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),

$$x = a\sin v\cos u$$ (1)

$$y = a\sin v\sin u$$ (2)

$$z = c\cos v,$$ (3)

with $a>c$, $u\in [0,2\pi)$, and $v\in[0, \pi]$. Its Cartesian equation is

$${x^2+y^2\over a^2}+{z^2\over c^2}=1.$$ (4)

The Ellipticity of an oblate spheroid is defined by

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

so that

$$1-e^2 = {c^2\over a^2}.$$ (6)

Then the radial distance from the rotation axis is given by

$$r(\delta) = a\left({1+{e^2\over 1-e^2}\sin^2\delta}\right)^{-1/2}$$ (7)

as a function of the Latitude $\delta$.

The Surface Area and Volume of an oblate spheroid are

$$S = 2\pi a^2+\pi {c^2\over e} \ln\left({1+e\over 1-e}\right)$$ (8)

$$V = {\textstyle{4\over 3}} \pi a^2 c.$$ (9)

An oblate spheroid with its origin at a Focus has equation

$$r={a(1-e^2)\over 1+e\cos\phi}.$$ (10)

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

$$k \equiv e^2(1-e^2)^{-1} = e^2(1+e^2-2e^4+6e^6+\ldots)$$

$$= e^2+e^4-2e^6+\ldots$$ (11)

$$k^2 = e^4+e^6+\ldots$$ (12)

$$k^3 = e^6+\ldots.$$ (13)

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

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

In terms of Legendre Polynomials,

$${r\over a} = (1 - {\textstyle{1\over 6}} e^2 - {\textstyle{11\over 20}} e^4 - {\textstyle{103\over 1680}} e^6)$$

$$\phantom{=} + (- {\textstyle{1\over 3}} e^2 - {\textstyle{5\over 42}} e^4 - {\textstyle{3\over 56}} e^6) P_2$$

$$\phantom{=} + ({\textstyle{3\over 35}} e^4 + {\textstyle{57\over 770}} e^6) P_4 - {\textstyle{5\over 231}} e^6 P_6+\ldots.$$ (15)

The Ellipticity may also be expressed in terms of the Oblateness (also called Flattening), denoted $\epsilon$ or $f$.

$$\epsilon \equiv {a-c\over a}$$ (16)

$$c = a(1-\epsilon)$$ (17)

$$c^2 = a^2(1-\epsilon)^2$$ (18)

$$(1-\epsilon)^2 = 1-e^2,$$ (19)

so

$$\epsilon=1-\sqrt{1-e^2}$$ (20)

and

$$e^2 = 1-(1-\epsilon )^2 = 1-(1-2\epsilon +\epsilon^2) = 2\epsilon -\epsilon^2$$ (21)

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

Define $k$ and expand up to Powers of $\epsilon^6$

$$k \equiv (2\epsilon-\epsilon)(1-\epsilon)^{-2} = (2\epsilon-\epsilon^2)(1+2\epsilon-6\epsilon^2+\ldots)$$

$$= 2\epsilon+4\epsilon^4-12\epsilon^3-\epsilon^2-2\epsilon^3+\ldots$$

$$= 2\epsilon+3\epsilon^2-14\epsilon^3+\ldots$$ (23)

$$k^2 = 4\epsilon^2+6\epsilon^3+\ldots$$ (24)

$$k^3 = 8\epsilon^3+\ldots.$$ (25)

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

$${r\over a} = 1-{\textstyle{1\over 2}}(2\epsilon+3\epsilon^2-... ...lon^2+6\epsilon^3)\sin^4\delta+8\epsilon^3\sin^6\delta+\ldots.$$ (26)

In terms of Legendre Polynomials,

$${r\over a} = (1-{\textstyle{1\over 3}}\epsilon-{\textstyle{2\over 5}}\epsilon^... ...\epsilon-{\textstyle{1\over 7}}\epsilon^2-{\textstyle{1\over 21}}\epsilon^3)P_2$$

$$\phantom{=} +({\textstyle{12\over 35}}\epsilon^2-{\textstyle{96\over 385}}\epsilon^3)P_4-{\textstyle{40\over 231}}\epsilon^3 P_6+\ldots.$$ (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

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

Define

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

and $x\equiv\sin\theta$. Then

$$r(\theta) =a[1+k(1-x^2)]^{-1/2} = a(1+k-kx^2)^{-1/2}.$$ (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

$$y = r(\theta)\cos(\theta-B) = r(\theta)(\cos\theta\cos B-\sin\theta\sin B)$$

$$= r(\theta)(\sqrt{1-x^2}\cos B+x\sin B).$$ (31)

To find the highest projected point,

$${dy\over d\theta} = {{a\sin(B-\theta)}\over (1+k\cos^2\theta... ...theta)\cos\theta\sin\theta \over (1+k\cos^2\theta)^{3/2}} = 0.$$ (32)

Simplifying,

$$\tan(B-\theta)(1+k\cos^2\theta)+k\cos\theta\sin\theta = 0.$$ (33)

But

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

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

Plugging (34) into (33),

$${\sqrt{1-x^2}\tan B-x \over\sqrt{1-x^2}+x\tan B} [1+k(1-x^2)]+kx\sqrt{1-x^2} = 0$$ (35)

and performing a number of algebraic simplifications

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

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

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

$$(1+k)^2\tan^2 B(1-x^2)=x^2$$ (40)

$$x^2[1+(1+k)^2\tan^2 B]=(1+k)^2\tan^2 B$$ (41)

finally gives the expression for $x$ in terms of $B$ and $k$,

$$x^2={\tan^2 B(1+k)^2 \over 1+(1+k)^2\tan^2 B}.$$ (42)

Combine (30) and (31) and plug in for $x$,

$$y = a {\sqrt{1-x^2}\cos B+x\sin B \over \sqrt{1+k-k x^2}}$$

$$= a {\cos B+(1+k){\sin^2 B\over\cos B}\over \sqrt{(1+k)[1+(1+k)\tan^2 B]}}$$

$$= a {\cos^2 B+(1+k)\sin^2 B\over\cos B\sqrt{(1+k)[1+(1+k)\tan^2 B]}}.$$ (43)

Now re-express $k$ in terms of $a$ and $c$, using $\epsilon \equiv 1-{c/a}$,

$$k \equiv {(2-\epsilon)\epsilon \over (1-\epsilon)^2} = {\left({1+{c\over a}}\right) \left({1-{c\over a}}\right)\over \left({c\over a}\right)^2}$$

$$= {{1-\left({c\over a}\right)^2 \over\left({c\over a}\right)^2} = \left({a\over c}\right)^2-1},$$ (44)

so

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

Plug (44) and (45) into (43) to obtain the Semiminor Axis of the projected oblate spheroid,

$$c' = a {\cos^2 B+\left({a\over c}\right)^2\sin^2 B \over \cos B \sqrt{\left({a\over c}\right)^2\left[{1+\left({a\over c}\right)^2\tan^2 B}\right]}}$$

$$= a {\cos^2 B+\left({a\over c}\right)^2\sin^2 B \over {a\over c}\sqrt{\cos^2 B+\left({a\over c}\right)^2\sin^2 B}}$$

$$= c\sqrt{\cos^2 B+\left({a\over c}\right)^2\sin^2 B} = \sqrt{c^2 \cos^2 B+a^2 \sin^2 B}$$

$$= a\sqrt{(1-\epsilon)^2\cos^2 B+\sin^2 B}.$$ (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 $P$

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

$$= \left[\begin{array}{ccc}\cos P & 0 & \sin P \\ -\sin B \sin P & ... ...s B\cos P\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right].$$ (47)

Now, in the original coordinates $(x', y', z')$, the spheroid is given by the equation

$${{x'}^2\over a^2}+{{y'}^2\over c^2}+{{z'}^2\over a^2} = 1,$$ (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,

$$Ax^2+By^2+Cz^2+Dxy+Exz+Fyz=1,$$ (50)

where

$$A \equiv {\cos^2 P+\sin^2 B\sin^2 P\over a^2}+{\cos^2 B\sin^2P\over c^2}$$ (51)

$$B \equiv {\sin^2 P+\sin^2 B\cos^2 P\over a^2}+{\cos^2 B\cos^2 P\over c^2}$$ (52)

$$C \equiv {\cos^2 B\over a^2}+{\sin^2 B\over c^2}$$ (53)

$$D \equiv 2\cos P\sin P\left({{1-\sin^2 B\over a^2}-{\cos^2 B\over c^2}}\right)$$

$$= 2\cos P\sin P\cos^2 B\left({{1\over a^2}-{1\over c^2}}\right)$$ (54)

$$E \equiv 2\sin B\cos B\sin P\left({{1\over b^2}-{1\over a^2}}\right)$$ (55)

$$F \equiv 2\sin B\cos B\cos P\left({{1\over a^2}-{1\over b^2}}\right).$$ (56)

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

$$Cz^2+(Ex+Fy)z+(Ax^2+By^2+Dxy-1) = Cz^2+G(x,y)z+H(x,y) = 0,$$ (57)

where

$$G(x,y) \equiv Ex+Fy$$ (58)

$$H(x,y) \equiv Ax^2+By^2+Dxy-1.$$ (59)

$z$ can now be computed using the quadratic equation when $(x,y)$ is given,

$$z={-G(x,y)\pm\sqrt{G^2(x,y)-4CG(x,y)}\over 2C}.$$ (60)

If $P=0$, then we have $\sin P=0$ and $\cos P=1$, so (51) to (56) and (58) to (59) become

$$A \equiv {1\over a^2}$$ (61)

$$B \equiv {\sin^2 B\over a^2}+{\cos^2 B\over b^2}$$ (62)

$$C \equiv {\cos^2 B\over a^2}+{\sin^2 B\over b^2}$$ (63)

$$D \equiv 0$$ (64)

$$E \equiv 0$$ (65)

$$F \equiv 2\sin B\cos B\left({{1\over a^2}-{1\over b^2}}\right)$$ (66)

$$G(x,y) \equiv Fy = 2y\sin B\cos B\left({{1\over a^2}-{1\over b^2}}\right)$$ (67)

$$H(x,y) \equiv Ax^2+By^2-1$$

$$= {x^2\over a^2}+y^2\left({{\sin^2 B\over a^2}+{\cos^2 B\over b^2}}\right)-1.$$ (68)

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

References

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