Page 43 - Basic Structured Grid Generation

32 Basic Structured Grid Generation

and
2
¨ r = r ˙s + r ¨s. (2.13)
Hence we obtain
1 ¨ s ¨ s
2
κ = r · r = ¨ r − ˙ r · ¨ r − ˙ r
˙ s 4 ˙ s ˙ s
2
1 ¨ s ¨ s
= ¨ r · ¨ r − 2 ˙ r · ¨ r + ˙ r · ˙ r .
˙ s 4 ˙ s ˙ s 2
2
But using the identity ˙ r · ˙ r =˙ together with its derivative ˙ r · ¨ r =˙s¨s reduces this
s
expression for κ to
1
2
κ = (˙ r · ˙ r)(¨ r · ¨ r) − (˙ r · ¨ r) . (2.14)
3
(˙ r · ˙ r) 2
By the Lagrange identity (1.40) this is equivalent to
1 |˙ r × ¨ r|
κ = (˙ r × ¨ r) · (˙ r × ¨ r) = . (2.15)
3 3
(˙ r · ˙ r) 2 |˙ r|
For curves in two dimensions (the Oxy plane) eqn (2.15) reduces to the well-
known formula
|˙x ¨y −˙y ¨x|
κ = , (2.16)
2 2 3
(˙x +˙y ) 2
and if x is used as a parameter in place of t, this becomes
2 2
|d y/dx |
κ = 3 . (2.17)
2
[1 + (dy/dx) ] 2
The radius of curvature ρ at a given point is given by
1
ρ = . (2.18)
κ
Exercise 1. Show that for the twisted curve given by the parametric form x = ln cos t,
√
1
y = ln sin t, z = 2t(0 <t <π/2),we have κ = √ sin 2t.
2
2.2 The Serret-Frenet equations

Given a unit tangent vector t and a unit principal normal n at a point on a curve in
3
E , we can define a third unit vector b, called the unit binormal vector, orthogonal to
both of them, such that
b = t × n. (2.19)
The system of vectors (t, n, b) then forms a right-handed set of unit vectors, which
forms the moving trihedron as s varies, i.e. as we move along the curve. By definition,
we also have t = n × b and n = b × t. The planes at a point of the curve containing
the directions n and b, and the directions b and t, are called the normal plane and
the rectifying plane, respectively, (see Fig. 2.1) at that point. Differentiating eqn (2.19)