On Control Polygons of Planar Sextic Pythagorean Hodograph Curves

Abstract

In this paper, we analyze planar parametric sextic curves to determine conditions for Pythagorean hodograph (PH) curves. By expressing the curves to be analyzed in the complex form, the analysis is conducted in algebraic form. Since sextic PH curves can be classified into two classes according to the degrees of their derivatives’ factors, we introduce auxiliary control points to reconstruct the internal algebraic structure for both classes. We prove that a sextic curve is completely characterized by the lengths of legs and angles formed by the legs of their Bézier control polygons. As such conditions are invariant under rotations and translations, we call them the geometric characteristics of sextic PH curves. We demonstrate that the geometric characteristics form the basis for an easy and intuitive method for identifying sextic PH curves. Benefiting from our results, the computations of the parameters of cusps and/or inflection points can also be simplified.

1. Introduction

This paper describes an intuitive geometric method for determining whether a planar parametric sextic curve is a Pythagorean hodograph (PH) curve [1]. We call this determination the geometric characteristics of sextic PH curves. A curve is a PH curve if the length of its hodograph coincides with the square of a real polynomial. PH curves have wide-ranging applications due to the special algebraic structures of hodographs. For instance, in numerical control (NC) machining operation, many algorithms rely on the computation of arc lengths or offsets of curves. The arc length of a PH curve can be computed without numerical integration, which speeds up the algorithms for NC machining [2]. The offset of a PH curve can be represented exactly instead of being approximated [3].

PH curves in both polynomial and rational forms have been extensively studied in 2- or 3-dimensional Euclidean spaces [4,5,6,7,8]. Among different representations of PH curves, planar polynomial PH curves are the most widely used due to their simplicity. Hence, the identification of planar polynomial PH curves becomes an important problem in most related applications, e.g., reverse engineering [9].

A fair amount of literature has been focused on developing methods for identifying planar polynomial PH curves of different degrees. They can be classified into algebraic methods and geometric methods. In these methods, PH curves are usually expressed in the Bézier form due to its popularity in CAD systems. Algebraic methods provide conditions for a curve to be a PH curve according to the algebraic properties of PH curves. For example, Farouki et al. determined whether or not a polynomial curve is a PH curve based on the satisfaction of a system of algebraic constraints on its Bézier control polygon legs [9]. Although algebraic methods are straightforward, they are less intuitive in identifying PH curves in practical applications. Geometric methods provide a more flexible and efficient alternative construction or identification of PH curves according to the geometric properties of PH curves. For example, an intuitive geometric condition on the control polygon for a cubic Bézier curve to be a PH curve was proposed in [1]. To obtain more geometric freedoms for practical design problems, the study of geometric constraints on control polygons has proceeded to PH curves of higher degrees [10,11,12,13]. The conditions for control polygons of cubics and quintics in Euclidean and Minkowski spaces have been well studied [14], and the relationship between these curves in Euclidean and Minkowski spaces have been further investigated [15]. Šír [16] studied the classification of planar PH curves with respect to similarities and linear reparameterizations, and he gave general expressions for planar PH curves of low degrees. The work was then employed to discuss the classification of polynomial minimal surfaces [17].

The most-relevant previous work is [13], in which necessary and sufficient conditions for sextic PH curves were provided. Based on these conditions, three methods for the identification of planar sextic PH curves are proposed, where the first method relies on solving a system of degree eleven, the second method can only deal with convex control polygons, and the third method is limited to some special cases of planar sextic PH curves. In this paper, we provide a more succinct and complete description of the geometric characteristics of sextic PH curves. Following the geometric characteristics, we propose an efficient method for the identification of sextic PH curves. The key technique for making our method both intuitive and flexible is the introduction of auxiliary control points. In particular, we geometrically compute auxiliary control points for given sextic curves and obtain auxiliary control polygons. The geometric characteristics are then described using the leg lengths of the auxiliary control polygons and angles formed by the legs. Our method is suitable for general sextic polynomial curves, and we only resort to solve lower-degree (quadratic) systems.

The remainder of this paper is organized as follows. Section 2 introduces some fundamental aspects of sextic PH curves in the Bézier form. Section 3 presents some geometric properties of sextic PH curves. In Section 4, sextic PH curves are classified into two classes and their geometric characteristics are presented, respectively. Finally, in Section 5, we conclude the paper.

2. Bézier Representation for Sextic PH Curves

In this section, we recall the Bézier representation for planar parametric sextic PH curves. We first introduce some notations and, then, use the complex representation of ${\mathbb{R}}^2$ to facilitate the computation in the later analysis of planar PH curves [4,11].

Let $\mathbb{R}$ and $\mathbb{C}$ denote the sets of all real numbers and complex numbers, respectively. A complex number is denoted by a single bold character, e.g., $\mathit{z}$. The complex norm, the real part, the imaginary part, and the argument of $\mathit{z}$ in $\mathbb{C}$ are denoted by $\parallel \mathit{z}\parallel$, $\mathrm{Re}\mathit{z}$, $\mathrm{Im}\mathit{z}$, and $arg\mathit{z}\in [0,2\pi )$, respectively. A planar point $(x,y)$ is interchangeably represented as a complex number $x+\mathbf{i}y$. Similarly, a planar parametric curve $\mathit{P}\left(t\right)=\left(x\right(t),y(t\left)\right)$, $t\in [0,1]$, can be identified with a complex-valued function $\mathit{P}\left(t\right)=x\left(t\right)+\mathbf{i}y\left(t\right)$, and vice versa.

The Bézier representation for a planar sextic polynomial curve is defined by seven control points in the plane, ${\left\{{\mathit{P}}_i\right\}}_{i=0}^6$, as

$$\mathit{P}\left(t\right)=(x\left(t\right),y\left(t\right))=\sum _{i=0}^6{B}_i^6\left(t\right){\mathit{P}}_i,0\le t\le 1,$$

(1)

where $B_i^n\left(t\right)=\left(\genfrac{}{}{0pt}{}{n}{i}\right){(1-t)}^{n-i}{t}^i$, $i=0,\dots ,n$, are the Bernstein polynomials. The polygon formed by consecutive control points ${\left\{{\mathit{P}}_i\right\}}_{i=0}^6$ is called the control polygon. Let $\Delta {\mathit{P}}_i$ denote the first forward-difference of the i-th control point, i.e., $\Delta {\mathit{P}}_i={\mathit{P}}_{i+1}-{\mathit{P}}_i$; thus, the derivative of the curve (1) can be represented as

$${\mathit{P}}^{\prime }\left(t\right)=(x^{\prime }\left(t\right),y^{\prime }\left(t\right))=6\sum _{i=0}^5{B}_i^5\left(t\right)\Delta {\mathit{P}}_i.$$

(2)

A planar sextic Bézier curve (1) is called a Pythagorean hodograph (PH) curve if and only if its hodograph ${\mathit{P}}^{\prime }\left(t\right)=x^{\prime }\left(t\right)+\mathbf{i}{y}^{\prime }\left(t\right)$ satisfies the Pythagorean condition:

$$x^{\prime 2}\left(t\right)+y^{\prime 2}\left(t\right)={\sigma }^2\left(t\right),$$

for some real polynomial $\sigma \left(t\right)$ [1]. Equivalently, a planar sextic Bézier curve (1) is a PH curve if and only if

$${\mathit{P}}^{\prime }\left(t\right)=w\left(t\right){[u\left(t\right)+\mathbf{i}v\left(t\right)]}^2,$$

(3)

for some real polynomials $u\left(t\right)$, $v\left(t\right)$, and $w\left(t\right)$, where $u\left(t\right)$ and $v\left(t\right)$ are relatively prime [12].

3. Properties of Sextic PH Curves

In this section, we investigate the presence of the cusps, inflection points, and curvature of a sextic PH curve. According to the sufficient and necessary condition (3) for a sextic curve to be PH, the degrees of $u\left(t\right)+\mathbf{i}v\left(t\right)$ and $w\left(t\right)$ are either 2 and 1 or 1 and 3, which are mutually exclusive. The sextic PH curves corresponding to these two cases are referred to as Class I and Class II sextic PH curves, respectively, and will be discussed in detail in this section.

Class I

Polynomials $u\left(t\right)+\mathbf{i}v\left(t\right)$ and $w\left(t\right)$ are quadratic and linear, respectively. Regardless of the symmetry of control points, we rewrite the factors of ${\mathit{P}}^{\prime }\left(t\right)$ in Bernstein form as

$$\begin{array}{cc}\hfill & w\left(t\right)=a_0(1-t)+t,\hfill \\ \hfill & u\left(t\right)+\mathbf{i}v\left(t\right)=\sum _{i=0}^2{\mathit{z}}_i{B}_i^2\left(t\right),\hfill \end{array}$$

(4)

where ${\mathit{z}}_0,{\mathit{z}}_2\ne 0$. Remark: if ${\mathit{z}}_0=0$ or ${\mathit{z}}_2=0$, the curve is indeed a Class II sextic PH curve, which will be discussed later. If $a_0\ne 1$, then $w\left(t\right)$ is a linear polynomial with real coefficients, which has exactly one real root. Denoting by ${\xi }_0$ the root of $w\left(t\right)$, we have

$$a_0=\frac{{\xi }_0}{{\xi }_0-1}.$$

(5)

That is to say, a sextic PH curve with the hodograph of the form (4) has a cusp occurring at ${\xi }_0$. Otherwise, if $a_0=1$ (i.e., $w\left(t\right)\equiv 1$), the corresponding sextic curve degenerates to a Class I quintic PH curve [10]. On the contrary, a quintic Class I PH curve can also be considered as a sextic PH curve of this class using one step of degree elevation.

Class II

Polynomials $u\left(t\right)+\mathbf{i}v\left(t\right)$ and $w\left(t\right)$ are linear and cubic, respectively. We rewrite the factors of ${\mathit{P}}^{\prime }\left(t\right)$ in Bernstein form as

$$\begin{array}{cc}\hfill & w\left(t\right)=\sum _{i=0}^3{a}_i{B}_i^3\left(t\right),\hfill \\ \hfill & u\left(t\right)+\mathbf{i}v\left(t\right)={\mathit{z}}_0(1-t)+{\mathit{z}}_{1}t,\hfill \end{array}$$

(6)

for some $a_i\in \mathbb{R}$, $i=0,\dots ,3$, and ${\mathit{z}}_j\in \mathbb{C}$, $j=0,1$. Let ${\mathit{\xi }}_i$, $i=0,1,2$, be roots of the cubic polynomial $w\left(t\right)$ in Equation (6). According to the fundamental theorem of algebra, there is at least one real root. We assume ${\mathit{\xi }}_0\in \mathbb{R}$, then ${\mathit{\xi }}_1$ and ${\mathit{\xi }}_2$ are either real numbers or a pair of conjugate complex numbers. Hence, a Class II sextic PH curve presents one or three cusps, depending on the real root number of $w\left(t\right)$. Note that, when the cubic real polynomial $w\left(t\right)$ degenerates to a quadratic, linear, or constant polynomial, the sextic PH curve degenerates to a Class II quintic, quartic, or cubic PH curve, respectively. Equally, cubic, quartic, and Class II quintic PH curves can be converted to Class II sextic PH curves using three steps to one step of degree elevation. A Class II sextic PH curve degenerates to a line segment when ${\mathit{z}}_0=0$ or ${\mathit{z}}_1=0$, which is not discussed in this paper.

Let us now consider the curvature of sextic PH curves. In the following, the variables of functions in the notations are omitted for simplicity, e.g., u means $u\left(t\right)$. Following (3), the second derivative ${\mathit{P}}^{″}=(x^{″},y^{″})$ is

$$\begin{array}{cc}\hfill & x^{″}=w^{\prime }(u^2-v^2)+2w(u{u}^{\prime }-v{v}^{\prime }),\hfill \\ \hfill & y^{″}=2(u^{\prime }vw+u{v}^{\prime }w+uv{w}^{\prime }).\hfill \end{array}$$

(7)

Therefore,

$$\begin{array}{cc}\hfill {\mathit{P}}^{\prime }\times {\mathit{P}}^{″}& =2w(u^2-v^2)(u^{\prime }vw+u{v}^{\prime }w+uv{w}^{\prime })-2uvw[w^{\prime }(u^2-v^2)+2w(u{u}^{\prime }-v{v}^{\prime })]\hfill \\ \hfill & =2u^2{u}^{\prime }v{w}^2-2u^{\prime }{v}^3{w}^2+2u^3{v}^{\prime }{w}^2-2u{v}^2{v}^{\prime }{w}^2-2u^2{u}^{\prime }v{w}^2+2u{v}^2{v}^{\prime }{w}^2\hfill \\ \hfill & =2v^2{w}^2(u{v}^{\prime }-u^{\prime }v)+2u^2{w}^2(u{v}^{\prime }-u^{\prime }v)\hfill \\ \hfill & =2w^2(u^2+v^2)(u{v}^{\prime }-u^{\prime }v).\hfill \end{array}$$

Thus, its curvature k can be computed by

$$\begin{array}{cc}\hfill k& =\frac{|{\mathit{P}}^{\prime }\times {\mathit{P}}^{″}|}{|{\mathit{P}}^{\prime }{|}^3}=\frac{u{v}^{\prime }-u^{\prime }v}{w\sqrt{u^2+v^2}}.\hfill \end{array}$$

(8)

Note that linear polynomials u, v lead to a constant numerator of curvature $u^{\prime }v-u{v}^{\prime }$, which means that the curvature does not vanish, i.e., no inflections occur on Class II sextic PH curves. Furthermore, we may claim that there are no inflection points on cubic [1], quartic [12], Class II quintic [10], and Class III septic [18] PH curves, i.e., they either are convex curves or have cusps.

4. Geometric Characteristics of Sextic PH Curves

In this section, we describe the geometric characteristics of sextic PH curves using auxiliary control points, based on which we can identify whether a given sextic polynomial curve is a PH curve or not. We introduce auxiliary control points, which are linear combinations of the original control points. Following that, we discuss the sufficient and necessary conditions for a planar sextic Bézier curve to possess a PH based on the auxiliary control points.

4.1. Class I Sextic PH Curve

For a Class I sextic curve in Bézier representation, we introduce eight auxiliary control points ${\mathit{Q}}_i$ for $i=0,\cdots ,7$, by the following four steps:

• Let $\alpha =\frac{arg\Delta {\mathit{P}}_0+arg\Delta {\mathit{P}}_5}{2}$, and

  $$\begin{array}{cc}\hfill & {\mathit{Q}}_0={\mathit{P}}_1+k_0\Delta {\mathit{P}}_0,\hfill \\ \hfill & {\mathit{Q}}_7={\mathit{P}}_5-k_1\Delta {\mathit{P}}_5,\hfill \end{array}$$

  (9)

  where

  $$\begin{array}{cc}\hfill & k_0=\frac{tan\alpha ·\mathrm{Re}\Delta {\mathit{P}}_1-\mathrm{Im}\Delta {\mathit{P}}_1}{tan\alpha ·\mathrm{Re}\Delta {\mathit{P}}_0-\mathrm{Im}\Delta {\mathit{P}}_0},\hfill \\ \hfill & k_1=\frac{tan\alpha ·\mathrm{Re}\Delta {\mathit{P}}_4-\mathrm{Im}\Delta {\mathit{P}}_4}{tan\alpha ·\mathrm{Re}\Delta {\mathit{P}}_5-\mathrm{Im}\Delta {\mathit{P}}_5}.\hfill \end{array}$$

  (10)
• Let

  $$\begin{array}{cc}\hfill & c=\frac{\parallel {\mathit{P}}_4-{\mathit{Q}}_7{\parallel }^2}{\parallel \Delta {\mathit{P}}_0\parallel ·\parallel \Delta {\mathit{P}}_5\parallel },\hfill \\ \hfill & d=\frac{\parallel {\mathit{P}}_2-{\mathit{Q}}_0{\parallel }^2}{\parallel \Delta {\mathit{P}}_0\parallel ·\parallel \Delta {\mathit{P}}_5\parallel };\hfill \end{array}$$

  (11)

  thus, we obtain

  $$\begin{array}{cc}\hfill & {\mathit{Q}}_1={\mathit{P}}_1+\frac{\Delta {\mathit{P}}_0}{5x},\hfill \\ \hfill & {\mathit{Q}}_6={\mathit{P}}_5-\frac{x}{5}\Delta {\mathit{P}}_5,\hfill \end{array}$$

  (12)

  where x is a solution of the system of cubic equations as follows:

  $$\begin{array}{cc}\hfill & x^3-10k_1{x}^2+25k_1^{2}x-25d=0,\hfill \\ \hfill & 25c{x}^3-25k_0^2{x}^2+10k_{0}x-1=0,\hfill \\ \hfill & 5k_0{x}^2+(\pm 25\sqrt{cd}-25k_0{k}_1-1)x+5k_1=0.\hfill \end{array}$$

  (13)
  We defer deriving this system until the proof of Theorem 1. Note that the roots of the third equation of System (13) come in pairs:

  $$\begin{array}{cc}\hfill & x_1=\frac{-(\pm 25\sqrt{cd}-25k_0{k}_1-1)+\sqrt{{(\pm 25\sqrt{cd}-25k_0{k}_1-1)}^2-100k_0{k}_1}}{10k_0},\hfill \\ \hfill & x_2=\frac{-(\pm 25\sqrt{cd}-25k_0{k}_1-1)-\sqrt{{(\pm 25\sqrt{cd}-25k_0{k}_1-1)}^2-100k_0{k}_1}}{10k_0}.\hfill \end{array}$$

  (14)
  Hence, we solve the system (13) by verifying if the roots of the third equation of Equation (13) satisfy the first two equations. Although there are four candidate solutions due to different signs, the number of roots is no more than three for the cubic system. For each real solution, Steps 3–4 are further performed, respectively. If there are no real roots for the above system, the curve is not a PH curve, and the procedure is terminated immediately.
• Let ${\mathit{Q}}_2$ and ${\mathit{Q}}_5$ be points on lines ${\mathit{Q}}_1{\mathit{P}}_2$ and ${\mathit{Q}}_6{\mathit{P}}_4$, respectively, such that

  $$\begin{array}{cc}\hfill & {\mathit{Q}}_2={\mathit{P}}_2+\frac{{\mathit{P}}_2-{\mathit{Q}}_1}{2x},\hfill \\ \hfill & {\mathit{Q}}_5={\mathit{P}}_4-\frac{x}{2}({\mathit{Q}}_6-{\mathit{P}}_4).\hfill \end{array}$$

  (15)
• Finally, let $\beta =\frac{arg({\mathit{P}}_2-{\mathit{Q}}_1)+arg({\mathit{Q}}_6-{\mathit{P}}_4)}{2}$, and

  $$\begin{array}{cc}\hfill & {\mathit{Q}}_3={\mathit{P}}_3+\frac{\mathrm{Im}({\mathit{Q}}_2-{\mathit{P}}_3)·cos\beta -\mathrm{Re}({\mathit{Q}}_2-{\mathit{P}}_3)·sin\beta }{sin(\alpha -\beta )}·e^{\mathbf{i}\alpha }\hfill \\ \hfill & ={\mathit{Q}}_2+\frac{\mathrm{Im}({\mathit{Q}}_2-{\mathit{P}}_3)·cos\alpha -\mathrm{Re}({\mathit{Q}}_2-{\mathit{P}}_3)·sin\alpha }{sin(\alpha -\beta )}·e^{\mathbf{i}\beta },\hfill \\ \hfill & {\mathit{Q}}_4={\mathit{P}}_3+\frac{\mathrm{Im}({\mathit{Q}}_5-{\mathit{P}}_3)·cos\beta -\mathrm{Re}({\mathit{Q}}_5-{\mathit{P}}_3)·sin\beta }{sin(\alpha -\beta )}·e^{\mathbf{i}\alpha }\hfill \\ \hfill & ={\mathit{Q}}_5+\frac{\mathrm{Im}({\mathit{Q}}_5-{\mathit{P}}_3)·cos\alpha -\mathrm{Re}({\mathit{Q}}_5-{\mathit{P}}_3)·sin\alpha }{sin(\alpha -\beta )}·e^{\mathbf{i}\beta }.\hfill \end{array}$$

  (16)

Now, we give a necessary and sufficient condition for Class I sextic PH curves.

Theorem 1.

Given a planar sextic curve with Bézier control points ${\mathit{P}}_i$, $i=0,\dots ,6$, it is a Class I sextic PH curve if and only if there are auxiliary points ${\mathit{Q}}_j$, $j=1,\dots ,7$, constructed following (12)–(16), such that ${\mathit{Q}}_2{\mathit{Q}}_3⫽{\mathit{Q}}_4{\mathit{Q}}_5$, and they further satisfy

$$\begin{array}{cc}\hfill & 5\parallel {\mathit{P}}_2-{\mathit{Q}}_1{\parallel }^2=8\parallel \Delta {\mathit{P}}_0\parallel ·\parallel \Delta {\mathit{Q}}_2\parallel ,\hfill \\ \hfill & 5\parallel {\mathit{Q}}_6-{\mathit{P}}_4{\parallel }^2=8\parallel \Delta {\mathit{Q}}_4\parallel ·\parallel \Delta {\mathit{P}}_5\parallel ,\hfill \\ \hfill & \parallel {\mathit{Q}}_2-{\mathit{P}}_2{\parallel }^2=2\parallel {\mathit{Q}}_1-{\mathit{P}}_1\parallel ·\parallel \Delta {\mathit{Q}}_4\parallel ,\hfill \\ \hfill & \parallel {\mathit{P}}_4-{\mathit{Q}}_5{\parallel }^2=2\parallel \Delta {\mathit{Q}}_2\parallel ·\parallel {\mathit{P}}_5-{\mathit{Q}}_6\parallel ,\hfill \\ \hfill & 5\parallel {\mathit{P}}_3-{\mathit{Q}}_3{\parallel }^2=\parallel \Delta {\mathit{P}}_0\parallel ·\parallel {\mathit{P}}_5-{\mathit{Q}}_6\parallel ,\hfill \\ \hfill & 5\parallel {\mathit{Q}}_4-{\mathit{P}}_3{\parallel }^2=\parallel {\mathit{Q}}_1-{\mathit{P}}_1\parallel ·\parallel \Delta {\mathit{P}}_5\parallel .\hfill \end{array}$$

(17)

Proof. 

A planar sextic curve with Bézier control points ${\mathit{P}}_i$, $i=0,\dots ,6$, is a Class I sextic PH curve if and only if its derivative with factors in Equation (4) satisfies Equation (3), in other words, if and only if there are $a_0$, ${\mathit{z}}_0$, ${\mathit{z}}_1$, ${\mathit{z}}_2$ satisfying:

$$\begin{array}{cc}\hfill {\mathit{P}}^{\prime }\left(t\right)=& a_0{\mathit{z}}_0^2{(1-t)}^5\hfill \\ \hfill & +({\mathit{z}}_0^2+4a_0{\mathit{z}}_0{\mathit{z}}_1){(1-t)}^{4}t\hfill \\ \hfill & +(4{\mathit{z}}_0{\mathit{z}}_1+4a_0{\mathit{z}}_1^2+2a_0{\mathit{z}}_0{\mathit{z}}_2){(1-t)}^3{t}^2\hfill \\ \hfill & +(4{\mathit{z}}_1^2+2{\mathit{z}}_0{\mathit{z}}_2+4a_0{\mathit{z}}_1{\mathit{z}}_2){(1-t)}^2{t}^3\hfill \\ \hfill & +(4{\mathit{z}}_1{\mathit{z}}_2+a_0{\mathit{z}}_2^2)(1-t)t^4.\hfill \\ \hfill & +{\mathit{z}}_2^2{t}^5.\hfill \end{array}$$

By matching the coefficients of Bernstein polynomials in the above equations with Equation (2), we obtain the decompositions of the control polygon legs:

$$\begin{array}{cc}\hfill \frac{1}{6}{a}_0{\mathit{z}}_0^2=& \Delta {\mathit{P}}_0,\hfill \\ \hfill \frac{1}{30}({\mathit{z}}_0^2+4a_0{\mathit{z}}_0{\mathit{z}}_1)=& \Delta {\mathit{P}}_1,\hfill \\ \hfill \frac{1}{60}(4{\mathit{z}}_0{\mathit{z}}_1+4a_0{\mathit{z}}_1^2+2a_0{\mathit{z}}_0{\mathit{z}}_2)=& \Delta {\mathit{P}}_2,\hfill \\ \hfill \frac{1}{60}(4{\mathit{z}}_1^2+2{\mathit{z}}_0{\mathit{z}}_2+4a_0{\mathit{z}}_1{\mathit{z}}_2)=& \Delta {\mathit{P}}_3,\hfill \\ \hfill \frac{1}{30}(4{\mathit{z}}_1{\mathit{z}}_2+a_0{\mathit{z}}_2^2)=& \Delta {\mathit{P}}_4,\hfill \\ \hfill \frac{1}{6}{\mathit{z}}_2^2=& \Delta {\mathit{P}}_5.\hfill \end{array}$$

(18)

We define six auxiliary control points ${\tilde{\mathit{Q}}}_i$, $i=1,\cdots ,6$:

$$\begin{array}{cc}\hfill & {\tilde{\mathit{Q}}}_1={\mathit{P}}_1+\frac{1}{30}{\mathit{z}}_0^2={\mathit{P}}_2-\frac{2a_0}{15}{\mathit{z}}_0{\mathit{z}}_1,\hfill \\ \hfill & {\tilde{\mathit{Q}}}_2={\mathit{P}}_2+\frac{1}{15}{\mathit{z}}_0{\mathit{z}}_1,\hfill \\ \hfill & {\tilde{\mathit{Q}}}_3={\tilde{\mathit{Q}}}_2+\frac{a_0}{15}{\mathit{z}}_1^2={\mathit{P}}_3-\frac{a_0}{30}{\mathit{z}}_0{\mathit{z}}_2,\hfill \\ \hfill & {\tilde{\mathit{Q}}}_4={\mathit{P}}_3+\frac{1}{30}{\mathit{z}}_0{\mathit{z}}_2,\hfill \\ \hfill & {\tilde{\mathit{Q}}}_5={\tilde{\mathit{Q}}}_4+\frac{1}{15}{\mathit{z}}_1^2={\mathit{P}}_4-\frac{a_0}{15}{\mathit{z}}_1{\mathit{z}}_2,\hfill \\ \hfill & {\tilde{\mathit{Q}}}_6={\mathit{P}}_4+\frac{2}{15}{\mathit{z}}_1{\mathit{z}}_2={\mathit{P}}_5-\frac{a_0}{30}{\mathit{z}}_2^2.\hfill \end{array}$$

(19)

Let ${\tilde{\mathit{Q}}}_0$, ${\tilde{\mathit{Q}}}_7$ be points on lines ${\mathit{P}}_0{\mathit{P}}_1$ and ${\mathit{P}}_5{\mathit{P}}_6$, respectively, such that

$$\begin{array}{cc}\hfill & arg\frac{{\mathit{P}}_2-{\tilde{\mathit{Q}}}_0}{\Delta {\mathit{P}}_0}=arg\frac{\Delta {\mathit{P}}_5}{{\tilde{\mathit{Q}}}_7-{\mathit{P}}_4}=\frac{1}{2}arg\frac{\Delta {\mathit{P}}_5}{\Delta {\mathit{P}}_0},\hfill \\ \hfill & \mathrm{Re}({\tilde{\mathit{Q}}}_0-{\mathit{P}}_2)\mathrm{Im}({\mathit{P}}_4-{\tilde{\mathit{Q}}}_7)=\mathrm{Im}({\tilde{\mathit{Q}}}_0-{\mathit{P}}_2)\mathrm{Re}({\mathit{P}}_4-{\tilde{\mathit{Q}}}_7),\hfill \end{array}$$

(20)

that is ${\tilde{\mathit{Q}}}_0{\mathit{P}}_2⫽{\mathit{P}}_4{\tilde{\mathit{Q}}}_7$. By computing the intersecting point of lines ${\mathit{P}}_0{\mathit{P}}_1$ and ${\tilde{\mathit{Q}}}_0{\mathit{P}}_2$, we obtain ${\tilde{\mathit{Q}}}_0={\mathit{Q}}_0$; similarly, we have ${\tilde{\mathit{Q}}}_7={\mathit{Q}}_7$; see Figure 1a. Note that triangles $▵{\mathit{P}}_2{\mathit{Q}}_0{\tilde{\mathit{Q}}}_1$ and $▵{\tilde{\mathit{Q}}}_6{\mathit{Q}}_7{\mathit{P}}_4$ are similar triangles, as their edges are parallel with ${\mathit{z}}_2^2$, ${\mathit{z}}_0{\mathit{z}}_2$, ${\mathit{z}}_1{\mathit{z}}_2$, respectively. Hence, the ratios of the lengths of their corresponding edges are equal:

$$\frac{\parallel {\tilde{\mathit{Q}}}_1-{\mathit{Q}}_0\parallel }{\parallel {\mathit{P}}_4-{\mathit{Q}}_7\parallel }=\frac{\parallel {\mathit{P}}_2-{\mathit{Q}}_0\parallel }{\parallel {\mathit{Q}}_7-{\tilde{\mathit{Q}}}_6\parallel }=\frac{\parallel {\mathit{P}}_2-{\tilde{\mathit{Q}}}_1\parallel }{\parallel {\mathit{P}}_4-{\tilde{\mathit{Q}}}_6\parallel }.$$

(21)

Following Equations (18) and (19), we have

$$\begin{array}{cc}\hfill & \frac{\Delta {\mathit{P}}_0}{5({\tilde{\mathit{Q}}}_1-{\mathit{P}}_1)}=\frac{5({\mathit{P}}_5-{\tilde{\mathit{Q}}}_6)}{\Delta {\mathit{P}}_5}=a_0,\hfill \\ \hfill & \frac{\parallel {\mathit{P}}_2-{\tilde{\mathit{Q}}}_1{\parallel }^2}{\parallel {\mathit{P}}_4-{\tilde{\mathit{Q}}}_6{\parallel }^2}=\frac{a_0^2{\parallel {\mathit{z}}_0\parallel }^2}{\parallel {\mathit{z}}_2{\parallel }^2}=\frac{|a_0|·\parallel \Delta {\mathit{P}}_0\parallel }{\parallel \Delta {\mathit{P}}_5\parallel }.\hfill \end{array}$$

Using the notations from Equation (10), we have

$$\begin{array}{cc}\hfill & \parallel {\tilde{\mathit{Q}}}_1-{\tilde{\mathit{Q}}}_0\parallel =\parallel {\tilde{\mathit{Q}}}_1-{\mathit{P}}_1+{\mathit{P}}_1-{\tilde{\mathit{Q}}}_0\parallel =|k_0-\frac{1}{5a_0}|·\parallel \Delta {\mathit{P}}_0\parallel ,\hfill \\ \hfill & \parallel {\tilde{\mathit{Q}}}_7-{\tilde{\mathit{Q}}}_6\parallel =\parallel {\tilde{\mathit{Q}}}_7-{\mathit{P}}_5+{\mathit{P}}_5-{\tilde{\mathit{Q}}}_6\parallel =|k_1-\frac{a_0}{5}|·\parallel \Delta {\mathit{P}}_5\parallel .\hfill \end{array}$$

Substituting Equation (11) into Equation (21) and, then, squaring both sides of Equation (21), we obtain

$$\frac{|k_0-\frac{1}{5a_0}{|}^2}{c}=\frac{d}{|k_1-\frac{a_0}{5}{|}^2}=\left|a_0\right|.$$

Thus, the cubic equations in Equation (13) can be derived by considering every two items to be equal. In other words, there is a real root of $a_0$ such that ${\tilde{\mathit{Q}}}_i={\mathit{Q}}_i$, $i=1,2,5,6$, as shown in Figure 1b,c.

From Equation (19), we have $arg({\mathit{P}}_3-{\tilde{\mathit{Q}}}_3)=arg({\tilde{\mathit{Q}}}_4-{\mathit{P}}_3)=arg{\mathit{z}}_0{\mathit{z}}_2$, and

$$\begin{array}{cc}\hfill & arg\frac{{\mathit{P}}_3-{\tilde{\mathit{Q}}}_3}{\Delta {\mathit{P}}_0}=arg\frac{\Delta {\mathit{P}}_5}{{\tilde{\mathit{Q}}}_4-{\mathit{P}}_3}=arg\frac{{\mathit{z}}_2}{{\mathit{z}}_0},\hfill \\ \hfill & arg\frac{\Delta {\tilde{\mathit{Q}}}_2}{{\mathit{P}}_2-{\mathit{Q}}_1}=arg\frac{{\mathit{P}}_2-{\mathit{Q}}_1}{\Delta {\mathit{P}}_0}=arg\frac{{\mathit{z}}_1}{{\mathit{z}}_0},\hfill \\ \hfill & arg\frac{\Delta {\mathit{P}}_5}{{\mathit{Q}}_6-{\mathit{P}}_4}=arg\frac{{\mathit{Q}}_6-{\mathit{P}}_4}{\Delta {\tilde{\mathit{Q}}}_4}=arg\frac{{\mathit{z}}_2}{{\mathit{z}}_1};\hfill \end{array}$$

thus, we know the points ${\tilde{\mathit{Q}}}_3$, ${\mathit{P}}_3$, ${\tilde{\mathit{Q}}}_4$ are co-linear, and we have ${\tilde{\mathit{Q}}}_3{\tilde{\mathit{Q}}}_4⫽{\mathit{Q}}_0{\mathit{P}}_2⫽{\mathit{P}}_4{\mathit{Q}}_7$. Since non-parallel lines intersect at a unique point, we have ${\tilde{\mathit{Q}}}_i={\mathit{Q}}_i$, $i=3,4$, where ${\mathit{Q}}_i$ are defined in Equation (16); see Figure 1d. Thus far, we have proven that ${\tilde{\mathit{Q}}}_i$, $i=1,\dots ,6$, coincident with the auxiliary control points ${\mathit{Q}}_i$ constructed in Equations (12)–(16); see Figure 1d. Moreover, since $arg\Delta {\mathit{Q}}_2=arg\Delta {\mathit{Q}}_4=arg{\mathit{z}}_1^2$, we have ${\mathit{Q}}_2{\mathit{Q}}_3⫽{\mathit{Q}}_4{\mathit{Q}}_5$. Following Equation (19), Equation (17) can be computed as follows:

$$\begin{array}{cc}\hfill & 5\parallel {\mathit{P}}_2-{\mathit{Q}}_1{\parallel }^2=8\parallel \Delta {\mathit{P}}_0\parallel ·\parallel \Delta {\mathit{Q}}_2\parallel =\frac{4}{45}{a}_0^2\parallel {\mathit{z}}_0{\parallel }^2{\parallel {\mathit{z}}_1\parallel }^2,\hfill \\ \hfill & 5\parallel {\mathit{Q}}_6-{\mathit{P}}_4{\parallel }^2=8\parallel \Delta {\mathit{Q}}_4\parallel ·\parallel \Delta {\mathit{P}}_5\parallel =\frac{4}{45}\parallel {\mathit{z}}_1{\parallel }^2{\parallel {\mathit{z}}_2\parallel }^2,\hfill \\ \hfill & \parallel {\mathit{Q}}_2-{\mathit{P}}_2{\parallel }^2=2\parallel {\mathit{Q}}_1-{\mathit{P}}_1\parallel ·\parallel \Delta {\mathit{Q}}_4\parallel =\frac{1}{225}\parallel {\mathit{z}}_0{\parallel }^2{\parallel {\mathit{z}}_1\parallel }^2,\hfill \\ \hfill & \parallel {\mathit{P}}_4-{\mathit{Q}}_5{\parallel }^2=2\parallel \Delta {\mathit{Q}}_2\parallel ·\parallel {\mathit{P}}_5-{\mathit{Q}}_6\parallel =\frac{1}{225}{a}_0^2\parallel {\mathit{z}}_1{\parallel }^2{\parallel {\mathit{z}}_2\parallel }^2,\hfill \\ \hfill & 5\parallel {\mathit{P}}_3-{\mathit{Q}}_3{\parallel }^2=\parallel \Delta {\mathit{P}}_0\parallel ·\parallel {\mathit{P}}_5-{\mathit{Q}}_6\parallel =\frac{1}{180}{a}_0^2\parallel {\mathit{z}}_0{\parallel }^2{\parallel {\mathit{z}}_2\parallel }^2,\hfill \\ \hfill & 5\parallel {\mathit{Q}}_4-{\mathit{P}}_3{\parallel }^2=\parallel {\mathit{Q}}_1-{\mathit{P}}_1\parallel ·\parallel \Delta {\mathit{P}}_5\parallel =\frac{1}{180}\parallel {\mathit{z}}_0{\parallel }^2{\parallel {\mathit{z}}_2\parallel }^2.\hfill \end{array}$$

On the contrary, given a planar sextic curve with Bézier control points ${\mathit{P}}_i$, $i=0,\dots ,6$, if Equation (17) holds for auxiliary control points ${\mathit{Q}}_i$, $i=0,\dots ,7$, we have

$$\begin{array}{cc}\hfill a_0=& \frac{\Delta {\mathit{P}}_0}{5({\mathit{Q}}_1-{\mathit{P}}_1)}=\frac{{\mathit{P}}_2-{\mathit{Q}}_1}{2({\mathit{Q}}_2-{\mathit{P}}_2)}\hfill \\ \hfill =& \frac{\Delta {\mathit{Q}}_2}{\Delta {\mathit{Q}}_4}=\frac{{\mathit{P}}_3-{\mathit{Q}}_3}{{\mathit{Q}}_4-{\mathit{P}}_3}\hfill \\ \hfill =& \frac{2({\mathit{P}}_4-{\mathit{Q}}_5)}{{\mathit{Q}}_6-{\mathit{P}}_4}=\frac{5({\mathit{P}}_5-{\mathit{Q}}_6)}{\Delta {\mathit{P}}_5}.\hfill \end{array}$$

(22)

Note that, although the denominators and numerators are all complex numbers, the quotient $a_0$ is well-defined because the denominator and numerator in each item have the same arguments. Moreover, the complex coefficients ${\mathit{z}}_i$, $i=0,1,2$, can be solved by Equation (19), that is

$$\begin{array}{cc}\hfill & {\mathit{z}}_0=\pm \sqrt{30({\mathit{Q}}_1-{\mathit{P}}_1)},\hfill \\ \hfill & {\mathit{z}}_1=\pm \frac{15({\mathit{Q}}_2-{\mathit{P}}_2)}{\sqrt{30({\mathit{Q}}_1-{\mathit{P}}_1)}},\hfill \\ \hfill & {\mathit{z}}_2=\pm \frac{30({\mathit{Q}}_4-{\mathit{P}}_3)}{\sqrt{30({\mathit{Q}}_1-{\mathit{P}}_1)}}.\hfill \end{array}$$

By substituting $\Delta {\mathit{P}}_i$, $i=0,\dots ,5$, into Equation (2), we can verify that the curve is a Class I sextic PH curve because its hodograph is actually of the form (3), which proves the sufficiency. □

The construction of auxiliary control points depends only on the leg lengths and angles of the Bézier control polygon, which can be obtained by measurement; hence, our method is geometric-invariant.

An Example of a Class I Sextic PH Curve with a Concave Control Polygon

Now, we give an example of a Class I sextic PH curve with a concave control polygon, as shown in Figure 2. In this example, we assign the control points as follows:

$$\begin{array}{cc}\hfill & {\mathit{P}}_0=0+0\mathbf{i},\hfill \\ \hfill & {\mathit{P}}_1=1.228732694000000+2.054457192000000\mathbf{i},\hfill \\ \hfill & {\mathit{P}}_2=1.024962887533333+5.016989645599999\mathbf{i},\hfill \\ \hfill & {\mathit{P}}_3=0.519266461400000+7.371566599600000\mathbf{i},\hfill \\ \hfill & {\mathit{P}}_4=2.371552422866667+10.372970071533333\mathbf{i},\hfill \\ \hfill & {\mathit{P}}_5=7.268585548066668+11.789158331266666\mathbf{i},\hfill \\ \hfill & {\mathit{P}}_6=12.868452536400000+4.155075291266666\mathbf{i}.\hfill \end{array}$$

Following Equation (10), we compute

$$\begin{array}{cc}\hfill & k_0=1.488475973647499,\hfill \\ \hfill & k_1=-0.150203163391209,\hfill \end{array}$$

which immediately gives

$$\begin{array}{cc}\hfill & {\mathit{Q}}_0=3.057671787054164+5.112467361179306\mathbf{i}\hfill \\ \hfill & {\mathit{Q}}_7=1.433521476333333+2.396866724000000\mathbf{i}.\hfill \end{array}$$

By solving Equation (13), we obtain the unique solution:

$$a_0=x=1.200000000000002;$$

thus, we have

$$\begin{array}{cc}\hfill & {\mathit{Q}}_1=1.433521476333333-2.396866724000000\mathbf{i},\hfill \\ \hfill & {\mathit{Q}}_2=0.854730142200000-6.108707529599998\mathbf{i},\hfill \\ \hfill & {\mathit{Q}}_3=-0.522605614600000-7.322629161599998\mathbf{i},\hfill \\ \hfill & {\mathit{Q}}_4=1.387493191400000-7.412347797933333\mathbf{i},\hfill \\ \hfill & {\mathit{Q}}_5=0.239713394066667-8.423949157933333\mathbf{i},\hfill \\ \hfill & {\mathit{Q}}_6=5.924617470866667-13.621338260866665\mathbf{i}.\hfill \end{array}$$

We numerically verify that the leg lengths and angles of control polygon with auxiliary control points defined above satisfy the conditions in Theorem 1 within tolerance ${10}^{-13}$ [9]; thus, the corresponding curve is a Class I sextic PH curve. Moreover, we can further compute the complex coefficients of Equation (4):

$$\begin{array}{cc}\hfill & {\mathit{z}}_0=\pm (3.009399999999999-1.706700000000001\mathbf{i}),\hfill \\ \hfill & {\mathit{z}}_1=\pm (1.693000000000000-4.481399999999998\mathbf{i}),\hfill \\ \hfill & {\mathit{z}}_2=\pm (6.723300000000002+3.406399999999999\mathbf{i}),\hfill \end{array}$$

which can facilitate algebraic computations of the curves, e.g., arc lengths, curvatures (8), etc.

4.2. Class II Sextic PH Curve

In this section, we propose the geometric characteristics of Class II sextic PH curves in the same fashion as we did in the previous section for Class I sextic PH curves. The auxiliary control points ${\mathit{Q}}_i$, $i=1,\dots ,6$, for Class II sextic PH curves are constructed according to the following steps:

• Let $\alpha =\frac{arg\Delta {\mathit{P}}_0+arg\Delta {\mathit{P}}_5}{2}$; we obtain

  $$\begin{array}{cc}\hfill & {\mathit{Q}}_1={\mathit{P}}_1+k_0\Delta {\mathit{P}}_0,\hfill \\ \hfill & {\mathit{Q}}_6={\mathit{P}}_5-k_1\Delta {\mathit{P}}_5,\hfill \end{array}$$

  (23)

  where

  $$\begin{array}{cc}\hfill & k_0=\frac{\mathrm{Re}\Delta {\mathit{P}}_1·tan\alpha -\mathrm{Im}\Delta {\mathit{P}}_1}{\mathrm{Re}\Delta {\mathit{P}}_0·tan\alpha -\mathrm{Im}\Delta {\mathit{P}}_0},\hfill \\ \hfill & k_1=\frac{\mathrm{Re}\Delta {\mathit{P}}_4·tan\alpha -\mathrm{Im}\Delta {\mathit{P}}_4}{\mathrm{Re}\Delta {\mathit{P}}_5·tan\alpha -\mathrm{Im}\Delta {\mathit{P}}_5}.\hfill \end{array}$$

  (24)
• Compute

  $$\begin{array}{cc}\hfill & {\mathit{D}}_{\Delta {\mathit{Q}}_2}=\frac{5({\mathit{Q}}_1-{\mathit{P}}_1)}{2\Delta {\mathit{P}}_0}·({\mathit{P}}_2-{\mathit{Q}}_1),\hfill \\ \hfill & {\mathit{D}}_{\Delta {\mathit{Q}}_4}=\frac{5({\mathit{P}}_5-{\mathit{Q}}_6)}{2\Delta {\mathit{P}}_5}·({\mathit{Q}}_6-{\mathit{P}}_4),\hfill \\ \hfill & {\mathit{D}}_{{\mathit{P}}_2{\mathit{Q}}_2}=\frac{{\mathit{D}}_{\Delta {\mathit{Q}}_4}}{5({\mathit{P}}_2-{\mathit{Q}}_1)}·\Delta {\mathit{P}}_0,\hfill \\ \hfill & {\mathit{D}}_{{\mathit{P}}_3{\mathit{Q}}_4}=\frac{({\mathit{Q}}_6-{\mathit{P}}_4)}{10({\mathit{P}}_2-{\mathit{Q}}_1)}·\Delta {\mathit{P}}_0,\hfill \\ \hfill & {\mathit{D}}_{{\mathit{Q}}_3{\mathit{P}}_3}=\frac{({\mathit{P}}_2-{\mathit{Q}}_1)}{10({\mathit{Q}}_6-{\mathit{P}}_4)}·\Delta {\mathit{P}}_5,\hfill \\ \hfill & {\mathit{D}}_{{\mathit{Q}}_5{\mathit{P}}_4}=\frac{{\mathit{D}}_{\Delta {\mathit{Q}}_2}}{5({\mathit{Q}}_6-{\mathit{P}}_4)}·\Delta {\mathit{P}}_5.\hfill \end{array}$$

  (25)
• Let ${\mathit{Q}}_i$, $i=2,\dots ,5$, be points such that

  $$\begin{array}{cc}\hfill & {\mathit{Q}}_2={\mathit{P}}_2+{\mathit{D}}_{{\mathit{P}}_2{\mathit{Q}}_2},\hfill \\ \hfill & {\mathit{Q}}_3={\mathit{Q}}_2+{\mathit{D}}_{\Delta {\mathit{Q}}_2},\hfill \\ \hfill & {\mathit{Q}}_4={\mathit{P}}_3+{\mathit{D}}_{{\mathit{P}}_3{\mathit{Q}}_4},\hfill \\ \hfill & {\mathit{Q}}_5={\mathit{Q}}_4+{\mathit{D}}_{\Delta {\mathit{Q}}_4};\hfill \end{array}$$

  (26)

  thus, we have ${\mathit{P}}_0{\mathit{P}}_1⫽{\mathit{P}}_2{\mathit{Q}}_2⫽{\mathit{P}}_3{\mathit{Q}}_4$, ${\mathit{Q}}_1{\mathit{P}}_2⫽{\mathit{Q}}_2{\mathit{Q}}_3⫽{\mathit{Q}}_4{\mathit{Q}}_5⫽{\mathit{P}}_4{\mathit{Q}}_6$, ${\mathit{Q}}_3{\mathit{P}}_3⫽{\mathit{Q}}_5{\mathit{Q}}_4⫽{\mathit{P}}_5{\mathit{P}}_6$, as shown in Figure 3.

The following theorem describes the necessary and sufficient conditions for a sextic curves to be a Class II PH curve.

Theorem 2.

Let ${\mathit{P}}_i$, $i=0,\dots ,6$, be Bézier control points of a planar sextic curve and ${\mathit{Q}}_i$, $i=1,\dots ,6$, be auxiliary control points constructed following Equations (23)–(26), then it is a Class II sextic PH curve if and only if

$$\begin{array}{cc}\hfill & 5\parallel {\mathit{P}}_2-{\mathit{Q}}_1{\parallel }^2=8\parallel \Delta {\mathit{P}}_0\parallel ·\parallel {\mathit{P}}_3-{\mathit{Q}}_3\parallel ,\hfill \\ \hfill & \parallel \Delta {\mathit{Q}}_2{\parallel }^2=2\parallel {\mathit{Q}}_1-{\mathit{P}}_1\parallel ·\parallel {\mathit{P}}_4-{\mathit{Q}}_5\parallel ,\hfill \\ \hfill & \parallel \Delta {\mathit{Q}}_4{\parallel }^2=2\parallel {\mathit{Q}}_2-{\mathit{P}}_2\parallel ·\parallel {\mathit{P}}_5-{\mathit{Q}}_6\parallel ,\hfill \\ \hfill & 5\parallel {\mathit{Q}}_6-{\mathit{P}}_4{\parallel }^2=8\parallel {\mathit{Q}}_4-{\mathit{P}}_3\parallel ·\parallel \Delta {\mathit{P}}_5\parallel .\hfill \end{array}$$

(27)

Proof. 

For a Class II sextic PH curve, there are $a_i\in \mathbb{R}$, $i=0,\dots ,3$, ${\mathit{z}}_0$, and ${\mathit{z}}_1$ such that its hodograph has the form (3), where $u\left(t\right)+\mathbf{i}v\left(t\right)$, $w\left(t\right)$ are polynomials of the form (6). Since all control points are distinct, we assume that $a_3=1$; thus, Equation (3) can be rewritten as

$$\begin{array}{cc}\hfill {\mathit{P}}^{\prime }\left(t\right)=& a_0{\mathit{z}}_0^2{(1-t)}^5\hfill \\ \hfill & +(2a_0{\mathit{z}}_0{\mathit{z}}_1+3a_1{\mathit{z}}_0^2){(1-t)}^{4}t\hfill \\ \hfill & +(a_0{\mathit{z}}_1^2+6a_1{\mathit{z}}_0{\mathit{z}}_1+3a_2{\mathit{z}}_0^2){(1-t)}^3{t}^2\hfill \\ \hfill & +(3a_1{\mathit{z}}_1^2+6a_2{\mathit{z}}_0{\mathit{z}}_1+{\mathit{z}}_0^2){(1-t)}^2{t}^3\hfill \\ \hfill & +(3a_2{\mathit{z}}_1^2+2{\mathit{z}}_0{\mathit{z}}_1)(1-t)t^4\hfill \\ \hfill & +{\mathit{z}}_1^2{t}^5.\hfill \end{array}$$

By matching the coefficients of Bernstein polynomials in the above equation with Equation (2), we have

$$\begin{array}{cc}\hfill \frac{1}{6}{a}_0{\mathit{z}}_0^2=& \Delta {\mathit{P}}_0,\hfill \\ \hfill \frac{1}{30}(2a_0{\mathit{z}}_0{\mathit{z}}_1+3a_1{\mathit{z}}_0^2)=& \Delta {\mathit{P}}_1,\hfill \\ \hfill \frac{1}{60}(a_0{\mathit{z}}_1^2+6a_1{\mathit{z}}_0{\mathit{z}}_1+3a_2{\mathit{z}}_0^2)=& \Delta {\mathit{P}}_2,\hfill \\ \hfill \frac{1}{60}(3a_1{\mathit{z}}_1^2+6a_2{\mathit{z}}_0{\mathit{z}}_1+{\mathit{z}}_0^2)=& \Delta {\mathit{P}}_3,\hfill \\ \hfill \frac{1}{30}(3a_2{\mathit{z}}_1^2+2{\mathit{z}}_0{\mathit{z}}_1)=& \Delta {\mathit{P}}_4,\hfill \\ \hfill \frac{1}{6}{\mathit{z}}_1^2=& \Delta {\mathit{P}}_5.\hfill \end{array}$$

(28)

Let points ${\tilde{\mathit{Q}}}_i$, $i=1,\dots ,6$, be defined by

$$\begin{array}{cc}\hfill & {\tilde{\mathit{Q}}}_1={\mathit{P}}_1+\frac{a_1}{10}{\mathit{z}}_0^2={\mathit{P}}_2-\frac{a_0}{15}{\mathit{z}}_0{\mathit{z}}_1,\hfill \\ \hfill & {\tilde{\mathit{Q}}}_2={\mathit{P}}_2+\frac{a_2}{20}{\mathit{z}}_0^2,\hfill \\ \hfill & {\tilde{\mathit{Q}}}_3={\tilde{\mathit{Q}}}_2+\frac{a_1}{10}{\mathit{z}}_0{\mathit{z}}_1={\mathit{P}}_3-\frac{a_0}{60}{\mathit{z}}_1^2,\hfill \\ \hfill & {\tilde{\mathit{Q}}}_4={\mathit{P}}_3+\frac{1}{60}{\mathit{z}}_0^2,\hfill \\ \hfill & {\tilde{\mathit{Q}}}_5={\tilde{\mathit{Q}}}_4+\frac{a_2}{10}{\mathit{z}}_0{\mathit{z}}_1={\mathit{P}}_4-\frac{a_1}{20}{\mathit{z}}_1^2,\hfill \\ \hfill & {\tilde{\mathit{Q}}}_6={\mathit{P}}_4+\frac{1}{15}{\mathit{z}}_0{\mathit{z}}_1={\mathit{P}}_5-\frac{a_2}{10}{\mathit{z}}_1^2.\hfill \end{array}$$

(29)

Thus, points ${\tilde{\mathit{Q}}}_1$ and ${\tilde{\mathit{Q}}}_6$ are located on lines ${\mathit{P}}_0{\mathit{P}}_1$ and ${\mathit{P}}_5{\mathit{P}}_6$, respectively, and satisfy

$$\begin{array}{cc}\hfill & arg\frac{{\mathit{P}}_2-{\tilde{\mathit{Q}}}_1}{\Delta {\mathit{P}}_0}=arg\frac{\Delta {\mathit{P}}_5}{{\tilde{\mathit{Q}}}_6-{\mathit{P}}_4}=arg\frac{{\mathit{z}}_1}{{\mathit{z}}_0}=\frac{1}{2}arg\frac{\Delta {\mathit{P}}_5}{\Delta {\mathit{P}}_0},\hfill \\ \hfill & \mathrm{Re}({\mathit{P}}_2-{\mathit{Q}}_1)\mathrm{Im}({\mathit{P}}_4-{\mathit{Q}}_6)=\mathrm{Im}({\mathit{P}}_2-{\mathit{Q}}_1)\mathrm{Re}({\mathit{P}}_4-{\mathit{Q}}_6),\hfill \end{array}$$

that is ${\mathit{P}}_2{\mathit{Q}}_1⫽{\mathit{P}}_4{\mathit{Q}}_6$.

Note that points ${\tilde{\mathit{Q}}}_i$ and ${\mathit{Q}}_i$ are coincident with each other for $i=1,6$ as the lines intersect at a unique point in a Euclidean plane. From Equation (29), we have ${\mathit{P}}_0{\mathit{P}}_1⫽{\mathit{P}}_2{\tilde{\mathit{Q}}}_2⫽{\mathit{P}}_3{\tilde{\mathit{Q}}}_4$, ${\tilde{\mathit{Q}}}_1{\mathit{P}}_2⫽{\tilde{\mathit{Q}}}_2{\tilde{\mathit{Q}}}_3⫽{\tilde{\mathit{Q}}}_4{\tilde{\mathit{Q}}}_5⫽{\mathit{P}}_4{\tilde{\mathit{Q}}}_6$, ${\tilde{\mathit{Q}}}_3{\mathit{P}}_3⫽{\tilde{\mathit{Q}}}_5{\mathit{P}}_4⫽{\mathit{P}}_5{\mathit{P}}_6$, and

$$\begin{array}{c}3\Delta {\mathit{P}}_0:5({\tilde{\mathit{Q}}}_1-{\mathit{P}}_1):10({\tilde{\mathit{Q}}}_2-{\mathit{P}}_2):30({\tilde{\mathit{Q}}}_4-{\mathit{P}}_3)\hfill \\ =3({\mathit{P}}_2-{\tilde{\mathit{Q}}}_1):2\Delta {\tilde{\mathit{Q}}}_2:2\Delta {\tilde{\mathit{Q}}}_4:3({\tilde{\mathit{Q}}}_6-{\mathit{P}}_4)\hfill \\ =30({\mathit{P}}_3-{\tilde{\mathit{Q}}}_3):10({\mathit{P}}_4-{\tilde{\mathit{Q}}}_5):5({\mathit{P}}_5-{\tilde{\mathit{Q}}}_6):3\Delta {\mathit{P}}_5\hfill \\ =a_0:a_1:a_2:1.\hfill \end{array}$$

We can verify that ${\tilde{\mathit{Q}}}_i$ coincides with ${\mathit{Q}}_i$ defined in Equation (26) for $i=2,\dots ,5$. In other words, points ${\tilde{\mathit{Q}}}_i$, $i=1,\dots ,6$, are exactly the points constructed by Steps 1–3 in this section. Moreover, we have

$$\begin{array}{cc}\hfill & 5\parallel {\mathit{P}}_2-{\mathit{Q}}_1{\parallel }^2=8\parallel \Delta {\mathit{P}}_0\parallel ·\parallel {\mathit{P}}_3-{\mathit{Q}}_3\parallel =\frac{1}{45}{a}_0^2\parallel {\mathit{z}}_0{\parallel }^2{\parallel {\mathit{z}}_1\parallel }^2,\hfill \\ \hfill & \parallel \Delta {\mathit{Q}}_2{\parallel }^2=2\parallel {\mathit{Q}}_1-{\mathit{P}}_1\parallel ·\parallel {\mathit{P}}_4-{\mathit{Q}}_5\parallel =\frac{1}{100}{a}_1^2\parallel {\mathit{z}}_0{\parallel }^2{\parallel {\mathit{z}}_1\parallel }^2,\hfill \\ \hfill & \parallel \Delta {\mathit{Q}}_4{\parallel }^2=2\parallel {\mathit{Q}}_2-{\mathit{P}}_2\parallel ·\parallel {\mathit{P}}_5-{\mathit{Q}}_6\parallel =\frac{1}{100}{a}_2^2\parallel {\mathit{z}}_0{\parallel }^2{\parallel {\mathit{z}}_1\parallel }^2,\hfill \\ \hfill & 5\parallel {\mathit{Q}}_6-{\mathit{P}}_4{\parallel }^2=8\parallel {\mathit{Q}}_4-{\mathit{P}}_3\parallel ·\parallel \Delta {\mathit{P}}_5\parallel =\frac{1}{45}\parallel {\mathit{z}}_0{\parallel }^2{\parallel {\mathit{z}}_1\parallel }^2.\hfill \end{array}$$

Thus, we have proven the necessity.

On the contrary, if the conditions in Theorem 2 hold for a given planar sextic curve, we can obtain a group of solutions for $a_i$, $i=0,1,2,3$, ${\mathit{z}}_0$ and ${\mathit{z}}_1$ from Equation (29), e.g.,

$$\begin{array}{cc}\hfill & a_0=\frac{10({\mathit{P}}_3-{\mathit{Q}}_3)}{\Delta {\mathit{P}}_5},\hfill \\ \hfill & a_1=\frac{10({\mathit{P}}_4-{\mathit{Q}}_5)}{3\Delta {\mathit{P}}_5},\hfill \\ \hfill & a_2=\frac{5({\mathit{P}}_5-{\mathit{Q}}_6)}{3\Delta {\mathit{P}}_5},\hfill \\ \hfill & a_3=1,\hfill \\ \hfill & {\mathit{z}}_0=\pm \frac{15({\mathit{Q}}_6-{\mathit{P}}_4)}{\sqrt{6\Delta {\mathit{P}}_5}},\hfill \\ \hfill & {\mathit{z}}_1=\pm \sqrt{6\Delta {\mathit{P}}_5};\hfill \end{array}$$

therefore, the curve is a Class II sextic PH curve. □

An Example of a Class II Sextic PH Curve with a Concave Control Polygon

In this section, we show an example of a Class II sextic PH curve with a concave control polygon given by (see Figure 4)

$$\begin{array}{cc}\hfill & {\mathit{P}}_0=0+0\mathbf{i},\hfill \\ \hfill & {\mathit{P}}_1=-2.000009978666667-5.999907840000000\mathbf{i},\hfill \\ \hfill & {\mathit{P}}_2=-0.915235119476400-8.304442981248000\mathbf{i},\hfill \\ \hfill & {\mathit{P}}_3=-2.922077810388467-19.175025906393600\mathbf{i},\hfill \\ \hfill & {\mathit{P}}_4=14.418872723033649-18.001271454000001\mathbf{i},\hfill \\ \hfill & {\mathit{P}}_5=10.749764840044783-5.749685950089601\mathbf{i},\hfill \\ \hfill & {\mathit{P}}_6=8.749765161711450-1.749702910089601\mathbf{i}.\hfill \end{array}$$

Following Equation (24), we have

$$\begin{array}{cc}\hfill & k_0=0.406575000000000,\hfill \\ \hfill & k_1=3.020760000000000.\hfill \end{array}$$

Then, from Equations (23) and (26), we have

$$\begin{array}{cc}\hfill & {\mathit{Q}}_1=-2.813164035743067-8.439320370048000\mathbf{i},\hfill \\ \hfill & {\mathit{Q}}_2=-4.691203958949600-19.632118985472001\mathbf{i},\hfill \\ \hfill & {\mathit{Q}}_3=-2.762077836121800-19.495024549593602\mathbf{i},\hfill \\ \hfill & {\mathit{Q}}_4=-3.172079057721800-19.925014386393599\mathbf{i},\hfill \\ \hfill & {\mathit{Q}}_5=14.744132670720999-18.651788695795201\mathbf{i},\hfill \\ \hfill & {\mathit{Q}}_6=16.791283868366982-17.832674718000000\mathbf{i}.\hfill \end{array}$$

We can numerically verify that Theorem 2 holds; thus, the curve shown in Figure 4 is a Class II sextic PH curve. Moreover, we obtain the coefficients of Equation (6) as follows:

$$\begin{array}{cc}\hfill & a_0=0.800000000000004,\hfill \\ \hfill & a_1=0.542100000000000,\hfill \\ \hfill & a_2=5.034599999999999,\hfill \\ \hfill & {\mathit{z}}_0=\pm (4.027000000000003+5.587199999999996\mathbf{i}),\hfill \\ \hfill & {\mathit{z}}_1=\pm (2.723300000000000-4.406400000000000\mathbf{i}).\hfill \end{array}$$

5. Conclusions

In this paper, we developed a simple and intuitive geometric method for identifying sextic PH curves according to the geometric characteristics of sextic PH curves. The geometric characteristics were described by the leg lengths and angles of the Bézier control polygon with auxiliary control points. Different from previous work, which was limited to convex control polygons, our method applies to general sextic polynomial curves. While we focused on planar polynomial curves here, we believe that this work could be extended to Minkowski PH (MPH) curves [14,15], rational PH curves, and indirect PH curves [19]. These will be considered as research directions in our future work.