| Author | Message | ||
     Kevin Ellwood (ikevin) Member Username: ikevin Post Number: 7 Registered: 06-2003 |
Hi I have a set of data points that represents a smooth curve. While meshing the geomtry, the mesh is artificially concentrated at the points along the curve. This is due to the fact the I used "line to", making the curve a bunch of staight line segements. Is there a spline function in flexpde that I can't find or is there another way to work around this issue. Thanks Kevin | ||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 49 Registered: 06-2003 |
FlexPDE does not at present have a capability for spline boundaries (we have done some preliminary work on this and hope to include it in a future release). The best I can suggest is to define the curve with enough line segments that there are no serious breaks in slope. An alternative is to laboriously construct some circular or elliptical arcs that match the boundary reasonably well. | ||
| Kevin Ellwood (ikevin) Member Username: ikevin Post Number: 8 Registered: 06-2003 |
Thanks for the answer. I can do the spline interpolation myself, however, doesn't the fact that I am adding more boundary points mean that I will be adding more anchored nodes? I assume that all boundary segment points are connected to nodes. Thanks Kev | ||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 50 Registered: 06-2003 |
Yes, you will get at least one mesh vertex at each explicit boundary point. But on an arc, or on the proposed spline boundary, FlexPDE would add nodes to resolve the shape anyway. And I assume that the mesh concentration you referred to earlier is due to mesh refinement triggered by slope breaks. How many boundary points are we talking about here? With the speed of todays computers, a few hundred points is trivial. Cells constructed along a line boundary have linear geometry. Cells constructed along an arc, or along the proposed spline boundary, are interpolated with the same basis as the finite element interpolation - quadratic by default. We must insert enough nodes on the boundary so that the quadratic interpolated position is a reasonable representation of the specified curve. So if you can match arcs you will get a quadratic representation of the boundary shape, and the only slope breaks will be those due to mismatch of the arcs. | ||
| Kevin Ellwood (ikevin) Member Username: ikevin Post Number: 9 Registered: 06-2003 |
The mesh concentration is due to refinement triggered by slope breaks. I will write a program this weekend to do the spline interpolation and let you know how it goes. |