Page 67 - Basic Structured Grid Generation

56 Basic Structured Grid Generation

The first equation shows that circular cross-sections (parallels) of the surface u =

constant can be geodesics only when f (u) = 0, i.e. when the radius of the cross-section
is stationary with respect to u. The second equation shows that meridians v = constant
2
are geodesics. More generally, if the second equation is multiplied through by f it
can then be integrated directly to give
dv
2
f = h
ds
where h is a constant of integration. An explicit form for a geodesic then follows,
since by eqn (3.33)
2
2
4
2
2
2
2
2
2
f dv = h ds = h [(f 2 + g ) du + f dv ],
and then, re-arranging,
2 2 2 2 2 2 2 2
f (f − h ) dv = h (f + g ) du ,
so that

h f + g
2 2
v = C ± du (3.74)

2
f f − h 2
in terms of two constants of integration C, h. For example, in the case of a circular
cylinder, with f(u) = a,and g(u) = cu,where a and c are constants, we see that the
relationship between u and v must be linear. It follows that geodesics on a circular
cylinder are helices.
2
1
If we put u = u, u = v, the two eqns (3.71) may be combined into one by
assuming that the solution can be represented as a relation between the parameters u,

v. Since, with d/ds denoted by () ,
2

dv v d v d v 1 u v − v u
= and = = ,
du u du 2 du u u u 3

substituting for u and v from eqn (3.71) leads directly to
2
d v 1 dv 3 2 1 dv 2 2 1 dv 2
− 22 + ( 22 − 2 ) + (2 12 − ) + 11 = 0. (3.75)
11
12
du 2 du du du
For a planar surface on which we take surface co-ordinates to be rectangular carte-
sians u = X, v = Y, the Christoffel symbols are all zero, and eqn (3.75) reduces to
2
d Y
= 0,
dX 2
with straight line solutions Y = mX + C as expected.
Another, non-intrinsic, feature of geodesics arises when we consider their curvature
1
2
3
as viewed from E . For any curve on the surface with points r = r(u ,u ),there are
unit tangent vectors
dr ∂r du 1 ∂r du 2 du 1 du 2
= + = a 1 + a 2 . (3.76)
1
2
ds ∂u ds ∂u ds ds ds