Question #698365 on Yade changed:
https://answers.launchpad.net/yade/+question/698365
Status: Open => Answered
Jan Stránský proposed the following answer:
> I would expect incidentVel for the contact to give almost zero
why almost zero, why not strict zero? :-)
If almost zero, what is the limit where it is almost and what not?
> I would expect incidentVel for the contact to give almost zero, but it
gives 6.8847e-04, one order of magnitude less than the linear velocities
of the spheres in contact.
incidentVel is independent of absolute values of velocities of the particles.
Only their difference matters.
You could have two particles at the speed of light (or even more in a model
:-), still they can have arbitrary incidentVel (depending also significantly on
rotation).
You could have two particles with fixed position, still they can have arbitrary
incidentVel (bacause of rotation).
> Is it normal that there is relative shear displacement at a contact
that does not satisfy the sliding criterion?
yes, absolutely
> According to the contact force they don't slide, according to the
relative velocity at the contact they do.
depends on the definition of "slide".
They do not slide from friction point of view (there is no energy dissipation
by friction).
But they still have mutual shear displacement/velocity, although in an elastic
regime Fs=G*us, as explained by Karol.
As discussed, (under general loading conditions, not like nothing
happens or free fall etc.) there is no reason why two "free" (non-clump
etc.) spheres should have no contact "sliding" (zero mutual
displacement/velocity, normal or shear).
The contact geometry (sliding, penetrationDepth etc.) influences (through
computed forces) PARTIALLY the motion of particles. Actual motion of particles
is SUM of ALL contacts.
The motion of particles FULLY determines the contact geometry (sliding etc.).
Therefore, you cannot expect certain exact value of contact geometry based only
on that one contact.
Cheers
Jan
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