32354544 L9 Trajectory Planning 1 V1

7/28/2019 32354544 L9 Trajectory Planning 1 V1

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Trajectory Planning

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niversitiKualaLumpurMalaysiaFranceInstitute

Originally prepared by: Prof Engr Dr Ishkandar Baharin

Head of Campus & Dean

UniKL MFI


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Path and trajectory planning is the way that a robot is moved from onelocation to another in a controlled manner.

It requires the use of both kinematics and dynamics of robots.

Path : A sequence of robot configurations in a particular orderwithout regard to the timing of these configurations.

Trajectory: It concerned about when each part of the path must

be attained in a certain constraint, thus specifying timing.

Sequential robot movements in a path

Here the robotconfiguration is more

important than the speed.

Path and Trajectory Planning

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J OINT-SPACE VS. CARTESIAN-SPACE DESCRIPTIONS

J oint-space description:- The description of the motion to be made by the robot by its joint values.- The motion between the two points is unpredictable.

Cartesian space description:

- The motion between the two points is known at all times and controllable.- It is easy to visualize the trajectory, but is difficult to ensure thatsingularity.

Sequential motions of a robot to follow a straight lineU



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Cartesian-space trajectory (a) The trajectory specified in Cartesiancoordinates may force the robot to run into itself, and (b) the trajectory may

requires a sudden change in the joint angles.

Problem: A robot must not harm itself !!!

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J oint-space non-normalized movements of a robot with two degrees of freedom.

Move the robot from A to B, to run both jointsat their maximum angular velocities. After 2 [sec], the lower link will have finished its

motion, while the upper link continues for another3 [sec].

The path is irregular and the distances traveledby the robots end are not uniform.U



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J oint-space, normalized movements of a robot with two degrees of freedom.

Both joints move at different speeds, but movecontinuously together.

The resulting trajectory will be different.

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niversitiKualaLumpurMalaysiaFrance

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Cartesian-space movements of a two-degree-of-freedom robot.

Divide the line into five segments and solve fornecessary angles and at each point.

The joint angles are not uniformly changing.

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Trajectory planning with an acceleration-deceleration regiment.

It is assumed that the robots actuators are strong enough to provide largeforces necessary to accelerate and decelerate the joints as needed.Divide the segments differently.

The arm move at smaller segments as we speed up at the beginning.Go at a constant cruising rateDecelerate with smaller segments as approaching point B.

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Trajectory Planning

It is a planning of how to move an object from its initial location to the finallocation in the work space with a consideration of time, i.e. it describes thepositions, orientations, linear velocities, angular velocities and accelerations ofthe joint movements.

There are generally two types of trajectory planning:

Polynomial trajectories - workspace without obstaclesPolynomial trajectories via points - workspace with obstacles

Polynomials TrajectoriesIf there is no obstacle in the work-space, the trajectory for a robot manipulator canbe easily planned using the specified initial position, initial velocity, final positionand final velocity in the joint space. We may use inverse kinematics and motionkinematics to find the corresponding information of all joints in the joint space.

Generally, we can use third-order (cubic) polynomial function as the positiontrajectory.U


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(1) (1) (1) 2(1) 1 2 2

( ) 2 3t a a t a t = + +&

(1) (1)(1) 2 3( ) 2 6t a a t = +

&&

(2) (2) (2) 2

( 2) 1 2 3( ) 2 ( 1) 3 ( 1)t a a t a t = + + &

(2) (2)(2) 2 3( ) 2 6 ( 1)t a a t = +

&&

0 1t 0 1t 1 2t 1 2t

for

for

for

for

The constraints of the trajectory, from the question, are listed as follow:

0 0(1) (1) (1)

0

(2) (1) (2) (1) (2)

0

(2) (2)

(0) 10 , (0) 0, (1) 5 ,

(0) 5 , (1) (1), (1) (1)(2) 50 , (2) 0

= = =

= = == =

&

& & && &&

&&

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Using the constraints in the two cubic polynomials and their derivatives,we obtain the following equations:

(1 )0

(1 )

1

(1 ) (1 ) (1 ) (1 )

0 1 2 3( 2 )

0

( 2 ) (1 ) (1 ) (1 )

1 1 2 3

( 2 ) (1 ) (1 )2 2 3

( 2 ) ( 2 ) ( 2 ) ( 2 )

0 1 2 3

( 2 ) ( 2 ) ( 2 )

1 2 3

1 0

0

5

5

2 3

2 2 6

5 0

2 3 0

a

a

a a a a

a

a a a a

a a a

a a a a

a a a

=

=

+ + + =

=

= + +

= +

+ + + =

+ + =U


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Solving the above 8 equations, we obtain the polynomial parameters asfollow:

(1)0

(1)

1

(1)

2

(1)

3

(2)

0

(2)

1

(2)

2

(2)

3

10

0

45

40

5

30

75

60

a

a

a

a

a

a

a

a

=

=

=

=

=

=

=

=

2 3(1) ( ) 10 45 40t t t = +

2 3(2) ( ) 5 30( 1) 75( 1) 60( 1)t t t t = + +

0 1t 1 2t

Therefore, the detailed expressions of the two cubic segments are:

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2(1) ( ) 90 120t t t = +

&

(1)

( ) 90 240t t = +&&

2(2) ( ) 30 150( 1) 180( 1)t t t = +

&

(2) ( ) 150 360( 1)t t =

&&

0 1t

0 1t

1 2t

1 2t

And the velocity profiles and the acceleration profiles of the two trajectorysegments are also given below:

for

for

for

for

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Homework Study the chapter on Trajectory Planning,

chapter 7. Try to understand and follow the

examples given, Ex 7.2-1, 7.3-1, 7.3-2. Try out the exercises by yourself.

Read and understand the reference

slides given.

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32354544 L9 Trajectory Planning 1 V1