DIFF GEOM: Geodesics on the Sphere   

If r( u,v) is a parameterization of a sphere of radius R centered at the origin, then its unit normal n is parallel to r.

LiveGraphics3d Applet

Thus, a curve g( t) = r( u(t) ,v( t) ) is a geodesic on the sphere only if its acceleration is parallel to g( t) itself. However, if g^¢¢( t) = lg( t) for all t (and for some number l), then

d dt ( g( t) × g' ( t) ) = g' ( t) × g' ( t) + g(t) × g'' ( t) = 0+ g( t) ×l g(t) = 0

Consequently, g( t) × g' ( t) is constant and as a result, g(t) must lie in a plane through the origin. Thus, a geodesic on a sphere is a great circle, which is a circle formed by the intersection of the sphere with a plane through the sphere's center.

Suppose that P = á x₁,y₁,z₁ ñ  and Q = á x₂,y₂,z₂ ñ are two points ( = position vectors ) on a sphere of radius R centered at the origin. Let's develop a method for finding the great circle that passes through P and Q, as well as the shortest distance between P and Q on that sphere. 

Since P·P = R², a vector w that is perpendicular to P is given by

w = Q - projP( Q) = Q -   P·Q P·P   P = Q - R-2( P·Q) P

We rescale to obtain a vector Q_p perpendicular to P that has a length of R :

Qp =    R | | w| |   w

The great circle through P and Q is subsequently given by

g( t) = cos(t) P + sin(t) Qp

(4)

since g^¢¢ = - g and || g( t) || = R by construction. 

Drag the points to change the circle. (Adapted from an applet by Martin Kraus)

Moreover, if we define the angle a to be  

a = cos-1 æ è  P·Q R2 ö ø

then it can be shown that g(0) = P and g( a) = Q. Thus, g( t) is the great circle through P and Q.  Moreover, the shortest distance s  from P to Q on the surface of the sphere is 

s = Ra

since this is the shortest distance from P to Q along the great circle g( t).      

EXAMPLE 7    What is the parameterization of the geodesic that passes through the points P = ( 2,0,2) and Q = ( 0,2,2) on the sphere of radius 2Ö2 centered at the origin? What is the shortest distance between P and Q?        

Solution: Since R = 2Ö2 and P·Q = 4 the vector w is given by

w = Q-R-2( P·Q) P = á 2,0,2 ñ -( 2Ö2) -2(4) á 0,2,2 ñ = á 2,0,2 ñ -  1 2 á0,2,2 ñ

Thus, w = á 2,-1,1 ñ , so that ||w|| = Ö4+1+1 = Ö6 and

Qp =  R || w || w =  2Ö2 Ö6 á2,-1,1 ñ = á ñ

so that the parametrization of the great circle passing through P and Q is

g( t) = á 2,0,2 ñ cos(t) +  4 Ö3 ,  -2 Ö3 ,  2 Ö3 sin( t)

Since a = cos^-1( 4/8) , the shortest distance from P to Q is

s = 2Ö2 cos-1 æ è  1 2 ö ø =  p2Ö2 3 = 2.9619

as opposed to a distance of p = 3.14159 calculated in example 4.

       

If distances are not too great, then it is safe to assume the earth is a sphere (i.e., less than a thousand miles, for example). In such instances, great circles can be used to determine shortest paths between two points on the earth's surface. First, however, latitudes j must be converted into spherical angles f using one of the following formulas:

f =  p 2 -j    or    f = 90°-j

EXAMPLE 5    Johnson City, TN, is located at 82^°22^¢07^¢¢ W and 36^°19^¢53^¢¢ N.     Memphis, TN, is located at 90^°00^¢25^¢¢ W and 35^°6^¢20^¢¢ N. Assuming that the earth is a sphere of radius R = 3963 miles, what is the parametric equation of the great circle which passes through both Johnson City and Memphis?       

Solution: Johnson City in decimal degrees is located at latitude j = 36.331389^° and longitude q = 82.368611^°, while Memphis is at j = 35.105556^° and q = 90.006944^°. Thus, their spherical coordinates are given by

Johnson City : ( 3963,90°-36.331389°,82.368611°) = ( 3963, 53.668611°,82.368611°) Memphis :  ( 3963,90°-35.105556°,90.006944°) = ( 3963, 54.894444°,90.006944°)

Conversion to Cartesian coordinates yield the points P and Q, respectively, which rounded to the nearest mile are

Johnson City :     P = á424,-3164,2348 ñ Memphis :     Q = á0,-3242,2279 ñ

It follows that

w = Q-R-2( P·Q) P = á 0,-3242,2279 ñ -( 3963) -2(15608780) á 424,-3164,2348 ñ = á -421. 3924,-97. 4588,-54. 55965 ñ

so that Q_p is given by

Qp =  3963 435.9432 á-421. 3924,-97. 4588,-54. 55965 ñ = á-3831,-886,-496 ñ

Thus, the great circle passing through Memphis and Johnson City is

r( t) = cos( t) á424,-3164,2348 ñ +sin( t) á-3831,-886,-496 ñ

In addition, a = cos^-1( ( 3963)^-2 (15,608,780) ) =  0.11096 radians, so that the shortest distance from Memphis to Johnson City is

s = Ra = 3963·0.11096 = 439.73  miles