Isotropie des manipulateurs parallèles de la classe H4 Isotropy of

Transcription

Isotropie des manipulateurs parallèles de
la classe H4
Isotropy of the H4 class of parallel
manipulators
Benoit Rousseau
Département de génie mécanique,
École Polytechnique de Montréal,
[email protected]
Luc Baron
Département de génie mécanique,
École Polytechnique de Montréal,
[email protected]
About the H4 • Loop closure equation • Rigidity condition • Jacobian matrix • Isotropy condition • Design procedure • Example • Future work • Questions
About the H4
1
• Paralel manipulator with 4 DOF (x, y, z, θ)
• Four arms
• Articulated traveling plate
• Revolute and spherical joints
1
O. Company, S. Krut and F. Pierrot. Internal singularity analysis of a class of lower mobility parallel
manipulators with articulated traveling plate. IEEE Transactions on Robotics, 2006
CCTOMM 2009, Isotropy of the H4 class of parallel manipulators, Benoit Rousseau & Luc Baron
2
Loop closure equation
Bi
pi
Ai
ri
qi
Ci
ûi
si
Di
θ
ti
P
One of the four legs
Ai : Revolute joint at the base
Bi and Ci : Spherical joints
Di : Revolute joint
P : Effector
3
Bi
pi
Ai
ri
qi
ûi
Ci
si
Di
θ
ti
P
bi = ai + pi (qi )
4
Bi
pi
Ai
ri
qi
ûi
Ci
si
Di
θ
ti
P
bi = ai + pi (qi )
5
Bi
pi
Ai
ri
qi
ûi
Ci
si
Di
θ
ti
P
bi = ai + pi (qi )
6
Bi
pi
Ai
ri
qi
ûi
Ci
si
Di
θ
ti
P
bi = ai + pi (qi )
ci = p − ti (θ) − si
7
Bi
pi
Ai
ri
qi
ûi
Ci
si
Di
θ
ti
P
bi = ai + pi (qi )
8
Bi
pi
Ai
ri
qi
ûi
Ci
si
Di
θ
ti
P
bi = ai + pi (qi )
9
Bi
pi
Ai
ri
qi
ûi
Ci
si
Di
θ
ti
P
bi = ai + pi (qi )
10
Bi
pi
Ai
ri
qi
ûi
Ci
si
Di
θ
ti
P
bi = ai + pi (qi )
ri = ci − bi
11
Bi
pi
Ai
ri
qi
ûi
Ci
si
Di
θ
ti
P
bi = ai + pi (qi )
ri = ci − bi
12
Bi
pi
Ai
ri
qi
ûi
Ci
si
Di
θ
ti
P
bi = ai + pi (qi )
ri = ci − bi
13
Bi
pi
Ai
ri
qi
Ci
si
ûi
Di
θ
ti
P
bi = ai + pi (qi )
ri = ci − bi
ri = p − ti (θ) − si − ai − pi (qi )
14
Rigidity condition
The length or ri is constant :
ri T ri = ri 2
(1)
Knowing :
ri = p − ti (θ) − si − ai − pi (qi )
(6)
We can write :
ri 2 = ri T (p − ti (θ) − si − ai − pi (qi ))
(7)
15
Jacobian matrix
Derivating with respect to time :
ri T ṗ + ti × k̂ θ̇ − (pi × ûi ) q̇i = 0
(8)
ri T ṗ + ri T ti × k̂ θ̇ = ri T (pi × ûi ) q̇i
(9)
Using the distributivity property of the scalar product, we can separate the terms
ṗ, θ̇ and q˙i :
Writing this equation for each of the four legs :
Aẋ = B q̇
(10)
where :
2
r1 T
6
6 T
6r2
A≡6
6 T
6r3
4
r4 T
“
”3
r1 T t1 × k̂
2 3
2 T
“
”7
7
T
r1 (p1 × û1 )
r2 t2 × k̂ 7
6ṗ7
“
”7 ẋ ≡ 6 7 B ≡ 4
0
7
4 5
r3 T t3 × k̂ 7
0
“
”5
θ̇
T
r4 t4 × k̂
0
...
0
2 3
q˙1
6 7
5 q̇ ≡ 6q˙2 7
4q˙3 5
r4 T (p4 × û4 )
q˙4
0
0
3
16
Adimensionalisation
Choosing λ as the natural length we then get :

r1 T /λ

 T
r2 /λ
A≡
 T
r3 /λ

r4 T /λ

 T
r1 T t1 × k̂ /λ2
r1 (p1 × û1 )



λ2
r2 T t2 × k̂ /λ2 

 B≡
0


r3 T t3 × k̂ /λ2 


0
r4 T t4 × k̂ /λ2
0
0
...
0
T
0
r4 (p4 × û4 )
λ2






17
Isotropy condition
Fixing the effector speed manifold to an hyper-spherical shape :
ẋT ẋ = 1
(12)
Subtituing Aẋ = B q̇ in the last equation :
q̇ T (A−1 B)T A−1 B q̇ = 1
Defining C T ≡ B −1 A we then get :
C T C = α2 1


λ r1 T /g1 h1 /g1
λ r2 T /g2 h2 /g2 

CT ≡ 
λ r3 T /g3 h3 /g3 
λ r4 T /g4 h4 /g4
(13)
where gi ≡ ri T (pi × ûi ) and hi ≡ ri T (ti × k̂)
18
Orthogonality condition
The scalar dot product between two rows of the C matrix must be null :
λ ri T /gi
hi /gi
λ rj T /gj
hj /gj
T
=0
We can write this equation for every pairs of rows of the C matrix :
t1,4
t2,3
µ1
µ2
λ
λ
t2,3
t2,3
=
µ2
µ3
λ
λ
t2,3
t1,4
µ1
µ3
λ
λ
t1,4
t2,3
=
µ2
µ4
λ
λ
t1,4
t1,4
µ1
µ4
λ
λ
t1,4
t2,3
=
µ3
µ4
λ
λ
− σ1,2 =
− σ1,3 =
− σ1,4 =
− σ2,3
− σ2,4
− σ3,4
r̂
where µi ≡ cos ∠ tr̂ii×k̂ ≡ r̂iT (t̂i × k̂) and σi,j ≡ cos ∠ r̂ij ≡ r̂iT r̂j .
ti
µi we get :
λ
− σ1,2 = β1 β2
− σ1,3 = β1 β3
− σ1,4 = β1 β4
− σ2,3 = β2 β3
− σ2,4 = β2 β4
− σ3,4 = β3 β4
Defining βi ≡
(17)
From which we can find the following constraint :
σ1,2 σ3,4 = σ1,3 σ2,4 = σ1,4 σ2,3
(18)
19
Normality condition
All rows of the C matrix must have the same norm :
λ ri T /gi
hi /gi
λ ri T /gi
hi /gi
T
= α2
We can write this equation for every rows of the C matrix :
t1,4 2 2
p1 2
µ1 = α2 2 η1 2
2
λ
λ
2
p
t2,3 2 2
3
1 + 2 µ3 = α2 2 η3 2
λ
λ
1+
t2,3 2 2
p2 2
µ2 = α2 2 η2 2
2
λ
λ
2
t1,4 2 2
p
4
1 + 2 µ4 = α2 2 η4 2
λ
λ
1+
p ×ûi
where µi ≡ cos ∠ tr̂ii×k̂ ≡ r̂iT (t̂i × k̂) and ηi ≡ cos ∠ r̂ii
(20)
≡ r̂iT (p̂i × ûi )
20
Relation between the σi,j
Using the angles between the ri instead of the ri themselves, the σi,j , which
represent the angle between the ri , must then respect the following condition :
arccos
σ1,3 − σ1,4 σ3,4
σ2,3 − σ2,4 σ3,4
σ1,2 − σ1,4 σ2,4
±arccos
±arccos
=0
κ1,4 κ2,4
κ1,4 κ3,4
κ2,4 κ3,4
(21)
r̂
r̂
where σi,j ≡ cos ∠ r̂ij et κi,j ≡ sin ∠ r̂ij .
For the solutions not to be complex we must have the following conditions :
−1 ≤
σ1,2 − σ1,4 σ2,4
σ1,3 − σ1,4 σ3,4
σ2,3 − σ2,4 σ3,4
≤ 1 −1 ≤
≤ 1 −1 ≤
≤1
κ1,4 κ2,4
κ1,4 κ3,4
κ2,4 κ3,4
21
Design procedure
Choice of σ1,4
Choice of σ2,4
Choice of σ1,2
The σi,j are cossines :
Calculus of σ1,3
r̂
σi,j ≡ cos ∠ r̂ij
Calculus of σ2,3 and σ3,4
Calculus of βi
We must then choose σ1,4 between −1 and 1 :
Calculus of µi and of r̂i
−1 ≤ σ1,4 ≤ 1
Calculus of ti /λ
Choice of α
Choice of ηi
Calculus of pi /λ
22
Design procedure
Choice of σ1,4
From (21) we have this condition :
Choice of σ2,4
Choice of σ1,2
Calculus of σ1,3
Calculus of βi
−1 ≤
σ1,2 − σ1,4 σ2,4
≤1
κ1,4 κ2,4
(22a)
for σ1,2 = +1 :
−κ1,4 κ2,4 ≤ (+1) − σ1,4 σ2,4 ≤ κ1,4 κ2,4
for σ1,2 = −1 :
Calculus of ti /λ
Choice of α
Choice of ηi
Calculus of pi /λ
−κ1,4 κ2,4 ≤ (−1) − σ1,4 σ2,4 ≤ κ1,4 κ2,4
Then, σ2,4 must p
be in the smallest
p interval between
[−1, +1] and [− 1 − σ1,4 2 , + 1 − σ1,4 2 ].
23
Design procedure
Knowing
Choice of σ1,4
β1 = −
Choice of σ2,4
Choice of σ1,2
Calculus of σ1,3
Calculus of βi
σ1,4
β4
β2 = −
σ2,4
β4
Then
−σ1,2 =
σ1,4 σ2,4
β4 2
σ1,2
=−
|σ1,2 |
σ1,4
|σ1,4 |
σ2,4
|σ2,4 |
The sign of σ1,2 must be the inverse of the sign of
the product of σ1,4 and σ2,4 .
Calculus of ti /λ
Choice of α
Choice of ηi
Calculus of pi /λ
We must then choose σ1,2 accordingly in one of the
following intervals :
p
p
1 − σ1,4 2 1 − σ2,4 2 ]
p
p
[0, σ1,4 σ2,4 − 1 − σ1,4 2 1 − σ2,4 2 ]
[0, σ1,4 σ2,4 +
24
Design procedure
Choice of σ1,4
Choice of σ2,4
Expressing σ2,3 and σ3,4 as fonction of σ1,2 , σ1,3 ,
σ1,4 et σ2,4 :
σ2,3 =
Choice of σ1,2
Calculus of σ1,3
Calculus of βi
Calculus of ti /λ
Choice of α
Choice of ηi
Calculus of pi /λ
σ1,3 σ2,4
σ1,4
σ1,3 σ2,4
σ1,2
σ3,4 =
(23)
Then substituing in (21) and isolating σ1,3 :
σ1,3 = ±(σ1,4 2 (−σ2,4 4 +σ2,4 2 −σ1,4 4 −2σ1,4 2 σ2,4 2 −σ1,4 2 σ1,2 2
−2σ1,2 σ1,4 σ2,4 −σ2,4 2 σ1,2 2 +2σ1,2 3 σ1,4 σ2,4 −4σ1,2 2 σ1,4 2 σ2,4 2
3
3
2
+ 4σ1,2 σ1,4 σ2,4 + 4σ1,2 σ1,4 σ2,4 + σ1,4 ))
2
(1/2)
2
σ1,2 ((−σ1,4
2
3
+ 2σ1,2 σ1,4 σ2,4 − σ2,4 )(σ1,2 − σ1,4 σ2,4 ) /(2σ1,2 σ1,4 σ2,4
− 3σ1,2 2 σ1,4 2 σ2,4 2 − σ1,4 2 σ1,2 2 − σ2,4 2 σ1,2 2 + 2σ1,2 σ1,4 3
σ2,4 + 2σ1,2 σ1,4 σ2,4 3 − σ1,4 2 σ2,4 2 ))(1/2) /(2σ1,2 2 σ1,4 σ2,4
−2σ1,2 σ1,4 2 σ2,4 2 −σ1,2 σ2,4 2 −σ1,2 σ1,4 2 +σ1,4 3 σ2,4 +σ1,4 σ2,4 3 )
The sign of σ1,3 can be choosen freely.
25
Design procedure
Choice of σ1,4
Choice of σ2,4
Choice of σ1,2
Calculus of σ1,3
From the precedent step we know that :
Calculus of βi
σ2,3 =
σ1,3 σ2,4
σ1,4
σ3,4 =
σ1,3 σ2,4
σ1,2
(23)
Calculus of ti /λ
Choice of α
Choice of ηi
Calculus of pi /λ
26
Design procedure
Choice of σ1,4
Choice of σ2,4
From (17) we can express β1 as a function of σi,j :
Choice of σ1,2
Calculus of σ1,3
β1 = ±
Calculus of βi
−σ1,2 σ1,3
σ2,3
The sign of β1 can be choosen freely. The other βi
can then be calculated from β1 :
Calculus of ti /λ
s
β2 = −
σ1,2
β1
β3 = −
σ1,3
β1
β4 = −
σ1,4
β1
Choice of α
Choice of ηi
Calculus of pi /λ
27
Design procedure
Choice of σ1,4
Choice of σ2,4
Choice of σ1,2
Calculus of σ1,3
Because β1 and β4 share the same t1,4 , and because β2 and β3 share t2,3 , we can then obtaint :
β4
β1
=
µ1
µ4
β2
β3
=
µ2
µ3
Also knowing :
Calculus of βi
Calculus of ti /λ
r̂i r̂j = βi βj
µi = r̂i (t̂i × k̂)
We then have a system of equation which we can
solve for r̂i and µi .
Choice of α
Choice of ηi
Calculus of pi /λ
28
Design procedure
Choice of σ1,4
Choice of σ2,4
Choice of σ1,2
Calculus of σ1,3
Calculus of βi
From the definition of βi :
t1,4 /λ = B1/µ1
t2,3 /λ = B2/µ2
Calculus of ti /λ
Choice of α
Choice of ηi
Calculus of pi /λ
29
Design procedure
Choice of σ1,4
Choice of σ2,4
Choice of σ1,2
Calculus of σ1,3
Calculus of βi
The amplification factor α can be choosen freely but
must be positive.
Calculus of ti /λ
Choice of α
Choice of ηi
Calculus of pi /λ
30
Design procedure
Choice of σ1,4
Choice of σ2,4
Choice of σ1,2
The ηi are cossines :
Calculus of σ1,3
ηi ≡ cos ∠ r̂pii ×ûi
Calculus of βi
We must then choose the ηi between −1 and 1 :
−1 ≤ ηi ≤ 1
Calculus of ti /λ
Choice of α
Choice of ηi
Calculus of pi /λ
31
Design procedure
Choice of σ1,4
Choice of σ2,4
Choice of σ1,2
Calculus of σ1,3
Calculus of βi
Calculus of ti /λ
From equation (20) we can find the pi /λ :
1 + (t1,4 /λ)2 µ1 2
p1
p
=
λ
α2 η1 2
p3
1 + (t2,3 /λ)2 µ3 2
p
=
λ
α2 η3 2
p2
λ
p4
λ
1 + (t2,3 /λ)2 µ2 2
p
α2 η2 2
1 + (t1,4 /λ)2 µ4 2
p
=
α2 η4 2
=
Choice of α
Choice of ηi
Calculus of pi /λ
32
Example
! ! ! [Numerical Example : 3D view, Values, ... ] (one slide)
33
Future work
• Evaluate the contitionning of the jacobian matrix in the whole
workspace
• Determine if a given H4 manipulator has isotropic points
• Find the best isotropic H4 manipulator with respect to an other
criterion such as the size of the workspace
34
Questions ?
35

Isotropie des manipulateurs parallèles de la classe H4 Isotropy of

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