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 7, 17:38 +0000
From: Lu Abel
Date: 2015 Oct 7, 17:38 +0000
I have no doubt of these numbers, Gary -- and at the same time they seem counterintuitive. If I were to fit a perfect sphere to the curvature of the earth at the equator it would be - as you pointed out earlier - about 11 nm greater in diameter than a perfect sphere that matched the curvature of the earth at the poles. A greater diameter should imply a greater surface distance for one degree than for a smaller diameter. What am I missing?
From: Gary LaPook <NoReply_LaPook@fer3.com>
To: luabel@ymail.com
Sent: Tuesday, October 6, 2015 10:42 PM
Subject: [NavList] Re: The magical maths of Google Maps
That made me curious so I turned to table 6 of Bowditch which gives the length of a degree of latitude for each degree from 0 to 90. This length varies from 59.701 NM at the equator to 60.313 NM at the pole based on the Clarke Spheroid of 1866. I added all the distances up and found that half way, at 45 degrees of latitude the distance is 8.3 NM short of the 2,700 NM it would be on a spherical earth and then the remaining distance to the pole is 9.2 NM greater so the total I came up with was 5,400.5915 NM but I assume the discrepancy from the perfect 5,400 NM is probably due to rounding. So to test Google Earth it doesn't do to measure the distance from the pole to the equator since you will get 5400 either way but you should measure from the pole to 45 degree latitude.glgl
From: Frank Reed <NoReply_FrankReed@fer3.com>
To: garylapook---.net
Sent: Tuesday, October 6, 2015 10:12 AM
Subject: [NavList] Re: The magical maths of Google Maps
David Pike, you wrote:
"From this I conclude that the Google Maps projection is close to standard Mercator’s.". You’ll find you’ve got a “Globe” presentation, which you can tilt until the line between Boston and London is straight while the distance remains constant. You would appear to have hit the view in with the plane of the great circle."Yes, that's right. 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.Frank Reed
Conanicut Island USA
