"\frac{1}{R}\sqrt{\frac{L}{C}}" <bla@bla.org> wrote in message 
news:49a12de0$0$198$e4fe514c@news.xs4all.nl...
> Jon 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
>>
>>
>
> In GPS the positions and velocities and clocks of the satellites are
> normally known, they are determined by tracking the satellites from
> ground, and the information is uploaded to the satellites. The message
> from the satellite contains time and position information and the
> receiver on the ground knows it as well.
>
> In this application of GPS the observation equations are formed for
> so-called pseudo-distances, which are time differences between the
> receiver clock on ground and the clock information received from the
> satellite. The receiver clock is quartz oscillator just like your wrist
> watch, and it comes with an offset typically in the order of several
> seconds. The receiver clock error (I call it \Delta t_q) is a parameter
> to solve for together with the coordinates of the receiver.
>
> Normally you start with 4 satellites so that the equations to solve are:
>
> \rho_{aq} = s * (t_q - t_a) + s * \Delta t_q
> \rho_{bq} = s * (t_q - t_b) + s * \Delta t_q
> \rho_{cq} = s * (t_q - t_c) + s * \Delta t_q
> \rho_{dq} = s * (t_q - t_d) + s * \Delta t_q
>
> where
>
> \rho_{aq} etc are so-called pseudo-distances between receiver "q" and
> satellite "a", other satellites are "b" "c" and "d", "s" is the speed of
> light, t_q is the time in the receiver, and t_a is the time in satellite
> a, and \Delta t_q is the time bias or the clock error of the receiver
>
> The term s * (t_q - t_a) is the geometric distance between satellite "a"
> and the receiver, and \rho_{aq} \rho_{bq} \rho_{cq} \rho_{dq} are
> measured by the receiver. You can also write:
>
> \rho_{aq} = r_{aq} + s * \Delta t_q
> \rho_{bq} = r_{bq} + s * \Delta t_q
> \rho_{cq} = r_{cq} + s * \Delta t_q
> \rho_{dq} = r_{dq} + s * \Delta t_q
>
> where r_{aq} etc are the geometric distances. To solve these equations
> one normally linearizes the equations relative to an approximate
> position that is provided to the receiver, or that is memorized from
> earlier computations. Let us call this approximate location (x0,y0,z0).
>
> In that case the linearized equations become:
>
> \begin{vec}{c}
>  \Delta \rho_{aq}  \\
>  \Delta \rho_{bq}  \\
>  \Delta \rho_{cq}  \\
>  \Delta \rho_{dq}  \\
> \end{vec}
> =
> \begin{mat}{cccc}
>  \diff{r_{aq}}{x0}  \diff{r_{aq}}{y0} \diff{r_{aq}}{z0}  s   \\
>  \diff{r_{bq}}{x0}  \diff{r_{bq}}{y0} \diff{r_{bq}}{z0}  s   \\
>  \diff{r_{cq}}{x0}  \diff{r_{cq}}{y0} \diff{r_{cq}}{z0}  s   \\
>  \diff{r_{dq}}{x0}  \diff{r_{dq}}{y0} \diff{r_{dq}}{z0}  s   \\
> \end{mat}
> *
> \begin{vec}
> \Delta x0 \\
> \Delta y0 \\
> \Delta z0 \\
> \Delta t_q \\
> \end{vec}
>
> where \Delta \rho_{aq} etc are observed minus approximated pseudo range
> differences (approximated because we can estimate them once the
> approximated location x0 y0 z0 is provided), \diff{r_{aq}}{x0} etc are
> partial derivatives from geometric range to coordinates, and \Delta x0
> etc are shifts to apply to the approximate coordinates.
>
> The left hand side of these equations is known, the right hand side
> matrix is known (although it is an approximation) and the right hand
> side vector can be solved by inverting the equations. Within a couple of
> iterations you update the approximate location until there are no more
> coordinate shifts as discussed above.
>
> Now, the OP said he had only three pseudo ranges. In that case you must
> add another constraint equation to the system of equations because it
> becomes rank deficient. Normally one adds a constraint equation like a
> height of the GPS receiver.
>
> Hope this helps,
>
> Q
>
> -- 
> CO2 at 390 ppm and counting, put a tiger in your tank -- ESSO commercial

Thank you for your insight.  That means I only solved *part* of the GPS 
problem.  The triangulation portion of the solution is at,

http://www.geocities.com/jongiff2000/Triangulation_Solver.xls

For the rest of it, you proved it can be done.

Here's another problem I solved:

http://www.geocities.com/jongiff2000/Mach_Projectile_Solver.xls