1. Vibratory Bowl Feeder “The real problem is not part transfer but part orientation” -Frank Riley, Bodine Corp.

2. Parts Presentation accounts for 50% of assembly time. (Nevins and Whitney ‘78) Why not use Machine Vision? (Cost) Sensor Noise: Lighting, Pixel Resolution Gripper Interface: Calibration, Communication Why not use Vibratory Bowl Feeders? (Floorspace, Acoustic Noise) Part Damage, Contamination Design of Tracks is a Black Art Cost: $7K - 18K per track Set Up Time

3. Universal Turning Machine

4. Related Work Mechanical Parts Feeders Compliant Motion Planning: Geometric Backchaining Randomization and Stochastic Plans Boothroyd, Poli, and Murch ‘82 Mani and Wilson ‘85 Natarajan ‘86 Singer and Seering ‘87 Hitakawa ‘88 Epstein ‘90 Erdmann, Mason, Vanacek ‘91 Lozano-Perez, Mason, and Taylor ‘84 Erdmann ‘84 Erdmann and Mason ‘86 Buckley ‘87 Canny ‘87 Donald ‘87 Taylor, Mason, and Goldberg ‘87 Peshkin ‘86 Latmobe ‘89 Brost ‘91 Erdmann ‘89 Goldberg and Mason ‘90

5. Feeding Polygonal Parts Animation

7. Kinematically Yielding Gripper Animation

8. Given a list of n vertices describing a planar part. Find the shortest sequence of actions guaranteed to orient the part up to symmetry. Assumptions: 1. All motion in the plane. 2. Rigid part. 3. Initial orientation unknown. 4. Inertial forces are negligible. 5. Contacting surfaces are frictionless. We do not address: Means for isolating parts: “singulating.” Problem Statement

9. Width Function:

10. Transfer Function, s:   

12. PUSH GRASP ACTIONS:

13. 3-Step Push-Grasp Plan for House-Shaped Part Animation Experimental setupExperimental trial

14. Definition: Orienting a Part up to Symmetry S ( ) has period T if:  s (  + T) = s (  S ( ) always has T =  due to gripper symmetry. S ( ) can have T = 2  / r due to part’s rotational symmetry. Periodicity in s ( ) gives rise to aliasing: A plan that maps  to  ’ also maps  + T to  ’ + T. Definition: A plan orients a part up to symmetry if the set of final poses includes exactly 2 /T poses equally spaced on S 1. Problem Statement: Given a list of n vertices describing the convex hull of a polygonal part, find the shortest sequence of squeeze actions guaranteed to orient the part up to symmetry.

15. s-intervals Def. an interval is a connected subset of S 1. For interval , let |  | be its Lebesgue measure. Def. An s-interval is a semi-closed interval of the form: [a, b) such that a, b are points of discontinuity in s ( ). An n-sided part defines O (n 2 ) s-intervals. Def. The s-image of a set is the smallest interval containing the image of that set. 1. Compute the squeeze function. 2. Find the widest single step in the squeeze function and set 1 equal to the corresponding s-interval. Let I =1. 3. While there exists an s-interval such that |s (  )| < |  |, Set   equal to the widest such s-interval. Increment i. 4. Return the list (  1,  2, …,  i ). Algorithm

16. Recovering the Plan To allow for control error, let  j = ½(|  j | - |s(  j+1 )|). Thus we define the plan:    For j from i - 1 down to 1:  j = s(  j+1 ) -  j -  j +  j+1 Given the list (  1,  2, …,  i ), Find a plan, a sequence of i squeeze actions  i = (  i,  i-1,…,  1 ), that collapses  i to the point: s (  1 ). Consider the rectangular part. Initially pose is anywhere in  2. After  = 0, pose is constrained to s (  2 ). Since |s (  2 )| < |  1 |, we could collapse s (  2 ) if we could align it with 1: open and rotate gripper by:  = s(  2 ) - 1.

17. Proof of Completeness For any Piecewise-Constant Monotonic Step Function on S 1 and any h, Either we can find a larger pre-image:  s(  h) - s(  < h, (1) Or h is a period of symmetry:  s(  h) = s(  + h (2)

18. Proof of Completeness For any Piecewise-Constant Monotonic Step Function on S 1 and any h, Either we can find a larger pre-image:  s(  h) - s(  < h, (3) Or h is a period of symmetry:  s(  h) = s(  + h (4)

19. Results Theorem (Completeness): A sensorless plan exists for any polygonal part. Theorem (Correctness): The algorithm will always find the shortest plan. Theorem (Complexity): For a polygon of n sides, the algorithm runs in time O(n 2 ) and finds the plans of length O(n). Extensions Stochastically Optimal Plans Reduction from O(n 3 ) to O(n 2 ) (Chen and Ierardi) Parallel Implentation (Prasanna and Rao) Extension to Non-Zero Friction (with Rao)

20. Algebraic Parts (Rao + Goldberg, 1995)

21. A Complete Algorithm for Designing Fence Arrangements Jeff Wiegley and Ken Goldberg (USC) Mike Brokowski and Mike Peshkin (Northwestern U)

22. Analysis: Optimality Claim:  is the shortest such plan. Lemma: Any plan that collapses a set  S 1 the smallest interval containing . (due to monotonicity of s()). Let (  ,    i  correspond to plan . Suppose there is a shorter plan . Let (  ’ 1,  ’ 2, …,  ’ j ) correspond to plan  ’, j < i. |  1 | |  ’ 1 | by definition of the algorithm. Since  ‘ terminates before , |  j |<|  ’ j |. Then for some k, |  k | |  ’ k | and |  k+1 | < |  ’ k+1 |. Cannot occur by definition of algorithm and Lemma.

23. Analysis: Correctness By Completeness, |  i | = T, smallest period of symmetry. Plan  collapses an interval of length T to a point. Recall s (  + T ) - s (  ) + T Consider a 2-step plan,  2 = (  1,  2 ).  2 (  + T) = s ( s (  + T -  1 ) -  2 ) = s ( s (  -  1 ) -  2 ) + T =  2 (  ) + T Outcome will be one of 2  / T orientations equally spaced on S 1.