Page 45 - Basic Structured Grid Generation

34 Basic Structured Grid Generation

Thus
c
τ = , (2.22)
2
a + c 2
a constant, in this case. Now if c is positive, so is τ. This is the case for a ‘right-handed’
helix, which twists around the z-axis in the same sense as a right-handed screwdriver
advancing in the direction of increasing z. But negative values of c give a left-handed
helix with the opposite sense, and the corresponding value of the torsion is negative
by eqn (2.22 ). Thus the convention used here for the sign of τ associates positive τ
with right-handed twisting about some axis and negative τ with left-handed twisting.
Right or left-handed twisting of a space-curve is also associated with a particular sense
of rotation of the osculating plane with increasing s. If the osculating plane does not
rotate, then b is a constant vector, the torsion is zero at every point, and the curve
must lie in a plane.
To evaluate dn/ds, we can differentiate n = b × t to obtain
dn dt db
= b × + × t = κb × n − τn × t =−κt + τb.
ds ds ds
Summarising, we have the Serret-Frenet formulas:
dt
= κn
ds
dn
=−κt + τb (2.23)
ds
db
= −τn,
ds
where the skew-symmetric nature of the array of scalars on the right-hand side serves
as an aide-memoire.
If at a point in space initial values of t and n (and hence of b also, since b = t × n)
are given, and if functions κ(s) and τ(s) of arc-length parameter s are also specified,
then in principle eqns (2.23) can be integrated to determine t, n,and b as functions
of s, and thus determine the space-curve in its entirety. The fundamental theorem
of space-curves states that specification of the functions κ(s) and τ(s) determines a
space-curve uniquely apart from its precise position in space. A proof is not difficult,
though we do not include one here.
A useful formula for torsion may be derived as follows. We begin with the identities

t = r , t = r ,and t = r , where the prime again stands for d/ds. Now, regarding
everything as a function of s,we have
2

t = (t ) = (κn) = κn + κ n = κ(−κt + τb) + κ n =−κ t + κ n + κτb.
Hence
2
3
2

r × r = t × t = (κn) × (−κ t + κ n + κτb) = κ b + κ τt.
Finally
3
2
2

r · (r × r ) = t · (κ b + κ τt) = κ τ,
so that

1 r · (r × r )

2
τ = r · (r × r ) = ρ r · (r × r ) = , (2.24)
κ 2 (r · r )