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Generate and plot 3D solution, given 2D solution
Posted 08.04.2013, 09:38 GMT-4 Version 4.2a 4 Replies
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Hi all,
I'm using weak-form modeling to solve for a vector field u(x,y,z). The vector field has 3 scalar components u1, u2 and u3.
I know the functional form of the solution in the z-direction; it is given by
u(x,y,z) = u(x,y)*cos(Kz)
So I only solve for u(x,y) in a 2D-model. All I want to do now is take this solution u(x,y) and use it to plot u(x,y,z) in 3D displacement plot. However, I have been unable to do so. I can't find any buit-in way to do this.
Does anyone have an idea of how to do this?
Thanks!
I'm using weak-form modeling to solve for a vector field u(x,y,z). The vector field has 3 scalar components u1, u2 and u3.
I know the functional form of the solution in the z-direction; it is given by
u(x,y,z) = u(x,y)*cos(Kz)
So I only solve for u(x,y) in a 2D-model. All I want to do now is take this solution u(x,y) and use it to plot u(x,y,z) in 3D displacement plot. However, I have been unable to do so. I can't find any buit-in way to do this.
Does anyone have an idea of how to do this?
Thanks!
4 Replies Last Post 17.04.2013, 07:38 GMT-4
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Posted:
1 decade ago
You can do in different ways depending on you preferences.
1. Use the Parametric Extrusion dataset: Create a parameter z under Global Definitions, then add a Parametric sweep under the Study. Parameterize over z and solve. Then go to Results->Datasets and add the Parametric Extrusion 2D dataset. Add a 3D plot group and create your plots, using Dataset: Parametric Extrusion.
2. Add a 3D model with a mesh but no physics, of which the sole purpose is to visualize your solution: Right click the top node in the model tree and select Add model, 3D, and finish. Draw and mesh a geometry that fits your 2D model. Go to the 2D model and under Definitions, select Model Couplings->Linear Extrusion to set up the solution mapping from the 2D to 3D model.
kind regards
Niklas
1. Use the Parametric Extrusion dataset: Create a parameter z under Global Definitions, then add a Parametric sweep under the Study. Parameterize over z and solve. Then go to Results->Datasets and add the Parametric Extrusion 2D dataset. Add a 3D plot group and create your plots, using Dataset: Parametric Extrusion.
2. Add a 3D model with a mesh but no physics, of which the sole purpose is to visualize your solution: Right click the top node in the model tree and select Add model, 3D, and finish. Draw and mesh a geometry that fits your 2D model. Go to the 2D model and under Definitions, select Model Couplings->Linear Extrusion to set up the solution mapping from the 2D to 3D model.
kind regards
Niklas
Hi all,
I'm using weak-form modeling to solve for a vector field u(x,y,z). The vector field has 3 scalar components u1, u2 and u3.
I know the functional form of the solution in the z-direction; it is given by
u(x,y,z) = u(x,y)*cos(Kz)
So I only solve for u(x,y) in a 2D-model. All I want to do now is take this solution u(x,y) and use it to plot u(x,y,z) in 3D displacement plot. However, I have been unable to do so. I can't find any buit-in way to do this.
Does anyone have an idea of how to do this?
Thanks!
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Posted:
1 decade ago
Hi Niklas,
thanks for your response. I've tried both ways, without success.
(1) In this case I solve the model again for each value of "z", which takes a long time. This is unnecessary computational time, since I know the analytical connection between u(x,y,z) and u(x,y). But maybe I missed something...
(2) This seems to be the most promising suggestion, but I'm not able to implement it. The problem is that I want to connect a 2D domain to a 3D domain. The "linear extrusion" only allows me to connect the 2D domain to a particular 3D boundary. Furthermore, I don't see how I could implement my analytical z-dependency:
u1(x,y,z) = Real(u1(x,y)*exp(-i*K*z))
u2(x,y,z) = Real(u2(x,y)*exp(-i*K*z))
u3(x,y,z) = Real(u3(x,y)*exp(-i*K*z))
Any more ideas? My model is attached.
Your help is greatly appreciated!
thanks for your response. I've tried both ways, without success.
(1) In this case I solve the model again for each value of "z", which takes a long time. This is unnecessary computational time, since I know the analytical connection between u(x,y,z) and u(x,y). But maybe I missed something...
(2) This seems to be the most promising suggestion, but I'm not able to implement it. The problem is that I want to connect a 2D domain to a 3D domain. The "linear extrusion" only allows me to connect the 2D domain to a particular 3D boundary. Furthermore, I don't see how I could implement my analytical z-dependency:
u1(x,y,z) = Real(u1(x,y)*exp(-i*K*z))
u2(x,y,z) = Real(u2(x,y)*exp(-i*K*z))
u3(x,y,z) = Real(u3(x,y)*exp(-i*K*z))
Any more ideas? My model is attached.
Your help is greatly appreciated!
Attachments:
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Posted:
1 decade ago
Dear Raphael,
I attached a version of your model where I implemented the solution number 2 as explained by my colleague.
It is your file with few adaptations:
0. removed "z" from the parameter list; cleared the parametric solver sequence
1. described the linear extrusion with only 3 points; and filled in the destination vertices as well
2. changed the 3D model so that the z axis was the elongated one
3. solve study
4. add a new solution; now with the 3D model as shape
5. created a new 3D plot group; with a surface plot
6. used the expression mod1.linext1(u1)*exp(-i*K*z)
This works, but note that the K value is equal to 0, therefore you do not see anything "happening" in the z direction.
If you replace K by a value 2e7; you see that it nicely varies over the height.
Best regards,
Frank
ps. I also answered you directly via support.
I attached a version of your model where I implemented the solution number 2 as explained by my colleague.
It is your file with few adaptations:
0. removed "z" from the parameter list; cleared the parametric solver sequence
1. described the linear extrusion with only 3 points; and filled in the destination vertices as well
2. changed the 3D model so that the z axis was the elongated one
3. solve study
4. add a new solution; now with the 3D model as shape
5. created a new 3D plot group; with a surface plot
6. used the expression mod1.linext1(u1)*exp(-i*K*z)
This works, but note that the K value is equal to 0, therefore you do not see anything "happening" in the z direction.
If you replace K by a value 2e7; you see that it nicely varies over the height.
Best regards,
Frank
ps. I also answered you directly via support.
Attachments:
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Posted:
1 decade ago
Thanks a lot Frank, had a quick look and it seems to do exactly what I want! Will test it more extensively in the coming days.