Great Circle

The shortest path between two points on a Sphere, also known as an Orthodrome. To find the great circle (Geodesic) distance between two points located at Latitude $\delta$ and Longitude $\lambda$ of $(\delta_1, \lambda_1)$ and $(\delta_2, \lambda_2)$ on a Sphere of Radius $a$, convert Spherical Coordinates to Cartesian Coordinates using

$${\bf r}_i=a\left[{\matrix{\cos\lambda_i\cos\delta_i\cr \sin\lambda_i\cos\delta_i\cr \sin\delta_i\cr}}\right].$$ (1)

(Note that the Latitude $\delta$ is related to the Colatitude $\phi$ of Spherical Coordinates by $\delta=90^\circ-\phi$, so the conversion to Cartesian Coordinates replaces $\sin\phi$ and $\cos\phi$ by $\cos\delta$ and $\sin\delta$, respectively.) Now find the Angle $\alpha$ between ${\bf r}_1$ and ${\bf r}_2$ using the Dot Product,

$$\cos\alpha = \hat{\bf r}_1\cdot\hat{\bf r}_2$$

$$= \cos\delta_1\cos\delta_2(\sin\lambda_1\sin\lambda_2+\cos\lambda_1\cos\lambda_2)+\sin\delta_1\sin\delta_2$$

$$= \cos\delta_1\cos\delta_2\cos(\lambda_1-\lambda_2)+\sin\delta_1\sin\delta_2.$$ (2)

The great circle distance is then

$$d=a\cos^{-1}[\cos\delta_1\cos\delta_2\cos(\lambda_1-\lambda_2)+\sin\delta_1\sin\delta_2].$$ (3)

For the Earth, the equatorial Radius is $a\approx 6378$ km, or 3963 (statute) miles. Unfortunately, the Flattening of the Earth cannot be taken into account in this simple derivation, since the problem is considerably more complicated for a Spheroid or Ellipsoid (each of which has a Radius which is a function of Latitude).

The equation of the great circle can be explicitly computed using the Geodesic formalism. Writing

$$u = \lambda$$ (4)

$$v = \delta={\textstyle{1\over 2}}\pi-\phi$$ (5)

gives the $P$, $Q$, and $R$ parameters of the Geodesic (which are just combinations of the Partial Derivatives) as

$$P \equiv \left({\partial x\over \partial u}\right)^2+\left({\partial y\over \partial u}\right)^2 + \left({\partial z\over \partial u}\right)^2 = a^2\sin^2 v$$ (6)

$$Q \equiv {\partial x\over \partial u}{\partial x\over \partial v} +{\parti... ...\over \partial v} +{\partial z\over \partial u}{\partial z\over \partial v} = 0$$ (7)

$$R \equiv \left({\partial x\over \partial v}\right)^2+\left({\partial y\over \partial v}\right)^2 +\left({\partial z\over \partial v}\right)^2 = a^2.$$ (8)

The Geodesic differential equation then becomes

$$\cos v\sin^4 v+2\cos v\sin^2 v v'^2+\cos v v'^4-\sin v v''=0.$$ (9)

However, because this is a special case of $Q=0$ with $P$ and $R$ explicit functions of $v$ only, the Geodesic solution takes on the special form

$$v = c_1\int \sqrt{R\over P^2-{c_1}^2 P}\,dv = c_1\int{dv\over a^2\sin^4 v-{c_1}^2\sin^2 v}$$

$$= \int {dv\over\sin v\sqrt{\left({a\over c_1}\right)^2\sin^2 v-1}}$$

$$= -\tan^{-1}\left[{\cos v\over\sqrt{\left({a\over c_1}\right)^2-1}}\right]+ c_2$$ (10)

(Gradshteyn and Ryzhik 1979, p. 174, eqn. 2.599.6), which can be rewritten as

$$v = -\sin^{-1}\left({\cot v\over\sqrt{\left({a\over c_1}\right)^2-1}}\right)+ c_2.$$ (11)

It therefore follows that

$$(\sin c_2)a\sin v\cos u-(\cos c_2)a\sin v\sin u-{a\cos v\over\sqrt{\left({a\over c_1}\right)^2-1}}=0.$$ (12)

This equation can be written in terms of the Cartesian Coordinates as

$$x\sin c_2-y\cos c_2-{z\over\sqrt{\left({a\over c_1}\right)^2-1}}=0,$$ (13)

which is simply a Plane passing through the center of the Sphere and the two points on the surface of the Sphere.

See also Geodesic, Great Sphere, Loxodrome, Mikusinski's Problem, Orthodrome, Point-Point Distance--2-D, Sphere

References

Gradshteyn, I. S. and Ryzhik, I. M. Tables of Integrals, Series, and Products, 5th ed. San Diego, CA: Academic Press, 1979.

Weinstock, R. Calculus of Variations, with Applications to Physics and Engineering. New York: Dover, pp. 26-28 and 62-63, 1974.