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Osculatory interpolation
Computer Aided Geometric Design 18 (2001) 739–750 www.elsevier.com/locate/comaid
Osculatory interpolation M. Sakai Department of Mathematics, University of Kagoshima, Kagoshima, 890-0065 Japan Received May 2000; revised March 2001
Abstract We consider the problem of G2 two-point Hermite interpolation by rational cubics. Given two points with unit tangent vectors and consistent signed curvatures, the necessary and sufficient conditions are placed on the weights of the rational cubic curve which ensures that (i) if the data (control polygon) suggest a C-shaped curve, the the rational cubic interpolates a C-shaped curve without loops, cusps, or inflections, and (ii) if the data suggest an S-shaped curve, the the rational cubic interpolates an S-shaped curve with a single inflection, no loops and no cusps. 2001 Elsevier Science B.V. All rights reserved. Keywords: Rational cubics; Curvature; G2 Hermite interpolation; C-shaped; S-shaped; Inflections; Loops; Cusps; Mathematica
1. Introduction and description of the method Many efforts and proposals recently have been made for an approximation (interpolation) theory of parametric curves or planar data by geometric splines with polynomial or rational segments from the viewpoints of approximation orders and shape preserving properties; refer to (Degen, 1993; Degen, 1995) and therein for a unified theory of the general geometric Hermite interpolation. We consider G2 two-point Hermite interpolation by a planar rational cubic curve z(t) = (x(t), y(t)), 0 t 1, with four Bézier control vertices pi , 0 i 3, of the form; 3 3 3 i z(t) = pi wi Bi (t) wi Bi (t), Bi (t) = t (1 − t)3−i , (1.1) i i=0
i=0
where we assume without loss of generality that w0 = w3 = 1. E-mail address: [email protected] (M. Sakai). 0167-8396/01/$ – see front matter 2001 Elsevier Science B.V. All rights reserved. PII: S 0 1 6 7 - 8 3 9 6 ( 0 1 ) 0 0 0 5 2 - 8
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M. Sakai / Computer Aided Geometric Design 18 (2001) 739–750
The main object of this paper is to determine the regions of non-existence of inflections and singularities (loops and cusps) with respect to the weight (w1 , w2 ) according to the shapes of the data (control polygons). The signed curvature κ(t) is given by κ(t) = (z × z )(t)/z (t)3 ,
0 t 1,
(1.2)
where “×” means the cross product of the two planar vectors. The following osculatory interpolation problem is well-known: given the four control vertices and a curvature value at each end point, find weights w1 , w2 such that the resulting rational cubic of the form (1.1) assumes the given curvatures at the end points (Farin, 1993; Goodman, 1988). de Boor et al. (1987) considered another osculatory cubic polynomial interpolation (i.e., wi = 1, 0 i 3) passing through given two points and satisfying the corresponding tangent directions and consistent curvatures at the end points for a C-shaped control polygon. Then the curvatures κi , i = 0, 1, can be viewed as shape parameters while our results enable us to view the weights wi , i = 1, 2, as the shape ones for the given (fixed) curvatures, though the weights can be also used to achieve a high order approximation or a high order of contact at the end points, i.e., use of rational cubic approximations instead of polynomial yields considerable improvement in accuracy: The approximation order is raised to eight instead of six (Degen, 1993). For simplicity, we assume that at the end points (1, 0), (−1, 0), the tangent directions (the counterclockwise angle from the x-axis to the tangent vectors) π − θ, π + ψ (0 < θ, |ψ| < π ) and the curvatures κ0 , κ1 are given where C-and S-shaped polygons mean ψ > 0 and ψ < 0, respectively. Let p0 = (1, 0), p1 = (−b cos θ + 1, b sin θ ), p2 = (c cos ψ − 1, c sin ψ) and p3 = (−1, 0) with the positive parameters b, c. We remark that the “consistent” curvatures at the end points p0 , p 3 with respect to the corresponding tangent directions mean sign(z (i) × (p3 − p 0 )) = (−1)i sign(κi ), i = 0, 1, i.e., κ0 > 0 and κ1 > 0 (or < 0) for ψ > 0 (or ψ < 0). Then the rational cubic segment z(t) = ((x(t), y(t)) of the form (1.1) is given by (1 − t)3 + 3w1 (1 − b cos θ )(1 − t)2 t + 3w2 (c cos ψ − 1)(1 − t)t 2 − t 3 , (1 − t)3 + 3w1 (1 − t)2 t + 3w2 (1 − t)t 2 + t 3 3bw1 (1 − t)2 t sin θ + 3cw2 (1 − t)t 2 sin ψ y(t) = . (1.3) (1 − t)3 + 3w1 (1 − t)2 t + 3w2 (1 − t)t 2 + t 3
x(t) =
Fig. 1. C-shaped (θ > 0, ψ > 0) and and S-shaped (θ > 0, ψ < 0) polygons.
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Then, note that (i)
z(0) = (1, 0),
z(1) = (−1, 0),
(ii)
z (0)/z (0) = (− cos θ, sin θ ),
z (1)/z (1) = −(cos ψ, sin ψ).
(1.4)
Here we remark that once the above z(t)(= (x(t), y(t))) of the form (1.3) has been determined, we easily obtain the rational interpolation Z(t) with the end points (α, β) and (γ , δ) not necessarily equal to (1, 0) and (−1, 0). Consider a transformation: z → Z as 1 α − γ −(β − δ) 1 α+γ Z(t) = z(t) + , 0 t 1. (1.5) 2 β − δ α − γ cr 2 β +δ Then, note that for the curvature K(t) of Z(t) and Z (= Z(1) − Z(0)), (i)
Z(0) = (α, β),
(ii)
K(0) = (2/ Z)κ0 ,
(iii)
Z(1) = (γ , δ), K(1) = (2/ Z)κ1 ,
Z (0) × Z = Z (0) Z sin θ,
Z × Z (1) = ZZ (1) sin ψ,
i.e., the counterclockwise angle from Z (0) to Z be θ ; the counter-clockwise angle from Z to Z (1) be ψ. From (1.3), the prescribed curvatures κ0 , κ1 at the end points yield a quadratic system of equations in b, c: 3κ0 w12 2 b + 2c sin (θ + ψ) = 4 sin θ, w2 3κ1 w22 2 c + 2b sin (θ + ψ) = 4 sin ψ. w1
(1.6)
As in (Farin, 1993; p. 257), if p1 , p2 are also prescribed in addition to p0 , p3 , then (1.6) can be equivalently rewritten as 4w2 4w2 area[p0 , p1 , p2 ] = c κ0 = 0 , 3w12 dist3 [p0 , p1 ] 3w12 (1.7) 4w1 4w1 area[p1 , p2 , p3 ] κ1 = = c1 , 3w22 dist3 [p2 , p3 ] 3w22 where then solve (1.7) for w1 , w2 to obtain 2 1/3 c0 c1 c0 c12 1/3 4 (w1 , w2 ) = . , 3 κ02 κ1 κ0 κ12
(1.8)
First we consider the case when θ + ψ = π, 0, i.e., p 0 p 1 is not parallel to p 2 p3 . Introduce the following two positive quantities R0 , R1 which are different from the ones in (de Boor et al., 1987) by constant multipliers: κ0 w12 sin2 ψ κ1 w22 sin2 θ 3 (R0 , R1 ) = 2 , . (1.9) w2 sin θ w1 sin ψ sin (θ + ψ)
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We derive the necessary and sufficient condition on (R0 , R1 ), strictly speaking, on the weight (w1 , w2 ) for G2 Hermite interpolation: given the two fixed points p0 , p 3 (note that p1 , p2 are movable) and corresponding tangent directions and consistent curvatures where “consistent” means R0 , R1 > 0. As in (de Boor et al., 1987), an introduction of the new parameter (ρ0 , ρ1 ) as b sin (θ + ψ) c sin (θ + ψ) , , (1.10) (ρ0 , ρ1 ) = 2 sin ψ 2 sin θ reduces the quadratic system (1.6) to the quadratic one of equations in (ρ0 , ρ1 ) for (b, c) as ρ0 = 1 − R1 ρ12 ,
ρ1 = 1 − R0 ρ02
(1.11)
from which we obtain a quartic equation f (ρ) = 0 of ρ0 : 1 2 1 f (ρ) = ρ 2 − − 2 (1 − ρ). R0 R0 R1
(1.12)
Depending on the C- and S- shapes of the control polygons, we note the following requirements on the solution(s) (ρ0 , ρ1 ) of the quadratic system (1.11) or the zero(s) ρ0 of the quartic equation (1.12): Requirements on solution(s) ρ0 of quartic equation f (ρ) = 0 For a C-shaped polygon:
then 0 < ρ0 , ρ1 < 1 ⇒ ρ0 ∈ (0, 1/ R0 ) ∩ (0, 1) Case 2: if π < θ + ψ < 2π , then ρ0 , ρ1 < 0 ⇒ ρ0 ∈ (−∞, −1/ R0 )
Case 1: if
0 < θ + ψ < π,
For an S-shaped polygon:
√ Case 3: if −π < θ + ψ < 0, then ρ1 < 0 < ρ0 < 1 ⇒ ρ0 ∈ (1/ R0 , ∞) ∩ (0, 1) √ Case 4: if 0 < θ + ψ < π , then ρ0 < 0 < ρ1 < 1 ⇒ ρ0 ∈ (−1/ R0 , 0) ∗ ∗ ρ ∈ (−1/√R , 0) is equivalent to ρ ∈ (1/√R , ∞) ∩ (0, 1). 0 0 1 1
First, we give the following two theorems for the C-shaped polygon. Theorem 1.1 (Case 1)(de Boor et al., 1987). The number of the solutions (ρ0 , ρ1 ) of the quadratic system (1.11) (the number of rational cubic curves of the form (1.3)) is given with respect to (R0 , R1 ) in Fig. 2 where the curve given by R0 R1 {256R0R1 − 256(R0 + R1 ) + 288} = 27 has a cusp at (3/4, 3/4). Proof. The number of the positive solutions (ρ0 , ρ1 ) of the quadratic system (1.11) has been completely determined; for R0 R1 {256R0R1 − 256(R0 + R1 ) + 288} = 27, refer to Appendix. ✷ For example, Theorem 1.1 (Eq. (1.9)) would give a choice of the weight (w1 , w2 ) which comes from (R0 , R1 ) = (1, 1) giving the unique positive solution of the system (1.11) as w1 =
sin2 (θ + ψ) 2/3 1/3
3κ0 κ1
sin ψ
,
w2 =
sin2 (θ + ψ) 1/3 2/3
3κ0 κ1
sin θ
.
(1.13)
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Fig. 2. Number of solutions of quadratic system (1.11) (de Boor et al., 1987; p. 272).
Then, the quadratic system of Eq. (1.6) nicely has the unique solution: √ √ ( 5 − 1) sin ψ ( 5 − 1) sin θ b= , c= . sin (θ + ψ) sin (θ + ψ)
(1.14)
Theorem 1.2 (Case 2). The number of the solution (ρ0 , ρ1 ) of the quadratic system (1.11) (the number of rational cubic curve of the form (1.3)) is just one for R0 , R1 > 0. Proof. Note f (−d − 1/r) = r = R0 .
r 5 R1 d 4 + 4r 4 R1 d 3 + 4r 3 R1 d 2 − rd − (r + 1) , r 5 R1
(1.15)
Use Decartes’ rule of signs (concerning the number of positive roots) and intermediate value of theorem to give the desired result, i.e., f (ρ) has always a unique negative zero in (−∞, −1/r). ✷ Next, we consider the following two theorems for the S-shaped polygon. Theorem 1.3 (Case 3). The number of the solution (ρ0 , ρ1 ) of the quadratic system (1.11) (the number of rational cubic curve of the form (1.3)) is just one for R0 > 1, R1 > 0 and 0 for R0 1, R1 > 0. √ √ Proof. First, (1/ R0 , ∞) ∩ (0, 1) = ∅ requires r > 1 with r = R0 . Then, 2 f (1/r) = −(1 − 1/r)/ r 4 R1 < 0, f (1) = 1 − 1/r 2 > 0, f (ρ) = 4 ρ 2 − 1/r 2 ρ + 1/ r 4 R1 > 0 (ρ > 1/r), from which f (ρ) has a unique positive zero in (1/r, ∞) if and only if r > 1.
(1.16) ✷
Theorem 1.4 (Case 4). The number of the solution (ρ0 , ρ1 ) of the quadratic system (1.11) (the number of rational cubic curve of the form (1.3)) is just one for R0 > 0, R1 > 1 and 0 for R0 > 0, 0 < R1 1.
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Proof. Instead of ρ1 , eliminate ρ0 from (1.11) to get a quartic equation in ρ1 and note √ ρ1 ∈ (1/ R1 , ∞) to obtain the desired result as in the proof of Theorem 1.3. ✷ Remark. For the special cases when θ + ψ = π, 0, i.e., p 0 p1 is parallel to p 2 p 3 , the quadratic system of (1.6) reduces to 3κ0 w12 2 b = 4 sin θ, w2
3κ1 w22 2 c = 4 sin ψ. w1
(1.17)
Hence, no restrictions on the weights w1 , w2 are placed and positive unknowns b, c are given as follows: for a C-shaped polygon (θ + ψ = π ), “consistent” (with respect to the fixed points, tangent directions and curvatures; κ0 > 0, κ1 > 0) and for an S-shaped polygon (ψ = −θ ), “consistent” (κ0 > 0, κ1 < 0 ) give
2 w1 sin ψ 2 w2 sin θ , c= (0 < θ < π). (1.18) b= w1 3κ0 w2 3κ1
2. Shape preserving properties We shall prove the absence of singularities (loop and cusp) for both C- and S-shaped polygons. Assume that the rational cubic z(t)(= (x(t), y(t))) of the form (1.3) would have a loop, i.e., there exists (m, n) = (1/(u + 1), 1/(v + 1)), 0 < u = v < 1, such that x(m) − x(n) = 0,
y(m) − y(n) = 0.
(2.1)
With help of Mathematica, let (r, s) = (u + v, uv) to obtain a linear system of equations in b, c: w1 b 3w2 s + r − s 2 cos θ + w2 c(3w1 s + rs − 1) cos ψ (2.2) = 2(w1 r + 3w1 w2 s + w2 rs) + 2 r 2 − s /3, w1 b 3w2 s + r − s 2 sin θ − w2 c(3w1s + rs − 1) sin ψ = 0. First, we assume that θ + ψ = π, 0. Solve (2.2) for c to have c=
2{3(w1r + w2 rs + 3w1 w2 s) + r 2 − s} sin θ , 3w2 (3w1 s + rs − 1) sin (θ + ψ)
(2.3)
from which 3(w1 r + w2 rs + 3w1 w2 s) + r 2 − s > 0, 3w2 (3w1 s + rs − 1) 3(w1 r + w2 ) + r 2 − s > 0. 1 − ρ1 = R0 ρ02 = − 3w2 (3w1 s + rs − 1) ρ1 =
(2.4)
Note r 2 − 4s(= (u − v)2 ) > 0 and 0 < ρ1 < 1 to see that Eqs. (2.4) cannot be valid at the same time, i.e., the rational cubic z of the form (1.3) has no loops. Note that the system (2.2) cannot be valid since (cos ψ, sin ψ) = (− cos θ, sin θ ) for θ + ψ = π and
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(cos ψ, sin ψ) = (cos θ, − sin θ ) for θ + ψ = 0. This completes the proof of the absence of a loop. Next, assume that the rational cubic z(t)(= (x(t), y(t))) of the form (1.3) would have a cusp at t = m, 0 < m < 1 for which the necessary condition is x (m) = y (m) = 0 (the curvature becomes unbounded). With help of Mathematica, the condition gives with m = 1/(d + 1), 0 < d < ∞ w1 db −d 3 + 3w2 d + 2 cos θ + w2 c 2d 3 + 3w1 d 2 − 1 cos ψ = 4w2 d 3 + 2(3w1w2 + 1)d 2 + 4w1 d, w1 db −d 3 + 3w2 d + 2 sin θ − w2 c 2d 3 + 3w1 d 2 − 1 sin ψ = 0.
(2.5)
First, we assume that θ + ψ = π, 0. Solve (2.5) for c to obtain c=
2d(2w2d 2 + d + 3w1 w2 d + 2w1 ) sin θ w2 (2d 3 + 3w1 d 2 − 1) sin(θ + ψ)
(2.6)
from which follow d(2w2 d 2 + d + 3w1 w2 d + 2w1 ) > 0, w2 (2d 3 + 3w1 d 2 − 1) −(d 2 + 2w1 d + w2 ) > 0. 1 − ρ1 = w2 (2d 3 + 3w1 d 2 − 1)
ρ1 =
(2.7)
Note that Eqs. (2.7) cannot be valid at the same time, i.e., the rational cubic z of the form (1.3) has no cusps. When θ + ψ = 0, π , as in the treatment of the loop we have the absence of the cusp from (2.5). Theorem 2.1. The rational cubics of the form (1.3) in Theorems 1.1, 1.2 and Remark have neither loops nor cusps for both C- and S-shaped polygons. Next, we shall consider whether inflection points exist or not for both C- and S-shaped polygons. With help of Mathematica, (z × z )(t) =
9(d + 1)6 ci d i , 3 2 3 (d + 3w1 d + 3w2 d + 1) 3
t = 1/(d + 1) (d > 0), (2.8)
i=0
with
(c0 , c1 , c2 , c3 ) = 3κ1 w23 c3 , 4w2 c sin ψ, 4w1 b sin θ, 3κ0 w13 b3
where Eqs. (1.6) reduce c0 , c3 as c0 (= w1 w2 c{4 sin ψ − 2b sin (θ + ψ)}) = 3κ1 w23 c3 , c3 (= w1 w2 b{4 sin θ − 2c sin (θ + ψ)}) = 3κ0 w13 b3 . First note that b, c > 0. For the “consistent” C-shaped polygon (θ, ψ > 0), κ0 , κ1 > 0, from which ci > 0, 0 i 3, are all positive. For the “consistent” S-shaped polygon (θ > 0, ψ < 0), κ0 > 0, κ1 < 0, from which ci , 0 i 3, are (−, −, +, +). Use Decartes’
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rule of signs and intermediate value of theorem to prove the absence of inflection points for a C-shaped polygon, and the existence of just one inflection point for an S-shaped polygon. Theorem 2.2. The rational cubics of the form (1.3) in Theorems 1.1, 1.2 and Remark have no inflection point for a C-shaped polygon and just one inflection point for an S-shaped polygon.
3. Numerical examples We consider some numerical examples. Note that a choice of the weight (w1 , w2 ) = (1, 1) (if possible) ensures the 6th order accurate cubic polynomial interpolation (de Boor et al., 1987). Example 1 (C-shaped polygon). (θ, ψ, κ0 , κ1 ) = (π/8, π/6, 0.2, 0.6). Example 2 (C-shaped polygon). (θ, ψ, κ0 , κ1 ) = (π/8, π/4, 2, 1). Fig. 3.1 gives the number of rational cubics with respect to (w1 , w2 ). Note that a cubic polynomial curve (equivalent to (w1 , w2 ) = (1, 1)) is not available. Figs. 3.2 and 3.3 give the rational cubic curves, control polygons and curvatures where “thick curves” mean the curves, polygons and curvatures of rational cubic curves of the form (1.3) with the weights (w1 , w2 ) ≈ (1.45469, 1.31784) and (0.253477, 0.590101) by (1.13). Note that the abscissa in the curvatures is the chord-length distance. Example 3 (S-shaped polygon). (θ, ψ, κ0 κ1 ) = (π/4, −π/8, 0.5, −0.2). Fig. 4.1 shows a possible choice of the weight (w1 , w2 ) = (1, 1) ensuring the 6th order accurate cubic polynomial interpolation. Example 4 (de Boor et al., 1987). We give an example of the interpolation for a circular arc with radius 1 with 4 interpolation points. Then (1.13) gives (w1 , w2 ) = (2/3, 2/3)
Fig. 3.1. Number of solutions with respect to (w1 , w2 ) for Examples 1 and 2.
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Fig. 3.2. Curves, polygons and curvatures with (w1 , w2 ) ≈ (1.454, 1.317), (1, 1) for Example 1.
Fig. 3.3. Curves, polygons and curvatures with (w1 , w2 ) ≈ (0.253, 0.590), (1, 1.5) for Example 2.
Fig. 4.1. Number of solutions with respect to (w1 , w2 ) for Example 3.
Fig. 4.2. Curve, polygon and curvature with (w1 , w2 ) = (1, 1) for Example 3.
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Fig. 5. Curves, polygons and curvatures with (w1 , w2 ) = (2/3, 2/3), (1, 1), (1/2, 1/2) for Example 4.
Fig. 6. Curves, polygons and curvatures with (w1 , w2 ) = (0.157635, 0.219432) for Example 5.
(then the curve, polygon and curvature are denoted by “thick” ones. For w1 = w2 (= π(4 + π)/24) ≈ 0.9348 (equivalent to R0 = R1 (= π/(4 + π)/16) ≈ 1.40225), the derivative of the rational cubic of the form (1.3) does equal to the exact value of the unit circular arc at the end points (the curve, control polygon and the curvature are denoted in “dashed lines”). Note the change in vertical scale in Fig. 5. Example 5 (Degen, 1995). Finally we consider the logarithmic spiral: r = r0 ecφ with r0 = 30, c = 0.46, φ ∈ [0, 2.35]. In this example, (1.13) gives (w1 , w2 ) = (0.157635, 0.219432) with which the curve, polygon and curvature are denoted by “thick” ones while the original spiral and curvature are denoted by “thin” ones.
Acknowledgements The author gratefully acknowledges the comments and suggestions of the anonymous referees which helped to improve the paper.
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Appendix (Proof of Theorem 1.1) We assume R0 R1 > 0 since the other case R1 R0 > 0 can be similarly treated by deleting ρ0 from the quadratic system (1.11) to obtain a quartic equation √ in ρ1 . We have to note count the number of the roots of the quartic equation (1.12) in (0, 1/ R0 ) where we√ that the root(s) (if they exist) satisfy ρ0 < 1. To do so, let ρ = 1/{r(d + 1)} with r = R0 to obtain: f (ρ) =
r(R1 − 1)d 4 + (4rR1 − 4r + 1)d 3 r 5 R1 (d + 1)4 (4rR1 − 6r + 3)d 2 − (4r − 3)d − (r − 1) + . r 5 R1 (d + 1)4
Then, for the numerator g(d) of the above equation, g (d) = (d + 1) 4r(R1 − 1)d 2 + (8rR1 − 8r + 3)d − (4r − 3) . Depending on values of√Ri , i = 0, 1, we count the number N of the positive root(s) of g(d) = 0 where note r = R0 . Case 1. R0 R1 1: Combine Decartes’ rule of signs and intermediate value of theorem to give N = 1. Case 2. R1 R0 9/16: g (d) has just one positive zero, g(0) = −(r − 1) > 0 and g(∞) = −∞. Hence N = 1. Case 3. R0 > 9/16, R1 < Min{R0 , 1}: Assume that g (d) = 0 has two positive roots α, β, i,e., 64R0 R12 − 64R0 R1 + 9 > 0. Then, with help of Mathematica, we obtain g(α)g(β) =
R1 [R0 R1 {256R0R1 − 256(R0 + R1 ) + 288} − 27] . 256R0 (R1 − 1)3
Then, if
R0 R1 256R0R1 − 256(R0 + R1 ) + 288 > 27,
N = 2 (or 3) for g(0) > 0 (or < 0). Next, either if g(α)g(β) > 0 or if g (d) = 0 has no positive roots, g(0) > 0 (or < 0) gives N = 1 (or 0). Note that the curve represented by R0 R1 256R0R1 − 256(R0 + R1 ) + 288 = 27 is over the one by 64R0 R12 − 64R0 R1 + 9 = 0 where these two curves intersect only at (R0 , R1 ) = (3/4, 3/4). References de Boor, C., Höllig, K., Sabin, M., 1987. High accuracy geometric Hermite interpolation. Computer Aided Geometric Design 14, 269–278. Degen, W., 1992. Best approximation of parametric curves by splines, in: Lyche, T., Schumaker, L.L. (Eds.), Mathematical Methods in Computer Aided Geometric Design II. Academic Press, New York, pp. 171–184. Degen, W., 1993. High accurate rational approximation of parametric curves. Computer Aided Geometric Design 10, 293–313.
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Degen, W., 1995. High accuracy approximation of parametric curves, in: Daehlen, M., Lyche, T., Schumaker, L.L. (Eds.), Mathematical Methods for Curves and Surfaces. Vanderbilt University Press, Nashville, pp. 83–98. Farin, G., 1993. Curves and Surfaces for Computer Aided Geometric Design, A Practical Guide. Academic Press, New York. Floater, M., 1995. Rational cubic implicitization, in: Dæhlen, M., Lyche, T., Schumaker, L.L. (Eds.), Mathematical Methods for Curves and Surfaces. Vanderbilt University Press, Nashville, pp. 151– 159. Goodman, T.N.T., 1988. Shape preserving interpolation by parametric rational cubic splines. Technical Report, U. of Dundee, Department of Mathematics, and Computer Science. Li, Y.-M., Cripps, R., 1992. Identification of inflection points and cusps on rational curves. Computer Aided Geometric Design 9, 1–24. Mørken, K., 1995. Parametric interpolation by quadratic polynomial in the plane, in: Daehlen, M., Lyche, T., Schumaker, L.L. (Eds.), Mathematical Methods for Curves and Surfaces. Vanderbilt University Press, Nashville, pp. 385–402. Seymour, C., Unsworth, K., 1999. Interactive shape preserving interpolation by curvature continuous rational cubic splines. J. Comput. Appl. Math. 102, 87–117.
Osculatory interpolation M. Sakai Department of Mathematics, University of Kagoshima, Kagoshima, 890-0065 Japan Received May 2000; revised March 2001
Abstract We consider the problem of G2 two-point Hermite interpolation by rational cubics. Given two points with unit tangent vectors and consistent signed curvatures, the necessary and sufficient conditions are placed on the weights of the rational cubic curve which ensures that (i) if the data (control polygon) suggest a C-shaped curve, the the rational cubic interpolates a C-shaped curve without loops, cusps, or inflections, and (ii) if the data suggest an S-shaped curve, the the rational cubic interpolates an S-shaped curve with a single inflection, no loops and no cusps. 2001 Elsevier Science B.V. All rights reserved. Keywords: Rational cubics; Curvature; G2 Hermite interpolation; C-shaped; S-shaped; Inflections; Loops; Cusps; Mathematica
1. Introduction and description of the method Many efforts and proposals recently have been made for an approximation (interpolation) theory of parametric curves or planar data by geometric splines with polynomial or rational segments from the viewpoints of approximation orders and shape preserving properties; refer to (Degen, 1993; Degen, 1995) and therein for a unified theory of the general geometric Hermite interpolation. We consider G2 two-point Hermite interpolation by a planar rational cubic curve z(t) = (x(t), y(t)), 0 t 1, with four Bézier control vertices pi , 0 i 3, of the form; 3 3 3 i z(t) = pi wi Bi (t) wi Bi (t), Bi (t) = t (1 − t)3−i , (1.1) i i=0
i=0
where we assume without loss of generality that w0 = w3 = 1. E-mail address: [email protected] (M. Sakai). 0167-8396/01/$ – see front matter 2001 Elsevier Science B.V. All rights reserved. PII: S 0 1 6 7 - 8 3 9 6 ( 0 1 ) 0 0 0 5 2 - 8
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The main object of this paper is to determine the regions of non-existence of inflections and singularities (loops and cusps) with respect to the weight (w1 , w2 ) according to the shapes of the data (control polygons). The signed curvature κ(t) is given by κ(t) = (z × z )(t)/z (t)3 ,
0 t 1,
(1.2)
where “×” means the cross product of the two planar vectors. The following osculatory interpolation problem is well-known: given the four control vertices and a curvature value at each end point, find weights w1 , w2 such that the resulting rational cubic of the form (1.1) assumes the given curvatures at the end points (Farin, 1993; Goodman, 1988). de Boor et al. (1987) considered another osculatory cubic polynomial interpolation (i.e., wi = 1, 0 i 3) passing through given two points and satisfying the corresponding tangent directions and consistent curvatures at the end points for a C-shaped control polygon. Then the curvatures κi , i = 0, 1, can be viewed as shape parameters while our results enable us to view the weights wi , i = 1, 2, as the shape ones for the given (fixed) curvatures, though the weights can be also used to achieve a high order approximation or a high order of contact at the end points, i.e., use of rational cubic approximations instead of polynomial yields considerable improvement in accuracy: The approximation order is raised to eight instead of six (Degen, 1993). For simplicity, we assume that at the end points (1, 0), (−1, 0), the tangent directions (the counterclockwise angle from the x-axis to the tangent vectors) π − θ, π + ψ (0 < θ, |ψ| < π ) and the curvatures κ0 , κ1 are given where C-and S-shaped polygons mean ψ > 0 and ψ < 0, respectively. Let p0 = (1, 0), p1 = (−b cos θ + 1, b sin θ ), p2 = (c cos ψ − 1, c sin ψ) and p3 = (−1, 0) with the positive parameters b, c. We remark that the “consistent” curvatures at the end points p0 , p 3 with respect to the corresponding tangent directions mean sign(z (i) × (p3 − p 0 )) = (−1)i sign(κi ), i = 0, 1, i.e., κ0 > 0 and κ1 > 0 (or < 0) for ψ > 0 (or ψ < 0). Then the rational cubic segment z(t) = ((x(t), y(t)) of the form (1.1) is given by (1 − t)3 + 3w1 (1 − b cos θ )(1 − t)2 t + 3w2 (c cos ψ − 1)(1 − t)t 2 − t 3 , (1 − t)3 + 3w1 (1 − t)2 t + 3w2 (1 − t)t 2 + t 3 3bw1 (1 − t)2 t sin θ + 3cw2 (1 − t)t 2 sin ψ y(t) = . (1.3) (1 − t)3 + 3w1 (1 − t)2 t + 3w2 (1 − t)t 2 + t 3
x(t) =
Fig. 1. C-shaped (θ > 0, ψ > 0) and and S-shaped (θ > 0, ψ < 0) polygons.
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Then, note that (i)
z(0) = (1, 0),
z(1) = (−1, 0),
(ii)
z (0)/z (0) = (− cos θ, sin θ ),
z (1)/z (1) = −(cos ψ, sin ψ).
(1.4)
Here we remark that once the above z(t)(= (x(t), y(t))) of the form (1.3) has been determined, we easily obtain the rational interpolation Z(t) with the end points (α, β) and (γ , δ) not necessarily equal to (1, 0) and (−1, 0). Consider a transformation: z → Z as 1 α − γ −(β − δ) 1 α+γ Z(t) = z(t) + , 0 t 1. (1.5) 2 β − δ α − γ cr 2 β +δ Then, note that for the curvature K(t) of Z(t) and Z (= Z(1) − Z(0)), (i)
Z(0) = (α, β),
(ii)
K(0) = (2/ Z)κ0 ,
(iii)
Z(1) = (γ , δ), K(1) = (2/ Z)κ1 ,
Z (0) × Z = Z (0) Z sin θ,
Z × Z (1) = ZZ (1) sin ψ,
i.e., the counterclockwise angle from Z (0) to Z be θ ; the counter-clockwise angle from Z to Z (1) be ψ. From (1.3), the prescribed curvatures κ0 , κ1 at the end points yield a quadratic system of equations in b, c: 3κ0 w12 2 b + 2c sin (θ + ψ) = 4 sin θ, w2 3κ1 w22 2 c + 2b sin (θ + ψ) = 4 sin ψ. w1
(1.6)
As in (Farin, 1993; p. 257), if p1 , p2 are also prescribed in addition to p0 , p3 , then (1.6) can be equivalently rewritten as 4w2 4w2 area[p0 , p1 , p2 ] = c κ0 = 0 , 3w12 dist3 [p0 , p1 ] 3w12 (1.7) 4w1 4w1 area[p1 , p2 , p3 ] κ1 = = c1 , 3w22 dist3 [p2 , p3 ] 3w22 where then solve (1.7) for w1 , w2 to obtain 2 1/3 c0 c1 c0 c12 1/3 4 (w1 , w2 ) = . , 3 κ02 κ1 κ0 κ12
(1.8)
First we consider the case when θ + ψ = π, 0, i.e., p 0 p 1 is not parallel to p 2 p3 . Introduce the following two positive quantities R0 , R1 which are different from the ones in (de Boor et al., 1987) by constant multipliers: κ0 w12 sin2 ψ κ1 w22 sin2 θ 3 (R0 , R1 ) = 2 , . (1.9) w2 sin θ w1 sin ψ sin (θ + ψ)
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We derive the necessary and sufficient condition on (R0 , R1 ), strictly speaking, on the weight (w1 , w2 ) for G2 Hermite interpolation: given the two fixed points p0 , p 3 (note that p1 , p2 are movable) and corresponding tangent directions and consistent curvatures where “consistent” means R0 , R1 > 0. As in (de Boor et al., 1987), an introduction of the new parameter (ρ0 , ρ1 ) as b sin (θ + ψ) c sin (θ + ψ) , , (1.10) (ρ0 , ρ1 ) = 2 sin ψ 2 sin θ reduces the quadratic system (1.6) to the quadratic one of equations in (ρ0 , ρ1 ) for (b, c) as ρ0 = 1 − R1 ρ12 ,
ρ1 = 1 − R0 ρ02
(1.11)
from which we obtain a quartic equation f (ρ) = 0 of ρ0 : 1 2 1 f (ρ) = ρ 2 − − 2 (1 − ρ). R0 R0 R1
(1.12)
Depending on the C- and S- shapes of the control polygons, we note the following requirements on the solution(s) (ρ0 , ρ1 ) of the quadratic system (1.11) or the zero(s) ρ0 of the quartic equation (1.12): Requirements on solution(s) ρ0 of quartic equation f (ρ) = 0 For a C-shaped polygon:
then 0 < ρ0 , ρ1 < 1 ⇒ ρ0 ∈ (0, 1/ R0 ) ∩ (0, 1) Case 2: if π < θ + ψ < 2π , then ρ0 , ρ1 < 0 ⇒ ρ0 ∈ (−∞, −1/ R0 )
Case 1: if
0 < θ + ψ < π,
For an S-shaped polygon:
√ Case 3: if −π < θ + ψ < 0, then ρ1 < 0 < ρ0 < 1 ⇒ ρ0 ∈ (1/ R0 , ∞) ∩ (0, 1) √ Case 4: if 0 < θ + ψ < π , then ρ0 < 0 < ρ1 < 1 ⇒ ρ0 ∈ (−1/ R0 , 0) ∗ ∗ ρ ∈ (−1/√R , 0) is equivalent to ρ ∈ (1/√R , ∞) ∩ (0, 1). 0 0 1 1
First, we give the following two theorems for the C-shaped polygon. Theorem 1.1 (Case 1)(de Boor et al., 1987). The number of the solutions (ρ0 , ρ1 ) of the quadratic system (1.11) (the number of rational cubic curves of the form (1.3)) is given with respect to (R0 , R1 ) in Fig. 2 where the curve given by R0 R1 {256R0R1 − 256(R0 + R1 ) + 288} = 27 has a cusp at (3/4, 3/4). Proof. The number of the positive solutions (ρ0 , ρ1 ) of the quadratic system (1.11) has been completely determined; for R0 R1 {256R0R1 − 256(R0 + R1 ) + 288} = 27, refer to Appendix. ✷ For example, Theorem 1.1 (Eq. (1.9)) would give a choice of the weight (w1 , w2 ) which comes from (R0 , R1 ) = (1, 1) giving the unique positive solution of the system (1.11) as w1 =
sin2 (θ + ψ) 2/3 1/3
3κ0 κ1
sin ψ
,
w2 =
sin2 (θ + ψ) 1/3 2/3
3κ0 κ1
sin θ
.
(1.13)
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Fig. 2. Number of solutions of quadratic system (1.11) (de Boor et al., 1987; p. 272).
Then, the quadratic system of Eq. (1.6) nicely has the unique solution: √ √ ( 5 − 1) sin ψ ( 5 − 1) sin θ b= , c= . sin (θ + ψ) sin (θ + ψ)
(1.14)
Theorem 1.2 (Case 2). The number of the solution (ρ0 , ρ1 ) of the quadratic system (1.11) (the number of rational cubic curve of the form (1.3)) is just one for R0 , R1 > 0. Proof. Note f (−d − 1/r) = r = R0 .
r 5 R1 d 4 + 4r 4 R1 d 3 + 4r 3 R1 d 2 − rd − (r + 1) , r 5 R1
(1.15)
Use Decartes’ rule of signs (concerning the number of positive roots) and intermediate value of theorem to give the desired result, i.e., f (ρ) has always a unique negative zero in (−∞, −1/r). ✷ Next, we consider the following two theorems for the S-shaped polygon. Theorem 1.3 (Case 3). The number of the solution (ρ0 , ρ1 ) of the quadratic system (1.11) (the number of rational cubic curve of the form (1.3)) is just one for R0 > 1, R1 > 0 and 0 for R0 1, R1 > 0. √ √ Proof. First, (1/ R0 , ∞) ∩ (0, 1) = ∅ requires r > 1 with r = R0 . Then, 2 f (1/r) = −(1 − 1/r)/ r 4 R1 < 0, f (1) = 1 − 1/r 2 > 0, f (ρ) = 4 ρ 2 − 1/r 2 ρ + 1/ r 4 R1 > 0 (ρ > 1/r), from which f (ρ) has a unique positive zero in (1/r, ∞) if and only if r > 1.
(1.16) ✷
Theorem 1.4 (Case 4). The number of the solution (ρ0 , ρ1 ) of the quadratic system (1.11) (the number of rational cubic curve of the form (1.3)) is just one for R0 > 0, R1 > 1 and 0 for R0 > 0, 0 < R1 1.
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Proof. Instead of ρ1 , eliminate ρ0 from (1.11) to get a quartic equation in ρ1 and note √ ρ1 ∈ (1/ R1 , ∞) to obtain the desired result as in the proof of Theorem 1.3. ✷ Remark. For the special cases when θ + ψ = π, 0, i.e., p 0 p1 is parallel to p 2 p 3 , the quadratic system of (1.6) reduces to 3κ0 w12 2 b = 4 sin θ, w2
3κ1 w22 2 c = 4 sin ψ. w1
(1.17)
Hence, no restrictions on the weights w1 , w2 are placed and positive unknowns b, c are given as follows: for a C-shaped polygon (θ + ψ = π ), “consistent” (with respect to the fixed points, tangent directions and curvatures; κ0 > 0, κ1 > 0) and for an S-shaped polygon (ψ = −θ ), “consistent” (κ0 > 0, κ1 < 0 ) give
2 w1 sin ψ 2 w2 sin θ , c= (0 < θ < π). (1.18) b= w1 3κ0 w2 3κ1
2. Shape preserving properties We shall prove the absence of singularities (loop and cusp) for both C- and S-shaped polygons. Assume that the rational cubic z(t)(= (x(t), y(t))) of the form (1.3) would have a loop, i.e., there exists (m, n) = (1/(u + 1), 1/(v + 1)), 0 < u = v < 1, such that x(m) − x(n) = 0,
y(m) − y(n) = 0.
(2.1)
With help of Mathematica, let (r, s) = (u + v, uv) to obtain a linear system of equations in b, c: w1 b 3w2 s + r − s 2 cos θ + w2 c(3w1 s + rs − 1) cos ψ (2.2) = 2(w1 r + 3w1 w2 s + w2 rs) + 2 r 2 − s /3, w1 b 3w2 s + r − s 2 sin θ − w2 c(3w1s + rs − 1) sin ψ = 0. First, we assume that θ + ψ = π, 0. Solve (2.2) for c to have c=
2{3(w1r + w2 rs + 3w1 w2 s) + r 2 − s} sin θ , 3w2 (3w1 s + rs − 1) sin (θ + ψ)
(2.3)
from which 3(w1 r + w2 rs + 3w1 w2 s) + r 2 − s > 0, 3w2 (3w1 s + rs − 1) 3(w1 r + w2 ) + r 2 − s > 0. 1 − ρ1 = R0 ρ02 = − 3w2 (3w1 s + rs − 1) ρ1 =
(2.4)
Note r 2 − 4s(= (u − v)2 ) > 0 and 0 < ρ1 < 1 to see that Eqs. (2.4) cannot be valid at the same time, i.e., the rational cubic z of the form (1.3) has no loops. Note that the system (2.2) cannot be valid since (cos ψ, sin ψ) = (− cos θ, sin θ ) for θ + ψ = π and
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(cos ψ, sin ψ) = (cos θ, − sin θ ) for θ + ψ = 0. This completes the proof of the absence of a loop. Next, assume that the rational cubic z(t)(= (x(t), y(t))) of the form (1.3) would have a cusp at t = m, 0 < m < 1 for which the necessary condition is x (m) = y (m) = 0 (the curvature becomes unbounded). With help of Mathematica, the condition gives with m = 1/(d + 1), 0 < d < ∞ w1 db −d 3 + 3w2 d + 2 cos θ + w2 c 2d 3 + 3w1 d 2 − 1 cos ψ = 4w2 d 3 + 2(3w1w2 + 1)d 2 + 4w1 d, w1 db −d 3 + 3w2 d + 2 sin θ − w2 c 2d 3 + 3w1 d 2 − 1 sin ψ = 0.
(2.5)
First, we assume that θ + ψ = π, 0. Solve (2.5) for c to obtain c=
2d(2w2d 2 + d + 3w1 w2 d + 2w1 ) sin θ w2 (2d 3 + 3w1 d 2 − 1) sin(θ + ψ)
(2.6)
from which follow d(2w2 d 2 + d + 3w1 w2 d + 2w1 ) > 0, w2 (2d 3 + 3w1 d 2 − 1) −(d 2 + 2w1 d + w2 ) > 0. 1 − ρ1 = w2 (2d 3 + 3w1 d 2 − 1)
ρ1 =
(2.7)
Note that Eqs. (2.7) cannot be valid at the same time, i.e., the rational cubic z of the form (1.3) has no cusps. When θ + ψ = 0, π , as in the treatment of the loop we have the absence of the cusp from (2.5). Theorem 2.1. The rational cubics of the form (1.3) in Theorems 1.1, 1.2 and Remark have neither loops nor cusps for both C- and S-shaped polygons. Next, we shall consider whether inflection points exist or not for both C- and S-shaped polygons. With help of Mathematica, (z × z )(t) =
9(d + 1)6 ci d i , 3 2 3 (d + 3w1 d + 3w2 d + 1) 3
t = 1/(d + 1) (d > 0), (2.8)
i=0
with
(c0 , c1 , c2 , c3 ) = 3κ1 w23 c3 , 4w2 c sin ψ, 4w1 b sin θ, 3κ0 w13 b3
where Eqs. (1.6) reduce c0 , c3 as c0 (= w1 w2 c{4 sin ψ − 2b sin (θ + ψ)}) = 3κ1 w23 c3 , c3 (= w1 w2 b{4 sin θ − 2c sin (θ + ψ)}) = 3κ0 w13 b3 . First note that b, c > 0. For the “consistent” C-shaped polygon (θ, ψ > 0), κ0 , κ1 > 0, from which ci > 0, 0 i 3, are all positive. For the “consistent” S-shaped polygon (θ > 0, ψ < 0), κ0 > 0, κ1 < 0, from which ci , 0 i 3, are (−, −, +, +). Use Decartes’
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rule of signs and intermediate value of theorem to prove the absence of inflection points for a C-shaped polygon, and the existence of just one inflection point for an S-shaped polygon. Theorem 2.2. The rational cubics of the form (1.3) in Theorems 1.1, 1.2 and Remark have no inflection point for a C-shaped polygon and just one inflection point for an S-shaped polygon.
3. Numerical examples We consider some numerical examples. Note that a choice of the weight (w1 , w2 ) = (1, 1) (if possible) ensures the 6th order accurate cubic polynomial interpolation (de Boor et al., 1987). Example 1 (C-shaped polygon). (θ, ψ, κ0 , κ1 ) = (π/8, π/6, 0.2, 0.6). Example 2 (C-shaped polygon). (θ, ψ, κ0 , κ1 ) = (π/8, π/4, 2, 1). Fig. 3.1 gives the number of rational cubics with respect to (w1 , w2 ). Note that a cubic polynomial curve (equivalent to (w1 , w2 ) = (1, 1)) is not available. Figs. 3.2 and 3.3 give the rational cubic curves, control polygons and curvatures where “thick curves” mean the curves, polygons and curvatures of rational cubic curves of the form (1.3) with the weights (w1 , w2 ) ≈ (1.45469, 1.31784) and (0.253477, 0.590101) by (1.13). Note that the abscissa in the curvatures is the chord-length distance. Example 3 (S-shaped polygon). (θ, ψ, κ0 κ1 ) = (π/4, −π/8, 0.5, −0.2). Fig. 4.1 shows a possible choice of the weight (w1 , w2 ) = (1, 1) ensuring the 6th order accurate cubic polynomial interpolation. Example 4 (de Boor et al., 1987). We give an example of the interpolation for a circular arc with radius 1 with 4 interpolation points. Then (1.13) gives (w1 , w2 ) = (2/3, 2/3)
Fig. 3.1. Number of solutions with respect to (w1 , w2 ) for Examples 1 and 2.
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Fig. 3.2. Curves, polygons and curvatures with (w1 , w2 ) ≈ (1.454, 1.317), (1, 1) for Example 1.
Fig. 3.3. Curves, polygons and curvatures with (w1 , w2 ) ≈ (0.253, 0.590), (1, 1.5) for Example 2.
Fig. 4.1. Number of solutions with respect to (w1 , w2 ) for Example 3.
Fig. 4.2. Curve, polygon and curvature with (w1 , w2 ) = (1, 1) for Example 3.
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Fig. 5. Curves, polygons and curvatures with (w1 , w2 ) = (2/3, 2/3), (1, 1), (1/2, 1/2) for Example 4.
Fig. 6. Curves, polygons and curvatures with (w1 , w2 ) = (0.157635, 0.219432) for Example 5.
(then the curve, polygon and curvature are denoted by “thick” ones. For w1 = w2 (= π(4 + π)/24) ≈ 0.9348 (equivalent to R0 = R1 (= π/(4 + π)/16) ≈ 1.40225), the derivative of the rational cubic of the form (1.3) does equal to the exact value of the unit circular arc at the end points (the curve, control polygon and the curvature are denoted in “dashed lines”). Note the change in vertical scale in Fig. 5. Example 5 (Degen, 1995). Finally we consider the logarithmic spiral: r = r0 ecφ with r0 = 30, c = 0.46, φ ∈ [0, 2.35]. In this example, (1.13) gives (w1 , w2 ) = (0.157635, 0.219432) with which the curve, polygon and curvature are denoted by “thick” ones while the original spiral and curvature are denoted by “thin” ones.
Acknowledgements The author gratefully acknowledges the comments and suggestions of the anonymous referees which helped to improve the paper.
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Appendix (Proof of Theorem 1.1) We assume R0 R1 > 0 since the other case R1 R0 > 0 can be similarly treated by deleting ρ0 from the quadratic system (1.11) to obtain a quartic equation √ in ρ1 . We have to note count the number of the roots of the quartic equation (1.12) in (0, 1/ R0 ) where we√ that the root(s) (if they exist) satisfy ρ0 < 1. To do so, let ρ = 1/{r(d + 1)} with r = R0 to obtain: f (ρ) =
r(R1 − 1)d 4 + (4rR1 − 4r + 1)d 3 r 5 R1 (d + 1)4 (4rR1 − 6r + 3)d 2 − (4r − 3)d − (r − 1) + . r 5 R1 (d + 1)4
Then, for the numerator g(d) of the above equation, g (d) = (d + 1) 4r(R1 − 1)d 2 + (8rR1 − 8r + 3)d − (4r − 3) . Depending on values of√Ri , i = 0, 1, we count the number N of the positive root(s) of g(d) = 0 where note r = R0 . Case 1. R0 R1 1: Combine Decartes’ rule of signs and intermediate value of theorem to give N = 1. Case 2. R1 R0 9/16: g (d) has just one positive zero, g(0) = −(r − 1) > 0 and g(∞) = −∞. Hence N = 1. Case 3. R0 > 9/16, R1 < Min{R0 , 1}: Assume that g (d) = 0 has two positive roots α, β, i,e., 64R0 R12 − 64R0 R1 + 9 > 0. Then, with help of Mathematica, we obtain g(α)g(β) =
R1 [R0 R1 {256R0R1 − 256(R0 + R1 ) + 288} − 27] . 256R0 (R1 − 1)3
Then, if
R0 R1 256R0R1 − 256(R0 + R1 ) + 288 > 27,
N = 2 (or 3) for g(0) > 0 (or < 0). Next, either if g(α)g(β) > 0 or if g (d) = 0 has no positive roots, g(0) > 0 (or < 0) gives N = 1 (or 0). Note that the curve represented by R0 R1 256R0R1 − 256(R0 + R1 ) + 288 = 27 is over the one by 64R0 R12 − 64R0 R1 + 9 = 0 where these two curves intersect only at (R0 , R1 ) = (3/4, 3/4). References de Boor, C., Höllig, K., Sabin, M., 1987. High accuracy geometric Hermite interpolation. Computer Aided Geometric Design 14, 269–278. Degen, W., 1992. Best approximation of parametric curves by splines, in: Lyche, T., Schumaker, L.L. (Eds.), Mathematical Methods in Computer Aided Geometric Design II. Academic Press, New York, pp. 171–184. Degen, W., 1993. High accurate rational approximation of parametric curves. Computer Aided Geometric Design 10, 293–313.
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Degen, W., 1995. High accuracy approximation of parametric curves, in: Daehlen, M., Lyche, T., Schumaker, L.L. (Eds.), Mathematical Methods for Curves and Surfaces. Vanderbilt University Press, Nashville, pp. 83–98. Farin, G., 1993. Curves and Surfaces for Computer Aided Geometric Design, A Practical Guide. Academic Press, New York. Floater, M., 1995. Rational cubic implicitization, in: Dæhlen, M., Lyche, T., Schumaker, L.L. (Eds.), Mathematical Methods for Curves and Surfaces. Vanderbilt University Press, Nashville, pp. 151– 159. Goodman, T.N.T., 1988. Shape preserving interpolation by parametric rational cubic splines. Technical Report, U. of Dundee, Department of Mathematics, and Computer Science. Li, Y.-M., Cripps, R., 1992. Identification of inflection points and cusps on rational curves. Computer Aided Geometric Design 9, 1–24. Mørken, K., 1995. Parametric interpolation by quadratic polynomial in the plane, in: Daehlen, M., Lyche, T., Schumaker, L.L. (Eds.), Mathematical Methods for Curves and Surfaces. Vanderbilt University Press, Nashville, pp. 385–402. Seymour, C., Unsworth, K., 1999. Interactive shape preserving interpolation by curvature continuous rational cubic splines. J. Comput. Appl. Math. 102, 87–117.