oblate spheroid

[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:

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!!

[adriank]

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.

[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.

[tony873004]

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.

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.

[adriank]

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.

[D H]

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.

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
EGM96, http://cddis.nasa.gov/926/egm96/egm96.html, to degree and order 360
GGM02C, http://www.csr.utexas.edu/grace/gravity
EGM2008, http://earth-info.nga.mil/GandG/wgs84/gravitymod/egm2008/index.html, to degree 2190 and order 2159.

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)

[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?

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.