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Non Uniform Rational B-Spline (NURBS)

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disediakan oleh Suriati Bte Sadimon GMM, FSKSM, UTM, 2004. Non Uniform ... equations can exactly represent the conic curves (circles, ellipse, parabola... – PowerPoint PPT presentation

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Title: Non Uniform Rational B-Spline (NURBS)


1
Non Uniform Rational B-Spline (NURBS)
2
NURBS
n
  • P(u) ? wiNi,k(u)pi
  • ? wiNi,k(u)
  • NURBS curve degree k-1
  • P0, P1,..Pn control points
  • Knot vector u (u0, u1um)
  • W (w0, w1wm) weight ? non negative
  • The number of control points and the number of
    weights must agree
  • Use homogenous coordinates

i0
n
i0
3
NURBS
  • If all wi are set to the value 1 or all wi have
    the same value ? we have the standard B-Spline
    curve
  • NURBS curve equation is a general form that can
    represent both B-Spline and NURBS curves. A
    Bezier curve is a special case of a B-Spline
    curve, so the NURBS equation can also represent
    Bezier and rational Bezier curves.

4
NURBS important properties
  • NURBS have all properties of B-Spline.(pls refer
    to the notes)
  • Extra properties
  • More versatile modification of a curve becomes
    possible if the curve is represented by a NURBS
    equation. It is due to ? B-spline curve is
    modified by changing the x, y, z coord, but NURBS
    curve use homogenous coord (x, y, z, h)
  • ?Or B-Spline (degree, control points knots)
    but NURBS (degree, control points, knots
    weights)

5
NURBS important properties
  • Extra properties (cont)
  • NURBS equations can exactly represent the conic
    curves (circles, ellipse, parabola)
  • Projective invariance ? If a projective
    transformation is applied to a NURBS curve, the
    result can be constructed from the projective
    images of its control points. Therefore, we do
    not have to transform the curve. We can obtain
    the correct view (no distortion).
  • Bézier curves and B-spline curves only
    satisfy the affine invariance property rather
    than this projective invariance property. This is
    because only NURBS curves involve projective
    transformations.

6
NURBS modifying weights
  • increasing the value of wi will pull the curve
    toward control point Pi. In fact, all affected
    points on the curve will also be pulled in the
    direction to Pi
  • When wi approaches infinity, the curve will pass
    through control point Pi
  • decreasing the value of wi will push the curve
    away from control point Pi

7
NURBS modifying weights
8
NURBS circle
  • NURBS equations represents the circle
  • Consider a half circle
  • Split the half circle into two circular arcs -1
    and 2
  • (actually many ways)

P2
P1
P3
1
2
P4
P0
9
NURBS circle
  • Consider arc 1
  • Conic section (quadratic) degree 2
  • 3 control points, P0 (1,0) ,P1 (1,1), P2
    (0,1)
  • Knot value (0,0,0,1,1,1)
  • Weights w0 1 w1 cos ? cos 45 1/?2 w21

P2
P1
?
1
P0
10
NURBS circle
  • Follow the same procedure for arc 2
  • We obtain
  • P2 (0,1) ,P3 (-1,1), P4 (-1,0)
  • Weights w2 1, w3 cos 45 1/?2, w41
  • Knot value (0, 0, 0,1,1,1) shifted to (1, 1, 1,
    2, 2, 2) for the composition

P2
P3
2
P4
11
NURBS circle
  • Composite curve (arc 1 and arc 2)
  • P0 (1,0) ,P1 (1,1), P2 (0,1),P3 (-1,1), P4
    (-1,0)
  • Weights w0 1 w1 1/?2 w21, w3 1/?2, w41
  • k 3, n 4 number of knot 8
  • Knot value (0, 0, 0,1,1,1) (1, 1, 1, 2, 2, 2)
  • Knot value (0, 0, 0, 1, 1, 1, 2, 2, 2) shift
    1 value
  • ? (0, 0, 0,1, 1, 2, 2, 2)

P2
P1
P3
1
2
P4
P0
12
NURBS circle
P2
  • So, for a circle
  • P0 P8 (1,0) ,P1 (1,1), P2 (0,1),P3 (-1,1),
    P4 (-1,0), P5 (-1,-1), P6
    (0,-1), P7 (1,-1)
  • Weights w0 1 w1 1/?2 w21, w3 1/?2, w41
  • k 3, n 8 number of knot 12
  • ? (0, 0, 0,1, 1, 2, 2, 3, 3, 4, 4, 4)

P1
P3
1
2
P4
P8
P0
P5
P7
P6
13
NURBS circle
  • Other way ? split into 3 arcs
  • ? 60?

P3
P2
P4
P1
P5
?
P0
14
NURBS circle
  • show that NURBS equations exactly represent the
    circle
  • Consider arc 1
  • NURBS equation is
  • P(u) ? wiNi,k(u)pi
  • ? wiNi,k(u)
  • P(u) w0P0N0, 3(u) w1P1N1, 3(u) w2P2N2, 3(u)
  • w0N0, 3(u) w1N1, 3(u) w2N2,
    3(u)

15
NURBS circle
  • w0 w2 1, w1 ?2 /2
  • N0, 3 (1-u) 2, N1, 3 2u(1-u), N2, 3 u2 ? pls
    refer to last topic (Curve1.pdf, slide 30)
  • P(u) P0 (1-u) 2 ?2 /2P1 2u(1-u) P2 u2
  • (1-u) 2 ?2 /2 2u(1-u) u2
  • P0 1,0,1 P1 1,1,1, P2 0,1,1 ?
    homogenous
  • x(u) (1- ?2) u2 ?2 (1-?2)u 1
  • (2- ?2) u2 (?2-2)u 1
  • y(u) ( 1- ?2) u2 ?2u .
  • (2- ?2) u2 (?2-2)u 1
  • x(u) 2 y(u) 2 1 -? circle radius 1 ?
    prove!!

16
Knot Insertion NURBS
  • three steps
  • (1) converting the given NURBS curve in 3D to a
    B-spline curve in 4D
  • (2) performing knot insertion to this four
    dimensional B-spline curve
  • (3) projecting the new set of control points
    back to 3D to form the the new set of control
    points for the given NURBS curve.

17
Knot Insertion NURBS
  • EXAMPLE
  • a NURBS curve of degree 3 with a knot vector as
    follows
  • 5 control points in the xy-plane and weights
  • Insert new knot t 0.4

u0 to u3 u4 u5 to u8
0 0.5 1
P0 P1 P2 P3 P4
X -70 -70 74 74 -40
Y -76 75 75 -77 -76
w 1 0.5 4 5 1
18
Knot Insertion NURBS
  • t 0.4 lies in knot span u3 ,u4)
  • the affected control points are P3, P2, P1 and
    P0.
  • 1) convert to 3D B-Spline (homogenous coord) ?
    multiply all control points with their
    corresponding weights

Pw0 Pw1 Pw2 Pw3
X -70 -35 296 370
Y -76 37.5 300 -385
w 1 0.5 4 5
19
Knot Insertion NURBS
  • 2) Perform knot insertion
  • compute a3, a2 and a1
  • a3 t - u3 0.4 0 0.4
  • u6 -u3 1 0
  • a2 t - u2 0.4 0 0.4
  • u5 -u2 1 0
  • a1 t u1 0.4 0 0.8
  • u4 -u1 0.5 0

20
Knot Insertion NURBS
  • 2) Perform knot insertion (cont)
  • The new control points Qw3, Qw2 and Qw1 are
  • Qw3 (1-a3)Pw2 a3Pw3 (325.6, 26, 4.4)
  • Qw2 (1-a2)Pw1 a2Pw2 (97.4, 142.5, 1.9)
  • Qw1 (1-a1)Pw0 a1Pw1 (-42, 14.8, 0.6)

21
Knot Insertion NURBS
  • 3) Projecting these control points back to 2D by
    dividing the first two components with the third
    (the weight),
  • Q3 (74, 5.9 ) with weight 4.4
  • Q2 (51.3, 75 ) with weight 1.9
  • Q1 (-70, 24.6) with weight 0.6

22
Knot Insertion NURBS
  • control points in the xy-plane and weights after
    knot insertion
  • new knot vector

P0 Q1 Q2 Q3 P3 P4
X -70 74 51.3 -70 74 -40
Y -76 5.9 75 24.6 -77 -76
w 1 4.4 1.9 0.6 5 1
u0 to u3 u4 u5 u6 to u9
0 0.4 0.5 1
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