Transfinite Interpolation Based Analysis Nathan Collier 1 V.M. Calo 1 Javier Principe 2 1 King Abdullah University of Science and Technology 2 International Center for Numerical Methods in Engineering 7 June 2010 Collier (KAUST) Transfinite Element Analysis 7 June 2010 1 / 26
Motivation - www.freedesign3d.com Collier (KAUST) Transfinite Element Analysis 7 June 2010 2 / 26
Overview 1 Transfinite Element Analysis Basis Function Construction Sample Basis Functions 2 Application 1: Terrain Modeling Automatic Refinement Insert Features 3 Application 2: Finite Elements with Simplified Meshing Current Challenges 4 Conclusions Collier (KAUST) Transfinite Element Analysis 7 June 2010 3 / 26
Interpolate Features 1 u1 d1 u d2 u2 2 1 f1(u1) F (u) f2(u2) 2 d3 3 u3 f3(u3) 3 i 1 d 1 F (u) = f 1 (u 1 ) 1 d 1 + 1 d 2 + 1 + f 2 (u 2 ) 1 d 3 d 1 + 1 d 2 + 1 + f 3 (u 3 ) 1 d 3 d 1 + 1 d 2 + 1 d 3 n f W i n f = f i (u i ) nf = f i (u i )Ŵ i (d 1, d 2, d 3 ) j W j i 1 d 2 1 d 3 Collier (KAUST) Transfinite Element Analysis 7 June 2010 4 / 26
Basis Function Construction n f F (u) = f i (u i )Ŵi(d) Each footprint can then be parameterized by its own one-dimensional basis, B i,k n b f i (u) = B i,k (u)c i,k which substituted into the surface approximation becomes n f n b F (u) = B i,k (u)c i,k Ŵ i (d) i k n f n b i k ( ) = B i,k (u)ŵ i (d) C i,k = N j C j i k }{{} j N i,k Collier (KAUST) Transfinite Element Analysis 7 June 2010 5 / 26
Sample Basis Functions Linear Footprints, Convex Polygons Collier (KAUST) Transfinite Element Analysis 7 June 2010 6 / 26
Sample Basis Functions Gradients N N x N y Node 0 Node 1 Node 2 Collier (KAUST) Transfinite Element Analysis 7 June 2010 7 / 26
Derivative Control: Ribbons Not only can we construct functions which interpolate features, but we can also control the derivative as the feature is crossed. 1 u1 d1 u d2 u2 2 1 f1(u1) F (u) f2(u2) 2 d3 3 u3 f3(u3) 3 Collier (KAUST) Transfinite Element Analysis 7 June 2010 8 / 26
Bi-cubic: 28 vs 16 Collier (KAUST) Transfinite Element Analysis 7 June 2010 9 / 26
Non-standard Basis Functions Collier (KAUST) Transfinite Element Analysis 7 June 2010 10 / 26
Overview 1 Transfinite Element Analysis Basis Function Construction Sample Basis Functions 2 Application 1: Terrain Modeling Automatic Refinement Insert Features 3 Application 2: Finite Elements with Simplified Meshing Current Challenges 4 Conclusions Collier (KAUST) Transfinite Element Analysis 7 June 2010 11 / 26
Projection onto Function Space Currently using L 2 projections. Find U A such that argmin U A Z H(U A ) 2 where Z is a function to approximate, and H is the interpolation which is controlled by parameters U A. Collier (KAUST) Transfinite Element Analysis 7 June 2010 12 / 26
Footprint Refinement Each footprint can be refined locally due to the surface interpolating each footprint. Collier (KAUST) Transfinite Element Analysis 7 June 2010 13 / 26
Footprint Refinement Tensor Product (Old way) Transfinite (New way) Collier (KAUST) Transfinite Element Analysis 7 June 2010 14 / 26
Footprint Refinement Other degrees of freedom can then be obtained via a surface L 2 projection G 1 Cubic G 2 Quintic Collier (KAUST) Transfinite Element Analysis 7 June 2010 15 / 26
Footprint Insertion Combine image processing with projective interpolation Collier (KAUST) Transfinite Element Analysis 7 June 2010 16 / 26
Footprint Insertion Combine image processing with projective interpolation Collier (KAUST) Transfinite Element Analysis 7 June 2010 17 / 26
Footprint Insertion C 0 Continuity C 1 Continuity Collier (KAUST) Transfinite Element Analysis 7 June 2010 18 / 26
Footprint Insertion Collier (KAUST) Transfinite Element Analysis 7 June 2010 19 / 26
Overview 1 Transfinite Element Analysis Basis Function Construction Sample Basis Functions 2 Application 1: Terrain Modeling Automatic Refinement Insert Features 3 Application 2: Finite Elements with Simplified Meshing Current Challenges 4 Conclusions Collier (KAUST) Transfinite Element Analysis 7 June 2010 20 / 26
Motivation - www.freedesign3d.com Collier (KAUST) Transfinite Element Analysis 7 June 2010 21 / 26
Laplace Equation Find θ(x) such that with boundary conditions θ = 0 x Ω θ x=0 = 0, n θ x=1 = 1, n θ y=0,1 = 0 Collier (KAUST) Transfinite Element Analysis 7 June 2010 22 / 26
Integration Check the relative change in the stiffness matrix diagonal terms as refining the integration mesh. Gaussian Quadrature 4x4 rule used with uniform refinements. Collier (KAUST) Transfinite Element Analysis 7 June 2010 23 / 26
Discontinuity B 5 (x, y) d dx B 5(x, y) d dy B 5(x, y) Collier (KAUST) Transfinite Element Analysis 7 June 2010 24 / 26
Discontinuity - Removed by Assembly B 5 (x, y) d dx B 5(x, y) d dy B 5(x, y) Pre-Assembly Post-Assembly Collier (KAUST) Transfinite Element Analysis 7 June 2010 25 / 26
Conclusion Transfinite Interpolation is a powerful, flexible way to define a basis Grants the freedom to place degrees of freedom at will without propagating them to unintentional regions of the domain Ties in directly to CAD programs whose geometries are watertight and requires no meshing in the traditional sense Current work is focused in accurate integration techniques Collier (KAUST) Transfinite Element Analysis 7 June 2010 26 / 26