DESIGN OF CURVES 91(pi,qi,ri) and (pi+1,qi+1,ri+1). Also, r(2) (u 2 ) is modeled with end points (xi+1,yi+1,zi+1), (xi+2,yi+2,zi+2)
and end slopes (pi+1,qi+1,ri+1) and (pi+2,qi+2,ri+2). From Eq. (4.7) for continuity of the two curves
at the common joint:
C^0 continuity: r(1)(1) = r(2)(0) = (xi+1,yi+1,zi+1)C^1 continuity: d
dud
du
= (0)rr(1) (2) (pi+1,qi+1,rr+1)Thus, from the way the Ferguson segments result, the composite Ferguson curve is always C^1
continuous at the junction points. C^1 continuity is guaranteed at other intermediate points on the
curve as well since the segments are differentiable indefinitely, the segments being cubic in degree.
It is often difficult to conjecture the geometric interpretation of the slope dr(u)/du or (pi,qi,ri) at
the junction points in terms of their use as design parameters. A designer would desire to specify only
data points in curve design and evade the specifications pertaining to the slope, curvature or higher
order information. To avoid the slope specification from the user at intermediate data points, we can
additionally impose curvature or C^2 continuity at the junction points for which it would require that
κ(1)(1) = κ(2)(0)or
d
dud
du
d
dud
dud
du
d
durrrrrr(1)^2
2(1)(1)3(2)^2
2(2)(2)3(1) (1)(1)=(0) (0)(0)××
(4.12)Eq. (4.12) on substituting the conditions for position and slope continuity from Eqs. (4.10) and (4.11)
becomes
tr (1) = tr(0)2
2(1)^2
1(^22)
2
× ⎛ (2)
⎝
⎜
⎞
⎠
⎟ ×
d
du
d
du
α
α (4.13)
An equation that satisfies the condition above is
d
du
d
du
d
du
2
2
(1)
21
2 2
2
rrr(1) = (αα/ ) (2)(0) + μ (2)(0)
(4.14)
whereμ is some arbitrary scalar. Note that d
du
d
du
2
2
rr(1)(0) and (2)(0) have the same direction for
which theis class product is zero. For Ferguson’s composite curve, d
du
d
du
rr(1)(1) = (2)(0) for which
α 1 = α 2 from Eq. (4.11). Assuming μ = 0, Eq. (4.14) becomes
d
du
d
du
2
2
(1)^2
2
rr(1) = (2)(0) (4.15)
The second derivative of r(u) from Eq. (4.9) is given by
d
du
uu uu ui i i i
2
2 ( ) = (–6 + 12 ) + rP P+1(6 – 12 ) + (– 4 + 6 ) + T T+1(–2 + 6 ) (4.16)
From Eqs. (4.15) and (4.16), we have
6 Pi – 6Pi+1 + 2Ti + 4Ti+1 = –6Pi+1 + 6Pi+2 – 4Ti+1 – 2Ti+2
or Ti + 4Ti+1 + Ti+2 = 3Pi+2 – 3Pi, i = 0, 1,... , n – 2 (4.17)