Hello

[Mike1843]

Mike,

If the heavenly body's declination is Contrary name to your DR Latitude enter a negative sign before it.

Pretty easy.

Clen

Hello again,

Apologies for the tardy response, I have just got around to having a proper look at the "proper formula" and as I think I have got it sorted I thought I would post my calculations.

As I indicated above I do have a dozen practice sights already worked out using sight reduction tables AP3270.

First one as follows

DR  49 17.4N     Adjusted to 49 degrees N
    017 14.3W

Dec 15 38.2N     Converted to 15.6366 degrees

LHA 302 09.5     Adjusted to 302 degrees

As I am using a calculator I am aware I do not need to adjust DR Lat and LHA but to get a proper comparison with AP3270 I will use the same figures.

Hc using AP3270 is 32 degrees 33 minutes

Using proper formula------

Hc = sin -1       sin dec x sin lat          +                cos lat  x cos dec x cos LHA

Hc = sin -1       0.2695 x 0.7547          +               0.6560 x 0.9629 x 0.5299

Hc = sin -1           0.2033                    +                         0.3347

Hc = sin -1                                    0.538

Hc =                                       32.5475 degrees      (inverse sin) ? On my calculator (2nd Function) (sin) 

                                                  0.5475 x 0.6 = 32.8

Hc = 32 degrees 32.8 minutes (Just 2 tenths different to AP3270)

Having adjusted DR Lat to 49 degrees this will become the Lat of my AP. but what is the longitude? As I have adjusted LHA by 09.5 minutes surely I should also adjust DR longitude by 09.5 minutes. In the moderate amount of study I have undertaken into celestial navigation I cannot remember this being made clear, it just seems logical. Therefore my DR longitude of 017 14.3 needs 09.5 adding = 17 degrees 23.8 minutes. Is this correct???. 

Final AP ready to plot is 

 49 degrees N

017 degrees 23.8 minutes W

Delighted to have any of my misunderstandings corrected.

Thanks

Mike