1. Trajectory Planning

2.  Goal: to generate the reference inputs to the motion control system which ensures that the manipulator executes the planned trajectory Motion control system Robot Trajectory planning system torques Position, velocity, acceleration

3. Path and Trajectory  Path: the locus of points in the joint space or in the operational space  Trajectory: a path on which a time law is specified in terms of velocities and/or accelerations t

4. Path and Trajectory Trajectory planning algorithm Path description path constraints(obstacles) constraints imposed by robot dynamics (smooth) (limits, not modeled resonant modes) Joint (end-effector) trajectories in terms of position, velocity and acceleration

5. Path and Trajectory  Specification of geometric path  Extremal points, possible intermediate points,geometric primitives interpolating the points  Specification of motion time law  Total trajectory time, maximum velocity and acceleration, velocity and acceleration at points of interests

6. Joint Space Trajectory Inverse kinematics algorithm Trajectory parameters in operation space Joint (end-effector) trajectories in terms of position, velocity and acceleration Trajectory parameters in joint space Trajectory planning algorithm Initial and final end-effector location, traveling time, etc.

7. Joint Space Trajectory  Requirements for joint space trajectory planning algorithm  The generated trajectory be easy to compute  Position and velocity (acceleration) be continuous function of time  Undesirable effects be minimized  Point-to-point motion  Moving from an initial to a final joint configuration in a given time t f

8. Point-to-point Motion Figure 4-2 from Fu, Gonzalez and Lee  (t f )  (t 2 )  (t 0 )  (t 1 ) “Lift-off” “Set-down” Final Initial

9.  Polynomial interpolation  Example: initial and final position and velocity be given. Point-to-point Motion

12.  Example: initial and final acceleration also be given.  Six constraints (initial and final position, velocity and acceleration  Order at least five Point-to-point Motion

13.  Trapezoidal velocity profile  Directly verifying whether the velocity and acceleration violate the mechanical limits Point-to-point Motion

14.  Area enclosed by the velocity profile given acceleration Point-to-point Motion

16.  Given the following conditions:  initial and final position are given  initial and final velocity are set to zero  Maximum velocity and acceleration are given  What’s the minimum traveling time?

17. Path Motion  Disadvantages of single high order polynomial  A suitable number of low order polynomials

18. Operation Space Trajectory  Not easy to predict end-effector motion due to kinematics nonlinearity  Path motion planning similar to joint space  Different method if the end-effector motion has to follow a prescribed trajectory of motion such as line, circle, etc.

19. Two-link Planar Arm Parameters

20. Cams  Motion programming historically associated with mechanical cams  Constant speed rotation of camshaft converted to variable linear displacement of valve (or other device attached to cam follower) – Camshafts in auto engines (all 4 strokes) – Sewing machine (older mechanical style)

21. Pictures of Cams http://www.howstuffworks.com/camshaft1.htm Industrial Car Engines

22. Cam Motion Profiles - DRD  Dwell – Rise – Dwell – initial period of no motion (“dwell”) – “rise” to a maximum displacement – final period of no motion (“dwell”) “dwell” “rise” “dwell” Time, t Displacement, s s=s max, v=0, a=0 s=0, v=0, a=0

23. Cam Motion Profiles - DRRD  Dwell – Rise – Return – Dwell – initial period of no motion (“dwell”) – “rise” to a maximum displacement – Immediately “return” to origin – final period of no motion (“dwell”) “dwell” “rise” “dwell” Time, t Displacement, s s=s max, v=0, a  0 s=0, v=0, a=0 “return” s=0, v=0, a=0

24. Cam Motion Profiles - RR  Rise – Return – “rise” to a maximum displacement – Immediately “return” to origin – No “dwell” – do same thing over again “rise” Time, t Displacement, s s=s max, v=0, a  0 “return” s=0, v=0, a  0

25. Accel.-Vel.-Disp. #1 Time, t Acceleration, A Zero order, A = constant Time, t Velocity, V Time, t Displacement, S T T T First order, V=k 1 t Second order, S=k 2 t 2

26. Accel.-Vel.-Disp. #1a Time, t Acceleration, A Time, t Velocity, V Time, t Displacement, S T T T this area equals this value this area equals this value

27. Accel.-Vel.-Disp. #1b Time, sec Acceleration, A Velocity, V Displacement, S T 0 0.1 0.2 0.3 T 25 m/s 2 V1 V2 V3 S1 S2 S3 Find numerical values for V1, V2, and V3 Find numerical values for S1, S2, and S3 Suitable for a “rise”

28. General Curve Shape: y=Kx n Area under the curve y=Kx n between x=0 and x=x 1 is Note that y 1 =Kx 1 n, so

29. Accel.-Vel.-Disp. #2 Time, sec Acceleration, A Velocity, V Displacement, S T 0 0.1 0.2 0.3 T 25 m/s 2 V1 V2 V3 S1 S2 S3 Find numerical values for V1, V2, and V3 Find numerical values for S1, S2, and S3 Suitable for a “dwell - rise”

30. Accel.-Vel.-Disp. #3 Time, sec Acceleration, A Velocity, V Displacement, S T 0 0.15 0.3 T 25 m/s 2 V1 V2 S1 S2 Find numerical values for V1 and V2 Find numerical values for S1 and S2 Suitable for a “dwell - rise”

31. Accel.-Vel.-Disp. #4 Time, sec Acceleration, A Velocity, V Displacement, S T 0 0.1 0.2 0.3 T 25 m/s 2 V1 V2 V3 S1 S2 S3 Find numerical values for V1, V2, and V3 Find numerical values for S1, S2, and S3 Suitable for a “dwell - rise”

32. Time, sec A V S T T 25 m/s 2 V1 V2 V3 S1 S2 S3 Find numerical values for V4 and V5 Find numerical values for S4, S5, and S6 Suitable for a “rise-return” -25 m/s 2 0.1 0.2 0.3 0.4 0.5 0.6 V5 V4 S4 S5 S6

33. Analytical Solution  Solve the previous problem analytically: Hint – solve first parts (for t<0.3 sec), find boundary conditions for 2 nd parts

34. Solve Numerically  Use Excel and trapezoidal integration

35. A Numerical Gimmick  Use this gimmick to improve accuracy when you have abrupt changes in acceleration “double up” at any sharp transitions

36. Motion Programming #2 Robot Joint Motions

37. Typical Robot Motion Figure 4-2 from Fu, Gonzalez and Lee  (t f )  (t 2 )  (t 0 )  (t 1 ) “Lift-off” “Set-down” Final Initial

38. Position Constraints  Initial position,  1 – initial velocity and acceleration (normally=0)  Lift off position,  2 – velocity and acceleration must match here  Set-down position,  3 – velocity and acceleration must match here  Final position,  f – final velocity and acceleration (normally=0)

39. Typical Solution  “4-3-4” trajectory 4 th order polynomial from initial to lift-off 4 th order polynomial from initial to lift-off 3 rd order polynomial from lift-off to set- down 3 rd order polynomial from lift-off to set- down 4 th order polynomial from set-down to final 4 th order polynomial from set-down to final  “3-5-3” trajectory same as above but 3 rd and 5 th order poly same as above but 3 rd and 5 th order poly  5-cubic trajectory Cubic splines used for 5 segments Cubic splines used for 5 segments  lift-off to set-down split into 3 segments

40. “4-3-4” Trajectory  1 st segment:  2 nd segment:  3 rd segment:  14 unknowns – need 14 equations!

41. Boundary Conditions #1- #3 1.Initial position,  0 =  (t 0 ) (set t 0 =0)1.Initial position,  0 =  (t 0 ) (set t 0 =0)  2.Initial velocity =  0 (typically 0) 2. 3.Initial acceleration =  0 (typically 0)

42. Boundary Conditions #4- #5 4.Lift-off position,  (t 1 )4.Lift-off position,  (t 1 ) 4. 5.Lift-off position,  (t 1 )

43. Boundary Conditions #4- #5 4.Lift-off position,  (t 1 )4.Lift-off position,  (t 1 ) 4. 5.Lift-off position,  (t 1 )

44. Boundary Conditions #6- #7 6.Lift-off velocity match from both sides6.Lift-off velocity match from both sides 2. 7.Lift-off acceleration match both sides

45. Boundary Conditions #8- #9 8.Set-down position,  (t 2 )8.Set-down position,  (t 2 )  9.Set-down position,  (t 2 )

46. Boundary Conditions #10 - #11 10. Set-down velocity match from both sides10. Set-down velocity match from both sides 2. 11.Set-down acceleration match both sides

47. Boundary Conditions #12- #14 12.Final position,  (t f )12.Final position,  (t f )  13. Final velocity, v f (typically 0) 2. 14.Final acceleration, a f (typically 0)

48. Solution Format  14 simultaneous linear equations with 14 unknowns:  11 values required to find solution:

49. Solution Format  Solve using Matlab (or possibly Maple)

50. After Matlab Solution  Once we find the 14 coefficients, how do we find velocities and accelerations? – Take derivative of h 1 (t), h 2 (t), h 3 (t) to get velocity – Take derivative of velocity to get accel. – Both are easily plotted in Matlab or Excel

51. Problems  What if the velocities or accelerations we find are too large? – Increase times t 1, t 2, t f – Move the pick-up point closer to  (t 0 ) – Move the drop-off point closer to  (t f )

52. Motion Programming #3 Friday, October 19, 2001

53. Plan Motion Program

54. Inverse Kinematics  What is the minimum time T that meets velocity and acceleration constraints for both joints? from Inverse Kinematics:

55. Motion Programming #1 Work in groups of 2:Work in groups of 2:  Determine an “S” curve for each joint  Need something suitable for DRD motion  Turn in solution will have two groups show solution will have two groups show solution Use T = 1.0 second as a “reasonable” time

56. Straight-Line Motion  Define 10 “knot” points along the straight line  Find inverse kinematic solution at each point:

57. Motion Programming #2  Select T = 1 second as a “reasonable” time for complete straight line motion   Work in groups of 2:  Are there any constraints that need to be considered? What do you recommend to determine desired positions at intermediate times?