1. Path Control in Robotics ME 4135, F 2012 Richard R. Lindeke, Ph. D.

2. Motion Types of Interest Point – to – Point Motion:  All Axes start and end simultaneously  All Geometry is computed for targets and relevant Joint changes which are then forced to be followed during program execution Path or Trajectory Controller Motion  Here the motion is performed through a time sequence of intermediate configurations computed ahead of time (like above but without stop-start operation) or in real time  Paths are “Space Curves” for the n-Frame to follow  This motion is a continuous scheme to move the TCP from one location to the next along a desired (straight or curved) line under direct operational control

3. Path Control and Motion Types: We will explore the following types of Motion:  Lead Through Path Creation  (Cubic) Polynomial Paths w/ Via Points  Minimum Time Trajectory w/ controlled Acceleration  Lower order Path-Poly Control  LSPB Paths  Craig’s Method for acceleration smoothing  Strict Velocity Control  Joint Interpolated Control  Full Cartesian Control

4. Lead Through Path Creation Basically this was a technique whereby a skilled operator took a robot arm (for welding or painting) and used it like his/her weld tool or paint sprayer and performed the required process at reasonable speed The robot is equipped with a position recording device and memorizes a large number of points during the teaching session These learned points then would be “played back” to replicate the skilled operators motions

5. Lead Through Path Creation Advantages:  Simple way to create complex paths  All points are sure to be physically attainable  Playback speed can be controlled by an external device Disadvantages:  Precision placements are required (program must be replayed at exactly the initial placement)  Major concern with operator safety: robot is powered and operator is physically touching it (OSHA rules it unsafe practice!)

6. Modern Path Control: (Lets look at a simple example) Dr. D’s new ‘Self-powered Automated Coffee Drinker’ Robot It is a simple cantilevered Cartesian device equipped with a spherical wrist that responds to eye movement and thoughts to help the overworked design engineer get coffee while designing and drafting & typing of Reports It follows a straight line path from the cup’s point on a table to the worker’s mouth in ½ second

7. Lets look at a simple example:

8. We see that the ‘Bot must travel a space path of 16.45” which can be decomposed into a movement of 9.5” along each of the prismatic joints For ‘accuracy’ lets divide each of the these joint paths into 100 segments From Physics: V joint   d/  t = (9.5/100)/(.5/100) = 19in/second (a reasonable speed!)

9. Lets look at a simple example: During the 1 st Step then:  Joint 1 starts at 0 and moves to 0.095”  Moves there in 0.005 seconds  How will it do it?  Of course by Accelerating from a stop to 19 in/sec in 0.005 seconds  Compute Acc reqr   V/  t Darn – this says that the acceleration is 3800in/sec 2 – this is  10G!!!!!

10. Lets look at a simple example: So this will certainly be difficult to accomplish! (more likely it will not work) OSHA would be just as upset as when we had the worker holding on to the powered robot – what should we do? I think our approach is too naïve! If we examine the Pos vs. Time, Vel Vs. Time and Acc vs. Time plots we may see why:

11. Look at a simple example’s Trajectory Curves:

12. This is Physically Impossible (or rather ‘very energy intensive’) Can we build a reasonable solution that keeps the acceleration to an achievable level? What this would mean is we wouldn’t “instantly” – in one time step – go from stopped to full speed This can be achieved with a “time polynomial” model of motion

13. Building a ‘Path Polynomial’ Motion Set These are the ‘trajectory’ equations for a joint (Position, Velocity and Acceleration)

14. Solving the ‘Path Polynomial’ means finding a i ’s for SPECIFIC PATHS We would have “boundary” conditions for position and velocity at both ends of the path We would have the desired total time of travel Using these conditions we can solve for a 0, a 1, a 2 and a 3 to build a 3 rd order path polynomial for the required motion

15. Solving the ‘Path Polynomial’ means finding a i ’s for SPECIFIC PATHS ‘Poly’s’ holding at starting time and position ‘Poly’s’ holding at ending time and position

16. Solving the ‘Path Polynomial’ means finding a i ’s for SPECIFIC PATHS Writing these as Matrix Forms:

17. Solving the ‘Path Polynomial’ means finding a i ’s for SPECIFIC PATHS If we set t 0 = 0 (starting time is when we start counting motion!) then: By examination, a 0 = q 0 & a 1 = q 0 (dot)

18. Solving the ‘Path Polynomial’ means finding a i ’s for SPECIFIC PATHS Completing the solution consists of forming relationships for: a 2 & a 3 Done by substituting a 0 & a 1 values and solving the last two equation simultaneously: Be Careful and note the order of the positions and velocities!

19. Applying it to the Coffee ‘Bot Start: X = 0; v = 0 @ time = 0 End: X = 9.5”; v = 0 @ time =.5 sec a 0 = 0 ; a 1 = 0 a 2 = (3 * 9.5)/(0.5 2 ) = 114 a 3 = (2 *(- 9.5))/(0.5 3 ) = -152

20. Applying it to the Coffee ‘Bot Here (specifically):

21. Applying it to the Coffee ‘Bot Simplifying:

22. Applying it to the Coffee ‘Bot: Position Position vs. Time

23. Applying it to the Coffee ‘Bot: Velocity Joint Velocity Vs. Time

24. Applying it to the Coffee ‘Bot: Acceleration Acceleration Vs. Time

25. Applying it to the Coffee ‘Bot Using the Path Polynomial Approach: Max Velocity is: 28.5 in/sec (compares to 19 in/sec) Max Acceleration is: 228 in/sec 2 (.6 g) compared to  10 g But, in this method, I require a 100% duty cycle motor since throughout the entire path, the motor is accelerating (either with positive or negative orientation) Can we make a path solution where we accelerate for only part of the path? Turns out we can and we will call it LSPB!

26. Studying the LSPB model In this model, we will carry forward with a ‘parabolic model’– that is 2 nd order In this model, we will determine a time where we will blend from startup until we reach a constant velocity – (and its greater than 1/100 of the total time!) Here we will see an acceleration followed by a period of “coasting” and then deceleration (often called a trapezoidal velocity model)

27. Model Building: We must define an acceleration constraint (minimum value) such that the acceleration is guaranteed to be completed within half of the allocated time period of the travel: This assures that there is no overlap for the ‘BLEND Regions’

28. Looking at the motion over the various regions: During Region 1 (while the joint is Accelerating) (time interval 0 to t blend [t b ]) the Joint moves: q = (V/2)*t b  t b is the acceleration time During the region of ‘Constant Velocity’ the Joint moves: q = V*(t – 2t b ) During Region 3 -- while the joint is decelerating the joint moves: q = (V/2)*t b Total travel distance is q B - q A

29. Writing a motion equation:

30. Substituting and Isolating the Unknown (which is blend time) : This is a quadratic equation in t b

31. Solving for t b : Note: Acceleration is subject to constraint determined above

32. Applying it to the Coffee ‘Bot Acceleration constraint: Blend time:

33. Applying it to the Coffee ‘Bot t linear =.5 – 2*0.159 = 0.181 s Linear Velocity: Positions:  By t b, the Joint has moved: (27.89/2)*.159 = 2.222”  During linear velocity joint moves: 0.181*27.89 = 5.055 in (thus the pos = 7.277”)  During deceleration joint travels 2.222”  Adding them gives full travel distance: 2.222 + 5.055 + 2.222 in  9.5in

34. Plotting the Path trajectory: Notice: the accelerator is ‘off’ during the linear travel segment

35. The 2 Previous Path Control Methods focused on Start/Stopping Approaches What can we do if we desire to travel by continuing along a path w/o stopping at each point? Here we will focus on a method called ‘dog-tracking’ after the lead and follow techniques employed in dog racing Essentially we would have a situation where the path is laid out (as a series of Via Points) and the joints smoothly maneuver through and between them

36. ‘Craig’s’ Dog Tracking Method

37. Craig’s Dog Tracking Method Upon Examination of the motion, we find that there are three ‘regimes’ in the motion These are:  Start up regime  Intermediate regimes  Stopping regime Starting and Stopping are similar to LSPB in the way they compute blend time and acceleration During an Intermediate regime we compute acceleration by comparing incoming and outgoing velocities about each point

38. Craig’s Dog Tracking Method Starting regime Equations: Start Acceleration Start Blend Time Linear Velocity 1→2 Time @ linear Velocity

39. Craig’s Dog Tracking Method Stopping Equations: Stop ‘acceleration’ Stop Blend Time L. Velocity to stop Time @ L. Velocity

40. Craig’s Dog Tracking Method Intermediate Equations: Linear Velocity Acceleration Blend time Time @ L. Velocity

41. Craig’s Dog Tracking Method Upon examination of the set of equation on the previous 3 slides several point should be noted:  Start and Stop are essentially the same but very important differences must be noted  One can’t complete any of the regimes without looking ahead – actually looking ahead to the 2 nd point beyond to see if a joint is stopping or continuing  Start/Stop require  position/  time relationships  Intermediate regimes require  velocity/  time relationships

42. Craig’s Dog Tracking Method Step 1: Calculate Global ‘usable’ acceleration (magnitude) constraint based on LSPB model applied Pairwise (1 → 2; 2 → 3; etc) Step 2: Focus on Start and Stop Segments Step 3: Complete the table of accelerations, blend times, linear velocity and time at linear velocity

43. Point Name Position (inch) Delta Position (inch) Arrival Time (sec) Delta Time (sec) A00------- B9.5 0.5.5 C111.53.53 D9.5-1.56.53 E0-9.57.51 Lets Expand on Dr. D’s Coffee ‘Bot:

44. Step 1: Global Acc. Constraint This is largest – should work globally – but lets make sure it doesn’t miss so choose 200ips 2 ‘ cause it’s easier to calculate and is only about.6g

45. Lets Expand on Dr. D’s Coffee ‘Bot: Next we focus on the start & stop equations: ◦ Starting ◦ Stopping NOTE: can’t compute t lij yet – we lack the data!

46. Lets Expand on Dr. D’s Coffee ‘Bot: Considering Intermediate B → C:

47. Lets Expand on Dr. D’s Coffee ‘Bot: On to “C → D” Segment

48. Lets Expand on Dr. D’s Coffee ‘Bot: Now for Segment D → E:

49. Summarizing PtPOS T. Time  POS  time T i (blend)AccL. Vel Time @L. Vel A00---------.106+200 21.3.342 B9.5.59.5.5.103-200.52.946 C113.51.53.005-200 -.52.974 D9.56.5-1.53.046-200 -9.74.928 E07.5-9.51.049+200

50. A Final Thought on Dog-Tracking: But if we must travel over or through a certain point, we can define ‘Pseudo-Via’ points that flank the desired target and force the arm to pass the ‘pseudos’ and drive right over the original desired target point

51. Looking at Velocity Control Achievable Path Desired Path

52. With this “Velocity Control” The acceleration is set and the new velocity for an upcoming segment is inserted at the appropriate time (place) Over the segment, the arm (joints) lags the desired path We extend the next path velocity curve to intersect the actual achievable path (to a point that is earlier in time than when we would have expected to change) this will mean that the joint can ‘catch up’ to the desired plan for the travel as it ‘blends’ to the new velocity

53. 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

54. 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:

55. 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

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

57. 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

58. Cartesian Control At Time 1 (P 1 ): At Time 2 (P 2 ):

59. Cartesian Control 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

60. 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

61. 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!

62. 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:

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

64. 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:

65. 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’

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

67. 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 measured at point of closest approach between the robot and part!

68. 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

69. 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

70. 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! What if we equally space the Via’s?