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
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