Principles of Computer-Aided Design and Manufacturing Second Edition 2004 ISBN 0-13-064631-8

Principles of Computer-Aided Design and Manufacturing

Second Edition 2004

ISBN 0-13-064631-8

Author: Prof. Farid. Amirouche

University of Illinois-Chicago

University of Illinois-Chicago

Chapter 4

Description of Curves and Surfaces

Principles of Computer-Aided Design and Manufacturing Second Edition 2004 – ISBN 0-13-064631-8

Author: Prof. Farid. Amirouche, University of Illinois-Chicago

CHAPTER 4 4.1 Line Fitting

4.1 LINE FITTING • Suppose we desire to fit a linear function to

the data set, as illustrated in Table 4.1.

i x y

1 xi yi

2 xi+1 yi+1

3 xi+2 yi+2

Table 4.1




21 cxcg(x)

1,2....Li,cxcyxgyr 2i1iiii

22i1i

L

1i

2i

L

1icxcyrR

0cxcyx2c

R2i1ii

L

1i1

0cxcy2c

R2i1i

L

1n2

2

1

2

1

2,22,1

1,21,1

z

z

c

c

aa

aa

We have two equations and two unknowns and the coefficient are given by :

(4.1)

(4.2)

(4.3)

(4.4)

(4.5)

(4.6)

points. data ofnumber total theis L where



2i

L

1i1,1 xa

i

L

1i2,11,2 xaa

LaL

1i2,2

0ix

ii

L

1i1 yxz

L

1ii2 yz

2,11,22,21,121,212,21 aaaa/zazac

2,11,22,21,112,221,12 aaaa/zazac

(4.7)

(4.8)

(4.9)

(4.10)

(4.11)

(4.12)

(4.13)

The solution to equation (4.6) is found by Cramer’s rule




Example 4.1

Determine the regression line for the data in Table 4.2 by solving Equation (4.6). After the regression line is obtained, examine the deviation error of the line from the data. Table 4.2

i xi yi x2i xiyi

1 0.1 0.22 0.01 0.022

2 0.2 0.39 0.04 0.078

3 0.3 0.57 0.09 0.171

4 0.4 0.81 0.16 0.324

5 0.5 1.02 0.25 0.51

6 0.6 1.18 .36 0.708

Total 2.1 4.19 0.91 1.813

a21 z2 a11 z1




Solution:

19.4,6,1.2

813.1,1.2,91.0

22,21,2

12,11,1

zaa

zaa

19.4

813.1

61.2

1.291.0

2

1

c

c

0053.,98.1 21 cc

0.00531.98xxg

i xi yi g=c1x+c2 Deviation (error)

1 0.1 0.22 0.2033 0.0167

2 0.2 0.39 0.4013 -0.0113

3 0.3 0.57 0.5993 -0.0293

4 0.4 0.81 0.7973 0.0127

5 0.5 1.02 0.9953 0.0247

6 0.6 1.18 1.1933 -0.0753

TABLE 4.3


x

0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

x

y

y=1.98x+.0053

Figure 4.1 The line fitted to the data






CHAPTER 4 4.2 Nonlinear Curve Fitting

4.2 NONLINEAR CURVE FITTING WITH A POWER

FUNCTION αβxxg βlogxαlogglog

xlogX

log(βoc

αc

log(g)G

2

1

21 cXcG where

(4.14)

(4.15)

(4.16)

(4.17)



Example 4.2

A following data set is used to demonstrate how curve fitting of a power function can be carried out making use of the regression line technique. Consider Table 4.4, when x, y represent experimental data between force (lbs) and displacement (mm). We need to find a mathematical function to describe the data and it is perceived that a power function is most suitable.

i 1 2 3 4 5 6 7 8 9 10 11 12 Total

x 0.1 0.25 0.39 0.60 1.03 1.32 1.78 2.13 2.45 3.07 3.98 4.64

y 3.21 3.81 4.09 5.21 7.97 8.32 8.88 9.27 9.97 10.8 11.34 13.08

X=log(x) -1 -.602 -.408 -.22 .0128 .1205 0.25 0.328 .389 0.487 .60 0.666 .6233

Y=log(y) 0.506 .580 0.611 0.716 0.9014 0.920 0.948 .967 .998 1.033 1.054 1.116 10.35

X2 1 .3624 .1664 .0484 0.0001 0.014 .0625 .1075 0.151 .2371 .36 .443 2.9524

XY -.506 -.349 -.249 -.1575 .0115 .1108 .237 .3171 .388 .5030 0.6324 .7432 1.6815

Table 4.4


x x x

2 4 6 8 10 12 14 0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

x

g(x)



35.10,12,6233.0

6815.1,6233.0,9524.2

22,21,2

12,11,1

zaa

zaa

9405.1)log(,3917.0 21 cc

0.3197α 6.9622xβxxg

Figure 4.2 The curve fitted to the data

C2 = 0.8422

β =2.3215




4.3 CURVE FITTING WITH A HIGHER-ORDER

POLYNOMIAL

11

21 ... n

nn cxcxcxg

CHAPTER 4 4.3 Higher order Curve Fitting

Considering a set of data (xi, yi).

Let us try to interpolate the data with a polynomial of order n :

xi yi

x1 y1

x2 y2

…. …..

xL yL




LnnL

n

nn

y

y

y

y

c

c

c

c

x

x

xx

A.

,.

,

1..

....

1..

1.

2

1

1

2

1

2

111

** ycAoryAAcA tt ** 1

yAc

yAyandAAA tt **

11

21

11

32313

11

22212

11

12111

...

...........................................

...

...

...

nn

Ln

LL

nnn

nnn

nnn

cxcxcy

cxcxcy

cxcxcy

cxcxcyThe system can be written :

In a matrix form :



In order to find the best fit, the error needs to be minimized :

11

21 ... n

nn cxcxcxg

Lixgyr iii ,...2,1,

L

iirR

1

21,...,2,1,0

njc

R

j

1,...2,1,1

11

1 1

22

nkyxcxL

ii

knij

n

j

L

i

kjni

L

ii

L

ii

ni

L

ii

ni

nL

ii

L

i

ni

L

i

ni

L

i

ni

L

i

ni

L

i

ni

L

i

ni

y

yx

yx

c

c

c

xx

xx

xxx

1

1

1

1

1

2

1

1

0

1

1

1

1

12

11

12

1

2

..

..

....

..

.

yAc

(4.18)

(4.19)

(4.21)(4.20)

(4.22)

(4.23)(4.24)




Example 4.3

A data set of a biomechanical experiment is provided in Table 4.5. Find a polynomial of order 12 that best fits the data.




Solution:

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 2

4

6

8

10

12

14

X

Y

Figure 4.3 Plot of the quadratic polynomial fitted




CHAPTER 4 4.4 Chebyshev Polynomial Fit

4.4 CHEBYSHEV POLYNOMIAL FIT

1. A Chebyshev polynomial is defined over the interval [-1,1].

2. The range of the independent variable must then be

3. The zeroth-order Chebyshev polynomial is

4. The first-order Chebyshev polynomial is

5. The second-order Chebyshev polynomial is

The definition of a Chebyshev polynomial is contained in the following rules:

.1...1 10 nxxx

.1)(0 xT

.)(1 xxT

.12)( 22 xxT



)()(2)( 21 xTxxTxT kkk

n

kkk xTaxf

0

).()(

Example 4.4

Figure 4.4 Free Body Analysis of a Vehicle on a Road

(4.29)

(4.30)




56.0448.0336.0224.0112.00]56.0,0[],[ ba

.12

ab

axI

,112

02

xx

I

,16.02.02.06.01 xI

xTaxTaxTaxTaxTaxTaxf o 55443322110 )()()()(

,1)( xTo

,)(1 xxT

,12 22 xxT

,342 3123 xxxTxTxxT

,188)( 244 xxxT

.52016)( 355 xxxxT

where

(4.31)

(4.32)

(4.33)




The approximating function becomes

1883412 244

33

2210 xxaxxaxaxaaxf

xxxa 52016 355

0

176.1

902.1

902.1

176.1

0

111111

0788.0

845.0

843.0

693.0

936.0

568.0

28.0

92.0

6.0

2.0

1

1

84512.06928.0568.092.02.01

07584.08432.0936.028.06.01

111111

5

4

3

2

1

0

a

a

a

a

a

a

533.203490.001844.20 a

543210 533.20023.03490.00004.01844.20019.0)( TTTTTTxf

(4.34)

(4.35)

(4.36)




VALUE OF X IN THE FUNCTION Y=2*SIN X

RESULTS FROM APPROXIMAT

ION

DESIRED RESULTS

0.1 1.6071 0.1997

0.6 4.1959 1.1293

1.1 2.0987 1.7824

1.6 -0.6599 1.9991

2.1 -2.0871 1.7264

2.6 -1.7240 1.0310

3.1 -0.1475 0.0832

3.6 1.5274 -0.8850

4.1 2.1458 -1.6366

4.6 1.0104 -1.9874

5.1 -1.6205 -1.8516

5.6 -4.0349 -1.2625

6.1 -2.5692 -0.3643

TABLE 4.6




CHAPTER 4 4.5 Fourier Series

4.5 FOURIER SERIES OF DISCRETE SYSTEMS

• By performing a variable transformation, we can transform the physical interval by using a new independent variable that has the range from some given interval . We, then subdivide this interval into 2N equally spaced parts by using . The function is then known at the points . There are 2N known values of the function through which the series will be fitted. Then we have



),0(0 fy ),(1 fy

].)1(2[12 Nfy N

.

.

.

m

jjj jbja

af

1

0 sincos2

)cos...2coscos(2 21

0 maaaa

f m )sin...2sinsin( 21 mbbb m

0

cos.2

djfa j .,....,2,1,0 mj

0

sin.2

djfb j .,....,2,1,0 mj

where is the Time Period.

j

(4.38)

(4.39)

(4.41)

(4.42)




12

0

2])([N

kkyfE

kkj jfN

a cos1 ,,.....,2,1,0 mj

kkj jfN

b sin)(1

,,.....,2,1,0 mj

where

)./( Nkkk

Figure 4.5 Mass M with Support Motion

(4.43)

(4.44)

(4.45)

3

23

4 2

f 3 3 0

0

0




)2sinsin()2coscos(2 2121

0 bbaaa

f

2sinsin 21 bbf

kkfN

b sin1

1

2sin*0

3

4sin*3

3

2sin*30sin*0

2

1

5.1

kkfN

b 2sin1

2

4sin*0

3

8sin*3

3

4sin*30sin*0

2

1

5.1 2sin5.1sin5.1 f

(4.46)

(4.47)

(4.48)

(4.49)

(4.50)

We apply Fourier series method to the data and use two-term Fourier series.

Because the function is odd all a’s are zeros.




0 1 2 3 4 5 6 7-3

-2

-1

0

1

2

3

X

Y

Figure 4.6 Graph for 2sin5.1sin5.1 f

f(y=2sin




2sinsin 21 bbf

kkfN

b sin1

1

29.154sin*8676.086.102sin*95.143.51sin*56.10sin*0(4

1

)360sin*058.308sin*56.115.257sin*95.172.205sin*868.0

75.1 kkf

Nb 2sin

12

58.308sin*8676.072.205sin*95.186.102sin*56.10sin*0(4

1

)720sin*016.617sin*56.13.514sin*95.144.411sin*868.0

sin75.1f

0

(4.52)

(4.53)(4.54)

2N=8


0 1 2 3 4 5 6 7-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

X

Y



Figure 4.7 Graph for sin75.1f

732.1176.1416.0416.0176.1732.1989.1902.1486.1813.00

240216192168144120967248240

f 0813.0486.1902.1989.1

360336312288264

y=2sin

f(




kkfN

b sin1

1

2sinsin 21 bbf

120sin*732.196sin*989.172sin*902.148sin*486.124sin*813.00sin*0(8

1

240sin*732.1216sin*176.1192sin*416.0168sin*416.0144sin*176.1

)360sin*0336sin*813.0312sin*486.1288sin*902.1264sin*989.1

876.1

kkfN

b 2sin1

2

240sin*732.1192sin*989.1144sin*902.196sin*486.148sin*813.00sin*0(8

1

480sin*732.1432sin*176.1384sin*416.0336sin*416.0288sin*176.1 )720sin*0672sin*813.0624sin*486.1576sin*902.1528sin*989.1

0

,sin876.1 f


0 1 2 3 4 5 6 7-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

X

Y



Figure 4.8 Graph for sin876.1f

f(

y=2sin




2sinsin 21 bbf




kkfN

b sin1

1




b2


0 1 2 3 4 5 6 7-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

X

Y

sin96.1fFigure 4.9 Graph for



f(

y=2sin


The benefits of using cubic splines are as follows:11. They reduce computational requirements and numerical instabilities that arise from higher-order curves.

2. They have the lowest degree space curve that allows inflection points.33. They have the ability to twist in space.



CHAPTER 4 4.6 Cubic Splines

4.6 CUBIC SPLINES

4

1

1)(i

iixaxy (1<i<4)

A spline is a smooth curve that can be generated by computer to go through a set of data points. The mathematical spline derives from its physical counterpart - the thin elastic beam. Because the beam is supported at specified points (we call them knots), it can be shown that its deflection (assumed small) is characterized by a polynomial of order three, hence a cubic spline. It is not a mere coincidence that the principle of explaining the deflection of beams under different loads results into a function of a third order.

(4.55)



4.7 PARAMETRIC CUBIC SPLINES

33,

22,1,,)( tatataatS iiioii

niforYXPSa iiiii .,.,.1),(0,

iii yxa 0,

niforYYXXt iiiii .,..,2121

211

.1,..,.10

,,,,,)(

1

33,3,

22,2,1,1,0,0,

niandttwhere

taataataaaatStStS

i

yixiyixiyixiyixiyixii

Consider a set of data points described in the x-y plane by (xi yi) with i=1,

…,n. Our objective is to pass a parametric cubic spline between all these points. A parametric cubic spline is a curve that is represented as a function of one or more parameters.

(4.56)

(4.57)

(4.58)

(4.59)

CHAPTER 4 4.7 Parametric Cubic Splines



UnknownsnaS

PtsControlnKnownsYXPaSTherefore

atataatd

tSdtSS

atatataatSS

i

iiiii

itiii

t

iii

itiiiiii

1,1

0,

1,0

23,2,1,

0

0,0

33,

22,1,0,

'

:),(

32)(

)0(''

)0(

3,3

3'"

3,2,2

2

23,2,1,

'

6)(

)(

62)(

)("

32)(

)(

ii

i

iii

i

iiii

i

atd

tSdtS

taatd

tSdtS

tataatd

tSdtS

33,

22,11 ')( tatatSStS iii

111

1111

')0(')('

)0()(

iiii

iiiii

StSttS

PStSttS

(4.60)

(4.61)

(4.62)

(4.63)

(4.64)

(4.65)

(4.66)

(4.67)




12

13,12,

13

13,2

12,1

''

'

iiiiii

iiiiiiii

StataS

StatatSS

)''(1

)(2

)2(13

121

131

3,

''1

112

12,

iii

iii

i

iii

iii

i

SSt

SSt

a

SSt

SSt

a

322

'2

22

'1

32

)212

2

'2

2

'1

22

12'11

(22)(3)( t

t

S

t

S

t

SSt

t

S

t

S

t

SSSStSi

Therefore, the spline function between P1 & P2 could simply be expressed as

(4.68)

(4.69)

(4.70)

(4.71)




IIn the context of computer graphics and general-purpose algorithm development, we need to ask the following questions: 11. How can we generate a solution for and for all cubic functions S i(t), Si+1(t), .

. . Sn(t)?

22. How do we select t, t1, and t2 for a given set of data points?

3. How do we assure continuity between the splines at knots P1, P2,. . . ,

Pn?

32

1

'1

21

'1

31

12

1

'1

1

'

21

1' )(22)(3)( t

t

S

t

S

t

SSt

t

S

t

S

t

SStSStS

i

i

ii

ii

i

i

i

i

i

iiiii

taatS iii 3,2," 62

2," 20 ii aS

23,2,2" 62 taatS iii

0)( "12

" ii StS

(4.72)

(4.73)

(4.74)

(4.75)

(4.76)




)21()()/3(

)(2

12

2122

121

'21

'112

'2

niSStSSttt

StSttSt

iiiiiiii

iiiiiii

)()(3

)(33

3

(200

0)(200

0020

0002

212

12

11

23244

23

43

122323

22

32

'

'3

'2

'1

)1

4545

3434

2323

nnnnnnnn

nnnnn

SStSSttt

SStSSttt

SStSSttt

S

S

S

S

tttt

tttt

tttt

tttt

ti+2S’i


i



Boundary Conditions

a) Natural Spline:

00"11" tSS

0)("" 1 nnn ttSS

212'2

'1 /)(5.15.0 tSSSS

)(/642 1''

1 nnnnn SStSS

b) Clamped Spline:

The boundary conditions for this spline are such that the first derivatives (slope) at t=0 and t=tn are specified.

(4.79)

(4.80)

(4.81)

(4.82)


Adding Equations (4.81) and (4.82) to the n-2 equations given by Equation (4.78) we can solve for all the S’.



TThe parametric cubic spline between any two points is constructed as follows: 11. Find the maximum cord length and determine t1, t2, . . . ,tn. 22. Use Equation (4.78) together with the corresponding boundary conditions to solve for the , , . . .. , . 33. Solve for the coefficients that make up the parametric cubic splines using equations (4.62), (4.69) and (4.70).

Summary




Example 4.4 For following data set (1,1), (1.5,2), (2.5,1.75) & (3.0,3.25). Find the parametric cubic spline assuming a relaxed condition at both ends of the data.

Solution:

21

211 iiiii YYXXt


We first compute the cord length



3222313

122

21

''2')1

)(3

''2)0

StSttSti

SSt

SSi

122323

22

32

3SStSSt

tt

4333424 ''2')2 StSttSti 232434

23

43

3SStSSt

tt

)(3

'2')3 344

43 SSt

SSi

t32

t42

The above equations are found using boundary conditions given by equations(4.81), (4.82) and (4.77).

Equation (4.78) in notational form is siT CSC '




2100

031.1476.3707.00

0118.1298.4031.1

0012

TCwhere

4

34

232434

23

43

4

34

232434

23

43

122323

22

3212

2323

22

32

2

12

2

12

3

3

3

3

33

33

t

SS

SStSSttt

t

SS

SStSSttt

SStSSttt

SStSSttt

t

SS

t

SS

C

YY

yyyy

XX

xxxx

YYYYXXXX

YYXX

s

121.2121.2

672.1245.4

952.1637.4

683.2342.1

(4.86)

(4.85)

Last Eqn

t2t3

t42

t42




sTi CCS 1'

1,4

1,3

1,2

1,1

9745.06217.0

1720.08776.0

0996.07836.0

2917.12792.0

a

aa

a

= (4.87)

Since we have three splines we need to compute three coefficientsof ai,2 and ai,3.

To solve for Si’ we multiply equation (4.84) by [CT]-1

to get the ai,1 constants .




3,2,1'2'1

)(

31

12

1

12,

iforSStt

SSa ii

ii

iii

135.1361.0

067.1452.0

00

2,3

2,2

2,1

a

a

a

ii

i

ii

i

i SSt

SSt

a ''12

121

131

3,

536.0171.0

713.0263.0

317.0135.0

3,3

3,2

3,1

a

a

a

(4.88)

(4.90)

(4.89)

(4.91)

Si+1

Using equation (4.69) to find ai,2

Using equation (4.70) to find ai,3

(ti+1)3




32

3

322

321

536.0171.0135.1361.0172.0878.075.15.2

713.0263.0067.1452.01.0784.025.1

317.0135.000292.1279.011

tttS

tttS

tttS

1 1.5 2 2.5 3 3.5 1

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

2.8

X

Y

Figure 4.10 Parametric cubic curve

(4.92)

S1

S2

S3




CHAPTER 4 4.8 Nonparametric Cubic Spline

4.8 NONPARAMETRIC CUBIC SPLINE

32 dxcxbxaS(x)

y)(xS ii

1i1i1i1i y)(xSxS

A nonparametric cubic spline is defined as a curve having a function of only one parameter. Non-parametric cubic splines allow a direct variable relationship between the parameter value x and the value of the cubic spline function to be determined.

(4.93)

(4.94)

(4.95)

Cubic spline S(x) is composed of (n-1) cubic segment splines.

Each point has an x and y value.

For the interval [xi,xi+1] we can write



3ii

2iiiiii xxd)x(xc)x(xba(x)S

2iiiii

'i )x(x3d)x(x2cbS

)x(x6d2cS iii"i

(4.98)

(4.99)

(4.100)

The non-parametric cubic spline can be expressed as:

Its first and second derivatives are

)(xS)(xS 1i'

1i1i'i

)(xS)(xS 1i"

1i1i"i

(4.96)

(4.97)

By considering the smoothness and continuity of the cubic splines the following conditions are derived:




iiii ya)(xS

3ii

2iiiii1i1ii hdhchbaa)(xS

ii'i b)(xS

2iiiii1i1i

'1i1i

'i h3dh2cbb)(xS)(xS

iii1i1i"

1i1i"i h6d2c2c)(xS)(xS

(4.101)

(4.102)

(4.103)

(4.104)

(4.105)




i1ii xxh

where

3

)c(2ch

h

aab 1iii

i

i1ii

3

)c(2ch

h

aab

as expressed be alsocan b

i1i1i

1i

1iii

i

1i

1ii

i

i1i1iiii1i1i1i h

aa

h

aa3chchh2ch

(4.106)

(4.107)

(4.108)




2n

2n1n

1n

1nn

0

01

1

12

n

2

1

0

1n1n2n2n

322

2211

1100

h

aa

h

aa

h

aa

h

aa

3

c

c

c

c

hhh2h000

0hh2h00

0hhh2h0

00hhh2h

1.n1,...,iforAcH hi

3

)c(2ch

h

aab i1i1i

1i

1iii

1.n0,...,ifor3h

ccd

i

i1ii

(4.109)

(4.110)

(4.111)

(4.112)




4.9 BOUNDARY CONDITIONS

4.9.1 Natural Splines

0)(xS")(xS" n0

0cc n0

4.9.2 Clamped Splines

)(xf')(xS' 00

)(xf')(xS' nn

S”(x0)

(4.113)

(4.114)

(4.115)

(4.116)

When substituted into equation (4.105) yields

CHAPTER 4 4.9 Boundary Conditions



Example 4.6Find the nonparametric cubic spline (natural spline) for the points shown in the Table below.

Solution: Step 1: Control points. Intervals, and ai

Step 2: Solve for c1: Natural Spline (c0=c2=0) using equation ( 4.109 )

25.2

225.03 3

5.0

12

1

275.1301 15.0205.0

3 2

1

1

1

0

01

1

122111000

c

c

c

h

aa

h

aachchhch

i xi yi hi

0 1 1 0.5

1 1.5 2 1

n=2 2.5 1.75 -




Step 3: Solve for bi and di from equation 4.106)

1.n0,...,iforh3

ccd

1.n0,...,ifor3

)c(2ch

h

aab

i

i1ii

1iii

i

i1ii

1.5h3

ccd

2.3753

)c(2ch

h

aab

0i

0

010

100

0

010

0.75h3

ccd

1.253

)c(2ch

h

aab

1i

1

121

211

1

121




The results are compiled in the following table:

2.5x1.5 )x-0.75(x )x-2.25(x-)x-1.25(x2(x)S

1.5x1.0 )x-1.5(x-)x-2.375(x1(x)S

:bygiven are (4.98)equation from calculated splines theHence

3i

2ii2

3ii1

i xi hi yi=ai bi ci di

0 1 0.5 1 2.375 0 -1.5

1 1.5 1.0 2 1.25 -2.25 0.75

n=2 2.5 - 1.75 - 0 -




Figure 4.11: Nonparametric cubic spline function

s1s2




CHAPTER 4 4.10 Bezier Curves

4.10 BEZIER CURVES

iniin, t1t

i

n(t)J

The shapes of Bezier curves are defined by the position of the points, and the curves may not intersect all the given points except for the endpoints.

i)!(ni!

n!

i

n

where

*2)(n*1)(n*nn!

1)t(0(t)JStS in,i

n

1i

The curve points are defined by

where i=1 to n, and the Si contain the vector components of the various

points.

(4.117)

(4.118)

(4.119)



n

i

in, n

ini)(ni

i

n

n

iJ

The following example illustrates the Bezier curve method of curve fitting.

Example 4.7Define the Bezier Curve that passes through the following points:

5210 10 PP

1654 32 PP

Find the Bezier curve space that passes through these points.

Solution

33,3

23,2

23,1

3303,0

t(t)J

t)(13t(t)J

t)3(1(t)J

t)(1t)(1(1)t(t)J

3,323,223,113,00 JPJPJPJPS(t)

(4.120)

(4.121)

(4.122)




100 S

529.29.015.0 S

The resulting S (t) function is then found as

733.3102.2)35.0( S

43)5.0( S

733.3904.365.0 S

529.2099.5)85.0( S

16)1( S

TABLE 4.8 Evaluation of the Bezier function J3,1(I=0,1,2,3) in

terms of the parameter t.

t J3,0 J3,1 J3,2 J3,3

0 1 0 0 0

0.15 0.614 0.325 0.0574 0.0034

0.35 0.275 0.444 0.239 0.043

0.5 0.125 0.375 0.375 0.125

0.65 0.043 0.239 0.444 0.275

0.85 0.0034 0.0574 0.325 0.614

1 0 0 0 1


X

3.5 4 4.5 5 5.5 6 0

2

4

6

8

10

12

14

16

x

y



Figure 4.12 Bezier curve




4.11 DIFFERENTIATION OF BEZIER CURVE

EQUATION )10()(,

1

ttJStS ini

n

i

dt

tJStJStdS ini

n

iini

n

i)()( ,

1,

1

ini

in tti

n

dt

dtJ )1()(,

11 )1()1()1(

iniini tt

i

nntt

i

ni

)100.4()1()1()1()( 11

iini

iini Stt

i

nnStt

i

ni

dt

tdS

(4.123)

(4.124)

(4.125)

(4.126)




j

nn

j

nj

1

1)1(

i

nn

i

nin

1)(

iijni

n

i

SStti

nn

dt

tdS

1

11

0

)1(1)(

iini

n

ij

jnjn

j

Stti

ninStt

j

nj

dt

tdS 11

01

11

0

)1()()1(1

)1()(

(4.128)




CHAPTER 4 4.12 B-Spline Curve

4.12 B-SPLINE CURVE B-Splines were introduced to overcome some weaknesses in the Bezier curve. It seems that the number of control points affect the degree of the curve. Furthermore the properties of the blending functions used in the Bezier curve do not allow for an easier way to modify the shape of the curve locally.

)()( 11,0

nkkii

n

ittttNStS

1

1,1

1

1,,

)()()()()(

iki

kiki

iki

kiiki tt

tNtt

tt

tNtttN

where

0

1)(1, tN i rest theall

1 ii ttt

(4.129)

(4.130)

(4.131)



knin 2

(4.133) n ik 1

ki0 0

t

:knots periodicNon b)

(4.132) )k ni(0 k -i T

:knots Periodic a)

:knots of types twoare There

i

i

kn

ki




Example 4.8 Define the B-spline curve of order 3 for non-periodic uniform knots. The control points for the curve are given by P0, P1 and P2

Solution:

We obtain the (n+k+1) knot values as follows:

t0 = 0, t1 = 0, t2 = 0, t3 = 1, t4 = 1 and t5 = 1

(Note that n = 2 and k = 3)

Order 1. Let us compute all possible functions.




0

1)(

0

1)(

0

1)(

0

1)(

0

1)(

1,4

1,3

1,2

1,1

1,0

tN

tN

tN

tN

tN

else

else

else

else

else

ttt

ttt

ttt

ttt

ttt

5

4

3

2

1

4

3

2

1

0

(4.134)




(4.138) Stt)S2t(1St)(1S(t)

t(t)N

(4.137) t)2t(1(t)N

t)(1(t)N

(4.136) t

t)(t

tt

)Nt(t(t)N

and

t)(1

(4.135) N t)(1

tt

t)N(t

tt

)Nt(t(t)N

22

102

32,3

1,3

20,3

4

23

2,122,2

2,1

23

2,13

12

1,111,2

We obtain order 2 Ni,2 function as follows:

In a similar fashion, we obtain the Ni,3(t) functions for order 3.

Where S0, S1 and S2 correspond to control points P0,P1 and P2, respectively.




CHAPTER 4 4.13 Non-Uniform B-Spline Curve

4.13 NON-UNIFORM RATIONAL B-SPLINE CURVE (NURBS)

n

0iki,ii (t))N.x(hx.h

n

0iki,ii (t))N.y(hy.h

n

0iki,ii (t))N.z(hz.h

n

0iki,i (t)Nhh

n

0iki,i

n

0iki,ii

(t)Nh

(t)NShS(t)The equation for NURBS curve S(t) is given by:

(4.139)

(4.140)



Example 4.9Derive a NURBS representation of a quarter circle of radius 1. Let the arc be

defined in the (x, y) plane. Determine the corresponding coordinates of the control points, and the knot values.

Solution:




t0 = 0, t1 = 0, t2 = 0, t3 = 1, t4 = 1 and t5 = 1 h0 = 1,

2

2

2

11 h 12 h

t)2t(1(t)N

t)(1(t)N

t(t)N

t(t)N

t1(t)N

0

1(t)N

where

(t)Nh(t)Nh(t)Nh

(t)NSh(t)NSh(t)NShS(t)

1,3

20,3

22,3

2,2

1,2

2,1

2,321,310,30

2,3221,3110,300

(4.141)

(4.142)

(4.143)




with S0 = P0, S1 = P1 and S2 = P2 ; after substitution the NURBS equation is then found to be :

22

22

1tt)2t(12

2t)1.(1

t

0

1

0

1t)2t(1

0

1

1

2

2t)(1

0

0

1

1.

S(t)

(4.144)




CHAPTER 4 4.15 Plane Surface

4.15 PLANE SURFACE

Figure 4.14 Plane surface formed by intersecting lines



Figure 4.15 Plane surface formed by intersecting curves

CHAPTER 4 4.15 Plane Surface



CHAPTER 4 4.16 Ruled Surface

4.16 RULED SURFACE

Figure 4.16 Ruled surface formed by 2 Curves



CHAPTER 4 4.17 Rectangular Surface

4.17 RECTANGULAR SURFACE

Figure 4.17 Rectangular surface formed by 4 curves



CHAPTER 4 4.18 Surface of Revolution

4.18 SURFACE OF REVOLUTION

Figure 4.18 Revolved Surface



CHAPTER 4 4.19 Application Software

4.19 APPLICATION SOFTWARE

Different Ways to Create a Surface

• Extrude-Create

Figure 4.19 Plane surface



•Revolve-Create

Figure 4.20 Revolved surface




•Sweep-Create

Figure 4.21 Sweep surface




•Blend-Create

Figure 4.22 Blend surface




•Flat-Create

Figure 4.23 Flat surface




•Offset-Create

Figure 4.24 Offsetting of a surface




•Copy-Create

Figure 4.25 Copying of a surface by selection method


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Principles of Computer-Aided Design and Manufacturing Second Edition 2004 ISBN 0-13-064631-8

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