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    Re: Azimuth Formula Questions
    From: George Huxtable
    Date: 2005 Oct 30, 00:58 +0100

    .
    Bill wrote-
    
    
    > George wrote:
    >> Here's the expression to use to obtain azimuth from its Tan. It presumes
    >> that southerly latitudes and declinations are negative, that local hour
    >> angle is measured positively Westwards 0 to 360 degrees, and that
    >> azimuths
    >> are measured clockwise from North, 0 to 360 degrees.
    >>
    >> az = arctan (sin (hour angle) / (cos (hour angle) sin lat - cos lat tan
    >> dec))
    >>
    >> Then to put az into the right quadrant, apply the following rules-
    >>
    >> if tan az was negative, add 180 degrees to az.
    >> if hour angle was less than 180 degrees (i.e., to the West of the
    >> observer),
    >> add another 180 degrees to az
    >
    > I notice another (longer) formula in the HO229 background section that
    > also
    > does not require Hc:
    >
    > tanZ = (cosd sinLHA)/(cosL sind - sinL cosd cosLHA)
    >
    > L is latitude
    > d is declination
    >
    > Zn is determined as per 229 usual rules:
    > North lat
    > Z > 180, Zn = Z
    > Z < 180, Zn = 360-Z
    > South lat
    > Z > 180, Zn = 180-Z
    > Z < 180, Zn = 180-Z
    >
    > For the mathematically inclined, are there pros and cons for either
    > formula?
    >
    > Bill.
    
    Reply from George.
    
    The two are the same, basically.
    
    Take the expression Bill quoted;
    
    tanZ = (cosd sinLHA)/(cosL sind - sinL cosd cosLHA)
    
    Divide the numerator and the denominator both by cos d,  remember that sin d
    / cos d = tan d, and you get;
    
    tan z = sin LHA / (cos L tan d - sin L cos LHA)    (equation A)
    
    That expression has two fewer trig terms in it, so it's that much simpler.
    
    compare it with the expression I gave, of  -
    
    az = arctan (sin (hour angle) / (cos (hour angle) sin lat - cos lat tan
    dec))
    
    which can just as well be written-
    
     tan az  = sin (hour angle) / (cos (hour angle) sin lat - cos lat tan dec)
    (equation B),
    
    Compare the two, A and B, allowing for the fact that they use slightly
    different symbols for the same quantities, Az for z, hour angle for LHA, lat
    for L, dec for d, and you will see that they are exactly the same, except
    that the order of the terms, subtracted in the denominator, is reversed.
    That's the same as multiplying by -1.
    
    In the case of equation A, I made it clear that hour angle was taken
    positive westwards. Bill doesn't state whether HO 229 also takes LHA to be
    positive Westwards, and I don't have a copy to check it, but let's assume
    that it does. Then equation A would give the opposite sign for tan az ,
    compared with equation B. In that case, the resulting azimuth from equation
    A would be 180 - (the az that resulted from equation B).
     Fundamentally, then, the two expressions are mathematically the same,
    except for that negative sign.
    That discrepancy between the two could perhaps be corrected by using
    different.selection rules to put the azimuth into the right quadrant.
    
    However, I'm a bit puzzled where Bill quotes-
    
    > Zn is determined as per 229 usual rules:
    > North lat
    > Z > 180, Zn = Z
    > Z < 180, Zn = 360-Z
    > South lat
    > Z > 180, Zn = 180-Z
    > Z < 180, Zn = 180-Z
    
    If you deduce an angle from its tan, you get a result, from a calculator,
    that varies between -90 degrees (when tan Z approaches minus infinity),
    through zero (when tan Z = 0), to +90 degrees (when tan Z approaches  plus
    infinity). So I'm rather puzzled about how you apply those tests, comparing
    az with 180 degrees, because it will never exceed 90 degrees either way. Can
    Bill (or anyone) explain how that works?
    
    For an observer at lat +45, compare the azimuths of two bodies, one with dec
    = +1 deg, and another, with dec = -1 deg, both with the same LHA of +90.
    The azimuths of those  3 bodies should be nearly the same, differing only by
    a degree or so, which is obvious if you look at a globe. However, expression
    A, using the the selection rules as Bill stated, seems to give very
    different azimuths in the two cases.. It just doesn't seem right.
    I suspect strongly, then, that those selection rules, as quoted by Bill, are
    unworkable and wrong. Have they been copied verbatim from HO 229? If so, the
    matter needs investigating further.
    
    George.
    ===============================================
    contact George Huxtable at george@huxtable.u-net.com
    or at +44 1865 820222 (from UK, 01865 820222)
    or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK
    
    
    

       
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