new g.irving et al. siggraph 2007 ki-hoon...

Ki-hoon Kim

2013-12-27

Computer Graphics @ Korea University

G.Irving et al.SIGGRAPH 2007

Ki-hoon Kim | 2013-12-27 | # 2Computer Graphics @ Korea University

• Treating both object contact and self-contact as linear constraints during the incompressible solve.

• Conserves the volume locally near each node in a finite element mesh.

Abstract


• “The most important rule to squash and stretch is that, no matter how squashed or stretched out a particular object gets, its volume remains constant.”

Principles of Traditional Animation Applied to 3D Computer Animation

• John LasseterSIGGRAPH 1987

IntroductionVolume Conserving


• Sweep-based Freeform Deformations

Yoon and KimEUROGRAPHICS 2006

• Swirling-sweepers: Constant-volume modeling

A. Angelidis et al.Graphical Models 2006

• Vector field based shape deformations.

W.Funck et al.SIGGRAPH 2006

IntroductionVolume Conserving: Shape Modeling


• Fast Volume Preservation for a Mass-Spring System

Hong et al.IEEE Computer Graphics and Applications 2006

• Localized volume preservation for simulation and animation

Punak and PetersPoster, SIGGRAPH 2006

IntroductionVolume Conserving: Only Total Volume


• A versatile and robust model for geometrically complex deformable solids.

M. Teschner et al.CGI 2004

• Limitation

Volume is not preserved in the presence of competing forces

IntroductionVolume Conserving: Approximate Local Volume(with simple spring-like forces)


• Modified Semi-implicit Newmark scheme.

Simulation of clothing with folds and wrinkles.

• R. Bridson et al.SCA 2003

Time Discretization


Time DiscretizationOverview


• Step 2 and 4 combine to form(w/o collision)

𝑥𝑛+1 = 𝑥𝑛 + Δ𝑡𝑣∗𝑛+

12 + Δ𝑡𝛾𝑥

• The final volumes should equal the initial volumes𝑉 𝑥𝑛+1 = 𝑉 𝑥0

𝑉 𝑥𝑛 + Δ𝑡𝐝𝐢𝐯𝑣∗𝑛+

12 + Δ𝑡𝐝𝐢𝐯𝛾𝑥 = 𝑉 𝑥0

where, 𝐝𝐢𝐯 is volume-weighted divergence.

• Similar to the typical pressure correction in fluids,

𝑣𝑛+1/2 = 𝑣∗𝑛+1/2

−𝛻 𝑝

𝜌

𝛾𝑥 = −𝛻 𝑝

𝜌

Time Discretizationstep 2. Pressure: Volume Conserving


• Contd.

𝑉 𝑥𝑛 + Δ𝑡𝐝𝐢𝐯𝑣∗𝑛+

12 + Δ𝑡𝐝𝐢𝐯𝑀−1𝐠𝐫𝐚𝐝 𝑝 = 𝑉 𝑥0

𝐠𝐫𝐚𝐝 is the volume-weighted gradient and 𝑀 is the mass matrix.

• Rearranging into standard Poisson equation form,

−𝐝𝐢𝐯𝑀−1𝐠𝐫𝐚𝐝 𝑝 = 𝐝𝐢𝐯𝑣∗𝑛+

12 −

𝑉 𝑥𝑛 − 𝑉 𝑥0

Δ𝑡⋯ (1)

which can be solved for 𝑝, and then 𝛾𝑥 = −𝑀−1𝐠𝐫𝐚𝐝 𝑝.

Time Discretizationstep 2. Pressure: Volume Conserving(Contd.)


• We correct the velocity to be divergence free in step 6, although this can be executed at any point in the algorithm since it is a static projection.

• Taking the divergence of step 6 and setting 𝛻 ∙ 𝑣𝑛 = 𝛻 ∙ 𝑣∗

𝑛 + 𝛻 ∙ 𝛾𝑣 = 0

𝛾𝑣 = −𝛻 𝑝

𝜌

• Similar to (1) we obtain

−𝐝𝐢𝐯𝑀−1𝐠𝐫𝐚𝐝 𝑝 = 𝐝𝐢𝐯𝑣∗𝑛+

12⋯ (2)

which can be solved for 𝑝, and then 𝛾𝑣 = −𝑀−1𝐠𝐫𝐚𝐝 𝑝.

Time Discretizationstep 6. Pressure: Divergence free velocity field


• Mesh with 𝑛 nodes has more than 4𝑛 tetrahedrons.

Over-constrained system makes locking.

• We avoid locking by enforcing incompressibility on one-rings.

Spatial DiscretizationOverview


• Standard finite volume discretization with all information coloacted on the nodes of the mesh.

Finite Volume Methods for the Simulation of Skeletal Muscle

• J. Teran et al.SCA 2003

• Define

𝐃 = 𝐱𝟏 − 𝐱𝟎, 𝐱𝟐 − 𝐱𝟎, 𝐱𝟑 − 𝐱𝟎

𝐃 = 𝐯𝟏 − 𝐯𝟎, 𝐯𝟐 − 𝐯𝟎 , 𝐯𝟑 − 𝐯𝟎

𝐁 = 𝑉𝐃−𝐓 = −(𝐚𝐧𝟏, 𝐚𝐧𝟐, 𝐚𝐧𝟑)/3

• 𝑉 is volume of the tetrahedron.

• 𝐚𝐧𝒌 is area-weighted normal opposite vertex 𝑘.

Spatial DiscretizationFinite volume method


• Linearly interpolated velocity field𝐯 𝐱 = 𝐃𝐃−1 𝐱 − 𝐱0 𝐯0

and

𝛻 ∙ 𝐯 𝐱 = 𝐭𝐫 𝐃𝐃−1 = 𝐃−T: 𝐃

volume-weighted divergence at node 𝑘 is

• is the set of tetrahedra incident on 𝑘.

Spatial DiscretizationVolume-weighted Divergence


• Gradient operator: negative transpose of the divergence operator.

(1) and (2) result in symmetric positive definite system.

• We want 𝛁𝑝, 𝐯 = 𝑝,−𝛁 ∙ 𝐯 , that is ( 𝐚, 𝐛 = 𝐚𝑻𝐛 and 𝑎, 𝑏 = 𝑎𝑏)

Ω

𝛻𝑝 ∙ 𝐯 𝑑𝐱 + Ω

𝑝𝛻 ∙ 𝐯 𝑑𝐱 = Ω

𝛻 ∙ (𝑝𝐯) 𝑑𝐱 = 𝜕Ω

𝑝𝐯 ∙ 𝑑𝐒 = 𝟎

Dirichlet boundary condition. (𝑝 = 0 on the boundary)

Spatial DiscretizationVolume-weighted Gradient and Pressure: Define


• We partition the pressure field into

𝑝 =

𝑡∈𝑹(𝑘)

𝑝𝑡

𝑝𝑡 is a pressure field that agrees with 𝑝 in 𝑡.

• The linearity of the gradient operator gives

𝐠𝐫𝐚𝐝 𝑝 𝑘 =

𝑡∈𝑹(𝑘)

𝐠𝐫𝐚𝐝 𝑝𝑡 𝑘

Spatial DiscretizationVolume-weighted Gradient and Pressure: Pressure


• From divergence operator on a single tetrahedron

𝐝𝐢𝐯 𝐯 𝑘 =1

4𝐁: 𝐃 = −

1

12

𝑗=1

3

𝐯𝑗 − 𝐯0 ∙ 𝐚𝐧𝑗 = −1

12

𝑗=0

3

𝐯𝑗 ∙ 𝐚𝐧𝑗

−𝐝𝐢𝐯 =1

12𝐍 𝐍 𝐍 𝐍 T where 𝐍 = 𝐚𝐧0

T 𝐚𝐧1T 𝐚𝐧2

T 𝐚𝐧3T T

.

• So, gradient operator on a single tetrahedron is

𝐠𝐫𝐚𝐝 =1

12𝐍 𝐍 𝐍 𝐍

,and gradient pressure at a node 𝑘 is

𝐠𝐫𝐚𝐝 𝑝 𝑘 =1

12𝐍𝑘

𝑗=0

3

𝑝𝑗 =1

3𝐚𝐧𝑘 𝑝

Spatial DiscretizationVolume-weighted Gradient and Pressure: On a single tetrahedron


• Summing over 𝑹(𝑘) gives

𝐠𝐫𝐚𝐝 𝑝 𝑘 =1

3

𝑡∈𝑹(𝑘)

𝐚𝐧𝑡,𝑘 𝑝𝑘

This equation is exactly standard FVM force for stress of 𝑝.

Spatial DiscretizationVolume-weighted Gradient and Pressure: On multi tetrahedrons


Collision and ContactRemind Time Discretization


• Problem

Over-constraints can cause serious artifacts

• Competing constraints

• Resulting in unusably tangled surface

• Solution Propose

Incorporate collision constraints into our Poisson equation.

Collision and ContactRemind Time Discretization: Conclusion


• Step 5.

Set the position and velocity of particles to respect collision.

𝐜𝑇𝐱 ≥ 0

• 𝐧𝑇 𝐱𝑝 − 𝑤1𝐱1 − 𝑤2𝐱2 − 𝑤3𝐱3 ≥ 0

Point-Triangle pair collision

𝑤𝑖 are barycentric weights.

𝐧 is triangle’s normal

• 𝒔𝑇 1 − 𝛼1 𝐱1 + 𝛼1𝐱2 − (1 − 𝛼2)𝐱3 − 𝛼2𝐱3 ≥ 0

Edge-Edge pair collision

𝛼𝑖 are position of the interacting points

𝒔 is shortest vector between the interacting segments

• Step 6.

Solve volume conserving problem

Solver change the velocity

• Δ𝐯 is the change in velocity.

Collision and ContactProblem Analysis


• Step 7.

Maintain the correct normal velocity for colliding particles.

i.e. 𝐜TΔ𝐯 = 0

• 𝐧𝑇 𝚫𝐯𝑝 − 𝑤1𝚫𝐯1 − 𝑤2𝚫𝐯2 −𝑤3𝚫𝐯3 = 0

• 𝒔𝑇 1 − 𝛼1 𝚫𝐯1 + 𝛼1𝚫𝐯2 − (1 − 𝛼2)𝚫𝐯3 − 𝛼2𝚫𝐯3 = 0

• How to?

Project out any constraint by redefinding𝛾 = −ℙ𝑀−1 𝐠𝐫𝐚𝐝 𝑝

Collision and ContactProblem Solution


• ℙ projects a change in velocity using an impulse 𝐣ℙΔ𝐯 = Δ𝐯 +𝑀−1𝐣

• Find impulse 𝐣 by

Minimize : 1

2𝐣𝑇𝑀−1𝐣

Subject to: 𝐜𝑇ℙΔ𝐯 = 0

Objective function:1

2𝐣𝑇𝑀−1𝐣 + 𝜆(𝐜𝑇Δ𝐯 + 𝐜𝑇𝑀−1𝐣)

𝐣 = −𝐜𝜆 and 𝜆 = 𝐜𝑇𝑀−1𝐜 −1𝐜𝑇Δ𝐯

• ℙ = 𝕀 −𝑀−1𝐜 𝐜𝑇𝑀−1𝐜 −1𝐜𝑇

ℙ𝑀−1 is symmetric positive semidefinite.

Collision and ContactProject out any constraint


• In the case of many constraints𝐂𝑇Δ𝐯 = 𝐜1⋯𝐜𝑛 Δ𝐯 = 0

• 1. This can apply the projection in simple Gaussian-Seidel order.ℙ𝑀−1 = ℙ𝑛⋯ℙ1𝑀

−1

This is only symmetric if none of the constraints overlap.

• 2. Apply all constraints at once.ℙ = 𝕀 −𝑀−1𝐂 𝐂𝑇𝑀−1𝐂 −1𝐂𝑇

Require inversion of 𝑛 × 𝑛 matrix 𝐂𝑇𝑀−1𝐂 −1

This is prohibitively expensive for complex scenarios.

Collision and ContactProject out many constraints : Attempt


• Apply the projection in alternating forward and backward sweeps.ℙ𝑀−1 = ℙ1⋯ℙ𝑛⋯ℙ1

𝑞𝑀−1

This is positive semidefinite matrix.

𝑞 is small integer, we use 𝑞 = 4.

ℙ𝑖 strictly reduces energy, so this iteration is stable.

Collision and ContactProject out many constraints : Last Solution


• Using our method

(Top) maintains correct volume within 1%.

• Using standard finite element forces

(Middle) with a Poisson’s ratio of 0.45 results in a maximum volume loss of over 15%.

(Bottom) with a Poisson’s ratio to 0.499 reduces the maximum volume loss to 2%

ExamplesVolume Conserving


ExamplesSelf-Collision


ExamplesMany Objects Collision


• Proposed a novel technique from fluid dynamics.

• Simplicity and flexibility of tetrahedra.

• Avoiding the locking by one-rings volume preservation

• Incorporate collision constraints into the incompressible solver.

• The method is trivially adapted for triangles and thin shells.

Conclusion

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