A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Antoine Couëtte
Date: 2023 Nov 12, 12:24 -0800
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:
2.1.5 - 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 :
22.214.171.124 - This updated "enhanced" PA rate should be significantly closer to our upcoming Formula (1.2) determination. And :
126.96.36.199 - 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.