Henri Wilson wrote:
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


Does this mean that the BaTh predicts that clocks on
a higher gravitational potential will run faster?

Paul

On Aug 26, 3:50 am, "q-bit" wrote:
"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

Kookfite coming up !

"Paul B. Andersen" wrote
Henri Wilson wrote:
On Tue, 07 Aug 2007 11:45:54 GMT, "Androcles" wrote:
"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.

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


Does this mean that the BaTh predicts that clocks on
a higher gravitational potential will run faster?

As I see it, yes, obviously; one just needs to
reverse the direction in his calculations...

"Paul B. Andersen" wrote in message
...
: Henri Wilson wrote:
: 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
:
:
: Does this mean that the BaTh predicts that clocks on
: a higher gravitational potential will run faster?
:
: Paul

It means the kooks haven't be booked for cooking the books.

'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' because I SAY SO and you have to
agree because I'm the great genius, STOOOPID, don't you
dare question it. -- Albert Einstein,
who in 1895 failed an examination that would have allowed
him to study for a diploma as an electrical engineer at
the Eidgenössische Technische Hochschule in Zurich
(couldn't even pass the SATs).

http://www.androcles01.pwp.blueyonde...rt/tAB=tBA.gif
"Counterfactual assumptions yield nonsense.
If such a thing were actually observed, reliably and reproducibly, then
relativity would immediately need a major overhaul if not a complete
replacement." -- Tom Roberts.

On Aug 7, 4:05 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

Nobody ever taught you about significant figures, did they?
All but a few of those digits are meaningless, and can be
replaced with output of any random number generator without
making the number less correct.

Same for most of the rest of your numbers.

- Randy

Henri Wilson wrote:
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?

How so?
I bet you don't even know what the author is talking about
Where is the "automated bang-bang controller"?
Which frequency is adjusted with this controller?
I know the answer.
Do you?
If you read the document from where you quoted the above, you might find out.
(But probably not. It is not well explained.)
http://www.agi.com/downloads/support...k_Analysis.pdf
or:
http://tinyurl.com/yoru3y

I will however admit that the quoted part is easy to misunderstand.
(And you read it out of context, didn't you?)
It is indeed very confusing.
What the hell is "frequency-rate"? :-)

If you really want to learn what is adjusted with a "bang-bang controller", read this:
ftp://igscb.jpl.nasa.gov/igscb/resou...97tech_rpt.pdf

BTW, I can quote too:
http://tycho.usno.navy.mil/ptti/1995/Vol%2027_10.pdf

Because the GPS frequency standards are physically inaccessible,
reliability is very important for system integrity. Each GPS-satellite
contains four frequency standards (two cesium and two rubidium).
In order to meet the required mission lifetime of 7.5 years,
each of the four clocks should be expected to operate within
stability specifications for approximately two years. ...


This is among my "previous claims".
Does it look pretty stupid? :-)

I will throw you a straw, though.
The above is true for Block II and Block IIA satellites,
but is not strictly true for Block IIR satellites.
The first IIR was launched in 1997, so the GPS was operated
for 20 years with physically inaccessible frequency standards.
Now about half of the satellites are II/IIA and half are IIR.

Paul

"Paul B. Andersen" wrote in message
...
: Henri Wilson wrote:
: 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?
:
: How so?
: I bet you don't even know what the author is talking about
: Where is the "automated bang-bang controller"?
: Which frequency is adjusted with this controller?
: I know the answer.
: Do you?
: If you read the document from where you quoted the above, you might find
out.
: (But probably not. It is not well explained.)
:
http://www.agi.com/downloads/support...k_Analysis.pdf
: or:
: http://tinyurl.com/yoru3y
:
: I will however admit that the quoted part is easy to misunderstand.
: (And you read it out of context, didn't you?)
: It is indeed very confusing.
: What the hell is "frequency-rate"? :-)

Ticks per second per second, same units as 38 usecs per day.


: If you really want to learn what is adjusted with a "bang-bang
controller", read this:
: ftp://igscb.jpl.nasa.gov/igscb/resou...97tech_rpt.pdf
:
: BTW, I can quote too:
: http://tycho.usno.navy.mil/ptti/1995/Vol%2027_10.pdf
:
: Because the GPS frequency standards are physically inaccessible,
: reliability is very important for system integrity. Each GPS-satellite
: contains four frequency standards (two cesium and two rubidium).
: In order to meet the required mission lifetime of 7.5 years,
: each of the four clocks should be expected to operate within
: stability specifications for approximately two years. ...
:
:
: This is among my "previous claims".
: Does it look pretty stupid? :-)

Yes.

:
: I will throw you a straw, though.
: The above is true for Block II and Block IIA satellites,
: but is not strictly true for Block IIR satellites.
: The first IIR was launched in 1997, so the GPS was operated
: for 20 years with physically inaccessible frequency standards.
: Now about half of the satellites are II/IIA and half are IIR.
:
: Paul

On Sun, 26 Aug 2007 22:09:06 +0200, "Paul B. Andersen"
wrote:

Henri Wilson wrote:
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?

How so?
I bet you don't even know what the author is talking about
Where is the "automated bang-bang controller"?
Which frequency is adjusted with this controller?
I know the answer.
Do you?
If you read the document from where you quoted the above, you might find out.
(But probably not. It is not well explained.)
http://www.agi.com/downloads/support...k_Analysis.pdf
or:
http://tinyurl.com/yoru3y

I will however admit that the quoted part is easy to misunderstand.
(And you read it out of context, didn't you?)
It is indeed very confusing.
What the hell is "frequency-rate"? :-)

If you really want to learn what is adjusted with a "bang-bang controller", read this:
ftp://igscb.jpl.nasa.gov/igscb/resou...97tech_rpt.pdf

BTW, I can quote too:
http://tycho.usno.navy.mil/ptti/1995/Vol%2027_10.pdf

Because the GPS frequency standards are physically inaccessible,
reliability is very important for system integrity. Each GPS-satellite
contains four frequency standards (two cesium and two rubidium).
In order to meet the required mission lifetime of 7.5 years,
each of the four clocks should be expected to operate within
stability specifications for approximately two years. ...


This is among my "previous claims".
Does it look pretty stupid? :-)

I will throw you a straw, though.
The above is true for Block II and Block IIA satellites,
but is not strictly true for Block IIR satellites.
The first IIR was launched in 1997, so the GPS was operated
for 20 years with physically inaccessible frequency standards.
Now about half of the satellites are II/IIA and half are IIR.

Don't try to wriggle out. The article clearly states that the clock rates are
adjusted regularly in orbit. You have been wrong for as long as I can remember.

Please apologise for being such a fool.

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.