Non-Rigid Shape and Motion Recovery: Degenerate Deformations - PowerPoint PPT Presentation

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Non-Rigid Shape and Motion Recovery: Degenerate Deformations

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Ambiguity in non-rigid SFM if only Rotation Constraints used (ECCV 2004) ... K1 are rank 1. K2 are rank 2. K3 are rank 3. Kd = K1 2xK2 3xK3 ... – PowerPoint PPT presentation

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Title: Non-Rigid Shape and Motion Recovery: Degenerate Deformations


1
Non-Rigid Shape and Motion Recovery Degenerate
Deformations
  • Jing Xiao and Takeo Kanade
  • CVPR 2004

2
Problem Addressed
  • Ambiguity in non-rigid SFM if only Rotation
    Constraints used (ECCV 2004)
  • SFM recovery in Degenerate Deformations

3
Basis Formulation for Non-Rigid Shape
  • Shape at any time instant

4
With Degenerate Deformation
  • Of the K bases
  • K1 are rank 1
  • K2 are rank 2
  • K3 are rank 3
  • Kd K1 2xK2 3xK3
  • WMB is not a unique decomposition
  • Essentially the problem is to find G

5
Degenerate contd
  • First K3 triple columns of M correspond to
    non-degenerate basis and the rest to degenerate
    basis
  • rj is a 3x1 eigen vector that corresponds to the
    degenerate basis shape
  • Arises cause Bi (degenerate) can be factored as

6
Rotation Constraints
  • Denote by Qi, we have
  • Qi has unknowns, so given enough frames
    we can find a solution?
  • This is not true in general
  • Qi can be written as where H satisfies

7
Rotation cont
  • is an arbitrary scalar and is a skew
    symmetric matrix
  • The solution has degrees of
    freedom

8
Basis Constraints
  • Ambiguity cause any non-singular transformation
    on the bases gives another valid set of bases.
  • To handle this choose the set of K3 frames that
    have smallest condition number (ECCV paper)
  • Denote the chosen frames as the first K3 frames
  • Coefficients will be

9
Basis contd
  • New constraints
  • Finally combining Rotation
  • and Basis constraints

10
Constraints contd
  • degrees of
    freedom and linear
    constraints.
  • Therefore solution space has
    degrees of freedom.
  • When ND is 0, there is a unique solution.

11
Finding Qi
  • If K2gt0 then
  • If K20, a unique solution exists using the
    constraints.

12
Results
  • Synthetic data

13
Results
  • Real Data
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