On Feb 23, 11:22 am, koobee.wub...@gmail.com wrote:
> 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.

No, it was earlier than that. Why do you care?

>
> 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>

Really?

Show us how an arbitrary scaling in time does not change the position.
It is just algebra and it would prove everyone here wrong, so it would
be a worthwhile exercise for you.