Oblate spheroid

1. Dec 21, 2008

tony873004

oblate spheriod
From http://en.wikipedia.org/wiki/Oblate_spheroidal_coordinates I'm given the formulas to compute cartesian coordinates from oblate spheroidal coordinates.

$$\begin{array}{l} x = a\cosh \mu \,\,\cos \nu \,\,\cos \phi \\ y = a\cosh \mu \,\,\cos \nu \,\,\sin \phi \\ z = a\cosh \mu \,\,\sin \nu \\ \end{array}$$

But I've got a few questions:

I've written a short block of code to plot an oblate spheroid:
Code (Text):

mu = 0.5
For a = 0 To 100
For nu = 0 To (2*pi) Step 0.02
For phi = 0 To (2*pi) Step 0.02
x = a * cosh(mu) * Cos(nu) * Cos(phi)
y = a * cosh(mu) * Cos(nu) * Sin(phi)
z = a * sinh(mu) * Sin(nu)
Call PlotPoint(x, z)
Next phi
Next nu
Next a

What is μ ? I don't see it defined on Wikipedia's page. I image it is a number that describes how oblate the object is. So I imagine there should be a number I can set it to that gives me a sphere. My guess would be that mu can range from 0 to 1, with one of those limits giving me a sphere.

So I tried it. I see that μ = 0 gives me a straight line, meaning completely oblate, and the larger I set this number, the more spherical the shape becomes. But it seems to become a sphere at μ = 2, contrary to my guess. Any value larger than 2 gives me a larger sphere. Does anyone know exactly what μ is, and what it's range is?

Also, I see Earth described as an oblate spheroid. What is its μ value?

Thanks!!

2. Dec 21, 2008

I think you have μ and a backwards: μ is one of the coordinates, and a is a scale parameter. The surfaces of constant μ are oblate spheroids; they are never spheres, since cosh μ and sinh μ never coincide. For large μ they may appear to be very close, though, so it would look quite like a sphere. If you're plotting using this coordinate system, you want to keep a constant and vary μ. The reason the coordinate system is defined that way is to make it orthogonal. You will only get a sphere if μ is very large (and a is accordingly small).

If you're just trying to plot an oblate spheroid, you might consider using stretched spherical coordinates like such:
x = r sin θ cos φ
y = r sin θ sin φ
z = kr cos θ.​
where your coordinates are (r, θ, φ) and k is a parameter.

Last edited: Dec 21, 2008
3. Dec 21, 2008

tony873004

Thanks.

The Wikipedia article says "Thus, the two foci are transformed into a ring of radius a in the x-y plane." That's why I was assuming that a defined the size and μ defined the shape.

I just tried swapping 'a' and μ in my code. Now as it draws concentric shells, they start out oblate and become more spherical the larger μ becomes. Before, the oblateness was consistent between the concentric shells, and their size varied from 0 to max as 'a' got larger.

4. Dec 21, 2008

tony873004

Ulitmatey, what I'd like to do is a triple integral so I can compute the strength of Earth's gravity at a defined point in space, without approximating Earth as a point mass.

5. Dec 21, 2008

In that case, either coordinate system would work in principle, but it's probably a mess either way. I'm thinking it's probably better to use the stretched spherical coordinates, since in that case the volume element is just dV = kr2 sin θ dr dθ dφ.

6. Dec 21, 2008

D H

Staff Emeritus
Use spherical harmonics instead. You can get spherical harmonics coefficients to more accuracy than you get ever envision using off the web. The gold standards
For a brief description of spherical harmonics as used in the realm of gravity modeling, see http://www.cdeagle.com/pdf/gravity.pdf. Also see Vallado, "Fundamentals of Astrodynamics and Applications".

For low accuracy (but better than a point mass model), you can just use the second-order zonal harmonic, J2 component, the negative of the unnormalized C2,0 component: J2 = 0.00108263.

The Earth's potential with this second-order zonal harmonic term is

$$U(r) = \frac {GM_{\text{earth}}} {r} \left(1+\left(\frac a r)^2 P_{2,0}(sin\phi)C_{2,0}\cos 2 \lambda+\cdots\right)$$

where
• $P_{2,0}(x) = \frac 1 2 (3x^2 - 1)[/tex] is the second associated Legendre polynomial • [itex]a=6,378.135\,\text{km}$ is the Earth's equatorial radius
• $r$ is the radial distance from the center of the Earth to the point in question
• $\lambda$ is the longitude of the point in question
• $\phi$ is the geocentric (not geodetic) latitude (i.e. simple spherical geometry)

Last edited: Dec 21, 2008
7. Dec 22, 2008

tony873004

Thanks Adriank and DH. How does this formula know how oblate Earth is? Is it in C 2,0? For C 2,0, do I use .00108263, or the negative of that? Is this the number that describes how oblate Earth is?
Code (Text):

Private Sub Command1_Click()
Dim U As Double, G As Double, M As Double, a As Double, r As Double, C20 As Double
Dim U2 As Double
G = 6.6725985E-11: M = 5.97369125232006E+24: a = 6378135: r = 100000000#: lambda = 0.1: phi = 0.1: C20 = 0.00108263

U = (G * M / r ^ 2) * (1 + (a / r) ^ 2 * P20(Sin(phi)) * C20 * Cos(2 * lambda))
U2 = G * M / r ^ 2 'point mass
End Sub

Private Function P20(x As Double) As Double
P20 = 0.5 * (3 * x ^ 2 - 1)
End Function

In the above code, I squared the first instance of r to get acceleration, rather than potential. Is that valid to do in this equation? It seems to behave like I want, with the answer approaching the point mass approximation as r becomes large.