"qbit" wrote in message
...
: For the well known relativistic correction of 38 microseconds per day
: they do a frequence shifting in the transmitters of the GPS satellites.
: They shift the frequence down from 10.23 MHz to 10.22999999543 MHz
instead.

Shifting frequency doesn't change time, it changes N where N = t.f
Example:
t = 1 second, f = 1 Hz, number of cycles counted N = 1
t = 1 second, f = 2 Hz, number of cycles counted N =2
t = 1 second, f = 1 MHz, number of cycles counted N =1,000,000
t = 60 seconds, f = 10 MHz, number of cycles counted N =600,000,000
N = t.f

Questions:
What does the term "4.465 parts in 10^10" mean?

The number of extra cycles that suddenly appeared out of nowhere.
They were not sent but they did arrive.
If you send 10,229,999,995.43 cycles in a second (or an hour or a day) and
10,230,000,000 arrive in a second (or an hour or day) then 45.7 cycles
magically appeared out of thin aether. The number of cycles sent has to
match the number of cycles received no matter what time frame they are
sent in.

What does it represent? The 38 us/d ? How?
How is this value calculated?

From this:
"we establish by definition that the time required by light to travel from A
to B equals the time it requires to travel from B to A." -- Albert ****wit
Einstein
St. Einstein the Divine ****wit can change the passage of time because St.
Einstein the Divine ****wit said so.




Is this 446.5 x 10^-12, ie. 446.5 pico ?

I'm not going to teach Ashby to count, he should have learnt that at 6 years
old.

How is the 10.22999999543 MHz calculated?

From this:
"we establish by definition that the time required by light to travel from A
to B equals the time it requires to travel from B to A." -- Albert ****wit
Einstein
St. Einstein the Divine ****wit can change the passage of time because St.
Einstein the Divine ****wit said so.

And last but not least:
Why can Professor Ashby not use a standard mathematical language?

Because he wants to sound important. All Einstein Dingleberries are
pompous prigs and illogical fools.



Why is he not consistent: sometimes he uses 4.465 and sometimes 4.4647.
How come?
I have the feeling he is doing much "curve-fitting".
Where in his document at
http://relativity.livingreviews.org/...3-1/index.html
is the calculation of the above numbers shown?

It doesn't matter, it's wrong.

"qbit" wrote in message
...
For the well known relativistic correction of 38 microseconds per day
they do a frequence shifting in the transmitters of the GPS satellites.
They shift the frequence down from 10.23 MHz to 10.22999999543 MHz
instead.

Citation from http://en.wikipedia.org/wiki/Global_Positioning_System
"[...] For the GPS satellites, general relativity predicts that the
atomic clocks at GPS orbital altitudes will tick more rapidly,
by about 45,900 nanoseconds (ns) per day, because they are in a
weaker gravitational field than atomic clocks on Earth's surface.
Special relativity predicts that atomic clocks moving at GPS orbital
speeds will tick more slowly than stationary ground clocks by
about 7,200 ns per day.
When combined, the discrepancy is 38 microseconds per day;
a difference of 4.465 parts in 10^10. To account for this,
the frequency standard onboard each satellite is given a rate offset
prior to launch, making it run slightly slower than the desired frequency
on Earth; specifically, at 10.22999999543 MHz instead of 10.23 MHz.
[...]"

Questions:
What does the term "4.465 parts in 10^10" mean?
What does it represent? The 38 us/d ? How?
How is this value calculated?
Is this 446.5 x 10^-12, ie. 446.5 pico ?
How is the 10.22999999543 MHz calculated?
And last but not least:
Why can Professor Ashby not use a standard mathematical language?
Why is he not consistent: sometimes he uses 4.465 and sometimes 4.4647.
How come?
I have the feeling he is doing much "curve-fitting".
Where in his document at
http://relativity.livingreviews.org/...3-1/index.html
is the calculation of the above numbers shown?

Perhaps the following will answer your questions.

1. Light travels at a constant speed
of 299 792.458 meters per second
in the absence of matter,
and in media with sparse matter,
such as the Earth's atmosphere.

2. Time interval measurements of E-M waves in air, and space,
are equivalent to distance measurements.
distance = time interval * C

3. Synchronized clocks can be used to
quantize the distance between the points
by measuring the time it takes light/radio waves
to travel from one point to another.

Clock(A) sends a message that it is time(X).
Clock(B) notes that it is time(X) + I1 on its' clock.

The distance between the clocks is
I1 * C

In other words, systems of synchronized clocks
can quantize the distances between the clocks,
by transmitting the time at each clock's location.

Any clock can determine the distances
between it and other clocks,
by simply determining time(I) for all of the other clocks.

For example,
if one measures a time delay of "I1" of a radio wave
from New York, they must be somewhere on
the surface of a sphere, with a distance radius of I1 * C,
centered about New York

If they also measure a time delay of "I2" of a radio wave
from San Francisco, they must be somewhere on
the surface of a sphere, with a distance radius of I2 * C,
centered about San Francisco.

If they measure both,
they must be on a circle represented by the
intersection of the two spheres.

As can be seen, the measurement of a third point,
would be the intersection of the circle with
another sphere, and would let tell the observer that
they are on one of two points.

A fourth measurement would resolve the situation,
and tell them at which of the two points they are
located.

4. As the GPS satellites are moving,
whereas New York and San Francisco are located
at fixed points (With respect to Earth bound observers.),
it is necessary that GPS receivers know where
the satellites were when they transmitted the time.

This is handled, by having each satellite
transmit its' position in space, along with
the time data.

Each satellite not only transmits where it is ("ephemeris data"),
it transmits its' orbital data ("almanac data"),
along with its' time.

The "ephemeris data"
serves the same purpose to the GPS receiver,
as the Sun does is to a sailor with a sextant.

5. Ground stations continuously monitor
the satellites' orbits and transmissions,
and when changes exceed certain amounts,
signals are sent to the offending satellites,
updating their "almanac data", their "ephemeris data",
their time settings, and drift in their oscillators
with respect to the master clock and oscillator on Earth.

In other words, the ground station monitors the data
transmitted by the satellites and when necessary
sends them signals that tells them, that their clock is x nano-seconds fast,
that their oscillator is running y x10^12 too fast, their orbit has changed
to
such and such (Perhaps because of dust drag, etc.),
that their "ephemeris data" should be xxx, etc.

The GPS clocks are set,
to some reference time,
just as your digital watch is,
the only difference being that
the ticks are far more stable, and much finer,
nanoseconds, rather than tenths of seconds.

Drifts in oscillators are set and corrected by
inserting or ignoring "ticks", and by adjusting
divider circuits to divide by the desired count.

6. As portable GPS receivers do not have
extremely stable oscillators, they must
derive precision times from the satellites.

As the satellites are at an altitude of about 11,000 miles,
and radio waves travel 186,000 miles in one second,
it takes about .006 seconds for the
time, ephemeris, and almanac data
to reach a sea level receiver.

This means that in a typical transmission,
the GPS receiver must subtract about .006 seconds
from its' clock, in order to set its' clock.
GPS receivers receive and average the times
from several satellites, and recursively
home in on the master time, and make an adjustment
for recursively computed position of the satellite.

In other words, at the reception of the first data,
the GPS receiver knows the master time to about .006 seconds
higher than the first time it receives,
and as it picks up signals from other satellites,
and recursively computes the distances to the
satellites, and averages out multi-path signal variations,
its' own clock homes in on the master clock time.

Also, the GPS receiver knows that it is in ONE place,
and as it calculates its' "place" using the data from
each satellite, it can recursively shift its' clock time
slightly in order to make the locations come out
as closely as possible to ONE spot, rather than
four spots several meters or kilometers apart.

As the satellites take about 12 hours
(43200 seconds) to orbit the Earth,
and the ephemeris data takes about .006 seconds
to reach the receiver, this means that
the GPS receiver knows where the
satellite is to an accuracy of about one part in
43000 / .006 = 71600000 parts,
even without clock and ephemeris corrections.

Considering that the Earth is about
24,000 miles or 126,000,000 feet in circumference,
this amounts to a sphere of uncertainty of about
1.76 feet at sea level.

7. The clocks used in the GPS system are extremely stable.
They have a long term and short term stability
of about 1 part in 10^14 over one day and even months.

As there are about 3 x 10^13 MICROseconds in a year,
this means that the GPS clocks can maintain microsecond
agreement for over a year, even if no corrections are made.

But of course, corrections ARE made to the clocks
on a regular basis by a ground clock,
that all of the GPS clocks are referenced to.

8. As the satellites have a life expectancy of about 10 years,
their orbits are very stable.
In other words, when ground stations get a fix on a satellite's orbit,
we know pretty much where the satellite will be for a long time, and
GPS receivers on the ground have an extremely dependable target to sight on

9. There is some variation in the time it takes the
signal to reach the receiver due to multi paths
taken by the radio wave to the GPS receiver,
so GPS receivers are programmed to compute out the
multi-path variations, and to compute the time,
using the most reliable data it gets from several satellites.

10. The GPS satellites broadcast on two carrier frequencies:
L1 at 1575.42 MHz and L2 at 1227.6 MHz.
They transmit a "coarse acquisition code" at 1.0 bits per nanosecond and
a "precision code" at a bit rate of 10.230 bits per nanosecond.

As light travels at about 300,000,000 meters per second,
or 300 meters in one micro-second,
a one nano second error would result in an error sphere of about .3 meters
( One foot), and a 10 nanosecond error would
result in an error of about 3 meters or ten feet.

By averaging data from multiple satellites,
a receiver can reduce the timing uncertainty
due to multipaths, and can reduce the error sphere
by only averaging where the error spheres
of several satellites overlap.

The single largest contributor to time transfer uncertainty is path delay,
the delay introduced as the signal travels from the satellite
to the receiver.

In order to recognize the tiny desired signals from the satellites,
which would be obscured by all kinds of noise,
the satellite transmissions are preceded with an 12 bit code
that produces a large cross-correlation.
(Various sequences of 12 bits produce various levels
of cross-correlation. The best sequence is used.
I don't recall what it is, but if I remember, or can find the
name of the code sequence on the Internet, I will post it.)

A copy of this 12 code, plus copies of quasi-random
codes that identify each satellite are stored in GPS receivers.

The digital computer in the GPS receiver
first cross-correlates frames of the incoming signal
with the 12 bit code transmitted by all satellites,
and after a high cross-correlation indicates a lock
on a signal, the quasi-random numbers associated
with the active satellites are appended to the
12 bit "recognition" code so that each particular satellite can
be identified, and the data from each extracted.

The largest contributor
to time transfer uncertainty is caused by
variations path delay, due to signals reflected
off mountains, buildings, etc., and as note,
much of the path delay errors can be averaged out,
because the satellites are moving, and signals
are received from several satellites.

The best GPS receivers can,
by using the methods addressed above,
reduce the uncertainty in time to about one nanosecond,
which amounts to a sphere of uncertainty of about one foot.

Regarding your question:
"How is the 10.22999999543 MHz calculated?"

As Galileo discover over 300 years ago,
oscillators are affected by acceleration,
and the acceleration "g" at the altitude where the
GPS satellites operate would allow an oscillator
to oscillate slightly faster, so when the atomic oscillators
are on the ground being tested,
the output frequency slaved to the atomic clock,
is set to 10.22999999543 MHz so that when the oscillator is
in orbit, it will oscillate at the same rate as the master
oscillator on the surface of the Earth (10.3 MHz).

Although the General Relativity Gurus,
who are on the taxpayer dole,
make a big production about how GTR, with about
13 Classical Physics hacks, comes up a little closer to
this value than Galileo's 300 year old equation,
the fact of the matter is that the satellites could
operate on slightly differences, as long as they all stayed
within the bandpass of the GPS receivers,
and the system would work fine.

Note that time does not change in orbit.
What changes is the frequency of the oscillator.
But all times must be referenced to a master
tick accumulator (Clock).

The important factor is that the satellite clocks
should stay synchronized as close as possible to the
master clock in order to prevent frequent time updates.
and by making the oscillators run slower on
the Earth they will maintain time sync better while in orbit,

but as can be seen, this could be easily taken care of by
having the ground station monitor each satellite oscillator,
and send each satellite a command to adjust
the frequency synthesizers and dividers to
minimize the drift of the satellite oscillator
with respect to the master oscillator.

--
Tom Potter

*** Time Magazine Person of the Year 2006 ***
*** May 2007 Anti-Bigot Award ***
http://home.earthlink.net/~tdp
http://tdp1001.googlepages.com/home
http://no-turtles.com
http://www.frappr.com/tompotter
http://spaces.msn.com/tdp1001
http://www.flickr.com/photos/tom-potter
http://tom-potter.blogspot.com





--
Posted via a free Usenet account from http://www.teranews.com

"Tom Potter" wrote in message
...
:
: "qbit" wrote in message
: ...
: For the well known relativistic correction of 38 microseconds per day
: they do a frequence shifting in the transmitters of the GPS satellites.
: They shift the frequence down from 10.23 MHz to 10.22999999543 MHz
: instead.
:
: Citation from http://en.wikipedia.org/wiki/Global_Positioning_System
: "[...] For the GPS satellites, general relativity predicts that the
: atomic clocks at GPS orbital altitudes will tick more rapidly,
: by about 45,900 nanoseconds (ns) per day, because they are in a
: weaker gravitational field than atomic clocks on Earth's surface.
: Special relativity predicts that atomic clocks moving at GPS orbital
: speeds will tick more slowly than stationary ground clocks by
: about 7,200 ns per day.
: When combined, the discrepancy is 38 microseconds per day;
: a difference of 4.465 parts in 10^10. To account for this,
: the frequency standard onboard each satellite is given a rate offset
: prior to launch, making it run slightly slower than the desired
frequency
: on Earth; specifically, at 10.22999999543 MHz instead of 10.23 MHz.
: [...]"
:
: Questions:
: What does the term "4.465 parts in 10^10" mean?
: What does it represent? The 38 us/d ? How?
: How is this value calculated?
: Is this 446.5 x 10^-12, ie. 446.5 pico ?
: How is the 10.22999999543 MHz calculated?
: And last but not least:
: Why can Professor Ashby not use a standard mathematical language?
: Why is he not consistent: sometimes he uses 4.465 and sometimes 4.4647.
: How come?
: I have the feeling he is doing much "curve-fitting".
: Where in his document at
: http://relativity.livingreviews.org/...3-1/index.html
: is the calculation of the above numbers shown?
:
: Perhaps the following will answer your questions.
:
: 1. Light travels at a constant speed
: of 299 792.458 meters per second
: in the absence of matter,
: and in media with sparse matter,
: such as the Earth's atmosphere.
:
: 2. Time interval measurements of E-M waves in air, and space,
: are equivalent to distance measurements.
: distance = time interval * C

No no no, Tom.
St. Einstein the Divine ****wit wrote a papal bull that said 'we establish
BY DEFINITION that the "time" required by light to travel from A to B
equals the "time" it requires to travel from B to A', and another that
said "But the ray moves relatively to the initial point of k, when measured
in
the stationary system, with the velocity c-v, so that t = x'/(c-v)".
Therefore the distance one-way, A to B, is different to the distance back
again,
B to A, it taking light the same "time" in either direction.

"Tom Potter" wrote in message ...

As Galileo discover over 300 years ago,
oscillators are affected by acceleration,
and the acceleration "g" at the altitude where the
GPS satellites operate would allow an oscillator
to oscillate slightly faster, so when the atomic oscillators
are on the ground being tested,
the output frequency slaved to the atomic clock,
is set to 10.22999999543 MHz so that when the oscillator is
in orbit, it will oscillate at the same rate as the master
oscillator on the surface of the Earth (10.3 MHz).

Do you have some handy references/links to texts of
the classical prediction methods, and/or the equations?
I like the classical methods, ie. where no "c" is present in the equations.
We should grab out the history of the classical methods.

Although the General Relativity Gurus,
who are on the taxpayer dole,
make a big production about how GTR, with about
13 Classical Physics hacks, comes up a little closer to
this value than Galileo's 300 year old equation,
the fact of the matter is that the satellites could
operate on slightly differences, as long as they all stayed
within the bandpass of the GPS receivers,
and the system would work fine.

Tom Potter wrote:

[..]

As the satellites take about 12 hours
(43200 seconds) to orbit the Earth,
and the ephemeris data takes about .006 seconds
to reach the receiver, this means that
the GPS receiver knows where the
satellite is to an accuracy of about one part in
43000 / .006 = 71600000 parts,
even without clock and ephemeris corrections.

Considering that the Earth is about
24,000 miles or 126,000,000 feet in circumference,
this amounts to a sphere of uncertainty of about
1.76 feet at sea level.

!!!! :-O

Paul, out of words

On Tue, 7 Aug 2007 07:06:44 +0200, "qbit"
wrote:

For the well known relativistic correction of 38 microseconds per day
they do a frequence shifting in the transmitters of the GPS satellites.
They shift the frequence down from 10.23 MHz to 10.22999999543 MHz instead.

Citation from http://en.wikipedia.org/wiki/Global_Positioning_System
"[...] For the GPS satellites, general relativity predicts that the
atomic clocks at GPS orbital altitudes will tick more rapidly,
by about 45,900 nanoseconds (ns) per day, because they are in a
weaker gravitational field than atomic clocks on Earth's surface.
Special relativity predicts that atomic clocks moving at GPS orbital
speeds will tick more slowly than stationary ground clocks by
about 7,200 ns per day.
When combined, the discrepancy is 38 microseconds per day;
a difference of 4.465 parts in 10^10. To account for this,
the frequency standard onboard each satellite is given a rate offset
prior to launch, making it run slightly slower than the desired frequency
on Earth; specifically, at 10.22999999543 MHz instead of 10.23 MHz.
[...]"

Questions:
What does the term "4.465 parts in 10^10" mean?
What does it represent? The 38 us/d ? How?
How is this value calculated?
Is this 446.5 x 10^-12, ie. 446.5 pico ?
How is the 10.22999999543 MHz calculated?
And last but not least:
Why can Professor Ashby not use a standard mathematical language?
Why is he not consistent: sometimes he uses 4.465 and sometimes 4.4647. How come?
I have the feeling he is doing much "curve-fitting".
Where in his document at http://relativity.livingreviews.org/...3-1/index.html
is the calculation of the above numbers shown?
Ashby wrote a simple GPS paper as I recall in which the time delays
came down to Lorentz transforms so I was surprised to see this
tortured set of equations that are extreme overkill, as if intended to
conceal rather than reveal.
But if your scientific proficiency is so limited as to ask
"What does the term "4.465 parts in 10^10" mean?", there's no point in
spending a lot of time on the finer points. You're not ready for it.

On Tue, 07 Aug 2007 11:45:54 GMT, "Androcles"
wrote:


"qbit" wrote in message
...
: For the well known relativistic correction of 38 microseconds per day
: they do a frequence shifting in the transmitters of the GPS satellites.
: They shift the frequence down from 10.23 MHz to 10.22999999543 MHz
instead.

Shifting frequency doesn't change time, it changes N where N = t.f
Example:
t = 1 second, f = 1 Hz, number of cycles counted N = 1
t = 1 second, f = 2 Hz, number of cycles counted N =2
t = 1 second, f = 1 MHz, number of cycles counted N =1,000,000
t = 60 seconds, f = 10 MHz, number of cycles counted N =600,000,000
N = t.f

Questions:
What does the term "4.465 parts in 10^10" mean?

The number of extra cycles that suddenly appeared out of nowhere.
They were not sent but they did arrive.
If you send 10,229,999,995.43 cycles in a second (or an hour or a day) and
10,230,000,000 arrive in a second (or an hour or day) then 45.7 cycles
magically appeared out of thin aether. The number of cycles sent has to
match the number of cycles received no matter what time frame they are
sent in.

What does it represent? The 38 us/d ? How?
How is this value calculated?

As a matter of interest, Ballistic theory comes up with exactly the same figure
of 45.7.
BaTh says photons accelerate as they fall to Earth, just like anything else.
The fractional change in speed multiplied by the number of seconds in a day
just happens to produce the magic figure of 45.7

**********************************
G=6.67E-11
Solar Mass = 5.983E24 kgm
Equatorial Radius = 6378388 m

GPS : velocity of 3.874 km/s at an altitude of 20,184 km = 26562388 m from
centre

______________________

For unit mass starting at velocity c:

F= -GM/r^2

O---------o 26600

d(PE)=$Fdr= [GMm/r] from r1 to r2
_____________
GM=6.67E-11 * 5.983E24 kgm

= 3.9907 E14
_____________

Determine 1/r1-1/r2

1/Re= 15.68E-8
1/Rs= 3.76E-8
1/r1-1/r2= 1.192E-7

(altrnatively: 20,184,000/169e12 = 1.19E-7)

Lost PE= 4.757E7

This equals gained KE = 1/2[(c+v)^2] - 1/2(c^2)

~cv

velocity gained = 4.757E7/2.997E8

= 1.59E-1 m/sec

as proportion of c: v/c = 5.3E-10

multiply by seconds per day= 457920E-10

= 45.7E-6








www.users.bigpond.com/hewn/index.htm

The difference between a preacher and a used car salesman is that the latter at least has a product to sell.

On Tue, 07 Aug 2007 22:52:25 +0200, "Paul B. Andersen"
wrote:

Tom Potter wrote:

[..]

As the satellites take about 12 hours
(43200 seconds) to orbit the Earth,
and the ephemeris data takes about .006 seconds
to reach the receiver, this means that
the GPS receiver knows where the
satellite is to an accuracy of about one part in
43000 / .006 = 71600000 parts,
even without clock and ephemeris corrections.

Considering that the Earth is about
24,000 miles or 126,000,000 feet in circumference,
this amounts to a sphere of uncertainty of about
1.76 feet at sea level.

!!!! :-O

Paul, out of words

Here is a quote I found:

"Inspection of the differences
between UTC/TAI and the phase of each NAVSTAR
GPS clock enables the USNO to identify GPS clocks that
require particular frequency-rate control corrections. Use of
this knowledge enables the USAF to adjust frequency rates of
selected GPS clocks. Currently the USAF uses an automated
bang-bang controller(3) on frequency-rate."

.....seems to make your previous claims look pretty stupid, eh Paul?



www.users.bigpond.com/hewn/index.htm

The difference between a preacher and a used car salesman is that the latter at least has a product to sell.

"Henri Wilson" HW@.... wrote

What does the term "4.465 parts in 10^10" mean?

As a matter of interest, Ballistic theory comes up with exactly the
same figure of 45.7.

Any reference or link?

BaTh says photons accelerate as they fall to Earth, just like anything else.
The fractional change in speed multiplied by the number of seconds
in a day just happens to produce the magic figure of 45.7

You presented IMO an interessting calculation.
Some questions:

G=6.67E-11
Solar Mass = 5.983E24 kgm

This should be "Earth Mass", isn't it?

For unit mass starting at velocity c:
F= -GM/r^2
O---------o 26600

What does this 26600 represent? How did you get it?

(altrnatively: 20,184,000/169e12 = 1.19E-7)

What does the value 169e12 represent?

Thx

"Henri Wilson" HW@.... wrote

What does the term "4.465 parts in 10^10" mean?

As a matter of interest, Ballistic theory comes up with exactly the
same figure of 45.7.

Any reference or link?

BaTh says photons accelerate as they fall to Earth, just like anything else.
The fractional change in speed multiplied by the number of seconds
in a day just happens to produce the magic figure of 45.7

You presented IMO an interessting calculation.
Some questions:

G=6.67E-11
Solar Mass = 5.983E24 kgm

This should be "Earth Mass", isn't it?

For unit mass starting at velocity c:
F= -GM/r^2
O---------o 26600

What does this 26600 represent? How did you get it?

(altrnatively: 20,184,000/169e12 = 1.19E-7)

What does the value 169e12 represent?

Thx

BTW, your website brings an error when trying to download the following files:
moonrelay.jpg and msvbrm50.zip