yade-dev team mailing list archive

[Vincent Richefeu <richefeu@xxxxxxxxxxxxxxx>]

Le 9 févr. 2011 à 17:45, Bruno Chareyre a écrit :

> 
>> Oh simply cubic. Say that you divide the main cell into 8 cubic parts
>> and then you get the average of stress in each of them (I simply want
>> to compare them). 
> Here is the algorithm :
> 
> foreach sphere:
>    cubeStress+=bodyStress(sphere.id)*volumeInsideCube
> cubeStress/=cubeVolume
> 
> Then if you have N cubes, sum_N(cubeStress)/N will be exactly the stress
> in the cell.
> 
> Probably similar to what Vincent suggests. I don't think it is
> equivalent with weighting by branch length, since volumeInsideCube is a
> non linear function of position.
> 
> Writing the volumeInsideCube function will be the boring part...
> 
> Bruno

I agree with you Bruno. Your 3-line algorithm is what I suggest in a more readable way (less words = easier to understand)

Chiara, I have no reference about it. Just some discussions with JJ Moreau when I was in Montpellier.
I can scan and send you some explanations if you want, but I think Bruno's mail is clear enough.

I don't know if you use sphere bodies. If yes, computing a local mean stress could be easier with a function volumeSphereInsideSphere (instead of volumeInsideCube), but sum_N(sphereStress)/N will NOT be the mean stress in the cell...

Vincent