On Wed, Jul 15, 2009 at 7:17 PM, <alejandro.marti...@utoronto.ca> wrote: > The jacobian for a 2d triangular element is a constant. I heard that for a > quadratic 2d triangular element the jacobian should be linear but I'm > getting a quadratic one (I'm using classical coordinate transformations: x = > ax' + by' + cx'y' + dx'^2 + ey'^2). It should be the same in natural > coordinates. Should the final result be linear?
The 2D Jacobian is a 2x2 matrix given by [dx/dxi, dx/deta] [dy/dxi, dy/deta] Expressing x (resp y.) in terms of the quadratic basis functions phi: x = \sum_i x_i \phi_i and differentiating, you should be able to convince yourself that the entries of the Jacobian for the quadratic triangular element are linear in xi and eta, the coordinates of the reference element. Give it a try, the basis functions for the TRI6 can be found in src/fe/fe_lagrange_shape_2D.C starting at line 158. -- John ------------------------------------------------------------------------------ Enter the BlackBerry Developer Challenge This is your chance to win up to $100,000 in prizes! For a limited time, vendors submitting new applications to BlackBerry App World(TM) will have the opportunity to enter the BlackBerry Developer Challenge. See full prize details at: http://p.sf.net/sfu/Challenge _______________________________________________ Libmesh-users mailing list Libmesh-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/libmesh-users
