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    Re: The magical maths of Google Maps
    From: Gary LaPook
    Date: 2015 Oct 6, 20:52 -0700

    OK, 21,600 NM/ Pi = 6,875.49354157 NM.
    
    But, for the Clarke spheroid if 1866 the polar diameter is 6,864.564 NM;
    Clarke spheroid of 1880 it is 6,864.49 NM;
    International spheroid it is 6,864.918 NM.
    So the spheroid is about 11 NM shorter in the pole to pole distance than the spherical earth.
    
    gl
    --------------------------------------------
    On Tue, 10/6/15, Lu Abel  wrote:
    
     Subject: [NavList] Re: The magical maths of Google Maps
     To: garylapook---.net
     Date: Tuesday, October 6, 2015, 5:57 PM
     
     That's the
     distance along the surface.   Sorry if I didn't make
     it clear, but I was asking about the diameter.
     
       
           From: Gary LaPook
     
      To: luabel{at}ymail.com
     
      Sent: Tuesday, October
     6, 2015 4:22 PM
      Subject: [NavList] Re:
     The magical maths of Google Maps
      
      
     The pole
     to pole distance on a spherical earth is simple, 10,800 NM.
     (180 X 60.)
     gl
       
     
           From: Lu Abel
     
      To: garylapook---.net
     
      Sent: Tuesday, October
     6, 2015 3:27 PM
      Subject: [NavList] Re:
     The magical maths of Google Maps
      
      
     Hmmm, that's very
     interesting that the geodesic distance is greater than the
     great circle distance.  I'm traveling right now with
     very limited internet access, so I must ask a question that
     I might look up for myself were I at home:  What is the
     pole-to-pole distance of the "spherical" earth as
     opposed to the oblate spheroid (ie, geodesic) earth.  Or,
     in a similar vein, where do the two intersect -- at the
     poles, at the equator, or at some intermediate point??
     
      
     
      
        
       From: David Pike
     
      To: luabel{at}ymail.com
     
      Sent: Tuesday, October
     6, 2015 2:37 PM
      Subject: [NavList] Re:
     The magical maths of Google Maps
      
      
     Frank you
     wrote.  The only thing you have to double-check with an
     earth-based mapping product is that they are not being too
     clever by including ellipsoidal corrections to the distance
     calculation. If that happens then it's not quite a great
     circle but awafully close. I haven't checked, but I
     assume that Google Maps products use the standard spherical
     great circle calculation.Yes, if
     spherical geometry’s hard, ellipsoidal geometry’s almost
     impossible (Well for me it is).  I
     see that in November 2000 (that’s only 15 years ago) I
     completed an MSc assignment to work out and compare: the
     bearing from London to New York; the bearing of New York to
     London; the Great Circle distance; and the longitude for the
     GC lying E/W using spherical trig c.f. the angles and the
     geodesic using ellipsoidal geometry. 
     All I had to do it with was a hand-held Casio fx-992s
     with sin, cos, tan, and 1/x, and it took ten sides of narrow
     lined A4.  Between 51° 30’ N
     000° 05’W and 40° 43’N 073° 59’W I
     got the Great Circle distance = 5,577.886km and the Geodesic
     distance = 5,586.662km (I suppose I could now check it out
     on Google Maps, but it’s cocoa time). 
     I see that at the time the assessor was kind enough
     to give me 100/100, but today I can’t understand a word of
     it.  DaveP
     
     
     
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