# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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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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