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[Jan Stránský <honzik.stransky@xxxxxxxxx>]

Public bug reported:

Hi all,

consider

O.forces.addF(id,(2,0,0),True)
... # some code, a few steps or whatever
O.forces.addF(id,(4,0,0),True)

what should be the result? Intuitively (according to addF name), I would
expect applied force (6,0,0). However, the permanent force is rewritten
with (4,0,0), so applied force (4,0,0) is the result..

I suggest to split addF to 2 cases:
1) addF without permanent parameter
2) setPermanentForce ()or something similar), substituting addF(permanent=True), with the name reflecting more accurately what it really does..

any suggestions or comments?
cheers
Jan

** Affects: yade
     Importance: Low
         Status: New

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https://bugs.launchpad.net/bugs/1476159

Title:
  several ForceContainer.addF behaves non-intuitively with
  permanent=True

Status in Yade:
  New

Bug description:
  Hi all,

  consider

  O.forces.addF(id,(2,0,0),True)
  ... # some code, a few steps or whatever
  O.forces.addF(id,(4,0,0),True)

  what should be the result? Intuitively (according to addF name), I
  would expect applied force (6,0,0). However, the permanent force is
  rewritten with (4,0,0), so applied force (4,0,0) is the result..

  I suggest to split addF to 2 cases:
  1) addF without permanent parameter
  2) setPermanentForce ()or something similar), substituting addF(permanent=True), with the name reflecting more accurately what it really does..

  any suggestions or comments?
  cheers
  Jan

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