Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can someone help to describe some possible parametrizations for the ellipsoid:

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1?$$

I am thinking polar coordinates, but there may be the concept of steographic projecting (not sure how to apply it here), and not sure how many ways I can provide such possible parametrizations for the ellipsoid.

Thanks

share|improve this question
    
For ellipsis it looks like this $x = a\cos(\varphi), y = b\sin(\varphi)$ –  nikita2 Oct 2 '12 at 10:05

3 Answers 3

The answer to your question will depend on what you want to do with the ellipsoid. The Wikipedia page on geodesics on ellipsoids gives three possible parametrizations of the surface: (1) geographic latitude and longitude (useful if you're determining your position by astronomical observations); (2) parametric coordinates, probably the simplest to deal with computationally; (3) ellipsoidal coordinates, nice in that they are lines of curvature and hence orthogonal.

share|improve this answer

You mention polar coordinates, which allow you to parameterise a sphere.

Let $X = \frac{x}{a}$, $Y = \frac{y}{b}$, and $Z = \frac{z}{c}$. Then the equation becomes

$$X^2 + Y^2 + Z^2 = 1.$$

This is the equation of a sphere of radius $1$, so you can parameterise it using polar coordinates. Once you have done that, use the fact that $x = aX$, $y = bY$, and $z = cZ$ to obtain a parameterisation of the ellipsoid.

share|improve this answer

I think in this way: $x = a\sin(\theta)\cos(\varphi),\quad y = b\sin(\theta)\sin(\varphi),\quad z = c\cos(\theta)$

share|improve this answer
    
See also en.wikipedia.orgwiki/Ellipsoid#Parameterization –  nikita2 Mar 22 '13 at 6:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.