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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@pacbell.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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