path control: linear and near-linear solutions

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Path Control: Linear and Near-Linear Solutions Slide Set 9: ME 4135 R. Lindeke, PhD

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Path Control: Linear and Near-Linear Solutions. Slide Set 9: ME 4135 R. Lindeke, PhD. Linear Path Control. Sometimes thought of as Cartesian Control It is based on the idea of Transitions between consecutive required geometries These transitions are based on the solution of a Drive Matrix:. - PowerPoint PPT Presentation

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Page 1: Path Control: Linear and Near-Linear Solutions

Path Control: Linear and Near-Linear Solutions

Slide Set 9:ME 4135R. Lindeke, PhD

Page 2: Path Control: Linear and Near-Linear Solutions

Linear Path Control

Sometimes thought of as Cartesian Control It is based on the idea of Transitions between

consecutive required geometries These transitions are based on the solution of a

Drive Matrix:

a oD r T r R r R r

The matrices T(r), Ra(r), and Ro(r) are the translation, rotation wrt Z and rotation wrt Y in transitioning from Pinitial to Pfinal

Page 3: Path Control: Linear and Near-Linear Solutions

Developing the Drive Matrix:

Given: P1 is (n1, o1, a1, d1) And: P2 is (n2, o2, a2, d2) Then:

1 2 1 2 1 2 1 2 1

1 2 1 2 1 2 1 2 1

1 2 1 2 1 2 1 2 1

0 0 0 1

n n n o n a n d do n o o o a o d d

D ra n a o a a a d d

*Derivation is found in Paul’s Reference: pgs 139 - 151

Page 4: Path Control: Linear and Near-Linear Solutions

Cartesian Control

It is used when very exact interaction is required

It ‘guaranties’ accurate tool placement at all times

It is typically used in time dependent solutions – like interaction while a product is moving

Page 5: Path Control: Linear and Near-Linear Solutions

Cartesian Control

NOTE: On the conveyor, the H-frame is a time dependent pose in C (conveyor space)

Page 6: Path Control: Linear and Near-Linear Solutions

Cartesian Control

We desire to attach the “Quality Tag” to the part as it moves by the robot station

Requires that the part and robot tool must be in exact contact throughout the attachment process

This becomes a ‘Time-based’ Mapping problem

Page 7: Path Control: Linear and Near-Linear Solutions

Cartesian Control

At Time 1 (P1):

At Time 2 (P2):

1 1 1

1 1n o R C H P Tar MWo R U U C H P nT t T T T T t T T T

1 1 1

2 2n o R C H P Tar MWo R U U C H P nT t T T T T t T T T

Page 8: Path Control: Linear and Near-Linear Solutions

Using these two (time-dependent) Poses, we can build the desired drive matrix

We can compute the accuracy of the path then as a series of changes to the three control vectors: a, o and d

These are updated in real time

Cartesian Control

Page 9: Path Control: Linear and Near-Linear Solutions

Cartesian Control

Problems that can result (and must be accounted for): Intermediate points that are unreachable – After we

compute the initial and final points (that prove to be reachable as individuals), we request the tracking of a, o and d vectors but they exceed joint capabilities or require positions outside the work envelope during the driving action

In certain situations where only certain solutions are possible for the robot, like being near singularities, the desired linear velocity may require very high joint velocities – exceeding capabilities – and the path actually followed will deviate from the one desired as the joints run at their velocity limits

Page 10: Path Control: Linear and Near-Linear Solutions

Near Cartesian (Joint Interpolated) Control

This is a semi-precise control method developed as a compromise between full-Cartesian and point-to-point motion

Basically it is used when a process needs to be held within a ‘band’ about an ideal linear path – for example during painting or bar-code scanning

The path is designed to ‘track’ the work as it moves and maintains no more than a given “focal distance” separation between the tool and work surface

It is a path that is close to the target path at all poses but exact only at a few!

Page 11: Path Control: Linear and Near-Linear Solutions

Joint-Interpolated Control

Step 1: determine the desired path Step 2: Compute the tolerable error and the

number of points (‘VIAs’) needed to maintain tool–to–work distances

Step 3:Compute IKS’s at each of the VIAs Step 4: Determine “Move time” for each segment:

2 1

each joint i

note: is joint velocity

K

i iSEG

i

i

q qT MAX

Page 12: Path Control: Linear and Near-Linear Solutions

Joint-Interpolated Control

Step 5: Divide the Tseg into ‘m’ equal time intervals:

is the controller positional sampling rate

1

:

segsampling

sampling

T f

where

m T fseg sampling

f

Page 13: Path Control: Linear and Near-Linear Solutions

Joint-Interpolated Control

Step 6: For each joint, determine angular distance during each time segment tseg:

Step 7: at the beginning of the nth step over a path, joint i servo control receives a target point:

2 1

/i i

i time

q qq

m

target = i seg startn q q

Page 14: Path Control: Linear and Near-Linear Solutions

Joint-Interpolated Control

Implementing this method begins with determination of the distance between and ultimately the number of Via Points needed

This is (really!) a simple trigonometry problem based on the offset distance and error tolerance () at ‘closest approach’

Page 15: Path Control: Linear and Near-Linear Solutions

Joint-Interpolated Control -- Model

Robot Base

2

1

2

Note: R = R1 = R2

Page 16: Path Control: Linear and Near-Linear Solutions

Joint-Interpolated Control -- Model

Notice line #1, #2 and R1 form a right triangle

2 2 21

2 221

221

2

1 2

2 1

: 1 2

2 2

2 4 4

R

R

But R

R R

R

#2 is half the distance

between Via Points!

R (= R1) is taken at point

of closest approach

between the robot and

part!

Page 17: Path Control: Linear and Near-Linear Solutions

Lets try one:

A Part 6m long moves by a stationary robot on a conveyor moving at 0.04mps (counter flowing compared to painting direction) If we desire that the robot complete its spraying in 15 seconds

– Then, the robot must travel 5.4 m to spray the side of the part nearest it since the part moves during the painting operation.

At closest approach, the robot is 1.5 m from the part and needs to have its sprayer 20 cm ( 5 cm) from the part.

From this data, the R value is: 1.5 - .20 + .05 = 1.35 m

Page 18: Path Control: Linear and Near-Linear Solutions

Lets try one: Distance between Via’s is found using:

Therefore: The number of Via’s – = 5.4/1.020 = 5.29 so round up to 6– plus the initial point = 6 + 1 = 7

2

2

2 4 4

4 1.35 .05 4 .05 .26

2 0.510

2 2 2 .510 1.020

R

m

m

distance between Via's is:

Page 19: Path Control: Linear and Near-Linear Solutions

Follow-up

Distance between is: 5.4/6 = .9m (rather than 1.020m)

What is actual Error band? Here we see it is: 3.86cm

Typically, we find that Joint Interpolated solutions provide better than required (or expected) process control!

.52

2 2

2

!

9. 4 1.35 42.45 5.4 4

1.35 .0506 0.0772

2.0386 3.86

m

m cm

A Quadratic eqn. in

What if we equally space the Via’s?