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Thanks Antoine for this interesting issue.

Solution for the example in "231217-LAN-Part-I-Near-Noon-methods-comparison-.pdf"

• EX1.Kermit.obs input file for Nautical Astronomy.exe (I did it quickly and I do not know if I interpret Hmoon data correctly)
• EX1.Kermit.current.png
• EX1.Kermit.png

231219-LAN-Part-II-Culmination-to-Meridan-Passage-fully-solved-.pdf

• LoogBook.xlsx (in zip file) simulated data, B1 example, for anyone who wants to play with it.

It is also interesting to compare with the SR solution as described on paper Celestial Fix - n LoPs - by Least squares Sight Reduction algorithm. Paper & Excel (B1solution.MP.png, B1solution.SR) 

Back in 2009 I made a lot of simulation,  huge amount of data, to see how time of culmination differs from time of meridian passage: simulacion V.xls, RV 135 100.7z 

And conclude that Jim Wilson's method gives bad results when SOG is greater than say 10 kn.

Background

At culmination celestial body altitude H = maximum/minimum -> dH/dt = 0

Meridian passage:

• LHA = 0  (Upper Meridian of Place) or LHA = 180º (Lower Meridian of a Place)
• Azimuth Zn = 180º/0º 

References (attached files):

Pages from A1 - Algoritmos.pdf

• Latitude by meridian transit (Latitud al paso por el meridiano del lugar)
• Position by meridian transit (Hs and time of meridian transit)
• Ex-meridian sight (Latitud por altura circunmeridiana)

Pages from Position&Time.pdf

• The equations of the observations: Altitude
• Time derivatives
• Stationary observer 

To solve this problem, I use a matrix integration of all differential equations for each observation, (see Position&Time.pdf).

Fair winds and following seas!

--

Andrés Ruiz
Navigational Algorithms
https://sites.google.com/site/navigationalalgorithms/

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