1. Limited time only! Sign up for a free 30min personal 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!

Oblate spheroid equations

  1. Aug 18, 2012 #1
    I'm a little confused by what appears to me to be a paradox. I understand that the equation for an oblate spheroid is given by
    \frac{x ^ 2+y^2}{a^2} + \frac{z ^ 2}{c^2}= 1, c < a
    where a and c are the semi-major and semi-minor axes respectively.
    However, the definitions for the WGS 84 oblate spheroid model of the Earth using the ECEF <--> geodetic latitude/longitude/height are (Wiki Link) the following equations
    x = N(\phi) cos\phi cos\lambda \\
    y = N(\phi) cos\phi sin\lambda \\
    z = N(\phi)(1-e^2) sin \phi \\
    N(\phi) = \frac{a}{\sqrt{1-e^2 sin^2\phi}}
    where [itex]\phi[/itex] is the geodetic latitude, [itex]\lambda[/itex] is the longitude, e is the eccentricity and a is the semi-major axis. I removed the height parameter as I am only concerned with the spheroid surface.

    I have also read a paper which states the equation relating the ECEF X,Y,Z is
    \frac{x ^ 2+y^2}{N(\phi)^2} + \frac{z ^ 2}{[N(\phi)(1-e^2)]^2}= 1

    The last equation has denominators that depends on the geodetic latitude while the first equation for the oblate spheroid has denominators which are constant. Is there something wrong?
    Last edited: Aug 19, 2012
  2. jcsd
  3. Aug 19, 2012 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The equation you quote for an oblate spheroid uses geocentric latitude. I believe the equations on the Wiki page use geodetic latitude. There is a comment about that on the page you linked.
  4. Aug 19, 2012 #3
    Well yes, but isn't ECEF independent of geodetic/geocentric coordinates? Then in the last equation, this seems to suggest that x,y,z must fit a different equation for different geodetic latitudes.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook