[Gmsh] The volume of the second order tetrahedron

[Christophe Geuzaine]

On 08/03/11 04:46, Geordie McBain wrote:
> 2011/3/7 Mikhail Artemiev<artemiev.mikhail at ngs.ru>:
>>>> Hello, Geordie.
>>>> Thank you for reply.
>>>> I used "Tools - Visibility - Numeric - Mesh - Hide all elements - Show
>>>> Element (for instance, 1365)" to draw that figure.
>>>
>>> Oh right, O.K.  It's a funny looking shape, isn't it.  Perhaps that's
>>> just how Gmsh depicts nonlinear elements?
>>
>> It's a very important question!
>
> Indeed.  I don't know much about the internals of the Gmsh, but let's
> see if we can't make some progress.
>
>> Please, look at 2 figures:
>> this is a shpere that was approximated by first order tets
>> http://saveimg.ru/show-image.php?id=47cdfa0e6f7a01f5dee8880c7081e151
>> this is a sphere that was approximated by second order tets
>> http://saveimg.ru/show-image.php?id=fa66051392be4e36e2262055272170e1
>> These 2 meshes was created by gmsh from one geo file (and with the same
>> characteristic lengths).
>
> It looks as though all the nodes on the outer six-node triangular
> faces of the ten-node tetrahedra lie on the geometric sphere.  Is that
> right?  If so, that's good, and your quadratic mesh is a better
> representation of the geometry than the linear one.
>
>> I will wonder if it is a feature of visualization.
>
> Why?  I don't know how the visualizer works internally but if (say)
> all it can do is depict triangles, then what it's showing is a
> reasonable representation of a quadratic tetrahedron, isn't it?
>

Hello - The visualization of high-order mesh elements can be enhanced 
with the following "Mesh.NumSubEdges" option (default=2),. For example:

gmsh demos/sphere.geo -clscale 4 -order 4 -string "Mesh.NumSubEdges=10;"




>> Standard 10-node second order tetrahedron differs from 4-node first order
>> tetrahedron by adding 6 node in the middles of the edges of tetrahedron.
>
> No, the additional six nodes don't have to be on the midpoints of the
> edges of the tetrahedron.  They can be, and probably will be if you're
> only meshing a polyhedron, but in general no.  There are some
> restrictions, they can't be just anywhere (or the Jacobian of the
> transformation from the canonical element will change sign within the
> element), but they can move around a bit, and indeed this is most
> desirable when meshing a curved geometry.
>
>> I think that gmsh not only adds new 6 nodes but changes the coordinates of
>> these nodes too.
>
> Perhaps to make them lie on the geometric sphere?
>
>> Therefore we have nonstandard 10-node quadratic tetrahedron and the formulae
>> of the shape functions defined on the standard one don't work.
>
> I think they will. I'm still hopeful this is a standard quadratic
> tetrahedron.  In terms of figure 4.3.1 on p. 228 of Ciarlet's book,
> referred to earlier, I trust that these are are `isoparametric'
> tetrahedra `of type (2)'; you'll also find drawings of `three
> isoparametric tetrahedra of type (2)' in figure 4.4.2 on p. 251.
>
>> Am I wrong?
>
> I'm not sure, but I'm hopeful Gmsh is correctly approximating `a
> curved boundary with isoparametric finite element' as described by
> Ciarlet pp. 248 ff.
>
> I presume the same thing happens in two-dimensions, e.g. using 6-node
> triangles to mesh a sector.  I've tried this, as attached.  It looks
> good.  In sector.png, I've gotten Gmsh to number the nodes, as they
> appear in sector.msh, which was generated by "gmsh -2 -order 2
> sector.geo". The edges of each triangle along the geometric perimeter
> aren't drawn as curves (parabolas), but I think that's just economy of
> depiction, and we're free to treat the elements as isoparametric
> triangles of type (2), no?
>
>
>
> _______________________________________________
> gmsh mailing list
> gmsh at geuz.org
> http://www.geuz.org/mailman/listinfo/gmsh


-- 
Prof. Christophe Geuzaine
University of Liege, Electrical Engineering and Computer Science
http://www.montefiore.ulg.ac.be/~geuzaine