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On control polygons of quartic Pythagorean–hodograph curves
Computer Aided Geometric Design 26 (2009) 1006–1015
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Computer Aided Geometric Design www.elsevier.com/locate/cagd
On control polygons of quartic Pythagorean–hodograph curves Guozhao Wang, Lincong Fang ∗ Institute of Computer Graphics and Image Processing, Department of Mathematics, Zhejiang University, Hangzhou 310027, China
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 16 November 2008 Received in revised form 18 August 2009 Accepted 23 August 2009 Available online 26 August 2009 Keywords: Pythagorean–hodograph Bézier curves Control polygon Hermite interpolation
In this paper, we study a necessary and sufficient condition for a planar quartic Bézier curve to possess a Pythagorean–hodograph (PH). Based on the definition of PH curve and complex representation of planar curve, we deduce geometric conditions in terms of the legs of the control polygon which guarantee the PH property. We also discuss the problem of G 1 Hermite interpolation by planar PH quartics. © 2009 Elsevier B.V. All rights reserved.
1. Introduction
Pythagorean–hodograph (PH) curves introduced by Farouki and Sakkalis (1990) have their roots in the rational parameterization of curves and surfaces in the practical field of computer aided geometric design. These curves provide a number of advantages in applications, such as CNC (computer-numerical-control) machining and control of digital motion along curved paths. For example, arc-lengths of PH curves can simply be computed by evaluating polynomials (avoiding the need of numerical integrations), which speed up the algorithms for numerically controlled (NC) machining (Farouki et al., 1998). Also, the fixed-distance offsets of PH curves are rational curves and can exactly be represented in CAD systems (Farouki and Sakkalis, 1990; Farouki, 1992). PH curves have been exhaustively studied (Choi and Kwon, 2008; Farouki, 1996, 1997; Farouki et al., 2008; Kosinka and Jüttler, 2006; Pelosi et al., 2007; Moon et al., 2001; Walton and Mekk, 2004) by Farouki and others since their first introduction. For more details, the reader may consult the excellent surveys of Farouki (2002, 2008) and the references therein. Farouki and Sakkalis (1990) have formulated geometric constraints on control polygon which guarantee the PH property of cubic curves, but for PH quartics, they just proposed a simple way for constructing curves by substituting chosen polynomials without geometric constraints. In this paper, we present a necessary and sufficient geometric characterization of PH quartics in terms of Bernstein– Bézier forms. We study the geometric relationship among the Bézier control points of PH quartics. Based on these, we provide a novel geometric approach for constructing PH quartics. The rest of the paper is organized as follows. Section 2 introduces some fundamental aspects of PH curves. In Section 3 we derive geometry properties of control polygons of PH quartics. In the last section we analyze the problem of G 1 Hermite interpolation by PH quartics.
*
Corresponding author. E-mail address: [email protected] (L. Fang).
0167-8396/$ – see front matter doi:10.1016/j.cagd.2009.08.003
© 2009 Elsevier B.V.
All rights reserved.
G. Wang, L. Fang / Computer Aided Geometric Design 26 (2009) 1006–1015
1007
2. Preliminaries This sectionsummarizes some basic concepts and some results concerning planar PH curves. n Let B nj (t ) = j (1 − t )n− j t j be a Bernstein basis function. The Bézier curve is defined by
P (t ) =
n
B nj (t ) P j ,
(1)
j =0
where { P j }nj=0 is a set of given (distinct) points in the plane. The ordered sequence of control points P 0 , P 1 , . . . , P n determines the control polygon of the curve (1). Let P j = P j +1 − P j be the directed leg of the control polygon. The hodograph of (1) is
P (t ) = n
n −1
B nj −1 (t ) P j .
(2)
j =0
Lemma 1. (See Kubota, 1972.) Three real polynomials a(t ), b(t ), and c (t ) satisfy the Pythagorean condition a2 (t ) + b2 (t ) ≡ c 2 (t ) if and only if they can be expressed in terms of real polynomials u (t ), v (t ), and w (t ) as
a(t ) = w (t ) u 2 (t ) − v 2 (t ) , b(t ) = 2w (t )u (t ) v (t ),
c (t ) = w (t ) u 2 (t ) + v 2 (t ) . Definition 2. A planar curve P (t ) = (x(t ), y (t )) is called a Pythagorean–hodograph (PH) curve if there exists a real polynomial σ (t ) such that (x (t ), y (t ), σ (t )) is a Pythagorean polynomial triple, where x (t ) and y (t ) are the components of the derivative of P (t ). The complex representation (Farouki, 1994; Kong et al., 2008) of a planar curve P (t ) = (x(t ), y (t )) is the complex function P (t ) = x(t ) + iy (t ). On making use of the complex representation of planar curve, it follows immediately from Lemma 1 that the theorem below holds if we identify x (t ) with a(t ) and y (t ) with b(t ). Theorem 3. In the complex representation, a planar curve P (t ) = x(t ) + iy (t ) is a PH curve if and only if P (t ) = w (t )(u (t ) + i v (t ))2 for some real polynomials u (t ), v (t ), and w (t ). In this paper, we focus on the control polygons of PH quartics, that is, for n = 4. As Farouki and Sakkalis (1990) proposed, PH quartics are cuspidal curves corresponding to the case when deg( w ) = max{deg(u ), deg( v )} = 1. Without loss of generality, we assume that gcd(u , v ) = 1 in the following sections. 3. The control polygons of PH quartics We confine here to PH quartics with control points P 0 , . . . , P 4 . Our goal is to formulate geometric constraints on the control polygon that will guarantee the PH property. Let t = ξ be the location of the cusp. If ξ = 0 or ξ = 1, the control polygon is degenerate ( P 1 = P 0 in the former case, and P 4 = P 3 in the latter), hence the curve is irregular. Thus, for regular curves, we have ξ = αα −1 for some α ∈ R \ {0, 1}. We see that if α > 0, the parameter value of the cusp is not in the interval [0, 1], and we call such a curve a curve without cusps in what follows. On the other hand, if α < 0, the parameter value of the cusp is in the interval (0, 1), and we call such a curve a curve with cusp. From Theorem 3, we have P (t ) = w (t )(u (t ) + i v (t ))2 , and we see that ξ is the root of w (t ). If we write the linear polynomial w (t ) and the complex linear polynomial u (t ) + i v (t ) in Bernstein–Bézier form, then
P (t ) =
α (1 − t ) + t β(1 − t ) + γ t
2
for some β, γ ∈ C. From (2), the hodograph of a planar PH quartic is
P (t ) = 4
3 j =0
Hence we get
B 3j (t ) P j .
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G. Wang, L. Fang / Computer Aided Geometric Design 26 (2009) 1006–1015
4 P 0 = α β 2 , 4 P 1 = 4 P 2 =
(3a)
2α β γ + β
2
3 2β γ + αγ 2 3
,
(3b)
,
(3c)
4 P 3 = γ 2 .
(3d)
Let · be the usual complex norm on C and denote P 1 P 2 = P 2 − P 1 . Let L j be the length of a leg of the control polygon. We may choose a coordinate system in which the legs of the control polygon can be expressed in polar form
P 0 = L0, P 1 = L 1 eiψ , P 2 = L 2 eiϕ , P 3 = L 3 e i2θ , with ψ ∈ (0, π ). Obviously we have
√
γ = 2 L 3 eiθ .
3.1. PH quartics without cusps The parameter value ξ of the cusp is not in the interval [0, 1] if α > 0. To avoid the degenerate cases, we assume that β ∈ R \ {0} and γ ∈ C \ R. Without loss of generality, we may assume 2θ ∈ (0, 2π ), that is, θ ∈ (0, π ). Since (3b) is a complex equation, by considering the real part and imaginary part separately, it is equivalent to
4L 1 cos ψ = L 1 sin ψ =
4 3
1 3
L 3 cos θ · α β +
1 3
· β 2,
L 3 sin θ · α β.
(4)
Solving the system of linear equations (4),
3L 1 sin ψ
αβ = √ β2 =
L 3 sin θ
,
12L 1 sin(θ − ψ) sin θ
(5)
.
From β = 0 we have θ − ψ ∈ (0, π ). From (3a) and (5), we have
α= β=
αβ 2 β2
=
L 0 sin θ 3L 1 sin(θ − ψ)
,
√
α β 2 4L 0 L 3 sin θ = . αβ 3L 1 sin ψ
(6)
We see that β > 0. In addition, from (5) and (6) we have
12L 1 sin(θ − ψ) sin θ
=
4L 0
√
L 3 sin θ
2
3L 1 sin ψ
,
that is,
L 20 L 3 sin3 θ =
27 4
L 31 sin2 ψ sin(θ − ψ).
(7)
Similarly, (3c) can be written as
L 2 cos ϕ = L 2 sin ϕ =
1 3 1 3
L 3 cos θ · β +
1
L 3 cos 2θ · α , 3 1 L 3 sin θ · β + L 3 sin 2θ · α . 3
Solving the system of linear equations (8),
(8)
G. Wang, L. Fang / Computer Aided Geometric Design 26 (2009) 1006–1015
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Fig. 1. The control polygon of a planar quartic Bézier curve without cusps.
α= β= Since
3L 2 sin(ϕ − θ)
,
L 3 sin θ 3L 2 sin(2θ − ϕ )
√
L 3 sin θ
(9)
.
α , β > 0, we get ϕ − θ , 2θ − ϕ ∈ (0, π ). From (6) and (9) we have L 0 L 3 sin2 θ = 9L 1 L 2 sin(ϕ − θ) sin(θ − ψ)
(10)
and
L 0 L 3 sin2 θ =
9 4
L 1 L 2 sin(2θ − ϕ ) sin ψ.
(11)
Furthermore, by substituting (9) into (3a), we have
4L 0 L 3 sin θ 3L 2 sin(ϕ − θ)
=
3L 2 sin(2θ − ϕ )
√
2 ,
L 3 sin θ
which simplifies to
L 0 L 23 sin3 θ =
27 4
L 32 sin2 (2θ − ϕ ) sin(ϕ − θ).
(12)
Now (7), (10), (11), and (12) are equivalent to (3a)–(3d). Hence we have following lemma if a coordinate system is chosen as stated in Section 3. Lemma 4. A planar quartic P (t ) with Bézier control points P 0 , . . . , P 4 is a PH curve if and only if
L 20 L 3 sin3 θ =
27 4
L 31 sin2 ψ sin(θ − ψ),
L 0 L 3 sin2 θ = 9L 1 L 2 sin(ϕ − θ) sin(θ − ψ), 9 L 0 L 3 sin2 θ = L 1 L 2 sin(2θ − ϕ ) sin ψ, 4 27 3 3 2 L 0 L 3 sin θ = L sin2 (2θ − ϕ ) sin(ϕ − θ). 4 2
(13)
All of the angles θ, ψ, θ − ψ, ϕ − θ , and 2θ − ϕ are in the interval (0, π ), and from Eqs. (3b) and (3c) we have Re( P 1 ) · Im( P 2 ) − Im( P 1 ) · Re( P 2 ) > 0, that is, ϕ − ψ ∈ (0, π ). Thus there exists a line which passes through P 2 and intersects the lines given by P 0 , P 1 and P 3 , P 4 at A and B respectively such that P 1 A P 2 = P 2 B P 3 = π − θ , as shown in Fig. 1. Let R 0 = A P 1 , R 1 = A P 2 , R 2 = B P 2 , and R 3 = B P 3 . Note that
sin P 1 A P 2 = sin θ,
sin A P 2 P 1 = sin(θ − ψ),
sin A P 1 P 2 = sin ψ,
sin P 2 B P 3 = sin θ,
sin B P 2 P 3 = sin(ϕ − θ),
sin B P 3 P 2 = sin(2θ − ϕ ).
By the law of sines we may write
L1 sin P 1 A P 2 L2 sin P 2 B P 3
= =
R0 sin A P 2 P 1 R3 sin B P 2 P 3
= =
R1 sin A P 1 P 2 R2 sin B P 3 P 2
, .
(14)
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G. Wang, L. Fang / Computer Aided Geometric Design 26 (2009) 1006–1015
Fig. 2. Elevation of a planar cubic Bézier curve.
By substituting (14) into (13), we get
L 20 · L 3 =
27
R 0 · R 21 , 4 L 0 · L 3 = 9R 0 · R 3 , 9 L0 · L3 = R 1 · R 2, 4 27 2 2 L0 · L3 = R · R3, 4 2 which are equivalent to
L 0 · R 2 = 3R 0 · R 1 , L 3 · R 1 = 3R 2 · R 3 , R 1 · R 2 = 4R 0 · R 3 .
(15)
Here the result for those quartics which are elevated from cubics will be given. Let P (t ) be a planar cubic with Bézier control points P 0 , . . . , P 3 , and let P −1 = P 4 = 0 + 0i. For j = 0, . . . , 4, let
Pˆ j = 1 −
j
4
Pj+
j 4
P j −1 .
P (t ) can be elevated to a quartic Bézier curve
P (t ) =
3
B 3j (t ) P j =
j =0
4
B 4j (t ) Pˆ j .
j =0
Farouki and Sakkalis (1990) showed that P (t ) is a PH curve if and only if P 1 P 2 2 = P 0 P 1 · P 3 P 4 and P 0 P 1 P 2 = { Pˆ j }4j=0 satisfies the conditions in (15) if we identify P 1 with A and P 2 with B here. P 1 P 2 P 3 , as shown in Fig. 2. Obviously the control polygon of
3.2. PH quartics with cusp If the parameter value ξ of the cusp is in the interval (0, 1), we have α < 0, β 2 < 0, and γ ∈ C. As β is purely imaginary, to avoid degenerate cases, we assume that Re(γ ) = 0. Without loss of generality, we may choose 2θ ∈ [π , 3π ). From Re(γ ) = 0, we have θ ∈ (π /2, 3π /2), and get similar results as in (13):
L 20 L 3 cos3 θ = −
27 4
L 31 sin2 ψ cos(θ − ψ),
L 0 L 3 cos2 θ = 9L 1 L 2 cos(ϕ − θ) cos(θ − ψ), 9 L 0 L 3 cos2 θ = L 1 L 2 sin(2θ − ϕ ) sin ψ, 4 27 2 3 L 0 L 3 cos θ = − L 32 sin2 (2θ − ϕ ) cos(ϕ − θ). 4 We replace θ with θ + π /2 such that θ ∈ (0, π ) and deduce that
L 20 L 3 sin3 θ =
27 4
L 31 sin2 ψ sin(ψ − θ ),
L 0 L 3 sin2 θ = 9L 1 L 2 sin(ϕ − θ ) sin(ψ − θ ),
G. Wang, L. Fang / Computer Aided Geometric Design 26 (2009) 1006–1015
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Fig. 3. The control polygon of a planar quartic Bézier curve with cusp.
L 0 L 3 sin2 θ =
9
L 1 L 2 sin(ϕ − 2θ ) sin ψ, 4 27 3 L 0 L 23 sin3 θ = L sin2 (ϕ − 2θ ) sin(ϕ − θ ). 4 2 As in the previous case, all the interior angles of the control polygon are in (0, π ). There exists a line which passes through P 2 and intersects the lines given by P 0 , P 1 and P 3 , P 4 at A and B respectively such that P 1 A P 2 = P 2 B P 3 = π − θ , as shown in Fig. 3. Similarly, we have (15) by the law of sines. Thus we have a geometric characterization of PH quartics which is valid for both classes of curves (with or without the cusp). Theorem 5. For a planar quartic P (t ) with Bézier control points P 0 , . . . , P 4 , it is a PH curve if and only if there is a line A B which passes through P 2 and intersects with P 0 P 1 and P 3 P 4 at A and B respectively, such that P1AP2
= P 2 B P 3,
L 0 · R 2 = 3R 0 · R 1 , L 3 · R 1 = 3R 2 · R 3 , R 1 · R 2 = 4R 0 · R 3 .
(16)
4. G 1 Hermite interpolation Theorem 5 provides simple geometric construction procedures for interpolating G 1 Hermite data by PH quartics. Given points P 0 and P 4 with associated tangent directions d0 and d4 , we first choose P 2 on some curve segments, and then find the inner control points P 1 and P 3 which satisfy the conditions in Theorem 5 with
P 1 = P 0 + λd0 , P 3 = P 4 − μd4 for some real constants λ and μ. The curve depends on the choice of the control point P 2 if G 1 Hermite data is given. If P 2 is chosen, we get P 1 and P 3 from (16) and the quartic curve is completely determined. In the rest of this section, we see that P 2 is not arbitrarily chosen. The locus of P 2 can be evaluated. For convenience, we write
Lˆ 0 = A P 0 ,
Lˆ 1 = A B ,
Lˆ 2 = B P 4
in the following subsections. 4.1. PH quartics without cusps For a curve without cusps, its control polygon satisfies
Lˆ 0 = L 0 + R 0 , Lˆ 1 = R 1 + R 2 , Lˆ 2 = L 3 + R 3 . Let q =
R1 . R2
(17)
From (16) we see that q = 0 will lead to degenerate control polygon, thus we have
4 Lˆ 0 3q + 1
·
Lˆ 2 q 3+q
=
Lˆ 21 q
(q + 1)2
.
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G. Wang, L. Fang / Computer Aided Geometric Design 26 (2009) 1006–1015
Fig. 4. Control polygon of a PH quartic without cusps.
Hence
3 Lˆ 21 − 4 Lˆ 0 Lˆ 2 q2 + 10 Lˆ 21 − 8 Lˆ 0 Lˆ 2 q + 3 Lˆ 21 − 4 Lˆ 0 Lˆ 2 = 0,
(18)
which is a quadratic equation. Let q1 and q2 be the two solutions of Eq. (18). From Viète’s theorem we have q1 · q2 = 1. Therefore, Eq. (18) have real roots if and only if
10 Lˆ 21 − 8 Lˆ 0 Lˆ 2
2
2 − 4 3 Lˆ 21 − 4 Lˆ 0 Lˆ 2 0
and
3 Lˆ 21 − 4 Lˆ 0 Lˆ 2 · 10 Lˆ 21 − 8 Lˆ 0 Lˆ 2 < 0.
It follows that
Lˆ 0 Lˆ 2 Lˆ 21 <
4 3
Lˆ 0 Lˆ 2 .
The PH quartic degenerates to a cubic curve with control points P 0 , A , B, and P 4 if
Lˆ 0 Lˆ 2 = Lˆ 21 , which implies R 1 = R 2 . Given two points P 0 and P 4 with associated tangent vectors d0 and d4 , we have lines A B and A B which satisfy P 0 A B
= P 4 B A = P 0 A B = P 4 B A , 4
A B 2 = A P 0 · B P 4 , 3
2
A B = A P 0 · B P 4 . Let C be the midpoint of the segment A B. It is obvious that
4 A P 2 · B P 2 = A B − 2 C P 2 · A B + 2 C P 2 .
(19)
A P
By substituting q = B P 2 , Eqs. (17) and (19) into Eq. (18), we get 2
C P 2 2 = A B 2 − A P 0 · B P 4 .
(20)
Now we choose the coordinates such that the origin is located at the midpoint O of the segment A B , while the line A B is aligned with the x-axis, as shown in Fig. 4. Let H be the midpoint of the segment A B . Further let a = A P 0 , b = B P 4 , c = A B , d = A B , h = O H , and l = A A . If P 2 = x + yi, then it follows from Eq. (20) that
h2 x2 − (c − d)2 − l2 y 2 + 2c (c − d) + l(a + b) hy + ab − c 2 h2 = 0. Hence P 2 is on a quadratic curve segment with ends A and B . In Fig. 5, given P 0 = 0 + 0i and P 4 = 5 + 0i, with the tangent vectors d0 = √1 + √2 i and d4 = − √1 + √2 i, the locus of 5
5
5
5
P 2 is a part of an ellipse. In the figure, to determine the curve, we choose P 2 to be the intersection of the line x = 2 and the curve segment. We have a PH quartic without cusps.
G. Wang, L. Fang / Computer Aided Geometric Design 26 (2009) 1006–1015
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Fig. 5. G 1 Hermite interpolation by a PH quartic without cusps.
Fig. 6. Control polygon of a PH quartic with cusp and with
d0 d4
∈ / R.
4.2. PH quartics with cusp For a curve with cusp, the control polygon satisfies
Lˆ 0 = L 0 − R 0 , Lˆ 1 = | R 1 − R 2 |, Lˆ 2 = L 3 − R 3 . Similar to Eq. (18), we have the quadratic equation
3 Lˆ 21 + 4 Lˆ 0 Lˆ 2 x2 − 10 Lˆ 21 + 8 Lˆ 0 Lˆ 2 x + 3 Lˆ 21 + 4 Lˆ 0 Lˆ 2 = 0.
This equation always has two positive solutions x1 , x2 ∈ ( 13 , 3). Further we also have an equation similar to (20):
C P 2 2 = A B 2 + A P 0 · B P 4 . d0 d4
∈ / R, let O be the intersection of P 0 P 1 and P 3 P 4 and choose the coordinates such that the origin is located at O while the angle bisector of P 0 O P 4 is aligned with the y-axis, as shown in Fig. 6. Let φ ∈ (0, π2 ) be the argument of d0 , and let m = O P 0 and n = O P 4 . We see that the locus of P 2 = x + yi is If
Since
3 tan2 φ − 1 x2 − (m − n)x sec φ + mn − y 2 = 0.
R1 R2
∈ ( 13 , 3), we have x ∈ (−m cos φ, n cos φ).
In Fig. 7, given P 0 = 0 + 0i and P 4 = 5 + 0i with d0 = √1 + √2 i and d4 = − √1 + √2 i, we see that the locus of P 2 is a 5
5
5
5
hyperbola with two branches. On the left-hand side of the figure, P 2 is chosen as the intersection of the line x = 3 and the top branch, and a PH quartic is generated. On the right-hand side of the figure, P 2 is chosen as the intersection of the line x = 3 and the bottom branch, and another PH quartic is constructed. d If d0 ∈ R, the line P 0 P 1 is parallel to the line P 3 P 4 . We cannot choose a coordinate system as above. Instead we choose 4
the line which is equidistance to both P 0 P 1 and P 3 P 4 as x-axis, and the line A B which satisfies A P 0 = B P 4 as y-axis, as shown in Fig. 8. Let r = A P 0 = B P 4 and s = A B . The locus of P 2 = x + yi is
x2 + y 2 = r 2 + s2 . Since RR 1 ∈ ( 13 , 3), we have |x| < r. 2 In Fig. 9, given P 0 = 0 + 0i and P 4 = 5 + 3i with d0 = 1 + 0i and d4 = −1 + 0i, we see that the locus of P 2 are two arcs of a circle. In the figure, to determine the curve, P 2 is chosen as the intersection of the line x = 3 and the top arc. A PH quartic with cusp is then constructed.
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G. Wang, L. Fang / Computer Aided Geometric Design 26 (2009) 1006–1015
Fig. 7. G 1 Hermite interpolation by a PH quartic with cusp and with
Fig. 8. Control polygon of a PH quartic with cusp and with
d0 d4
d0 d4
∈ / R.
∈ R.
Fig. 9. G 1 Hermite interpolation by a PH quartic with cusp and with
d0 d4
∈ R.
Acknowledgements The authors thank the anonymous referees for their helpful suggestions and comments. Also the authors wish to thank Dr. Zhaozhen Huang for his great help in improving the language. This work is supported by the Natural Science Foundation of China (No. 60773179) and Foundation of State Key Basic Research 973 Development Programming Item of China (No. G2004CB318000). References Choi, H.I., Kwon, S.-H., 2008. Absolute hodograph winding number and planar PH quintic splines. Computer Aided Geometric Design 25, 230–246. Farouki, R.T., 1992. Pythagorean hodograph curves in practical use. In: Barnhill, R.E. (Ed.), Geometry Processing for Design and Manufacturing. SIAM, pp. 3– 33. Farouki, R.T., 1994. The conformal map z → z2 of the hodograph plane. Computer Aided Geometric Design 11, 363–390. Farouki, R.T., 1996. The elastic bending energy of Pythagorean curves. Computer Aided Geometric Design 13, 227–241. Farouki, R.T., 1997. Pythagorean–hodograph quintic transition curves of monotone curvature. Computer-Aided Design 29, 606–610. Farouki, R.T., 2002. Pythagorean hodograph curves. In: Farin, G., Hoschek, J., Kim, M.-S. (Eds.), Handbook of Computer Aided Geometric Design. NorthHolland, pp. 405–427. Farouki, R.T., 2008. Pythagorean–Hodograph Curves: Algebra and Geometry Inseparable. Springer. Farouki, R.T., Giannelli, C., Manni, C., Sestini, A., 2008. Identification of spatial PH quintic Hermite interpolants with near-optimal shape measures. Computer Aided Geometric Design 25, 274–297. Farouki, R.T., Manjunathaiah, J., Nichlas, D., Yuan, G.F., Jee, S., 1998. Variable-feedrate CNC interpolators for constant material removal rates along Pythagorean–hodograph curves. Computer-Aided Design 30, 631–640.
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Farouki, R.T., Sakkalis, T., 1990. Pythagorean hodographs. IBM Journal of Research and Development 34, 736–752. Kong, J.H., Jeong, S.P., Lee, S., Kim, G.I., 2008. C 1 Hermite interpolation with simple planar PH curves by speed reparametrization. Computer Aided Geometric Design 25, 214–229. Kosinka, J., Jüttler, B., 2006. G 1 Hermite interpolation by Minkowski Pythagorean hodograph cubics. Computer Aided Geometric Design 23, 401–418. Kubota, K.K., 1972. Pythagorean triples in unique factorization domain. American Mathematical Monthly 79, 503–505. Moon, H.P., Farouki, R.T., Choi, H.I., 2001. Construction and shape analysis of PH quintic Hermite interpolants. Computer Aided Geometric Design 18, 93–115. Pelosi, F., Sampoli, M.L., Farouki, R.T., Manni, C., 2007. A control polygon scheme for design of planar C 2 PH quintic spline curves. Computer Aided Geometric Design 24, 28–52. Walton, D.J., Mekk, D.S., 2004. A generalisation of the Pythagorean hodograph quintic spiral. Journal of Computational and Applied Mathematics 172, 271– 287.
Contents lists available at ScienceDirect
Computer Aided Geometric Design www.elsevier.com/locate/cagd
On control polygons of quartic Pythagorean–hodograph curves Guozhao Wang, Lincong Fang ∗ Institute of Computer Graphics and Image Processing, Department of Mathematics, Zhejiang University, Hangzhou 310027, China
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 16 November 2008 Received in revised form 18 August 2009 Accepted 23 August 2009 Available online 26 August 2009 Keywords: Pythagorean–hodograph Bézier curves Control polygon Hermite interpolation
In this paper, we study a necessary and sufficient condition for a planar quartic Bézier curve to possess a Pythagorean–hodograph (PH). Based on the definition of PH curve and complex representation of planar curve, we deduce geometric conditions in terms of the legs of the control polygon which guarantee the PH property. We also discuss the problem of G 1 Hermite interpolation by planar PH quartics. © 2009 Elsevier B.V. All rights reserved.
1. Introduction
Pythagorean–hodograph (PH) curves introduced by Farouki and Sakkalis (1990) have their roots in the rational parameterization of curves and surfaces in the practical field of computer aided geometric design. These curves provide a number of advantages in applications, such as CNC (computer-numerical-control) machining and control of digital motion along curved paths. For example, arc-lengths of PH curves can simply be computed by evaluating polynomials (avoiding the need of numerical integrations), which speed up the algorithms for numerically controlled (NC) machining (Farouki et al., 1998). Also, the fixed-distance offsets of PH curves are rational curves and can exactly be represented in CAD systems (Farouki and Sakkalis, 1990; Farouki, 1992). PH curves have been exhaustively studied (Choi and Kwon, 2008; Farouki, 1996, 1997; Farouki et al., 2008; Kosinka and Jüttler, 2006; Pelosi et al., 2007; Moon et al., 2001; Walton and Mekk, 2004) by Farouki and others since their first introduction. For more details, the reader may consult the excellent surveys of Farouki (2002, 2008) and the references therein. Farouki and Sakkalis (1990) have formulated geometric constraints on control polygon which guarantee the PH property of cubic curves, but for PH quartics, they just proposed a simple way for constructing curves by substituting chosen polynomials without geometric constraints. In this paper, we present a necessary and sufficient geometric characterization of PH quartics in terms of Bernstein– Bézier forms. We study the geometric relationship among the Bézier control points of PH quartics. Based on these, we provide a novel geometric approach for constructing PH quartics. The rest of the paper is organized as follows. Section 2 introduces some fundamental aspects of PH curves. In Section 3 we derive geometry properties of control polygons of PH quartics. In the last section we analyze the problem of G 1 Hermite interpolation by PH quartics.
*
Corresponding author. E-mail address: [email protected] (L. Fang).
0167-8396/$ – see front matter doi:10.1016/j.cagd.2009.08.003
© 2009 Elsevier B.V.
All rights reserved.
G. Wang, L. Fang / Computer Aided Geometric Design 26 (2009) 1006–1015
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2. Preliminaries This sectionsummarizes some basic concepts and some results concerning planar PH curves. n Let B nj (t ) = j (1 − t )n− j t j be a Bernstein basis function. The Bézier curve is defined by
P (t ) =
n
B nj (t ) P j ,
(1)
j =0
where { P j }nj=0 is a set of given (distinct) points in the plane. The ordered sequence of control points P 0 , P 1 , . . . , P n determines the control polygon of the curve (1). Let P j = P j +1 − P j be the directed leg of the control polygon. The hodograph of (1) is
P (t ) = n
n −1
B nj −1 (t ) P j .
(2)
j =0
Lemma 1. (See Kubota, 1972.) Three real polynomials a(t ), b(t ), and c (t ) satisfy the Pythagorean condition a2 (t ) + b2 (t ) ≡ c 2 (t ) if and only if they can be expressed in terms of real polynomials u (t ), v (t ), and w (t ) as
a(t ) = w (t ) u 2 (t ) − v 2 (t ) , b(t ) = 2w (t )u (t ) v (t ),
c (t ) = w (t ) u 2 (t ) + v 2 (t ) . Definition 2. A planar curve P (t ) = (x(t ), y (t )) is called a Pythagorean–hodograph (PH) curve if there exists a real polynomial σ (t ) such that (x (t ), y (t ), σ (t )) is a Pythagorean polynomial triple, where x (t ) and y (t ) are the components of the derivative of P (t ). The complex representation (Farouki, 1994; Kong et al., 2008) of a planar curve P (t ) = (x(t ), y (t )) is the complex function P (t ) = x(t ) + iy (t ). On making use of the complex representation of planar curve, it follows immediately from Lemma 1 that the theorem below holds if we identify x (t ) with a(t ) and y (t ) with b(t ). Theorem 3. In the complex representation, a planar curve P (t ) = x(t ) + iy (t ) is a PH curve if and only if P (t ) = w (t )(u (t ) + i v (t ))2 for some real polynomials u (t ), v (t ), and w (t ). In this paper, we focus on the control polygons of PH quartics, that is, for n = 4. As Farouki and Sakkalis (1990) proposed, PH quartics are cuspidal curves corresponding to the case when deg( w ) = max{deg(u ), deg( v )} = 1. Without loss of generality, we assume that gcd(u , v ) = 1 in the following sections. 3. The control polygons of PH quartics We confine here to PH quartics with control points P 0 , . . . , P 4 . Our goal is to formulate geometric constraints on the control polygon that will guarantee the PH property. Let t = ξ be the location of the cusp. If ξ = 0 or ξ = 1, the control polygon is degenerate ( P 1 = P 0 in the former case, and P 4 = P 3 in the latter), hence the curve is irregular. Thus, for regular curves, we have ξ = αα −1 for some α ∈ R \ {0, 1}. We see that if α > 0, the parameter value of the cusp is not in the interval [0, 1], and we call such a curve a curve without cusps in what follows. On the other hand, if α < 0, the parameter value of the cusp is in the interval (0, 1), and we call such a curve a curve with cusp. From Theorem 3, we have P (t ) = w (t )(u (t ) + i v (t ))2 , and we see that ξ is the root of w (t ). If we write the linear polynomial w (t ) and the complex linear polynomial u (t ) + i v (t ) in Bernstein–Bézier form, then
P (t ) =
α (1 − t ) + t β(1 − t ) + γ t
2
for some β, γ ∈ C. From (2), the hodograph of a planar PH quartic is
P (t ) = 4
3 j =0
Hence we get
B 3j (t ) P j .
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4 P 0 = α β 2 , 4 P 1 = 4 P 2 =
(3a)
2α β γ + β
2
3 2β γ + αγ 2 3
,
(3b)
,
(3c)
4 P 3 = γ 2 .
(3d)
Let · be the usual complex norm on C and denote P 1 P 2 = P 2 − P 1 . Let L j be the length of a leg of the control polygon. We may choose a coordinate system in which the legs of the control polygon can be expressed in polar form
P 0 = L0, P 1 = L 1 eiψ , P 2 = L 2 eiϕ , P 3 = L 3 e i2θ , with ψ ∈ (0, π ). Obviously we have
√
γ = 2 L 3 eiθ .
3.1. PH quartics without cusps The parameter value ξ of the cusp is not in the interval [0, 1] if α > 0. To avoid the degenerate cases, we assume that β ∈ R \ {0} and γ ∈ C \ R. Without loss of generality, we may assume 2θ ∈ (0, 2π ), that is, θ ∈ (0, π ). Since (3b) is a complex equation, by considering the real part and imaginary part separately, it is equivalent to
4L 1 cos ψ = L 1 sin ψ =
4 3
1 3
L 3 cos θ · α β +
1 3
· β 2,
L 3 sin θ · α β.
(4)
Solving the system of linear equations (4),
3L 1 sin ψ
αβ = √ β2 =
L 3 sin θ
,
12L 1 sin(θ − ψ) sin θ
(5)
.
From β = 0 we have θ − ψ ∈ (0, π ). From (3a) and (5), we have
α= β=
αβ 2 β2
=
L 0 sin θ 3L 1 sin(θ − ψ)
,
√
α β 2 4L 0 L 3 sin θ = . αβ 3L 1 sin ψ
(6)
We see that β > 0. In addition, from (5) and (6) we have
12L 1 sin(θ − ψ) sin θ
=
4L 0
√
L 3 sin θ
2
3L 1 sin ψ
,
that is,
L 20 L 3 sin3 θ =
27 4
L 31 sin2 ψ sin(θ − ψ).
(7)
Similarly, (3c) can be written as
L 2 cos ϕ = L 2 sin ϕ =
1 3 1 3
L 3 cos θ · β +
1
L 3 cos 2θ · α , 3 1 L 3 sin θ · β + L 3 sin 2θ · α . 3
Solving the system of linear equations (8),
(8)
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Fig. 1. The control polygon of a planar quartic Bézier curve without cusps.
α= β= Since
3L 2 sin(ϕ − θ)
,
L 3 sin θ 3L 2 sin(2θ − ϕ )
√
L 3 sin θ
(9)
.
α , β > 0, we get ϕ − θ , 2θ − ϕ ∈ (0, π ). From (6) and (9) we have L 0 L 3 sin2 θ = 9L 1 L 2 sin(ϕ − θ) sin(θ − ψ)
(10)
and
L 0 L 3 sin2 θ =
9 4
L 1 L 2 sin(2θ − ϕ ) sin ψ.
(11)
Furthermore, by substituting (9) into (3a), we have
4L 0 L 3 sin θ 3L 2 sin(ϕ − θ)
=
3L 2 sin(2θ − ϕ )
√
2 ,
L 3 sin θ
which simplifies to
L 0 L 23 sin3 θ =
27 4
L 32 sin2 (2θ − ϕ ) sin(ϕ − θ).
(12)
Now (7), (10), (11), and (12) are equivalent to (3a)–(3d). Hence we have following lemma if a coordinate system is chosen as stated in Section 3. Lemma 4. A planar quartic P (t ) with Bézier control points P 0 , . . . , P 4 is a PH curve if and only if
L 20 L 3 sin3 θ =
27 4
L 31 sin2 ψ sin(θ − ψ),
L 0 L 3 sin2 θ = 9L 1 L 2 sin(ϕ − θ) sin(θ − ψ), 9 L 0 L 3 sin2 θ = L 1 L 2 sin(2θ − ϕ ) sin ψ, 4 27 3 3 2 L 0 L 3 sin θ = L sin2 (2θ − ϕ ) sin(ϕ − θ). 4 2
(13)
All of the angles θ, ψ, θ − ψ, ϕ − θ , and 2θ − ϕ are in the interval (0, π ), and from Eqs. (3b) and (3c) we have Re( P 1 ) · Im( P 2 ) − Im( P 1 ) · Re( P 2 ) > 0, that is, ϕ − ψ ∈ (0, π ). Thus there exists a line which passes through P 2 and intersects the lines given by P 0 , P 1 and P 3 , P 4 at A and B respectively such that P 1 A P 2 = P 2 B P 3 = π − θ , as shown in Fig. 1. Let R 0 = A P 1 , R 1 = A P 2 , R 2 = B P 2 , and R 3 = B P 3 . Note that
sin P 1 A P 2 = sin θ,
sin A P 2 P 1 = sin(θ − ψ),
sin A P 1 P 2 = sin ψ,
sin P 2 B P 3 = sin θ,
sin B P 2 P 3 = sin(ϕ − θ),
sin B P 3 P 2 = sin(2θ − ϕ ).
By the law of sines we may write
L1 sin P 1 A P 2 L2 sin P 2 B P 3
= =
R0 sin A P 2 P 1 R3 sin B P 2 P 3
= =
R1 sin A P 1 P 2 R2 sin B P 3 P 2
, .
(14)
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Fig. 2. Elevation of a planar cubic Bézier curve.
By substituting (14) into (13), we get
L 20 · L 3 =
27
R 0 · R 21 , 4 L 0 · L 3 = 9R 0 · R 3 , 9 L0 · L3 = R 1 · R 2, 4 27 2 2 L0 · L3 = R · R3, 4 2 which are equivalent to
L 0 · R 2 = 3R 0 · R 1 , L 3 · R 1 = 3R 2 · R 3 , R 1 · R 2 = 4R 0 · R 3 .
(15)
Here the result for those quartics which are elevated from cubics will be given. Let P (t ) be a planar cubic with Bézier control points P 0 , . . . , P 3 , and let P −1 = P 4 = 0 + 0i. For j = 0, . . . , 4, let
Pˆ j = 1 −
j
4
Pj+
j 4
P j −1 .
P (t ) can be elevated to a quartic Bézier curve
P (t ) =
3
B 3j (t ) P j =
j =0
4
B 4j (t ) Pˆ j .
j =0
Farouki and Sakkalis (1990) showed that P (t ) is a PH curve if and only if P 1 P 2 2 = P 0 P 1 · P 3 P 4 and P 0 P 1 P 2 = { Pˆ j }4j=0 satisfies the conditions in (15) if we identify P 1 with A and P 2 with B here. P 1 P 2 P 3 , as shown in Fig. 2. Obviously the control polygon of
3.2. PH quartics with cusp If the parameter value ξ of the cusp is in the interval (0, 1), we have α < 0, β 2 < 0, and γ ∈ C. As β is purely imaginary, to avoid degenerate cases, we assume that Re(γ ) = 0. Without loss of generality, we may choose 2θ ∈ [π , 3π ). From Re(γ ) = 0, we have θ ∈ (π /2, 3π /2), and get similar results as in (13):
L 20 L 3 cos3 θ = −
27 4
L 31 sin2 ψ cos(θ − ψ),
L 0 L 3 cos2 θ = 9L 1 L 2 cos(ϕ − θ) cos(θ − ψ), 9 L 0 L 3 cos2 θ = L 1 L 2 sin(2θ − ϕ ) sin ψ, 4 27 2 3 L 0 L 3 cos θ = − L 32 sin2 (2θ − ϕ ) cos(ϕ − θ). 4 We replace θ with θ + π /2 such that θ ∈ (0, π ) and deduce that
L 20 L 3 sin3 θ =
27 4
L 31 sin2 ψ sin(ψ − θ ),
L 0 L 3 sin2 θ = 9L 1 L 2 sin(ϕ − θ ) sin(ψ − θ ),
G. Wang, L. Fang / Computer Aided Geometric Design 26 (2009) 1006–1015
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Fig. 3. The control polygon of a planar quartic Bézier curve with cusp.
L 0 L 3 sin2 θ =
9
L 1 L 2 sin(ϕ − 2θ ) sin ψ, 4 27 3 L 0 L 23 sin3 θ = L sin2 (ϕ − 2θ ) sin(ϕ − θ ). 4 2 As in the previous case, all the interior angles of the control polygon are in (0, π ). There exists a line which passes through P 2 and intersects the lines given by P 0 , P 1 and P 3 , P 4 at A and B respectively such that P 1 A P 2 = P 2 B P 3 = π − θ , as shown in Fig. 3. Similarly, we have (15) by the law of sines. Thus we have a geometric characterization of PH quartics which is valid for both classes of curves (with or without the cusp). Theorem 5. For a planar quartic P (t ) with Bézier control points P 0 , . . . , P 4 , it is a PH curve if and only if there is a line A B which passes through P 2 and intersects with P 0 P 1 and P 3 P 4 at A and B respectively, such that P1AP2
= P 2 B P 3,
L 0 · R 2 = 3R 0 · R 1 , L 3 · R 1 = 3R 2 · R 3 , R 1 · R 2 = 4R 0 · R 3 .
(16)
4. G 1 Hermite interpolation Theorem 5 provides simple geometric construction procedures for interpolating G 1 Hermite data by PH quartics. Given points P 0 and P 4 with associated tangent directions d0 and d4 , we first choose P 2 on some curve segments, and then find the inner control points P 1 and P 3 which satisfy the conditions in Theorem 5 with
P 1 = P 0 + λd0 , P 3 = P 4 − μd4 for some real constants λ and μ. The curve depends on the choice of the control point P 2 if G 1 Hermite data is given. If P 2 is chosen, we get P 1 and P 3 from (16) and the quartic curve is completely determined. In the rest of this section, we see that P 2 is not arbitrarily chosen. The locus of P 2 can be evaluated. For convenience, we write
Lˆ 0 = A P 0 ,
Lˆ 1 = A B ,
Lˆ 2 = B P 4
in the following subsections. 4.1. PH quartics without cusps For a curve without cusps, its control polygon satisfies
Lˆ 0 = L 0 + R 0 , Lˆ 1 = R 1 + R 2 , Lˆ 2 = L 3 + R 3 . Let q =
R1 . R2
(17)
From (16) we see that q = 0 will lead to degenerate control polygon, thus we have
4 Lˆ 0 3q + 1
·
Lˆ 2 q 3+q
=
Lˆ 21 q
(q + 1)2
.
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Fig. 4. Control polygon of a PH quartic without cusps.
Hence
3 Lˆ 21 − 4 Lˆ 0 Lˆ 2 q2 + 10 Lˆ 21 − 8 Lˆ 0 Lˆ 2 q + 3 Lˆ 21 − 4 Lˆ 0 Lˆ 2 = 0,
(18)
which is a quadratic equation. Let q1 and q2 be the two solutions of Eq. (18). From Viète’s theorem we have q1 · q2 = 1. Therefore, Eq. (18) have real roots if and only if
10 Lˆ 21 − 8 Lˆ 0 Lˆ 2
2
2 − 4 3 Lˆ 21 − 4 Lˆ 0 Lˆ 2 0
and
3 Lˆ 21 − 4 Lˆ 0 Lˆ 2 · 10 Lˆ 21 − 8 Lˆ 0 Lˆ 2 < 0.
It follows that
Lˆ 0 Lˆ 2 Lˆ 21 <
4 3
Lˆ 0 Lˆ 2 .
The PH quartic degenerates to a cubic curve with control points P 0 , A , B, and P 4 if
Lˆ 0 Lˆ 2 = Lˆ 21 , which implies R 1 = R 2 . Given two points P 0 and P 4 with associated tangent vectors d0 and d4 , we have lines A B and A B which satisfy P 0 A B
= P 4 B A = P 0 A B = P 4 B A , 4
A B 2 = A P 0 · B P 4 , 3
2
A B = A P 0 · B P 4 . Let C be the midpoint of the segment A B. It is obvious that
4 A P 2 · B P 2 = A B − 2 C P 2 · A B + 2 C P 2 .
(19)
A P
By substituting q = B P 2 , Eqs. (17) and (19) into Eq. (18), we get 2
C P 2 2 = A B 2 − A P 0 · B P 4 .
(20)
Now we choose the coordinates such that the origin is located at the midpoint O of the segment A B , while the line A B is aligned with the x-axis, as shown in Fig. 4. Let H be the midpoint of the segment A B . Further let a = A P 0 , b = B P 4 , c = A B , d = A B , h = O H , and l = A A . If P 2 = x + yi, then it follows from Eq. (20) that
h2 x2 − (c − d)2 − l2 y 2 + 2c (c − d) + l(a + b) hy + ab − c 2 h2 = 0. Hence P 2 is on a quadratic curve segment with ends A and B . In Fig. 5, given P 0 = 0 + 0i and P 4 = 5 + 0i, with the tangent vectors d0 = √1 + √2 i and d4 = − √1 + √2 i, the locus of 5
5
5
5
P 2 is a part of an ellipse. In the figure, to determine the curve, we choose P 2 to be the intersection of the line x = 2 and the curve segment. We have a PH quartic without cusps.
G. Wang, L. Fang / Computer Aided Geometric Design 26 (2009) 1006–1015
1013
Fig. 5. G 1 Hermite interpolation by a PH quartic without cusps.
Fig. 6. Control polygon of a PH quartic with cusp and with
d0 d4
∈ / R.
4.2. PH quartics with cusp For a curve with cusp, the control polygon satisfies
Lˆ 0 = L 0 − R 0 , Lˆ 1 = | R 1 − R 2 |, Lˆ 2 = L 3 − R 3 . Similar to Eq. (18), we have the quadratic equation
3 Lˆ 21 + 4 Lˆ 0 Lˆ 2 x2 − 10 Lˆ 21 + 8 Lˆ 0 Lˆ 2 x + 3 Lˆ 21 + 4 Lˆ 0 Lˆ 2 = 0.
This equation always has two positive solutions x1 , x2 ∈ ( 13 , 3). Further we also have an equation similar to (20):
C P 2 2 = A B 2 + A P 0 · B P 4 . d0 d4
∈ / R, let O be the intersection of P 0 P 1 and P 3 P 4 and choose the coordinates such that the origin is located at O while the angle bisector of P 0 O P 4 is aligned with the y-axis, as shown in Fig. 6. Let φ ∈ (0, π2 ) be the argument of d0 , and let m = O P 0 and n = O P 4 . We see that the locus of P 2 = x + yi is If
Since
3 tan2 φ − 1 x2 − (m − n)x sec φ + mn − y 2 = 0.
R1 R2
∈ ( 13 , 3), we have x ∈ (−m cos φ, n cos φ).
In Fig. 7, given P 0 = 0 + 0i and P 4 = 5 + 0i with d0 = √1 + √2 i and d4 = − √1 + √2 i, we see that the locus of P 2 is a 5
5
5
5
hyperbola with two branches. On the left-hand side of the figure, P 2 is chosen as the intersection of the line x = 3 and the top branch, and a PH quartic is generated. On the right-hand side of the figure, P 2 is chosen as the intersection of the line x = 3 and the bottom branch, and another PH quartic is constructed. d If d0 ∈ R, the line P 0 P 1 is parallel to the line P 3 P 4 . We cannot choose a coordinate system as above. Instead we choose 4
the line which is equidistance to both P 0 P 1 and P 3 P 4 as x-axis, and the line A B which satisfies A P 0 = B P 4 as y-axis, as shown in Fig. 8. Let r = A P 0 = B P 4 and s = A B . The locus of P 2 = x + yi is
x2 + y 2 = r 2 + s2 . Since RR 1 ∈ ( 13 , 3), we have |x| < r. 2 In Fig. 9, given P 0 = 0 + 0i and P 4 = 5 + 3i with d0 = 1 + 0i and d4 = −1 + 0i, we see that the locus of P 2 are two arcs of a circle. In the figure, to determine the curve, P 2 is chosen as the intersection of the line x = 3 and the top arc. A PH quartic with cusp is then constructed.
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G. Wang, L. Fang / Computer Aided Geometric Design 26 (2009) 1006–1015
Fig. 7. G 1 Hermite interpolation by a PH quartic with cusp and with
Fig. 8. Control polygon of a PH quartic with cusp and with
d0 d4
d0 d4
∈ / R.
∈ R.
Fig. 9. G 1 Hermite interpolation by a PH quartic with cusp and with
d0 d4
∈ R.
Acknowledgements The authors thank the anonymous referees for their helpful suggestions and comments. Also the authors wish to thank Dr. Zhaozhen Huang for his great help in improving the language. This work is supported by the Natural Science Foundation of China (No. 60773179) and Foundation of State Key Basic Research 973 Development Programming Item of China (No. G2004CB318000). References Choi, H.I., Kwon, S.-H., 2008. Absolute hodograph winding number and planar PH quintic splines. Computer Aided Geometric Design 25, 230–246. Farouki, R.T., 1992. Pythagorean hodograph curves in practical use. In: Barnhill, R.E. (Ed.), Geometry Processing for Design and Manufacturing. SIAM, pp. 3– 33. Farouki, R.T., 1994. The conformal map z → z2 of the hodograph plane. Computer Aided Geometric Design 11, 363–390. Farouki, R.T., 1996. The elastic bending energy of Pythagorean curves. Computer Aided Geometric Design 13, 227–241. Farouki, R.T., 1997. Pythagorean–hodograph quintic transition curves of monotone curvature. Computer-Aided Design 29, 606–610. Farouki, R.T., 2002. Pythagorean hodograph curves. In: Farin, G., Hoschek, J., Kim, M.-S. (Eds.), Handbook of Computer Aided Geometric Design. NorthHolland, pp. 405–427. Farouki, R.T., 2008. Pythagorean–Hodograph Curves: Algebra and Geometry Inseparable. Springer. Farouki, R.T., Giannelli, C., Manni, C., Sestini, A., 2008. Identification of spatial PH quintic Hermite interpolants with near-optimal shape measures. Computer Aided Geometric Design 25, 274–297. Farouki, R.T., Manjunathaiah, J., Nichlas, D., Yuan, G.F., Jee, S., 1998. Variable-feedrate CNC interpolators for constant material removal rates along Pythagorean–hodograph curves. Computer-Aided Design 30, 631–640.
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