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Transcript of Vibratory Bowl Feeder “The real problem is not part transfer but part orientation” -Frank Riley,...
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Vibratory Bowl Feeder
“The real problem is not part transfer but part orientation”-Frank Riley, Bodine Corp.
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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
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Universal
Turning
Machine
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Related Work
Mechanical Parts Feeders
Compliant Motion Planning:
Geometric Backchaining
Randomization and Stochastic Plans
Boothroyd, Poli, and Murch ‘82Mani and Wilson ‘85 Natarajan ‘86Singer and Seering ‘87 Hitakawa ‘88
Epstein ‘90 Erdmann, Mason, Vanacek ‘91
Lozano-Perez, Mason, and Taylor ‘84Erdmann ‘84 Erdmann and Mason ‘86
Buckley ‘87 Canny ‘87 Donald ‘87Taylor, Mason, and Goldberg ‘87
Peshkin ‘86 Latmobe ‘89 Brost ‘91
Erdmann ‘89 Goldberg and Mason ‘90
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Feeding Polygonal Parts
Animation
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Kinematically Yielding Gripper
Animation
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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
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Width Function:
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Transfer Function, s:
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PUSH GRASP ACTIONS:
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3-Step Push-Grasp Plan for House-Shaped Part
Animation •Experimental setup
•trial
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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 S1.
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.
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s-intervals
Def. an interval is a connected subset of S1.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 (n2) 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
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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.
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Proof of Completeness
For any Piecewise-Constant Monotonic Step Function on S1 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)
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Proof of Completeness
For any Piecewise-Constant Monotonic Step Function on S1 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)
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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(n2) and finds the plans of length O(n).
Extensions
• Stochastically Optimal Plans• Reduction from O(n3) to O(n2) (Chen and Ierardi)• Parallel Implentation (Prasanna and Rao)• Extension to Non-Zero Friction (with Rao)
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Algebraic Parts (Rao + Goldberg, 1995)
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A Complete Algorithm for Designing Fence Arrangements
Jeff Wiegley and Ken Goldberg (USC)Mike Brokowski and Mike Peshkin (Northwestern U)
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Analysis: Optimality
Claim: is the shortest such plan.
Lemma: Any plan that collapses a set S 1the smallest interval containing . (due to monotonicity of s()).
Let ( , icorrespond 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.
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Analysis: Correctness
By Completeness, | i| = T, smallest period of symmetry.
Plan collapses an interval of length T to a point.
Recalls ( + 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 S1.