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. 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 PresentationTRANSCRIPT
Path Control: Linear and Near-Linear Solutions
Slide Set 9:ME 4135R. 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:
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
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
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
Cartesian Control
NOTE: On the conveyor, the H-frame is a time dependent pose in C (conveyor space)
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
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
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
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
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!
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
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
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
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’
Joint-Interpolated Control -- Model
Robot Base
2
1
2
Note: R = R1 = R2
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!
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
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:
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?