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    Re: Interpolation of Meridional Part Table
    From: George Huxtable
    Date: 2009 Mar 24, 22:17 -0000

    Brad wrote- responding to a question about meridional parts-
    
    "= LOG(TAN(RADIANS(Latitude/2+45)))*7915.704468"
    
    =================
    
    Comment from George.
    
    Well, not quite. Not if you're working to high precision, as a constant such 
    as 7915.704468 implies.
    
    That would be right for a spherical Earth, but needs correcting for its 
    actual non-spherical shape. Taking the flattening to be 1 part in 298, a 
    more accurate answer is given by
    MP= log(tan(radians(lat/2 +45)))*7915.7 - 23*sin L, which is good to about a 
    tenth of a mile. A more precise expression can be found in Bowditch ((1981) 
    vol 2, page 4 (table 5 explanation)
    
    Slightly different answers, in different sets of tables, may reflect nothing 
    more than updating the model for the Earth's shape, which isn't a precise 
    spheroid anyway.
    
    In that edition of Bowditch the meridional parts is found in table 5, not 
    table 6 as John Parsons quotes from his 1972 edition, but the interpolated 
    value at 48� 23.5' is exactly the same as his, as  3309.5, interpolating for 
    the half-minute. There's no problem at all with making such a linear 
    interpolation.
    
    Most likely, that example in Dutton's was formulated many years earlier, 
    when a slightly different model had been adopted for the Earth's flattening.
    
    What intrigues me is why John is worrying about such small discrepancies. 
    Navigation isn't such an exact science as that.
    
    George.
    
    contact George Huxtable, at  george@hux.me.uk
    or at +44 1865 820222 (from UK, 01865 820222)
    or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
    
    -----Original Message-----
    From: NavList@fer3.com [mailto:NavList@fer3.com] On Behalf 
    Of NavList@aol.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
    
    
    ---------------------------------------
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