1. Trajectory Generation
Goal:
Turn a specified Cartesian-space trajectory of Pe into appropriate joint position reference values
Steps:
Use inverse kinematics of a robot manipulator arm to find joint values for any particular location of Pe
Use sampling and curve fitting to reduce computation
Output:
a series of joint position/velocity reference values to send to the controller

2. 5 Step Process for Trajectory Generation
Obtain function for workspace path
Sample function to get discrete joint pts
Apply IK & Jacobian calculations
Fit function to joint points
Sample to get discrete reference points C D
C=continuous
D=discrete

3. Step One: Continuous Fcn
Obtain an analytic function to describe motion with respect to the base frame
Obtain rate of change of location

4. Step 2: Sample
Sample the trajectory to obtain a finite number, m, of sample points on the continuous trajectory:
Sample rate of change

5. Step 3: IK & J
(a) Use inverse kinematics to convert each Cartesian trajectory sample point vector, into a corresponding joint space vector,
Handle multiple solutions, admissibility, etc.

6. Step 3: IK & J
(b) Use the inverse Jacobian relation to convert each velocity vector, into a corresponding joint speed vector,
Handle singular configurations

7. Step 4: Fit Continuous Curve to Joint Points
Use the sequence of vectors and i=1,…,m to generate continuous expressions for each joint and j=1,…,dof which pass through or sufficiently near to each of joint space sample points, and rate of change sample points, to produce continuous joint space trajectories for each joint.

8. Step 4: Fit Continuous Curve to Joint Points
Spline or
Polynomial Fit
& derivatives:

9. Step 4: Fit Continuous Curve to Joint Points
Let’s look at fitting a curve to one interval

10. Step 4. Fit Continuous Curve to Joint Points
Fit a continuous function, q(t) to the points:
Time info – from original sampling
For now use notation (get rid of subscripts i and i+1):
t=t0
t=tf

11. Step 4. Fit Continuous Curve to Joint Points
Splines, polynomials,…
To match position, velocity and acceleration at end points use a quintic polynomial (6 parameters to match the 6 unknowns):

12. Step 4. Fit Continuous Curve to Joint Points
Note: To match only position and velocity at end points use a cubic polynomial (4 parameters to match the 4 unknowns):

13. Step 4. Fit Continuous Curve to Joint Points
Use endpoints and time values in quintic polynomial (6 linear equations, 6 unknowns)

14. Step 4. Fit Continuous Curve to Joint Points
In matrix form:

15. Step 4. Fit Continuous Curve to Joint Points
In matrix form:

16. Step 4: Computational Thoughts
Need to perform fit for each joint but…
on each TIME interval, this matrix is the same for each joint – compute inverse only once

17. Step 4: Fit Continuous Curve to Joint Points
…..
one for each time interval
(i, i+1)
Piecewise polynomials: one polynomial for each joint for each timte interval (and we can easily take derivatives)

18. Step 5: Sample Joint Curve
Sample each continuous joint trajectory to generate a sequence of discrete reference values for each joint, where ttotal/N is the sampling period used.
Sample

19. Step 5: Sample Joint Curve
Sample joint speeds.
Sample

20. 5 Step Process for Trajectory Generation
Workspace path function
Sample -- discrete joint pts
Apply IK & Jacobian calculations
Joint function
Sample -- discrete reference points C
m pts
N pts
m ~ N ~ 1000+
Note: N >> m

21. Example: Linear Motion
Step One: Express line as continuous function (i.e. parameterize by time):
x(t), y(t), f(t)
Suppose we specify the line y= mx +b
Want to move along line with constant speed, v

22. Parameterize the Line Equation of line Differentiate Velocity vector
Substitute

23. Parameterize the Line Velocity Speed (magnitude)
Solve (pick appropriate sign)

24. Parameterize the Line
Velocity,

25. Sample Continuous Path Fcn
Use M samples for total time, ttotal time

26. Example Line Let’s take a 2-link planar arm with
Link lengths: l1 = 4, l2 = 3 m
Line y= - x + l1+l2/4 or y=-x with speed, v = 3m/s
Start point = (0, l1 + l2 /4)=(0,4.75)
endpoint = (l1 + l2 /4,0)=(4.75,0)
constant speed
Note, in practice, speed usually follows a trapezoidal profile with acceleration/deceleration at start/end of the motion

27. Example Line Using our 2-link planar data the path has
And for constant speed will take

28. Example Line We will use M=9 sample points
And for constant speed we will have

30. Step 3: IK & Jacobian
inverse kinematics:
where

31. Step 3: IK & Jacobian
Jacobian relationship:

32. Step 3: Discrete joint values

33. Step 3: Discrete Joint Velocity Values

34. Step 4: Fit Cubic To Positions/Speeds
rmse = rad ~ 1 deg

35. Curve Fitting Comments
Typically, a single cubic is not sufficient for the entire motion
A quintic polynomial can fit position, velocity and accelerations of end-points
What are the implications of fitting only position/velocity?

36. Reference Positions with 0.02s timesteps (sample cubic function)

38. Cartesian vs. Joint space
CONS
Actual robot position is sometimes unclear, particularly in presence of known obstacles
PROS
Actual control of the robot occurs in joint space
Simpler to plan trajectories in real-time; less computation
No problem with singularities

39. Cartesian vs. Joint space
CONS
Need Inverse kinematics (can be multi-valued)
Smooth trajectory in Cartesian space may not map to continuous joint space trajectory
Joint limits may prevent position from being realized
End-effector paths may not be safe for rest of manipulator
computational demand and analytic complexity
PROS
We usually desire a Cartesian path.
Easy to visualize the trajectory
Occurs in many robotic activities
Shortest Euclidean path
Straight line path minimizes inertial forces
Cartesian trajectory can be robot independent

41. Path not in workspace
Start & goal in different solution branches
May need to flip between configurations