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?

On Aug 6, 9:06 pm, "qbit" wrote:
[...]

I have the feeling he is doing much "curve-fitting".

Who cares what you feel?

Where in his document athttp://relativity.livingreviews.org/Articles/lrr-2003-1/index.html
is the calculation of the above numbers shown?

Why should anyone bother answering any of your questions when this
question alone shows that you didn't even LOOK for the answer?

On Aug 7, 2:06 am, "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 fromhttp://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.

Where does Special relativity predict this?
Special relativity says:

"The [ ] Incompatibility of the Law of Propagation of
Light with the Principle of Relativity [is only] Apparent"
http://www.bartleby.com/173/7.html

Sue...

"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?

Ok, finally I've managed to find the formula for calculating
the GPS frequency shift (see below).

Using the the documented frequence shift value of 10.22999999543 MHz
as the basis, the best matching RelEffect number (ie. a more accurate value
for the famous value "38 us/day") seems to be one of the following
values depending on what the GPS engineers used for DaySeconds,
ie. whether they used Sideral Day (86164.091 s) or the normal 86400 s:

GPS frequency shifting algorithm:
RelEffect s/day = 0.000038597067448680351906158357771260997067448680 35
DaySeconds = 86400
FreqBase MHz = 10.23
FreqBase Hz = 10230000
Factor = 0.000000000446725317693059628543499511241446725317 69
Value Hz = 0.004569999999999999999999999999999999999999999999 77
FreqResult Hz = 10229999.99543
FreqResult MHz = 10.22999999543

GPS frequency shifting algorithm:
RelEffect s/day = 0.000038491680925708699902248289345063538611925708 70
DaySeconds = 86164.091
FreqBase MHz = 10.23
FreqBase Hz = 10230000
Factor = 0.000000000446725317693059628543499511241446725317 69
Value Hz = 0.00457
FreqResult Hz = 10229999.99542999999999999999999999999999999999999 999999999
FreqResult MHz = 10.22999999543


The formula is:
Factor = RelEffect / DaySeconds
Value = Factor * FreqBase
Result = FreqBase - Value

Or in short form:
Result = FreqBase - (RelEffect / DaySeconds) * FreqBase


(- Also Wikipedia should correct its number from the 38700 ns to one of
the IMO more correct ones above, ie. to 38597.067 ns or 38491.681 us)

On Aug 7, 12:05 am, "qbit" wrote:
"qbit" wrote in ...

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 fromhttp://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 athttp://relativity.livingreviews.org/Articles/lrr-2003-1/index.html
is the calculation of the above numbers shown?

Ok, finally I've managed to find the formula for calculating
the GPS frequency shift (see below).

Using the the documented frequence shift value of 10.22999999543 MHz
as the basis, the best matching RelEffect number (ie. a more accurate value
for the famous value "38 us/day") seems to be one of the following
values depending on what the GPS engineers used for DaySeconds,
ie. whether they used Sideral Day (86164.091 s) or the normal 86400 s:

GPS frequency shifting algorithm:
RelEffect s/day = 0.000038597067448680351906158357771260997067448680 35

You do not have 50 significant figures of accuracy.

[snip]

"qbit" wrote in message ...
"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?

Ok, finally I've managed to find the formula for calculating
the GPS frequency shift (see below).

Using the the documented frequence shift value of 10.22999999543 MHz
as the basis, the best matching RelEffect number (ie. a more accurate value
for the famous value "38 us/day") seems to be one of the following
values depending on what the GPS engineers used for DaySeconds,
ie. whether they used Sideral Day (86164.091 s) or the normal 86400 s:

GPS frequency shifting algorithm:
RelEffect s/day = 0.000038597067448680351906158357771260997067448680 35
DaySeconds = 86400
FreqBase MHz = 10.23
FreqBase Hz = 10230000
Factor = 0.000000000446725317693059628543499511241446725317 69
Value Hz = 0.004569999999999999999999999999999999999999999999 77
FreqResult Hz = 10229999.99543
FreqResult MHz = 10.22999999543

GPS frequency shifting algorithm:
RelEffect s/day = 0.000038491680925708699902248289345063538611925708 70
DaySeconds = 86164.091
FreqBase MHz = 10.23
FreqBase Hz = 10230000
Factor = 0.000000000446725317693059628543499511241446725317 69
Value Hz = 0.00457
FreqResult Hz = 10229999.99542999999999999999999999999999999999999 999999999
FreqResult MHz = 10.22999999543


The formula is:
Factor = RelEffect / DaySeconds
Value = Factor * FreqBase
Result = FreqBase - Value

Or in short form:
Result = FreqBase - (RelEffect / DaySeconds) * FreqBase


(- Also Wikipedia should correct its number from the 38700 ns to one of
the IMO more correct ones above, ie. to 38597.067 ns or 38491.681 us)

Correction:
the IMO more correct ones above, ie. to 38597.067 ns or 38491.681 ns)

"Eric Gisse" wrote
On Aug 7, 12:05 am, "qbit" wrote:
"qbit" wrote

Ok, finally I've managed to find the formula for calculating
the GPS frequency shift (see below).

Using the the documented frequence shift value of 10.22999999543 MHz
as the basis, the best matching RelEffect number (ie. a more accurate value
for the famous value "38 us/day") seems to be one of the following
values depending on what the GPS engineers used for DaySeconds,
ie. whether they used Sideral Day (86164.091 s) or the normal 86400 s:

GPS frequency shifting algorithm:
RelEffect s/day = 0.000038597067448680351906158357771260997067448680 35

You do not have 50 significant figures of accuracy.

Maybe on the GPS computers not, but I can use any number
of significant decimal digits for accuracy; it's only up to available memory.
Actually internally I even used 100 :-), only the output was limited to 50.
I can choose whatever I like, even 10000 :-)

On Aug 7, 12:48 am, "qbit" wrote:
"Eric Gisse" wrote



On Aug 7, 12:05 am, "qbit" wrote:
"qbit" wrote

Ok, finally I've managed to find the formula for calculating
the GPS frequency shift (see below).

Using the the documented frequence shift value of 10.22999999543 MHz
as the basis, the best matching RelEffect number (ie. a more accurate value
for the famous value "38 us/day") seems to be one of the following
values depending on what the GPS engineers used for DaySeconds,
ie. whether they used Sideral Day (86164.091 s) or the normal 86400 s:

GPS frequency shifting algorithm:
RelEffect s/day = 0.000038597067448680351906158357771260997067448680 35

You do not have 50 significant figures of accuracy.

Maybe [...]

There is no "maybe".

"Eric Gisse" wrote
On Aug 7, 12:48 am, "qbit" wrote:
"Eric Gisse" wrote

On Aug 7, 12:05 am, "qbit" wrote:
"qbit" wrote

Ok, finally I've managed to find the formula for calculating
the GPS frequency shift (see below).

Using the the documented frequence shift value of 10.22999999543 MHz
as the basis, the best matching RelEffect number (ie. a more accurate value
for the famous value "38 us/day") seems to be one of the following
values depending on what the GPS engineers used for DaySeconds,
ie. whether they used Sideral Day (86164.091 s) or the normal 86400 s:

GPS frequency shifting algorithm:
RelEffect s/day = 0.000038597067448680351906158357771260997067448680 35

You do not have 50 significant figures of accuracy.

Maybe [...]

There is no "maybe".

You must be sick man! Why have you cut my reply?
A dumb message manipulator you are!
My reply was:

Maybe on the GPS computers not, but I can use any number
of significant decimal digits for accuracy; it's only up to available memory.
Actually internally I even used 100 :-), only the output was limited to 50.
I can choose whatever I like, even 10000 :-)

fup set

On Aug 7, 1:16 am, "qbit" wrote:
"Eric Gisse" wrote



On Aug 7, 12:48 am, "qbit" wrote:
"Eric Gisse" wrote

On Aug 7, 12:05 am, "qbit" wrote:
"qbit" wrote

Ok, finally I've managed to find the formula for calculating
the GPS frequency shift (see below).

Using the the documented frequence shift value of 10.22999999543 MHz
as the basis, the best matching RelEffect number (ie. a more accurate value
for the famous value "38 us/day") seems to be one of the following
values depending on what the GPS engineers used for DaySeconds,
ie. whether they used Sideral Day (86164.091 s) or the normal 86400 s:

GPS frequency shifting algorithm:
RelEffect s/day = 0.000038597067448680351906158357771260997067448680 35

You do not have 50 significant figures of accuracy.

Maybe [...]

There is no "maybe".

You must be sick man! Why have you cut my reply?
A dumb message manipulator you are!
My reply was:

Maybe on the GPS computers not, but I can use any number
of significant decimal digits for accuracy; it's only up to available memory.

Just as wrong as when you said it.

You might be able to get enough precision on a computer to _get_ that
many digits of accuracy but the number is _NOT_ correct to 50
significant figures.

Actually internally I even used 100 :-), only the output was limited to 50.
I can choose whatever I like, even 10000 :-)

fup set