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    Re: Azimuth Equation
    From: Bill B
    Date: 2005 Oct 19, 15:04 -0500

    >
    > I'm using the azimuth equation out of the Nautical Almanac:
    > cos-1[(sinDec cosLat)-(cosDec cosLHA sinLat)/cosHc]
    >
    > To find Hc: sin-1[(sinDec sinLat)+(cosDec cosLHA cos Lat)]
    > In entering the equation, if the Declination or latitude is South, or the
    > Longitude is West, the value is negative.
    
    Todd
    
    After a nights sleep, and looking at the almanac, have rethought the
    problem.  Your signing of Lat, West Lon, and declination is correct.
    Remember LHA must always be a positive number from 0 to 359.xxx.  If GHA+W
    Long is negative, add 360. If GHA + E Long is equal to or greater than 360,
    subtract 360.  Some will tell you when using a modern calculator, a negative
    LHA or LHA greater than 360 will work fine, the calculator knows the
    appropriate trig value of those angles.
    
    The formula for Hc looks correct: sinHC=(sinDec sinLat)+(cosDec cosLHA cos
    Lat)
    
    cos-1[(sinDec cosLat)-(cosDec cosLHA sinLat)/cosHc] seems strange to me.
    Perhaps it folds the formula for Hc in the formula below.  For Z and Zn
    (azimuth) I use:
    
    cosZ = (sinDec - (sinLat sinHc))/(cosLat cosHc)
    
    If sinLHA is negative, Z=Zn
    If sinLHA is positive, Zn = 360-Z
    
    Purely by inspection if, LHA is >0 to <180, the body will be in the SW or NW
    quadrant. If LHA is >180 to <360, the body will be in the NE or SE quadrant,
    which tells you whether to use Z as Zn, of subtract it from 360.
    
    Hope that helps,
    
    Bill
    
    
    

       
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