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Nurbs (1)
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M TECH students of [cad & cam] can be benefited .

M TECH students of [cad & cam] can be benefited .

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Nurbs (1) Nurbs (1) Presentation Transcript

  • NURBS n P(u) = ∑ wiNi,k(u)pi i=0 n  ∑ wiNi,k(u) i=0 NURBS curve degree = k-1 P0, P1,..Pn – control points Knot vector u = (u0, u1…um) W= (w0, w1…wm) – weight  non negative The number of control points and the number of weights must agree Use homogenous coordinates
  • 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.
  • 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)
  • 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.
  • 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
  • NURBS : modifying weights
  • 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 P3 P1 2 1 P4 P0
  • 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 θ = cos 45 = 1/√2 w2=1 P2 θ P1 1 P0
  • 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, w4=1Knot value =(0, 0, 0,1,1,1) shifted to (1, 1, 1, 2, 2, 2) for the composition P2 P3 2 P4
  • 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 w2=1, w3 = 1/√2, w4=1 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 P3 P1 2 1 P4 P0
  • NURBS : circle P2• So, for a circle P3 P1 2 1 P4 P0 P8 P5 P7 P6• 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 w2=1, w3 = 1/√2, w4=1 k = 3, n= 8 number of knot = 12  (0, 0, 0,1, 1, 2, 2, 3, 3, 4, 4, 4)
  • NURBS : circleOther way  split into 3 arcsθ = 60° P3 P2 P4 θ P1 P5 P0
  • 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)
  • 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 /2[P1 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!!
  • 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.
  • Knot Insertion : NURBS u0 to u3 u4 u5 to u8 0 0.5 1 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 P0 P1 P2 P3 P4 X -70 -70 74 74 -40 Y -76 75 75 -77 -76 w 1 0.5 4 5 1
  • 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
  • Knot Insertion : NURBS2) Perform knot insertioncompute a3, a2 and a1a3 = t - u3 = 0.4 –0 = 0.4 u6 -u3 1 – 0a2 = t - u2 = 0.4 –0 = 0.4 u5 -u2 1 – 0a1 = t – u1 = 0.4 –0 = 0.8 u4 -u1 0.5 – 0
  • Knot Insertion : NURBS2) 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)
  • 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
  • Knot Insertion : NURBS after *control points in the xy-plane and weights knot insertion 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 *new knot vector u0 to u3 u4 u5 u6 to u9 0 0.4 0.5 1