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    Pub.229 GC-Calculation distance and course.
    From: Guus Dekker
    Date: 2009 Nov 28, 12:48 -0800

    jm7857 Guus Dekker.
    I am exercising all kinds of nav problems. Can somebody explain me about 1 
    tiny problem I can not solve by my self.
    It deals about a sailing calculation problem finding GC-Distance and GC-Course 
    from Fremantle to Durban, with nonintegral degrees and using a graphical 
    method to simplify the DSD (Double-Second Difference) calculations.
    It's about the supplement of pub. 229 (vol.3 all the supplements are more or less the same).
    CHAPTER:  D. Other applications 2. Great-Circle Sailing page xx. Case 1 example. point (6).
    
    (6) The Azimuth angle interpolated for declination, LHA, and latitude 
    increments is S66.3W; the initial greatcircle course from the point of 
    departure is 246.3*
    How is this calcutalted ?! What is the 0.2* correction from 246.1*(S66.1*W)
    
    The Z part is calc. like this: Z=66.9* Zdiff.= 66.9-66.0 =0.9 corr.= -0.8* 
    distance = S66.3W - corr. = S66.1W
    Next: I think the 0.2* is the correction angle between  - AP to destination -  
    and  - Departure Point to destination. -
    Pub.229 says on page 172  LHA 85* lat. 32* Decl. 29* that Hc=18* 45.4' d= +27.0' Z=66.9*
    d (from decl. 28* just above) = +27.2'. Is this the 0.2* diff. between the outcome S66.3*W and S66.1W !?
    
    Please... somebody !!??
    
    
    
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