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    Re: Sight Reduction Formula
    From: Jan Kalivoda
    Date: 2003 Oct 19, 23:42 +0200

    Hello, list,
    
    I will return to Ralph's question again.
    
    First of all, his purported "standard" formula for the Azimuth in the form 
    "cos Z = (sin Dec [+ or abs. diff.] / (cos Hc * cos Lat)" seems erroneous. I 
    have never seen it, but above all, it produces wrong results. See the file 
    http://web.dkm.cz/kalivoda/Ralph.xls , where I tried the case with contrary 
    names for Dec and Lat, too. (You have my last trial after several previous 
    ones there.) Ralph's "standard" formula gives a wrong result. Maybe Ralph 
    made a mistake while transcribing it?
    
    But Ralph's "Almanac" formula gives the correct result, as you can see in the 
    same file. It even considers contrary names of Lat and Dec.
    
    There is a sort of trigonometric formulas that work with *four* known elements 
    of the spherical triangle to obtain the fifth unknown element. They are named 
    "formulas of five elements" accordingly. They weren't used in old times, as 
    the fifth element would have added substantially to the difficulty of 
    logarithmic solutions. But now, in the time of digital gadgets, it is not an 
    issue. Ralph's "Almanac" formula resembles those equations, if my memory 
    doesn't fail.
    
    But I cannot verify this my opinion, as I don't have a complete manual of 
    spherical trigonometry at home. If anybody has such handbook available, he 
    can confirm or refute my assertion.
    
    
    Jan Kalivoda
    
    
    ----- Original Message -----
    From: "Clampitt, Ralph" rclampitt{at}KUMIN.ALASKA.COM
    To: 
    Sent: Saturday, October 18, 2003 1:06 AM
    Subject: Sight Reduction Formula
    
    
    I am new to the list and have just recently started using a programmable 
    calculator for sight reductions.  I have found several useful formulae on 
    this list and in other references, however the one shown in the Nautical 
    Almanac for azimuth puzzles me.
    Can anyone explain the formula for azimuth from the Almanac?
    
    Summarizing from page 279 - 2002 Nautical Almanac
    
    6. The calculated altitude and azimuth.
    Step 1.  Calculate the local hour angle.
            LHA = GHA + Long.
    
    Step 2.  Calculate S, C and the altitude Hc from
            S = sin Dec
            C = cos Dec cos LHA
    Sin Hc = (S sin LAT + C cos LAT)
    
    (This appears to be analogous to the fundamental cosine formula of spherical 
    trigonometry and to be the classic formula for computing altitude.   Most 
    references however seem to use [+ or abs. difference]) i.e. sin Hc = sin LAT 
    * sin Dec [+ or absolute diff.] cos LAT * cos Dec * cos LHA.
    
    The trouble, for me, starts in using the formula for azimuth from the almanac, 
    which seems to be quite different from those most commonly shown in other 
    references, such as one that seems fairly standard
    
            Cos Z = (sin Dec  [+ or abs. diff.] sin Hc) / (cos Hc * cos Lat)
    
    The Almanac formula for azimuth show in Step 3 is as follows;
    
    Step 3. Calculate X and A from
            X= (S cos Lat - C sin LAT)/cos Hc
            Cos A = X
    
    Determine the azimuth Z
    If HA > 180 degrees than Z= A
    Otherwise Z = 360 degrees - A
    
    The formula than seems to be
    Cos A = [(sin Dec * cos LAT) - (cos Dec *cos LHA * sin LAT)]/cos Hc
    
    As I said, I have not seen this formula anywhere else and also I have had 
    trouble with it when Dec and Lat are opposite signs. Please understand that I 
    am not much of a mathimatician.
    
    Ralph Clampitt
    
    
    

       
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