Welcome to the NavList Message Boards.

NavList:

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

Compose Your Message

Message:αβγ
Message:abc
Add Images & Files
    or...
       
    Reply
    Re: The magical maths of Google Maps
    From: Gary LaPook
    Date: 2015 Oct 6, 23:15 +0000
    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






       
    Reply
    Browse Files

    Drop Files

    NavList

    What is NavList?

    Join NavList

    Name:
    (please, no nicknames or handles)
    Email:
    Do you want to receive all group messages by email?
    Yes No

    You can also join by posting. Your first on-topic post automatically makes you a member.

    Posting Code

    Enter the email address associated with your NavList messages. Your posting code will be emailed to you immediately.
    Email:

    Email Settings

    Posting Code:

    Custom Index

    Subject:
    Author:
    Start date: (yyyymm dd)
    End date: (yyyymm dd)

    Visit this site
    Visit this site
    Visit this site
    Visit this site
    Visit this site
    Visit this site