NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
From: Greg Rudzinski
Date: 2018 Jan 21, 06:12 -0800
Jeremey,
As a check on the Bowditch ex-meridian tables try a pocket trig calculator using this unique formula:
(t°)(t°).5235 (cos dec cos Lat / cos Ho) = + ex-meridian correction ' in minutes of arc
I use this formula with a 6" slide rule and get good results to the nearest tenth of a degree.
See this link to a custom table PDF to compare to the Bowditch XM tables :
http://fer3.com/arc/m2.aspx/Compact-ExMeridian-Table-PDF-Rudzinski-jan-2017-g37815
Greg Rudzinski
From: Jeremy C
Date: 2018 Jan 20, 20:52 -0800My 2nd mate is studying for her Chief Mate unlimited oceans test and she was having trouble with an ex-meridian (upper transit) of the sun problem. The app she was using had a solution, and there was another solution she and a friend had found online from a different source. I didn't agree with either methodology or answer when I reviewed them. There were 2 twists to the problem and each solution failed to address one of the twists.
The first twist was a given 1200 LT DR while traveling at 078 deg and 17 knots. Usually in USCG world, this means a mid-latitude sailing (or at least a parallel sailing) for the 12 minutes between the sight and the DR time must be determined to find the DR at sight time. This 3.5 minutes of longitude will affect the "t" angle entered into the second table.
The second twist was an Hs of about 81 degrees. To me, since this is near the limit of the tables, you would need to do a double interpolation of the table to obtain the correct "a" value. One online solution used the wrong number for "a" as it was was outside the range of the 4 tabular entries. The second one picked an "a" without any interpolation (nearest Lat and Dec) and also didn't account for the change in longitude between the DR and sight time.
The book answer was the one without the sailing, which makes sense if the person supplying the answer didn't do the sailing or interpolate the "a" table. The one that had the incorrect interpolation had a number between two of the answers. I did a problem combining a sailing and the double interpolation and came up with was 2 minutes less than the smallest and "correct" answer.
So the question is my methodology all wet? I am not one to double interpolate if the difference in "a" is a tenth or 2, but here the "a" varied over 3 units between the 4 tabular values. Also, since the "t" factor is squared in the formula, that 3.5 minutes of longitude can certainly affect the final latitude.
I'd be happy to hear some thoughts on this. Jeremy
