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    Re: Interpolation of Meridional Part Table
    From: Brad Morris
    Date: 2009 Mar 24, 17:00 -0400

    The constant 7915.704468 may be wrong.  That's from Bowditch, 1849.
    
    Sorry about that.  The technique is correct however.
    
    Best Regards
    Brad
    
    
    -----Original Message-----
    From: NavList@fer3.com [mailto:NavList@fer3.com] On Behalf Of Brad Morris
    Sent: Tuesday, March 24, 2009 4:57 PM
    To: NavList@fer3.com
    Subject: [NavList 7756] RE: Interpolation of Meridional Part Table
    
    
    = LOG(TAN(RADIANS(Latitude/2+45)))*7915.704468
    
    Best Regards
    Brad
    
    -----Original Message-----
    From: NavList@fer3.com [mailto:NavList@fer3.com] On Behalf Of NavList---.com
    Sent: Tuesday, March 24, 2009 3:42 PM
    To: NavList@fer3.com
    Subject: [NavList 7754] Interpolation of Meridional Part Table
    
    
    Anyone,
    I am looking at a sample problem in section 505 of the 1972 Dutton's.  It says 
    that the latitude of Cape Flattery Light is 48d23.5' and says that the 
    meridional parts (m) for that latitude are (is?) 3309.2.
    
    Using an on-line versuion of Bowditch Table 6, I located 48 deg latitude and 
    went down the minutes column to 23' to find m=3308.8, and to 24' to find 
    m=3310.3.  A linear interpolation would put 48d23.5' at 1/2 the difference, 
    or 3309.55 (round up or down to taste), and one would then have m=3309.5 or 
    3309.6.
    
    I recognize that a linear interpolation doesn't really fit the facts as far as 
    the way meridional parts change; so is there a technique I shold know about 
    that would have given me the result Dutton's got, 3309.2?  That is, how did 
    they do that?
    
    -John Parsons
    
    
    ---------------------------------------
    [Sent from archive by: jkp-AT-obec.com]
    
    
    
    
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