# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: Position Angles (PA) angular rates**

**From:**Antoine Couëtte

**Date:**2023 Nov 12, 12:24 -0800

I recently wrote : QUOTE " Our last topic on ** Alnair identification **gave me a very first opportunity to deal with Position Angles (

**PA**)

**" UNQUOTE.**

*angular rates*I was wondering whether there would exist some reasonably simple formula or 3D vector algorithm capable of computing such angular rate.

After devoting my research and efforts onto 3D vector solutions - to no avail until now - I ended up with an unexpected *mathematically exact* solution.

A close study of the attachment - handdrawn to best depict our initial Alnair identification puzzle - readily shows that** the Peacok-Alnair PA rate seen from the Observer is exactly equal to the Observer's changing Azimuth rate as seen from the Peacock sub-stellar point**. Such point is the intersection of the Earth surface with the straight line Peacock center to Earth center.

*******

**(1) - Let's first compute the Azimuth rate dAz/dUT of a celestial body** with equatorial coordinates (Dec, GHA) as seen from a [standard] Observer with given (Lat, Lon) coordinates

With *P = GHA - Lon* , our well known Azimuth Formula tells us :

**Formula (1.1) : sinP * (cosAz / sinAz) = tgDec * cosLat - sinLat*cosP**

1st order differentiaiton immediately gives : *dP * cosP * (cosAz / sin Az) + sinP * dAz * ( -1 / sin²Az) = sinLat * dP * sinP*

Start multiplying both sides by sin²Az, then divide both sides by sinP and get :* dP * (sinAz * cosAz / tgP ) = dP * sinLat * sin²Az*

Regroup terms and eventually obtain : *dAz/dP = (sinAz*cosAz/tgP - sinLat*sinLat*sin²Az).*

On the other hand we know that : *dAz/dUT = dAz/dP * dP/dUT *with *dP/dUT = +15.0410686 °/h*.

*With PA seen from Observer watching 2 Celestial Bodies and Za seen from the [lowest] Celestial Body towards the Observer*

we finally obtain :

**Formula (1.2) : dPA/dUT = (sinAz*cosAz/tgP - ****sinLat*sin²Az) * dP/dUT**

*Again Az is to be computed from Substellar point towards Observer.*** **

*******

**(2) - Let's compare ****our previous PA angular rate for UT = 05h54m**

**with the result to be obtained from Formula (1.2).**

**2.1 - Result from previous method**

In order to get a more meaningful comparison with ** Formula (1.2) **let us slightly refine our

*previous determination method*through 2 minor adjustments :

2.1.1 - Assume both no refraction and observer at sea-level.

2.1.2 - For an unchanged target UT at 05h54m, narrow the bracketing interval from +/- 15 min to +/- 5 min. And get:

**2.1.3 - At UT = 05h49m00.0s**

**Alnair **Height/Azimuth = 34.72672°/221.29267° and **Peacock **Height/Azimuth = 17.37065°/214.27908°, **PA**=18.48766°

**2.1.4 -** **At UT = 05h59m00.0s**

**Alnair **Height/Azimuth = 33.10401°/221.95652° and **Peacock** Height/Azimuth = 14.99405°/214.31987°, **PA**=20.59411°

2.1.5 - For a difference of 2.10644° over a 10 minute time span, we are deriving an **updated PA angular rate of 12.63878 °/h** (vs. our *earlier 12.6718 °/h *).

We should nonetheless remember the following :

2.1.5.1 - This updated "enhanced" PA rate should be significantly closer to our upcoming ** Formula (1.2) **determination. And :

2.1.5.2 - The following ** Formula (1.2)** determination is to be the best available one we can get since it is "

*mathematically*" exact.

**2.2 - Result from Formula (1.2) here-above computed for target UT = 05h54m**

*Once again Az and dAz/dUT are to be computed here from Substellar point towards Observer.*

Hence in **Formula (1)** , *Latitude* is to be taken as the *Body Declination*, and* P *is to be taken as : "*Lon - GHA" (vs. "GHA- Lon")*.

So for Peacok at UT = 05h54m with Dec = -56.67486° and GHA = 242.33072° aiming towards the Observer at Lat 13°S and Longitude 163° W, we get an Azimuth equal to 88.37609°

Entering** Formula (2)** with the appropriate values obtained just here-above yields a "true" mathematical PA rate :** dPA/dUT = 12.63799 °/h**, a value which differs from our previous "updated" approximation by les than 3" / h.

*******

**(3) - Conclusions**

3.1 - With **Formula (1.2)** a solid result has been obtained to solve our initial *PA angular rates* quizz. Most probably some similar result has already been earlier establishes if not published already. Therefore ** Formula (1.2) **is most likely an independent re-discovery.

3.2 -* I could not find any 3D vectorial algorithm yet which would have my strongest preference* since

**Formula (1.2)**, albeit a rigourous one, has some shortcomings due to the possibility of division by quantities equal to zero, requiring specific local limited developments in order to cancel out terms of the form 0/0, except when Observer's Latitude is exactly equal to Body Declination when P=0, a situation incurring a sharp discontinuity in the angular rate.

From the results of the introduction to this post, ** our problem to be solved boils down to determining Azimuths angular rates**. A 3D vector solution / algorithm might be easier to obtain here. I am to investigate.

3.3 - And finally, this current PA rate topic was triggered by our previous Alnair identification quizz by Frank E. Reed.

I just see that Peter Blaskett has come up there with some *extremely interesting insights there*, and I will revert to you Peter asap.

Kermit