Computer Aided Engineering Design

DESIGN OF CURVES 85

slopes via interpolation or curve fitting. Since the aim is to use low order parametric segments, two
models of cubic segments are discussed in detail in this chapter, namely (a) Ferguson’s or Hermite
cubic segments and (b) Bézier segments. Cubic segments are usually a good compromise for form
representation in most engineering applications. Differential properties like the tangents, normals and
curvatures are easy to compute. A cubic form of the segment ensures continuity of a composite curve
up to second order. It is computationally more efficient than the higher degree polynomials. Linear
or quadratic forms of curve segments, on the other hand, are incapable of modeling inflexions in the
curve.
We may commence with the parametric representation of a three-dimensional curve r(u) in
Eq. (6), that is

rijk() = () + () + () [(), (), ()]u xu yu zu≡ xuyuzu

Forr(u) to be cubic in u, the scalar polynomials x(u),y(u) and z(u) can be correspondingly expanded
as

x(u) = a 0 x + a 1 xu + a 2 xu^2 + a 3 xu^3

y(u) = a 0 y + a1yu + a 2 yu^2 + a 3 yu^3 (4.1)

z(u) = a 0 z + a 1 zu + a 2 zu^2 + a 3 zu^3

wherea 0 x,a 1 x,a 2 x and a 3 x are all unknowns in x(u) and likewise for y(u) and z(u). Alternatively, r(u)
may be written in the matrix form [x(u),y(u),z(u)] as

ru xu yu zu u u u

aaa

aaa

aaa

aaa

xyz

xyz

xyz

xyz

( ) [ ( ), ( ), ( )] = [^32 1] =

333

222

111

000

≡

⎡

⎣

⎢

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

⎥

UA (4.2)

whereU is the power basis vector andAthealgebraic coefficient matrix determined by the conditions
imposed on the curve segment to acquire a desired shape.

Example 4.1. Find a parametric cubic curve that starts at P 0 (–1, 2), ends at P 3 (8, 4) and passes
through two prescribed points P 1 (2, 4) and P 2 (6, 6).
We use Eq. (4.2) for this 2-D curve, ignoring the third column in A. Let u 0 = 0 at P 0 and
u 3 = 1 at P 3. Different ways may be employed to determine parameter values u 1 and u 2 corresponding
to the points P 1 and P 2 , respectively. For instance, we may place them uniformly in [0, 1] by setting
u 1 = 0.33 and u 2 = 0.66. The other alternative is to determine them in a manner suggestive of the
relative position of data points. We can find the chord lengths P 0 P 1 ,P 1 P 2 and P 2 P 3 as

d 1 = P 0 P 1 = √{(2 – (–1))^2 + (4 – 2)^2 } = 3.61

d 2 = P 1 P 2 = √{(6 – 2)^2 + (6 – 4)^2 } = 4.47

d 3 = P 2 P 3 = √{(8 – 6)^2 + (4 – 6)^2 } = 2.83

and determine parameter values as

uu

d

d

j j

01

1

=1

= 0, = 3 =

3.61

3.61 + 4.47 + 2.83

= 0.33

Σ