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Position Angles (PA) angular rates
From: Antoine Couëtte
Date: 2023 Nov 8, 01:11 -0800

A puzzle from Kermit ...

Our last topic on Alnair identification gave me a very first opportunity to deal with Position Angles (PA) angular rates as follows :

QUOTE

2 - From S13°00.0' - W163°00.0' on Jan 01st, 2019 (Height of Eye = 15' , P=1013.25 mb and T = 25°C)

(2.1) - At UT = 05h39m00.0s

Al N'Aïr Height/Azimuth=36°24.6'/220.539° and Peacock Height/Azimuth=18°51.2'/214.185°, ...///... , PA=16.3681° ...///...

(2.2) - At UT = 06h09m00.0s3 ( i.e. 30 minutes later)

Al N'Aïr Height/Azimuth=31°33.0'/222.535° and Peacock Height/Azimuth=14°44.3'/214.308°, ...///... , PA=22.7040° ...///...

(3) - Through backwards linear interpolation on both PA and HR, we get :

For PA : UT = 05h39m + 0.5732 h : UT(PA) = 05h56m +/- 4 min

UNQUOTE

I totally unexpectedly ran into the quite surprising "efficiency" of such PA reverse interpolation.

Our PA angular rate was quite high indeed : 12.6718 °/h , not far from our Lady Earth [inertial] rotation angular rate of 15.0411... °/h.

By the way, this PA method to "date and time" our initial puzzle picture by Frank might very well be the most reliable one since it involves only measuring an angle and it is almost insensitive to the environmental conditions, and in particular to refraction and DIP.

Back to this new topic of ours ... is there a known formula or method (be it a 1st order one) to compute (or approximate) such PA angular rate ?

I have not studied this topic which has now become certainly enticing, to me at least.

I would surmise that this PA angular rate would depend on Latitude, Height and Azimuth ... Is it possible to solve it through simple additions of 3D rotation vectors, which would be absolutely great ?

Anything published here ? Probably so ...

Who knows more or better ?

Kermit

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