NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: The magical maths of Google Maps
From: Lu Abel
Date: 2015 Oct 6, 23:56 +0000
From: Lu Abel
Date: 2015 Oct 6, 23:56 +0000
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 <NoReply_LaPook@fer3.com>
To: luabel@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 <NoReply_LuAbel@fer3.com>
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 <NoReply_DavidPike@fer3.com>
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
