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Curves
Toni SellarèsUniversitat de Girona
2
Curves
• Many objects we want to model are not straight.
– Ex. Text, sketches, etc.
• How can we represent a curve?
– A large number of points on the curve.
– Approximate with connected line segments.• “piecewise linear approximation”
3
Accuracy/Space Trade-off
Problem
• Piecewise linear approximations require many pieces to look good (realistic, smooth, etc.).
• Set of individual curve points would take large amounts of storage.
Solution
• Higher-order formulae for coordinates on curve.
• If a simple formula won’t work, subdivide curve into pieces that can be represented by simple formulae.
• May still be an approx., but uses much less storage.
• Downside: harder to specify and render.
4
Curve Representations• Explicit: y = f(x)
• Example: y = x2
• The curve must be a function (only one value of y for each x)
• Parametric: (x(t),y(t)) = (f(t),g(t))
• Example: (x(t), y(t)) = (cos t, sin t)
• Easy to specify, modify, control
• Easy to join curve segments smoothly
• Implicit: F(x,y) = 0
• Example: x2-y2-r2=0
• Could exist several values of y for each x
• Need constraints to model just one part of a curve
•Joining curves together smoothly is difficult
• Hard to specify, modify, control
5
• Two Ways to Define a Circle
Parametric:
x(u) = r cos (u)y(u) = r sin (u)
Implicit:
F(x,y) = x² + y² - r² = 0
Curve examples
6
Parametric Equations
CTtQ )(
Cubic preferred
Balance between flexibility and complexity in specifying and computing shape.
Matricial expression
Q(t)=(x(t),y(t),z(t))
x(t) = axt3 + bxt2 + cxt + dx
y(t) = ayt3 + byt2 + cyt + dy
z(t) = azt3 + bzt2 + czt + dz
123 tttT
zyx
zyx
zyx
zyx
ddd
ccc
bbb
aaa
CWhere:
t [0,1]
7
Joining Curve Segments
G0 geometric continuity: Two curve segments join together.
G1 geometric continuity: The direction of the two segments’ tangent vectors are equal at the join point.
C0 continuity: Curves share the same point where they join.
C1 continuity: Tangent vectors of the two segments are equal in magnitude and direction (share the same parametric derivatives).
C2 continuity: Curves share the same parametric second derivatives where they join.
C1 G1, unless tangent vector = [0, 0, 0].
8
Joining Examples
C0
C1
C2
TV2
TV3
TV1
P1
P2
P3
Q1
Q2
Q3
Q1 and Q2 are C1 continuous because their tangents, TV1and TV2, are equal. Q1 and Q3 are only G1 continuous.
Join point
S joins C0, C1, and C2 with C0, C1, and C2 continuity, respectively.
9
Continuity examples
Continuous in position
Continuous in position and tangent vector
Continuous in position, tangent, and curvature
10
Curve Fitting: Interpolation and Approximation
Interpolation
curve must pass through control points
Approximation
curve is influenced by control points
11
Curve representations: polynomial bases
• Polynomials are easy to analyze, derivatives remain polynomial, etc.
• Monomial basis {1, x, x2, x3, …}.
– Coefficients are geometrically meaningless.
– Manipulation is not robust.
• Number of coefficients = polynomial rank.
12
Polynomial Interpolation
• An n-th degree polynomial fits a curve to n+1 points.
– Example: fit a second degree curve to three points:
• x(u)= au2 + bu + c.
• control points to interpolate: (u1, x1), (u2, x2), (u3, x3).
• solve for coefficients (a, b, c): 3 linear eqns, 3 unknowns.
– called Lagrange Interpolation.
– result is a curve that changes to any control point affects entire curve (non-local) [this method is poor].
• We usually want the curve to be as smooth as possible:
– minimize the wiggles.
– high-degree polynomials are bad.
13
Lagrange Interpolation
The Lagrange interpolating polynomial is the polynomial of degree n-1 that passes through the n points,
(x1, y1), (x2, y2), …, (xn, yn),
and is given by:
2 1 31 2
1 2 1 2 1 2 3 2
1
1 1
1
P x
n n
n n
nn
n n
nj
ii j i i j
x x x x x x x x x xy y
x x x x x x x x x x
x x x xy
x x x x
x xy
x x
14
Splines
A spline is a parametric curve defined by control points:
– term “spline” dates from engineering drawing, where a spline was a piece of flexible wood used to draw smooth curves.
– control points are adjusted by the user to control shape of curve.
– wood splines:
• have second-order continuity,
• pass through the control points.
Drawing curves with wood splines
15
Spline Curve Families
• Hermite cubic
– Defined by its 2 endpoints and tangent vectors at endpoints.– Interpolates all its control points.– Special case of Bezier and B-Spline.
• Bezier
– Interpolates first and last control points.– Curve is tangent to first and last segments of control polygon.– Easy to subdivide.– Curve segment lies within convex hull of control polygon.
• B-Spline
– Not guaranteed to interpolate control points.– Curve segment lies within convex hull of control polygon.– Greater local control than Bezier.
0
1
2
3
5
4
6
7
89
16
Linear Interpolation
• Linear interpolation (Lerp) is a common technique for generating a new value that is somewhere in between two other values.
• By interpolating between two points p0 and p1 by some parameter t we obtain a segment:
p0
p1
t=1.
. 0<t<1t=0
17
Three ways of writing a segment:
1. Weighted average of the control points: Q(t) = (1-t)p0 + tp1 .
Basis Functions:
B0(t)=1-t; B1(t)=t
B0(t)+B1(t)=1
Q(t) = B0(t)p0 + B1(t)p1
2. Polynomial in t: Q(t) = (p1-p0 )t + p0
3. Matrix form:
pp
tQ t1
0
01
111)(
Linear Interpolation
18
Hermite Curve Examples
P0P1
R1
R0
The set of Hermite curves that have the same values for the endpoints P0 and P1, tangent vectors R0 and R1 of the same direction, but with different magnitudes for R0. The magnitude of R1 remains fixed.
19
Hermite Curve Examples
All tangent vector magnitudes are equal, but the direction of the left tangent vector varies.
20
p
qDp
DqP(t) = at3 + bt2 +ct +d
t in [0,1]
P(0) = p
P(1) = q
P'(0) = Dp
P'(1) = Dq
Is a cubic curve for which the user provides:
The endpoints of the curve: p, q;The parametric derivatives of the curve at the endpoints: Dp, Dq.
Hermite Cubic Curves
21
Hermite Coefficients
For each coordinate, we have 4 linear equations in 4 unknowns
22
Boundary Constraint Matrix
23
Hermite Matrix
2 2 1 1
3 3 2 1
0 0 1 0
1 0 0 0
a
b
c
d
p
q
Dp
Dq
MH = (NH)-1 GH
P(t) = T MH GH
24
Hermite Basis Functions
0 1 2 3t H t H t H t H t P p q Dp Dq
25
23
3
232
231
230
2
32
132
tttH
ttttH
tttH
tttH
H0(t) H1(t)
Hermite Basis Functions
- Functions:
are called Hermite Basis or Blending Functions.
- Observe that: H0(t)+H1(t)+H2(t)+H3(t) 1.
- An Hermite cubic curve can be thought as a higher order extension of linear interpolation:
P(t)=H0(t)p + H1(t)q+H2(t)Dp+H3(t)Dq
26
• Hermite curves are difficult to use because we usually have control points but not derivatives.
• However, Hermite curves are the basis of the Bezier curves.
• Bezier curves are more intuitive, since we only need to specify control points.
From Hermite to Bezier curves
Displaying Hermite curves
• Evaluate the curve at a fixed set of parameter values and join the points with straight lines.
27
• Bezier curve is an approximation of given control points.
• Bezier curve of degree d is defined over d+1 control points {Pi}i=0,...,d.
• Have two formulations:
– Algebraic (higher order extension of linear interpolation).
– Geometric.
Bezier Curves of degree d
28
Bezier Curves: Algebraic Formulation
• The user supplies d+1 control points:
pi ; i=0, … , d.
• Write the curve of degree d as:
• The functions Bid(t) are the Bernstein basis polynomials or Bezier
blending functions of degree d.
pi
d
i
di tBtQ
0
ididi tt
i
dtB
1
0
0 )!!*(/!
otherwise
diwhenidid
i
d
29
Bezier Curves: Algebraic Formulation
• Q(0)=p0 and Q(1)=pd, that means the Bézier curve lies on p0 and pd .
• Q‘(0)=d(p1 - p0) and Q‘(1)=d(pd - pd-1) (tangents in start and end points).
pi
d
i
di tBtQ
0
=
=(1-t)d p0 + t(1-t)d-1p1+ …+td-1(1-t)pd-1+tdpd.
30
d
i
d
i
di
iid
d
tBtti
d
tt
0 0)(1
11
0,1t 1)(0
d
i
di tB
Proof:
• They are all positive in interval [0,1],
• Their sum is equal to 1:
• Symmetry:
• Recursion:
Bernstein Basis Polynomials
)1()( tBtB did
di
)()1()()( 11 tBttBttB id
idi
di
31
Bezier Curves
32
Some cubic Bezier curves
For cubic Bezier curves, two control points define endpoints, and two control the tangents at the endpoints in a geometric way.
Cubic Bezier Curves (d=3)
33
Bernstein Basis Polynomials for d=3
)1()(33
0 ttB
ttB33
3 )(
)1(3)(23
1 tttB
)1(3)( 23
2 tttB
)(3
0 tB )(3
3 tB
)(3
1 tB )(3
2 tB
34
zyx
zyx
zyx
zyx
ppp
ppp
ppp
ppp
ttttQ
333
222
111
000
23
0001
0033
0363
1331
1)(
Q(t) = T MB GB
)(3010
ppDp
)(3233
ppDp
Bézier Matrix for d=3
From:
We can deduce:
MB GB
35
Bezier Cubic Curves Properties
• The first and last control points are interpolated,
• The tangent to the curve at the first control point is along theline joining the first and second control points,
• The tangent at the last control point is along the line joining the second last and last control points,
• The curve lies entirely within the convex hull of its control points:
– Every point on curve is a linear combination of the control points,
– The weights of the combination are all positive,
– The sum of the weights is 1,
– Therefore, the curve is a convex combination of the control points.
• By specifying multiple coincident points at a vertex, we pull the curve in closer and closer to that vertex.
36
The properties of the Bernstein polynomials ensure that all Bezier curves lie in the convex hull of their control points.
Convex Hull Property
37
De Casteljau Algorithm
• Describes the curve as a recursive series of linear interpolations.
• Is useful for providing an intuitive understanding of the geometry involved.
• Provides a means for evaluating Bezier curves.
38
De Casteljau Algorithm
Q (1/3)
We take points 1/3 of the way
);(:)(
);()()1(:)(
do to:=For
do to1:For
;:)( do to0:For
][
]1[]1[1
][
]0[
tPtQ
ttPtPttP
nji
nj
PtPni
nn
ji
ji
ji
ii
Select t[0,1] value. Then:
d
d
);(][ tP dd
d
39
p0
q0
p1
p2
p3
q2
q1
322
211
100
,,
,,
,,
ppq
ppq
ppq
tLerp
tLerp
tLerp
De Casteljau Algorithm for d=3
tqptqptLerp 1,,Where:
We start with our original set of points p0, p1, p2 and p3.
40
q0
q2
q1
r1
r0
211
100
,,
,,
qqr
qqr
tLerp
tLerp
De Casteljau Algorithm
41
10 ,,)( rrtLerptQ
De Casteljau Algorithm
r1Q(t)r0 •
Q(t)•
p0
p1
p2
p3
Bezier curve
42
Recursive Linear Interpolation
322
211
100
,,
,,
,,
ppq
ppq
ppq
tLerp
tLerp
tLerp
211
100
,,
,,
qqr
qqr
tLerp
tLerp
10 ,,)( rrtLerptQ
3
2
1
0
p
p
p
p
2
1
0
q
q
q
1
0
r
r)(tQ
3
2
1
0
p
p
p
p
43
Expanding the Lerps
3221
211010
3221211
2110100
32322
21211
10100
111
1111,,)(
111,,
111,,
1,,
1,,
1,,
pppp
pppprr
ppppqqr
ppppqqr
ppppq
ppppq
ppppq
ttttttt
ttttttttLerptQ
tttttttLerp
tttttttLerp
tttLerp
tttLerp
tttLerp
44
Bernstein Polynomial Form
33
22
12
03
3221
2110
13131)(
111
1111)(
pppp
pppp
pppp
tttttttQ
ttttttt
ttttttttQ
45
Animations from Wikipedia
De Casteljau Algorithm
d=1
d=2
46
De Casteljau Algorithm
d=3
d=4
Animations from Wikipedia
47
Drawing Algorithms
Evaluate points and draw lines.
Possibilities:
1. Evaluate recursively Bernstein Polynomials.
2. Compute the geometric matrix.
3. Use the recursive algorithm of De Casteljau.
48
• Are hard to control and hard to work with.
• The intermediate points don’t have obvious effect on shape.
• For large sets of points – curve deviates far from the points.
• Degree corresponds to number of control points.
• Changing any control point can change the whole curve:
– We want local support: each control point only influences nearby portion of curve.
Bezier Curves Drawbacks
49
Piecewise Curves
• A single cubic Hermite o Bezier curve can only capture a small class of curves:
– At most 2 inflection points.
• One solution is to raise the degree:
– Allows more control, at the expense of more control points and higher degree polynomials.
– Control is not local, one control point influences entire curve.
• Alternate, most common solution is to join pieces of cubic curve together into piecewise cubic curves:
– Total curve can be broken into pieces, each of which is cubic.
– Local control: each control point only influences a limited part of the curve.
– Interaction and design is much easier.
50
Piecewise Bezier Curves
“knot”P0,0
P0,1 P0,2
P0,3
P1,0
P1,1
P1,2
P1,3
If complicated curves are to be generated, they can be formed
by piecing together several Bézier sections.
The curve segments join at knots.
51
Achieving Continuity
• For Hermite curves:
– the user specifies the derivatives, so C1 is achieved simply by sharing points and derivatives across the knot.
• For Bezier curves:
– They interpolate their endpoints, so C0 is achieved by sharing control points.
– The parametric derivative is a constant multiple of the vector joining the first/last 2 control points.
– So C1 is achieved by setting P0,3=P1,0=P, and making P0,2 and P and P1,1 collinear, with P-P0,2=P1,1-P.
– C2 comes from further constraints on P0,1 and P1,2
C0 continuity
52
Why more curves?
• Bezier and Hermite curves have global influence:
– One could create a Bezier curve that required 15 points to define the curve…
• Moving any one control point would affect the entire curve.
– Piecewise Bezier or Hermite don’t suffer from this, but they don’t enforce derivative continuity at join points.
• B-Splines consist of curve segments whose polynomial coefficients depend on just a few control points:
– Local control
53
Cubic B-Spline: Geometric Definition• Approximates n+1 control points: P0, P1, … , Pn, n >= 3.
• Consists of n –2 cubic polynomial curve segments: Q3, Q4, … , Qn.
• Defined using a non-decreasing sequence of n-1 parameters: t3,… ,tn+1(knots).
• It is uniform if the elements of the knot sequence are uniformly spaced: ti+1-ti=c , i=3,…,n (without loss of generality, we can assume t3=0 and ti+1-ti =1), otherwise it is non-uniform.
• Curve segment Qi is defined by the control points Pi-3, Pi-2, Pi-1 and Pi over the knot interval [ti,ti+1) (there exist local control).
P0
P1
P2
P3
P5
P4
P6
P7
P8P9
Q3
Q4
Q5
Q6
Q7
Q8
Q9
54
Uniform Cubic B-Splines
55
Uniform Cubic B-Splines
To determine the equation of the curve segment Q3(t), t in [t3,t4):
• Each control point affects 4 curve segments.
• Formulate 16 equations to solve the 16 unknowns, that correspond to 4 polynomial of degree 3.
• The 16 equations enforce the C0, C1, and C2 continuity between adjoining curve segments.
We obtain:
33
232
132
032
3
04,3
6
13331
6
1364
6
1331
6
1
)()(
PtPtttPttPttt
PtBt kk
kQ
for t in [t3 ,t4).
56
Uniform Cubic B-Splines
33
232
132
10
32
3
04,3
6
13331
6
1364
6
1331
6
1
)()(
PtPtttPttPttt
PtBt kk
kQ
From:
We obtain the matricial expression:
3
2
1
0
233
0141
0303
0363
1331
6
11
P
P
P
P
ttttQ
57
3
4,3
32
4,2
32
14,1
32
4,0
6
1)(
33316
1)(
3646
1)(
3316
1)(
tt
tttt
ttt
tttt
B
B
B
B
Uniform Cubic B-Splines
Basis or blending cubic functions for curve segment Q3(t):
0
0,1
0,2
0,3
0,4
0,5
0,6
0,7
0 0,1
0,2
0,3
0,4
0,5
0,6
0,7
0,8
0,9 1
t
B0,4
B1,4 B2,4
B3,4
Basis functions for Q3(t), t in [t3,t 4]=[0,1].
58
Uniform Cubic B-Splines
It is not difficult to see that for the curve segment Qi(t),
t in [ti,ti+1), i=3,…,n, we have:
i
ikkki
tBPtQ3
4, )()(
where:
Bi,4(t)=B0,4(t-i).
59
• The blending functions sum to one, and are positive everywhere.
• The curve lies inside the convex hull of the control points.
• The curve does not interpolate its endpoints.
• Uniform B-splines are C2.
• Moving a control point has a local effect.
Uniform Cubic B-Splines
60
i
n
idi PtBtQ
0
,
Uniform B-Splines: Algebraic Definition
A B-spline curve Q(t), is defined by:
where:
• The Pi, i = 0,1,...,n are the n+1 control points.
• d is the order of the polynomial segments of the B-Spline curve. That means that the curve is made up of piecewise polynomial segments of degree d-1.
• The order d is independent of the number of control points (n+ 1).
• Bi,d are the uniform B-Spline basis or blending functions of degree d-1.
61
B-Spline Basis FunctionsGiven a non-decreasing knot sequence of n+d+1 parameterst0, … , tn+d , we define:
otherwise 0
1 11,
iii
ttttB
tBtt
tttB
tt
tttB di
idi
didi
kdi
idi 1,1
11,
1,
for d >1 and t in [td-1,n+1).
If the denominator terms on the right hand side of the last equation are zero or the subscripts are out of the range of the summation limits, then the associated fraction is not evaluated and the term becomes zero (avoiding 0/0 expressions).
We will concentrate in uniform B-splines and, without loss of generality, we will assume t0= -3 and ti+1-ti =1.Observe that for a cubic (d=4) B-Spline now we have n+5 parameters: t0, … , tn+4 .
62
B-Splines Basis Functions Computation
B0,0(t) B1,0(t) B2,0(t) B3,0(t) …
B0,1(t) B1,1(t) B2,1(t)
B0,2(t) B1,2(t)
B0,3(t)
…
…
…
t0 t1 t2 t3 t4 …
63
Bi,1(t)
Bi,1(t) has support (is non zero) on interval [i-3,i-2).
Bi,1(t) is a constant function.
Bi+1,1(t) is just Bi,1 (t) shifted one unit to the right, because the uniform election of parameter knots. We can write: Bi,1(t)=B0,1(t-i).
Blending functions for d=1
B 0,1
0
0,2
0,4
0,6
0,8
1
1,2
-3
-2,8
-2,6
-2,4
-2,2 -2
-1,8
-1,6
-1,4
-1,2 -1
-0,8
-0,6
-0,4
-0,2 0
0,2
0,4
0,6
0,8 1
t
B0
,1(t
)
B 2,1
0
0,2
0,4
0,6
0,8
1
1,2
-3-2 ,8 -2 ,6 -2 ,4 -2 ,2 -2
-1 ,8 -1 ,6 -1 ,4 -1 ,2 -1-0 ,8 -0 ,6 -0 ,4 -0 ,2 0
0 ,2 0 ,4 0 ,6 0 ,8 1
t
B2
,1(t
)
B 3,1
0
0,2
0,4
0,6
0,8
1
1,2
-3-2 ,8 -2 ,6 -2 ,4 -2 ,2 -2
-1 ,8 -1 ,6 -1 ,4 -1 ,2 -1-0 ,8 -0 ,6 -0 ,4 -0 ,2 0
0 ,2 0 ,4 0 ,6 0 ,8 1
t
B3
,1(t
)
B 1,1
0
0,2
0,4
0,6
0,8
1
1,2
-3-2 ,8 -2 ,6 -2 ,4 -2 ,2 -2
-1 ,8 -1 ,6 -1 ,4 -1 ,2 -1-0 ,8 -0 ,6 -0 ,4 -0 ,2 0
0 ,2 0 ,4 0 ,6 0 ,8 1
t
B1
,1(t
)
64
Bi,2(t)B 0,2
0
0,2
0,4
0,6
0,8
1
1,2
-3
-2,8
-2,6
-2,4
-2,2 -2
-1,8
-1,6
-1,4
-1,2 -1
-0,8
-0,6
-0,4
-0,2 0
0,2
0,4
0,6
0,8 1
t
B0
,2(t
)
B 1,2
0
0,2
0,4
0,6
0,8
1
1,2
-3
-2,8
-2,6
-2,4
-2,2 -2
-1,8
-1,6
-1,4
-1,2 -1
-0,8
-0,6
-0,4
-0,2 0
0,2
0,4
0,6
0,8 1
t
B1,
2(t)
B 2,2
0
0,2
0,4
0,6
0,8
1
1,2
-3-2
,8-2
,6-2
,4-2
,2 -2-1
,8-1
,6-1
,4-1
,2 -1-0
,8-0
,6-0
,4-0
,2 00,2 0,4 0,6 0,8 1
t
B2
,2(t
)
12 1
23 3)(2,0 tt
tttB
Blending functions for d=2
Bi,2(t) has support on interval [i-3,i-1).
Bi,2(t) is piecewise linear, made up of two linear segments joined continuously.
We have: Bi,2(t)=B0,2(t-i).
65
Bi,3(t) B 0,3
0
0,1
0,2
0,3
0,4
0,5
0,6
0,7
0,8
-3
-2,8
-2,6
-2,4
-2,2 -2
-1,8
-1,6
-1,4
-1,2 -1
-0,8
-0,6
-0,4
-0,2 0
0,2
0,4
0,6
0,8 1
t
B0,
3(t)
B 1,3
00,10,20,30,40,50,60,70,8
-3-2 ,8 -2 ,6 -2 ,4 -2 ,2 -2
-1 ,8 -1 ,6 -1 ,4 -1 ,2 -1-0 ,8 -0 ,6 -0 ,4 -0 ,2 0
0,2 0,4 0,6 0,8 1
t
B1,
3(t)
01
12 362
23 3
2
1)(
2
2
2
3,0
tt
ttt
tt
tB
Blending functions for d=3
Bi,3(t) has support on interval [i-3,i).
Bi,3(t) is a piecewise quadratic curve made up of three parabolic segments that are joined continuously.
We have: Bi,3(t)=B0,3(t-i).
66
10 1
01 1333
12 521153
23 3
6
1)(
3
23
23
3
4,0
tt
tttt
tttt
tt
tB
Blending functions for d=4 (cubic)
Bi,4(t) has support on interval [i-3,i+1).
Bi,4(t) is a piecewise cubic curve made up of four cubic segments that are joined continuously.
B 0,4
0
0,1
0,2
0,3
0,4
0,5
0,6
0,7
-3
-2,8
-2,6
-2,4
-2,2 -2
-1,8
-1,6
-1,4
-1,2 -1
-0,8
-0,6
-0,4
-0,2 0
0,2
0,4
0,6
0,8 1
t
B0
,4(t
)
B0,4(t)
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Uniform Cubic B-spline Blending Functions
0
0,1
0,2
0,3
0,4
0,5
0,6
0,7
-3 -2,5 -2 -1,5 -1 -0,5 0 0,5 1 1,5 2 2,5 3 3,5 4 4,5 5
t
B0,4 B1,4 B2,4 B3,4 B4,4 B5,4 B6,4
We have: Bi,4(t)=B0,4(t-i).
Observe that at interval [0,1) justfour of the functions are non-zero: B0,4(t), B1,4(t), B0,4(t) and B1,4(t).
They are the four functions that determine the curve segment Q3.
68
Example of Uniform Cubic B-Spline (n=6)
0
0,05
0,1
0,15
0,2
0,25
-3
-2,7
-2,3 -2
-1,6
-1,3
-0,9
-0,6
-0,2 0,1
0,5
0,8
1,2
1,5
1,9
2,2
2,6
2,9
3,3
3,6 4
4,3
4,7
t
Pi
n
ii tBtQ
0
4,
B0,4P0
B1,4P1B2,4P2
B3,4P3
B4,4P4
B5,4P5
B6,4P6
The curve can’t start until there are 4 basis functions active.
69
Uniform Cubic B-spline at Arbitrary t
• The interval from an integer parameter value i to i+1 is essentially the same as the interval from 0 to 1:
– The parameter value is offset by i.– A different set of control points is needed.
• To evaluate a uniform cubic B-spline at an arbitrary parameter value t:
– Find the greatest integer less than or equal to t: i = floor(t)– Evaluate:
• Valid parameter range: 0 t <n-3, where n is the number of control points
kik
k PitBtQ
3
04,
70
Closed curve
• To obtain a closed curve, repeat the control points P0, P1, P2 at the end of the sequence: P0, P1, P2, ... , P0, P1, P2.
Point interpolation
• On can force the curve pass through a control point by giving that point a multiplicity 3: Pi = Pi+1 = Pi+2 (however, tangent discontinuity may result).
• In particular, to force endpoints interpolation, let: P0 = P1 = P2 and Pn-2 = Pn-1 = Pn.
Cubic B-Splines Shape Modification Using Control Points
71
NURBS: Non-Uniform Rational B-Splines
NURBS are defined as:
where every point control point has associated a weight .iP
)(
)()(
0,
0,
tBw
PtBwtQ n
jdjj
n
iidii
i
72
NURBS can be made to pass arbitrarily far or near to a control point by
changing the corresponding weight.
All conic sections (a circle for example) can be modelled exactly:
Constructing a 2D-circle with NURBS
NURBS
NURBS inherit all advantages of B-Splines (locality, ...) while extending the liberty of modelling.
73
• If all wi are set to the value 1, we obtain standard B-Splines.
• A Bezier curve is a special case of a B-Spline, so NURBS can also represent Bezier curves.
• Today NURBS are standard for modelling.
NURBS
Curves
Toni SellarèsUniversitat de Girona
