INTRODUCTION TO DYNAMICS ANALYSIS OF ROBOTS (Part 6)
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INTRODUCTION TO DYNAMICS ANALYSIS OF ROBOTS (Part 6)
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Transcript of INTRODUCTION TO DYNAMICS ANALYSIS OF ROBOTS (Part 6)
INTRODUCTION TO
DYNAMICS ANALYSIS
OF ROBOTS(Part 6)
This lecture continues the discussion on the analysis of the instantaneous motion of a rigid body, i.e. the velocities and accelerations associated with a rigid body as it moves from one configuration to another.
After this lecture, the student should be able to:•Analyze problems of evolution of rigid body configuration with time
Introduction to Dynamics Analysis of Robots (6)
Application Issues
In many industry applications (e.g. welding and assembly), the desired path of the gripper is known. In these applications, it is common to divide the path into many points that the gripper must successively pass through. These points are then used to get the required joint angles based on inverse kinematics solutions.This raises another issue. How fast the gripper traverses these points in succession will depend on the rate of change of the joint variables, i.e. the linear velocity of the gripper will depend on the angular velocities of the joints. This issue is important because for a robot to maneuver in real time, the joint angles and velocities have to be determined quickly for the robot control. The accuracy of the results will ultimately affect the accuracy and repeatability of the robot.
Application Issues
Point ‘A’ Point ‘B’
Robot path
Inverse kinematics can determine the angles 1A, 2A & 3A
at point ‘A’.
Inverse kinematics can also determine the angles 1B, 2B & 3B at point ‘B’.
How fast the arm travel from point ‘A’ to B’ will depend on the rate the corresponding angles 1A, 2A & 3A change to 1B, 2B & 3B.
f 0
Trajectory Planning
Consider the rotation of the following link:
The link starts from rest, i.e. at t0=0
0)( 00 t
The link must stop at tf, i.e.
0)( fft
The total time taken for the rotation is
At the time th where the rotation should have covered 2
Tth
20 f
h
The mid point!
0ttT f
Trajectory Planning
One way to accomplish the rotation is as follow:
t
t
ftht0t 1t
1T
From t0=0 to t1: constant acceleration Velocity will change from to 1
From t1 to t2: zero acceleration
Velocity held constant at 1
From t2 to tf constant deceleration at
Velocity will change from to 0
1
2t t
01h2
f
0
1
00
1T
0
Trajectory Planning
0 tdt
(constant)
Note: =0 (initially at rest)0
For t0 t t1:
02
21 ttdtdt
At t=t1:
02
11 21 T
11 T
11 T
0For t1 t t2:
)( 111 TtT
(constant) will change from 1 at t=t1 to 2 at t=t2 where
t
t
ftht0t 1t
11T
2t t
01h2
f
0
1T
Trajectory Planning
t
t
ftht0t 1t
11T
2t t
01h2
f
0
20 f
h
T
1T0
211 2
1 T
21
01
02
10
1
21
22
21
22
T
T
fh
fh
111
1
1
1
22 T
TTtth
h
hh
112
112
2
)(2
TTtth
Trajectory Planning
t
t
ftht0t 1t
11T
2t t
01h2
f
0
T
1T
21
011
111
11
1
21
222)2(
)(2)2(
22
TTTT
TTT
TTT
f
h
hh
0012
1 fTTT
Rearranging:
2
4 0
2
1
fTTT
Solving it to get:
Trajectory Planning
t
t
ftht0t 1t
11T
2t t
01h2
f
0
T
1T
2
4 0
2
1
fTTT
021
02
1 42
fhh
f
ttT
TTT
Obviously T1th:
02
1
fhh ttT
This is the time required for acceleration and deceleration
Trajectory Planning
t
t
ftht0t 1t
11T
2t t
01h2
f
0
T
1T
2
4 0
2
1
fTTT
To avoid any imaginary solutions:
20
0
2
4
4
T
T
f
f
This is the constraint on the value of the constant acceleration
Trajectory Planning
t
t
ftht0t 1t
11T
2t t
01h2
f
0
T
1T
Given 0 and f and total time “T”, the plan works as follow:•Select a value for acceleration
2
04
Tf
•Find the “blend time” T1 using:
02
1
fhh ttT
2
Tth where
•Implement the trajectory plan according to the diagrams on the left
Trajectory Planning
t For t0 t t1:
02
21 t
At t=T1:
02
11 21 T
111 T
11 T For t1 t t2:
)( 111 TtT (constant)
t
t
ftht0t 1t
11T
2t t
01h2
f
0
1T
At t=T2:
)2( 1112 TTT 112 T
T-2T1
(constant)
0
Trajectory Planning
)}({ 11 TTtT For t1 t tf:
21
112
)}({21
)}({
TTt
TTtT
At t=tf=T:
f
0
t
t
ftht0t 1t
11T
2t t
01h2
f
0
1T
T-2T1
(constant)
0
Point ‘A’ Point ‘B’
Robot path
Putting it Altogether
We want the gripper to move from point “A” to “B”. Given these points w.r.t. base frame, we can apply the inverse kinematics to derive the joint angles 0s at point “A” and angles fs at point “B”. We then have to decide on the time “T” to accomplish the motion. With these parameters, we can apply the trajectory planning discussed.
Other Trajectory Planning Approaches
All trajectory planning approaches basically involve generating a path w.r.t. between the given starting point (represented by 0) and the ending point (represented by f).
t
f
0
T
There are many possibilities!
0t
1t
The constraints are:
ffftt
tt
,
, 000
One common approach is to use a cubic polynomial to represent the path
Cubic Polynomial
The cubic polynomial has the form
Tttatataat 0,)( 33
2210
2321 32)( tataat
33
2210
0000
)(
)(0
TaTaTaaTTtt
attt
ff
2321
1000
32)(
)(0
TaTaaTTtt
attt
ff
2320 32 TaTaf
33
2200 TaTaTf
Cubic Polynomial
TTTaa
TaTaTaTa
f
ff
032
02
322
320
32
3232
TTTTaa
TTaTaTaTaT
f
ff
020
232
003
32
23
32
200
TTTaa f 0
20
32 2222
200
3 2TT
Ta ff
TTTaa f 0
32 32
30
20
3 2TT
a ff
TTTaa f 0
20
32 3333
TTTaa f 0
32 32
Cubic Polynomial
TTa ff 0
20
2
23
Summarizing: Tttatataat 0,)( 33
2210
00 a
01 a
TTa ff 0
20
2
23
30
20
3 2TT
a ff
This lecture continues the discussion on the analysis of the instantaneous motion of a rigid body, i.e. the velocities and accelerations associated with a rigid body as it moves from one configuration to another.
The following were covered:•Evolution of rigid body configuration with time
Summary
