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Anupam Saxena Associate Professor Indian Institute of Technology KANPUR 208016.

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Presentation on theme: "Anupam Saxena Associate Professor Indian Institute of Technology KANPUR 208016."— Presentation transcript:

1 Anupam Saxena Associate Professor Indian Institute of Technology KANPUR

2 Geometric/PARAMETRIC Modeling Solid Modeling Perception of Solids Topology and Solids Solid Modeling 1- 2 Transformations and Projections 1-2 Modeling of Curves Representation, Differential Geometry Ferguson Segments Bezier Segments 1-2 B-spline curves 1-5 NURBS Modeling of Surfaces (Patches) Differential Geometry Tensor Product Boundary Interpolating Composite NURBS

3 n P R d n is perpendicular to the tangent plane, r u.n = r v.n = 0 second fundamental matrix D Geometric/PARAMETRIC Modeling Solid Modeling Perception of Solids Topology and Solids Solid Modeling 1-2 Transformati ons and Projections 1-2 Modeling of Curves Representati on, Differential Geometry Ferguson Segments Bezier Segments 1- 2 B-spline curves 1-5 NURBS Modeling of Surfaces (Patches) Differential Geometry Tensor Product Boundary Interpolating Composite NURBS

4 tangent plane intersects the surface at all points where d = 0 Case 1:No real value of du P is the only common point between the tangent plane and the surface No other point of intersection P  ELLIPTICAL POINT

5 Case 2: L 2 +M 2 +N 2 > 0 du =  (M/L)dv u – u 0 =  (M/L)(v – v 0 ) tangent plane intersects the surface along this straight line P  PARABOLIC POINT Case 3: two real roots for du tangent plane at P intersects the surface along two lines passing through P P  HYPERBOLIC POINT Case 4: L = M = N = 0 P  FLAT POINT

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7 P t n ncnc ncnc nnnn gtggtg t g  t k n =  n n normal curvature k g =  g t g geodesic curvature Since n.t = 0 since k g and n are perpendicular k g.n = 0

8 decomposing dr and dn along parametric lengths du and dv Since r u and r v are both perpendicular to n

9 the expression for the normal curvature is where The above equation can be written as For an optimum value of normal curvature Differentiation yields

10 Thus This can be simplified to For a non trivial solution, the determinant of the coefficient matrix is zero

11 K is the Gaussian curvature… H is the mean curvature

12 parametric equation of a Monkey Saddle Compute the Gaussian and Mean curvatures

13 Monkey saddle maximum principal curvature minimum principal curvature Gaussian curvaturemean curvature

14 To identify a certain class of surface patches e.g. For developable surfaces, the Gaussian curvature is ZERO


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