[Gmsh] The volume of the second order tetrahedron

[Geordie McBain]

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?

> 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?
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