Original Research ARTICLE

Front. Phys., 09 June 2020 | https://doi.org/10.3389/fphy.2020.00175

Necessary and Sufficient Conditions for Expressing Quadratic Rational Bézier Curves

Chaoyu Yang1*, Jie Yang2, Ying Liu1 and Xianya Geng3
  • 1School of Economics and Management, Anhui University of Science and Technology, Huainan, China
  • 2School of Computing and Information Technology, Faculty of Engineering and Information Sciences, University of Wollongong, Wollongong, NSW, Australia
  • 3School of Mathematics and Physics, Anhui University of Science and Technology, Huainan, China

Quadratic rational Bézier curve transformation is widely used in the field of computational geometry. In this paper, we offer several important characteristics of the quadratic rational Bézier curve. More precisely, on the basis of proving its monotonicity, the necessary and sufficient conditions for transforming a quadratic rational Bézier curve into a point, line segment, parabola, elliptic arc, circular arc, and hyperbola are proved, respectively.

1. Introduction

Bézier curves have wide application in computer-aided geometric design, being used to provide precisely described points along a given curve [1]. Compared to other methods, such as the French curve, Bézier-based approaches are more computationally affordable and reliable. Additionally, the advantages of the Bézier curve in geometric design include its simple but clear mathematical function [2]. For instance, it is capable of incorporating both conic sections and parametric cubic curves as special cases [3]. As such, one can deal with two different curves simultaneously using one unique computational procedure. Some preliminary studies and applications of Bézier curves can be found in Lu et al. [4], Lee [5], and Han [6].

In this paper, to better understand the basic characteristics of Bézier curves, we conduct some fundamental research. In particular, we discuss the necessary and sufficient conditions for representing six different basic shapes, including a point, line segment, parabola, elliptic arc, circular arc, and hyperbola, using Bézier curves [7, 8]. These results play a fundamental role in the shape formulation and can help in facilitating any subsequent computer-based geometric design.

To begin with, we introduce the mathematical model of the quadratic rational Bézier curve [1].

Definition 1. The quadratic rational Bézier curve is defined as follows:

p(t)=(1t)2ω0P0+2t(1t)ω1P1+t2ω2P2(1t)2ω0+2t(1t)ω1+t2ω2,t[0,1],    (1)

where

t=ω0μω0μ+ω2(1-μ),μ[0,1],    (2)

and ω0 and ω2 are not zero values at the same time.

The monotonicity of Formula (1.2) is discussed below. Let μ1 ∈ [0, 1], μ2 ∈ [0, 1], and μ1 ≤ μ2. Accordingly, in the case of μ1 = 0, we have:

t1=ω0μ1ω0μ1+ω2(1-μ1)=0.    (3)

Note that 1 ≥ μ2 > μ1 ≥ 0, and t2=ω0μ2ω0μ2+ω2(1-μ2)0; then it is easy to have t2t1 = 0.

In the case of μ1 ≠ 0 and μ2 ≠ 0, according to Formula (1.2), we have:

t1t2=ω0μ1ω0μ1+ω2(1μ1) / ω0μ2ω0μ2+ω2(1μ2)  =(ω0+ω2(1μ21)) / (ω0+ω2(1μ11))1.    (4)

In other words, we have the conclusion that t is monotonically increasing [911]. Furthermore, if we apply linear transformation to Formula (1.1), it is easy to know

p(μ)=ω0ω2(1μ)2P0+2ω0ω2ω1μ(1μ)P1+ω0ω2μ2P2ω0ω2(1μ)2+2ω0ω2ω1μ(1μ)+ω0ω2μ2         =  ((1μ)2 Ρ0 +2ω12ω0ω2μ(1μ)Ρ1+μ2Ρ2)/((1μ)2          +2ω12ω0ω2μ(1μ)+μ2).    (5)

Let ω=ω12ω0ω2 and substitute μ with t in the standard form of the quadratic rational Bézier curve. To this end, we have the simplified version of the quadratic rational Bézier curve, which is expressed as follows:

p(t)=(1-t)2P0+2ωt(1-t)P1+t2P2(1-t)2+2ωt(1-t)+t2.    (6)

2. Sufficient and Necessary Conditions for a Quadratic Rational Bézier Curve to Degenerate into a Point

Theorem 1. A quadratic rational Bézier curve degenerates into a point if and only if three control points P0, P1, P2 coincide.

Proof. Assume that the quadratic rational Bézier curve degenerates to a point PA. That is,

p(t)=(1-t)2P0+2ωt(1-t)P1+t2P2(1-t)2+2ωt(1-t)+t2=PA(1-t)2(P0-PA)+2t(1-t)ω(P1-PA)+t2(P2-PA)=0.    (7)

As can be seen from Formula (7), when t ∈ (0, 1), we have (1−t)2 ≠ 0, t2 ≠ 0, 2t(1−t) ≠ 0, so P0 = PA, P1 = PA, P2 = PA. That is, when the quadratic rational Bézier curve degenerates into a point, P0, P1, P2 are the same point of PA.

On the other hand, when three control points coincide (say, the same point PA), we know that:

p(t)=(1-t)2PA+2t(1-t)ωPA+t2PA(1-t)2+2t(1-t)ω+t2=PA.    (8)

As can be seen from Formula (8), when three control points P0, P1, P2 coincide, the quadratic rational Bézier curve degenerates into a point [12, 13].

ALGORITHM 1
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Algorithm 1: To degenerate a Quadratic Rational Bézier Curve into a Point

3. Necessary and Sufficient Conditions for Degradation of a Quadratic Rational Bézier Curve into a Linear Section

Theorem 2. The quadratic rational Bézier curve degenerates into a straight line segment if and only if the control points P0, P2 do not coincide, the weight factor ω = 0, or the control point P1 is on the line segment [1416].

Proof. First, we assume that one point is with two coordinates; alternatively, we have P0 = (x0, y0), P1 = (x1, y1), and P2 = (x2, y2). As such, for an arbitrary point p(t) = (x, y), according to Formula (6) it is easy to have:

x=(1-t)2x0+2t(1-t)ωx1+t2x2(1-t)2+2t(1-t)ω+t2,y=(1-t)2y0+2t(1-t)ωy1+t2y2(1-t)2+2t(1-t)ω+t2,    (9)

On the other hand, a general form for a line function can be expressed as: y = ax + b, where a, and b is a constant [17]. Then, substituting x and y using Formula (12), we can get:

(1-t)2y0+2t(1-t)ωy1+t2y2(1-t)2+2t(1-t)ω+t2=a (1-t)2x0+2t(1-t)ωx1+t2x2(1-t)2+2t(1-t)ω+t2+b,    (10)

If we simplify the above formula, it is easy to know:

(y0-ax0-b+y2-ax2-b-2y1ω+2ax1ω+2bω)t2    (11)
-2(y0-ax0-b-y1ω+ax1ω+bω)t+y0-ax0-b=0.

First, we assume that one point is with two coordinates; alternatively, we have P0 = (x0, y0), P1 = (x1, y1), and P2 = (x2, y2). As such, for an arbitrary point p(t) = (x, y), according to Formula (6) it is easy to have:

x=(1-t)2x0+2t(1-t)ωx1+t2x2(1-t)2+2t(1-t)ω+t2,y=(1-t)2y0+2t(1-t)ωy1+t2y2(1-t)2+2t(1-t)ω+t2,    (12)

Now, control points P0 and P2 are the first and last points of the Bézier curve. As they are all on the Bézier curve, they will also be on the straight line [1820]. Alternatively, we have:

y0=ax0+b,y2=ax2+b.    (13)

Therefore, Formula (11) is further simplified:

(y1-ax1-b)(ωt-ωt2)=0.    (14)

Next, Formula (14) is analyzed in the following aspects:

1. If the control point P1 is also on the Bézier curve (or on the straight line), then y1ax1b = 0, and Formula (14) clearly holds.

2. If the control point P1 is not on the Bézier curve (or not on the straight line), then y1ax1b ≠ 0, and Formula (14) can be simplified as

-ωt2+ωt=0.    (15)

Therefore, when t ∈ [0, 1], in order to make Formula (15) hold, we have ω = 0.

As such, it is proved that when the quadratic rational Bézier curve degenerates into a straight line segment, two conditions are met: (1) the weight factor ω = 0, or (2) the control point P1 is on the line segment with the control point P0, P2 as the end point. In the following, we discuss these two conditions separately.

1. According to Formula (6), when the weight factor ω = 0, we have:

p(t)=(1-t)2P0+2ωt(1-t)P1+t2P2(1-t)2+2ωt(1-t)+t2=(1-t)2P0+t2P2(1-t)2+t2,    (16)

and

x=(1-t)2x0+t2x2(1-t)2+t2=(1-t)2x0(1-t)2+t2+t2x2(1-t)2+t2.    (17)
y=(1-t)2y0+t2y2(1-t)2+t2=(1-t)2y0(1-t)2+t2+t2y2(1-t)2+t2.    (18)

To simplify the calculation process, let us assume that:

α=(1-t)2(1-t)2+t2.    (19)

and

1-α=t2(1-t)2+t2.    (20)

Now the following formula holds:

x=αx0+(1-α)x2x-x2=α(x0-x2).    (21)
y=αy0+(1-α)y2y-y2=α(y0-y2).    (22)

As the control points P0, P2 do not coincide, x0x2, y0y2,

α=y-y2y0-y2=x-x2x0-x2    (23)
yy0-y2-xx0-x2=y2y0-y2-x2x0-x2,    (24)

where x0, x2, y0, y2 are constants. We assume that 1y0-y2=A, 1x0-x2=B, y2y0-y2-x2x0-x2=C (that is, A,B,C are all constants). Accordingly, we know that AyBx = C is a line segment [21].

2. Let the conditional control point P1 be the end point (on the line segment with the control points P0 and P2) [22]; thus, it can be seen that:

P1=(1-v)P0+vP2,v[0,1],    (25)

The Formula (25) can be substituted with Formula (6) to have:

p(t)=(1t)2+2t(1t)ω(1v)(1t)2+2t(1t)ω+t2P0          +(2t(1t)ωv+t2(1t)2+2t(1t)ω+t2P2.    (26)

Then we set:

u=(2t(1t)ωv+t2(1t)2+2t(1t)ω+t2.    (27)

Comparing Formula (26) with Formula (27), it is easy to find that

p(t)=(1-u)P0+uP2.    (28)

In conclusion, Formula (28) is the parametric formula of the line segment. When the control point P1 is on the line segment with the control point (P0, P2) as the end point, Formula (26) can be written as the parametric formula of the line segment of Formula (28). As such, it is proved that it degenerates into a line segment [23, 24].

ALGORITHM 2
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Algorithm 2: To Degenerate a Quadratic Rational Bézier Curve into a Linear section

ALGORITHM 3
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Algorithm 3: To degenerate a Quadratic Rational Bézier Curve into a Linear section

4. Necessary and Sufficient Conditions for a Quadratic Rational Bézier Curve to Represent a Section of Arc

Theorem 3. Quadratic rational Bézier curves can be used to represent an arc if and only if |P0P1| = |P2P1| and 0 ≤ ω ≤ 1 [25].

Proof. The equation of a circle passing through three collinear points Qi(xi, yi), (i = 1, 2, 3), on a rectangular coordinate plane is:

|x2+y2xy1x02+y02x0y01x12+y12x1y11x22+y22x2y21|=0.    (29)

Given three points that are not collinear, we have:

P0(x0,y0,0)=(-a,0,0),A(xA,yA,0),P2(x2,y2,0)=(a,0,0).    (30)

The arc curve starts from point P0 and passes through point A to point P2. Now, let us find another control vertex P1. To do so, P0, A, P2 are substituted into the three-point common-circle equation 29, and we get

|x2+y2xy1x02+y02x0y01xA2+yA2xAyA1x22+y22x2y21|=0.    (31)

From Formula (30) to Formula (31), we can find that x0 = −a, y0 = 0, x2 = a, y2 = 0. Furthermore, by expanding the determinant 31 in the first row, we have

(x2+y2)|a010yA1a01|+(1)x|a201yA2yA1a201|+y|a2a1yA201a2a1|                                                                                        +(1)|a2a0yA20yAa2a0|=0.    (32)

Among them,

|a010yA1a01|=2ayA,|a201yA2yA1a201|=0,      |a2a1yA201a2a1|=a3+ayA2+ayA2a3,|a2a0yA20yAa2a0|=a3yAa3yA.    (33)

Finally, the above formula can be simplified as follows:

(-2ayA)(x2+y2)+y(-a3+2ayA2-a3)+2a3yA=0.    (34)

Because yA ≠ 0, it is easy to know

x2+(y+a2-yA22yA)2=a2+(a2-yA2)24yA2.    (35)

On the other hand, as xA = 0, we can add xA to have

x2+(y+a2-(xA2+yA2)2yA)2=a2+(a2-(xA2+yA2))24yA2.    (36)

Summarizing the above formula, the coordinates of the center of the circle O are:

xO=0,yO=a2-(xA2+yA2)2yA.    (37)

The radius of the circle is:

r=a2+(a2-(xA2+yA2))24yA2.    (38)

The vertical lines of OP0 and OP2 are made from points P0 and P2, respectively. According to the symmetry, if two vertical lines intersect with the Y axis at point P1, then point P1 is the control vertex of the arc curve. That is,

y1=2a2yAa2-(xA2+yA2).    (39)

Accordingly, the coordinates of point P1 are:

x1=0,y1=2a2yAa2-(xA2+yA2).    (40)

From the definition of the Bézier Curve in Formula (1), we have:

x(t)=(1-t)2ω0x0+2t(1-t)ω1x1+t2ω2x2(1-t)2ω0+2t(1-t)ω1+t2ω2,y(t)=(1-t)2ω0y0+2t(1-t)ω1y1+t2ω2y2(1-t)2ω0+2t(1-t)ω1+t2ω2.    (41)

To simply Formula (41), we further introduce the Quadratic Bernstein Basis Function (Bi,2(t)), which can be expressed as follows:

B0,2(t)=(1-t)2,B1,2(t)=2t(1-t),B2,2(t)=t2.    (42)

As such, Formula (41) can be rewritten by applying Bi,2(t) in the following format:

x(t)=-aω0B0,2(t)+aω2B2,2(t)ω0B0,2(t)+ω1B1,2(t)+ω2B2,2(t),y(t)=2t(1-t)ω1y1ω0B0,2(t)+ω1B1,2(t)+ω2B2,2(t).    (43)

On the other hand, note that the standard equation of curve arc circle can be estimated as

x2(t)+(y(t)+acotθ)2=a2/sin2θ.    (44)

Consequently, by substituting Formulas (43) into Equation (44), the following results are obtained:

(aω0B0,2(t)+aω2B2,2(t)ω0B0,2(t)+ω1B1,2(t)+ω2B2,2(t))2        +(ω1y1B1,2(t)ω0B0,2(t)+ω1B1,2(t)+ω2B2,2(t)+acotθ)2=a2sin2θ,    (45)

Note that

a2sin2θ-a2cot2θ=a2.    (46)

As such, Formula (45) can be further simplified as

a2ω02B0,22(t)+a2ω22B2,22(t)2a2ω0ω2B0,2(t)B2,2(t)+ω12y12B1,22(t)                 +(2ω1y1B1,2(t)acotθ(ω0B0,2(t)+ω1B1,2(t)+ω2B2,2(t))                                                 = a2(ω0B0,2(t)+ω1B1,2(t)+ω2B2,2(t))2    (47)

Furthermore, according to Formula (38) and Formula (40), we can have

y1cotθ=2a2yAa2-(xA2+yA2)×(a2-(xA2+yA2))2ayA=a,    (48)

and then,

(y12+a2)ω12B1,22(t)-4a2ω0ω2B0,2(t)B2,2(t)=0.    (49)

Again, we consider the Quadratic Bernstein Basis Function, and then the above formula (in Formula 49) can be simplified as follows:

((y12+a2)ω12-a2ω0ω2)(1-t)2t2=0.    (50)

Next, according to Formula (40), we know

(ω12sec2θ-ω0ω2)(1-t)2t2=0,    (51)

and t ∈ (0, 1), t2(1 − t)2 ≠ 0. It is thus easy to know

ω12=ω0ω2cos2θ.    (52)

According to the standard form of the quadratic rational Bézier curve (see Formula 6), we can further estimate ω0 = ω2 = 1, ω1 = cosθ, and the value range of θ of the center angle of the semicircle should be 0 ≤ θ ≤ π/2 [26].

In summary, the rational quadratic Bézier expressions of arc curves passing through points P0, A, P2 are as follows,

C(t)=(1-t)2P0+2cos(θ)t(1-t)P1+t2P2(1-t)2+2cos(θ)t(1-t)+t2.    (53)

Compared with the standard formula of a rational quadratic Bézier, the following results are obtained,

ω=cos(θ),    (54)

where 0 ≤ θ ≤ π/2, 0 ≤ ω ≤ 1. Consequently, the necessary and sufficient conditions for a rational quadratic Bézier curve to represent a circular arc are expressed as follows:

|P0P1|=|P2P1|  and   0ω1.    (55)
ALGORITHM 4
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Algorithm 4: For a Quadratic Rational Bézier Curve to Represent a section of an Arc

5. Necessary and Sufficient Conditions for Quadratic Rational Bézier Curves to Represent a Parabola, Elliptic Arc and Hyperbola

Theorem 4. Quadratic rational Bézier curve represents a parabola, elliptic arc, and hyperbola if and only if ω = ±1, −1 < ω < 1, and ω < −1 or ω> 1, respectively [27].

Proof. According to the second order Bernstein basis function of Formula (42), Bézier curve from Formula (1) is written as follows,

p(t)=ω0B0,2(t)P0j=02Bj,2(t)ωj+ω1B1,2(t)P1j=02Bj,2(t)ωj+ω2B2,2(t)P2j=02Bj,2(t)ωj=i=02Ri,2(t)Pi,    (56)

where

Ri,2(t)=ωiBi,2(t)j=02Bj,2(t)ωj.    (57)

Next, we introduce the Local Oblique Coordinate System P1, S, T, so that S = P0P1, T = P2P1. Since point P(t) is within δP0P1P2 for arbitrary t ∈ [0, 1], P(t) can be rewritten as

P(t)=P1+u(t)S+v(t)T     =P1+u(t)(P0-P1)+v(t)(P2-P1)         =u(t)P0+[1-u(t)-v(t)]P1+v(t)P2.    (58)

Comparing the coefficients from both Formula (56) and Formula (58), we know that

R0,2(t)=u(t),R1,2(t)=1-u(t)-v(t),R2,2(t)=v(t).    (59)

Let k=ω0ω2/ω12, where k is the shape-invariant factor of a conic, so

u(t)v(t)=R0,2(t)2,2(t)=14k[1-u(t)-v(t)]2.    (60)

Formula (60) is an implicit equation of a quadratic curve in the local oblique coordinate system P1, S, T. The expansion of Formula (60) further indicates that:

ku2(t)+(2k-4)u(t)v(t)+kv2(t)-2ku(t)-2kv(t)+k=0.    (61)

In the Cartesian coordinate system, the image of a binary quadratic equation can represent a conic curve, and all conic curves can be derived in the aforementioned way [1]. The equation has the following forms [28]:

Ax2+Bxy+Cy2+Dx+Ey+F=0,A,B,C are not all zero,    (62)

where A, B, C, D, E, F are polynomial coefficients. If the following conditions are satisfied,

B2-4AC<0,    (63)

then Formula (62) represents an ellipse; furthermore, under the same condition, if the conic degenerates (that is, A = C, B = 0), the equation represents a circle. Additionally, if the following conditions are satisfied,

B2-4AC=0,    (64)

then Formula (62) represents a parabola [29]. Finally, if the following conditions are satisfied,

B2-4AC>0    (65)

then Formula (62) represents an hyperbola. The coefficients from Formula (61) and Formula (62) can be obtained as follows: A = k, B = k − 2, C = k, D = −2k, E = −2k, F = k. As such, we can get:

B2-4AC=1-k.    (66)
ALGORITHM 5
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Algorithm 5: For Quadratic Rational Bézier Curves to Represent a Parabola, Elliptic Arc and Hyperbola

We then provide the discussion and judgment of Formula (66). That is, from the condition of Formula (63), if the curve is an ellipse, then in Formula (66) we have B2 − 4AC = 1 − k < 0. Therefore, when k > 1, the curve is an ellipse. From the condition of (64), if the curve is a parabola, then B2 − 4AC = 1 − k = 0 (again see Formula 66). Therefore, when k = 1, the curve is a parabola. From the condition of 65, if the curve is a hyperbola, then B2 − 4AC = 1 − k > 0, so when k < 1, the curve is a hyperbola.

Note that k=ω0ω2/ω12. In summary, under the standard form of the quadratic rational Bézier curve, we have ω0 = ω2 = 1, and ω = ω1. Consequently, we prove that when −1 < ω < 1, the quadratic rational Bézier curve is a ellipse; when ω = ±1, the quadratic rational Bézier curve is a parabola; when ω < −1, or ω > 1, the quadratic rational Bézier curve is a hyperbola.

6. Conclusion

In this paper, we discuss the necessary and sufficient conditions for utilizing quadratic rational Bézier curves to represent different shapes, such as a point, line segment, parabola, elliptic arc, circular arc, and hyperbola. These results can be further used to facilitate other computer-aided geometric designs.

Data Availability Statement

The raw data supporting the conclusions of this article will be made available by the authors, without undue reservation.

Author Contributions

CY: conceptualization, methodology, software, validation, investigation, visualization, and writing original draft. JY: software, writing - review & editing, and supervision. YL: software, visualization, and writing - original draft. XG: writing - review & editing, validation, and visualization.

Funding

This work was supported by the National Natural Science Foundation of China (Grant No. 61873004, 51874003) and the Humanities and Social Sciences Foundation of Anhui Department of Education, China (Grant No. SK2017A0098). We would also like to thank JY from the University of Wollongong for his useful discussion.

Conflict of Interest

The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.

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Keywords: Bézier curve, quadratic rational, necessary and sufficient conditions, geometry, computer aided geometric design

Citation: Yang C, Yang J, Liu Y and Geng X (2020) Necessary and Sufficient Conditions for Expressing Quadratic Rational Bézier Curves. Front. Phys. 8:175. doi: 10.3389/fphy.2020.00175

Received: 25 March 2020; Accepted: 24 April 2020;
Published: 09 June 2020.

Edited by:

Jia-Bao Liu, Anhui Jianzhu University, China

Reviewed by:

Qin Zhao, Hubei University, China
Liangchen Li, Luoyang Normal University, China
Jianbing Liu, West Virginia University, United States

Copyright © 2020 Yang, Yang, Liu and Geng. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.

*Correspondence: Chaoyu Yang, yangchy@aust.edu.cn

These authors have contributed equally to this work