Particles center of mass
Hello everyone,
I have a problem with the calculation of the gyration radius in Python. (for postprocessing)
Agglomerates with three primary particles have a gyration radius that is smaller than the radius of the primary particles. It is not possible!
I double check the calculation and it is completely true. The only thing that I am not sure about it, is the particle's position.
I have a question regarding the particle position that EDEM provides. Is it exactly the particle's center of mass?
Please see the particle positions for an agglomerate containing 3 particles:
1:[-0.00021583 0.005800961 1.99864E-05 ]
2:[-0.00020695 0.00582015 6.49494E-05 ]
3:[-0.000205903 0.005802919 -2.86589E-05 ]
R1=2.5e-5 (primary particles radius)
Rg/R1=0.846!
The formula for gyration radius and shape on agglomerates have been attached.
Regards,
Fatemeh
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Hi,
Yes, the given particle position is its centre of mass. Whether the particle is centred at (0,0,0) though, is going to depend on whether you centred the particle during creation (you should do this unless you have a very specific reason not to). As long as the particle was centred during creation, the off-sets for the spheres will all be relative to the centre of mass / particle position. So with the particle position and orientation data, you can use that to determine the individual sphere locations and do your sum as needed.
Cheers,
Richard
Dear Richard,
Many thanks for your response.
I am a bit confused about this sentence:
Whether the particle is centred at (0,0,0) though, is going to depend on whether you centred the particle during creation (you should do this unless you have a very specific reason not to)
Could you please explain a bit more?
The geometry is a horizontal tube and particles are injected from the entrance. (it is a CFD-DEM simulation)
The position of injection is not(0,0,0).
I calculate r for each particle based on this formula:
r=sqrt(x^2+y^2+z^2).
and use this r for the calculation of the gyration radius.
Regards,
Fatemeh
Altair EmployeeDear Richard,
Many thanks for your response.
I am a bit confused about this sentence:
Whether the particle is centred at (0,0,0) though, is going to depend on whether you centred the particle during creation (you should do this unless you have a very specific reason not to)
Could you please explain a bit more?The geometry is a horizontal tube and particles are injected from the entrance. (it is a CFD-DEM simulation)
The position of injection is not(0,0,0).
I calculate r for each particle based on this formula:
r=sqrt(x^2+y^2+z^2).
and use this r for the calculation of the gyration radius.
Regards,
Fatemeh
I mean this tick box:
If the 'Automatically Center Particle' checkbox isn't checked, the particle centre will be somewhere other than the origin. The position reported by EDEM is the centre of mass, so if the particle isn't centred around the origin then any positions reported will include whatever off-set was included when setting the particle up in the Creator.
Long story short...you probably want the box ticked, so I'd double check that. The only reason I can think of to not tick this is if the particle has a non-homogenous density that you're trying to model.
Richard
Hi,
Yes, the given particle position is its centre of mass. Whether the particle is centred at (0,0,0) though, is going to depend on whether you centred the particle during creation (you should do this unless you have a very specific reason not to). As long as the particle was centred during creation, the off-sets for the spheres will all be relative to the centre of mass / particle position. So with the particle position and orientation data, you can use that to determine the individual sphere locations and do your sum as needed.
Cheers,
Richard