Hi Jean Paul, Thanks for this - for now this should be sufficient, though for the rest I will have to do a long blurb.
Do you have an example of how to use BoundingBox? I'm a little confused by the document. I don't know what i have to set for the argument before i do something (i am guessing) like mydomain.get_boundary_points(); Thanks! On Friday, March 15, 2019 at 11:30:36 AM UTC, Jean-Paul Pelteret wrote: > > Dear Jane, > > There is the BoundingBox > <https://www.dealii.org/current/doxygen/deal.II/classBoundingBox.html> class > that could tell you the overall extents of the domain. So if your geometry > is rectangular and not rotated then you could work out the difference > between the y-coordinate of the top vertex of the bounding box and your > point of interest. If you have an irregular domain then I think that you’re > going to have to work out the projection onto its upper surface yourself. I > don’t think that there’s any easy way to do this. > > I hope that this helps. > > Best, > Jean-Paul > > On 15 Mar 2019, at 11:14, [email protected] <javascript:> wrote: > > Hi Wolfgang, > > Yes, what you put into much better words than mine is exactly what I am > needing - For a given quadrature point at > (x,y), find how far the domain extends above (x,y) in y-direction? > > So I am looking to find the y-coordinate of the point which is directly > above the (x,y) in question, so that I can find how far it extends above it > (by subtracting it from what I am trying to find) > > On Friday, March 15, 2019 at 5:03:56 AM UTC, Wolfgang Bangerth wrote: >> >> 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] <javascript:>. > For more options, visit https://groups.google.com/d/optout. > > > -- 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.
