Discussion:
GPS question
(too old to reply)
Jon
2009-02-20 12:33:51 UTC
Permalink
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
dlzc
2009-02-20 14:32:29 UTC
Permalink
On Feb 20, 5:33 am, "Jon" <***@peoplepc.com> wrote:
...
Post by Jon
Is this how it is done?
In no way like the kindergarten method you show. Thanks for asking
your "question", sending traffic to your site, and promoting download
of your virus-injection-capable spreadsheet.

David A. Smith
Dave
2009-02-20 15:20:57 UTC
Permalink
Post by Jon
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 athttp://mypeoplepc.com/members/jon8338/math/id8.html
I suggest you read http://en.wikipedia.org/wiki/Gps, which explains
how four satellites are used. Start with the "postion calculation
introduction" section, and then move on to the "position calculation
advanced" section.

Dave
\\frac{1}{R}\\sqrt{\\frac{L}{C}}
2009-02-22 10:50:27 UTC
Permalink
Post by Jon
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
Jon
2009-02-22 17:38:13 UTC
Permalink
Post by \\frac{1}{R}\\sqrt{\\frac{L}{C}}
Post by Jon
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.
\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
\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).
\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
Tim Smith
2009-02-22 19:42:36 UTC
Permalink
Post by \\frac{1}{R}\\sqrt{\\frac{L}{C}}
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.
I'm curious. Has anyone made a GPS receiver that contains no clock? It
should be possible, by using the GPS signal itself for timing. The
frequency is very precisely controlled, so a cycle counter could be used
for timing.
--
--Tim Smith
Jim Heckman
2009-02-23 01:18:05 UTC
Permalink
Post by Tim Smith
Post by \\frac{1}{R}\\sqrt{\\frac{L}{C}}
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.
I'm curious. Has anyone made a GPS receiver that contains no clock? It
should be possible, by using the GPS signal itself for timing. The
frequency is very precisely controlled, so a cycle counter could be used
for timing.
Possible, but not particularly practical for most applications. You
usually want the receiver to know about what time it is when you
turn it off and later back on, both for apps that use it as a
clock, and so that it has an idea what satellites are likely to be
visible and what their Doppler shifts are likely to be, so you can
find them quickly. Of course, with today's receivers that have a
bazillion parallel channels, that last is becoming less and less
relevant -- just search simultaneously for all 32 PRNs in all
possible frequency bins.
--
Jim Heckman (former GPS receiver engineer)
k***@gmail.com
2009-02-23 18:49:52 UTC
Permalink
Post by Jon
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.
PD
2009-02-23 19:09:37 UTC
Permalink
Post by k***@gmail.com
Post by Jon
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.
Except that t1, t2, t3, and t4 are Earth-referenced times, not the
native time of the clocks on the satellites, which are t1', t2', t3',
t4'. To get from t1', t2', t3', t4' to t1, t2, t3, t4, the satellites
have programmed in a correction that comes from understanding
relativistic effects.

PD
k***@gmail.com
2009-02-23 19:57:18 UTC
Permalink
Post by PD
Post by k***@gmail.com
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.
Except that t1, t2, t3, and t4 are Earth-referenced times, not the
native time of the clocks on the satellites, which are t1', t2', t3',
t4'. To get from t1', t2', t3', t4' to t1, t2, t3, t4, the satellites
have programmed in a correction that comes from understanding
relativistic effects.
There are two ways to synchronize t1, t2, t3, and t4. Ground-
satellite synchronization introduces much more errors than satellite-
satellite synchronization. The latter does not have the ionosphere
and other atmospheric anomalies to consider. That latter also does
not have high speed to worry about since all satellites are either at
rest or moving very slowly relative to each other.

For precision GPS design, one must use the satellite-satellite
synchronization. In doing so, there is no relativistic correction
needed in the design. <shrug>
Dave
2009-02-23 21:39:20 UTC
Permalink
There are two ways to synchronize t1, t2, t3, and t4.  Ground-
satellite synchronization introduces much more errors than satellite-
satellite synchronization.  The latter does not have the ionosphere
and other atmospheric anomalies to consider.  That latter also does
not have high speed to worry about since all satellites are either at
rest or moving very slowly relative to each other.
Of course the last sentence above is complete nonsense. The satellites
lie in six orbital planes, with orbital inclinations of approximately
55 degrees. The planes are separated by 60 degrees right ascension of
the ascending node. The four satellites in each plane are nearly at
rest with each other, but the relative velocity of two satellites in
different planes can be as high as sqrt(3) times the orbital velocity
of the satellites, i.e., up to approximately 6.7 km/sec.

Dave
k***@gmail.com
2009-02-23 22:57:12 UTC
Permalink
Post by Dave
Post by k***@gmail.com
There are two ways to synchronize t1, t2, t3, and t4. Ground-
satellite synchronization introduces much more errors than satellite-
satellite synchronization. The latter does not have the ionosphere
and other atmospheric anomalies to consider. That latter also does
not have high speed to worry about since all satellites are either at
rest or moving very slowly relative to each other.
Of course the last sentence above is complete nonsense. The satellites
lie in six orbital planes, with orbital inclinations of approximately
55 degrees. The planes are separated by 60 degrees right ascension of
the ascending node. The four satellites in each plane are nearly at
rest with each other, but the relative velocity of two satellites in
different planes can be as high as sqrt(3) times the orbital velocity
of the satellites, i.e., up to approximately 6.7 km/sec.
Hmmm... I guess Dave never passed the first year physics course.
<shrug>

You want to try again?

Hint: All satellites have the same angular frequency and the same
altitude.

While you are trying that again, allow me to:

Ahahaha...
Dono
2009-02-23 23:16:08 UTC
Permalink
Post by Dave
There are two ways to synchronize t1, t2, t3, and t4.  Ground-
satellite synchronization introduces much more errors than satellite-
satellite synchronization.  The latter does not have the ionosphere
and other atmospheric anomalies to consider.  That latter also does
not have high speed to worry about since all satellites are either at
rest or moving very slowly relative to each other.
Of course the last sentence above is complete nonsense. The satellites
lie in six orbital planes, with orbital inclinations of approximately
55 degrees. The planes are separated by 60 degrees right ascension of
the ascending node. The four satellites in each plane are nearly at
rest with each other, but the relative velocity of two satellites in
different planes can be as high as sqrt(3) times the orbital velocity
of the satellites, i.e., up to approximately 6.7 km/sec.
Hmmm...  I guess Dave never passed the first year physics course.
<shrug>
You want to try again?
Hint:  All satellites have the same angular frequency and the same
altitude.
Umm, no, arsehole : http://en.wikipedia.org/wiki/Global_Positioning_System#Space_segment
PD
2009-02-23 23:23:16 UTC
Permalink
Post by Dave
There are two ways to synchronize t1, t2, t3, and t4.  Ground-
satellite synchronization introduces much more errors than satellite-
satellite synchronization.  The latter does not have the ionosphere
and other atmospheric anomalies to consider.  That latter also does
not have high speed to worry about since all satellites are either at
rest or moving very slowly relative to each other.
Of course the last sentence above is complete nonsense. The satellites
lie in six orbital planes, with orbital inclinations of approximately
55 degrees. The planes are separated by 60 degrees right ascension of
the ascending node. The four satellites in each plane are nearly at
rest with each other, but the relative velocity of two satellites in
different planes can be as high as sqrt(3) times the orbital velocity
of the satellites, i.e., up to approximately 6.7 km/sec.
Hmmm...  I guess Dave never passed the first year physics course.
<shrug>
You want to try again?
Hint:  All satellites have the same angular frequency and the same
altitude.
But not the same trajectory. Tell me, KW, if you have one satellite on
an equatorial orbit, and another on a polar orbit over the Greenwich
meridian, and a third on a polar orbit through the 90-degree
longitudinal line, all with the same angular frequency and altitude,
do you believe that they are all at rest or moving very slowly with
respect to each other?

Weren't you an aerospace engineer at some point?

Was there a head injury or something?
Ahahaha...
Dono
2009-02-23 23:27:57 UTC
Permalink
Post by PD
Was there a head injury or something?
What head? He's acepahlous.
k***@gmail.com
2009-02-24 02:08:16 UTC
Permalink
Post by PD
Post by k***@gmail.com
Hint: All satellites have the same angular frequency and the same
altitude.
But not the same trajectory. Tell me, KW, if you have one satellite on
an equatorial orbit, and another on a polar orbit over the Greenwich
meridian, and a third on a polar orbit through the 90-degree
longitudinal line, all with the same angular frequency and altitude,
do you believe that they are all at rest or moving very slowly with
respect to each other?
There are no polar orbits in GPS. There are 6 such orbits with all of
them inclined by 55 degrees or so from the equatorial plane and
shifted 60 degrees (6 x 60 = 360) from one orbit to another. <shrug>
Post by PD
Weren't you an aerospace engineer at some point?
That is irrelevant. The bottom line is that you as a professor are
totally clueless and irresponsibly ignorant about even Newtonian
mechanics. <shrug>
Post by PD
Was there a head injury or something?
You should not have passed your first year physics course. <shrug>

Of course, you are no professor but a children’s book writer. No
professor would agree with the likes of DAVE and Dono, the symbols of
gross ignorance. <shrug>
Dono
2009-02-24 02:37:40 UTC
Permalink
Post by k***@gmail.com
There are no polar orbits in GPS. There are 6 such orbits with all of
them inclined by 55 degrees or so from the equatorial plane and
shifted 60 degrees (6 x 60 = 360) from one orbit to another. <shrug>
Look at the animation, asshole:
http://en.wikipedia.org/wiki/Global_Positioning_System#Space_segment
k***@gmail.com
2009-02-24 06:07:58 UTC
Permalink
Post by Dono
Post by k***@gmail.com
There are no polar orbits in GPS. There are 6 such orbits with all of
them inclined by 55 degrees or so from the equatorial plane and
shifted 60 degrees (6 x 60 = 360) from one orbit to another. <shrug>
http://en.wikipedia.org/wiki/Global_Positioning_System#Space_segment
OK, this is the last time I am helping you out. Just concentrate on
any two dots regardless of either black or red. <shrug>
Dono
2009-02-24 06:15:56 UTC
Permalink
Post by k***@gmail.com
Post by Dono
Post by k***@gmail.com
There are no polar orbits in GPS. There are 6 such orbits with all of
them inclined by 55 degrees or so from the equatorial plane and
shifted 60 degrees (6 x 60 = 360) from one orbit to another. <shrug>
http://en.wikipedia.org/wiki/Global_Positioning_System#Space_segment
OK, this is the last time I am helping you out. Just concentrate on
any two dots regardless of either black or red. <shrug>
Dumbfuck, the satellites have non-zero relative speed. Get it,
asshole?
PD
2009-02-24 14:09:51 UTC
Permalink
Post by PD
Hint:  All satellites have the same angular frequency and the same
altitude.
But not the same trajectory. Tell me, KW, if you have one satellite on
an equatorial orbit, and another on a polar orbit over the Greenwich
meridian, and a third on a polar orbit through the 90-degree
longitudinal line, all with the same angular frequency and altitude,
do you believe that they are all at rest or moving very slowly with
respect to each other?
There are no polar orbits in GPS.  There are 6 such orbits with all of
them inclined by 55 degrees or so from the equatorial plane and
shifted 60 degrees (6 x 60 = 360) from one orbit to another.  <shrug>
That's right. And in both the case I mentioned (which should have been
exceptionally easy to visualize) and in the GPS orbits, it is the case
that the satellites are not slow moving or at rest relative to each
other.

I'm glad you finally gripped the point.
Post by PD
Weren't you an aerospace engineer at some point?
That is irrelevant.  The bottom line is that you as a professor are
totally clueless and irresponsibly ignorant about even Newtonian
mechanics.  <shrug>
Post by PD
Was there a head injury or something?
You should not have passed your first year physics course.  <shrug>
Of course, you are no professor but a children’s book writer.  No
professor would agree with the likes of DAVE and Dono, the symbols of
gross ignorance.  <shrug>
You're awfully proficient at indignant sniffing when you are
embarrassed.
Dave
2009-02-23 23:42:57 UTC
Permalink
Post by Dave
There are two ways to synchronize t1, t2, t3, and t4.  Ground-
satellite synchronization introduces much more errors than satellite-
satellite synchronization.  The latter does not have the ionosphere
and other atmospheric anomalies to consider.  That latter also does
not have high speed to worry about since all satellites are either at
rest or moving very slowly relative to each other.
Of course the last sentence above is complete nonsense. The satellites
lie in six orbital planes, with orbital inclinations of approximately
55 degrees. The planes are separated by 60 degrees right ascension of
the ascending node. The four satellites in each plane are nearly at
rest with each other, but the relative velocity of two satellites in
different planes can be as high as sqrt(3) times the orbital velocity
of the satellites, i.e., up to approximately 6.7 km/sec.
Hmmm...  I guess Dave never passed the first year physics course.
<shrug>
You guess wrong. I have had more physics courses than you probably can
name.
You want to try again?
No need to do that when I'm right the first time.
Hint:  All satellites have the same angular frequency and the same
altitude.
Perhaps you have not passed the first orbital mechanics class. Since
the orbits are not in the same plane, the orbital planes intersect.
Consider two satellites in two different planes that recently have
passed the orbital plane intersection.
Ahahaha...
You are laughing at your own ignorance.

Dave
\\frac{1}{R}\\sqrt{\\frac{L}{C}}
2009-02-24 04:16:45 UTC
Permalink
Post by k***@gmail.com
Post by PD
Post by k***@gmail.com
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.
Except that t1, t2, t3, and t4 are Earth-referenced times, not the
native time of the clocks on the satellites, which are t1', t2', t3',
t4'. To get from t1', t2', t3', t4' to t1, t2, t3, t4, the satellites
have programmed in a correction that comes from understanding
relativistic effects.
There are two ways to synchronize t1, t2, t3, and t4. Ground-
satellite synchronization introduces much more errors than satellite-
satellite synchronization. The latter does not have the ionosphere
and other atmospheric anomalies to consider. That latter also does
not have high speed to worry about since all satellites are either at
rest or moving very slowly relative to each other.
For precision GPS design, one must use the satellite-satellite
synchronization. In doing so, there is no relativistic correction
needed in the design. <shrug>
You don't need satellite to satellite synchronization in GPS, although
the newer satellites have this capability (block III i think they are
called, these guys could operate independently for a while)

Since the 70's any single GPS satellite in orbit is tracked with a few
stations on ground forming a ground segment.

In this way you get the orbit of the satellite, a master oscillator at
just one tracking station is good enough to synchronize all satellite
clocks to the same time system.

Q
--
CO2 at 390 ppm and counting, put a tiger in your tank -- ESSO commercial
x***@gmail.com
2009-02-24 02:22:56 UTC
Permalink
Post by PD
Post by k***@gmail.com
Post by Jon
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.
Except that t1, t2, t3, and t4 are Earth-referenced times, not the
native time of the clocks on the satellites, which are t1', t2', t3',
t4'. To get from t1', t2', t3', t4' to t1, t2, t3, t4, the satellites
have programmed in a correction that comes from understanding
relativistic effects.
PD- Hide quoted text -
- Show quoted text -
xxein: You can call them relativistic effects all you want but that
only satisfies how the theory is understood with it's conjugate math.
I can use a 50% different math and get the same (perhaps more
accurate) answer because of fewer "unnecessary" variables introduced.
This is an ideal formulation that still must be corrected with a
feedback because of the un-ideal reality of the orbits, but with a
regularity, it will produce a higher degree of precision.

There is still one issue that seemingly cancels out for either
method. The Shapiro delay and the bending of light in gravity are
factors that don't cancel out. What this means is that light does not
go from the Earth's surface in the same time it takes light to return
from the satellite. This comes from the Pound-Rebka experiments. It
is still not recognized as a logical flaw. But Carroll Alley pointed
out a flaw in the original development of GPS that might have had
something to do with that line of logical thought. It is sort of a
throw back to doubling light delay time of light from 1921 by
Eddington.

Somehow physicists don't see the conection. If light bends in gravity
with a delay, what is the "+bend" coming from a satellite directly
toward us? What is the "-bend" from us to the satellite? Now
consider that the satellites are never at zenith to GPS receivers on
Earth. Is that delay acceptable or can we work it out to more
accuracy?

If I had more calculus prowess,... But I don't. I can "see" it
clearly though with logic.
Eric Gisse
2009-02-23 20:01:51 UTC
Permalink
Post by k***@gmail.com
Post by Jon
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.
...even tnough t1-4 are wrong by roughly 50,000ns/day?
k***@gmail.com
2009-02-23 20:22:26 UTC
Permalink
Post by Eric Gisse
Post by k***@gmail.com
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>
Eric Gisse
2009-02-23 20:26:17 UTC
Permalink
Post by k***@gmail.com
Post by Eric Gisse
Post by k***@gmail.com
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?
Post by k***@gmail.com
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.
doug
2009-02-24 00:15:53 UTC
Permalink
Post by k***@gmail.com
Post by Eric Gisse
Post by k***@gmail.com
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>
Of course the relativitistic correction is needed. The satellites
get their location information from the ground. They need to know
the times for that. Read about how the gps works.
k***@gmail.com
2009-02-24 06:18:45 UTC
Permalink
[Toxic nonsense snipped]
It is truly amazing. The subject of GPS attracts Einstein
Dingleberries who knows nothing about GPS works. <shrug> The whole
subject can easily be explained by junior high mathematics. I guess
the trait among Einstein Dingleberries who worship Einstein the
nitwit, the plagiarist, and the liar as a messiah is that they do not
understand elementary algebra in the first place. Mysticism, however,
is just one small part of whatever nonsense the Einstein Dingleberries
are spoon-fed in the past 100 years:

** FAITH IS THEORY
** MYSTICISM IS WISDOM
** IGNORANCE IS KNOWLEDGE
** PLAGIARISM IS CREATIVITY
** CONJECTURE IS REALITY
** BELIEVING IS LEARNING
** LYING IS TEACHING

<shrug>
Eric Gisse
2009-02-24 06:25:31 UTC
Permalink
[Toxic nonsense snipped]
It is truly amazing.  The subject of GPS attracts Einstein
Dingleberries who knows nothing about GPS works.
Let's see you cite a reference that describes how the sattelites work.

[...]
RustyJames
2009-02-24 21:05:25 UTC
Permalink
Post by doug
Post by k***@gmail.com
Post by Eric Gisse
Post by k***@gmail.com
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>
Of course the relativitistic correction is needed. The satellites
get their location information from the ground. They need to know
the times for that.  Read about how the gps works.- Hide quoted text -
- Show quoted text -
Tracking Geosynchronous Satellites with GPS

The traditional method of orbit determination for a geosynchronous
spacecraft is to collect occasional range measurements from a tracking
station on the ground. More than one station might be used, but never
more than one at a time. Scientists have known this approach would be
improved by adding more tracking stations and positioning them as far
apart as possible along the circumference of Earth (as seen by the
satellite). Unfortunately, many factors make this arrangement
impossible. GPS offers a simpler solution.

Viewed from a geosynchronous spacecraft, GPS satellites (on the
opposite side of Earth) are positioned somewhat beyond the edge of the
planet—providing better tracking geometry than would be possible from
any location on the ground. Moreover, the orbital configuration of the
GPS constellation ensures that various satellites will be seen at
comparatively wide distances from each other, further improving the
tracking geometry. Also, in comparison to ground-based stations, GPS
satellites exhibit greater relative motion with respect to the
geosynchronous satellite.

Of course, a geosynchronous satellite will rarely, if ever, have four
GPS satellites in view. Thus, it cannot employ GPS signals the way a
user on Earth would. But geosynchronous satellites only use one
station in their traditional method of orbit determination anyway, so
an absence of more GPS signals is not a deficit.
This answers my question I understand the difference for applications
of them not being Geosynchronous GPS Satellites

C***@comcast.net
2009-02-23 20:33:09 UTC
Permalink
Post by k***@gmail.com
Post by Jon
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.
Currently, only Block IIR satellites are capable of autonomous
position monitoring and synchronization independent of the ground
control segment. This capability allows Block IIR satellites to
mutually correct their clocks and ephemerides information for a
period of up to several months should the ground segment be
knocked out by, say, nuclear war. For optimum performance under
non-wartime conditions, "ground truth" comparisons are still
performed on a daily basis to correct the satellites for drift.

GPS satellites before the Block IIR series are incapable of such
mutual correction, and require monitoring and daily corrections
by the Ground Segment.

Satellite control requires that the Space Segment and Ground
Segment clocks be synchronized. The only practical way of
synchronizing the world-wide network of ground stations is to
use the network of GPS satellites to disseminate time. The only
way to use the networks of GPS satellites to disseminate time is
if the satellite clocks are adjusted via the GR correction so
that they beat in time with Earth clocks.

Jerry
Michael Press
2009-02-23 21:28:16 UTC
Permalink
In article
Post by C***@comcast.net
Currently, only Block IIR satellites are capable of autonomous
position monitoring and synchronization independent of the ground
control segment. This capability allows Block IIR satellites to
mutually correct their clocks and ephemerides information for a
period of up to several months should the ground segment be
knocked out by, say, nuclear war. For optimum performance under
non-wartime conditions, "ground truth" comparisons are still
performed on a daily basis to correct the satellites for drift.
The Block IIR satellites use what references?
--
Michael Press
C***@comcast.net
2009-02-23 22:05:29 UTC
Permalink
Post by Michael Press
In article
Post by C***@comcast.net
Currently, only Block IIR satellites are capable of autonomous
position monitoring and synchronization independent of the ground
control segment. This capability allows Block IIR satellites to
mutually correct their clocks and ephemerides information for a
period of up to several months should the ground segment be
knocked out by, say, nuclear war. For optimum performance under
non-wartime conditions, "ground truth" comparisons are still
performed on a daily basis to correct the satellites for drift.
The Block IIR satellites use what references?
After decades of operation, the ground station positions are
known to within millimeters. So accurately have their positions
been characterized, that the effects of tides on the Earth's
crust need to be taken into account.

Jerry
Michael Press
2009-02-24 03:17:25 UTC
Permalink
In article
Post by C***@comcast.net
Post by Michael Press
In article
Post by C***@comcast.net
Currently, only Block IIR satellites are capable of autonomous
position monitoring and synchronization independent of the ground
control segment. This capability allows Block IIR satellites to
mutually correct their clocks and ephemerides information for a
period of up to several months should the ground segment be
knocked out by, say, nuclear war. For optimum performance under
non-wartime conditions, "ground truth" comparisons are still
performed on a daily basis to correct the satellites for drift.
The Block IIR satellites use what references?
After decades of operation, the ground station positions are
known to within millimeters. So accurately have their positions
been characterized, that the effects of tides on the Earth's
crust need to be taken into account.
Am I to infer that ground stations are reference points
for Block IIR satellites? Does the ground station have
to emit a signal? If so, what kind of signal?

Is it really spelled "Block IIR" with majuscule `B'?
--
Michael Press
Dono
2009-02-23 22:40:53 UTC
Permalink
Post by C***@comcast.net
Post by k***@gmail.com
Post by Jon
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.
Currently, only Block IIR satellites are capable of autonomous
position monitoring and synchronization independent of the ground
control segment. This capability allows Block IIR satellites to
mutually correct their clocks and ephemerides information for a
period of up to several months should the ground segment be
knocked out by, say, nuclear war. For optimum performance under
non-wartime conditions, "ground truth" comparisons are still
performed on a daily basis to correct the satellites for drift.
GPS satellites before the Block IIR series are incapable of such
mutual correction, and require monitoring and daily corrections
by the Ground Segment.
Satellite control requires that the Space Segment and Ground
Segment clocks be synchronized. The only practical way of
synchronizing the world-wide network of ground stations is to
use the network of GPS satellites to disseminate time. The only
way to use the networks of GPS satellites to disseminate time is
if the satellite clocks are adjusted via the GR correction so
that they beat in time with Earth clocks.
Jerry- Hide quoted text -
- Show quoted text -
Jerry

According to Dave's post (see above), the satellites in relative
motion (at about 6.7km/s) so the Kobee-Wublee idiot needs to
understand that, even with Block IIR, the relativistic corrections are
still needed.
torso
2009-02-23 22:54:42 UTC
Permalink
Post by Dono
Post by C***@comcast.net
Post by k***@gmail.com
Post by Jon
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.
Currently, only Block IIR satellites are capable of autonomous
position monitoring and synchronization independent of the ground
control segment. This capability allows Block IIR satellites to
mutually correct their clocks and ephemerides information for a
period of up to several months should the ground segment be
knocked out by, say, nuclear war. For optimum performance under
non-wartime conditions, "ground truth" comparisons are still
performed on a daily basis to correct the satellites for drift.
GPS satellites before the Block IIR series are incapable of such
mutual correction, and require monitoring and daily corrections
by the Ground Segment.
Satellite control requires that the Space Segment and Ground
Segment clocks be synchronized. The only practical way of
synchronizing the world-wide network of ground stations is to
use the network of GPS satellites to disseminate time. The only
way to use the networks of GPS satellites to disseminate time is
if the satellite clocks are adjusted via the GR correction so
that they beat in time with Earth clocks.
Jerry- Hide quoted text -
- Show quoted text -
Jerry
According to Dave's post (see above), the satellites in relative
motion (at about 6.7km/s) so the Kobee-Wublee idiot needs to
understand that, even with Block IIR, the relativistic corrections are
still needed.
i thought they were geostationary
Dono
2009-02-23 23:10:27 UTC
Permalink
Post by Dono
Post by C***@comcast.net
Post by k***@gmail.com
Post by Jon
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.
Currently, only Block IIR satellites are capable of autonomous
position monitoring and synchronization independent of the ground
control segment. This capability allows Block IIR satellites to
mutually correct their clocks and ephemerides information for a
period of up to several months should the ground segment be
knocked out by, say, nuclear war. For optimum performance under
non-wartime conditions, "ground truth" comparisons are still
performed on a daily basis to correct the satellites for drift.
GPS satellites before the Block IIR series are incapable of such
mutual correction, and require monitoring and daily corrections
by the Ground Segment.
Satellite control requires that the Space Segment and Ground
Segment clocks be synchronized. The only practical way of
synchronizing the world-wide network of ground stations is to
use the network of GPS satellites to disseminate time. The only
way to use the networks of GPS satellites to disseminate time is
if the satellite clocks are adjusted via the GR correction so
that they beat in time with Earth clocks.
Jerry- Hide quoted text -
- Show quoted text -
Jerry
According to Dave's post (see above), the satellites in relative
motion (at about 6.7km/s) so the Kobee-Wublee idiot needs to
understand that, even with Block IIR, the relativistic corrections are
still needed.
i thought they were geostationary- Hide quoted text -
- Show quoted text -
http://groups.google.com/group/sci.physics.relativity/msg/20b712a944ab8265?
Dono
2009-02-23 23:11:46 UTC
Permalink
Post by Dono
Post by C***@comcast.net
Post by k***@gmail.com
Post by Jon
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.
Currently, only Block IIR satellites are capable of autonomous
position monitoring and synchronization independent of the ground
control segment. This capability allows Block IIR satellites to
mutually correct their clocks and ephemerides information for a
period of up to several months should the ground segment be
knocked out by, say, nuclear war. For optimum performance under
non-wartime conditions, "ground truth" comparisons are still
performed on a daily basis to correct the satellites for drift.
GPS satellites before the Block IIR series are incapable of such
mutual correction, and require monitoring and daily corrections
by the Ground Segment.
Satellite control requires that the Space Segment and Ground
Segment clocks be synchronized. The only practical way of
synchronizing the world-wide network of ground stations is to
use the network of GPS satellites to disseminate time. The only
way to use the networks of GPS satellites to disseminate time is
if the satellite clocks are adjusted via the GR correction so
that they beat in time with Earth clocks.
Jerry- Hide quoted text -
- Show quoted text -
Jerry
According to Dave's post (see above), the satellites in relative
motion (at about 6.7km/s) so the Kobee-Wublee idiot needs to
understand that, even with Block IIR, the relativistic corrections are
still needed.
i thought they were geostationary- Hide quoted text -
- Show quoted text -
Ahh, the Australian fake idiot :-)
Dave
2009-02-23 23:31:06 UTC
Permalink
Post by torso
Post by Dono
Post by C***@comcast.net
Post by k***@gmail.com
Post by Jon
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.
Currently, only Block IIR satellites are capable of autonomous
position monitoring and synchronization independent of the ground
control segment. This capability allows Block IIR satellites to
mutually correct their clocks and ephemerides information for a
period of up to several months should the ground segment be
knocked out by, say, nuclear war. For optimum performance under
non-wartime conditions, "ground truth" comparisons are still
performed on a daily basis to correct the satellites for drift.
GPS satellites before the Block IIR series are incapable of such
mutual correction, and require monitoring and daily corrections
by the Ground Segment.
Satellite control requires that the Space Segment and Ground
Segment clocks be synchronized. The only practical way of
synchronizing the world-wide network of ground stations is to
use the network of GPS satellites to disseminate time. The only
way to use the networks of GPS satellites to disseminate time is
if the satellite clocks are adjusted via the GR correction so
that they beat in time with Earth clocks.
Jerry- Hide quoted text -
- Show quoted text -
Jerry
According to Dave's post (see above), the satellites in relative
motion (at about 6.7km/s) so the Kobee-Wublee idiot needs to
understand that, even with Block IIR, the relativistic corrections are
still needed.
i thought they were geostationary
You thought wrong. They orbit at about 20,200 km altitude, and have an
orbital period of 12 hours.

Dave
Dono
2009-02-23 23:39:46 UTC
Permalink
Post by Dave
Post by torso
Post by Dono
Post by C***@comcast.net
Post by k***@gmail.com
Post by Jon
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.
Currently, only Block IIR satellites are capable of autonomous
position monitoring and synchronization independent of the ground
control segment. This capability allows Block IIR satellites to
mutually correct their clocks and ephemerides information for a
period of up to several months should the ground segment be
knocked out by, say, nuclear war. For optimum performance under
non-wartime conditions, "ground truth" comparisons are still
performed on a daily basis to correct the satellites for drift.
GPS satellites before the Block IIR series are incapable of such
mutual correction, and require monitoring and daily corrections
by the Ground Segment.
Satellite control requires that the Space Segment and Ground
Segment clocks be synchronized. The only practical way of
synchronizing the world-wide network of ground stations is to
use the network of GPS satellites to disseminate time. The only
way to use the networks of GPS satellites to disseminate time is
if the satellite clocks are adjusted via the GR correction so
that they beat in time with Earth clocks.
Jerry- Hide quoted text -
- Show quoted text -
Jerry
According to Dave's post (see above), the satellites in relative
motion (at about 6.7km/s) so the Kobee-Wublee idiot needs to
understand that, even with Block IIR, the relativistic corrections are
still needed.
i thought they were geostationary
You thought wrong. They orbit at about 20,200 km altitude, and have an
orbital period of 12 hours.
Dave- Hide quoted text -
- Show quoted text -
You are dealing with "the Australian fake idiot" troll
RustyJames
2009-02-24 19:40:51 UTC
Permalink
Post by Dono
Post by C***@comcast.net
Post by k***@gmail.com
Post by Jon
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.
Currently, only Block IIR satellites are capable of autonomous
position monitoring and synchronization independent of the ground
control segment. This capability allows Block IIR satellites to
mutually correct their clocks and ephemerides information for a
period of up to several months should the ground segment be
knocked out by, say, nuclear war. For optimum performance under
non-wartime conditions, "ground truth" comparisons are still
performed on a daily basis to correct the satellites for drift.
GPS satellites before the Block IIR series are incapable of such
mutual correction, and require monitoring and daily corrections
by the Ground Segment.
Satellite control requires that the Space Segment and Ground
Segment clocks be synchronized. The only practical way of
synchronizing the world-wide network of ground stations is to
use the network of GPS satellites to disseminate time. The only
way to use the networks of GPS satellites to disseminate time is
if the satellite clocks are adjusted via the GR correction so
that they beat in time with Earth clocks.
Jerry- Hide quoted text -
- Show quoted text -
Jerry
According to Dave's post (see above), the satellites in relative
motion (at about 6.7km/s) so the Kobee-Wublee idiot needs to
understand that, even with Block IIR, the relativistic corrections are
still needed.- Hide quoted text -
- Show quoted text -
I thought they were geostationary

I believe their geosynchronous to the earths rotation
PD
2009-02-24 20:01:42 UTC
Permalink
Post by RustyJames
Post by Dono
Post by C***@comcast.net
Post by k***@gmail.com
Post by Jon
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.
Currently, only Block IIR satellites are capable of autonomous
position monitoring and synchronization independent of the ground
control segment. This capability allows Block IIR satellites to
mutually correct their clocks and ephemerides information for a
period of up to several months should the ground segment be
knocked out by, say, nuclear war. For optimum performance under
non-wartime conditions, "ground truth" comparisons are still
performed on a daily basis to correct the satellites for drift.
GPS satellites before the Block IIR series are incapable of such
mutual correction, and require monitoring and daily corrections
by the Ground Segment.
Satellite control requires that the Space Segment and Ground
Segment clocks be synchronized. The only practical way of
synchronizing the world-wide network of ground stations is to
use the network of GPS satellites to disseminate time. The only
way to use the networks of GPS satellites to disseminate time is
if the satellite clocks are adjusted via the GR correction so
that they beat in time with Earth clocks.
Jerry- Hide quoted text -
- Show quoted text -
Jerry
According to Dave's post (see above), the satellites in relative
motion (at about 6.7km/s) so the Kobee-Wublee idiot needs to
understand that, even with Block IIR, the relativistic corrections are
still needed.- Hide quoted text -
- Show quoted text -
I thought they were geostationary
I believe their geosynchronous to the earths rotation
Nope.
RustyJames
2009-02-24 20:08:25 UTC
Permalink
Post by RustyJames
Post by Dono
Post by C***@comcast.net
Post by k***@gmail.com
Post by Jon
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.
Currently, only Block IIR satellites are capable of autonomous
position monitoring and synchronization independent of the ground
control segment. This capability allows Block IIR satellites to
mutually correct their clocks and ephemerides information for a
period of up to several months should the ground segment be
knocked out by, say, nuclear war. For optimum performance under
non-wartime conditions, "ground truth" comparisons are still
performed on a daily basis to correct the satellites for drift.
GPS satellites before the Block IIR series are incapable of such
mutual correction, and require monitoring and daily corrections
by the Ground Segment.
Satellite control requires that the Space Segment and Ground
Segment clocks be synchronized. The only practical way of
synchronizing the world-wide network of ground stations is to
use the network of GPS satellites to disseminate time. The only
way to use the networks of GPS satellites to disseminate time is
if the satellite clocks are adjusted via the GR correction so
that they beat in time with Earth clocks.
Jerry- Hide quoted text -
- Show quoted text -
Jerry
According to Dave's post (see above), the satellites in relative
motion (at about 6.7km/s) so the Kobee-Wublee idiot needs to
understand that, even with Block IIR, the relativistic corrections are
still needed.- Hide quoted text -
- Show quoted text -
I thought they were geostationary
I believe their geosynchronous to the earths rotation
Nope.- Hide quoted text -
- Show quoted text -
PD what satellites are you referring two so I can get on the same page
then this would be helpful maybe I should not jump into a conversation
without knowing all the logistics of the satellites I apologize
PD
2009-02-24 20:23:04 UTC
Permalink
Post by RustyJames
Post by RustyJames
Post by Dono
Post by C***@comcast.net
Post by k***@gmail.com
Post by Jon
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.
Currently, only Block IIR satellites are capable of autonomous
position monitoring and synchronization independent of the ground
control segment. This capability allows Block IIR satellites to
mutually correct their clocks and ephemerides information for a
period of up to several months should the ground segment be
knocked out by, say, nuclear war. For optimum performance under
non-wartime conditions, "ground truth" comparisons are still
performed on a daily basis to correct the satellites for drift.
GPS satellites before the Block IIR series are incapable of such
mutual correction, and require monitoring and daily corrections
by the Ground Segment.
Satellite control requires that the Space Segment and Ground
Segment clocks be synchronized. The only practical way of
synchronizing the world-wide network of ground stations is to
use the network of GPS satellites to disseminate time. The only
way to use the networks of GPS satellites to disseminate time is
if the satellite clocks are adjusted via the GR correction so
that they beat in time with Earth clocks.
Jerry- Hide quoted text -
- Show quoted text -
Jerry
According to Dave's post (see above), the satellites in relative
motion (at about 6.7km/s) so the Kobee-Wublee idiot needs to
understand that, even with Block IIR, the relativistic corrections are
still needed.- Hide quoted text -
- Show quoted text -
I thought they were geostationary
I believe their geosynchronous to the earths rotation
Nope.- Hide quoted text -
- Show quoted text -
PD what satellites are you referring two so I can get on the same page
then this would be helpful maybe I should not jump into a conversation
without knowing all the logistics of the satellites I apologize
Geosynchronous satellites are at the altitude of 35,786 km.
GPS satellites are at the altitude of 20,200 km.
Completely different periods.

PD
k***@gmail.com
2009-02-23 22:54:47 UTC
Permalink
Post by C***@comcast.net
Post by k***@gmail.com
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.
[...]
Satellite control requires that the Space Segment and Ground
Segment clocks be synchronized. The only practical way of
synchronizing the world-wide network of ground stations is to
use the network of GPS satellites to disseminate time. The only
way to use the networks of GPS satellites to disseminate time is
if the satellite clocks are adjusted via the GR correction so
that they beat in time with Earth clocks.
These are what the mysticism the Einstein Dingleberries hypnotized
themselves in **believing in**. <shrug>

When given the truth, the Einstein Dingleberries still whining about
their mysticisms fed to them daily. Oh, well. Mysticism is only a
small part to the BS diet fed to Einstein Dingleberries where:

** FAITH IS THEORY
** MYSTICISM IS WISDOM
** IGNORANCE IS KNOWLEDGE
** PLAGIARISM IS CREATIVITY
** CONJECTURE IS REALITY
** BELIEVING IS LEARNING
** LYING IS TEACHING

<shrug>
torso
2009-02-23 22:57:51 UTC
Permalink
Post by k***@gmail.com
Post by C***@comcast.net
Post by k***@gmail.com
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.
[...]
Satellite control requires that the Space Segment and Ground
Segment clocks be synchronized. The only practical way of
synchronizing the world-wide network of ground stations is to
use the network of GPS satellites to disseminate time. The only
way to use the networks of GPS satellites to disseminate time is
if the satellite clocks are adjusted via the GR correction so
that they beat in time with Earth clocks.
These are what the mysticism the Einstein Dingleberries hypnotized
themselves in **believing in**. <shrug>
When given the truth, the Einstein Dingleberries still whining about
their mysticisms fed to them daily. Oh, well. Mysticism is only a
** FAITH IS THEORY
** MYSTICISM IS WISDOM
** IGNORANCE IS KNOWLEDGE
** PLAGIARISM IS CREATIVITY
** CONJECTURE IS REALITY
** BELIEVING IS LEARNING
** LYING IS TEACHING
<shrug>
in principle you are right

but they still need to add the missing ticks, nobody knows were they
are stored !!!saisc
Eric Gisse
2009-02-23 23:16:14 UTC
Permalink
Post by k***@gmail.com
Post by C***@comcast.net
Post by k***@gmail.com
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.
[...]
Satellite control requires that the Space Segment and Ground
Segment clocks be synchronized. The only practical way of
synchronizing the world-wide network of ground stations is to
use the network of GPS satellites to disseminate time. The only
way to use the networks of GPS satellites to disseminate time is
if the satellite clocks are adjusted via the GR correction so
that they beat in time with Earth clocks.
These are what the mysticism the Einstein Dingleberries hypnotized
themselves in **believing in**.  <shrug>
Why do the GPS designers and published specifications say otherwise,
and why are you unable to show otherwise?
Post by k***@gmail.com
When given the truth, the Einstein Dingleberries still whining about
their mysticisms fed to them daily.  Oh, well.  Mysticism is only a
**       FAITH IS THEORY
**   MYSTICISM IS WISDOM
**   IGNORANCE IS KNOWLEDGE
**  PLAGIARISM IS CREATIVITY
**  CONJECTURE IS REALITY
**   BELIEVING IS LEARNING
**       LYING IS TEACHING
<shrug>
C***@comcast.net
2009-02-24 00:16:57 UTC
Permalink
Post by k***@gmail.com
Post by C***@comcast.net
Satellite control requires that the Space Segment and Ground
Segment clocks be synchronized. The only practical way of
synchronizing the world-wide network of ground stations is to
use the network of GPS satellites to disseminate time. The only
way to use the networks of GPS satellites to disseminate time is
if the satellite clocks are adjusted via the GR correction so
that they beat in time with Earth clocks.
These are what the mysticism the Einstein Dingleberries hypnotized
themselves in **believing in**.  <shrug>
When given the truth, the Einstein Dingleberries still whining about
their mysticisms fed to them daily.  Oh, well.  Mysticism is only a
**       FAITH IS THEORY
**   MYSTICISM IS WISDOM
**   IGNORANCE IS KNOWLEDGE
**  PLAGIARISM IS CREATIVITY
**  CONJECTURE IS REALITY
**   BELIEVING IS LEARNING
**       LYING IS TEACHING
<shrug>
Here is an exercise for you. Go online and buy one of those neat
little Mars watches that are set for a day length of 24h 39m 35s.

Set your watch to accurate Earth time, go into a large office
building and using your Mars watch as a reference, synchronize
all of the office clocks over the course of a day so that they
show accurate Earth time to the nearest second.

That's sort of what the problem is like synchronizing the ground-
based clocks of the GPS Control Segment if you don't adjust the
rates of the satellite clocks with the GR correction.

Jerry
x***@gmail.com
2009-02-24 00:52:03 UTC
Permalink
Post by k***@gmail.com
Post by Jon
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 -
\\frac{1}{R}\\sqrt{\\frac{L}{C}}
2009-02-24 04:00:33 UTC
Permalink
Post by k***@gmail.com
Post by Jon
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.
The last statement is not true, a relativistic effect must be considered
in GPS because you're dealing with a space based clock that travels at
several km/s relative to an observer on Earth. Furthermore there is a
difference in the gravitational potential between Earth and the GPS
satellite so that you need the GR correction as well.

In reality both SR and GR corrections are deliberately applied to the
GPS satellite clock so that the thing appears to run at the proper
frequency for an Earth fixed observer. A nice homework exercise for you
is to compute the frequency shift required at the 10,something Mhz
reference oscillator in the satellite.

Sometimes these military guys are a bit smarter than you think.

Q
--
CO2 at 390 ppm and counting, put a tiger in your tank -- ESSO commercial
Dono
2009-02-24 05:14:30 UTC
Permalink
Post by \\frac{1}{R}\\sqrt{\\frac{L}{C}}
Post by k***@gmail.com
Post by Jon
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.
The last statement is not true, a relativistic effect must be considered
in GPS because you're dealing with a space based clock that travels at
several km/s relative to an observer on Earth. Furthermore there is a
difference in the gravitational potential between Earth and the GPS
satellite so that you need the GR correction as well.
In reality both SR and GR corrections are deliberately applied to the
GPS satellite clock so that the thing appears to run at the proper
frequency for an Earth fixed observer. A nice homework exercise for you
is to compute the frequency shift required at the 10,something Mhz
reference oscillator in the satellite.
Sometimes these military guys are a bit smarter than you think.
Q
--
CO2 at 390 ppm and counting, put a tiger in your tank -- ESSO commercial
You are dealing with the resident antirelativity, antiEinstein nutter,
you will not get anywhere.
Nice posts, though, you know your stuff.
k***@gmail.com
2009-02-24 06:05:57 UTC
Permalink
Post by \\frac{1}{R}\\sqrt{\\frac{L}{C}}
Post by k***@gmail.com
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.
The last statement is not true, a relativistic effect must be considered
in GPS because you're dealing with a space based clock that travels at
several km/s relative to an observer on Earth. Furthermore there is a
difference in the gravitational potential between Earth and the GPS
satellite so that you need the GR correction as well.
Hmmm... You never understood what I said and what the Wikipedia
article said below. <shrug>

http://en.wikipedia.org/wiki/Gps
Post by \\frac{1}{R}\\sqrt{\\frac{L}{C}}
[Rest of nonsense snipped]
\\frac{1}{R}\\sqrt{\\frac{L}{C}}
2009-02-24 07:21:12 UTC
Permalink
Post by k***@gmail.com
Post by \\frac{1}{R}\\sqrt{\\frac{L}{C}}
Post by k***@gmail.com
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.
The last statement is not true, a relativistic effect must be considered
in GPS because you're dealing with a space based clock that travels at
several km/s relative to an observer on Earth. Furthermore there is a
difference in the gravitational potential between Earth and the GPS
satellite so that you need the GR correction as well.
Hmmm... You never understood what I said and what the Wikipedia
article said below. <shrug>
http://en.wikipedia.org/wiki/Gps
Just look under the section relativity, it is explained there.
Post by k***@gmail.com
Post by \\frac{1}{R}\\sqrt{\\frac{L}{C}}
[Rest of nonsense snipped]
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CO2 at 390 ppm and counting, put a tiger in your tank -- ESSO commercial
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