Robotic Trajectory planning

1. Trajectory Planning

2. Trajectory Planning
 Path and trajectory planning means the way that a
robot is moved from one location to another in a
controlled manner.
 The sequence of movements for a controlled
movement between motion segment, in straight-line
motion or in sequential motions.
 It requires the use of both kinematics and dynamics of
robots.

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

4. Trajectory Planning
PATH VS. TRAJECTORY
 Path: A sequence of robot configurations in a
particular order without regard to the timing of
these configurations.
 Trajectory: It concerned about when each part of
the path must be attained, thus specifying timing.
Sequential robot movements in a path.

5. Trajectory planning
algorithm
Path description
path constraints(obstacles)
constraints imposed by
robot dynamics (smooth)
(
Joint (end-effector)
trajectories in terms
of position, velocity
and acceleration
TRAJECTORY PLANNING

6. TRAJECTORY PLANNING
Method 1
Plan a path in Cartesian Space
Use inverse kinematics for
finding the corresponding path
for the joints in the joint space.
Method 2
Plan a path in Joint Space
Use inverse kinematics for
finding the initial and final joint
position in joint space.
Without
obstacle
Cartesian Space
Techniques
Joint Space
Techniques

7. TRAJECTORY PLANNING
Method 1 Method 2
Computing the inverse
kinematics in real time,
time consuming and robot
efficiency may be affected.
When a specific end effector
trajectory is required, this
method is useful. E.g. in
the case of arc welding,
the electrode is required
to follow the seam
precisely.
Determine the initial and
final joint positions. The
path for all joints can be
easily planned in real time.
No need for forward and
inverse kinematics
recalculation.

8. Cartesian space
• We can track a shape
(for orientation : equivalent axes, Euler angles,…
• More expensive at run time
(after the path is calculated need joint angles
in a lot of points)
• Discontinuity problems
Joint space
• Easy to go through via points
(Solve inverse kinematics at all path points and plan)
• No problems with singularities
• Less calculations
• Can not follow straight line
TRAJECTORY PLANNING

9. Cartesian planning
difficulties :
Initial and Goal
Points are
reachable.
Intermediate points
(C) unreachable.
TRAJECTORY PLANNING
A
B
C

10. TRAJECTORY PLANNING
WITH POLYNOMIALS
No obstacle
The robot trajectory can be planned using initial
positions, initial velocity , final position and final velocity
of the joints in joint space. The third-order (cubic)
polynomial function is chosen as the position trajectory of
the ith joint:
3
3
2
210)( tatataat iiiii 
3210 ,,, iiii aaaa are constant parameters to be determined.

11. TRAJECTORY PLANNING
WITH POLYNOMIALS
No obstacle
Differentiate with respect to time t,
2
321 32)( tataat iiii 
we obtained the velocity trajectory of the
ith joint.

12. TRAJECTORY PLANNING
WITH POLYNOMIALS
No obstacle 2
321 32)( tataat iiii 
Assume that the initial position, final position, initial
velocity and final velocity are known, where the initial
time is t0=0 and final time is tf,
2
321
1
3
3
2
210
0
32)(
)0(
)(
)0(
fifiifi
ii
fififiifi
ii
tataat
a
tatataat
a











13. TRAJECTORY PLANNING
WITH POLYNOMIALS
No obstacle 2
321 32)( tataat iiii 
33
22
1
0
))()0(())()0((2
))()0(2())0()((3
)0(
)0(
f
ffiifii
i
f
ffiiifi
i
ii
ii
t
ttt
a
t
ttt
a
a
a













Four parameters of the cubic polynomial can be uniquely
determined by the four constraints: the initial position,
final position, initial velocity and final velocity.
Obtain this
equation

14. TRAJECTORY PLANNING
WITH POLYNOMIALS
No obstacle 2
321 32)( tataat iiii 
3
0
000
3
2
0
0
2
1
0
)(
)))(()(())()((2
)(
)))(()0(2())0()((3
)0(
)0(
tt
tttttt
a
tt
tttttt
a
ta
ta
f
ffiifii
i
f
ffiiifi
i
ii
ii















When t0 is not equal to 0,

15. TRAJECTORY PLANNING
WITH POLYNOMIALS
No obstacle
Consider a single link robot manipulator with a
rotary joint. Design a cubic trajectory, which starts
from the initial angular position and ends
at the final angular position , with zero
initial velocity and zero final velocity.
10)0(i
 90)2(i

16. TRAJECTORY PLANNING
WITH POLYNOMIALS
No obstacle
10)0(i  90)2(i
3
3
2
210)( tatataat iiiii 
0)0( i 0)2( i
20
2
)9010(2
60
2
)1090(3
0
10
33
22
1
0








a
a
a
a
32
206010)( ttt 
Final trajectory,

17. EXERCISE 1
The second joint of SCARA manipulator is required
to move from ϴ2 = 30° to 150° in 5 seconds. Find
the cubic polynomial to generate the smooth
trajectory for the joint. Assume that the initial and
final velocity of the joint is zero.

18. POLYNOMIALS TRAJECTORIES WITH
VIA POINTS
With obstacle
Some via points must be specified in order to avoid collisions.
A
B
CP(0)
P(tf)

19. POLYNOMIALS TRAJECTORIES WITH
VIA POINTS
With obstacle
Two methods fro trajectory planning with via point in space:
1. Use single high-order polynomial. The order of
polynomial depends on the number of the constraints.
2. With m via points, the whole trajectory could be divided
into m+1 segments, and each segment can be modeled
by using cubic polynomials.

20. POLYNOMIALS TRAJECTORIES WITH
VIA POINTSWith obstacle
Consider a single-link robot manipulator with a
rotary joint. Design a single polynomial
trajectory, which starts from the initial angular
position ϴ(0)=10°, passes the via point ϴ(1)=5°,
and then stops at the final angular position
ϴ(2)=50°, with zero initial velocity and zero final
velocity.

21. POLYNOMIALS TRAJECTORIES WITH
VIA POINTSWith obstacle
The initial angular position ,
initial velocity ,
passes the via point ,
the final angular position ,
final velocity .
0)0( 
10)0(
 5)1(
 50)2(
0)2( 

22. POLYNOMIALS TRAJECTORIES WITH
VIA POINTSWith obstacle
The order of polynomial depends on the number of
constraint. We have 5 constraints, fourth order polynomial
with five parameters as the candidate of the trajectory
function:
4
4
3
3
2
210)( tatatataat 
3
4
2
321 432)( tatataat 
2
432 1262)( tataat 
Velocity,
Acceleration,

23. POLYNOMIALS TRAJECTORIES WITH
VIA POINTSWith obstacle
100 a
543210  aaaaa
032124 4321  aaaa
5016842 43210  aaaaa
01 a
Using the five constraints, the following equation were
obtained,
25,90,70,0,10 43210  aaaaaThen,

24. POLYNOMIALS TRAJECTORIES WITH
VIA POINTSWith obstacle
Fourth order polynomial trajectory is formed,
25,90,70,0,10 43210  aaaaaThen,
432
25907010)( tttt 
32
100270140)( tttt 
2
300540140)( ttt 
Velocity,
Acceleration,

25. TRAJECTORY PLANNING
Previously, the velocity increase until peak and then
reduce until zero when the robot arm move
towards the final position.
There are cases where the robot arm are required
to move at a constant velocity. In that case, the
trajectory should be divided into three segments
i.e. acceleration segment, constant velocity
segment and deceleration segment.

26. POLYNOMIALS TRAJECTORIES WITH
VIA POINTSWith obstacle
Consider a single-link robot manipulator with a
rotary joint. Design a trajectory with following
cubic segments: the 1st segment connects initial
angular position ϴ(0)=10° to the via point
ϴ(1)=5°, and 2nd segment that connects via
point ϴ(1)=5° to the final angular position
ϴ(2)=50°. Assume that initial velocity and final
velocity are zero. At via point, the trajectory
should have continuous velocity and
acceleration

27. Quiz - 1
Consider a single-link robot manipulator with a
rotary joint. Design a trajectory with following cubic
segments: the 1st segment connects initial angular
position ϴ(0)=5° to the via point ϴ(5)=10°, and 2nd
segment connects via point ϴ(5)=10° to the final
angular position ϴ(20)=100°. Assume that initial
velocity and final velocity are zero. At via point, the
trajectory should have continuous velocity and
acceleration

28. TRAJECTORY PLANNING
Two parabolics model to define acceleration and
decelaration, one linear model to define
constant velocity segment, it is called Linear
Segment With Two Parabolic Blend (LPSB).