On 3/14/19 5:50 PM, [email protected] wrote: > > I've tried looking a cell->face(face_no)->at_boundary() type input within the > cell iterator and by using something like cell->vertex(v)(1) but I can't > quite > get my head around how to find the y coordinate of the top boundary for every > cell/quadrature point calculation.
Jane, if you've already identified that a face is at the boundary, then cell->face(face_no)->center() returns a Point<dim> object. So if you need the y-coordinate, it would be cell->face(face_no)->center()[1] Or are you asking the following question: "For a given quadrature point at (x,y), find how far the domain extends above (x,y) in y-direction"? Best W. -- ------------------------------------------------------------------------ Wolfgang Bangerth email: [email protected] www: http://www.math.colostate.edu/~bangerth/ -- The deal.II project is located at http://www.dealii.org/ For mailing list/forum options, see https://groups.google.com/d/forum/dealii?hl=en --- You received this message because you are subscribed to the Google Groups "deal.II User Group" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. For more options, visit https://groups.google.com/d/optout.
