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INTRODUCTION TO
DYNAMICS ANALYSIS
OF ROBOTS(Part 3)
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:•Derive the acceleration tensor and angular acceleration tensor•Derive the principles of relative motion between bodies in terms of acceleration analysis
Introduction to Dynamics Analysis of Robots (3)
Summary of previous lectures
333231
232221
131211
)()(
tRRt T
Velocity tensor and angular velocity vector
12
31
23
21
13
32
3
2
1
)(
t
)()()()()( tPtQttvtv PQ
)()()()()( tPtQttvtv PQ
bQabaobaQ PRPP ///
Velocity and moving FORs
bQabbQ
ababaobaQ vRPRvv /////
Relative Angular Velocity
Consider 3 FORs {a}, {b} and {c}. is the rotation of frame {b} w.r.t. frame {a}. Let
Rab
= relative angular velocity of frame {b} w.r.t. frame {a}aba
/
= relative angular velocity of frame {c} relative to frame {b} w.r.t. frame {a}bc
a/
= relative angular velocity of frame {c} w.r.t. frame {a}aca
/
)( // bcabbc
a R
)( ///
///
bcababac
bca
aba
aca
R
Example: Relative Angular Velocity
Example: The 3 DOF RRR Robot:
Y0, Y1
X0, X1
Z0, Z1
Z2
X2
Y2
Z3
X3
Y3
A=3 B=2 C=1
P
What is after 1 second if all the joints are rotating at
3,2,1,6
it
i
0/3
0
5236.0
0
1/2
100
0866.05.0
05.0866.001R
5236.0
0
0
0/1
Example: Relative Angular Velocity
5236.0
0
0
2/3
0866.05.0
100
05.0866.012R
100
0866.05.0
05.0866.023R
Solution: We re-used the following data obtained from the previous lecture
0
0472.1
0
5236.0
0
0
0866.05.0
100
05.0866.0
0
5236.0
0
)(
1/3
2/3121/21/3
R
Example: Relative Angular Velocity
5236.0
9068.0
5236.0
0
0472.1
0
100
0866.05.0
05.0866.0
5236.0
0
0
)(
0/3
1/3010/10/3
R
)(
)(
1/3010/10/3
2/3121/21/3
R
R
Example: Relative Angular Velocity
You should get the same answer from the overall rotational matrix and its derivative, i.e.
)1,2(
)3,1(
)2,3(
0/3
0/3
0/3
0/303
030/3
23
12
01
23
12
01
23
12
01
03
23
12
01
03
TRR
RRRRRRRRRR
RRRR
5236.0
9068.0
5236.0
0/3
05.0866.0
866.0433.025.0
5.075.0433.023
12
01
03 RRRR
09069.05236.0
2618.06545.02267.0
4534.02267.09162.023
12
01
23
12
01
23
12
01
03 RRRRRRRRRR
Example: Relative Angular Velocity
5236.0
9068.0
5236.0
05236.09068.0
5236.005235.0
9068.05236.00
05.0866.0
866.0433.025.0
5.075.0433.0
09069.05236.0
2618.06545.02267.0
4534.02267.09162.0
0/30/3
0/3
T
Acceleration tensor
Consider 2 points “P” and “Q” of a rigid body:
)()()()()()(
})()()(){()()()()()(
)()()()()()()()(
)()()()()(
2 tPtQtttPtQ
tPtQtttPtQttPtQ
tPtQttPtQttPtQ
tPtQttPtQ
Rearranging:
)()()()()( tPtQtAtata PQ
where)()()( 2 tttA
A(t) is called the acceleration tensor
00
00
000
)()(
tRRt T
Example: Acceleration tensor
Given
Find the acceleration tensor if =t2
Solution:
020
200
000
00
00
000
)(
t
222 tt
Example: Acceleration tensor
22
222
2
22
400
040
000
)(
00
00
000
00
00
000
00
00
000
)(
t
tt
t
22
22
22
222
420
240
000
)(
400
040
000
020
200
000
)(
t
ttA
t
ttA
Angular Acceleration vector
)()()()()()()()(
)()()()()()()()(
tvtvttPtQttPtQ
tPtQttPtQttPtQ
PQ
where
12
31
23
21
13
32
3
2
1
)(
t Angular
velocity vector
12
31
23
21
13
32
3
2
1
)(
t
Similarly:
Angular acceleration vector
Example: Angular Acceleration vector
00
00
000
)()(
tRRt T
Given
Find the angular acceleration vector if =t2
Solution:
020
200
000
00
00
000
)(
t
222 tt
0
0
2
)(
21
13
32
t
Acceleration and moving FORs
RRR
RRRRR
RRRR
PRPRPRPRPRvv
PRPRvv
T
TT
TT
bQabbQ
abbQ
abbQ
abbQ
abaobaQ
bQabbQ
abaobaQ
1
1111
1
///////
////
)()()(
)(
bQabbQ
abab
bQabababbQ
ababaobaQ
bQabbQ
abbQ
abbQ
abaobaQ
bQabbQ
abbQ
abbQ
abbQ
abaobaQ
bQabbQ
abbQ
abbQ
abbQ
abaobaQ
PRPR
PRPRaa
PRPRPRPRvv
PRPRPRPRPRvv
PRPRPRPRPRvv
///
///////
//////
///////
///////
)(2
)(
)(2)(
)()()(
)()()(
Acceleration and moving FORs
bQ
abbQ
abab
bQabababbQ
ababaobaQ
PRPR
PRPRaa
///
///////
)(2
)(
Let
bQabrel
bQabrel
bQabrel
PRa
PRV
PRP
/
/
/
)(
relrelabrelababrelabaobaQ aVPPaa )(2)( //////
Example: Acceleration and moving FORs
Example: The 3 DOF RRR Robot:
Y0, Y1
X0, X1
Z0, Z1
Z2
X2
Y2
Z3
X3
Y3
A=3 B=2 C=1
P
What is the acceleration of point “P” after 1 second if all the joints are rotating at
3,2,1,6
it
i
Example: Acceleration and moving FORs
We know from the previous lecture that at t=1
100
0866.05.0
05.0866.001R
000
05.0866.0
0866.05.0
11
1101
R
1000
0100
00)cos()sin(
00)sin()cos(
11
11
01
T
0000
0000
00)sin()cos(
00)cos()sin(
1111
1111
01
T
61
0000
0000
00)sin()cos(
00)cos()sin(
1111
1111
01
T
Example: Acceleration and moving FORs
0000
0000
00)sin()cos()cos()sin(
00)cos()sin()sin()cos(
112
11112
11
112
11112
11
01
T
000
0866.05.0
05.0866.02
12
1
21
21
01
R
Example: Acceleration and moving FORs
Similarly at t=162
1000
00)cos()sin(
0100
0)sin()cos(
22
22
12
A
T
0866.05.0
100
05.0866.012R
05.0866.0
000
0866.05.0
22
2212
R
0000
00)sin()cos(
0000
00)cos()sin(
2222
2222
12
T
Example: Acceleration and moving FORs
0000
00)sin()cos(
0000
00)cos()sin(
2222
2222
12
T
0000
00)sin()cos()cos()sin(
0000
00)cos()sin()sin()cos(
2222222
22
2222222
222
12
T
0866.05.0
000
05.0866.0
22
22
22
22
12
R
At t=1,63
1000
0100
00)cos()sin(
0)sin()cos(
33
33
23
B
T
000
05.0866.0
0866.05.0
33
3323
R
100
0866.05.0
05.0866.023R
0000
0000
00)sin()cos(
00)cos()sin(
3333
3333
23
T
Example: Acceleration and moving FORs
0000
0000
00)sin()cos(
00)cos()sin(
3333
3333
23
T
Example: Acceleration and moving FORs
0000
0000
00)sin()cos()cos()sin(
00)cos()sin()sin()cos(
3323333
233
3323333
233
23
T
000
0866.05.0
05.0866.023
23
23
23
23
R
TTT RRRRRRt )(
00 0/101
01
01
010/1
TT RRRR
Example: Acceleration and moving FORs
Substitute the matrices given into the equation, we get:
Similarly
00
00
2/323
23
23
232/3
1/212
12
12
121/2
TT
TT
RRRR
RRRR
TPP 001;6 3/321
We need to find 0/Pa
With
5236.0
0
0
0/1
5236.0
0
0
2/3
0
5236.0
0
1/2
For the data given, the following were determined in the previous lecture:
Example: Acceleration and moving FORs
0
4534.0
2618.0
2/Pv
4304.1
0
4304.1
1/Pv
4304.1
6571.1
6084.2
0/Pv
02/31/20/1
Example: Acceleration and moving FORs
3/
233/
232/3
3/232/32/33/
232/32/32/
)(2
)(
PP
PPoP
aRvR
PRPRaa
0
0
0
3/
3/
2/3
P
P
o
v
a
a
There is no translation acceleration between frames {3} and {2} and no translation velocity and acceleration of point “P” in frame {3}
3/232/32/32/ PP PRa
0
1371.0
2374.0
0
0
1
100
0866.05.0
05.0866.0
5236.0
0
0
5236.0
0
0
2/Pa
Example: Acceleration and moving FORs
2/
122/
121/2
2/121/21/22/
121/21/21/
)(2
)(
PP
PPoP
aRvR
PRPRaa
01/2 oa There is no translation acceleration between
frames {2} and {1}
TPoP PRPP 05.0866.23/232/32/
TPa 01371.02374.02/
01/2
TPP PRv 04534.02618.03/232/32/
2237.1
0
023.1
0
1371.0
2374.0
0866.05.0
100
05.0866.0
0
4534.0
2618.0
0866.05.0
100
05.0866.0
0
5236.0
0
2
0
5.0
866.2
0866.05.0
100
05.0866.0
0
5236.0
0
0
5236.0
0
1/
1/
P
P
a
a
Example: Acceleration and moving FORs
Substituting the values into the equation:
2/122/
121/22/
121/21/21/ )(2 PPPP aRvRPRa
Example: Acceleration and moving FORs
1/
011/
010/1
1/010/10/11/
010/10/10/
)(2
)(
PP
PPoP
aRvR
PRPRaa
00/1 oa There is no translation acceleration between
frames {1} and {0}
TPa 2237.10023.11/
0/ ab
TPoP PRPP 866.10232.52/121/21/
TPPP vRPRv 4304.104304.12/122/
121/21/
Example: Acceleration and moving FORs
Substituting the values into the equation:
1/011/
010/11/
010/10/10/ )(2 PPPP aRvRPRa
22.1
53.2
38.1
2237.1
0
023.1
100
0866.05.0
05.0866.0
4304.1
0
4304.1
100
0866.05.0
05.0866.0
5236.0
0
0
2
866.1
0
232.5
100
0866.05.0
05.0866.0
5236.0
0
0
5236.0
0
0
0/
0/
P
P
a
a
Example: Acceleration and moving FORs
We should get the same answer if we use transformation matrix method.
Try it at home and we’ll discuss this in the next lecture!
Summary
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:•The acceleration tensor and angular acceleration tensor•The principles of relative motion between bodies in terms of acceleration analysis