Barycentric coordinates and transfinite interpolation

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Coordinates for quadrilaterals v 01 v 11 x v 00 v 10 Can find λ, µ ∈ [0, 1] s.t. x = (1 − λ)(1 − µ)v 00 + λ(1 − µ)v 10 + (1 − λ)µv 01 + λµv 11 , giving an interpolant g(x) = (1 − λ)(1 − µ)f 00 + λ(1 − µ)f 10 + (1 − λ)µf 01 + λµf 11 .
To find λ <strong>and</strong> µ: let a = v 00 −x, b = v 10 −v 00 , c = v 01 −v 00 , d = v 00 −v 10 −v 01 +v 11 . Then need to solve a + bλ + cµ + dλµ = 0. Eliminating λ gives a quadratic in µ: (a + cµ) × (b + dµ) = 0. Similar for λ. The special cases c × d = 0 <strong>and</strong> b × d = 0 need to be treated separately.
Page 1 and 2: Barycentric coordinates and transfi
Page 3 and 4: Barycentric coordinates on a triang
Page 5 and 6: Properties ◮ Lagrange property:
Page 7: Properties ◮ Lagrange property:
Page 11 and 12: Proof: Meyer et al. 2002 Let A i =
Page 13 and 14: Application: barycentric mappings R
Page 15 and 16: Theorem (F. & Kosinka). Wachspress
Page 17 and 18: Proof: With e i = (v i − x)/‖v
Page 19 and 20: Transfinite interpolation, Warren e
Page 21 and 22: Examples of MV interpolants
Page 23 and 24: On arbitrary curves p ( x, θ) 2 p
Page 25 and 26: Application: mesh deformation, Ju,
Page 27 and 28: Method 1: weight function The cubic
Page 29 and 30: To use this construction we need to
Page 31 and 32: Method 2: radial minimization MV in
Page 33 and 34: Example of radial minimization

Barycentric coordinates and transfinite interpolation

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