ffc team mailing list archive

[Anders Logg <logg@xxxxxxxxx>]

Add -f quadrature_order=n when compiling for a high enough value of n.

-- 
Anders


On Tue, Jun 23, 2009 at 04:28:44PM -0400, Shawn Walker wrote:
> I've almost got this working.  When I compare the stiffness matrix from 
> my code to the one from dolfin the max difference in the entries is 
> machine precision.  This is when the mesh is straight (no curves).  If I 
> have a slightly curved mesh, the error is 0.0162, but I think this is 
> because the quad rule that FFC generated only has 4 points and I only 
> have two triangles with diameter approx 0.5 (not small enough resolve the 
> error).
>
> Is there a way to tell FFC to use more quad points than it `should'?  
> That way I can verify that I am computing this properly.
>
> - Shawn
>
> On Tue, 23 Jun 2009, Anders Logg wrote:
>
>> DOLFIN assumes that the mesh is ordered locally so that the local
>> vertices on each triangle are ordered increasingly. Take a look at the
>> XML files in data/meshes in DOLFIN and you'll see what I mean.
>>
>>
>>
>> On Tue, Jun 23, 2009 at 12:11:13PM -0400, Shawn Walker wrote:
>>> ok.  But I have one more question.
>>>
>>> Is there any kind of re-ordering of nodes that dolfin does internally?  I
>>> thought I remembered seeing this.  I have been trying to compare two
>>> matrices, and it looks like the entries are the same, except the ordering
>>> is a little different.
>>>
>>> - Shawn
>>>
>>> On Tue, 23 Jun 2009, Kristian Oelgaard wrote:
>>>
>>>> Quoting Shawn Walker <walker@xxxxxxxxxxxxxxx>:
>>>>
>>>>> Hello.  I need to know how FFC orders P2 Lagrange nodes.  I need to know
>>>>> this in order to compare a FEM matrix that I created in another program.
>>>>>
>>>>> For example, say you have one triangle with the following nodes:
>>>>>
>>>>> [0 1 2 3 4 5]
>>>>>
>>>>> Where is the #3 node in relation to the #0, #1, and #2 nodes?  Is #3
>>>>> opposite the #0 node?
>>>>
>>>> Yes.
>>>>
>>>> I usually look at the tabulate_dofs() function, for the P2 case:
>>>>
>>>> dofs[0] = c.entity_indices[0][0];
>>>> dofs[1] = c.entity_indices[0][1];
>>>> dofs[2] = c.entity_indices[0][2];
>>>> unsigned int offset = m.num_entities[0];
>>>> dofs[3] = offset + c.entity_indices[1][0];
>>>> dofs[4] = offset + c.entity_indices[1][1];
>>>> dofs[5] = offset + c.entity_indices[1][2];
>>>>
>>>> So you see that the first 3 dofs are the vertices, the next 3 are the midpoints
>>>> of the edges. Then you need to know the UFC ordering of edges to figure out
>>>> that e0 is the edge which does not contain vertex 0 (opposite to v0) etc.
>>>>
>>>> Kristian
>>>>
>>>>> - Shawn
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>>>>>
>>>>
>>>>
>>>>
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