1 / 14

Computer Aided Engineering Design

Computer Aided Engineering Design. Anupam Saxena Associate Professor Indian Institute of Technology KANPUR 208016. Surface from the tangent plane: Derivation. n. P. R. n is perpendicular to the tangent plane, r u . n = r v . n = 0. d. second fundamental matrix D.

jam
Télécharger la présentation

Computer Aided Engineering Design

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Computer Aided Engineering Design AnupamSaxena Associate Professor Indian Institute of Technology KANPUR 208016

  2. Surface from the tangent plane: Derivation n P R n is perpendicular to the tangent plane, ru.n= rv.n= 0 d second fundamental matrix D

  3. Classification of pointson the surface 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 P  ELLIPTICAL POINT No other point of intersection

  4. Classification of pointson the surface L2+M2+N2 > 0 du = (M/L)dv Case 2: u – u0 = (M/L)(v – v0) tangent plane intersects the surface along this straight line P  PARABOLIC POINT two real roots for du Case 3: 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

  5. Lecture # 35Gaussian and Mean Curvature of Surfaces

  6. Normal and geodesic curvatures kn = nn normal curvature kg = gtg geodesic curvature n t P nc Sincen.t= 0 nn tg t nc gtg since kg and n are perpendicular kg.n = 0

  7. Normal and geodesic curvatures decomposing drand dnalong parametric lengths du and dv Since ru and rv are both perpendicular to n

  8. Normal and geodesic curvatures the expression for the normal curvature is where The above equation can be written as For an optimum value of normal curvature Differentiation yields

  9. Normal and geodesic curvatures Thus This can be simplified to For a non trivial solution, the determinant of the coefficient matrix is zero

  10. Max and Min normal curvatures K is the Gaussian curvature… H is the mean curvature

  11. Example parametric equation of a Monkey Saddle Compute the Gaussian and Mean curvatures

  12. Curvature Plots of Monkey Saddle minimum principal curvature maximum principal curvature Monkey saddle Gaussian curvature mean curvature

  13. Why are these curvatures important ? To identify a certain class of surface patches e.g. For developable surfaces, the Gaussian curvature is ZERO

More Related