Learning Forum

[Jelte van Luenen]

For my project, I am extracting the B_vector information using the fields calculator from ansys Maxwell 2021 R1. In post-processing, I want to find the average B_vector (average for each of the vector elements) over a certain part of the geometry. Is there a way to find the size of the mesh elements given by the fields calculator such that I can take the weighted average to compensate for some mesh elements being larger than others? 

The data contains columns for [x-coordinate, y-coordinate, z-coordinate, b_x, b_y, b_z]. A small sample:

Vector data "SurfaceValue(Surface(AllObjects), B_Vector)"

NumElems 3656

9.6056089393257701e-02 5.3639681988800397e-02 0.0000000000000000e+00 1.7142683516020780e+00 9.9767139872533850e-01 0.0000000000000000e+00

9.4086033622851006e-02 5.3410703276034498e-02 0.0000000000000000e+00 1.7218518778208884e+00 9.8958311399298859e-01 0.0000000000000000e+00

9.5268805810850204e-02 5.5003246375166348e-02 0.0000000000000000e+00 1.7120729398496055e+00 9.8845994771479428e-01 0.0000000000000000e+00

Does anyone know of a way to do this averaging? I do not want to calculate the average inside the field calculator since this will limit my post-processing later on. 

 

Kind regards

 

[GLUO]

Hi,

No, you can only find the volume/size of object in field calculator, not a mesh element. Please calculate the size of mesh element using the exported data. You can click output>>write to get the node position and B vector.

GL

[Jelte van Luenen]

Hi GLUO,

Thanks for your reply. Do you know of any smart way to determine the size of the nodes? 
Since only the coordinates of a node and the quantity magnitude are given it is not trivial how dense the nodes are in a certain region and this is not constant over a single object. 

Kind regards

 

[GLUO]

No. There might be helpful script for similar type of problem on github.

GL

[Jelte van Luenen]

I have found an alternative solution by creating a regular grid on the surface and interpolating the data on that grid. 

Thanks for your help!