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
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.
I think in this way: $x = a\sin(\varphi)\cos(\theta),\quad y = b\sin(\varphi)\sin(\theta),\quad z = c\cos(\varphi)$
I know this is old but wanted to offer an answer. If you want to use spherical coordinates, you need to parameterize $r$ as a function of $\theta$ and $\phi$.
Start with
$x = r\cos(\phi)\sin(\theta)$
$y = r\sin(\phi)\sin(\theta)$
$z = r\cos(\theta)$
Plug into the equation for an ellipsoid and get
$$r = \frac{1}{\sqrt{\left ((\cos(\phi)/a)^2 + (\sin(\phi)/b)^2)\sin(\theta)^2 + (\cos(\theta)/c)^2 \right )}}$$
Given an angle pair $(\theta, \phi)$ the above equation will give you the distance from the center of the ellipsoid to a point on the ellipsoid corresponding to $(\theta, \phi)$. This may be a little more work than some of the other parameterizations but has its uses.
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.
