1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple (or should be) Trigonomic question

  1. Oct 23, 2009 #1

    Yay

    User Avatar

    1. The problem statement, all variables and given/known data

    I'm making a program that tracks an object moving along the surface of a sphere (earth). I'm using a set of Cartesian coordinates who origin is the center of the earth and Z directed through the north pole, X the greenwich meridian at the equator, and Y is 90 degrees to everyone else.

    Given a Lat/Long value I can convert into a position vector by doing:

    phi = (90 - lat)*pi/180
    theta = long*pi/180

    vector.x = sin(phi)*cos(theta)
    vector.y = sin(phi)*sin(theta)
    vector.z = cos(phi)

    After moving my object around I need to convert it back into lat / long. At the moment I'm just trying to take a position in Lat/long, convert it into my position vector, draw it on a sphere , then convert it back.

    I know my conversion TO the position vector is working properly since it points to the right spot on a sphere. What I can't get right is the conversion FROM the vector back to lat long.

    2. Relevant equations
    See above.


    3. The attempt at a solution

    I thought I could find phi from the Z component, then solve X and Y simultaneously for theta. So since Z = cos(phi), I thought easy phi = 1/cos(Z).

    Only that didn't work. Sat down, thought for a bit, played around with cos and sec, and realized cos and sec aren't inverse operations (which I thought for sure they were) so sec(cos(a)) != a.

    There's a simple little rule for finding going from A = cos(angle), to an expression for the angle, when you know A. Can someone please tell me what it is ? I've spent hours on something I thought I learnt back in Highschool !
     
  2. jcsd
  3. Oct 23, 2009 #2

    berkeman

    User Avatar

    Staff: Mentor

    Is there a reason you aren't using spherical coordinates instead of cartesian coordinates? Seems like you're making it extra hard on yourself.

    Here are some trig notes from wikipedia if they help:

    http://en.wikipedia.org/wiki/Trig

    .
     
  4. Oct 23, 2009 #3

    Yay

    User Avatar

    We were told to use Cartesian, not spherical. Gonna have another read through the wiki page, its gotta be there somewhere.
     
  5. Oct 23, 2009 #4

    Yay

    User Avatar

    I worked it out. Turns out sec is the reciprocal of cos, arccos is the inverse function. I'm annoyed I forgot something so basic, but at least its all working now.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook