time today.

In the meantime, if t1 is off by 50usec a day relative to the ground,
as long as t2, t3, and t4 are all off by the same amount, it does not
matter.  There is still no need for any relativistic correction.
<shrug>