NavList:

In the past 40 years or so, I never worried that much about computing Azimuths in the right quadrants since the dyadic functions P→R and R→P  do it all perfectly. They also have the wonderful advantage that they never have "cautious" value such as the Tangent / Cotangent and their related functions.

So ... this [self-]thread has given me the opportunity of again "digging into the subject of Azimuths" as derived through the "old methods" i.e. through Tables such as these ones taught in the French Navy then and still nowadays.

In my last post on Position Angles angular rates, I wrote :

QUOTE

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.

UNQUOTE

Well ... I have meanwhile investigated in detail and the previously uncertain situation then seems fully clarified now.

The enclosed attachment depicts the relationship between the Polar Angle "P" and the Local Hour Angle "LHA" of a Celestial Body, with "LHA" , most often abriged into "T".

And most importantly :

this Attachment reminds us of the difference between the Tabular Azimuth "Z" obtained from derivation of a formula using the Arc-Cotangent function (Acot) defined over [0°-180°] and its real world counterpart being the True Azimuth "Az" covering the entire [0°-360°] range.

Once this study was completed, it has been possible to improve 2 formulae recently listed on this thread.

(1) - Improvement of earlier Azimuth Formula (1.1) from Z to Az

As earlier indicated, our well known general Tabular Azimuth formula obtained from position and time is :

Formula (1.1) : SinPCotZ = TanDCosLat - SinLatCosDec

As an improvement to Formula (1.1) here-above, the Attachment gives the True Azimuth "Az" through the following :

Formula (1.1 bis) : Az = - Acot [(TanDCosLat - SinLatCosDec)/SinP] Sign(SinT)

(2) - Improvemenet of earlier Formula (1.2.0) on True Azimuth Angular Rate dAz/dP 

Through previous - although incomplete as seen today - term by term differentiation,

we have derived the previous Formula 1.2.0 : dZ/dP = SinZCosZ/TanP - SinLatSin²Z.

As shown in the Attachment, Formula 1.2.0 is incomplete since it requires further refinements in order to be used onto the entire [0°-360°] range of the LHA values.

The Attachment brings remedy to this shortcoming through first noticing that the required quantity to look for should not be "dZ/dP" but "dAz/dT" instead since "T" continuously varies in the same direction overtime throughout the entire [0°-360°] domain, while "P" abruptly changes directions at 180°. In other words its derivative dP/dT abruptly changes from +1 to -1 when T crosses the 180°mark.

As a result, we now have at our disposal the following Azimuth Angular Rate formula valid for all configurations :

Formula (1.2.0 bis) :  dAz/dUT = [SinLatSin²Az + (SinAzCosAz/TanP) Sign(SinT)] dT/dUT 

Note : Given our previous remark about dP/dT here-above, we could as well replace "Sign(SinT)" by the quantity "dP/dT".

(2) - Improvement of earlier Formula (1.2 bis) on Position Angles Angular Rates

We have seen earlier that for an Observer watching 2 celestial Bodies A & B having a Position Angle "PA" measured at Body A, the PA angular rate dPA/dUT is exactly equal to the Azimuth angular rate dAz*/dUT of the Observer as seen from Body A.

A seen earlier, the " * " postcript refers to measurements from the Celestial Body.

Earlier Formula (1.2 bis) therefore needs to be modifed accordingly in order to cover the entire [0°-360°] T (or LHA) range.

Also since T* (as seen from the Celestial Body) = -T (as seen from the Observer), then Sign(SinT*) = -Sign(SinT).

From all this we can derive the following updated Position Angle (PA) Angular Rate formula valid for all configurations:

Formula (1.2 ter) : dPA/dUT = - [SinLDecSin²Az* + (SinAz*CosAz*/TanP)Sign(SinT)] dT/dUT 

In this updated Formula (1.2 ter) and as earlier - with the only exception of Az* - all the other variables including T are our familiar ones applicable to the [real world] Observer.

Note : Again, and given our previous remark about dP/dT here-above, we could as well replace "Sign(SinT)" by the quantity "dP/dT".

(3) - Numerical examples :

3.1 - Celestial Body WEST of the Observer, i.e. LHA within the [0°-180°] interval

Let's consider again our previous Peacock - Alnair example, while exactly following this time the precepts and descriptions here-above.

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 :

(3.1.1) : μPA = 12.61725 °/h

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

Peacock : Dec = -56.67486 °

As seen from Peacock :

T* =  Observer's Longitude - (GHAγ  - Peacock ARA) = 163.10833 ° - (189.60035 ° - 306.76826 °) = - 79.72375°.

Hence P* = +79.72375°

From these data and through Formula (1.1 bis) we obtain : Tabular Z* = 88.85325 °, hence real world Az* = 88.85325 °

Formula (1.2 ter) yields :  dPA / dP = 0.83886 and with : dT / dUT = 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.

3.2 - Celestial Body EAST of the Observer, i.e. LHA within the [180°-360°] interval

From Beautiful Tasmania S41°14'/E146°18' let's compute the angular rate of the Betelgeuse to Rigel line at 12:26 UT on 22 Nov 2023.

To first approximate such PA rate we choose to evenly bracket our reference UT with a 3 minute time span interval this time.

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

At UT = 12h24m30.0s (Betelgeuse to Rigel) PA1= 7.85532° ° , and

at UT = 12h27m30s (i.e. 3 minutes later) (Betelgeuse to Rigel) PA2 = 7.55359 ° 

From which we deduce a PA angular rate equal to 20 * (PA2 - PA1) = -6.03459 °/h , or in other terms :

(3.2.1) : μPA = -6.03459 °/h

At mid-interval, i.e. at UT = 12h26m00.0s , let us use Formula (1.2 ter) with the following numerical values :

Betelgeuse Dec = 7.41263 °

As seen from Betelgeuse :

T*  =  Observer's Longitude - (GHAγ  - Betelgeuse ARA) = -146.30000 - (247.73537 ° - 89.11952 °) = 55.08416 ° ,

Hence P* = 55.08416 °

From these data and through Formula (1.1 bis) we obtain : Tabular Z* = 138.99057 ° , hence real world Az* = 221.00943 °

Formula (1.2 ter) yields :  dPA / dT = - 0.40118 and with : dT / dUT = dGHAγ / dUT = 15.04107 °/h . And we finally obtain :

dPA / dUT = - 6.03418 °/h , a result to which the approximate value given in (3.2.1) here-above  μPA = -6.03459 °/h compares quite well again.

4 - OVERALL CONCLUSION

I think that we have eventually got it all ... this time.

Kermit