NavList:

Hello Bill,

From your post and after looking at the Spherical Trigonometry Sine Rule and then at the Napier's Analogies (also here ), and further to reading Lars Bergman's and Andrés Ruiz posts, I became aware of the existence of at least one direct solution to Dave Walden's Problem submitted here.

As this problem looks like being a rather infrequent in the Classical CelNav environment, I decided to further investigate about it. If in fact the existence of at least one direct method can be inferred from previous publication(s), the explanation of such a method has not been given on NavList, at least recently to the best of my memories..

I am using here-under a, b, c for sides and A, B, C for angles opposite to respectively a, b, c.

Also excuse me for publishing so many digits, since this solution seems sensitive to some numerical instability .

I am therefore reconstructing here-after what I am guessing to be the Lars Bergman's Method which also seems to be among of the ones published by Andrés Ruiz, and which might also be implemented in Peter Hakel's Spreadsheet Solution.

From Dave's data, we know :

- One Angle : the Azimuth "A" (120.465 000 00 °). And 2 sides:

- The Sun-Pole Distance "a" (16.779 718 19 °) and the Observer's distance "c" to the Sun sub-solar point (35.065 910 58 °).

Hence from the Spherical Trigonometry Sine Rule  we can immediately compute the Sun LHA "C" = 31°146108 02 , which when added to the Sun GHA ( 45.852 021 14 °) yields the Observer's Longitude C = 76.998 129 16 ° = 76°59.888'

Then from the Napier's Analogies , since we now know 2 Angles (A and C ) and their opposite sides (a and c) , it possible to compute tan (b/2) with "b" being the Observer's co-Latitude (90° - Latitude). From the earlier values and subsequent results, we get b = 51.001 192 774 °.

Hence the Observer's Latitude is equal 90° - b = 38.998 072 26 = 38°59.884'

Yesterday I included a check on the WGS84 Ellipsoid, which actually was not really necessary.

WGS 84 checks are only useful for bodies with important Horizontal Parallax since in such cases (e.g. Lady Moon for us brave Celestial Navigators), both the Height and the Azimuth can be significantly modified by such Parallax effects.

I am considering direct methods as [far] superior to Iteration methods.

Dave, Bill, Andrés and Lars, thank you very much for having given me the opportunity to discover (or re-discover ?) the existence of such a direct method.

You have made my day, ...

... But you are now getting more than merely Thanks. Read here-after ...

Final Comment : I personnally hate solving angles from Sines since within the [0-180°] span, there are generally 2 solutions : namely the α  and ( π – α ) angles which may become quite cumbersome to single out for Azimuths near 90° or 270 °. You do always know when a body is W or E, or at least you should ..., but solving for an East Azimuth slightly to the North or to the South may not always be so easy or immediate.

Fortunately this remark does not apply to either Cosines, Tangents or Co-tangents which have only one determination within the [0-180°] span.

Hence my question to you Bill, Lars, Andrés or Peter, or others - if you are using the method here-above - how do you and can ensure that you consistently get the correct angle from the Sine Rule ?

Best Regards to all,

Antoine M. Couëtte