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Distance of a great circle

  1. Mar 11, 2009 #1
    I am the captain of my schools academic team. One of the main topics this year is Spherical trig and I can't find out how to find the shortest distance between two points on earth given the longitude and latitude of both points. I can easily convert the points from Degree- minute- second format to degree format but I don't know what to do from there. for example what is the distance between 53 09 02N ; 001 50 40W and 52 12 17N ; 000 08 26E?
  2. jcsd
  3. Mar 11, 2009 #2
    I'm not sure what level of math you're in. If you can make a triangle between the center of the earth and the two points, then you, can use trigonometric relationships to get the distance. Have you learned about the dot product yet? Or are you trying to do this based on basic geometry? Anyway, once you know the angle to getting the distance is basic trig if you go through, the earth. If you want the arc length then you [tex]S=r \theta[/tex] where [tex]\theta[/tex] is in radians.
  4. Mar 12, 2009 #3
    I'm in calculus 2 but the academic bowl is algebra, basic geometry and Spherical trig. i know how to find the dot product between two vectors from the little that i have done with vector algebra but i needed to find out the length by moveing on the earths serface. how could i do that given the infromation stated above??
  5. Mar 12, 2009 #4
    A great circle is also called a Geodesic. A Geodesic is the shortest distance between two points on a sphere. The airlines call these Great Circles and they like to fly on them to save fuel. Click on this link for more information.

    http://www.black-holes.org/relativity5.html [Broken]
    Last edited by a moderator: May 4, 2017
  6. Mar 12, 2009 #5


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    If [itex]\theta[/itex] is the difference between the latitudes and [itex]\phi[/itex] is the difference between the latitudes, then you have a spherical right triangle with legs of angular length [itex]\theta[/itex] and [itex]\phi[/itex]. The spherical version of the Pythagorean theorem is [itex]cos(\mu)= cos(\phi)cos(\theta)[/itex] where [itex]\mu[/itex] is the angular distance between the points. The actual distance between them is [itex]\mu[/itex] times the radius of the earth.

    http://www.math.uncc.edu/~droyster/math3181/notes/hyprgeom/node5.html [Broken]
    for more information.
    Last edited by a moderator: May 4, 2017
  7. Mar 13, 2009 #6
    so all that i would have to do is take 6 378.1km* ArcCos(Cos([itex]\phi[/itex])Cos([itex]\theta[/itex]) and that would give me the distance of my great circle?
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