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Reviewing this recent post I have come to the conclusion that its Formula (1.2) needs some explanations and some rewording in order to be more explicit.

(1) - First of all, for a stubborn Kermit sticking to West Longitudes as positive, and with:

P = Body Polar Angle, Lat = (Observer's Latitude) ,and Dec = (Body Declination),

the Azimuth Az of a Celestial Body can be computed through Formula (1.1) of this same post :

Formula (1.1) : SinP(CosAz / Sin Az)  = TanDecCosLat - SinLatCosP

from which we have derived the following 1st order Formula (1.2.0) : dAz/dP= (SinAzCosAz/TanP - SinLatSin²Az)

Important note : The Polar Angle "P" is not a nicely continuously increasing variable as it is only "authorized" to vary within [0° - 180°], i.e. exactly like the Acos function.

This had to be implemented by lack of the 2 variable functions P→R and R→P then.

Such [0° - 180°] restriction is sufficient for all practical Tabular Azimuth determination purposes since for both functions - i.e. Polar Angle and Acos - it is [almost] always possible to check whether a Body is East or West of its Observer in order to fully clarify the Tabular Azimuth uncertainty between Az and -Az . E.g. each of the two Tables d'Azimut de Bataille uses one of these variables.

Nonetheless and for this adverse mathematical reason of constraining P within [0° - 180°], carrying out the differentiation here-above is subject to strict limitations which I have not investigated in detail.

(2) - We have seen that for an Observer watching 2 celestial Bodies A & B with their Position Angle PA measured at Body A, the PA angular rate dPA/dUT as seen from the [steady] Observer is exactly equal to the Observer's Azimuth angular rate as seen from Body A subastral point [moving] on the Earth Surface.

In other words, we are to replace the initial actual Observer (Lat, Lon) by a fictitious Observer (Dec, GHAγ-ARA) and the actual Celestial Body (Dec, GHAγ-ARA) by a fictitious Celestial Body (Lat, Lon).

Accordingly for all computation purposes and for this Fictitious Observer :

- His Fictitious Latitude is equal to the Real World Body Declination. And :

- The Fictitious Celestial Body P* is equal to P since this angle is constrained to stay inside [0° - 180°]. And finally:

- The Fictitious Celestial Body Declination* is equal to the Real World Observer's Latitude.

Applied to our Fictitious Observer, Formula (1.2.0) is : dAz*/dP*= (SinAz*CosAz*/TanP* - SinLat*Sin²Az*).

Through substitution between Lat* and Dec, and substitution between P* and P, Formula (1.2.0) then becomes :

dAz*/dP*= (SinAz*CosAz*/TanP - SinDecSin²Az*)

On the other hand, we have just been reminded hereabove that dAz*/dP* = dPA/dP .

We can then conclude through rewriting our initial Formula (1.2) as :

Formula (1.2 bis) : dPA/dUT = (SinAz*CosAz*/TanP - SinDecSin²Az*) dP/dUT 

In this reworded Formula (1.2 bis) all variables, except Az* have become our familiar ones applicable to the [real world] Observer.

Numerical example :

In the preparation of my last post to our current Alnair identification thread, I earlier computed the following values of the (Peacock - Alnair) Position Angles :

From S13°15'/W163°06.5' , on 01 Jan 2019, HoE = 9.5 ft,  T = +25°C, P=29.92 "Hg ,

and for unrefracted values with HoE = 0', I am obtaining :

At UT = 05h54m00.0s (Peacock - Alnair) PA1= 19.59917 ° , and

at UT = 05h58m00.0s (i.e. 4 minutes later) (Peacock - Alnair) PA2 = 20.44032 ° 

From which we deduce a PA angular rate equal to 15 * (PA2 - PA1) = 12.61725 °/h , or in other terms :

(1.3) : μPA = 12.61725 °/h

At mid-interval, i.e. at UT = 05h56m00.0s , let us use Formula (1.2 bis) with the following numerical values :

Peacock : Dec = -56.67486 °

As seen from Peacock : Observer's Pole Angle P =  Observer's Longitude - (GHAγ  - Peacock ARA)=163.10833 ° - (189.60035 ° - 306.76826 °) = - 79.72375° , which we are to convert into P = +79.72375° since P is constrained to stay inside [0° - 180°], while remembering that the Observer is East of Peacock.

From these data and through Formula (1.1) we obtain : Az* = 88.85325 °

Formula (1.2 bis) yields :  dPA / dP = 0.83886 and with : dGHAγ / dUT = 15.04107 °/h we finally obtain :

dPA / dUT = 12.61734 °/h , a result to which the approximate value given in (1.3) here-above  μPA = 12.61725 °/h compares quite well.

This probably concludes this Position Angles (PA) angular rates thread too.

Kermit