# Elastic potential energy of polyhedra

Hi,

Following the discussion in [1], the normal stiffness "kn" in the Law2_PolyhedraG

When calculating the elastic potential energy, we currently use [2,3,4]:

scene->

"elastPote

where "kn" should be in [N/m] for the units to give energy values [Joules=N*m].

I think the energy calculation in the normal direction should change either as 0.5*(normalForc

where area the projection of the overlapping volume perpendicularly to the normal direction (this would work only if volumePower=1), or more safely, use the equivalentPenet

Cheers,

Vasileios

[1] https:/

[2] https:/

[3] https:/

[4] https:/

[5] https:/

[6] https:/

## Question information

- Language:
- English Edit question

- Status:
- Solved

- For:
- Yade Edit question

- Assignee:
- No assignee Edit question

- Solved by:
- Vasileios Angelidakis

- Solved:
- 2019-09-11

- Last query:
- 2019-09-11

- Last reply:
- 2019-09-06

Jan Stránský (honzik) said : | #1 |

Hi Vasileios,

good point. The equations [2,3,4] are wrong. In general, elastic energy is

int F(u) du

in case F=k*u, int F du = int k*u du = 0.5*k*u^2 = 0.5*F^2/k

But it is not the case of polyhedrons interaction, where the function F(u) is actually unknown.

But probably some estimation could be done, e.g. F=k*V and assume V being proportional to u^3.

Or the approaches mentioned by you (today is too late for me to investigate them :-)

> Btw, looking at the calculation of the equivalentPenet

estimation [7]

cheers

Jan

[7] https:/

Hi Jan,

Thanks for taking a look at this.

I think assuming that V is proportional to u^3 using a fixed estimation can be risky, since it depends highly on the geometry of the overlap region. :p

I like the idea of an equivalentPenet

Currently, this distance is estimated by dividing the penetrationVolume by the area perpendicularly to the contact normal direction.

I think it would be more straightforward if we calculated two opposite points of the overlap volume, along the normal direction, passing from the contact point, and calculate the equivalentPenet

Calculating the equivalentPenet

All the best,

Vasileios

Jan Stránský (honzik) said : | #3 |

> I think assuming that V is proportional to u^3 using a fixed estimation can be risky, since it depends highly on the geometry of the overlap region. :p

> the function F(u) is actually unknown.

this is what I meant :-)

> two opposite points of the overlap volume, along the normal direction, passing from the contact point, and calculate the equivalentPenet

then it would be needed to distinguish equivalentPenet

cheers

Jan

Good point, thanks Jan!

Regards,

Vasileios