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    Re: Zero-Altitude Shot
    From: Paul Hirose
    Date: 2019 Sep 6, 21:14 -0700

    On 2019-09-03 22:01, Bob Goethe wrote:
    > Image1 is a copy of my worksheet, with extraneous fields erased.  The key 
    field in question is field 8.5, the Hc of -1° 16.0'.
    > This seems to make sense to me when I work it out with Pub. 249.
    > My question has to do with direct calculation of this sort of problem with 
    the Bygrave equations.  In the nature of things, one will seemingly never get 
    a negative value for Hc with these equations as they stand.
    > My direct calculation values for Z and Zn agree with what I got from Pub. 
    249, seen in Image2.  But the Hc comes out positive instead of negative when 
    I reduce it with Bygrave.
    >
    > Attached File: http://fer3.com/arc/imgx/Image2.png
    > Attached File: http://fer3.com/arc/imgx/Image1.png
    
    I haven't looked in detail at your Bygrave implementation, but in the
    calculation of X you subtract W where I would add. I get Hc = 12.7486.
    
    My version of the Bygrave formulas:
    
    W = arctan(tan dec / cos LHA)
    
    If LHA is between 90 and 270, replace W with its supplement: W = 180 - W
    
    X = 90 - lat + W (if lat and dec same name)
    X = 90 - lat - W (contrary name)
    
    If X is not 0 to 180, add or subtract 180 to make it so. If the
    adjustment is necessary, altitude is negative.
    
    Z = arctan(tan LHA cos W / cos X)
    
    If (altitude < 0) or (X < 90), replace Z with its supplement:
    Z = 180 - Z
    Do the adjustment only when EXACTLY ONE of the conditions is true.
    
    Hc = arctan(tan X * cos Z)
    Apply negative sign if applicable.
    
    NOTES
    
    Ignore negative signs in trigonometric functions. All cosines and
    tangents are positive. The only possible negative number is X, and then
    it is immediately adjusted to a positive number.
    
    Latitude and declination are always positive, "same name" or "contrary
    name."
    
    Z is the conventional azimuth angle in the range 0 to 180, measured east
    or west from the elevated pole (north pole if assumed latitude is north).
    
    The cosine of 90 or 270 is zero, and therefore a forbidden "divide by
    zero" can occur in the W or Z formula. In that case, skip the formula
    and make the result 90.
    
    My Bygrave solution is valid for every combination of LHA, latitude, and
    declination, including negative altitude. For example, here are
    parameters for a sunset observation.
    
    95.3057 = LHA
    40 = lat
    5 = dec (same name)
    
    Bygrave solution:
    
    W = arctan(tan 5 / cos 95.3057)
    = 43.3154
    
    LHA is between 90 and 270, so adjust W.
    
    W = 180 - 43.3154 = 136.5855
    
    X = 90 - 40 + 136.5855
    = 186.5855
    
    That's outside the range 0 - 180, so subtract 180. The adjustment means
    altitude is negative.
    
    X = 6.5855
    
    Z = arctan(tan 95.3057 cos 136.5855 / cos 6.5855)
    = 82.7620
    
    No Z adjustment is necessary since (altitude < 0) and (x < 90).
    
    Hc = arctan(tan 6.5855 cos 82.7620)
    = .8333 = 50.00′
    
    Apply a negative sign as indicated in the computation of X. This is the
    nominal altitude of the Sun center at rise or set. The Bygrave
    calculation of this negative altitude was accurate.
    

       
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