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

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Presentation on theme: "Path Control: Linear and Near- Linear Solutions Slide Set 9: ME 4135 R. Lindeke, PhD."— Presentation transcript:

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

2 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: The matrices T(r), R a (r), and R o (r) are the translation, rotation wrt Z and rotation wrt Y in transitioning from P initial to P final

3 Developing the Drive Matrix: Given: P 1 is (n 1, o 1, a 1, d 1 ) And: P 2 is (n 2, o 2, a 2, d 2 ) Then: * Derivation is found in Paul’s Reference: pgs 139 - 151

4 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

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

6 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

7 Cartesian Control At Time 1 (P 1 ): At Time 2 (P 2 ):

8 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

9 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

10 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!

11 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:

12 Joint-Interpolated Control Step 5: Divide the T seg into ‘m’ equal time intervals:

13 Joint-Interpolated Control Step 6: For each joint, determine angular distance during each time segment  t seg : Step 7: at the beginning of the nth step over a path, joint i servo control receives a target point:

14 Joint-Interpolated Control Implementing this method begins with determination of the distance between and ultimately the number of Via’s This is (really!) a simple trigonometry problem based on the offset distance and error tolerance (  ) at ‘closest approach’

15 Joint-Interpolated Control -- Model Robot Base 22 1 2 Note: R = R1 = R2

16 Joint-Interpolated Control -- Model Notice line #1, #2 and R 1 form a right triangle #2 is half the distance between Via Points! R (= R 1 ) is taken at point of closest approach between the robot and part!

17 Lets try one: A Part 6m long moves by a stationary robot on a conveyor moving at 0.04mps (counter flowing) If we desire that the robot complete its spraying in 15 seconds – Note, 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 = 1.3 m

18 Lets try one: Distance between Via’s is: Therefore: The number of Via’s – = 5.4/1.015 = 5.32  6 – plus the initial point = 6 + 1 = 7

19 Follow-up Distance between is: 5.4/6 =.9m (rather than 1.015m) What is actual Error band? Here we see it is:  4.01cm Typically, we find that Joint Interpolated solutions provide better than desired (or expected) process control! What if we equally space the Via’s?


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