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    Re: Pub.229 GC-Calculation distance and course.
    From: Hewitt Schlereth
    Date: 2009 Dec 2, 08:28 -0400

    As I read it, what Guus seems to be asking is how they got the
    additional .2'  they added to the course from the AP (246.1�) to the
    course from the actual point of departure in Fremantle (246.3�).
    
    Hewitt
    
    On 12/1/09, John Karl  wrote:
    > Guus,
    >
    >  I don't understand the 0.2* that you're talking about.  The Z
    >  interpolation is for the dec between 29* and 30*, the desired Z being
    >  for dec = 29* 52'.  So the Z interpolation calculation is 52'/60' X
    >  0.9* = 0.8*, giving Z = 66.9* - 0.8* - 66.1* as in the book.
    >
    >  Make sense?
    >
    >  JK
    >
    >
    >  --
    >  NavList message boards: www.fer3.com/arc
    >  Or post by email to: NavList@fer3.com
    >  To , email NavList+@fer3.com
    
    -- 
    NavList message boards: www.fer3.com/arc
    Or post by email to: NavList@fer3.com
    To , email NavList+@fer3.com
    

       
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