On 11/07/2018 04:02 PM, Stogner, Roy H wrote:
On Wed, 7 Nov 2018, Michael Povolotskyi wrote:
On 11/07/2018 02:59 PM, Stogner, Roy H wrote:
On Wed, 7 Nov 2018, Michael Povolotskyi wrote:
if I use the trapezoidal rule:
QTrap qrule(dim)
Can I assume the quadrature points are ordered in the same way at the
element
nodes?
Doesn't look that way, in general, at first glance. We build the quad
and hex QTrap rules via tensor products, but that's not how we order
nodes.
---
Roy
Thank you,
then how to solve the following problem.
I have an element that has n_nodes;
It has the same number of quadrature points that are at the nodal positions
(trapezoidal rule is assumed).
How to find the node number that corresponds to a quadrature point?
Should I compare coordinates?
If you do that at runtime then it will throw a ton of unnecessary
calculation into an inner loop. You could just hardcode the
permutation vectors; looking over quadrature_trap*.C it looks like
(mapping from quadrature point numbers to node numbers, and don't
trust me on this without verifying it):
QUAD4: {0, 1, 3, 2}
HEX8: {0, 1, 3, 2, 4, 5, 7, 6}
And then edge/triangle/prism elements are all the identity
permutation... and we don't support QTrap on pyramids apparently?
---
Roy
Thank you,
I'll code a permutation vector inside my application.
Michael.
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