Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Equation (with polar coordinates) of circle on a sphere

  1. Mar 24, 2015 #1
    hi,
    i'm a newbie......
    i have this problem:
    i have a sphere with known and constant R (obvious),
    i have two point with spherical coordinates
    P1=(R,p_1,t_1) and P0=(R, p_0, t_0)
    p_x = phi x = latitude x
    t_x = theta x =longitude x
    the distance between point is
    D= R*arccos[cos(p_0)*cos(p_1)+sin(p_0)*sin(p_1)*cos(t_1-t_0)]
    source (http://mathforum.org/library/drmath/view/51882.html)
    or similar (http://www.movable-type.co.uk/scripts/latlong.html)

    but in spherical coordinates wich is explicit formula of the
    circle on sphere with center in P0 and radius P0 to P1 ?

    many thanks
    mario

    PS image on this link
    https://www.dropbox.com/s/807dqx0wovvw34v/2015-03-24_213647.png?dl=0

    PSS i cannot understand if this is the same problem ....
    https://www.physicsforums.com/threa...ven-two-points-on-circle.571535/#post-3732362
     
    Last edited: Mar 24, 2015
  2. jcsd
  3. Mar 25, 2015 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I take it P0 and P1 are both on a sphere radius R centered at the origin?
    You want the spherical-polar coordinates for a circle center P0 that passes through P1?

    There are a number of ways to construct it - i.e. you can rotate the axes to that the z axis goes through P0, work out the formula, then rotate back.
    What have you tried?
     
  4. Mar 25, 2015 #3
    i have tried many way (intersection plane A x + B y + C z = D with sphere x^2 + y^2 + z^2 = R^2, intersect cylinder (or cone) with sphere) but i'm searching the simpliest algebrically solution .....

    PS soultion with programs like Wolframalpha or Derive for R*arccos[cos(p_0)*cos(p_1)+sin(p_0)*sin(p_1)*cos(t_1-t_0)] - D = 0 are to complex and give not the correct solution
     
  5. Mar 26, 2015 #4

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    How about this:
    If P0 is at ##\vec r_0## and P1 is at ##\vec r_1## then these two vectors form a plane.
    The circle you want is in the plane perpendicular to ##\vec r_0## that contains the point P1.
    The radius of the circle in that plane is given by Pythagoras ... sketch out the vectors and you should see what I mean.
     
    Last edited: Mar 26, 2015
  6. Mar 27, 2015 #5
    Last edited by a moderator: Apr 18, 2017
  7. Mar 28, 2015 #6

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Technically that is not the equation you asked for, but you can certainly use it to find the one you asked for.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Equation (with polar coordinates) of circle on a sphere
  1. Equation of a circle (Replies: 2)

Loading...