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    Re:
    From: Millard Kirk
    Date: 1997 Jul 09, 12:42 PM

    Russell Sher wrote:
    >
    > Hi - I've just subscribed. Does anyone know the formula for calculating
    > meridional parts if given the Latitude ?
    > Cheers
    > Russell Sher
    
    >From "The Calculator Afloat"
    
    	M=7915.7045 x log tan (45d +L/2) - 23.2698 sin L
    
    Example:  Our position is Lat 32d 14.7'N, Lon 66d 28.9' W, and we wish
    to determine the course and distance to a point near Chesapeake Light,
    in Lat 36d 58.7'N, Lon 75d 42.2'S, be Mecator sailing.
    
    	L1      32d 14.7'N
    	L2	36d 58.7'N
                    ----------
    	l        284.0' n        l = difference of L1 and L2
    
    M1 = 7915.7045 x log tan(45d + 32.2450d/2) -23.2689 x sin 32.2450d x sin
    32.2450d
         2045.7409 - 12.4149 = 2033.3
    
    ~ = asbsolute value
    
    	L1      32d 14.7'N       M1	2033.3
            ~                        ~
    	L2	36d 58.7'N       M2     2377.0
                    ----------              -------
    	l        284.0' n 	         343.7
    
    next step difference in Lon in min of arc:
    
    	Lon1    66d 28.9' W
             ~
            Lon2    75d 42.2' W
                    -----------
             DLO       553.3' W
    
    	Tan C = 553.3'/343.7 = 1.6098
    	
    Therefore:  course is N 58.1522d W; the course to steer will be Cn
    301.8d
    	
    	Distance  D = 284.0'/cos 58.1522d  = 538.2206  or 538.2 miles
    
    Hope this will help
    
    --
    
    Learning the Hard Way!!
    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
     Millard Kirk KB8YQO 	Email - mkirk{at}marshall.edu
     116 Lewis Ave	     	Homepage- http://webpages.marshall.edu/~mkirk/
     Barboursville, WV      A West Virginia Blue Water Sailor
     25504                  (304) 736-6544
    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    
    
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