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Meaning of "node" vs. "vertex" vs. "degree of freedom" for linear finite elements.
Posted 01.04.2013, 13:52 GMT-4 Mesh, Studies & Solvers Version 4.2a 6 Replies
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Hi there,
can anybody tell me what COMSOL means by the word "node" in contrast to "vertex" or "degree of freedom" when it comes to the discretization? I have read
www.comsol.com/support/knowledgebase/875/
but am slightly confused about the fact that for linear elements both in 2D and 3D the number of nodes is smaller than the number of elements.
Let's assume we have only one dependend variable. Then the number of DOFs should equal the number of nodes, right?
Furthermore let's assume that we have linear Lagrange elements in 2D based on a triangle mesh. This means that the nodes coincide with the vertices of the triangle mesh. As each triangle has 3 vertices which are shared between surrounding triangles, I would have expected that we end up with about the same number of nodes as we have triangles.
Why does the knowledge base entry above then state the formula
(#nodes) = 0.5 * (#elements)
and even
(#nodes) = 0.3 * (#elements) for the 3D case?
How can there be less nodes/vertices than elements/triangles?
What is the relation between a "node" and a "vertex"?
Is it true that every DOF also is a node and that DOFs merely are collections of nodes with respect to the particular dependent variables?
I'm a bit confused here. I know my FEM theory from maths lectures at university, but it seems as if COMSOL uses the notions of "nodes", "vertices" and "DOFs" slightly differently.
Thanks a lot in advance,
Joerg
can anybody tell me what COMSOL means by the word "node" in contrast to "vertex" or "degree of freedom" when it comes to the discretization? I have read
www.comsol.com/support/knowledgebase/875/
but am slightly confused about the fact that for linear elements both in 2D and 3D the number of nodes is smaller than the number of elements.
Let's assume we have only one dependend variable. Then the number of DOFs should equal the number of nodes, right?
Furthermore let's assume that we have linear Lagrange elements in 2D based on a triangle mesh. This means that the nodes coincide with the vertices of the triangle mesh. As each triangle has 3 vertices which are shared between surrounding triangles, I would have expected that we end up with about the same number of nodes as we have triangles.
Why does the knowledge base entry above then state the formula
(#nodes) = 0.5 * (#elements)
and even
(#nodes) = 0.3 * (#elements) for the 3D case?
How can there be less nodes/vertices than elements/triangles?
What is the relation between a "node" and a "vertex"?
Is it true that every DOF also is a node and that DOFs merely are collections of nodes with respect to the particular dependent variables?
I'm a bit confused here. I know my FEM theory from maths lectures at university, but it seems as if COMSOL uses the notions of "nodes", "vertices" and "DOFs" slightly differently.
Thanks a lot in advance,
Joerg
6 Replies Last Post 04.04.2013, 12:09 GMT-4
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Posted:
1 decade ago
Hi
I read "vertex" as a geometrical vertex (intersection of two edges, or three planes non-parallel ...)
Vertex is for me asynonyme to Point when used for an geometrical Object (I distinguish a geometrical Object and a FEM Element = Domain, Boundary, possibly Edge, or Point which are made up once the geometry has been "analysed" = simplified)
Nodes are the limits (vertex or points in space) of the Mesh Elements.
While DoF is a combination of mesh elements, mesh nodes and discretization,
A 2nd order discretization, is close to saying you add a virtual node between each mesh physical node hence a triangle elements for one scalar depednent variable is of 3 DoF first order mesh elements and contain 6 Dofs for second order discretization
(pls check again, as I'm replying from my memory only, not having access to the docs from here, tonight ;)
--
Good luck
Ivar
I read "vertex" as a geometrical vertex (intersection of two edges, or three planes non-parallel ...)
Vertex is for me asynonyme to Point when used for an geometrical Object (I distinguish a geometrical Object and a FEM Element = Domain, Boundary, possibly Edge, or Point which are made up once the geometry has been "analysed" = simplified)
Nodes are the limits (vertex or points in space) of the Mesh Elements.
While DoF is a combination of mesh elements, mesh nodes and discretization,
A 2nd order discretization, is close to saying you add a virtual node between each mesh physical node hence a triangle elements for one scalar depednent variable is of 3 DoF first order mesh elements and contain 6 Dofs for second order discretization
(pls check again, as I'm replying from my memory only, not having access to the docs from here, tonight ;)
--
Good luck
Ivar
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Posted:
1 decade ago
Hi Ivar,
thanks for your answer. I totally agree with you and that is how I understand FEM from theory. But what confuses me is the formulas above from the knowledge base. They seem to be contradicting. Can you maybe explain those to me? That would be great.
Greetings,
Joerg
thanks for your answer. I totally agree with you and that is how I understand FEM from theory. But what confuses me is the formulas above from the knowledge base. They seem to be contradicting. Can you maybe explain those to me? That would be great.
Greetings,
Joerg
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Posted:
1 decade ago
Joerg-
Why can't you have more elements than nodes? You definitely can. See attached mesh with 24 elements and 17 nodes.
Also, I created a simple cube and meshed with an extra fine mesh resolution with the free tetrahedral mesher.
Results:
~400k linear elems
~69k DoF (nodes)
That is a ratio of 0.175 -- very close to the value given by the Knowledge Base post you referenced:
(#of DoF) = 0.2 * (#of Elems)
Best regards,
Josh Thomas
AltaSim Technologies
Why can't you have more elements than nodes? You definitely can. See attached mesh with 24 elements and 17 nodes.
Also, I created a simple cube and meshed with an extra fine mesh resolution with the free tetrahedral mesher.
Results:
~400k linear elems
~69k DoF (nodes)
That is a ratio of 0.175 -- very close to the value given by the Knowledge Base post you referenced:
(#of DoF) = 0.2 * (#of Elems)
Best regards,
Josh Thomas
AltaSim Technologies
Attachments:
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Posted:
1 decade ago
Hey Josh,
thank you so much for this picture. That was a real eye opener. I always assumed that there must be more vertices/nodes than tetrahedra. That is why outputs like
==========
Results:
~400k linear elems
~69k DoF (nodes)
==========
confused me in the first place. I just didn't have a proper counter example. Thank you so much. Now my FEM world view is back to normal again. :)
Btw, do you have any idea where the ratio 0.2 is coming from?
Greetings Joerg
thank you so much for this picture. That was a real eye opener. I always assumed that there must be more vertices/nodes than tetrahedra. That is why outputs like
==========
Results:
~400k linear elems
~69k DoF (nodes)
==========
confused me in the first place. I just didn't have a proper counter example. Thank you so much. Now my FEM world view is back to normal again. :)
Btw, do you have any idea where the ratio 0.2 is coming from?
Greetings Joerg
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Posted:
1 decade ago
You're welcome, Joerg. A picture is worth a thousand words!
Where exactly does the 0.2 ratio come from? I'm not sure. A round-off of 0.175 maybe? I tried a cube and a sphere and got the same ratio of 0.175 for Extra fine free tet mesher.
As is pointed out in the Knowledge Base article, the real ratio is going to depend on how much surface area per volume your geometry has. My intuition tells me that a sphere (lowest surface area to volume ratio of any object) is going to have the smallest node to element ratio possible.
Best regards,
Josh Thomas
AltaSim Technologies
Where exactly does the 0.2 ratio come from? I'm not sure. A round-off of 0.175 maybe? I tried a cube and a sphere and got the same ratio of 0.175 for Extra fine free tet mesher.
As is pointed out in the Knowledge Base article, the real ratio is going to depend on how much surface area per volume your geometry has. My intuition tells me that a sphere (lowest surface area to volume ratio of any object) is going to have the smallest node to element ratio possible.
Best regards,
Josh Thomas
AltaSim Technologies
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Posted:
1 decade ago
That sounds plausible. Thank you once again for your help.
Joerg
Joerg