1. Trajectory Generation
Cherevatsky Boris

2. Trajectory for single joint
Suppose we are given a simple robot
We want to move the joint from to in 4 seconds and the trajectory should be a cubic polynomial .

3. Trajectory for single joint
Let’s denote the initial velocity and the final velocity We also know that , so if
then we get that :
By solving this 4 equations we can extract the coefficients of the polynomial:

4. Using Matlab
The function [s,sd,sdd]=tpoly(start,end,time,start_v,end_v) generates a quantic polynomial trajectory.
Example:
tpoly(0,1,50) ; this means start_v=end_v=0

5. Problem with polynomials
Let’s try this example: tpoly(0,1,50,0.5,0)
What is the problem?

6. Problem with polynomials #2
Looking at the 1st example tpoly(0,1,50) we see that the velocity peaks at t=25, which means that for the most of the time the velocity is far less than the maximum. The mean velocity: mean(sd)/max(sd) = Only 52% of the peak.

7. Solution – Linear with bend!
S = lspb(0,1,50,0.025) Linear segment with parabolic blend
Mean(sd)/max(sd)=0.80

8. Multi-Dimensional Case
2] ,[1 -1], 50);
2] ,[1 -1], 50);

9. Multi-Dimensional with via points
mstraj(via,[2,1],[],[2,1],0.05,0);
The functions takes:
Via points
A vector of maximum speeds per joint
A vector of durations for each segment
Initial axis coordinates
Sample interval
Acceleration time

10. Acceleration time
mstraj(via,[2,1],[],[2,1],0.05,1);