Non Uniform Rational B-Spline (NURBS)

1
Non Uniform Rational B-Spline (NURBS)
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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
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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.

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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)

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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.

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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

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NURBS modifying weights
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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
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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
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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
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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
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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
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NURBS circle

• Other way ? split into 3 arcs
• ? 60?

P3
P2
P4
P1
P5
?
P0
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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)

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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!!

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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.

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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
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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
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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

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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)

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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

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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