A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Position-Finding
From: Tony Oz
Date: 2016 Nov 4, 07:06 -0700
LHA + EsLon 3° + (-66°30') = 3° - 66°30' = -63°30'.
LHA is always between 0° and 360° so you must "normalize" the negative LHA by adding 360° to it. ( If LHA was 425° you would subtract 360° )
LHA = -63°30' + 360° = 296°30'
LHA is Greater than 180° so the second rule would be used ie t = 360° - LHA
t = 360° - 296°30' = 63°30'
Thank you for further clarification. It confirmed that I did get your explanations correctly. I'll contact the author of that SR work-sheet to amend the instructions for the Step 1 accordingly.
Could Step 1 be simplified by not dealing with LHA at all? Could I get t directly from GHA and LonEP observing the "+ is East, - is West" rule? I think it would be less error-prone.