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[Jan Stránský <question691610@xxxxxxxxxxxxxxxxxxxxx>]

Question #691610 on Yade changed:
https://answers.launchpad.net/yade/+question/691610

    Status: Open => Answered

Jan Stránský proposed the following answer:
> the particle size of the paper I mentioned is larger than mine
> So I guess this is the reason that they have larger ForceN.

how much larger?

>> 2*a*a = (2*a*a*a) ^ (2/3) * 2^(1/3)
> Why use this formula to transfer from per volume to per area ? Sorry I am slow about this :(

This is a rough trick. You know how many interactions there is per volume. But pressure is related to area. So you need to estimate how many interactions you have in some area.
For a cube, volume is a*a*a and face area is a*a, so the volume->area formula is
a*a = (a*a*a)^(2/3)
A = V^(2/3)
You can relate (number of interactions per volume) to (number of interactions per area) with the same formula.

You can use this cube estimation for even more rough estimation, as the results are not too much different.
But you can use actual dimensions, in your case a=0.07, b=0.14
V = a*a*b = 2*a*a*a
A = a*b = 2*a*a
A = someFunction(V)
2*a*a = someFunction(2*a*a*a)
Turns out that the relation is:
2*a*a = (2*a*a*a) ^ (2/3) * 2^(1/3)

cheers
Jan

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