Strategy to lock the knee of exoskeleton stance leg: study in the framework of ballistic walking model Yannick Aoustin, Alexander Formalskii 1 Mesrob 2015, Nantes July 10, 2015 1 Supported Ministry of Education and Science of Russian Federation, Project No. 7.524.11.4012, and by Région des Pays de la Loire, Project LMA and Gérontopôle Autonomie Longévité des Pays de la Loire. Yannick Aoustin, Alexander Formalskii Strategy to lock the knee of exoskeleton stance leg: study in the fram
Motivation To design a wearable assist device for human. To improve daily life for patients or elderly To avoid the musculoskeletal disorders for industrial workers Desired performances: To be able to compensate a part of the loads due the human’s weight. To have an assist device with an energetic autonomy. Good adaptation to the shape of human’s body. Yannick Aoustin, Alexander Formalskii Strategy to lock the knee of exoskeleton stance leg: study in the fram
Statement of the problem To consider a five-link planar Biped with a strapped exoskeleton. There are not any actuators in our exoskeleton. To define a ballistic walking gait of the biped alone without any assist device for the transport of a load. During ballistic walking of the biped with exoskeleton the knee of the stance leg of the exoskeleton (and as a consequence of the biped) is locked. Walking of the biped consists of alternating single- and instantaneous double-support phases. At the instant of this phase, the knee of the previous swing leg is locked and the knee of the previous stance leg is unlocked. To compare the energy consumption of the biped alone and with its exoskeleton. Yannick Aoustin, Alexander Formalskii Strategy to lock the knee of exoskeleton stance leg: study in the fram
The geometrical structure of the biped and its wearable assist device. q 5 Γ 3 s T s t q 3 s p Γ 2 Γ 4 s s q 3 q 4 Γ 1 q 1 (b) (a) seven generalized coordinates x = [ q 1 , q 2 , q 3 , q 4 , q 5 , x , y ] ⊤ . Yannick Aoustin, Alexander Formalskii Strategy to lock the knee of exoskeleton stance leg: study in the fram
Physical parameters. Mass ( kg ) Length ( m ) Inertia moment center of ( kg . m 2 ) mass ( m ) I s = 0 . 0521 m s = 4 . 6 l s = 0 . 55 s s = 0 . 324 Human shin I t = 0 . 75 m t = 8 . 6 l t = 0 . 45 s t = 0 . 18 Human thigh I T = 11 . 3 m T = 48 . 6 l T = 0 . 75 s T = 0 . 386 Human trunk I 1 = 0 . 0260 m 1 = 1 . 0 l 1 = 0 . 497 s s = 0 . 324 Exoskeleton shin I 2 = 0 . 0354 m 2 = 2 . 0 l 2 = 0 . 41 s t = 0 . 18 Exoskeleton thigh I 3 = 0 . 3817 m 3 = 8 . 0 l 3 = 0 . 75 s T = 0 . 386 Exoskeleton trunk Yannick Aoustin, Alexander Formalskii Strategy to lock the knee of exoskeleton stance leg: study in the fram
Matrix equations Equations of the motion of the biped in single support x ) = D Γ + J ⊤ r 1 r 1 + J ⊤ A ( x )¨ x + h ( x , ˙ r 2 r 2 , (1) with the constraint equations, x + ˙ J r i ¨ J r i ˙ x = 0 for i = 1 or/and 2 . (2) The joint variables θ i for i = ( 1 , 2 , 3 , 4 ) as functions of the generalized coordinates are: θ 1 = q 2 − q 1 , θ 2 = q 5 − q 2 , θ 3 = q 5 − q 3 , θ 4 = q 3 − q 4 . (3) Yannick Aoustin, Alexander Formalskii Strategy to lock the knee of exoskeleton stance leg: study in the fram
Definition of the ballistic motion Statement of the problem Let x ( 0 ) be an initial configuration t = 0. Let x ( T ) be an final configuration t = T . Yannick Aoustin, Alexander Formalskii Strategy to lock the knee of exoskeleton stance leg: study in the fram
Definition of the ballistic motion ballistic motion be in single support: Γ = 0. � � � Γ 1 � + J ⊤ A ( x )¨ x + h ( x , ˙ x ) = D 1 D 2 D 3 D 4 r 1 r 1 , 0 3 × 1 with the constraint equation for the stance leg tip fixed on the ground: x + ˙ J r 1 ¨ J r 1 ˙ x = 0 , and with the constraint equation for the knee of the stance leg locked in the swing phase: D ⊤ 1 ¨ x = 0 . problem: Which velocity vector ˙ x ( 0 ) such that x ( t ) starting from x ( 0 ) reaches x ( T ) ⇒ A boundary value problem solved using a Newton method Yannick Aoustin, Alexander Formalskii Strategy to lock the knee of exoskeleton stance leg: study in the fram
Impulsive control decomposition of the impulsive impact. After impact the velocity of the biped has to be equal to the founded initial velocity. ⇒ impulsive impact. x b be the final velocity vector of the current ballistic swing, Let ˙ x a be the initial velocity vector of the next ballistic swing. Let ˙ Yannick Aoustin, Alexander Formalskii Strategy to lock the knee of exoskeleton stance leg: study in the fram
The structure of the instantaneous double support First sub-phase: Impulsive phase. � � I − � x − − ˙ x b ) = � + J ⊤ r 1 I − A [ x ( T )](˙ D 1 D 2 D 3 D 4 1 (4) r 1 . I − 3 × 1 x − = 0 2 × 1 J r 1 ˙ (5) The velocity of the inter-link angle of the stance (hind) leg after first sub-phase remains zero, therefore x − = 0 . D ⊤ 1 ˙ (6) I − 3 × 1 applied by human. Yannick Aoustin, Alexander Formalskii Strategy to lock the knee of exoskeleton stance leg: study in the fram
The structure of the instantaneous double support Second sub-phase: Passive impact. Knee of the leg 1 is unlocked, knee of the leg 2 is locked with I 4 . The second sub-phase: passive impact. The stance leg lifts off the ground (Hypothesis). � � � 0 3 × 1 x + − ˙ x − ) = � + J ⊤ A (˙ D 1 D 2 D 3 D 4 r 2 I r 2 (7) I 4 The associate equation: x + = 0 2 × 1 J r 2 ˙ (8) To take into account the locking of the knee of the leg 2, we have to complete equations (7) and (8) with equation: x + = 0 . D ⊤ 4 ˙ (9) Yannick Aoustin, Alexander Formalskii Strategy to lock the knee of exoskeleton stance leg: study in the fram
The structure of the instantaneous double support Third sub-phase: Impulsive impact. � � I + � x a − ˙ x + ) = � D 1 3 × 1 + J ⊤ r 2 I + A (˙ D 2 D 3 D 4 (10) I + r 2 4 There are 27 scalar equations to find 29 unknown variables, which are the components of the vectors and scalars: x − ( 7 × 1 ) , I − 1 , I − 3 × 1 ( I − 2 , I − 3 , I − ˙ 4 ) ⊤ , I − r 1 ( 2 × 1 ) (for the first sub-phase), x + ( 7 × 1 ) , I 4 , I r 2 ( 2 × 1 ) (for the second sub-phase), ˙ I + 3 × 1 ( I + 1 , I + 2 , I + 3 ) ⊤ , I + 4 and I + r 2 ( 2 × 1 ) (for the third sub-phase). Yannick Aoustin, Alexander Formalskii Strategy to lock the knee of exoskeleton stance leg: study in the fram
Cost functional 1/3 With the impulsive torques, the cost functional is T T + 4 3 � � � � � � � i ( t ) ˙ � i ( t ) ˙ � Γ − � Γ + W = θ i ( t ) � d t + θ i ( t ) � d t � � � � i = 2 i = 1 T − T 4 3 � � W − W + W = + (11) i i i = 2 i = 1 Yannick Aoustin, Alexander Formalskii Strategy to lock the knee of exoskeleton stance leg: study in the fram
Cost functional 2/3 The energy due to the impulsive torques becomes [Formal 82] 4 3 � W − � W + W = + with: i i i = 2 i = 1 Values W − ( i = 2 , 3 , 4 ) are calculated as follows: i � � ˙ i + ˙ θ b θ − � � ˙ i ˙ W − � I − i θ b θ − = if ≥ 0 i = 2 , 3 , and 4 , � � i i i 2 � � � � � i ) 2 + ( ˙ ( ˙ θ b θ − i ) 2 � � ˙ i ˙ W − � I − � θ b θ − = if < 0 i = 2 , 3 , and 4 , � � i i i � � ˙ i − ˙ θ − θ b � 2 � i � � ˙ ˙ ˙ θ b q b q b θ b q b q b θ b q b q b 2 = ˙ 5 − ˙ 3 = ˙ 5 − ˙ 4 = ˙ 3 − ˙ 2 , 3 , 4 , (12) θ − ˙ q − q − θ − ˙ q − q − θ − ˙ q − q − 2 = ˙ 5 − ˙ 3 = ˙ 5 − ˙ 4 = ˙ 3 − ˙ 2 , 3 , 4 Yannick Aoustin, Alexander Formalskii Strategy to lock the knee of exoskeleton stance leg: study in the fram
Cost functional 3/3 The energy due to the impulsive torques becomes [Formal 82] 4 3 � W − � W + W = + with: i i i = 2 i = 1 Values W + ( i = 1 , 2 , 3 ) are calculated using analogous formulas: i � � θ + ˙ i + ˙ θ a � � W + � I + θ + ˙ i ˙ i θ a = if i ≥ 0 i = 1 , 2 , and 3 , � � i i 2 � � � � � i ) 2 + ( ˙ ( ˙ θ + θ a i ) 2 � � ˙ i ˙ W + � I + � θ + θ a = if i < 0 i = 1 , 2 , and 3 . � � i i � � ˙ i − ˙ θ + � 2 θ a � � i � θ + ˙ q + q + θ + ˙ q + q + θ + ˙ q + q + 1 = ˙ 2 − ˙ 2 = ˙ 5 − ˙ 3 = ˙ 5 − ˙ 1 , 2 , 3 , (13) ˙ ˙ ˙ θ a q a q a θ a q a q a θ a q a q a 1 = ˙ 2 − ˙ 1 , 2 = ˙ 5 − ˙ 2 , 3 = ˙ 5 − ˙ 3 Yannick Aoustin, Alexander Formalskii Strategy to lock the knee of exoskeleton stance leg: study in the fram
Energy cost as a function of T . (1/2) 240 220 200 180 160 W 140 120 100 80 60 40 0.38 0.4 0.42 0.44 0.46 0.48 0.5 0.52 0.54 T For human with the exoskeleton (diamond), and for human alone (circle). The step length is 0.5 m. Yannick Aoustin, Alexander Formalskii Strategy to lock the knee of exoskeleton stance leg: study in the fram
Energy cost as a function of L . (2/2) 240 220 200 180 160 W 140 120 100 80 60 40 0.38 0.4 0.42 0.44 0.46 0.48 0.5 0.52 0.54 L For human with the exoskeleton (diamond), and for human alone (circle). The step time is 0 . 45 s Yannick Aoustin, Alexander Formalskii Strategy to lock the knee of exoskeleton stance leg: study in the fram
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