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: Interesting challenge
    From: Andrés Ruiz
    Date: 2012 Aug 10, 11:00 +0200
    Hi Andrew,  we have the same name!

    It seems that You have read my two papers, (available at my webpage):
    • Vector Solution for the Intersection of Two Circles of Equal Altitude
    • Sight Reduction with Matrices
    A summary of the methods used in Celestial Navigation.png

    regards,
    -- 
    Andrés Ruiz
    Navigational Algorithms
    http://sites.google.com/site/navigationalalgorithms/ 


    2012/8/6 Andrew Nikitin <nsg21@hotmail.com>

    > From: Lu Abel
    > Date: 5 Aug 2012 19:06


    > Right, but is there a closed-form solution?

    Lu,

    Solution to this problem does not require knowledge of spherical trigonomtry,
    just some linear algebra and the ability to convert between rectangular and
    spherical coordinates.

    You need to find point of intersection of 3 circles on a sphere.
    You know the centers of circles (defined by GP of bodies, or coordinates of the cities in your case) and the distance along the surface fromthe center to the edge of the circle. If you notice that each circle lies on a plane, which is perpendicular to the radius-vector connecting center of the sphere and GP, the point of intersection of the circles is the point of intersection of the planes.

    Equation for each plane give you one equation, with 3 circles you have 3
    planes, which gives you 3 linear eqautions to solve for for 3 coordiantes.
    Final step is to convert the rectangular x,y,z coordiantes into spherical.

    If some of the planes intersect at acute angles, you get the poorly conditioned
    system and there is not much you can do about it. Unless, you know more
    distances to more points with known GP. Then you have more than 3 linear
    equations and need to solve overdetermined system, which can be solved to find
    the point which is as close as possible to all planes.

    In your case, assuming coordinates (taken from wikipedia)
    San Jose CA 37°20′7″N 121°53′31″W
    Portland OR 45°31′12″N 122°40′55″W
    Houston TX 29°45′46″N 95°22′59″W
    and distances given in statute miles as 6003 5557 5420
    I got system of equations

    [ -0.420067 -0.675077 0.606478] [x] 0.0532418
    [ -0.378339 -0.589732 0.713495] * [y] = 0.165252
    [-0.0814387 -0.86426 0.49641] [z] 0.199303

    Solving for x,y,z get
    -0.16697 -0.723045 0.680436

    Convert to spherical
    lat=43.8835° lon=7.92476° R=0.99607

    which is close to Nice, France.

    With so many computers around, solving overdetermined system is only marginally
    harder, so if you have more distances to more points, the solution may be
    better, but I still doubt it will be far from Nice.

     

    File:


       
    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