research article a geometric modeling method based on th...

Research ArticleA Geometric Modeling Method Based onTH-Type Uniform B-Splines

Jin Xie12

1 Department of Mathematics and Physics Hefei University Hefei 230601 China2Department of Mathematics and Physics University of La Verne La Verne CA 91750 USA

Correspondence should be addressed to Jin Xie hfuuxiejin126com

Received 26 January 2014 Accepted 24 May 2014 Published 15 June 2014

Academic Editor Vassilios C Loukopoulos

Copyright copy 2014 Jin Xie This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

A geometric modeling method based on TH-type uniform B-splines which are composed of trigonometric and hyperbolicpolynomial with parameters is introduced in this paper The new splines possess many important properties of quadratic andcubic B-splines Taking different values of the parameters one can not only locally adjust the shape of the curves but also changethe type of some segments of a curve between trigonometric and hyperbolic functions as wellThe given curves can also interpolatedirectly control polygon locally by selecting special parameters Moreover the introduced splines can represent some quadraticcurves and transcendental curves with selecting proper control points and parameters

1 Introduction

B-splines are used as an important geometric modeling toolin computer aided geometric design (CAGD)However thereare still several limitations on B-splines in practical appli-cations [1] Firstly for the fixed control points and the knotsequences the shape of the curves and surfaces representedby B-splines is fixed Secondly the B-splines cannot representconics (except parabolas) and some known curves such as thecycloid and the helix exactly AlthoughNURBS can overcomethe shortcomings of B-splines as the complexity of its rationalbasis functions and its derivatives and integrals are hard tocompute it is not convenient to the user So in order to avoidtheir inconveniences recently several new splines defined indifferent space from the usual polynomial space have beenproposed for geometric modeling in CAGD [2ndash11] T-typesplines were introduced [2ndash7] which can exactly representthe ellipse the cycloid and the helix Pottmann and Wagner[8] and Koch and Lyche [9] presented a kind of exponentialsplines in tension in that space 1 119905 cosh 119905 sinh 119905 Lu et al[10] gave the explicit expressions for uniform splines Li andWang [11] generalized the curves and surfaces of exponentialforms to algebraic hyperbolic spline forms of any degreewhich can represent exactly some remarkable curves such as

the hyperbola and the catenary However H-type uniform B-splines in tension are not applicable to freeform polynomialcurves of high orders which severely restrict their applica-tions in CAGD

By comparing T-type uniform B-splines and H-typeuniform B-splines we found that T-type uniform B-splinesare located on one side of the B-spline andH-type uniformB-splines are located on the other side of the B-splineThereforeone thinks if the two different curves can be unified toproduce new blending splines then the new curve will havemore plentiful modeling power In order to construct moreflexible curves for curves and surface modeling Zhang et al[12 13] proposed a curve family named FB-spline that usesa unified basis 1 119905 cos 119905 sin 119905 and basis 1 119905 cosh 119905 sinh 119905FB-splines inherited nearly all the properties that the T-typeB-splines and the H-type B-splines have However the for-mulas for the FB-splines were rather complicated Wang andFang [14] unified and extended three types of splines by a newkind of spline (UE-spline for short) defined over the spacecos120596

119894119905 sin120596

119894119905 1 119905 119905

119897 where the type of a curve can

be switched by a frequency sequence 120596119894 However the

geometric meaning of the sequence 120596119894 is not obvious Over

the space span sin 119905 cos 119905 sinh 119905 cosh 119905 1 119905 119905119899minus5 119899 ge 5Xu and Wang [15] presented two new unified mathematics

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2014 Article ID 242469 7 pageshttpdxdoiorg1011552014242469

2 Mathematical Problems in Engineering

models of conics and polynomial curves called algebraichyperbolic trigonometric (AHT) Bezier curves and nonuni-form algebraic hyperbolic trigonometric (NUAHT) B-splinecurves of order 119899 which share most of the properties as thoseof the Bezier curves and B-spline curves in polynomial space

In this paper we present a new geometric modelingmethod based on two kinds of TH-type uniform B-splineswhich are composed of hyperbolic and trigonometric func-tions The introduced spline has the following features (1)the new spline curves can be adjusted totally or locally (2)The given curves can switch into T-type B-spline curves orH-type B-spline curves when the parameter is equal to 0 or 1(3) Without solving the system of equations the new curvescan interpolate certain control points directly (4) The TH-type B-spline curves can be used to represent some conicsand transcendental curves with the parameters and controlpoints chosen properly

The rest of this paper is organized as follows In Sections2 and 3 the TH-type basis functions and corresponding TH-type curves are established and the properties of the basisfunctions are proved In Section 4 some properties of theTH-type B-spline curves are discussed It is pointed out inSection 5 that some transcendental curves can be representedprecisely with the TH-type curves and the applications of thecurves are shown in Section 6

2 Quadratic TH-Type B-Spline

Definition 1 Given 119905 isin [0 1] the quadratic basis functionsbased on weighted trigonometric and hyperbolic polynomi-als are as follows

qth02

(119905 120582119894) =

1

2minus 120582119894

2119890 cosh (1 minus 119905) minus 1198902minus 1

2(119890 minus 1)2

+1

2(120582119894minus 1) sin 120587119905

2

qth12

(119905 120582119894 120582119894+1

) =1

2(1 minus 120582

119894) sin 120587119905

2

+ 120582119894

2119890 (cosh 1 minus cosh (1 minus 119905))

(119890 minus 1)2

+ 120582119894+1

2119890 (cosh 1 minus cosh (119905))(119890 minus 1)

2

+1

2(1 minus 120582

119894+1) cos 120587119905

2

qth22

(119905 120582119894+1

) = 120582119894+1

119890 (cosh (119905) minus 1)

2(119890 minus 1)2

+1

2(1 minus 120582

119894+1) (1 minus cos 120587119905

2)

(1)

which are named the basis functions of quadratic TH-typeB-spline

Theorem2 Theabove functions have the following properties(i) Partition of unity qth

02(119905 120582119894) + qth

12(119905 120582119894 120582119894+1

) +

qth22(119905 120582119894+1

) = 1

(ii) Symmetry qth02(119905 120582119894) = qth

22(1 minus 119905 120582

119894) qth12(119905 120582119894

120582119894+1

) = qth12(1 minus 119905 120582

119894+1 120582119894)

(iii) Nonnegativity if (1+radic119890)2((1minusradic119890)(1+radic119890)

2+radic2119890) le

120582119894 120582119894+1

le (119890minus 1)21205872((119890 minus 1)

21205872minus8119890) then qth

1198962(119905) ge

0 119896 = 0 1 2

Proof (i) and (ii) are easy to be proved by simple computa-tion Next we will prove (iii)

By direct computation we have qth02(0 120582119894) =

1 qth02(1 120582119894) = 0 And since 0 le 119905 le 1 120582

119894le

(119890 minus 1)21205872((119890 minus 1)

21205872minus 8119890) and qth1015840

02(119905 120582119894) le 0 then

we have qth02(119905 120582119894) ge 0 Evidenced by the same token we

have qth22(119905 120582119894+1

) ge 0From (ii) we have qth

12(119905 120582119894 120582119894+1

) = 1 minus qth02(119905 120582119894) minus

qth22(119905 120582119894+1

) Obviously if we can prove qth02(119905 120582119894) +

qth22(119905 120582119894+1

) le 1 we can prove qth12(119905 120582119894+1

) ge 0Let 119891(119905 120582

119894 120582119894+1

) = qth02(119905 120582119894) + qth

22(119905 120582119894+1

) we have119891(0 120582

119894 120582119894+1

) = 119891(1 120582119894 120582119894+1

) = 1 Thus when 120582119894 120582119894+1

ge (1 +

radic119890)2((1 minus radic119890)(1 + radic119890)

2+ radic2119890) we can get

1198911015840(119905 120582119894 120582119894+1

) =

lt 0 119905 isin [0 05)

= 0 119905 = 05

gt 0 119905 isin (05 1]

(2)

So the maximum value of the function 119891(119905 120582119894 120582119894+1

) equals1 That is qth

02(119905 120582119894) + qth

22(119905 120582119894+1

) le 1 which meansqth12(119905 120582119894+1

) ge 0

Definition 3 Given control points 119875119894isin 119877119889(119889 = 2 3 119894 =

0 1 119899) the curves

QTH1198942(119905 120582119894 120582119894+1

)

= 119875119894minus1

qth02

(119905 120582119894)

+ 119875119894qth12

(119905 120582119894 120582119894+1

) + 119875119894+1

qth22

(119905 120582119894+1

)

119905 isin [0 1] 119894 = 1 2 119899 minus 1

(3)

are defined quadratic TH-type B-spline curve segmentswith shape parameters 120582

119894and 120582

119894+1 where qth

02(119905 120582119894)

qth12(119905 120582119894 120582119894+1

) and qth22(119905 120582119894+1

) are the bases of quadraticTH-type B-spline

3 Cubic TH-Type B-Spline

By a similar method we may define the bases of cubic TH-type B-spline

Definition 4 For (119890 minus 1)2((119890 minus 1)

2minus 120587) le 120582

119894 le 120582119894+1

le (119890 minus

1)21205872((119890minus1)

21205872minus8119890) and 119905 isin [0 1] the following functions

cth03

(119905 120582119894) =

1

120587(120582119894minus 1) cos 120587

2119905 +

1

(119890 minus 1)2

times ((2(119890 minus 1)2minus (1 + 119890

2) 120582119894)

times (1 minus 119905) minus 2119890120582119894sinh (1 minus 119905))

Mathematical Problems in Engineering 3

cth13

(119905 120582119894 120582119894+1

) =(119890 minus 1)

2minus (1 + 119890

2) 120582119894+1

2(119890 minus 1)2

minus(119890 minus 1)

2minus (1 + 119890

2) (2120582119894+ 120582119894+1

)

2(119890 minus 1)2

times (1 minus 119905) +2

120587(1 minus 120582

119894) cos 120587119905

2

minus1

120587(1 minus 120582

119894+1) sin 120587119905

2

+(119890 + 1) 120582119894+1

2 (119890 minus 1)cosh (1 minus 119905)

minus(1 + 119890

2) 120582119894+1

+ 4119890120582119894

(119890 minus 1)2120587

sinh (1 minus 119905)

cth23

(119905 120582119894 120582119894+1

) =(119890 minus 1)

2minus (1 + 119890

2) 120582119894

2(119890 minus 1)2


2minus (1 + 119890

2) (120582119894+ 2120582119894+1

)

2(119890 minus 1)2

119905

+2

120587(1 minus 120582

119894+1) sin 120587119905

2

minus1

120587(1 minus 120582

119894) cos 120587119905

2+(119890 + 1) 120582119894

2 (119890 minus 1)cosh 119905

minus(1 + 119890

2) 120582119894+ 4119890120582

119894+1

(119890 minus 1)2

sinh 119905

cth33

(119905 120582119894+1

) =1

120587(120582119894+1

minus 1) sin 120587119905

2+

1

2(119890 minus 1)2

times (((119890 minus 1)2minus (1 + 119890

2) 120582119894+1

) 119905

+ 2119890120582119894+1

sinh 119905) (4)

are called basis functions of cubic TH-type B-spline withshape parameters 120582

119894and 120582

119894+1

It is easy to prove that the basis functions of cubicTH-type B-spline have the same properties nonnegativitypartition of unity and symmetry



CTH1198943(119905 120582119894 120582119894+1

)

= 119875119894minus1

cth03

(119905 120582119894) + 119875119894cth13

(119905 120582119894 120582119894+1

)

+ 119875119894+1

cth23

(119905 120582119894 120582119894+1

) + 119875119894+2

cth33

(119905 120582119894+1

)

119905 isin [0 1] 119894 = 1 2 119899 minus 1

(5)

are defined cubic TH-type B-spline curve segmentswith shape parameters 120582

119894and 120582

119894+1 where cth

03(119905 120582119894)

cth13(119905 120582119894 120582119894+1

) cth23(119905 120582119894 120582119894+1

) and cth33(119905 120582119894+1

) are thebasis functions of cubic TH-type B-spline

4 The Properties of the TH-TypeB-Spline Curves

According to the properties of the basis functions anddefinition it is easy to get the following properties of curves(3) and (5)

(i) Continuity

Theorem 6 For the uniform knots the curves (3) are 1198621

continuous and the curves (5) are 1198622 continuous

Proof For the curve (3) we can get

QTH1198942(0 120582119894 120582119894+1

) =1

2(119875119894minus1

+ 119875119894)

QTH1198942(1 120582119894 120582119894+1

) =1

2(119875119894+ 119875119894+1

)

QTH10158401198942(0 120582119894 120582119894+1

)

=(1 minus 119890) 120587 + (119890 (120587 minus 2) minus 2 minus 120587) 120582119894

4 (119890 minus 1)(119875119894minus 119875119894minus1

)

QTH10158401198942(1 120582119894 120582119894+1

)

=(1 minus 119890) 120587 + (119890 (120587 minus 2) minus 2 minus 120587) 120582119894+1

4 (119890 minus 1)(119875119894+1

minus 119875119894)

(6)

Thus we obtain QTH(119896)119894minus12

(1 120582119894 120582119894+1

) = QTH(119896)1198942(0 120582119894 120582119894+1

)

(119896 = 0 1) that is to say the curves (3) are 1198621 continuousFor the curves (5) we get

CTH1198943(0 120582119894 120582119894+1

)

=(119890 minus 1)

2(120587 + 2120582

119894minus 2) minus 2120587120582

119894

2(119890 minus 1)2120587

(119875119894minus1

+ 119875119894+1

)

+2(119890 minus 1)

2(120587 minus 120582

119894minus 2) + 2120587120582

119894

(119890 minus 1)2120587

119875119894

CTH1198943(1 120582119894 120582119894+1

)

=(119890 minus 1)

2(120587 + 2120582

119894+1minus 2) minus 2120587120582

119894+1

2(119890 minus 1)2120587

(119875119894+ 119875119894+2

)

+2(119890 minus 1)

2(120587 minus 120582

119894+1minus 2) + 2120587120582

119894+1

(119890 minus 1)2120587

119875119894+1

CTH10158401198943(0 120582119894 120582119894+1

) =1

2(119875119894+1

minus 119875119894minus1

)


CTH10158401198943(1 120582119894 120582119894+1

) =1

2(119875119894+2

minus 119875119894)

CTH101584010158401198943(0 120582119894 120582119894+1

)

=(119890 minus 1) 120587 + ((119890 minus 1) 120587 minus 2 (119890 + 1)) 120582119894

4 (119890 minus 1)

times (119875119894minus1

minus 2119875119894+ 119875119894+1

)

CTH101584010158401198943(1 120582119894 120582119894+1

)

=(119890 minus 1) 120587 + ((119890 minus 1) 120587 minus 2 (119890 + 1)) 120582119894+1

4 (119890 minus 1)

times (119875119894minus 2119875119894+1

+ 119875119894+2

)

(7)

So we have CTH(119896)119894minus13

(1 120582119894 120582119894+1

) = CTH(119896)1198943(0 120582119894 120582119894+1

) (119896 =

0 1 2) This implies that curves (5) are 1198622 continuousThis implies the theorem

(ii) Local Adjustable Properties From formulas (3) and (5) theparameter120582

119894only affects two curve segmentswithout altering

the remainder Figure 1 shows local adjustable quadraticuniform TH-type spline curves where all parameters 120582

119894=

05 in the solid curves and all parameters 120582119894= 05 except

1205823

= minus1 in the dotted curves The parameter only affectsthe 2th and the 3th curve segment Figure 2 shows the localadjustable cubic uniform TH-type spline curves where allthe parameters are equal to 05 in the solid curves and allthe parameters are equal to 05 except 120582

5= minus1 in the dotted

curves The parameter 1205825only affects the 4th and 5th curve

segmentObviously when all parameters 120582

119894are the same the

curves can be adjusted totally

(iii) Local Interpolating Properties For the curve (3) letting120582119894

= 120582119894+1

= (radic119890 + 1)2(1 minus radic2119890 + 119890) = 505952 then

qth1198942(05) = 119875

119894 that is the curve interpolates the point

119875119894 For the curve (5) when 120582

119894= (2 minus 120587)(119890 minus 1)

2(2(119890 minus

1)2minus 2120587) = 891206 cth

1198943(0) = 119875

119894 120582119894+1

= (2 minus 120587)(119890 minus

1)2(2(119890 minus 1)

2minus 2120587) = 891206 cth

1198943(1) = 119875

119894+1 that is the

curve interpolates the points 119875119894and 119875119894+1

Figure 3 shows localinterpolating quadratic TH-type spline curves where thecurve interpolates the point119875

5when the parameter120582

5= 1205826=

505952 The local interpolating cubic TH-type spline curvesare showed in Figure 4 where the curves interpolate the point1198755when the parameter 120582

5= 891206

5 The Representations of Some Known Curves

When the parameters 120582119894= 120582119894+1

= 0 the curves (3) and (5) areT-type uniformB-spline curves If the parameters 120582

119894= 120582119894+1

=

1 the curves (3) and (5) become H-type uniform B-splines

51 The Representation of the Conic Curves The ellipse andhyperbola are the most common in the conic curve If the

P1

P2P3

P4

P5

Figure 1 Local adjustable quadratic uniformTH-type spline curves

P1

P2P3

P4

P5

Figure 2 Local adjustable cubic uniform TH-type spline curves

P1

P2P3

P4

P5

Figure 3 Local interpolating quadratic TH-type spline curves


P1

P2P3

P4

P5

Figure 4 Local interpolating cubic uniform TH-type spline curves

control points and the parameters are selected properly thecurves (3) and (5) can represent them precisely

Given the uniform knots for the quadratic T-type B-spline curve we take the coordinates of the points119875

119894minus1 119875119894 and

119875119894+1

as follows

119875119894minus1

= (119898 minus 119886 119899 + 119887)

119875119894= (119898 + 119886 119899 + 119887)

119875119894+1

= (119898 + 119886 119899 minus 119887)

(119886119887 = 0)

(8)

For the cubic T-type B-spline curve we take

119875119894minus1

= (119898 119899 minus120587

2119887)

119875119894= (119898 +

120587

2119886 119899)

119875119894+1

= (119898 119899 minus120587

2119887)

119875119894+2

= (119898 minus120587

2119886 119899)

(119886119887 = 0)

(9)

Then when 119905 isin [0 1] and 120582119894= 120582119894+1

= 0 we obtain aparametric equation as follows

119909 (119905) = 119898 + 119886 cos 1205872119905

119910 (119905) = 119899 + 119887 sin 120587

2119905

(10)

It is the parametric form of the ellipse see Figure 5 In orderto represent the hyperbola for the quadratic H-type uniformB-spline curves the control points are taken as follows

119875119894minus1

= (119898 + 119886 119899 +1 minus 119890

1 + 119890119887)

119875119894= (119898 + 119886 119899

1 minus 119890

1 + 119890119887)

119875119894+1

= (119898 +1198902minus 119890 + 1

119890119886 119899 +

1198903minus 1

1198902 + 119890119887)

(119886119887 = 0)

(11)

For the cubic H-type uniform B-spline curves we take119875119894minus1

= (119898 + ((1198902+ 1)119890)119886 119899 minus ((119890

4+ 1)(119890

3minus 119890))119887) 119875

119894= (119898 +

119886 119899 + ((1198902+ 1)(119890

2minus 1))119887) 119875

119894+1= (119898 119899 + (2119890(119890

2minus 1))119887)

119875119894+2

= (119898 minus 119886 ((1198902+ 1)(119890

2minus 119890))119887) (119886119887 = 0) as control points

So we get a parametric equation as follows

119909 (119905) = 119898 + 119886 cosh 119905

119910 (119905) = 119899 + 119887 sinh 119905(12)

which represents an arc of the hyperbola see Figure 6

52 The Representation of the Transcendental Curves In thissection we can represent the transcendental curves with theuniform TH-type B-splines such as cycloid and catenary

When parameters 120582119894= 120582119894+1

= 0 control points are takenas follows

119875119894minus1

= (120587 minus 2

4119886

4 + 120587

4119886)

119875119894= (

2 minus 120587

4119886

4 minus 120587

4119886)

119875119894+1

= (6 minus 3120587

4119886

120587 + 4

4119886)

119875119894+2

= (10 minus 120587

4119886

4 + 3120587

4119886)

(119886 = 0)

(13)

So we obtain the parametric equation as follows

119909 (119905) = 119886 (119905 minus sin 120587

2119905)

119910 (119905) = 119886 (1 minus cos 1205872119905)

(14)

which represents an arc of a cycloid see Figure 7Similarly when taking 119875

119894minus1= (119898 + 2119886 119899 + ((119890

4+ 1)(119890

3minus

119890))119887)119875119894= (119898+119886 119899+((119890

2+1)(119890

2minus1)))119875

119894+1= (119898 119899+(2119890(119890

2minus

1))119887) and 119875119894+2

= (119898 minus 119886 119899 + ((1198902+ 1)(119890

2minus 1))119887) (119886119887 = 0) as

control points the parameters 120582119894= 120582119894+1

= 1 By formula (5)we have the following equation

119909 (119905) = 119898 + 119886119905

119910 (119905) = 119899 + 119887 cosh 119905(15)

which is the parametric equation of the catenary see Figure 8


Piminus1

Pi+1

Pi

(a)

Pi+1

Pi+2

Piminus1

Pi

(b)

Figure 5 The representation of ellipse with quadratic (a) and cubic (b) T-type B-spline curves

Piminus1Pi

Pi+1

(a)

Piminus1

Pi+2

Pi+1

Pi

(b)

Figure 6 The representation of hyperbola with quadratic (a) and cubic (b) H-type B-spline curves

Pi+2

Pi+1

Piminus1

Pi

Figure 7 The representation of cycloid with cubic T-type B-splinecurves

6 The Applications of the TH-Type Splines

From the last section we see letting the parameter be equalto 0 or 1 the types of the curves can be switched easily Soby selecting control points and parameters properly we canrepresent different type curve segments among a blendingcurve In Figure 9 a closed 119862

1 blending curve is composedof different type curves with the quadratic TH-type B-splineswhere the coordinates of control points are119875

0= 1198756= (minus3 (119890minus

1)(119890+1)) 1198751= 1198757= (minus3 (1minus 119890)(119890+1)) 119875

2= (3 (119890minus1)(119890+

1))1198753= (3 (1minus119890)(119890+1))119875

4= (4minus119890minus(1119890) (119890

3minus1)(119890

2+119890))

1198755

= (119890 minus 4 + (1119890) (1198903minus 1)(119890

2+ 119890)) and the parameters

Piminus1

Pi+2

Pi+1

Pi

Figure 8The representation of catenary with cubicH-type B-splinecurves

120582119894= (0 0 1 1 505952 505952 0) (119894 = 1 2 7) The 1st

segment is a trigonometric curve which is a quarter of aparabola The 3rd segment is a hyperbola arc The blendingcurve interpolates the point 119875

5in that the parameters 120582

5=

1205826= 505952Figure 10 shows an open 119862

2 blending curve representedby the cubic TH-type B-splines The control points are taken


P5

1

3

Figure 9 A closed 1198621 blending curve with quadratic TH-type B-

splines

1

10

6 5

P5

P4

Figure 10 An open 1198622 blending curve with cubic TH-type B-

splines

as follows 1198750= ((120587 minus 2)2 1) 119875

1= (0 (2 minus 120587)2) 119875

2= ((2 minus

120587)2 1) 1198753= (2 (2 +120587)2) 119875

4= (2 (119890

4+ 1198903minus 119890+ 1)(119890

3minus 119890))

1198755

= (1 21198902(1198902minus 1)) 119875

6= (0 (119890

2+ 2119890 minus 1)(119890

2minus 1))

1198757

= (minus1 21198902(1198902minus 1)) 119875

8= (minus(119890

2+ 1)119890 (119890

4+ 1198903minus

119890 + 1)(1198903minus 119890)) 119875

9= (minus(120587 + 4)2 0) 119875

10= (minus2 minus1205874)

11987511

= ((120587 minus 4)2 0) 11987512

= (minus2 1205874) where the parameters120582119894= (0 0 891206 891206 1 1 1 1 05 0) (119894 = 1 2 11)

The 1st segment of the bending curve is a trigonometric curvewhich is a part of the cycloidThe 5th and 6th segment are thecatenary and hyperbola respectively The 10th segment is theparabola arc Since the parameters 120582

3= 1205824= 120938 the

blending curve interpolates the points 1198754and 119875

5

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work was funded by the Natural Science Foundationof Anhui Province of China under Grant no 1208085MA15the Key Project Foundation of Scientific Research Edu-cation Department of Anhui Province under Grant noKJ2014ZD30 and Key Construction Disciplines Foundationof Hefei University under Grant no 2014XK08

References

[1] L Piegle and W Tiller The NURBS Book Springer BerlinGermany 1995

[2] J Zhang ldquoTwo different forms of C-B-splinesrdquo Computer AidedGeometric Design vol 14 no 1 pp 31ndash41 1997

[3] J Zhang ldquoC-curves an extension of cubic curvesrdquo ComputerAided Geometric Design vol 13 no 3 pp 199ndash217 1996

[4] H Wu and X Chen ldquoCubic non-uniform trigonometric poly-nomial curves with multiple shape parametersrdquo Journal ofComputer-Aided Design and Computer Graphics vol 18 no 10pp 1599ndash1606 2006 (Chinese)

[5] X Han ldquoCubic trigonometric polynomial curves with a shapeparameterrdquoComputer Aided Geometric Design vol 21 no 6 pp535ndash548 2004

[6] G Xu G Wang and W Chen ldquoGeometric construction ofenergy-minimizing Beezier curvesrdquo Science China InformationSciences vol 54 no 7 pp 1395ndash1406 2011

[7] W-T Wang and G-Z Wang ldquoTrigonometric polynomialuniform B-spline with shape parameterrdquo Chinese Journal ofComputers vol 28 no 7 pp 1192ndash1198 2005 (Chinese)

[8] H Pottmann and M G Wagner ldquoHelix splines as an exampleof affine Tchebycheffian splinesrdquo Advances in ComputationalMathematics vol 2 no 1 pp 123ndash142 1994

[9] P E Koch and T Lyche ldquoExponential B-splines in tensionrdquo inApproximationTheory VI C K Chui L L Schumaker and J DWard Eds pp 361ndash364 Academic Press New York NY USA1989

[10] Y Lu G Wang and X Yang ldquoUniform hyperbolic polynomialB-spline curvesrdquo Computer Aided Geometric Design vol 19 no6 pp 379ndash393 2002

[11] Y-J Li and G-Z Wang ldquoTwo kinds of B-basis of the algebraichyperbolic spacerdquo Journal of Zhejiang University Science A vol6 no 7 pp 750ndash759 2005

[12] J Zhang F-L Krause and H Zhang ldquoUnifying C-curves andH-curves by extending the calculation to complex numbersrdquoComputer Aided Geometric Design vol 22 no 9 pp 865ndash8832005

[13] J Zhang and F-L Krause ldquoExtending cubic uniform B-splinesby unified trigonometric and hyperbolic basisrdquo Graphical Mod-els vol 67 no 2 pp 100ndash119 2005

[14] G Wang and M Fang ldquoUnified and extended form of threetypes of splinesrdquo Journal of Computational and Applied Math-ematics vol 216 no 2 pp 498ndash508 2008

[15] G Xu and G-Z Wang ldquoAHT Bezier curves and NUAHT B-Spline curvesrdquo Journal of Computer Science and Technology vol22 no 4 pp 597ndash607 2007

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of


Mathematical PhysicsAdvances in

Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


Operations ResearchAdvances in

Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


models of conics and polynomial curves called algebraichyperbolic trigonometric (AHT) Bezier curves and nonuni-form algebraic hyperbolic trigonometric (NUAHT) B-splinecurves of order 119899 which share most of the properties as thoseof the Bezier curves and B-spline curves in polynomial space

In this paper we present a new geometric modelingmethod based on two kinds of TH-type uniform B-splineswhich are composed of hyperbolic and trigonometric func-tions The introduced spline has the following features (1)the new spline curves can be adjusted totally or locally (2)The given curves can switch into T-type B-spline curves orH-type B-spline curves when the parameter is equal to 0 or 1(3) Without solving the system of equations the new curvescan interpolate certain control points directly (4) The TH-type B-spline curves can be used to represent some conicsand transcendental curves with the parameters and controlpoints chosen properly

The rest of this paper is organized as follows In Sections2 and 3 the TH-type basis functions and corresponding TH-type curves are established and the properties of the basisfunctions are proved In Section 4 some properties of theTH-type B-spline curves are discussed It is pointed out inSection 5 that some transcendental curves can be representedprecisely with the TH-type curves and the applications of thecurves are shown in Section 6

2 Quadratic TH-Type B-Spline

Definition 1 Given 119905 isin [0 1] the quadratic basis functionsbased on weighted trigonometric and hyperbolic polynomi-als are as follows

qth02

(119905 120582119894) =

1

2minus 120582119894

2119890 cosh (1 minus 119905) minus 1198902minus 1

2(119890 minus 1)2

+1

2(120582119894minus 1) sin 120587119905

2

qth12

(119905 120582119894 120582119894+1

) =1

2(1 minus 120582

119894) sin 120587119905

2

+ 120582119894

2119890 (cosh 1 minus cosh (1 minus 119905))

(119890 minus 1)2

+ 120582119894+1

2119890 (cosh 1 minus cosh (119905))(119890 minus 1)

2

+1

2(1 minus 120582

119894+1) cos 120587119905

2

qth22

(119905 120582119894+1

) = 120582119894+1

119890 (cosh (119905) minus 1)

2(119890 minus 1)2

+1

2(1 minus 120582

119894+1) (1 minus cos 120587119905

2)

(1)

which are named the basis functions of quadratic TH-typeB-spline

Theorem2 Theabove functions have the following properties(i) Partition of unity qth

02(119905 120582119894) + qth

12(119905 120582119894 120582119894+1

) +

qth22(119905 120582119894+1

) = 1

(ii) Symmetry qth02(119905 120582119894) = qth

22(1 minus 119905 120582

119894) qth12(119905 120582119894

120582119894+1

) = qth12(1 minus 119905 120582

119894+1 120582119894)

(iii) Nonnegativity if (1+radic119890)2((1minusradic119890)(1+radic119890)

2+radic2119890) le

120582119894 120582119894+1

le (119890minus 1)21205872((119890 minus 1)

21205872minus8119890) then qth

1198962(119905) ge

0 119896 = 0 1 2

Proof (i) and (ii) are easy to be proved by simple computa-tion Next we will prove (iii)

By direct computation we have qth02(0 120582119894) =

1 qth02(1 120582119894) = 0 And since 0 le 119905 le 1 120582

119894le

(119890 minus 1)21205872((119890 minus 1)

21205872minus 8119890) and qth1015840

02(119905 120582119894) le 0 then

we have qth02(119905 120582119894) ge 0 Evidenced by the same token we

have qth22(119905 120582119894+1

) ge 0From (ii) we have qth

12(119905 120582119894 120582119894+1

) = 1 minus qth02(119905 120582119894) minus

qth22(119905 120582119894+1

) Obviously if we can prove qth02(119905 120582119894) +

qth22(119905 120582119894+1

) le 1 we can prove qth12(119905 120582119894+1

) ge 0Let 119891(119905 120582

119894 120582119894+1

) = qth02(119905 120582119894) + qth

22(119905 120582119894+1

) we have119891(0 120582

119894 120582119894+1

) = 119891(1 120582119894 120582119894+1

) = 1 Thus when 120582119894 120582119894+1

ge (1 +

radic119890)2((1 minus radic119890)(1 + radic119890)

2+ radic2119890) we can get

1198911015840(119905 120582119894 120582119894+1

) =

lt 0 119905 isin [0 05)

= 0 119905 = 05

gt 0 119905 isin (05 1]

(2)

So the maximum value of the function 119891(119905 120582119894 120582119894+1

) equals1 That is qth

02(119905 120582119894) + qth

22(119905 120582119894+1

) le 1 which meansqth12(119905 120582119894+1

) ge 0



QTH1198942(119905 120582119894 120582119894+1

)

= 119875119894minus1

qth02

(119905 120582119894)

+ 119875119894qth12

(119905 120582119894 120582119894+1

) + 119875119894+1

qth22

(119905 120582119894+1

)

119905 isin [0 1] 119894 = 1 2 119899 minus 1

(3)

are defined quadratic TH-type B-spline curve segmentswith shape parameters 120582

119894and 120582

119894+1 where qth

02(119905 120582119894)

qth12(119905 120582119894 120582119894+1

) and qth22(119905 120582119894+1

) are the bases of quadraticTH-type B-spline

3 Cubic TH-Type B-Spline

By a similar method we may define the bases of cubic TH-type B-spline

Definition 4 For (119890 minus 1)2((119890 minus 1)

2minus 120587) le 120582

119894 le 120582119894+1

le (119890 minus

1)21205872((119890minus1)

21205872minus8119890) and 119905 isin [0 1] the following functions

cth03

(119905 120582119894) =

1

120587(120582119894minus 1) cos 120587

2119905 +

1

(119890 minus 1)2

times ((2(119890 minus 1)2minus (1 + 119890

2) 120582119894)

times (1 minus 119905) minus 2119890120582119894sinh (1 minus 119905))


cth13

(119905 120582119894 120582119894+1

) =(119890 minus 1)

2minus (1 + 119890

2) 120582119894+1

2(119890 minus 1)2


2minus (1 + 119890

2) (2120582119894+ 120582119894+1

)

2(119890 minus 1)2


120587(1 minus 120582

119894) cos 120587119905

2

minus1

120587(1 minus 120582

119894+1) sin 120587119905

2

+(119890 + 1) 120582119894+1

2 (119890 minus 1)cosh (1 minus 119905)

minus(1 + 119890

2) 120582119894+1

+ 4119890120582119894

(119890 minus 1)2120587


cth23

(119905 120582119894 120582119894+1

) =(119890 minus 1)

2minus (1 + 119890

2) 120582119894

2(119890 minus 1)2


2minus (1 + 119890

2) (120582119894+ 2120582119894+1

)

2(119890 minus 1)2

119905

+2

120587(1 minus 120582

119894+1) sin 120587119905

2

minus1

120587(1 minus 120582

119894) cos 120587119905

2+(119890 + 1) 120582119894

2 (119890 minus 1)cosh 119905

minus(1 + 119890

2) 120582119894+ 4119890120582

119894+1

(119890 minus 1)2

sinh 119905

cth33

(119905 120582119894+1

) =1

120587(120582119894+1

minus 1) sin 120587119905

2+

1

2(119890 minus 1)2


2) 120582119894+1

) 119905

+ 2119890120582119894+1

sinh 119905) (4)


119894and 120582

119894+1




CTH1198943(119905 120582119894 120582119894+1

)

= 119875119894minus1

cth03

(119905 120582119894) + 119875119894cth13

(119905 120582119894 120582119894+1

)

+ 119875119894+1

cth23

(119905 120582119894 120582119894+1

) + 119875119894+2

cth33

(119905 120582119894+1

)

119905 isin [0 1] 119894 = 1 2 119899 minus 1

(5)


119894and 120582

119894+1 where cth

03(119905 120582119894)

cth13(119905 120582119894 120582119894+1

) cth23(119905 120582119894 120582119894+1

) and cth33(119905 120582119894+1




(i) Continuity




QTH1198942(0 120582119894 120582119894+1

) =1

2(119875119894minus1

+ 119875119894)

QTH1198942(1 120582119894 120582119894+1

) =1

2(119875119894+ 119875119894+1

)

QTH10158401198942(0 120582119894 120582119894+1

)



)

QTH10158401198942(1 120582119894 120582119894+1

)


4 (119890 minus 1)(119875119894+1

minus 119875119894)

(6)


(1 120582119894 120582119894+1

) = QTH(119896)1198942(0 120582119894 120582119894+1

)


CTH1198943(0 120582119894 120582119894+1

)

=(119890 minus 1)

2(120587 + 2120582

119894minus 2) minus 2120587120582

119894

2(119890 minus 1)2120587

(119875119894minus1

+ 119875119894+1

)

+2(119890 minus 1)

2(120587 minus 120582

119894minus 2) + 2120587120582

119894

(119890 minus 1)2120587

119875119894

CTH1198943(1 120582119894 120582119894+1

)

=(119890 minus 1)

2(120587 + 2120582

119894+1minus 2) minus 2120587120582

119894+1

2(119890 minus 1)2120587

(119875119894+ 119875119894+2

)

+2(119890 minus 1)

2(120587 minus 120582

119894+1minus 2) + 2120587120582

119894+1

(119890 minus 1)2120587

119875119894+1

CTH10158401198943(0 120582119894 120582119894+1

) =1

2(119875119894+1


)


CTH10158401198943(1 120582119894 120582119894+1

) =1

2(119875119894+2

minus 119875119894)

CTH101584010158401198943(0 120582119894 120582119894+1

)


4 (119890 minus 1)


minus 2119875119894+ 119875119894+1

)

CTH101584010158401198943(1 120582119894 120582119894+1

)


4 (119890 minus 1)

times (119875119894minus 2119875119894+1

+ 119875119894+2

)

(7)


(1 120582119894 120582119894+1

) = CTH(119896)1198943(0 120582119894 120582119894+1

) (119896 =





119894=


1205823








= 120582119894+1


qth1198942(05) = 119875



119894= (2 minus 120587)(119890 minus 1)

2(2(119890 minus

1)2minus 2120587) = 891206 cth

1198943(0) = 119875

119894 120582119894+1

= (2 minus 120587)(119890 minus

1)2(2(119890 minus 1)

2minus 2120587) = 891206 cth

1198943(1) = 119875





5= 1205826=


5= 891206




119894= 120582119894+1

=



P1

P2P3

P4

P5


P1

P2P3

P4

P5


P1

P2P3

P4

P5



P1

P2P3

P4

P5




119894minus1 119875119894 and

119875119894+1

as follows

119875119894minus1

= (119898 minus 119886 119899 + 119887)

119875119894= (119898 + 119886 119899 + 119887)

119875119894+1

= (119898 + 119886 119899 minus 119887)

(119886119887 = 0)

(8)


119875119894minus1

= (119898 119899 minus120587

2119887)

119875119894= (119898 +

120587

2119886 119899)

119875119894+1

= (119898 119899 minus120587

2119887)

119875119894+2

= (119898 minus120587

2119886 119899)

(119886119887 = 0)

(9)



119909 (119905) = 119898 + 119886 cos 1205872119905

119910 (119905) = 119899 + 119887 sin 120587

2119905

(10)


119875119894minus1

= (119898 + 119886 119899 +1 minus 119890

1 + 119890119887)

119875119894= (119898 + 119886 119899

1 minus 119890

1 + 119890119887)

119875119894+1

= (119898 +1198902minus 119890 + 1

119890119886 119899 +

1198903minus 1

1198902 + 119890119887)

(119886119887 = 0)

(11)


= (119898 + ((1198902+ 1)119890)119886 119899 minus ((119890

4+ 1)(119890

3minus 119890))119887) 119875

119894= (119898 +

119886 119899 + ((1198902+ 1)(119890

2minus 1))119887) 119875

119894+1= (119898 119899 + (2119890(119890

2minus 1))119887)

119875119894+2

= (119898 minus 119886 ((1198902+ 1)(119890



119909 (119905) = 119898 + 119886 cosh 119905

119910 (119905) = 119899 + 119887 sinh 119905(12)



When parameters 120582119894= 120582119894+1


119875119894minus1

= (120587 minus 2

4119886

4 + 120587

4119886)

119875119894= (

2 minus 120587

4119886

4 minus 120587

4119886)

119875119894+1

= (6 minus 3120587

4119886

120587 + 4

4119886)

119875119894+2

= (10 minus 120587

4119886

4 + 3120587

4119886)

(119886 = 0)

(13)


119909 (119905) = 119886 (119905 minus sin 120587

2119905)

119910 (119905) = 119886 (1 minus cos 1205872119905)

(14)


119894minus1= (119898 + 2119886 119899 + ((119890

4+ 1)(119890

3minus

119890))119887)119875119894= (119898+119886 119899+((119890

2+1)(119890

2minus1)))119875

119894+1= (119898 119899+(2119890(119890

2minus

1))119887) and 119875119894+2

= (119898 minus 119886 119899 + ((1198902+ 1)(119890

2minus 1))119887) (119886119887 = 0) as



119909 (119905) = 119898 + 119886119905

119910 (119905) = 119899 + 119887 cosh 119905(15)



Piminus1

Pi+1

Pi

(a)

Pi+1

Pi+2

Piminus1

Pi

(b)


Piminus1Pi

Pi+1

(a)

Piminus1

Pi+2

Pi+1

Pi

(b)


Pi+2

Pi+1

Piminus1

Pi





0= 1198756= (minus3 (119890minus

1)(119890+1)) 1198751= 1198757= (minus3 (1minus 119890)(119890+1)) 119875

2= (3 (119890minus1)(119890+

1))1198753= (3 (1minus119890)(119890+1))119875

4= (4minus119890minus(1119890) (119890

3minus1)(119890

2+119890))

1198755

= (119890 minus 4 + (1119890) (1198903minus 1)(119890


Piminus1

Pi+2

Pi+1

Pi


120582119894= (0 0 1 1 505952 505952 0) (119894 = 1 2 7) The 1st



5=




P5

1

3


splines

1

10

6 5

P5

P4


splines

as follows 1198750= ((120587 minus 2)2 1) 119875

1= (0 (2 minus 120587)2) 119875

2= ((2 minus

120587)2 1) 1198753= (2 (2 +120587)2) 119875

4= (2 (119890

4+ 1198903minus 119890+ 1)(119890

3minus 119890))

1198755

= (1 21198902(1198902minus 1)) 119875

6= (0 (119890

2+ 2119890 minus 1)(119890

2minus 1))

1198757

= (minus1 21198902(1198902minus 1)) 119875

8= (minus(119890

2+ 1)119890 (119890

4+ 1198903minus

119890 + 1)(1198903minus 119890)) 119875

9= (minus(120587 + 4)2 0) 119875


11987511

= ((120587 minus 4)2 0) 11987512



3= 1205824= 120938 the


5



Acknowledgments


References























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










cth13

(119905 120582119894 120582119894+1

) =(119890 minus 1)

2minus (1 + 119890

2) 120582119894+1

2(119890 minus 1)2


2minus (1 + 119890

2) (2120582119894+ 120582119894+1

)

2(119890 minus 1)2


120587(1 minus 120582

119894) cos 120587119905

2

minus1

120587(1 minus 120582

119894+1) sin 120587119905

2

+(119890 + 1) 120582119894+1

2 (119890 minus 1)cosh (1 minus 119905)

minus(1 + 119890

2) 120582119894+1

+ 4119890120582119894

(119890 minus 1)2120587


cth23

(119905 120582119894 120582119894+1

) =(119890 minus 1)

2minus (1 + 119890

2) 120582119894

2(119890 minus 1)2


2minus (1 + 119890

2) (120582119894+ 2120582119894+1

)

2(119890 minus 1)2

119905

+2

120587(1 minus 120582

119894+1) sin 120587119905

2

minus1

120587(1 minus 120582

119894) cos 120587119905

2+(119890 + 1) 120582119894

2 (119890 minus 1)cosh 119905

minus(1 + 119890

2) 120582119894+ 4119890120582

119894+1

(119890 minus 1)2

sinh 119905

cth33

(119905 120582119894+1

) =1

120587(120582119894+1

minus 1) sin 120587119905

2+

1

2(119890 minus 1)2


2) 120582119894+1

) 119905

+ 2119890120582119894+1

sinh 119905) (4)


119894and 120582

119894+1




CTH1198943(119905 120582119894 120582119894+1

)

= 119875119894minus1

cth03

(119905 120582119894) + 119875119894cth13

(119905 120582119894 120582119894+1

)

+ 119875119894+1

cth23

(119905 120582119894 120582119894+1

) + 119875119894+2

cth33

(119905 120582119894+1

)

119905 isin [0 1] 119894 = 1 2 119899 minus 1

(5)


119894and 120582

119894+1 where cth

03(119905 120582119894)

cth13(119905 120582119894 120582119894+1

) cth23(119905 120582119894 120582119894+1

) and cth33(119905 120582119894+1




(i) Continuity




QTH1198942(0 120582119894 120582119894+1

) =1

2(119875119894minus1

+ 119875119894)

QTH1198942(1 120582119894 120582119894+1

) =1

2(119875119894+ 119875119894+1

)

QTH10158401198942(0 120582119894 120582119894+1

)



)

QTH10158401198942(1 120582119894 120582119894+1

)


4 (119890 minus 1)(119875119894+1

minus 119875119894)

(6)


(1 120582119894 120582119894+1

) = QTH(119896)1198942(0 120582119894 120582119894+1

)


CTH1198943(0 120582119894 120582119894+1

)

=(119890 minus 1)

2(120587 + 2120582

119894minus 2) minus 2120587120582

119894

2(119890 minus 1)2120587

(119875119894minus1

+ 119875119894+1

)

+2(119890 minus 1)

2(120587 minus 120582

119894minus 2) + 2120587120582

119894

(119890 minus 1)2120587

119875119894

CTH1198943(1 120582119894 120582119894+1

)

=(119890 minus 1)

2(120587 + 2120582

119894+1minus 2) minus 2120587120582

119894+1

2(119890 minus 1)2120587

(119875119894+ 119875119894+2

)

+2(119890 minus 1)

2(120587 minus 120582

119894+1minus 2) + 2120587120582

119894+1

(119890 minus 1)2120587

119875119894+1

CTH10158401198943(0 120582119894 120582119894+1

) =1

2(119875119894+1


)


CTH10158401198943(1 120582119894 120582119894+1

) =1

2(119875119894+2

minus 119875119894)

CTH101584010158401198943(0 120582119894 120582119894+1

)


4 (119890 minus 1)


minus 2119875119894+ 119875119894+1

)

CTH101584010158401198943(1 120582119894 120582119894+1

)


4 (119890 minus 1)

times (119875119894minus 2119875119894+1

+ 119875119894+2

)

(7)


(1 120582119894 120582119894+1

) = CTH(119896)1198943(0 120582119894 120582119894+1

) (119896 =





119894=


1205823








= 120582119894+1


qth1198942(05) = 119875



119894= (2 minus 120587)(119890 minus 1)

2(2(119890 minus

1)2minus 2120587) = 891206 cth

1198943(0) = 119875

119894 120582119894+1

= (2 minus 120587)(119890 minus

1)2(2(119890 minus 1)

2minus 2120587) = 891206 cth

1198943(1) = 119875





5= 1205826=


5= 891206




119894= 120582119894+1

=



P1

P2P3

P4

P5


P1

P2P3

P4

P5


P1

P2P3

P4

P5



P1

P2P3

P4

P5




119894minus1 119875119894 and

119875119894+1

as follows

119875119894minus1

= (119898 minus 119886 119899 + 119887)

119875119894= (119898 + 119886 119899 + 119887)

119875119894+1

= (119898 + 119886 119899 minus 119887)

(119886119887 = 0)

(8)


119875119894minus1

= (119898 119899 minus120587

2119887)

119875119894= (119898 +

120587

2119886 119899)

119875119894+1

= (119898 119899 minus120587

2119887)

119875119894+2

= (119898 minus120587

2119886 119899)

(119886119887 = 0)

(9)



119909 (119905) = 119898 + 119886 cos 1205872119905

119910 (119905) = 119899 + 119887 sin 120587

2119905

(10)


119875119894minus1

= (119898 + 119886 119899 +1 minus 119890

1 + 119890119887)

119875119894= (119898 + 119886 119899

1 minus 119890

1 + 119890119887)

119875119894+1

= (119898 +1198902minus 119890 + 1

119890119886 119899 +

1198903minus 1

1198902 + 119890119887)

(119886119887 = 0)

(11)


= (119898 + ((1198902+ 1)119890)119886 119899 minus ((119890

4+ 1)(119890

3minus 119890))119887) 119875

119894= (119898 +

119886 119899 + ((1198902+ 1)(119890

2minus 1))119887) 119875

119894+1= (119898 119899 + (2119890(119890

2minus 1))119887)

119875119894+2

= (119898 minus 119886 ((1198902+ 1)(119890



119909 (119905) = 119898 + 119886 cosh 119905

119910 (119905) = 119899 + 119887 sinh 119905(12)



When parameters 120582119894= 120582119894+1


119875119894minus1

= (120587 minus 2

4119886

4 + 120587

4119886)

119875119894= (

2 minus 120587

4119886

4 minus 120587

4119886)

119875119894+1

= (6 minus 3120587

4119886

120587 + 4

4119886)

119875119894+2

= (10 minus 120587

4119886

4 + 3120587

4119886)

(119886 = 0)

(13)


119909 (119905) = 119886 (119905 minus sin 120587

2119905)

119910 (119905) = 119886 (1 minus cos 1205872119905)

(14)


119894minus1= (119898 + 2119886 119899 + ((119890

4+ 1)(119890

3minus

119890))119887)119875119894= (119898+119886 119899+((119890

2+1)(119890

2minus1)))119875

119894+1= (119898 119899+(2119890(119890

2minus

1))119887) and 119875119894+2

= (119898 minus 119886 119899 + ((1198902+ 1)(119890

2minus 1))119887) (119886119887 = 0) as



119909 (119905) = 119898 + 119886119905

119910 (119905) = 119899 + 119887 cosh 119905(15)



Piminus1

Pi+1

Pi

(a)

Pi+1

Pi+2

Piminus1

Pi

(b)


Piminus1Pi

Pi+1

(a)

Piminus1

Pi+2

Pi+1

Pi

(b)


Pi+2

Pi+1

Piminus1

Pi





0= 1198756= (minus3 (119890minus

1)(119890+1)) 1198751= 1198757= (minus3 (1minus 119890)(119890+1)) 119875

2= (3 (119890minus1)(119890+

1))1198753= (3 (1minus119890)(119890+1))119875

4= (4minus119890minus(1119890) (119890

3minus1)(119890

2+119890))

1198755

= (119890 minus 4 + (1119890) (1198903minus 1)(119890


Piminus1

Pi+2

Pi+1

Pi


120582119894= (0 0 1 1 505952 505952 0) (119894 = 1 2 7) The 1st



5=




P5

1

3


splines

1

10

6 5

P5

P4


splines

as follows 1198750= ((120587 minus 2)2 1) 119875

1= (0 (2 minus 120587)2) 119875

2= ((2 minus

120587)2 1) 1198753= (2 (2 +120587)2) 119875

4= (2 (119890

4+ 1198903minus 119890+ 1)(119890

3minus 119890))

1198755

= (1 21198902(1198902minus 1)) 119875

6= (0 (119890

2+ 2119890 minus 1)(119890

2minus 1))

1198757

= (minus1 21198902(1198902minus 1)) 119875

8= (minus(119890

2+ 1)119890 (119890

4+ 1198903minus

119890 + 1)(1198903minus 119890)) 119875

9= (minus(120587 + 4)2 0) 119875


11987511

= ((120587 minus 4)2 0) 11987512



3= 1205824= 120938 the


5



Acknowledgments


References























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










CTH10158401198943(1 120582119894 120582119894+1

) =1

2(119875119894+2

minus 119875119894)

CTH101584010158401198943(0 120582119894 120582119894+1

)


4 (119890 minus 1)


minus 2119875119894+ 119875119894+1

)

CTH101584010158401198943(1 120582119894 120582119894+1

)


4 (119890 minus 1)

times (119875119894minus 2119875119894+1

+ 119875119894+2

)

(7)


(1 120582119894 120582119894+1

) = CTH(119896)1198943(0 120582119894 120582119894+1

) (119896 =





119894=


1205823








= 120582119894+1


qth1198942(05) = 119875



119894= (2 minus 120587)(119890 minus 1)

2(2(119890 minus

1)2minus 2120587) = 891206 cth

1198943(0) = 119875

119894 120582119894+1

= (2 minus 120587)(119890 minus

1)2(2(119890 minus 1)

2minus 2120587) = 891206 cth

1198943(1) = 119875





5= 1205826=


5= 891206




119894= 120582119894+1

=



P1

P2P3

P4

P5


P1

P2P3

P4

P5


P1

P2P3

P4

P5



P1

P2P3

P4

P5




119894minus1 119875119894 and

119875119894+1

as follows

119875119894minus1

= (119898 minus 119886 119899 + 119887)

119875119894= (119898 + 119886 119899 + 119887)

119875119894+1

= (119898 + 119886 119899 minus 119887)

(119886119887 = 0)

(8)


119875119894minus1

= (119898 119899 minus120587

2119887)

119875119894= (119898 +

120587

2119886 119899)

119875119894+1

= (119898 119899 minus120587

2119887)

119875119894+2

= (119898 minus120587

2119886 119899)

(119886119887 = 0)

(9)



119909 (119905) = 119898 + 119886 cos 1205872119905

119910 (119905) = 119899 + 119887 sin 120587

2119905

(10)


119875119894minus1

= (119898 + 119886 119899 +1 minus 119890

1 + 119890119887)

119875119894= (119898 + 119886 119899

1 minus 119890

1 + 119890119887)

119875119894+1

= (119898 +1198902minus 119890 + 1

119890119886 119899 +

1198903minus 1

1198902 + 119890119887)

(119886119887 = 0)

(11)


= (119898 + ((1198902+ 1)119890)119886 119899 minus ((119890

4+ 1)(119890

3minus 119890))119887) 119875

119894= (119898 +

119886 119899 + ((1198902+ 1)(119890

2minus 1))119887) 119875

119894+1= (119898 119899 + (2119890(119890

2minus 1))119887)

119875119894+2

= (119898 minus 119886 ((1198902+ 1)(119890



119909 (119905) = 119898 + 119886 cosh 119905

119910 (119905) = 119899 + 119887 sinh 119905(12)



When parameters 120582119894= 120582119894+1


119875119894minus1

= (120587 minus 2

4119886

4 + 120587

4119886)

119875119894= (

2 minus 120587

4119886

4 minus 120587

4119886)

119875119894+1

= (6 minus 3120587

4119886

120587 + 4

4119886)

119875119894+2

= (10 minus 120587

4119886

4 + 3120587

4119886)

(119886 = 0)

(13)


119909 (119905) = 119886 (119905 minus sin 120587

2119905)

119910 (119905) = 119886 (1 minus cos 1205872119905)

(14)


119894minus1= (119898 + 2119886 119899 + ((119890

4+ 1)(119890

3minus

119890))119887)119875119894= (119898+119886 119899+((119890

2+1)(119890

2minus1)))119875

119894+1= (119898 119899+(2119890(119890

2minus

1))119887) and 119875119894+2

= (119898 minus 119886 119899 + ((1198902+ 1)(119890

2minus 1))119887) (119886119887 = 0) as



119909 (119905) = 119898 + 119886119905

119910 (119905) = 119899 + 119887 cosh 119905(15)



Piminus1

Pi+1

Pi

(a)

Pi+1

Pi+2

Piminus1

Pi

(b)


Piminus1Pi

Pi+1

(a)

Piminus1

Pi+2

Pi+1

Pi

(b)


Pi+2

Pi+1

Piminus1

Pi





0= 1198756= (minus3 (119890minus

1)(119890+1)) 1198751= 1198757= (minus3 (1minus 119890)(119890+1)) 119875

2= (3 (119890minus1)(119890+

1))1198753= (3 (1minus119890)(119890+1))119875

4= (4minus119890minus(1119890) (119890

3minus1)(119890

2+119890))

1198755

= (119890 minus 4 + (1119890) (1198903minus 1)(119890


Piminus1

Pi+2

Pi+1

Pi


120582119894= (0 0 1 1 505952 505952 0) (119894 = 1 2 7) The 1st



5=




P5

1

3


splines

1

10

6 5

P5

P4


splines

as follows 1198750= ((120587 minus 2)2 1) 119875

1= (0 (2 minus 120587)2) 119875

2= ((2 minus

120587)2 1) 1198753= (2 (2 +120587)2) 119875

4= (2 (119890

4+ 1198903minus 119890+ 1)(119890

3minus 119890))

1198755

= (1 21198902(1198902minus 1)) 119875

6= (0 (119890

2+ 2119890 minus 1)(119890

2minus 1))

1198757

= (minus1 21198902(1198902minus 1)) 119875

8= (minus(119890

2+ 1)119890 (119890

4+ 1198903minus

119890 + 1)(1198903minus 119890)) 119875

9= (minus(120587 + 4)2 0) 119875


11987511

= ((120587 minus 4)2 0) 11987512



3= 1205824= 120938 the


5



Acknowledgments


References























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










P1

P2P3

P4

P5




119894minus1 119875119894 and

119875119894+1

as follows

119875119894minus1

= (119898 minus 119886 119899 + 119887)

119875119894= (119898 + 119886 119899 + 119887)

119875119894+1

= (119898 + 119886 119899 minus 119887)

(119886119887 = 0)

(8)


119875119894minus1

= (119898 119899 minus120587

2119887)

119875119894= (119898 +

120587

2119886 119899)

119875119894+1

= (119898 119899 minus120587

2119887)

119875119894+2

= (119898 minus120587

2119886 119899)

(119886119887 = 0)

(9)



119909 (119905) = 119898 + 119886 cos 1205872119905

119910 (119905) = 119899 + 119887 sin 120587

2119905

(10)


119875119894minus1

= (119898 + 119886 119899 +1 minus 119890

1 + 119890119887)

119875119894= (119898 + 119886 119899

1 minus 119890

1 + 119890119887)

119875119894+1

= (119898 +1198902minus 119890 + 1

119890119886 119899 +

1198903minus 1

1198902 + 119890119887)

(119886119887 = 0)

(11)


= (119898 + ((1198902+ 1)119890)119886 119899 minus ((119890

4+ 1)(119890

3minus 119890))119887) 119875

119894= (119898 +

119886 119899 + ((1198902+ 1)(119890

2minus 1))119887) 119875

119894+1= (119898 119899 + (2119890(119890

2minus 1))119887)

119875119894+2

= (119898 minus 119886 ((1198902+ 1)(119890



119909 (119905) = 119898 + 119886 cosh 119905

119910 (119905) = 119899 + 119887 sinh 119905(12)



When parameters 120582119894= 120582119894+1


119875119894minus1

= (120587 minus 2

4119886

4 + 120587

4119886)

119875119894= (

2 minus 120587

4119886

4 minus 120587

4119886)

119875119894+1

= (6 minus 3120587

4119886

120587 + 4

4119886)

119875119894+2

= (10 minus 120587

4119886

4 + 3120587

4119886)

(119886 = 0)

(13)


119909 (119905) = 119886 (119905 minus sin 120587

2119905)

119910 (119905) = 119886 (1 minus cos 1205872119905)

(14)


119894minus1= (119898 + 2119886 119899 + ((119890

4+ 1)(119890

3minus

119890))119887)119875119894= (119898+119886 119899+((119890

2+1)(119890

2minus1)))119875

119894+1= (119898 119899+(2119890(119890

2minus

1))119887) and 119875119894+2

= (119898 minus 119886 119899 + ((1198902+ 1)(119890

2minus 1))119887) (119886119887 = 0) as



119909 (119905) = 119898 + 119886119905

119910 (119905) = 119899 + 119887 cosh 119905(15)



Piminus1

Pi+1

Pi

(a)

Pi+1

Pi+2

Piminus1

Pi

(b)


Piminus1Pi

Pi+1

(a)

Piminus1

Pi+2

Pi+1

Pi

(b)


Pi+2

Pi+1

Piminus1

Pi





0= 1198756= (minus3 (119890minus

1)(119890+1)) 1198751= 1198757= (minus3 (1minus 119890)(119890+1)) 119875

2= (3 (119890minus1)(119890+

1))1198753= (3 (1minus119890)(119890+1))119875

4= (4minus119890minus(1119890) (119890

3minus1)(119890

2+119890))

1198755

= (119890 minus 4 + (1119890) (1198903minus 1)(119890


Piminus1

Pi+2

Pi+1

Pi


120582119894= (0 0 1 1 505952 505952 0) (119894 = 1 2 7) The 1st



5=




P5

1

3


splines

1

10

6 5

P5

P4


splines

as follows 1198750= ((120587 minus 2)2 1) 119875

1= (0 (2 minus 120587)2) 119875

2= ((2 minus

120587)2 1) 1198753= (2 (2 +120587)2) 119875

4= (2 (119890

4+ 1198903minus 119890+ 1)(119890

3minus 119890))

1198755

= (1 21198902(1198902minus 1)) 119875

6= (0 (119890

2+ 2119890 minus 1)(119890

2minus 1))

1198757

= (minus1 21198902(1198902minus 1)) 119875

8= (minus(119890

2+ 1)119890 (119890

4+ 1198903minus

119890 + 1)(1198903minus 119890)) 119875

9= (minus(120587 + 4)2 0) 119875


11987511

= ((120587 minus 4)2 0) 11987512



3= 1205824= 120938 the


5



Acknowledgments


References























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Piminus1

Pi+1

Pi

(a)

Pi+1

Pi+2

Piminus1

Pi

(b)


Piminus1Pi

Pi+1

(a)

Piminus1

Pi+2

Pi+1

Pi

(b)


Pi+2

Pi+1

Piminus1

Pi





0= 1198756= (minus3 (119890minus

1)(119890+1)) 1198751= 1198757= (minus3 (1minus 119890)(119890+1)) 119875

2= (3 (119890minus1)(119890+

1))1198753= (3 (1minus119890)(119890+1))119875

4= (4minus119890minus(1119890) (119890

3minus1)(119890

2+119890))

1198755

= (119890 minus 4 + (1119890) (1198903minus 1)(119890


Piminus1

Pi+2

Pi+1

Pi


120582119894= (0 0 1 1 505952 505952 0) (119894 = 1 2 7) The 1st



5=




P5

1

3


splines

1

10

6 5

P5

P4


splines

as follows 1198750= ((120587 minus 2)2 1) 119875

1= (0 (2 minus 120587)2) 119875

2= ((2 minus

120587)2 1) 1198753= (2 (2 +120587)2) 119875

4= (2 (119890

4+ 1198903minus 119890+ 1)(119890

3minus 119890))

1198755

= (1 21198902(1198902minus 1)) 119875

6= (0 (119890

2+ 2119890 minus 1)(119890

2minus 1))

1198757

= (minus1 21198902(1198902minus 1)) 119875

8= (minus(119890

2+ 1)119890 (119890

4+ 1198903minus

119890 + 1)(1198903minus 119890)) 119875

9= (minus(120587 + 4)2 0) 119875


11987511

= ((120587 minus 4)2 0) 11987512



3= 1205824= 120938 the


5



Acknowledgments


References























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










P5

1

3


splines

1

10

6 5

P5

P4


splines

as follows 1198750= ((120587 minus 2)2 1) 119875

1= (0 (2 minus 120587)2) 119875

2= ((2 minus

120587)2 1) 1198753= (2 (2 +120587)2) 119875

4= (2 (119890

4+ 1198903minus 119890+ 1)(119890

3minus 119890))

1198755

= (1 21198902(1198902minus 1)) 119875

6= (0 (119890

2+ 2119890 minus 1)(119890

2minus 1))

1198757

= (minus1 21198902(1198902minus 1)) 119875

8= (minus(119890

2+ 1)119890 (119890

4+ 1198903minus

119890 + 1)(1198903minus 119890)) 119875

9= (minus(120587 + 4)2 0) 119875


11987511

= ((120587 minus 4)2 0) 11987512



3= 1205824= 120938 the


5



Acknowledgments


References























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









research article a geometric modeling method based on th...

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