Den 13.11.2023 20:36, skrev Lou:
> On Monday, 13 November 2023 at 18:36:49 UTC, Paul B. Andersen wrote:
>> Den 13.11.2023 14:36, skrev Lou:
>>>
>>> In your website table you have the following data.
>>>
>>> r rate
>>> ----------------------
>>> 10R 1.000000000626
>>> 9R 1.000000000618
>>> 8R 1.000000000608
>>> 7R 1.000000000596
>>> 6R 1.000000000579
>>> 5R 1.000000000556
>>> 4R 1.000000000522
>>> 3R 1.000000000464
>>> 2R 1.000000000348
>>> 1R 1.000000000000
>>>
>>> https://paulba.no/temp/ClockRate.pdf

The headline:
Gravitational blue shift of EM radiation from a stationary
clock in the non rotating Earth centred frame of reference.

>> Read the statement above the table:

Read it again:

>> “r" is the distance from the clock to the centre of the Earth.
>> "R" is the radius of the Earth.
>> The clock transmits a frequency 1 Hz,
>> "rate" is the frequency received on the ground.
>>
>> Gravitational Doppler shift, Gravitational blue shift.

>>> How would you calculate what frequency the rest frame
>>> (1R) c-133 atom frequency of 9192661770 hz would be at 2R?

>> It can't be "calculated".
>> The frequency of the photon associated with
>> the ground-state hyperfine transition of
>> the cesium-133 atom is 9192631770 Hz by definition.
>> Everywhere. Always.
>>
>> Fact! Nothing to discuss.
>>

> Yes I realise that under GR the rule is c-133 frequency never changes.
> Under GR its always everywhere the same.

Not "under GR". In the real world. Everywhere. Always. By definition!

> But you say a broadcast
> frequency can appear to be blueshifted to an observer on the ground
> and I’m trying to find out what that blueshifted f would be to an
> earth ground observer.

Not "can appear to be". IS blue shifted. Measured in the real world.

https://paulba.no/paper/Pound&Rebka.pdf

If the receiver is above the sender, it is gravitational red shift.

> So if it were possible to observe, from the ground, a clock orbiting at 2R.

Not orbiting! Stationary in the ECI frame at 2R.

> What frequency would 9192631770 Hz on a clock at 2R be blueshifted to
> an observer on the ground be? Would it appear to a ground observer
> to be 9192631773.48hz ?

The frequency of the received signal on the ground would
be gravitational blue shifted. That's a kind of Doppler shift.

That you hear the train-whistle change frequency when
the train passes you doesn't mean that the emitted
frequency has changed.

f_received = 1.000000000348⋅f_transmitted = 9192631773.1990 Hz

BECAUSE f_transmitted = 9192631770 Hz.

--
Paul

https://paulba.no/

On Tuesday, 14 November 2023 at 12:00:48 UTC+1, Paul B. Andersen wrote:

> > Yes I realise that under GR the rule is c-133 frequency never changes.
> > Under GR its always everywhere the same.
> Not "under GR". In the real world. Everywhere. Always. By definition!

Do you think sharks will start eating grass
if some idiot (just like you) define a shark as
4-leg grass eater? Definitions have a power,
indeed, but not that power, poor halfbrain,
sorry.

On Tuesday, November 14, 2023 at 12:10:16 AM UTC-5, Richard Hertz wrote:
> On Monday, November 13, 2023 at 3:36:49 PM UTC-3, Paul B. Andersen wrote:

> First, you pollute this thread arguing nonsense around my OP.
> Now you introduce more crap, playing with Einstein's 1911 crap. Why don't you take a rest?
> You introduce gravitational time dilation playing with 1911 Einstein's shit:

Off topic.

Stick to the OP topic.
Oh, true, you ran away from it. We have cornered you and you havent presented any counterargument whatsoever.
You havent answered one question psoed to you and havent answered one clarification posed to you.
You just run away or change subject. Now, isn't that the behavior of a crank?
Yes...you realize now you have become one...
Perhaps old age has crept up on you, but something is definitely wrong with you.
Even if you dont believe it, look into it.

> > Yes I realise that under GR the rule is c-133 frequency never changes.
> > Under GR its always everywhere the same.
> Not "under GR". In the real world. Everywhere. Always. By definition!
> > But you say a broadcast
> > frequency can appear to be blueshifted to an observer on the ground
> > and I’m trying to find out what that blueshifted f would be to an
> > earth ground observer.
> Not "can appear to be". IS blue shifted. Measured in the real world.
>
> https://paulba.no/paper/Pound&Rebka.pdf
>
> If the receiver is above the sender, it is gravitational red shift.
> > So if it were possible to observe, from the ground, a clock orbiting at 2R.
> Not orbiting! Stationary in the ECI frame at 2R.
> > What frequency would 9192631770 Hz on a clock at 2R be blueshifted to
> > an observer on the ground be? Would it appear to a ground observer
> > to be 9192631773.48hz ?
> The frequency of the received signal on the ground would
> be gravitational blue shifted. That's a kind of Doppler shift.
>
> That you hear the train-whistle change frequency when
> the train passes you doesn't mean that the emitted
> frequency has changed.
>
> f_received = 1.000000000348⋅f_transmitted = 9192631773.1990 Hz
>
> BECAUSE f_transmitted = 9192631770 Hz.
>
>
Interesting thanks. But are you sure it’s correct to convert the gps signal
frequency gain of 348 pico seconds at 10.23 Mhz to the gain of 3.19 Hz
for the recieved C-133 frequency using that method?

Because if I try your same formula for the transmitted f of 10.23MHz, I
get the following:
f_received = 1.000000000348 x 10.23Mhz =10230000.0036 hz
I would have thought it should be 10230000.0038 hz.

Lou wrote:

> Interesting thanks. But are you sure it’s correct to convert the gps
> signal frequency gain of 348 pico seconds at 10.23 Mhz to the gain of
> 3.19 Hz for the recieved C-133 frequency using that method?
> Because if I try your same formula for the transmitted f of 10.23MHz, I
> get the following:
> f_received = 1.000000000348 x 10.23Mhz =10230000.0036 hz I would have
> thought it should be 10230000.0038 hz.

that .0038 makes no sense. Ie, a photon with a frequency of 1 Hz, would
have a wavelength of 3E8 m, long radiowave already. GPS sats are closer
than this distance. (3E8/.0038 ~ 79E9). Please reconsider. Think on size
and accuracy of the antenna.

Lou wrote:

> Interesting thanks. But are you sure it’s correct to convert the gps
> signal frequency gain of 348 pico seconds at 10.23 Mhz to the gain of
> 3.19 Hz for the recieved C-133 frequency using that method?
> Because if I try your same formula for the transmitted f of 10.23MHz, I
> get the following:
> f_received = 1.000000000348 x 10.23Mhz =10230000.0036 hz I would have
> thought it should be 10230000.0038 hz.

that .0038 makes no sense. Ie, a photon with a frequency of 1 Hz, would
have a wavelength of 3E8 m, long radiowave already. GPS sats are closer
than this distance. (3E8/.0038 ~ 79E9). Please reconsider. Think on size
and accuracy of the antenna.

Anal-lena Baerbock's Grandfather was 𝗮 𝗡𝗮𝘇𝗶 𝗰𝗼𝗺𝗽𝗹𝗶𝗰𝗶𝘁 𝗶𝗻 𝗸𝗶𝗹𝗹𝗶𝗻𝗴 𝗥𝘂𝘀𝘀𝗶𝗮𝗻 𝗖𝗶𝘃𝗶𝗹𝗶𝗮𝗻𝘀
and she wants to go to war against Russia again - this nasty Slattern
should read some history and find out what Russian Soldiers 𝗱𝗶𝗱 𝘁𝗼 𝗽𝗲𝗼𝗽𝗹𝗲
𝗹𝗶𝗸𝗲 𝗵𝗲𝗿.

Den 14.11.2023 15:01, skrev Lou:

Who wrote what? DON'T SNIP THE ATTRIBUTIONS!

>>> Yes I realise that under GR the rule is c-133 frequency never changes.
>>> Under GR its always everywhere the same.

>> Not "under GR". In the real world. Everywhere. Always. By definition!

>>> But you say a broadcast
>>> frequency can appear to be blueshifted to an observer on the ground
>>> and I’m trying to find out what that blueshifted f would be to an
>>> earth ground observer.

>> Not "can appear to be". IS blue shifted. Measured in the real world.
>>
>> https://paulba.no/paper/Pound&Rebka.pdf
>>
>> If the receiver is above the sender, it is gravitational red shift.

>>> So if it were possible to observe, from the ground, a clock orbiting at 2R.
>> Not orbiting! Stationary in the ECI frame at 2R.
>>> What frequency would 9192631770 Hz on a clock at 2R be blueshifted to
>>> an observer on the ground be? Would it appear to a ground observer
>>> to be 9192631773.48hz ?

>> The frequency of the received signal on the ground would
>> be gravitational blue shifted. That's a kind of Doppler shift.
>>
>> That you hear the train-whistle change frequency when
>> the train passes you doesn't mean that the emitted
>> frequency has changed.
>>
>> f_received = 1.000000000348⋅f_transmitted = 9192631773.1990 Hz
>>
>> BECAUSE f_transmitted = 9192631770 Hz.
>>

> Interesting thanks. But are you sure it’s correct to convert the gps signal
> frequency gain of 348 pico seconds at 10.23 Mhz to the gain of 3.19 Hz
> for the recieved C-133 frequency using that method?

This statement is meaningless.

A GPS satellite is at r = 4.187883R
and is moving in the ECI frame at the speed 3.874 km/s.

https://paulba.no/pdf/GPS_clock_rate.pdf

>
> Because if I try your same formula for the transmitted f of 10.23MHz, I
> get the following:
> f_received = 1.000000000348 x 10.23Mhz =10230000.0036 hz

A more precise value is:
f_received = 1.000000000347674 x 10.23Mhz = 10230000.0035567 Hz

> I would have thought it should be 10230000.0038 hz.

OK, so that's what you believe.

--
Paul

https://paulba.no/

On Monday, November 13, 2023 at 9:10:16 PM UTC-8, Richard Hertz wrote:
> snip fresh imbecilities

Do you need a shovel?

On Tuesday, 14 November 2023 at 14:38:15 UTC+1, rotchm wrote:

> Oh, true, you ran away from it. We have cornered you and you havent presented any counterargument whatsoever.
> You havent answered one question psoed to you and havent answered one clarification posed to you.
> You just run away or change subject. Now, isn't that the behavior of a crank?

Sure it is! Relativistic idiots behave exactly the
same way.

On Tuesday, 14 November 2023 at 19:33:59 UTC, Paul B. Andersen wrote:
> Den 14.11.2023 15:01, skrev Lou:
>
> Who wrote what? DON'T SNIP THE ATTRIBUTIONS!
> >>> Yes I realise that under GR the rule is c-133 frequency never changes..
> >>> Under GR its always everywhere the same.
>
> >> Not "under GR". In the real world. Everywhere. Always. By definition!
>
> >>> But you say a broadcast
> >>> frequency can appear to be blueshifted to an observer on the ground
> >>> and I’m trying to find out what that blueshifted f would be to an
> >>> earth ground observer.
>
> >> Not "can appear to be". IS blue shifted. Measured in the real world.
> >>
> >> https://paulba.no/paper/Pound&Rebka.pdf
> >>
> >> If the receiver is above the sender, it is gravitational red shift.
>
> >>> So if it were possible to observe, from the ground, a clock orbiting at 2R.
> >> Not orbiting! Stationary in the ECI frame at 2R.
> >>> What frequency would 9192631770 Hz on a clock at 2R be blueshifted to
> >>> an observer on the ground be? Would it appear to a ground observer
> >>> to be 9192631773.48hz ?
>
> >> The frequency of the received signal on the ground would
> >> be gravitational blue shifted. That's a kind of Doppler shift.
> >>
> >> That you hear the train-whistle change frequency when
> >> the train passes you doesn't mean that the emitted
> >> frequency has changed.
> >>
> >> f_received = 1.000000000348⋅f_transmitted = 9192631773.1990 Hz
> >>
> >> BECAUSE f_transmitted = 9192631770 Hz.
> >>
>
> > Interesting thanks. But are you sure it’s correct to convert the gps signal
> > frequency gain of 348 pico seconds at 10.23 Mhz to the gain of 3.19 Hz
> > for the recieved C-133 frequency using that method?
> This statement is meaningless.
>
> A GPS satellite is at r = 4.187883R
> and is moving in the ECI frame at the speed 3.874 km/s.
>
> https://paulba.no/pdf/GPS_clock_rate.pdf
> >
> > Because if I try your same formula for the transmitted f of 10.23MHz, I
> > get the following:
> > f_received = 1.000000000348 x 10.23Mhz =10230000.0036 hz
> A more precise value is:
> f_received = 1.000000000347674 x 10.23Mhz = 10230000.0035567 Hz
> > I would have thought it should be 10230000.0038 hz.
> OK, so that's what you believe.

That was a typo. It should have been 10230000.00348hz.
But you have clarified that point thanks. It is as you
say10230000.0035567 Hz

Using your formula I get approximately 10230000.0054hz
recieved on earth for a gps signal of 10.23 Mhz from 4.12 R.
I can’t find any reference to confirm this. Do you know if that
approx frequency is observed at earth ground stations?

Den 15.11.2023 14:29, skrev Lou:
> On Tuesday, 14 November 2023 at 19:33:59 UTC, Paul B. Andersen wrote:
>>
>> A GPS satellite is at r = 4.187883R
>> and is moving in the ECI frame at the speed 3.874 km/s.

You can find the answer to your question below here.

>> https://paulba.no/pdf/GPS_clock_rate.pdf

>
> Using your formula I get approximately 10230000.0054hz
> recieved on earth for a gps signal of 10.23 Mhz from 4.12 R.
> I can’t find any reference to confirm this. Do you know if that
> approx frequency is observed at earth ground stations?

The received frequency from a stationary satellite
at GPS altitude which transmits the frequency 10.23 MHz
would be 10.230000005408 MHz.
But a GPS satellite is moving, so the frequency received from
a satellite in GPS orbit which transmits the frequency 10.23 MHz
would be 10.230000004567 MHz.

It is however impossible to directly measure this
frequency on the ground, because the Doppler shift due
to the orbital speed of the satellite is order of
magnitudes greater than the shift due to GR, and it
is changing all the time.

But the ground stations can read what the satellite clock shows,
and can compare it to clocks on the ground (GPS-time, UTC).
The difference will increase with time.

This was done in the first GPS-satellite with a clock
which was not corrected for relativistic effects.

https://paulba.no/paper/Initial_results_of_GPS_satellite_1977.pdf

--
Paul

https://paulba.no/

On Wednesday, November 15, 2023 at 3:20:28 PM UTC-3, Paul B. Andersen wrote:

<snip>

> But a GPS satellite is moving, so the frequency received from
> a satellite in GPS orbit which transmits the frequency 10.23 MHz
> would be 10.230000004567 MHz.
>
> It is however impossible to directly measure this frequency on the ground, because the Doppler shift due
> to the orbital speed of the satellite is order of magnitudes greater than the shift due to GR, and it
> is changing all the time.
>
> But the ground stations can read what the satellite clock shows, and can compare it to clocks on the ground (GPS-time, UTC).
> The difference will increase with time.

<snip>

Relativist, you are so full of shit that it's extremely disgusting.

So, relativistic effects can't be measured in any way because Doppler effect bury them 1,000 times under its values.

But EVERYONE have to trust you about the readings of the ground clock, because YOU HAD A JOB measuring it.

LIAR, DECEIVER, HYPOCRITE!

Den 15.11.2023 21:53, skrev Richard Hertz:
> On Wednesday, November 15, 2023 at 3:20:28 PM UTC-3, Paul B. Andersen wrote:
>
> <snip>
>
>> But a GPS satellite is moving, so the frequency received from
>> a satellite in GPS orbit which transmits the frequency 10.23 MHz
>> would be 10.230000004567 MHz.
>>
>> It is however impossible to directly measure this frequency on the ground, because the Doppler shift due
>> to the orbital speed of the satellite is order of magnitudes greater than the shift due to GR, and it
>> is changing all the time.
>>
>> But the ground stations can read what the satellite clock shows, and can compare it to clocks on the ground (GPS-time, UTC).
>> The difference will increase with time.

>> https://paulba.no/pdf/GPS_clock_rate.pdf

>
> Relativist, you are so full of shit that it's extremely disgusting.
>
> So, relativistic effects can't be measured in any way because Doppler effect bury them 1,000 times under its values.

The satellite is moving, the ground is moving, so the Doppler
shift of the carrier signals from a satellite will be anything
in the range (1 ± 1E-7). The receiver will obviously not measure
this frequency. Why should it?
The bandwidth of the receiver is bigger that that.
Receivers can receive the signals of up to 12 satellites at
the same time, and all the signals will be differently Doppler shifted.
They are separated by their PRN code. Phase locked loops.

Didn't you know this, Richard?
Did you believe that the receiver could measure the frequency
to a precision of 1E-12?

>
> But EVERYONE have to trust you about the readings of the ground clock, because YOU HAD A JOB measuring it.
>
> LIAR, DECEIVER, HYPOCRITE!
>

Mind the blood pressure, Richard. :-D

--
Paul, enjoying Richard's helpless rage

https://paulba.no/

Den 15.11.2023 22:34, skrev Paul B. Andersen:
> Den 15.11.2023 21:53, skrev Richard Hertz:
>> On Wednesday, November 15, 2023 at 3:20:28 PM UTC-3, Paul B. Andersen
>> wrote:
>>
>> <snip>
>>
>>> But a GPS satellite is moving, so the frequency received from
>>> a satellite in GPS orbit which transmits the frequency 10.23 MHz
>>> would be 10.230000004567 MHz.
>>>
>>> It is however impossible to directly measure this frequency on the
>>> ground, because the Doppler shift due
>>> to the orbital speed of the satellite is order of magnitudes greater
>>> than the shift due to GR, and it
>>> is changing all the time.
>>>
>>> But the ground stations can read what the satellite clock shows, and
>>> can compare it to clocks on the ground (GPS-time, UTC).
>>> The difference will increase with time.

This was the link I meant to send:

https://paulba.no/paper/Initial_results_of_GPS_satellite_1977.pdf

>
>
>>
>> Relativist, you are so full of shit that it's extremely disgusting.
>>
>> So, relativistic effects can't be measured in any way because Doppler
>> effect bury them 1,000 times under its values.
>
> The satellite is moving, the ground is moving, so the Doppler
> shift of the carrier signals from a satellite will be anything
> in the range (1 ± 1E-7). The receiver will obviously not measure
>  this frequency. Why should it?
> The bandwidth of the receiver is bigger that that.
> Receivers can receive the signals of up to 12 satellites at
> the same time, and all the signals will be differently Doppler shifted.
> They are separated by their PRN code. Phase locked loops.
>
> Didn't you know this, Richard?
> Did you believe that the receiver could measure the frequency
> to a precision of 1E-12?
>
>>
>> But EVERYONE have to trust you about the readings of the ground clock,
>> because YOU HAD A JOB measuring it.
>>
>> LIAR, DECEIVER, HYPOCRITE!
>>
>
> Mind the blood pressure, Richard. :-D
>

--
Paul, still enjoying Richard's helpless rage

https://paulba.no/

On Wednesday, 15 November 2023 at 19:20:28 UTC+1, Paul B. Andersen wrote:

> This was done in the first GPS-satellite with a clock
> which was not corrected for relativistic effects.

Of course, Paul is a piece of lying shit and this
first GPS satellite was made perfectly proper way,
exactly as it should be according to The Shit.

On Wednesday, 15 November 2023 at 18:20:28 UTC, Paul B. Andersen wrote:
> Den 15.11.2023 14:29, skrev Lou:
> > On Tuesday, 14 November 2023 at 19:33:59 UTC, Paul B. Andersen wrote:
> >>
> >> A GPS satellite is at r = 4.187883R
> >> and is moving in the ECI frame at the speed 3.874 km/s.
> You can find the answer to your question below here.
>
> >> https://paulba.no/pdf/GPS_clock_rate.pdf
> >
> > Using your formula I get approximately 10230000.0054hz
> > recieved on earth for a gps signal of 10.23 Mhz from 4.12 R.
> > I can’t find any reference to confirm this. Do you know if that
> > approx frequency is observed at earth ground stations?
> The received frequency from a stationary satellite
> at GPS altitude which transmits the frequency 10.23 MHz
> would be 10.230000005408 MHz.
> But a GPS satellite is moving, so the frequency received from
> a satellite in GPS orbit which transmits the frequency 10.23 MHz
> would be 10.230000004567 MHz.
>
> It is however impossible to directly measure this
> frequency on the ground, because the Doppler shift due
> to the orbital speed of the satellite is order of
> magnitudes greater than the shift due to GR, and it
> is changing all the time.
>
> But the ground stations can read what the satellite clock shows,
> and can compare it to clocks on the ground (GPS-time, UTC).
> The difference will increase with time.
>
> This was done in the first GPS-satellite with a clock
> which was not corrected for relativistic effects.
>
> https://paulba.no/paper/Initial_results_of_GPS_satellite_1977.pdf
>

Interesting facts thanks for clarifying things. Still a bit of discrepancy
though.The current NIST reference says the gps sat clock f is designated
as 10229999.99543Hz.
If I assume gps at 4.12R ** then it should have an approx picosecond gain
of 1.000000000542s
So using your formula I get:
10229999.99543 × 1.000000000542 = 10230000.001hz.
Small error but maybe still significant.
To get exactly 10.23 Mhz for GR clock gains at receiver I need a calculation
using closer to a 3R altitude clock gain of only 450 picoseconds.
Not 542 picoseconds

10229999.99543 × 1.00000000045 = 10.23 Mhz.

That’s equivelent to one whole radius r distance based on your table
below

As you can see below the sat clock f setting isn’t
matching GR predictions. Is that because the sat f
also has to take account the various Doppler speeds
you mentioned? I would have thought that the average
Doppler shift over time would be zero seeing as 1/2 the
time it’s(gps sat) coming towards reciever, 1/2 time away.

**
10R 1.000000000626 0.0100
9R 1.000000000618 0.0123
8R 1.000000000608 0.0156
7R 1.000000000596 0.0204
6R 1.000000000579 0.0278
5R 1.000000000556 0.0400
4.12 1.000000000542. *
4R 1.000000000522 0.0625
3R 1.000000000464 0.1111 *
2R 1.000000000348 0.2500
1R 1.000000000000 1.0000
https://paulba.no/temp/ClockRate.pdf

Den 16.11.2023 13:58, skrev Lou:
> On Wednesday, 15 November 2023 at 18:20:28 UTC, Paul B. Andersen wrote:
>> Den 15.11.2023 14:29, skrev Lou:
>>>
>>> Using your formula I get approximately 10230000.0054hz
>>> recieved on earth for a gps signal of 10.23 Mhz from 4.12 R.
>>> I can’t find any reference to confirm this. Do you know if that
>>> approx frequency is observed at earth ground stations?

>> The received frequency from a stationary satellite
>> at GPS altitude which transmits the frequency 10.23 MHz
>> would be 10.230000005408 MHz.

As I told you and you have ignored:
THE SATELLITE IS MOVING IN THE ECI-FRAME
========================================
>> But a GPS satellite is moving, so the frequency received from
>> a satellite in GPS orbit which transmits the frequency 10.23 MHz
>> would be 10.230000004567 MHz.

Note that I didn't say a GPS satellite, I said a satellite
in GPS orbit which transmits a 10.23 Mhz signal.
Then the signal received on the ground would be:
f = (1 + 4.4647E-10)⋅10.23 MHz = 10.230000004567389 MHz

>
> Interesting facts thanks for clarifying things. Still a bit of discrepancy
> though.The current NIST reference says the gps sat clock f is designated
> as 10229999.99543Hz.

Yes, the GPS satellite is NOT transmitting a 10.23 MHz signal.
The frequency is adjusted (the famous "GR-correction") to:
f₀ = (1 - 4.4647E-10)⋅10.23 MHz = 10.229999995432612 MHz

This frequency would observed on the ground be:
f₁ = (1 + 4.4647E-10)⋅10.229999995432612 MHz = 10.2300000000000000 MHz

(The frequency f₀ isn't transmitted at all, but it is used as reference
for the satellite clock, to make it stay in sync with GPS-time (UTC).)

> If I assume gps at 4.12R ** then it should have an approx picosecond gain
> of 1.000000000542s
> So using your formula I get:
> 10229999.99543 × 1.000000000542 = 10230000.001hz.

The satellite is moving!!!!
https://paulba.no/pdf/GPS_clock_rate.pdf
equation (8)

Read it!

> Small error but maybe still significant.
> To get exactly 10.23 Mhz for GR clock gains at receiver I need a calculation
> using closer to a 3R altitude clock gain of only 450 picoseconds.
> Not 542 picoseconds
>
> 10229999.99543 × 1.00000000045 = 10.23 Mhz.
>
> That’s equivelent to one whole radius r distance based on your table
> below
>
> As you can see below the sat clock f setting isn’t
> matching GR predictions. Is that because the sat f
> also has to take account the various Doppler speeds
> you mentioned? I would have thought that the average
> Doppler shift over time would be zero seeing as 1/2 the
> time it’s(gps sat) coming towards reciever, 1/2 time away.
>
> **
> 10R 1.000000000626 0.0100
> 9R 1.000000000618 0.0123
> 8R 1.000000000608 0.0156
> 7R 1.000000000596 0.0204
> 6R 1.000000000579 0.0278
> 5R 1.000000000556 0.0400
> 4.12 1.000000000542. *
> 4R 1.000000000522 0.0625
> 3R 1.000000000464 0.1111 *
> 2R 1.000000000348 0.2500
> 1R 1.000000000000 1.0000
> https://paulba.no/temp/ClockRate.pdf

The GPS-satellites are moving!
================================

Got it now?

--
Paul

https://paulba.no/

On Thursday, 16 November 2023 at 19:19:09 UTC, Paul B. Andersen wrote:
> Den 16.11.2023 13:58, skrev Lou:
> > On Wednesday, 15 November 2023 at 18:20:28 UTC, Paul B. Andersen wrote:
> >> Den 15.11.2023 14:29, skrev Lou:
> >>>
> >>> Using your formula I get approximately 10230000.0054hz
> >>> recieved on earth for a gps signal of 10.23 Mhz from 4.12 R.
> >>> I can’t find any reference to confirm this. Do you know if that
> >>> approx frequency is observed at earth ground stations?
>
> >> The received frequency from a stationary satellite
> >> at GPS altitude which transmits the frequency 10.23 MHz
> >> would be 10.230000005408 MHz.
> As I told you and you have ignored:
> THE SATELLITE IS MOVING IN THE ECI-FRAME
> =======================================> >> But a GPS satellite is moving, so the frequency received from
> >> a satellite in GPS orbit which transmits the frequency 10.23 MHz
> >> would be 10.230000004567 MHz.
> Note that I didn't say a GPS satellite, I said a satellite
> in GPS orbit which transmits a 10.23 Mhz signal.
> Then the signal received on the ground would be:
> f = (1 + 4.4647E-10)⋅10.23 MHz = 10.230000004567389 MHz
> >
> > Interesting facts thanks for clarifying things. Still a bit of discrepancy
> > though.The current NIST reference says the gps sat clock f is designated
> > as 10229999.99543Hz.
> Yes, the GPS satellite is NOT transmitting a 10.23 MHz signal.

Yes. I just told you it was 10.229999995432612 MHz I got that from NIST.

> The frequency is adjusted (the famous "GR-correction") to:
> f₀ = (1 - 4.4647E-10)⋅10.23 MHz = 10.229999995432612 MHz
>
> This frequency would observed on the ground be:
> f₁ = (1 + 4.4647E-10)⋅10.229999995432612 MHz = 10.23 Mhz
>
Where did you get 1 + 4.4647E-10 ?
That’s only 3R in your table.
According to you and your table if the gps satellite is at 4.12 R, then
your “famous correction” from broadcast to receive should include approx
5.42 Not 4.46. At 4.12 R the correction formula should be, *according
to you*:
f₁ = (1 + 5.52E-10)⋅10.229999995432612 MHz = 10230000.001hz
Notice it doesn’t work. There shouldn’t be an extra .001Hz
To correct that one needs to assume the satelitte is only at 3R to use
4.46 in your equation.
Which it isn’t. It’s at 4R.
You say that the reason why you pretend the satelitte is at 3 R is
“Because it is moving in the ECI frame”
Are you referring to kinematic velocity SR etc reducing the
total picoseconds gained from GR?

> (The frequency f₀ isn't transmitted at all, but it is used as reference
> for the satellite clock, to make it stay in sync with GPS-time (UTC).)
> > If I assume gps at 4.12R ** then it should have an approx picosecond gain
> > of 1.000000000542s
> > So using your formula I get:
> > 10229999.99543 × 1.000000000542 = 10230000.001hz.
> The satellite is moving!!!!
> https://paulba.no/pdf/GPS_clock_rate.pdf
> equation (8)
>
> Read it!
> > Small error but maybe still significant.
> > To get exactly 10.23 Mhz for GR clock gains at receiver I need a calculation
> > using closer to a 3R altitude clock gain of only 450 picoseconds.
> > Not 542 picoseconds
> >
> > 10229999.99543 × 1.00000000045 = 10.23 Mhz.
> >
> > That’s equivelent to one whole radius r distance based on your table
> > below
> >
> > As you can see below the sat clock f setting isn’t
> > matching GR predictions. Is that because the sat f
> > also has to take account the various Doppler speeds
> > you mentioned? I would have thought that the average
> > Doppler shift over time would be zero seeing as 1/2 the
> > time it’s(gps sat) coming towards reciever, 1/2 time away.
> >
> > **
> > 10R 1.000000000626 0.0100
> > 9R 1.000000000618 0.0123
> > 8R 1.000000000608 0.0156
> > 7R 1.000000000596 0.0204
> > 6R 1.000000000579 0.0278
> > 5R 1.000000000556 0.0400
> > 4.12 1.000000000542. *
> > 4R 1.000000000522 0.0625
> > 3R 1.000000000464 0.1111 *
> > 2R 1.000000000348 0.2500
> > 1R 1.000000000000 1.0000
> > https://paulba.no/temp/ClockRate.pdf
> The GPS-satellites are moving!
> ================================
Yes OK
“They are moving”. Could you elaborate.

On Thursday, 16 November 2023 at 20:19:09 UTC+1, Paul B. Andersen wrote:
> Den 16.11.2023 13:58, skrev Lou:
> > On Wednesday, 15 November 2023 at 18:20:28 UTC, Paul B. Andersen wrote:
> >> Den 15.11.2023 14:29, skrev Lou:
> >>>
> >>> Using your formula I get approximately 10230000.0054hz
> >>> recieved on earth for a gps signal of 10.23 Mhz from 4.12 R.
> >>> I can’t find any reference to confirm this. Do you know if that
> >>> approx frequency is observed at earth ground stations?
>
> >> The received frequency from a stationary satellite
> >> at GPS altitude which transmits the frequency 10.23 MHz
> >> would be 10.230000005408 MHz.
> As I told you and you have ignored:
> THE SATELLITE IS MOVING IN THE ECI-FRAME
> =======================================> >> But a GPS satellite is moving, so the frequency received from
> >> a satellite in GPS orbit which transmits the frequency 10.23 MHz
> >> would be 10.230000004567 MHz.
> Note that I didn't say a GPS satellite, I said a satellite
> in GPS orbit which transmits a 10.23 Mhz signal.

You didn't say but it is, measured both by
ground and satellite clocks.

> The satellite is moving!!!!

Was sure that you don't even believe this rrelative
motion idiocies.

Den 16.11.2023 21:16, skrev Lou:
> On Thursday, 16 November 2023 at 19:19:09 UTC, Paul B. Andersen wrote:
>> Den 16.11.2023 13:58, skrev Lou:
>>>
>>> Interesting facts thanks for clarifying things. Still a bit of discrepancy
>>> though.The current NIST reference says the gps sat clock f is designated
>>> as 10229999.99543Hz.

>> Yes, the GPS satellite is NOT transmitting a 10.23 MHz signal.

> Yes. I just told you it was 10.229999995432612 MHz I got that from NIST.
>

>> The frequency is adjusted (the famous "GR-correction") to:
>> f₀ = (1 - 4.4647E-10)⋅10.23 MHz = 10.229999995432612 MHz
>>
>> This frequency would observed on the ground be:
>> f₁ = (1 + 4.4647E-10)⋅10.229999995432612 MHz = 10.23 Mhz
>>

> Where did you get 1 + 4.4647E-10 ?

The correction is (1 - 4.4647E-10)
so f₁ = (1/(1 - 4.4647E-10))⋅f₀ ≈ (1 + 4.4647E-10)⋅f₀

The correction is specified in the
"INTERFACE SPECIFICATION DOCUMENT" for GPS:

https://www.gps.gov/technical/icwg/IS-GPS-200N.pdf

From 3.3.1.1 Frequency Plan:
"The carrier frequencies for the L1 and L2 signals shall be
coherently derived from a common frequency source within the SV.
The nominal frequency of this source -- as it appears to an observer
on the ground -- is 10.23 MHz. The SV carrier frequency and clock
rates -- as they would appear to an observer located in the SV --
are offset to compensate for relativistic effects. The clock rates
are offset by Δf/f = -4.4647E-10, equivalent to a change in the
P-code chipping rate of 10.23 MHz offset by a Δf = -4.5674E-3 Hz.
This is equal to 10.2299999954326 MHz."

SV = space vehicle, satellite.

<snip>

>>
>> The GPS-satellites are moving!
>> ================================
>>
>> Got it now?

Obviously not!

> “They are moving”. Could you elaborate.

I have given this link several times now:
https://paulba.no/pdf/GPS_clock_rate.pdf

The blue shift observed on the ground is given in equation (8).
READ IT!

Equation (8) is of the form:
dτ₂/dτ₁ = 1 + gravitational term + kinematic term

You keep ignoring the kinematic term.

--
Paul

https://paulba.no/

On Friday, 17 November 2023 at 12:10:04 UTC+1, Paul B. Andersen wrote:
> Den 16.11.2023 21:16, skrev Lou:
> > On Thursday, 16 November 2023 at 19:19:09 UTC, Paul B. Andersen wrote:
> >> Den 16.11.2023 13:58, skrev Lou:
> >>>
> >>> Interesting facts thanks for clarifying things. Still a bit of discrepancy
> >>> though.The current NIST reference says the gps sat clock f is designated
> >>> as 10229999.99543Hz.
>
> >> Yes, the GPS satellite is NOT transmitting a 10.23 MHz signal.
>
> > Yes. I just told you it was 10.229999995432612 MHz I got that from NIST..
> >
>
> >> The frequency is adjusted (the famous "GR-correction") to:
> >> f₀ = (1 - 4.4647E-10)⋅10.23 MHz = 10.229999995432612 MHz
> >>
> >> This frequency would observed on the ground be:
> >> f₁ = (1 + 4.4647E-10)⋅10.229999995432612 MHz = 10.23 Mhz
> >>
>
> > Where did you get 1 + 4.4647E-10 ?
> The correction is (1 - 4.4647E-10)
> so f₁ = (1/(1 - 4.4647E-10))⋅f₀ ≈ (1 + 4.4647E-10)⋅f₀
>
> The correction is specified in the
> "INTERFACE SPECIFICATION DOCUMENT" for GPS:
>
> https://www.gps.gov/technical/icwg/IS-GPS-200N.pdf
>
> From 3.3.1.1 Frequency Plan:
> "The carrier frequencies for the L1 and L2 signals shall be
> coherently derived from a common frequency source within the SV.
> The nominal frequency of this source -- as it appears to an observer
> on the ground -- is 10.23 MHz. The SV carrier frequency and clock
> rates -- as they would appear to an observer located in the SV --

Paul, poor halfbrain, who cares what "would appear" to a
non-existing gedanken person? It's the measurement result
that counts, andthis is 10.23, both measured with
ground or satellite clock.

> t' = γ (t - vx/c)

The relativistic velocity v is not constant.

On 11/17/2023 2:14 PM, Maciej Wozniak wrote:
> On Friday, 17 November 2023 at 12:10:04 UTC+1, Paul B. Andersen wrote:
>> Den 16.11.2023 21:16, skrev Lou:
>>> On Thursday, 16 November 2023 at 19:19:09 UTC, Paul B. Andersen wrote:
>>>> Den 16.11.2023 13:58, skrev Lou:
>>>>>
>>>>> Interesting facts thanks for clarifying things. Still a bit of discrepancy
>>>>> though.The current NIST reference says the gps sat clock f is designated
>>>>> as 10229999.99543Hz.
>>
>>>> Yes, the GPS satellite is NOT transmitting a 10.23 MHz signal.
>>
>>> Yes. I just told you it was 10.229999995432612 MHz I got that from NIST.
>>>
>>
>>>> The frequency is adjusted (the famous "GR-correction") to:
>>>> f₀ = (1 - 4.4647E-10)⋅10.23 MHz = 10.229999995432612 MHz
>>>>
>>>> This frequency would observed on the ground be:
>>>> f₁ = (1 + 4.4647E-10)⋅10.229999995432612 MHz = 10.23 Mhz
>>>>
>>
>>> Where did you get 1 + 4.4647E-10 ?
>> The correction is (1 - 4.4647E-10)
>> so f₁ = (1/(1 - 4.4647E-10))⋅f₀ ≈ (1 + 4.4647E-10)⋅f₀
>>
>> The correction is specified in the
>> "INTERFACE SPECIFICATION DOCUMENT" for GPS:
>>
>> https://www.gps.gov/technical/icwg/IS-GPS-200N.pdf
>>
>> From 3.3.1.1 Frequency Plan:
>> "The carrier frequencies for the L1 and L2 signals shall be
>> coherently derived from a common frequency source within the SV.
>> The nominal frequency of this source -- as it appears to an observer
>> on the ground -- is 10.23 MHz. The SV carrier frequency and clock
>> rates -- as they would appear to an observer located in the SV --
>
> Paul, poor halfbrain, who cares what "would appear" to a
> non-existing gedanken person?

NIST does, obviously, since they wrote that. They use "would appear"
since nobody is going to send an astronaut up to check, especially since
it was already set to 10.229999995432612 MHz on the ground before launch.

> It's the measurement result
> that counts, andthis is 10.23, both measured with
> ground or satellite clock.

So now it is you invoking the "would appear" since you are talking about
the satellite clock without an astronaut checking. But regardless of
whether there is an astronaut there or not, it still runs at
10.229999995432612 MHz because that is what it was set to run.

On Sunday, 19 November 2023 at 19:49:53 UTC+1, Volney wrote:
> On 11/17/2023 2:14 PM, Maciej Wozniak wrote:
> > On Friday, 17 November 2023 at 12:10:04 UTC+1, Paul B. Andersen wrote:
> >> Den 16.11.2023 21:16, skrev Lou:
> >>> On Thursday, 16 November 2023 at 19:19:09 UTC, Paul B. Andersen wrote:
> >>>> Den 16.11.2023 13:58, skrev Lou:
> >>>>>
> >>>>> Interesting facts thanks for clarifying things. Still a bit of discrepancy
> >>>>> though.The current NIST reference says the gps sat clock f is designated
> >>>>> as 10229999.99543Hz.
> >>
> >>>> Yes, the GPS satellite is NOT transmitting a 10.23 MHz signal.
> >>
> >>> Yes. I just told you it was 10.229999995432612 MHz I got that from NIST.
> >>>
> >>
> >>>> The frequency is adjusted (the famous "GR-correction") to:
> >>>> f₀ = (1 - 4.4647E-10)⋅10.23 MHz = 10.229999995432612 MHz
> >>>>
> >>>> This frequency would observed on the ground be:
> >>>> f₁ = (1 + 4.4647E-10)⋅10.229999995432612 MHz = 10.23 Mhz
> >>>>
> >>
> >>> Where did you get 1 + 4.4647E-10 ?
> >> The correction is (1 - 4.4647E-10)
> >> so f₁ = (1/(1 - 4.4647E-10))⋅f₀ ≈ (1 + 4.4647E-10)⋅f₀
> >>
> >> The correction is specified in the
> >> "INTERFACE SPECIFICATION DOCUMENT" for GPS:
> >>
> >> https://www.gps.gov/technical/icwg/IS-GPS-200N.pdf
> >>
> >> From 3.3.1.1 Frequency Plan:
> >> "The carrier frequencies for the L1 and L2 signals shall be
> >> coherently derived from a common frequency source within the SV.
> >> The nominal frequency of this source -- as it appears to an observer
> >> on the ground -- is 10.23 MHz. The SV carrier frequency and clock
> >> rates -- as they would appear to an observer located in the SV --
> >
> > Paul, poor halfbrain, who cares what "would appear" to a
> > non-existing gedanken person?
> NIST does, obviously, since they wrote that.

Too bad for NIST, stupid Mike.
Anyway, the measurement result gives a different number.

> > It's the measurement result
> > that counts, andthis is 10.23, both measured with
> > ground or satellite clock.
> So now it is you invoking the "would appear" since you are talking about
> the satellite clock without an astronaut checking. But regardless of
> whether there is an astronaut there or not, it still runs at
> 10.229999995432612 MHz

No, stupid Mike, it isn't,. I know it's the prediction of yourr
moronic Shit, but the measurement gives a different
result.