Question about fix rigid

Submitted by ericparteli on Wed, 08/29/2012 - 12:21

Hi,
I have got a question about the fix rigid.
When two clumps made of spherical particles (hybrid molecular granular) collide with each other, is the deformation on the particles that are participating at the collision calculated using the masses "m1 and m2" of each colliding sphere or the total masses "M1 and M2" of each clump?
I know that the motion of the clumps after the collision is computed by considering the moment of inertia "I1, I2" and total masses "M1, M2" of the clumps, but I am not sure whether "m1 and m2" (only the masses of the colliding spheres) or rather "M1 and M2" (the total masses of the clumps) are considered to compute the deformation during the collision and to solve the pair/gran/ (hertz or hooke) sytle equations then.
Thanks in advance for the help!
Best
Eric

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ckloss's picture

ckloss | Wed, 08/29/2012 - 14:04

Hi Eric,

for calculating the dissipation, the mass of the clump is used rather than the mass of the single particle. this is coded explicitly in the pair style

Cheers, Christoph

ericparteli | Thu, 08/30/2012 - 11:24

Dear Christoph,

thanks so much for your prompt reply!
Indeed this is clear in the pair_gran_history_history.cpp file:
if (fix_rigid) {
if(body[i] >= 0) mi = masstotal[body[i]];
if(body[j] >= 0) mj = masstotal[body[j]];
}

I have a final question: the effective radius is Ri*Rj / (Ri+Rj) where Ri and Rj are the particle radii of the colliding spheres, right? The only reference made to the size of the clump is through the total mass (masstotal[body[i,j]]) is it correct? Indeed I find following lines in the aforementioned file,
double ri = atom->radius[ip];
double rj = atom->radius[jp];
double reff=ri*rj/(ri+rj);
but I am not sure if I am missing some code typed elsewhere.

Thanks again in advance for clarifying this final issue.

Best
Eric
P.S. Oh, and I forgot to mention, the reason for these questions is that I have to give a talk next week about all this and therefore I would like to be sure that I didn't miss anything in the code!

ckloss's picture

ckloss | Thu, 08/30/2012 - 11:59

Hi Eric,

>>I have a final question: the effective radius is Ri*Rj / (Ri+Rj) where Ri and Rj are the particle radii of the colliding spheres, right?
yes - the sphere radii are used for that as they describe the geometric deformation that is happening at the contact point

Cheers, Christoph