introduction to dynamics analysis of robots (part 3)
TRANSCRIPT
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INTRODUCTION TO
DYNAMICS ANALYSIS
OF ROBOTS(Part 3)
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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)
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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 /////
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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)(
)()()(
)()()(
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Acceleration and moving FORs
bQ
abbQ
abab
bQabababbQ
ababaobaQ
PRPR
PRPRaa
///
///////
)(2
)(
Let
bQabrel
bQabrel
bQabrel
PRa
PRV
PRP
/
/
/
)(
relrelabrelababrelabaobaQ aVPPaa )(2)( //////
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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/
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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
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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/
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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
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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!
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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