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

Radius of Ellipsoid

  1. Aug 21, 2008 #1
    The one formula I need I cannot find.

    I have an Oblate Ellipsoid which I can describe as a=b=2c. The values of a, b, and c are known to me.

    I've encountered multiple representations of symbols here, so...

    let Phi = angle from positive z-axis: 0<= Phi <=180.
    let Theta = angle from positive x-axis: 0<= Theta <=360

    Phi and Theta are also known to me.

    How can I determine the radius where the vector described by Phi and Theta pass through the surface of the Ellipsoid?
  2. jcsd
  3. Aug 21, 2008 #2


    User Avatar
    Science Advisor

    To save writing, let p=phi and t = theta.


    Ellipse: (x/a)2+(y/b)2+(z/c)2=1

    Plug x, y, and z into the ellipse equation and solve for r.
  4. Aug 22, 2008 #3
    I must be missing something, because I cannot see how to solve the first three equations; it appears to me that each still has 2 unknowns. How can I determine x, y, or z without knowing r?

    I realize that, being an Oblate Ellipsoid, I could probably use some formula for an ellipse, since any value for t should not change the value of r for a given p. Unfortunately, every formula discussing r I could find was to do with a chord from one of the foci, not from the center.

    I did find one rather hefty equation for r that looked like I could solve it, except that it uses a "flattening" constant; perhaps if someone could tell me how to determine this constant for a given Ellipsoid I would be able to proceed and find r.......http://www.mathworks.com/access/hel...=org.mozilla:en-US:official&client=firefox-a"
    Last edited by a moderator: Apr 23, 2017
  5. Aug 22, 2008 #4


    User Avatar
    Science Advisor

    You don't determine x, y, or z. Since you have looking for the "radius" in some specific direction, you are given [itex]\theta[/itex] and [itex]\phi[/itex]. Put [itex]x= r cos(\theta)sin(\phi)[/itex], [itex]y= r sin(\theta)sin(\phi)[/itex], [itex]z= r cos(\phi)[/itex] into the equation of the ellipsoid giving you an equation depending only on r. Then solve for r.

    For example, suppose [itex]\theta= 0[/itex], [itex]\phi= \pi/2[/itex] (so we are on the positive x-axis). Then x= r(1)(1)= r, y= r(0)(1)= 0, z= r(0)= 0. Your equation for the ellipsoid, with a= b= 2c is
    [tex]\frac{x^2}{4c^2}+ \frac{y^2}{4c^2}+ \frac{z^2}{c^2}= 1[/itex]
    which with x= r, y= 0, z= 0 becomes r2/4c2= 1 so r= 2c, exactly what we would expect. Taking [itex]\theta= \pi/2[/itex], [itex]\phi= \pi/2[/itex] so we are on the positive y axis would also give r= 2c, and taking [itex]\phi= 0[/itex], [itex]\theta[/itex] anything, gives z= 2c.

    For a more difficult example, take [itex]\theta= \pi/4[/itex], [itex]\phi= \pi/4[/itex]. Then [itex]x= r(\sqrt{2}/2)(\sqrt{2}/2)= r/2[/itex], [itex]y= r(\sqrt{2}/2)(\sqrt{2}/2)= r/2[/itex], [itex]z= r(\sqrt{2}/2)[/itex] and the equation of the ellipse becomes
    [tex]\frac{r^2}{4(2c)}+ \frac{r^2}{4(2c)}+ \frac{r^2}(2c}= 6r^2/8c= 1[/tex]
    so [itex]r= 2c/\sqrt{3}= 2\sqrt{3}c/3[/itex].

    Last edited by a moderator: Apr 23, 2017
  6. Aug 22, 2008 #5
    Since I am after a general equation for r, I have attempted to take your post and find said equation. Does this look correct...

  7. Aug 22, 2008 #6


    User Avatar
    Science Advisor

    You can greatly simplify to r=2c/(1+3cos2(p))1/2
  8. Aug 22, 2008 #7
    Where does "p" come from? Is that just shorthand for phi?

    Also, I am fuzzy on exponents. Please clarify - am I taking the square root of (1+3cos^2(p))

    ...or does the 1/2 exponent mean something else?
  9. Aug 23, 2008 #8

    I appreciate your reply. I expected [tex]\theta[/tex] to disappear from the equation since it has no effect on r, however I have two problems with your equation. The first is that I am not certain how to read it. Is this the form you mean...


    Second, how did you manipulate my equation to reach this simplified form?
  10. Aug 23, 2008 #9
    UPDATE - Re: Radius of Ellipsoid

    Folks, I appreciate the help so far. However, I hold a BS in Computer Engineering, not mathematics, and I am attempting to write a program, not proof any theorems or become an expert in trigonometry. After Mathman's last posting I tried to discover some rules about manipulating trig equations; I found some such as http://en.wikipedia.org/wiki/Trigonometric_identity" [Broken], but at this point such information is overwhelming. I will have to learn this stuff at some point, and much more since I must learn how to manipulate homogenous matrices (or whatever they are called) so as to be able to perform the computer graphic functions necessary to rotate and zoom my 3D galaxy of stars - which is where all this is leading.

    After some reconsideration, I believe I should define my spheroid (galaxy) as merely oblate, that is, with the terms: a=b>c. My reasoning is that I may wish to alter the eccentricity of the spheroid at some point, so my first request was probably too limited.

    Furthermore, this thread has taught me something important: I am able to find the information I need on the Internet, but without interaction with those like you who understand these matters, the information is, at least initially, meaningless to me. For example, I was aware of the conversions from Spherical to Cartesian, but was ignorant of their applicability to my problem until HallsofIvy poitned me in the right direction. Further, I did not even realize the equation could be simplified until Mathman presented me with his equation.

    Applying what I have learned so far, and using a=b>c to describe my spheroid, I have come up with...

    [tex]r=\sqrt {\frac{a^2c^2}{c^2(cos^2 \theta sin^2 \phi + sin^2 \theta sin^2 \phi) + a^2 cos^2 \phi}}[/tex]

    Two questions:

    1. Is my equation correct?

    2. Could someone simplify that using latex format. I found the form Mathman used to be difficult to read, and since I am unfamiliar with these equations, I may easily mistake what I see if it is not clear in form.

    It is logically clear to me that [tex]\theta[/tex] should disappear from the equation, because in an oblate spheroid, changing [tex]\theta[/tex] should not alter r, and Mathman's equation bears this out. I simply do not know how to proceed with the simplification. If I keep reading and studying and learning I've no doubt I'll figure it out eventually, but my priority right now is to progress with my code, not become expert in trig.

    Thanking you in advance.
    Last edited by a moderator: May 3, 2017
  11. Aug 23, 2008 #10


    User Avatar
    Science Advisor

    p is my shorthand for phi (I haven't learned Latex)

    1/2 power is same as square root.

    To get rid of theta, observe that sin2(a) + cos2(a) = 1 for any angle a.
  12. Aug 23, 2008 #11


    User Avatar
    Science Advisor

    Re: UPDATE - Re: Radius of Ellipsoid

    [tex]r=\sqrt {\frac{a^2c^2}{c^2sin^2 \phi (cos^2 \theta + sin^2 ) + a^2 cos^2 \phi}}[/tex]
    [tex]r= \sqrt{\frac{a^2c^2}{c^2sin^2\phi+ a^2 cos^2\phi}}[/tex]
    Notice that if a= c, then becomes
    [tex]r= \sqrt{\frac{a^4}{a^2(sin^2\phi+ cos^2\phi}}= a[/tex]
    as we would expect.

    Last edited by a moderator: May 3, 2017
  13. Aug 24, 2008 #12
    Thank you all. That is working fine now. I am having programmatic issues converting to cartesian coordinates, but that is a problem for another forum.

    I truly appreciate all the help.
  14. Mar 5, 2011 #13
    Hi all, I am working on a similar problem,

    the above equation

    r= \sqrt{\frac{a^2c^2}{c^2sin^2\phi+ a^2 cos^2\phi}}

    can be used also if a=b<c (Prolate spheroid)?

    And, also, how must it be updated if I have an oblate spheroid which is rotated 45° on the x axis?

    Thank you very much!
  15. Mar 8, 2011 #14


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    For a quadric surface: (x/a)^2 + (y/b)^2 + (z/c)^2 = 1, using a given longitude (L) and latitude (B) of a point on the surface (See wiki for 'Ellipsoid'), the parametric representation of the Cartesian coordinates is: x = a * cos B * cos L; y = b * cos B * sin L; z = c * sin B

    The, the position vector r(x,y,z) has radius (x^2+y^2+z^2)^(1/2) Note: r = constant only for the sphere (a = b = c = 1)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook