Post-pro: Deformed Shape + scale factor

Hello,

I have difficulties selecting the right scale factor.

I want to compare the Deformed Shapes of an object, under different boundary conditions (computed with Code Aster). I am doing two simulations and importing two mesh files (med format) within Post-pro module of Salomé.

Then, I display the Deformed Shapes. Here, both different deformed shapes from the two results can be displayed with different default scale factor. If I leave as it is, then both of them have the same geometry. Also, I am quite sure the resulting geometries should be really different (elastic case with a pressure difference of 400%). However, if I choose a scale factor of 1 for both, the deformed shapes are the same as the initial mesh. Deformed shapes are at least different from the initial geometry.

I thought that Deformed shape superimposed with Scalar map (applied on a RESU__DEPL field computed from Code_Aster) could give a visual on the difference between initial and final step. Am I right?

I do not understand what the scale factor stands for here. Documentation says:

Scale Factor: visualization scaling of all vector values applied to the cells of the mesh

Does it mean it is a scaling multiplier applied to the displacement vector before displaying?

Therefore, how to choose it properly?

Thank you very much

I do it and didn't see any issue.

Try to place two meshes one over the other and use 1 as amplification factor on boths. After this, try to zoom very strictly and see if nodes are displaced on the two representations.

May be your problem is that the deformation is very little if compared to the dimensions of the model so you don't see relevant differences when the pressure varies of 400%.

In my opnion, an amplification factor = 1 menas that the post processor sows the mesh with the real deformation.

Example: a point of coordinates 0,0,50 and a deformation of 1%, will move to 0,0,50.5. If you choose ampl.factor = 1, you will find it in 0,0,50.5; if you chose ampl.factor =2 you will find it in 0,0,51.

You were right. The structure was locked and an increase of 400% in pressure did not change much. 

The amplification factor works correctly.

Thanks again.