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

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Vibratory Bowl Feeder real problem is not part transfer but part orienta -Frank Riley, Bodine Corp.

Transcript of Vibratory Bowl Feeder “The real problem is not part transfer but part orientation” -Frank Riley,...

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

Vibratory Bowl Feeder

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

Page 2: Vibratory Bowl Feeder “The real problem is not part transfer but part orientation” -Frank Riley, Bodine Corp.

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.