Trajectory Generation Cherevatsky Boris

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 .

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:

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

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

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) = 0.5231 Only 52% of the peak.

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

Multi-Dimensional Case X=mtraj(@tpoly,[0 2] ,[1 -1], 50); X=mtraj(@lspb,[0 2] ,[1 -1], 50);

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

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