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    Re: Azimuth formula for Great Circle sailings- having problems
    From: Paul Hirose
    Date: 2014 Dec 25, 16:37 -0800

    On 2014-12-23 14:51, Samuel L wrote:
    > I use the following formula to obtain Azimith-
    > Z = tan-1 (sin (*LHA*) / (cos (*LHA*) x sin(*AP Latitude*) - cos(*AP Latitude*) x tan(*Declination*))
    > If LHA is greater than 180 it's treated as a negative quantity.
    > If the Azimith angle as calculated is negative, add 180 to it.
    > In one Great Circle sailing probelm I'm working on to determine Z the answer 
    is negative unless the LHA is entered into the formula as a negative number. 
    The answer should be about 3d 55min. If I add 180 to to the negative answer 
    the wrong Z is obtained.
    > tan-1((sin(*LHA*) */ *(cos(*Present Latitude*) x tan(*Destination Latitude*) 
    – (sin(*Present Latitude*) x cos(*LHA*))
    > Here's the figures;
    > LHA= 0d 9m 40.9sec
    > Present Latitude- N 39d 58m
    > Destination Latitude= N 41d 43m 31.8sec
    > The answer I get is -3d 55min (notice the negative sign.
    Sam, the correct spelling is "azimuth". The 5th letter is u not i.
    We should be clear on the difference between azimuth angle (Z) and
    azimuth (Zn). Azimuth angle is zero in the direction of the elevated
    pole. That is, the zero direction is south for an observer in south
    latitude. Azimuth angle increases left and right to 180.
    On the other hand, azimuth is 0 at north, and increases right to 360.
    Now let's look at the two formulas. Assume present position latitude
    10N, destination 11N, and LHA = 1. Those easy numbers were chosen to put
    the destination to the northwest, at about azimuth angle 45 and azimuth 315.
    The first formula is clearly wrong since it yields -45, which becomes
    135 according to rule #2.
    What if rule #2 is changed to add 360 instead of 180? That fixes my
    first example. But if we move the destination south to 9N, the result is
    then 45, though the correct azimuth angle is obviously about 135.
    Applying the same rules to the second formula does give correct azimuth
    angle. Result is 45 for destination latitude 11N and LHA 1 or 359.
    Result is 135 for destination latitude 9N and LHA 1 or 359. By the way,
    the rule about treating an LHA greater than 180 as a negative quantity
    means that you simply put a negative sign on LHA.
    A third rule is that the present position latitude is always positive,
    even in the southern hemisphere. The destination latitude is negative if
    it's in the opposite hemisphere.
    After you have azimuth angle (Z), it's an easy matter to compute azimuth
    (Zn). In your example, present position is in the northern hemisphere.
    Since LHA is less than 180, azimuth of the destination is west of north.
    It follows that Z = N 4° W, and Zn = 356°.
    It's possible to get azimuth (not azimuth angle) from the second formula
    if your calculator has a rectangular to polar coordinate conversion
    function. First compute x and y coordinates. (Note that they are
    respectively the denominator and negative numerator in the second formula.)
    x = cos(Present Latitude) x tan(Destination Latitude) – sin(Present
    Latitude) x cos(LHA)
    y = -sin(LHA)
    Then convert x and y to polar form to obtain azimuth. The only rules are
    that south latitude is negative, and add 360 to azimuth if the result is

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