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Oblate Spheroid

A ``squashed'' Spheroid for which the equatorial radius $a$ is greater than the polar radius $c$, so $a>c$. To first approximation, the shape assumed by a rotating fluid (including the Earth, which is ``fluid'' over astronomical time scales) is an oblate spheroid. The oblate spheroid can be specified parametrically by the usual Spheroid equations (for a Spheroid with z-Axis as the symmetry axis),

$\displaystyle x$ $\textstyle =$ $\displaystyle a\sin v\cos u$ (1)
$\displaystyle y$ $\textstyle =$ $\displaystyle a\sin v\sin u$ (2)
$\displaystyle z$ $\textstyle =$ $\displaystyle c\cos v,$ (3)

with $a>c$, $u\in [0,2\pi)$, and $v\in[0, \pi]$. Its Cartesian equation is
{x^2+y^2\over a^2}+{z^2\over c^2}=1.
\end{displaymath} (4)

The Ellipticity of an oblate spheroid is defined by

e\equiv\sqrt{a^2-c^2\over a^2},
\end{displaymath} (5)

so that
1-e^2 = {c^2\over a^2}.
\end{displaymath} (6)

Then the radial distance from the rotation axis is given by
r(\delta) = a\left({1+{e^2\over 1-e^2}\sin^2\delta}\right)^{-1/2}
\end{displaymath} (7)

as a function of the Latitude $\delta$.

The Surface Area and Volume of an oblate spheroid are

$\displaystyle S$ $\textstyle =$ $\displaystyle 2\pi a^2+\pi {c^2\over e} \ln\left({1+e\over 1-e}\right)$ (8)
$\displaystyle V$ $\textstyle =$ $\displaystyle {\textstyle{4\over 3}} \pi a^2 c.$ (9)

An oblate spheroid with its origin at a Focus has equation
r={a(1-e^2)\over 1+e\cos\phi}.
\end{displaymath} (10)

Define $k$ and expand up to Powers of $e^6$,

$\displaystyle k$ $\textstyle \equiv$ $\displaystyle e^2(1-e^2)^{-1} = e^2(1+e^2-2e^4+6e^6+\ldots)$  
  $\textstyle =$ $\displaystyle e^2+e^4-2e^6+\ldots$ (11)
$\displaystyle k^2$ $\textstyle =$ $\displaystyle e^4+e^6+\ldots$ (12)
$\displaystyle k^3$ $\textstyle =$ $\displaystyle e^6+\ldots.$ (13)

Expanding $r$ in Powers of Ellipticity to $e^6$ therefore yields

{r\over a} = 1-{\textstyle{1\over 2}}(e^2+e^4-2e^4+6e^6)\sin...
...6)\sin^4\delta-{\textstyle{15\over 8}} e^6\sin^6\delta+\ldots.
\end{displaymath} (14)

In terms of Legendre Polynomials,
$\displaystyle {r\over a}$ $\textstyle =$ $\displaystyle (1 - {\textstyle{1\over 6}} e^2 - {\textstyle{11\over 20}} e^4 - {\textstyle{103\over 1680}} e^6)$  
  $\textstyle \phantom{=}$ $\displaystyle + (- {\textstyle{1\over 3}} e^2 - {\textstyle{5\over 42}} e^4 - {\textstyle{3\over 56}} e^6) P_2$  
  $\textstyle \phantom{=}$ $\displaystyle + ({\textstyle{3\over 35}} e^4 + {\textstyle{57\over 770}} e^6) P_4 - {\textstyle{5\over 231}} e^6 P_6+\ldots.$ (15)

The Ellipticity may also be expressed in terms of the Oblateness (also called Flattening), denoted $\epsilon$ or $f$.
\epsilon \equiv {a-c\over a}
\end{displaymath} (16)

c = a(1-\epsilon)
\end{displaymath} (17)

c^2 = a^2(1-\epsilon)^2
\end{displaymath} (18)

(1-\epsilon)^2 = 1-e^2,
\end{displaymath} (19)

\end{displaymath} (20)

e^2 = 1-(1-\epsilon )^2 = 1-(1-2\epsilon +\epsilon^2) = 2\epsilon -\epsilon^2
\end{displaymath} (21)

r = a\left[{1 + {2\epsilon -\epsilon^2\over (1-\epsilon)^2} \sin^2\delta}\right]^{-1/2}.
\end{displaymath} (22)

Define $k$ and expand up to Powers of $\epsilon^6$
$\displaystyle k$ $\textstyle \equiv$ $\displaystyle (2\epsilon-\epsilon)(1-\epsilon)^{-2} = (2\epsilon-\epsilon^2)(1+2\epsilon-6\epsilon^2+\ldots)$  
  $\textstyle =$ $\displaystyle 2\epsilon+4\epsilon^4-12\epsilon^3-\epsilon^2-2\epsilon^3+\ldots$  
  $\textstyle =$ $\displaystyle 2\epsilon+3\epsilon^2-14\epsilon^3+\ldots$ (23)
$\displaystyle k^2$ $\textstyle =$ $\displaystyle 4\epsilon^2+6\epsilon^3+\ldots$ (24)
$\displaystyle k^3$ $\textstyle =$ $\displaystyle 8\epsilon^3+\ldots.$ (25)

Expanding $r$ in Powers of the Oblateness to $\epsilon^3$ yields

{r\over a} = 1-{\textstyle{1\over 2}}(2\epsilon+3\epsilon^2-...
\end{displaymath} (26)

In terms of Legendre Polynomials,
$\displaystyle {r\over a}$ $\textstyle =$ $\displaystyle (1-{\textstyle{1\over 3}}\epsilon-{\textstyle{2\over 5}}\epsilon^...
...\epsilon-{\textstyle{1\over 7}}\epsilon^2-{\textstyle{1\over 21}}\epsilon^3)P_2$  
  $\textstyle \phantom{=}$ $\displaystyle +({\textstyle{12\over 35}}\epsilon^2-{\textstyle{96\over 385}}\epsilon^3)P_4-{\textstyle{40\over 231}}\epsilon^3 P_6+\ldots.$ (27)

To find the projection of an oblate spheroid onto a Plane, set up a coordinate system such that the z-Axis is towards the observer, and the $x$-axis is in the Plane of the page. The equation for an oblate spheroid is

r(\theta)=a\left[{1+{2\epsilon-\epsilon^2 \over (1-\epsilon)^2}\cos^2\theta}\right]^{-1/2}.
\end{displaymath} (28)

k\equiv {2\epsilon-\epsilon^2 \over (1-\epsilon)^2},
\end{displaymath} (29)

and $x\equiv\sin\theta$. Then
r(\theta) =a[1+k(1-x^2)]^{-1/2} = a(1+k-kx^2)^{-1/2}.
\end{displaymath} (30)

Now rotate that spheroid about the $x$-axis by an Angle $B$ so that the new symmetry axes for the spheroid are $x'\equiv
x$, $y'$, and $z'$. The projected height of a point in the $x=0$ Plane on the $y$-axis is
$\displaystyle y$ $\textstyle =$ $\displaystyle r(\theta)\cos(\theta-B) = r(\theta)(\cos\theta\cos B-\sin\theta\sin B)$  
  $\textstyle =$ $\displaystyle r(\theta)(\sqrt{1-x^2}\cos B+x\sin B).$ (31)

To find the highest projected point,
{dy\over d\theta} = {{a\sin(B-\theta)}\over (1+k\cos^2\theta...
...theta)\cos\theta\sin\theta \over (1+k\cos^2\theta)^{3/2}} = 0.
\end{displaymath} (32)

\tan(B-\theta)(1+k\cos^2\theta)+k\cos\theta\sin\theta = 0.
\end{displaymath} (33)

$\displaystyle \tan(B-\theta)$ $\textstyle =$ $\displaystyle {\tan B-\tan\theta \over 1+\tan B\tan\theta} = {\tan B-{\sin\thet...
...r\sqrt{1-\sin^2\theta}}\over 1+\tan B {\sin\theta\over \sqrt {1-\sin^2\theta}}}$  
  $\textstyle =$ $\displaystyle {\sqrt{1-\sin^2\theta}\tan B-\sin\theta \over\sqrt{1-\sin^2\theta}+\tan B\sin\theta}.$ (34)

Plugging (34) into (33),
{\sqrt{1-x^2}\tan B-x \over\sqrt{1-x^2}+x\tan B} [1+k(1-x^2)]+kx\sqrt{1-x^2} = 0
\end{displaymath} (35)

and performing a number of algebraic simplifications

(\sqrt{1-x^2}\tan B-x) (1+k-kx^2)+kx\sqrt{1-x^2}(\sqrt{1-x^2}+x\tan B) = 0
\end{displaymath} (36)

$[(1+k)\sqrt{1-x^2}\tan B-k x^2\sqrt{1-x^2}\tan B-x-kx+kx^3]$
$ +[kx(1-x^2)+kx^2\sqrt{1-x^2}\tan B]\quad$ (37)

(1+k)\tan B\sqrt{1-x^2}-kx(1-x^2)-x+kx(1-x^2) = 0
\end{displaymath} (38)

(1+k)\tan B\sqrt{1-x^2}=x
\end{displaymath} (39)

(1+k)^2\tan^2 B(1-x^2)=x^2
\end{displaymath} (40)

x^2[1+(1+k)^2\tan^2 B]=(1+k)^2\tan^2 B
\end{displaymath} (41)

finally gives the expression for $x$ in terms of $B$ and $k$,
x^2={\tan^2 B(1+k)^2 \over 1+(1+k)^2\tan^2 B}.
\end{displaymath} (42)

Combine (30) and (31) and plug in for $x$,
$\displaystyle y$ $\textstyle =$ $\displaystyle a {\sqrt{1-x^2}\cos B+x\sin B \over \sqrt{1+k-k x^2}}$  
  $\textstyle =$ $\displaystyle a {\cos B+(1+k){\sin^2 B\over\cos B}\over \sqrt{(1+k)[1+(1+k)\tan^2 B]}}$  
  $\textstyle =$ $\displaystyle a {\cos^2 B+(1+k)\sin^2 B\over\cos B\sqrt{(1+k)[1+(1+k)\tan^2 B]}}.$ (43)

Now re-express $k$ in terms of $a$ and $c$, using $\epsilon \equiv 1-{c/a}$,
$\displaystyle k$ $\textstyle \equiv$ $\displaystyle {(2-\epsilon)\epsilon \over (1-\epsilon)^2} = {\left({1+{c\over a}}\right)
\left({1-{c\over a}}\right)\over \left({c\over a}\right)^2}$  
  $\textstyle =$ $\displaystyle {{1-\left({c\over a}\right)^2 \over\left({c\over a}\right)^2} = \left({a\over c}\right)^2-1},$ (44)

1+k = \left({a\over c}\right)^2.
\end{displaymath} (45)

Plug (44) and (45) into (43) to obtain the Semiminor Axis of the projected oblate spheroid,
$\displaystyle c'$ $\textstyle =$ $\displaystyle a {\cos^2 B+\left({a\over c}\right)^2\sin^2 B \over \cos B \sqrt{\left({a\over c}\right)^2\left[{1+\left({a\over c}\right)^2\tan^2 B}\right]}}$  
  $\textstyle =$ $\displaystyle a {\cos^2 B+\left({a\over c}\right)^2\sin^2 B \over {a\over c}\sqrt{\cos^2 B+\left({a\over c}\right)^2\sin^2 B}}$  
  $\textstyle =$ $\displaystyle c\sqrt{\cos^2 B+\left({a\over c}\right)^2\sin^2 B} = \sqrt{c^2 \cos^2 B+a^2 \sin^2 B}$  
  $\textstyle =$ $\displaystyle a\sqrt{(1-\epsilon)^2\cos^2 B+\sin^2 B}.$ (46)

We wish to find the equation for a spheroid which has been rotated about the $x\equiv x'$-axis by Angle $B$, then the $z$-axis by Angle $P$

$\displaystyle \left[\begin{array}{c}x'\\  y'\\  z'\end{array}\right]$ $\textstyle =$ $\displaystyle \left[\begin{array}{ccc}1 & 0 & 0 \\  0 & \cos B & \sin B \\  0 &...
... & \cos P \end{array}\right]\left[\begin{array}{c}x\\  y\\  z\end{array}\right]$  
  $\textstyle =$ $\displaystyle \left[\begin{array}{ccc}\cos P & 0 & \sin P \\  -\sin B \sin P & ...
...s B\cos P\end{array}\right]\left[\begin{array}{c}x\\  y\\  z\end{array}\right].$  

Now, in the original coordinates $(x', y', z')$, the spheroid is given by the equation
{{x'}^2\over a^2}+{{y'}^2\over c^2}+{{z'}^2\over a^2} = 1,
\end{displaymath} (48)

which becomes in the new coordinates,

${(x\cos P+y\sin P)^2\over a^2}+{(-x\sin B\sin P+z\cos B+y\sin B\cos P)^2\over a^2}$
${}+ {(-x\cos B\sin P-z\sin B+y\cos B\cos P)^2\over c^2} = 1.\quad$ (49)
Collecting Coefficients,

\end{displaymath} (50)

$\displaystyle A$ $\textstyle \equiv$ $\displaystyle {\cos^2 P+\sin^2 B\sin^2 P\over a^2}+{\cos^2 B\sin^2P\over c^2}$ (51)
$\displaystyle B$ $\textstyle \equiv$ $\displaystyle {\sin^2 P+\sin^2 B\cos^2 P\over a^2}+{\cos^2 B\cos^2 P\over c^2}$ (52)
$\displaystyle C$ $\textstyle \equiv$ $\displaystyle {\cos^2 B\over a^2}+{\sin^2 B\over c^2}$ (53)
$\displaystyle D$ $\textstyle \equiv$ $\displaystyle 2\cos P\sin P\left({{1-\sin^2 B\over a^2}-{\cos^2 B\over c^2}}\right)$  
  $\textstyle =$ $\displaystyle 2\cos P\sin P\cos^2 B\left({{1\over a^2}-{1\over c^2}}\right)$ (54)
$\displaystyle E$ $\textstyle \equiv$ $\displaystyle 2\sin B\cos B\sin P\left({{1\over b^2}-{1\over a^2}}\right)$ (55)
$\displaystyle F$ $\textstyle \equiv$ $\displaystyle 2\sin B\cos B\cos P\left({{1\over a^2}-{1\over b^2}}\right).$ (56)

If we are interested in computing $z$, the radial distance from the symmetry axis of the spheroid ($y$) corresponding to a point

Cz^2+(Ex+Fy)z+(Ax^2+By^2+Dxy-1) = Cz^2+G(x,y)z+H(x,y) = 0,
\end{displaymath} (57)

$\displaystyle G(x,y)$ $\textstyle \equiv$ $\displaystyle Ex+Fy$ (58)
$\displaystyle H(x,y)$ $\textstyle \equiv$ $\displaystyle Ax^2+By^2+Dxy-1.$ (59)

$z$ can now be computed using the quadratic equation when $(x,y)$ is given,
z={-G(x,y)\pm\sqrt{G^2(x,y)-4CG(x,y)}\over 2C}.
\end{displaymath} (60)

If $P=0$, then we have $\sin P=0$ and $\cos P=1$, so (51) to (56) and (58) to (59) become
$\displaystyle A$ $\textstyle \equiv$ $\displaystyle {1\over a^2}$ (61)
$\displaystyle B$ $\textstyle \equiv$ $\displaystyle {\sin^2 B\over a^2}+{\cos^2 B\over b^2}$ (62)
$\displaystyle C$ $\textstyle \equiv$ $\displaystyle {\cos^2 B\over a^2}+{\sin^2 B\over b^2}$ (63)
$\displaystyle D$ $\textstyle \equiv$ $\displaystyle 0$ (64)
$\displaystyle E$ $\textstyle \equiv$ $\displaystyle 0$ (65)
$\displaystyle F$ $\textstyle \equiv$ $\displaystyle 2\sin B\cos B\left({{1\over a^2}-{1\over b^2}}\right)$ (66)
$\displaystyle G(x,y)$ $\textstyle \equiv$ $\displaystyle Fy = 2y\sin B\cos B\left({{1\over a^2}-{1\over b^2}}\right)$ (67)
$\displaystyle H(x,y)$ $\textstyle \equiv$ $\displaystyle Ax^2+By^2-1$  
  $\textstyle =$ $\displaystyle {x^2\over a^2}+y^2\left({{\sin^2 B\over a^2}+{\cos^2 B\over b^2}}\right)-1.$ (68)

See also Darwin-de Sitter Spheroid, Ellipsoid, Oblate Spheroidal Coordinates, Prolate Spheroid, Sphere, Spheroid


Beyer, W. H. CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, p. 131, 1987.

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© 1996-9 Eric W. Weisstein