Generating planar spiral by geometry driven subdivision scheme

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Generating planar spiral by geometry driven subdivision scheme DENG ChongYang1,2† & WANG GuoZhao1 1

Department of Mathematics, Zhejiang University, Hangzhou 310027, China;

2

Institute of Applied Mathematics and Engineering Computations, Hangzhou Dianzi University, Hangzhou 310018, China

Spirals are curves with one-signed, monotone increasing or decreasing curvature. They are commonly useful in a variety of applications, either for aesthetic or for engineering requirements. In this paper we propose a new iterative subdivision scheme for generating planar spiral segments from two points and their tangent vectors. The subdivision process consists of two main steps, computing new points and adjusting tangent vectors adaptively for each iteration. We categorize this iterative scheme as geometry driven because we utilize the old points and their tangent vectors whereas most other iterative schemes rely only on the old points. Some numerical examples are presented to show the advantageous properties of the new scheme. computer aided geometric design, spiral, geometry driven subdivision, circular arc splines

1 Introduction Spirals are curves with one-signed, monotone increasing or decreasing curvature and they are considered as fair curves because they have no internal curvature extrema and no cusps[1] . Spiral design is an interesting mathematical problem with importance for both physical[2] and aesthetic applications[3] . Much research focuses on designing spirals using (rational) B´ezier curves and Pythagorean Hodograph (PH) curves. Mineur et al.[4] show that certain B´ezier curves are guaranteed to be spiral. These curves are defined by the requirement that each leg of their B´ezier polygon is obtained by a rotation and scale of the previous leg. Frey and

Field[5] generate B´ezier conic segments with monotone curvature through confining the control points to well-defined regions bounded by circular arcs. Dietz and Piper et al.[6,7] use numerical techniques to design (rational) cubic B´ezier curves with monotonic curvature. Generating PH curve with monotone curvature is discussed extensively[8−10] , and they are used to design transition curves[11−15] . In this paper we present a geometry driven subdivision scheme for designing planar spirals. Though the curve generated by our scheme is not convenient for incorporation in traditional non-uniform rational B-spline (NURBS) based computer-aided design (CAD) packages, our scheme inherits the advantages of subdivision scheme, such as simplicity, efficiency, multi-resolution, and so on.

Received January 6, 2008; accepted October 28, 2008 doi: 10.1007/s11432-009-0160-3 † Corresponding author (email: [email protected]) Supported partially by the National Natural Science Foundation of China (Grant Nos. 60673032, 60773179), and the National Basic Research Program of China (Grant No. 2004CB318000), and the Scientific Starting Foundation of Hangzhou Dianzi University

Citation: Deng C Y, Wang G Z. Generating planar spiral by geometry driven subdivision scheme. Sci China Ser F-Inf Sci, 2009, 52(10): 1821–1829, doi: 10.1007/s11432-009-0160-3

Curve interpolation by repeated subdivision is an efficient shape design method in the field of computer aided geometric design (CAGD)[16] . Traditional linear subdivision schemes, such as the four-point subdivision scheme[17] and its extensions[18,19] , are simple to implement and easy to analyze, but it is difficult to control the shape of the limit curves accurately with linear subdivision schemes[20] . Recently, several nonlinear[21] and geometry driven schemes[16,20] have emerged, these schemes introduce some new properties, such as easy generalization to surface[21] and easy preservation of shape[16,21] , and so on, but the limit curves are only G1 continuous with these schemes. In this paper, a new scheme joining two points and matching their unit tangent vectors is proposed. The limit curve is a spiral segment which interpolates the given G1 Hermite data. When applying our algorithm to an array of G1 Hermite data, we obtain a G1 continuous spiral spline. For two points of a planar spiral segment, by Kneser’s theorem[22] , the osculating circle (circle of curvature) of the large curvature point is inside that of the small curvature point. Guided by this simple fact, in each iteration we introduce new points to make the radius of osculating circle (approximated by the circle interpolating three adjacent points) of a point bounded by those of its adjacent points. Though it is easy to calculate the radii of the circle interpolate three adjacent points, it is complex to compare the radii of the curvature circles. However, the radii of circular arc spline interpolating the vertices and their tangent vectors are comparatively easy to compute and compare. So in each iteration, we select the added points as if there is a G1 continuous circular arc spline with monotone radii array interpolating the vertices. By adjusting the tangent vectors adaptively, the sequence of circular arc splines converges to a spiral.

2 The subdivision scheme 2.1

Notations and conventions

In this paper, we define a planar angle to be positive if it is counter-clockwise and negative if it is clockwise. For a vector V = (Vx , Vy )T , the rotated 1822

vector with angle α from V can be obtained as α − sin α R(α)V , where R(α) = ( cos ) is the rotation sin α cos α matrix. In the following text we indicate the Euclidean distance between pki and pki+1 by pki+1 −pki  or pki pki+1 . The angle from Tik to pki pki+1 and the k are denoted as αki and angle from pki pki+1 to Ti+1 βik , respectively. We begin with two points p00 , p01 and their tangent vectors T00 , T10 . Then at iteration k, the point array has 2k + 1 points. To make sure that the initial data can define a spiral, we only consider the initial data which satisfy 0 < β00 < α00 < π and α00 + β00 < π. 2.2

Adding new points

The sequence of points at each iteration defines a control polygon which is refined in the following manner. Points in each new iteration which have even indices are inherited from the previous itera= pki , while new tion and are renumbered as pk+1 2i points with odd indices are computed to be the k angle bisectors of ∠Tik pki pki+1 , ∠pki pki+1 T i+1 , where k

k T i+1 = −Ti+1 (see Figure 1). In fact, pk+1 2i+1 can be computed as k pk+1 2i+1 = pi+1

+

sin(βik /2) R(−αki /2)(pki+1 − pki ). (1) sin(αki /2 + βik /2) k+1

For the new sequence of points {pk+1 }2i=0 , we i choose a set of temporary tangents as follows. k+1 = For points with even indices, we define U2i Tik , and for points with odd indices, we define pk −pk k+1 i = pi+1 (see Figure 1). Note that the U2i+1 k −pk  i+1

i

first and last points of each iteration, U0k+1 , U2k+1 k+1 are inherited from the initial conditions. k+1 and β i be the angles from Uik+1 to Let αk+1 i k+1 k+1 k+1 pk+1 pi+1 to Ui+1 , repk+1 i i+1 and the angle from pi spectively, we have k+1

αk+1 = β 2i 2i αk+1 2i+1

=

= 0.5αki ;

k+1 β 2i+1

=

0.5βik .

(2) (3)

From eqs. (2) and (3) we know that there exists k+1 satisfying pk+1 a circular arc segment Λk+1 2i 2i , p2i+1 k+1 k+1 and matching U2i , U2i+1 , and there is another cirk+1 k+1 cular arc segment Λk+1 2i+1 satisfying p2i+1 , p2i+2 with k+1 k+1 end tangents U2i+1 , U2i+2 . With simple computa-

DENG C Y et al. Sci China Ser F-Inf Sci | Oct. 2009 | vol. 52 | no. 10 | 1821-1829

Figure 1

Adding new points.

k+1 tion we have the radii for Λk+1 and Λk+1 = 2i+1 as r2i 2i k+1 pk+1 2i p2i+1 

2 sin |αk+1 2i |

and

k+1 r2i+1

=

k+1 pk+1 2i+1 p2i+2 

2 sin |αk+1 2i+1 |

, respectively.

So in this paper we designate the values of these coefficients.

From eqs. (2) and (3) it is easy to see that all ank+1 , β i } are positive if all angles {αki , βik } gles {αk+1 i are positive, then Proposition 1. There is a G1 spline of circular arcs Ψk+1 interpolating the Hermite data k+1 , Uik+1 }i and all angles {αk+1 , β i } are posi{pk+1 i i tive if all angles {αki , βik } are positive. 2.3

Calculating tangent vectors

We compute new tangent vectors Tik s from the temporary tangent vectors Uik s. To interpolate the initial tangent vectors, we define T0k = T00 , T2kk = T10 . The tangent vector Tik (0 < i < 2k ) at pki is defined as Tik

=

R(−δik )Uik (0

k

< i < 2 ),

(4)

where δik = min{1.02mki−1 αki−1 , 0.25mki αki−1 , (5) mki αki , 1.02mki+1 αki }, ⎧ k k ⎪ ⎨ ri − ri−1 0 < i < 2k , pk pk  k rik = i i+1k mki = rik + ri−1 ⎪ 2 sin |αi | ⎩ 0.5 i = 0, 2k ,

Figure 2

Calculating tangent vector.

From eq. (4) and Figure 2 we can see that αki = αki + δik ,

(6)

βik

(7)

=

αki

−

k δi+1 .

By the definition of mki , we know |mki | < 1, then δik < min{0.25|αki−1 |, 0.25|α ki |}.

(8)

(see Figure 2).

By eqs. (6)–(8) we know that all angles {αki , βik }i k are positive if all angles {αki , β i }i are positive. Combining with Proposition 1 and the initial condition α00 , β00 > 0 we have the following proposition.

Remark. The coefficients 1.02, 0.25 in eq. (5) seem to be strange. In fact, to make the limit curve be a spiral, the variaion ranges of the coefficients are very narrow and their relations are complex.

Proposition 2. All angles {αki , βik }i , {αki , β i }i are positive and there is no inflexion in {pki , Tik }i . For point pki , by the law of sine, we can express k+1 k+1 k in terms of ri−1 : r2i in terms of rik and r2i

DENG C Y et al. Sci China Ser F-Inf Sci | Oct. 2009 | vol. 52 | no. 10 | 1821-1829

k

1823

k+1 r2i = k+1 = r2i−1

k+1 k rik sin αki sin[(αki − δi+1 pki pki+1  sin(βik /2) sin(αki /2) )/2] pk+1 2i p2i+1  , = = k+1 k k k k k k k sin[(αi + βi )/2] sin[(αi + δi )/2] sin[αi + (δi − δi+1 )/2] 2 sin α2i k k ri−1 sin αki−1 sin[(αki−1 + δi−1 )/2] . k k k k sin[(αi−1 − δi )/2) sin[αi−1 + (δi−1 − δik )/2]

k k k > rik > ri−1 > ri−2 , by eq. (5), δik has If ri+1 k k the same sign as αi−1 , αi . From eqs. (9) and (10) we know that the effects of δik > 0 are to decrease k+1 k+1 k+1 and to increase r2i−1 . Then the ratio of r2i r2i k+1 over r2i−1 is expected to be less than that of rik over k ri−1 (for detailed proof see Lemma 6 of Appendix). This leads to the curvature continuity in the end.

3 Smoothness analysis k

Let θk = max{αki , βik }, θ k = max{αki , β i }, first we i i prove the following: Theorem 1. For the subdivision scheme defined in section 2, limk→∞ θk = limk→∞ θk = 0. Proof. By eqs. (6) and (8), we have αki = αki +δik  1.25αki . Similarly, we have βik  1.25αki . So θk  1.25θ k . From eqs. (2) and (3), we have θk = 0.5θk−1 , which implies that θk  1.25θ k = 1.25(0.5θk−1 ) = 0.625θk−1 .

(11)

So θk  (0.625)k θ0 . Thus limk→∞ θk = 0 and limk→∞ θk = limk→∞ 0.5θk−1 = 0. Theorem 2. For the subdivision scheme defined in section 2, the polygon series converge to a continuous curve. Proof. Let Γk be the polygon after kth subdivision. We compute the distance dk between Γk+1 and Γk . From the initial condition π > α00 > β00 > 0 and 0 α0 + β00 < π it is easy to see that θ1 < π2 . For k  1, let pk+1 2i+1 be a new added point corresponding to edge pki pki+1 , by law of sine, we have pki pki+1  sin(βik /2)  pki pki+1 , sin[(αki + βik )/2] pki pki+1  sin(αki /2) k  pki pki+1 . pk+1 2i+1 pi+1  = sin[(αki + βik )/2] pki pk+1 2i+1  =

k k pk+1 So maxi {pk+1 i i+1 }  maxi {pi pi+1 }  · · ·  maxi {p1i p1i+1 } := L.

1824

(9) (10)

k k Now, the distance from pk+1 2i+1 to the edge pi pi+1 can be computed as k Lθk αki k k+1 αi < p  .  sin p  dki = pki pk+1 2i+1 i i+1 2 2 2 Let dk = maxi {dki }, by eq. (11), we have k−1 dk  Lθ2 k  (0.625)2 Lθ1 . This means that polygons {Γk } form a Cauchy sequence and this sequence of polygons converges uniformly. Since each polygon is a piecewise linear curve, the limit curve is continuous. By Proposition 1, there is a G1 continuous arc spline Ψk interpolating {pki , Uik }i (k  1). With simple computation we can see that the distance from Ψk to Γk is no larger than the distance from Γk+1 to Γk . Then, {Ψk } converges to the same limit curve as {Γk } when k → ∞.

Theorem 3. For the subdivision scheme defined in section 2, the limit curve is G1 continuous. Proof. We first prove that the tangent for every pki on the limit curve exists. k+1 Since the angle between Tik and T2ik+1 , i.e. U2i and T2ik+1 , is δik , by eqs. (8) and (11), k

δik  0.25 min{αki , β i−1 }  0.25θ k = 0.25θk−1  0.125(0.625)k−1 θ0 .

(12)

Then {T2k+l l i }l is a Cauchy sequence and the tangent vector at pki exists. On the other hand, the subdivision curve is also the limit curve of the sequence of arc splines {Ψk }. As the tangent along every arc spline Ψk is continuous and tangent along every arc segment is bounded by tangents at end points, the sequence of tangent curves converge to a continuous curve too. So the limit curve of {Γk } is also G1 continuous. Theorem 4. For the subdivision scheme defined in section 2, the limit curve is spiral. Proof. For a point pki , in the Appendix we 1 1 = liml→∞ rk+l (Lemmas prove that liml→∞ rk+l 2l i

2l i−1

7 and 8 of Appendix). Noting that the curvature

DENG C Y et al. Sci China Ser F-Inf Sci | Oct. 2009 | vol. 52 | no. 10 | 1821-1829

for every arc spline is piecewise constant we conclude that the limit curve of {Ψk } is G2 continuous. On the other hand, for any k, in the Appendix we k (i = 0, 1, . . . , 2k − 1) (Lemma 4 prove that rik > ri−1 of Appendix). This means that the limit curve of has monotone curvature. From these two conclusions we conclude that the limit curve is a spiral.

4 Examples We present 6 examples to show the advantageous

Figure 3 Generating spiral segments with different initial tangent vectors. (a) Example 1: α00 = 40◦ , β00 = 10◦ ; (b) example 2: α00 = 70◦ , β00 = 20◦ ; (c) example 3: α00 = 90◦ , β00 = 30◦ ; (d) example 4: α00 = 120◦ , β00 = 40◦ ; (e) example 5: α00 = 120◦ , β00 = 50◦ ; (f) fxample 6: α00 = 80◦ , β00 = 60◦ . Table 1

properties of the subdivision scheme introduced in section 2. In these 6 examples, we compute spiral segments with different initial tangent vectors, i.e. with different initial angles α00 , β00 . To illustrate the qualities of these subdivision curves more clearly, we compute and plot the discrete curvatures with the curves after specified numbers of times of subdivisions (see Figure 3). The curvature plots of the curves are also shown (see Figure 4). Finally, we present the maximum curvature difference Dk = maxi { rk1 − r1k } and the convergence i−1

Figure 4

i

The curvature plot of the spiral segments.

The maximum curvature difference and the ratios after each subdivision

k

1

2

3

4

5

6

7

8

9

Example 1

51.692

46.042

39.979

32.595

24.692

17.236

10.988

6.393

3.647

0.8907

0.8683

0.8153

0.7576

0.6980

0.6375

0.5818

0.5705

61.945

52.460

40.981

29.712

19.841

12.187

6.926

3.884

0.8759

0.8469

0.7812

0.7250

0.6678

0.6142

0.5683

0.5608

58.473

47.780

35.704

24.695

15.808

9.418

5.273

2.900

0.8561

0.8171

0.7473

0.6917

0.6401

0.5958

0.5599

0.5500

65.923

55.217

42.234

30.110

20.041

12.525

7.419

4.208

0.8736

0.8376

0.7649

0.7129

0.6656

0.6250

0.5923

0.5671

42.797

32.810

22.953

14.970

9.191

5.368

3.014

1.641

0.8255

0.7666

0.6996

0.6522

0.6139

0.5841

0.5615

0.5445

11.195

7.060

4.161

2.339

1.270

0.673

0.352

0.183

0.7039

0.6307

0.5894

0.5621

0.5430

0.5297

0.5235

0.5195

Example 2

Example 3

Example 4

Example 5

Example 6

70.719

68.301

75.4614

51.8417

15.903

DENG C Y et al. Sci China Ser F-Inf Sci | Oct. 2009 | vol. 52 | no. 10 | 1821-1829

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ratios DDk+1 after each subdivision step for the exk amples (see Table 1). From Table 1 we can see that after a few steps of subdivisions, the convergence ratios run to a constant which is about 0.5 and the maximum curvature differences converge quickly though the convergence rate is conversely estimated as in Appendix.

5 Conclusions In this paper we propose a geometric driven subdivision scheme for generating spiral segment from two points and their tangent vectors. The spiral 1 Farin G. Curves and Surfaces for Computer Aided Geometric Design: A practical Guide. 4 ed. San Diego: Academic Press, 1997 2 Gibreel G M, Easa S M, Hassan Y, et al. State of the art of highway geometric design consistency. J Transp Eng, 1999, 125(4): 305–313 3 Burchard H G, Ayers J A, Frey W H, et al. Approximation with aesthetic constraints. In: Designing Fair Curves and Surfaces, SIAM, Philadelphia, 1994 4 Mineur Y, Lichah T, Castelain J M, et al. A shape controlled fitting method for B´ezier curves. Comput Aided Geom D, 1998, 15(9): 879–891 5 Frey W H, Field D A. Designing B´ezier conics segments with monotone curvature. Comput Aided Geom D, 2000, 17(6): 457–483 6 Dietz D A, Piper B. Interpolation with cubic spirals. Comput Aided Geom D, 2004, 21(2): 165–180 7 Dietz D A, Piper B, Sebe E. Rational cubic spirals. Comput Aided Design, 2008, 40(1): 3–12 8 Farouki R T. Pythagorean hodograph quintic transition curves of monotone curvature. Comput Aided Design, 1997, 29(9): 601–606 9 Walton D J, Meek D S. A Pythagorean hodograph quintic spiral. Comput Aided Design, 1996, 28(12): 943–950 10 Walton D J, Meek D S. A generalization of the Pythagorean hodograph quintic spiral. J Comput Appl Math, 2004, 172(2): 271–287 11 Habib Z, Sakai M. G2 PH quintic spiral transition curves and their applications. Sci Math Japon, 2005, 61(2): 207–217 12 Habib Z, Sakai M. On PH quintic spirals joining two circles with one circle inside the other. Comput Aided Design, 2007, 39(2): 125–132

Appendix

13 Habib Z, Sakai M. G2 Pythagorean hodograph quintic transition between two circles with shape control. Comput Aided Geom D, 2007, 24(5): 252–266 14 Walton D J, Meek D S. Planar G2 transition with a fair Pythagorean Hodograph quintic curve. J Comput Appl Math, 2002, 138(1): 109–126 15 Walton D J, Meek D S. G2 curve design with a pair of Pythagorean Hodograph quintic spiral segments. Comput Aided Geom D, 2007, 24(5): 267–285 16 Yang X. Normal based subdivision scheme for curve design. Comput Aided Geom D, 2006, 23(3): 243–260 17 Dyn N, Levin D, Gregory J A. A 4-point interpolatory subdivision scheme for curve design. Comput Aided Geom D, 1987, 4(4): 257–268 18 Dyn N, Floater M S, Hormann K. A C 2 four-point subdivision scheme with fourth order accuracy and its extensions. In: Daehlen M, Mørken K, Schumaker L L, eds. Mathematical Methods for Curves and Surfaces, Tromso 2004, 2005. 145– 156 19 Hassan M F, Ivrissimitzis I P, Dodgson N A, et al. An interpolating 4-point C 2 ternary stationary subdivision scheme. Comput Aided Geom D, 2002, 19(1): 1–18 20 Marinov M, Dyn N, Levin D. Geometrically controlled 4-point interpolatory schemes. In: Dodgson N A, Floater M S, Sabin M A, eds. Advances in Multiresolution for Geometric Modeling. Berlin: Springer-Verlag, 2005 21 Aspert N, Ebrahimi T, Vandergheynst P. Non-linear subdivision using local spherical coordinates. Comput Aided Geom D, 2003, 20(3): 165–187 22 Guggenheimer H. Differential Geometry. New York: McGrawHill, 1963

If 0 < α  β  arccos λ, we have sin β β β  ; (A1) (1 − λ) + λ  α sin α α

Lemma 1.

At first we propose Lemma 1 which is used to replace the ratio of sines by the ratio of angles.

1826

segment interpolates the initial G1 Hermite data. However, the curvature of the start and end points is determined by the initial G1 Hermite data and the subdivision scheme, then when we use the scheme to interpolate arbitrary point array, the subdivision curve is only G1 in the joint points. We conjecture that we can interpolate any admissible G2 Hermite data with spiral segments by modifying the subdivision rule slightly, and by selecting appropriate initial tangent vectors we can interpolate arbitrary ordered discrete point arrays by spiral segments joined in G2 manner.

DENG C Y et al. Sci China Ser F-Inf Sci | Oct. 2009 | vol. 52 | no. 10 | 1821-1829

sin α α α   (1 − λ) + λ . β sin β β

k Let g(δi+1 ) =

(A2)

Proof. We prove (A1), the proof of (A2) is similar. sin x − αx , For α  x  arccos λ, let f1 (x) = sin α sin x − λ αx − (1 − λ), by the well known f2 (x) = sin α inequality sin x < x < tan x(0 < x < π2 ), we have 1 cos α 1 cos x − < − < 0, sin α α sin α α λ λ cos x λ  − > − > 0. f2 (x) = sin α α sin α α

f1 (x) =

So f1 (β)  f1 (α) = 0, f2 (β)  f2 (α) = 0, i.e. (A1) holds. Lemma 2.

For k  2, θk < π4 .

Proof. The initial conditions are π > α00 > β00 > 0 and α00 + β00 < π. Because T00 = T01 = T02 , T10 = T21 = T42 , we have 1 0 1 0 α20 = α20 = α40 < π4 , β42 = β22 = β41 < π4 . By eqs. (2), (3), (6) and (7), 5β 1 5α21 = 0 4 8 1 1 1 5β 0 5(β 0 − δ1 ) < = 8 8 5(α00 + β00 ) π 5β00 < < , = 16 32 4 1 π α α1 α0 2 2 β02 = β 0 − δ12 < β 0 = 0 = 0 = 0 < . 2 2 4 4

α21 = α21 + δ12
2, by eq. (11), θk < (0.625)k−2 θ2 < π4 . Lemma 3. For k  2,    1 − 0.113mki 1 − mki k ri 1 − 0.385mki 1 + 0.25mki   δik k+1 k (A3) < r2i < ri 1 − 0.75 k ; αi   1 − 0.25mki  1 k 1 + 0.272m i k ri−1 1 + mki   1 δk 1 1 − 0.75 ki . (A4) < k+1 < k ri−1 αi−1 r2i−1 Proof. We only prove the left side inequality of (A3), the proof of other inequalities are similar.

k+1 r2i

k k rik sin αk i sin[(αi −δi+1 )/2] k )/2] sin[αk +(δ k −δ k sin[(αk +δ i i i i i+1 )/2]

=

(see eq. (9)), because k )=− g (δi+1

rik sin2 [αki − δik /2] < 0. k 2 sin αki sin2 [αki + δi+1 /2]

δk

< 1.02mki , and αi+1 k i g(1.02mki αki ), i.e. k+1 r2i 

:=

sin[αki f (δik ).

k we have g(δi+1 )



rik sin αki sin[(αki − 1.02mki αki )/2] + (δik − 1.02mki αki )/2] sin[(αki + δik )/2]

Because k  2, by Lemma 2 we have θk < π4 , then √ on the left side of inequality (A1) we can set λ = 22 . It is easy to see that f (δik ) is monotone decreasing function of δik , then by δik  0.25mki αki and the left side of inequalities (A1) and (A2), k+1 r2i  f (δik )

 f (0.25mki αki ) sin αki sin[(αki − 1.02mki αki )/2] sin(αki − 0.385mki αki ) sin[(αki + 0.25mki αki )/2]

  √ k √  2αi 2 k 1− + > ri 2 2(αki − 0.385mki αki ) 

αki − 1.02mki αki · αki + 0.25mki αki



 1 − 0.113mki 1 − 1.02mki k . > ri 1 − 0.385mki 1 + 0.25mki = rik

Lemma 4. Let rik (k  1) be the radius of Λki k which interpolate pki , pki+1 and match Uik , Ui+1 , we k k k have ri+1 > ri (i = 0, 1, · · · , 2 − 1). Proof. We prove Lemma 4 by induction. For k = 1, obviously Lemma 4 holds. Assume that for k = n, Lemma 4 holds, i.e. n ri+1 > rin (i = 0, 1, · · · , 2n − 1), then by eq. (5), δin > 0(i = 0, 1, · · · , 2n − 1). For each i, by eqs. (9) and (10) we have 2  n+1 r2i+1 sin[(αni + δin )/2] > 1, (A5) n+1 = n sin[(αni − δi+1 )/2] r2i and by the left side inequalities of (A3) and (A4) we have   n+1 rin 1 − 0.113mni r2i n+1 > n ri−1 1 − 0.385mni r2i−1   1 − 0.25mni (1 + 0.272mni ) · 1 + mni

DENG C Y et al. Sci China Ser F-Inf Sci | Oct. 2009 | vol. 52 | no. 10 | 1821-1829

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  rin 1 − mni > n > 1. (A6) ri−1 1 + mni By (A5) and (A6), Lemma 4 also holds for k = n + 1. So Lemma 4 holds. Lemma 5. We can find an integer k0 , such that for k > k0 , 0.999δik > ; (A7) mk+1 2i−1 αki−1 0.999δik > . (A8) mk+1 2i+1 αki + 0.999δik Proof. We only prove (A7), the proof of (A8) is similar. By Theorem 1, there exists an integer k0 , such that for k > k0 , θk < arccos 0.999. Then for k > k0 , by eqs. (9), (10) and the left side of inequality (A1), 2  k+1 k sin[(αki−1 + δi−1 )/2] r2i−1 = k+1 sin[(αki−1 − δik )/2] r2i−2  2 k αki−1 + δi−1 > 0.001 + 0.999 k αi−1 − δik 1.998δik >1+ k . αi−1 − δik (y > 1) is monotone increasBecause f (y) = y−1 y+1 ing function of y, we have k+1 k+1 k+1 k+1 r2i−1 r2i−1 − r2i−2 /r2i−2 −1 = = mk+1 2i−1 k+1 k+1 k+1 k+1 r2i−1 + r2i−2 r2i−1 /r2i−2 + 1 1 + 1.998δik /(αki−1 − δik ) − 1 1 + 1.998δik /(αki−1 − δik ) + 1 0.999δik . > αki−1

1 r2k+k k1 i−1

Proof.

1.02 min{mki−1 , mki+1 } > 0.25mki . x−1 , by (A11), x+1 k+1 r2i < x(1 − 0.75 k+1 r2i−1

Since mki =

< 1 + 0.9925

Denote x =

rik

k ri−1

 rik −1 . k ri−1 , then k

mki δk

=

× 0.25mki )

< 1 + 0.6(x − 1).

(A12)

So in this case we can let k1 = 1. (2) δik = min{1.02mki−1 αki−1 , 1.02mki+1 αki }, i.e., 1.02 min{mki−1 , mki+1 } < 0.25mki . Without loss of generality we can assume δik = 1.02mki+1 αki . If mki+1  0.01, by (A11), k+1 r2i < x(1 − 0.75 × 0.01) k+1 r2i−1 < 1 + 0.9925(x − 1),

(A13)

else mki+1 < 0.01, first we prove that k+1 k+1 k+1 min{1.02mk+1 2i−1 α2i−1 , 1.02m2i+1 α2i } k+1 min{αk+1 2i−1 , α2i }

(A14)

By (A8), 0.999δik αk+1 αki + 0.999δik 2i 0.999 × 1.02 = mk αk+1 1 + 0.999 × 1.02mki+1 i+1 2i

k+1 > 1.02 1.02mk+1 2i+1 α2i

> mki+1 αk+1 2i .

(A9)

(A15)

By Eqs. (2), (3), (6) and (7), x−1 . x+1

< 0.25mki , By Lemma 4 and eq. (8), 0 < αδik , αi+1 k i i then by (A1), we have  k 2  2 k+1 r2i+1 1 + 0.25mki αi + δik < < k+1 k 1 − 0.25mki αki − δi+1 r2i

28x + 4 = 1 + (x − 1) (3x + 5)2 1828

By the right side of inequalities (A3), (A4),    k+1 r2i rik 0.75δik 0.75δik < k 1− 1− k k+1 ri−1 αki αi−1 r2i−1   k k 0.75δi r . (A11) 1− < ki ri−1 min{αki−1 , αki } In the following we classify two cases to prove (A9). (1) δik = 0.25mki min{αki−1 , αki }, i.e.

For a point pki , we can find k1 such 

(A10)

> mki+1 .

>

Lemma 6. that 1 r2k+k k1 i

< 1 + 0.5(x − 1).

k k k αk+1 2i−1 = 0.5βi−1 = 0.5(α i−1 − δi ),

αk+1 = 0.5αki = 0.5(αki + δik ). 2i If αki−1  αki , by mki+1 < 0.01, we have δik < 0.0102α ki < 0.0102α ki−1 , then by (A7), k+1 1.02mk+1 2i−1 α2i−1

> 1.02

2 0.999δik αki−1 − δik · k · αk+1 k 2 αi−1 αi + δik 2i

DENG C Y et al. Sci China Ser F-Inf Sci | Oct. 2009 | vol. 52 | no. 10 | 1821-1829

αki−1 − δik k k+1 αki · mi+1 α2i αki + δik αki−1 1.022 × 0.999 × (1 − 0.0102) k k+1 mi+1 α2i > 1 + 0.0102 > mki+1 αk+1 (A16) 2i ; k k else, αi−1 < αi , also by (A7), 1.022 × 0.999α ki k k+1 k+1 mi+1 α2i−1 1.02mk+1 2i−1 α2i−1 > αki−1 = 1.022 × 0.999

> mki+1 αk+1 (A17) 2i−1 . By (A15)–(A17), (A14) holds. Because k+1 k+1 k+1 k+1 = min{1.02mk+1 δ2i 2i−1 α2i−1 , 0.25m2i α2i−1 , k+1 k+1 k+1 0.25mk+1 2i α2i , 1.02m2i+1 α2i }, by (A11) and (A14), k+2 r4i

k+2 r4i−1


2,   r2k+l li k l < max x(1 − m ) , 1 + 0.9925 (x − 1) . i+1 r2k+l l i−1 Since liml→∞ (1− mki+1 )l = 0, we can find k1 such that for k > k1 , (A9) holds. By the proof of the two cases we know that Lemma 6 holds. limk→∞

1 r0k

and limk→∞

1 r kk

exist.

2

Proof. We only prove that limk→∞ r1k exists, 0 the proof of that of limk→∞ rk1 is similar. 2k

mk+1 = 1

Because T0k+1 = T0k , δ0k = 0. Then by eqs. (9), (10) and the right side of inequality (A1), 2   2 sin(αk0 /2) αk0 r1k+1 = < . sin[(αk0 − δ1k )/2] αk0 − δ1k r0k+1

r1k − r0k δ1k < < 0.25mk1 . r1k + r0k αk0 + δ1k

< (0.25)k m11 . By induction we have mk+1 1 By eq. (9) and the right side of inequality (A2), sin αk0 sin[(αk0 − δ1k )/2] αk0 − δ1k r0k+1 > = r0k sin(αk0 /2) sin(αk0 − δ1k /2) αk0 > 1 − 0.25mk1 . So r0k r02 r0k = · · · · · r01 r01 r0k−1 > (1 − 0.25mk−1 ) · · · · · (1 − 0.25m11 ) 1 > (1 − (0.25)k−1 m11 ) · · · · · (1 − 0.25m11 )



k+1 min{αk+1 2i−1 , α2i } 

k+1 0.75δ2i k < x(1 − 0.75mi+1 ) 1 − k+1 min{αk+1 2i−1 , α2i } 

 k+1 r2i k 2 < max x(1 − mi+1 ) , 1 + 0.9925 −1 k+1 r2i−1   < max x(1 − mki+1 )2 , 1 + 0.9925 (x − 1) .

Lemma 7.

Then

> 1 − [((0.25)k−1 + · · · · · +0.25]m11 > 2/3. So {r0k }k is a monotone decreasing number array and have low bound (2r01 )/3, then limk→∞ r1k 0 exists. Lemma 8. liml→∞

1 r k+l l

For a point pki , liml→∞

1 r k+l l

=

2 i

.

2 i−1

Proof. For a point pki , by (A3), (A4) and k+1 k+1 k < r2i < rik , then {r2k+l Lemma 5, ri−1 < r2i−1 li } is a monotone decreasing number array and have 1 k , then liml→∞ rk+l exists. Similarly, low bound ri−1 liml→∞

2l i

1

exists.

r k+l 2l i−1

By Lemma 6, for arbitrary integer k2 , we can find an integer l1 such that   k 1 r2k+l ri l1 i k2 < 1 + 0.9925 −1 . k 1 ri−1 r2k+l l1 i−1 This leads to liml→∞ liml→∞

1 r k+l l

r k+l l 2 i

r k+l l

2 i−1

= 1, so liml→∞

1 r k+l l

=

2 i

.

2 i−1

DENG C Y et al. Sci China Ser F-Inf Sci | Oct. 2009 | vol. 52 | no. 10 | 1821-1829

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