Computer Aided Engineering Design

DIFFERENTIAL GEOMETRY OF CURVES 77

Example 3.3. Find the length of a portion of the helix x=a cos u,y=a sin u,z=bu.
To use Eq. (3.21)

xauyauzbuauaubu ̇ ̇ = – sin , = cos , = , ( ) = cos + sin + ̇ rijk

Therefore, xyzab ̇ ̇222 2 2 + + = ̇ +

Hence sabduabu

u

= + = ( + )

0

22 22

∫

Since the length s is independent of the co-ordinate axis chosen, another representation of the helical
curve in terms of natural parameter would be

rijk( ) = cos

+

+ sin

+

+

(^222222) +
sa
s
ab
a
s
ab
b
s
ab
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎤
⎦
⎥
The normal at a point on a two-dimentional curve is unique. However, in three-dimensions, there
exists a plane of vectors perpendicular to the slope T. This plane is often referred to as the normal
plane (Figure 3.8 (a)). To span the vectors that are orthogonal to T, two unique vectors are identified
in the normal plane. The first is the principal normal N, while the second is the binormalB. To
determineN, consider
T() =
()
s = ( ) and ( + ) = ( + )
ds
ds
sssss
r
rT r′′ΔΔ
Figure 3.8 (a) The normal plane and (b) definition of a unit normal N
T(s + Δs)N
T(s)
N P(s)
r(s)
r(s

• Δ
  s)
  P(s + Δs)
  s = s 0
  (b)
  Osculating circle
  (a)
  Osculating plane
  Normal plane
  Rectifying plane
  B
  T
  N
  The net change in the direction of unit tangent in moving from P to a neighboring point Q
  (Figures 3.6 and 3.8b) is given by
  ΔΔTT T T ΔΔ Δ
  T T
  T
  T
  ( ) = ( + ) – ( ) = ( ) + +^1
  2!
  ( ) +... – ( )
  (^2) ()
  2
  ssss sd^2
  ds
  s d
  ds
  ss
  ds
  ds
  s
  ⎧
  ⎨
  ⎩
  ⎫
  ⎬
  ⎭
  ≈ (3.24)