vibratory bowl feeder “the real problem is not part transfer but part orientation” -frank riley,...

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

Universal

Turning

Machine

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

Feeding Polygonal Parts

Animation

Kinematically Yielding Gripper

Animation

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

Width Function:

Transfer Function, s:

PUSH GRASP ACTIONS:

3-Step Push-Grasp Plan for House-Shaped Part

Animation •Experimental setup

•trial

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.

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

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.

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)

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)

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)

Algebraic Parts (Rao + Goldberg, 1995)

A Complete Algorithm for Designing Fence Arrangements

Jeff Wiegley and Ken Goldberg (USC)Mike Brokowski and Mike Peshkin (Northwestern U)

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.

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.