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Why does 3D simulation cost so much memory?
Posted 30.08.2012, 07:00 GMT-4 Studies & Solvers 2 Replies
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Hi, all
For 2D, several millions DOF cost less 50G, however for 3D only 1 million DOF will cost more than 100G. I understand the consumed memory is proportional to DOF. But the coefficient should be the same for 2D and 3D, isn't it?
Thanks!
For 2D, several millions DOF cost less 50G, however for 3D only 1 million DOF will cost more than 100G. I understand the consumed memory is proportional to DOF. But the coefficient should be the same for 2D and 3D, isn't it?
Thanks!
2 Replies Last Post 30.08.2012, 18:28 GMT-4
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Posted:
1 decade ago
Hello Pu,
Memory usage is much greater in 3D than in 2D, in COMSOL as in any other FEA software.
Here's a quick-and-dirty explanation why:
Consider a regular arrangement of four square-shaped elements that share one central node. For quadratic elements, that central node is in the same element as all the other 24 nodes, so that node will contribute 25x25=625 non-zero entries to the stiffness matrix.
Now consider a regular arrangement of 8 cube-shaped elements elements that share one central node. For quadratic elements, that central node is in the same element as all the other 124 nodes, so that node will contribute 125x125=15,625 non-zero entries to the stiffness matrix.
So you see from the example above that in the 3D case each node can contribute many more non-zero entries to the stiffness matrix. Those entries need to be stored, and that takes memory.
Jeff
PS: This is a quick-and-dirty explanation because in reality memory is not used only when you assemble the stiffness matrix but also when you solve the problem, so things are certainly more complex than what the above may suggest.
Memory usage is much greater in 3D than in 2D, in COMSOL as in any other FEA software.
Here's a quick-and-dirty explanation why:
Consider a regular arrangement of four square-shaped elements that share one central node. For quadratic elements, that central node is in the same element as all the other 24 nodes, so that node will contribute 25x25=625 non-zero entries to the stiffness matrix.
Now consider a regular arrangement of 8 cube-shaped elements elements that share one central node. For quadratic elements, that central node is in the same element as all the other 124 nodes, so that node will contribute 125x125=15,625 non-zero entries to the stiffness matrix.
So you see from the example above that in the 3D case each node can contribute many more non-zero entries to the stiffness matrix. Those entries need to be stored, and that takes memory.
Jeff
PS: This is a quick-and-dirty explanation because in reality memory is not used only when you assemble the stiffness matrix but also when you solve the problem, so things are certainly more complex than what the above may suggest.
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Posted:
1 decade ago
Thanks Jeff!
Now I have a rough understanding. The proportional coefficients of 2D and 3D are very different in value...
Now I have a rough understanding. The proportional coefficients of 2D and 3D are very different in value...