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Simulation dynamique de perte d’équilibre : Applicationaux passagers debout de transport en commun
Zohaib Aftab
To cite this version:Zohaib Aftab. Simulation dynamique de perte d’équilibre : Application aux passagers debout detransport en commun. Médecine humaine et pathologie. Université Claude Bernard - Lyon I, 2012.Français. �NNT : 2012LYO10243�. �tel-00982026�
Thèse
Dynamic Simulation of Balance
Recovery: Application to the
standing passengers of
public transport
Présentée devant L’UNIVERSITE CLAUDE BERNARD LYON I
Pour l'obtention
du DIPLÔME DE DOCTORAT
Formation doctorale : Biomécanique École doctorale : École doctorale MEGA
Par
M. Zohaib AFTAB
Soutenue le 21 Novembre 2012 devant la Commission d’examen
Jury
Rapporteur M. Patrick LACOUTURE Professeur (Université de Poitiers)
Rapporteur M. Philippe FRAISSE Professeur (Université Montpellier 2)
Examinateur Mme. Laurence CHÈZE Professeur (Université Lyon 1)
Directeur de thèse M. Bernard BROGLIATO Directeur de recherche (INRIA Grenoble)
Co-directeur M. Thomas ROBERT Chargé de recherche (IFSTTAR, Bron)
Co-directeur M. Pierre-Brice WIEBER Chargé de recherche (INRIA Grenoble)
Laboratoire de Biomécanique et Mécanique des Chocs,
Ifsttar, 25 av. Francois Mitterrand, case 24, 69 675 BRON Cedex
−
−
−
−
−
−2 −3
−2
−2
−2
−2>
−2>
−2
−2 −2
K
u x
wst
ust
W i = W g +W c
W i W g W c
≃
m
l
j
α
cx
px cx
mg(px − cx) = jα ≈ −ml2cx
l
cx = ω20(cx − px)
ω0 =√
g/l
cx(t) = px + (cx(0)− px)cosh(ω0t) +cx(0)
ω0
sinh(ω0t)
cx(t) ≤ px
cx +cx
ω0
≤ px
cx +cxω0
≃
≃∫ tk+T
tk‖c(n)x ‖2dt
cx +cx
ω0
≤ px
px
cx +cxω0
cx = ω2(cx − px)
ω =√
g/h
ELIP =1
2c2x −
g
2h(cx − px)
2
px = xcapt = cx +cx
ω
xcapt
xcapt
xcapt
xcapt = cx +cx
ω± τmax
mg[eωTR2 − 2eω(TR2−TR1) + 1
eωTR2
]
τmax TR1 TR2
l
l
cx = −l cosθ θ, cz = −l sinθ θ
c+x − c−xc+z − c−z
= tanθ
c−x c+x
c−z c+z
c+z = 0
c+x = c−x − c−z tanθ
cx +c+xω0
= 0
θ = ωsinθcos1/2θ
cos2θ
θ
2θ
θc
θ0 tcont
θc = θ0cosh(
√
3g
2l− 3ka
ml2tcont)
θc
Fmax
α = 2θc
tcont
√gH
st
nd
st
τmax = 190N.m
◦
1st
st
20.8 ± 6.4 cm
nd 13±4.6 cm
1st
tk LT
∫ tk+T
tk
L(q(t), q(t), q(t), u(t), λ(t))dt
u(t)
tk+1
N
T c
ck =
⎡
⎢
⎢
⎣
ck
ck
ck
⎤
⎥
⎥
⎦
,
tk
Ck+1 =
⎡
⎢
⎢
⎣
ck+1
ck+N
⎤
⎥
⎥
⎦
, Ck+1 =
⎡
⎢
⎢
⎣
ck+1
ck+N
⎤
⎥
⎥
⎦
, Ck+1 =
⎡
⎢
⎢
⎣
ck+1
ck+N
⎤
⎥
⎥
⎦
Ck+1 =
⎡
⎢
⎢
⎣
c k+1
c k+N
⎤
⎥
⎥
⎦
Ck+1 = Spck + UpCk,
Ck+1 = Sv ck + UvCk,
Ck+1 = Sack + UaCk.
zx = cx −h
gcx
px zx
Zk+1 =
⎡
⎢
⎢
⎣
zk+1
zk+N
⎤
⎥
⎥
⎦
Ck
Zk+1 = Sz ck + UzCk,
Sz = Sp −h
gSa,
Uz = Up −h
gUa.
f ′
i
‖f ′
i‖ ≤ f ′ .
zx
D(zi − fi) ≤ b,
D b
fi
Ck+1
Crefk+1
Ck
Zk+1 Fk+1
1
c21‖Ck+1 − C
refk+1‖2 +
1
c22‖Ck‖2 +
1
c23‖Zk+1 − Fk+1‖2,
c1 c2 c3
u =
[
Ck
Fk+1
]
Ck
Fk+1
Ck+1 = 0
T = 1
T = 25
m
T
T
T
T = T + T + T T
T
T
T T + T
l ×lf ×
a × lf×
θ H Treac Tprep
Tstep
H θ Treac Tprep Tstep
m.s−1
c1 c3
c1 = 1 m.s−1
c2 c3 100
×
c1−1
c2−3
c2lf
vmax−1
T
±
±
±
mh cx + j θ = mg(cx − zx)
cx zx θ
j g
N T
c θ
θk =
⎡
⎢
⎢
⎣
θk
θk
θk
⎤
⎥
⎥
⎦
tk
Θk+1 =
⎡
⎢
⎢
⎣
θk+1
θk+N
⎤
⎥
⎥
⎦
, Θk+1 =
⎡
⎢
⎢
⎣
θk+1
θk+N
⎤
⎥
⎥
⎦
, Θk+1 =
⎡
⎢
⎢
⎣
θk+1
θk+N
⎤
⎥
⎥
⎦
Θk+1 =
⎡
⎢
⎢
⎣
θ k+1
θ k+N
⎤
⎥
⎥
⎦
Θk+1 = Spθk + UpΘk,
Θk+1 = Svθk + UvΘk,
Θk+1 = Saθk + UaΘk.
Sp, Up
z = cx −h
gcx −
j
mgθ.
Zk+1 =
⎡
⎢
⎢
⎣
zk+1
zk+N
⎤
⎥
⎥
⎦
Ck Θk
Zk+1 = Sz
[
ck
θk
]
+ Uz
[
Ck
Θk
]
,
Sz =[
Sp − hgSa − j
mgSa
]
,
Uz =[
Up − hgUa − j
mgUa
]
.
i ∈ [k + 1, . . . k +N ]
θ ≤ θi ≤ θ .
j|θi| ≤ τ
fi ti
‖cix − fi‖ ≤ l .
f ′
i
‖f ′
i‖ ≤ f ′
zx
D(zi − fi) ≤ b,
D b
fi
Fk+1 =
⎡
⎢
⎢
⎣
fk+1
fk+N
⎤
⎥
⎥
⎦
fk
Fk+1
Vk+1
Vk+1 ti
Fk+1 = Vk+1fk + Vk+1Fk+1.
Vk+1 =
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
1
1
0
0
0
0
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
, Vk+1 =
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
0 0
0 0
1 0
1 0
0 1
0 1
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
Ck+1
Θk+1
F ′
k+1
1
c21‖Ck+1‖2 +
1
c22‖Θk+1‖2 +
1
c23‖F ′
k+1‖2.
Ck Θk
Zk+1 Fk+1
1
c21‖Ck+1‖2 +
1
c22‖Θk+1‖2 +
1
c23‖F ′
k+1‖2
+1
c24‖Ck‖2 +
1
c25‖Θk‖2 +
1
c26‖Zk+1 − Fk+1‖2,
u =[
Ck Θk Fk+1
]T
c1
c6
H
m
h = 0.575×H
lf = 0.152×H
0.81× lf0.19× lfθ π/2θ −π/2j 2
τmax
1
c21‖Ck+1‖2 +
1
c22‖Θk+1‖2.
u =[
Ck Θk
]T
π2
N.m
τmax θmax
−1
−1
τmax = 190N.m θmax = π2
−1 −1
−1
‖Ck‖2
−1
c4 c5 c4 c5−3 −3
−1
‖Ck+1‖2
F ′
k+1
‖F ′
k+1‖2
1c21
‖Ck+1‖2 + 1c23
‖F ′
k+1‖2 c3c1
−2
c3
c4 c5
c1−1 c4
−3
c5−3
c2 c3 c2
c2−1
c3
‖F ′
k+1‖2 103 104
‖Ck+1‖2
c1−1
c2−1
c3−2
c4−3
c5−3
c6
c1 c3 ≈ −2
1
c21‖Ck+1‖2 +
1
c22‖Θk+1‖2 +
1
c23‖F ′
k+1‖2
+1
c24‖Ck‖2 +
1
c25‖Θk‖2 +
1
c26‖Zk+1 − Fk+1‖2
c2−1 c5
−3
−1
c2 c5
c2−1 c5
−3
c2 c5
c2−1
−1
◦
◦
c2 c5 c2
c4
c6
c4
c4 = 1000 m.s−2
c6
c6
c2 c6
F ′′
k+1
c3 = −2
c3 = −2
c3F ′
k+1
Tsampling
Treac Tprep
Tstep
◦
×
lf
f ′ −2
θ π/2τmax
c1−1
c2−1
c3−2
c4−3
c5−3
c6
Tland
Tland = Treac + Tprep + Tstep
Tstep
±
±
nd
±
Tstep
c3−2
◦
±
Tland
Tstep
Tstep
±
st
nd
st
nd
Treac
Tprep
Tstep
± ± ± ±± ±± ±
± ± ± ±± ±
1
c21‖Ck+1‖2+
1
c22‖Θk+1‖2+
1
c23‖F ′
k+1‖2+1
c24‖Ck‖2+
1
c25‖Θk‖2+
1
c26‖Zk+1−Fk+1‖2
c1 c6
Treac
Tprepst
nd
Tstep
×
f ′ −2
θ π/2τmax
c1−1
c2−1
c3−2
c4−3
c5−3
c6
‖F ′
k+1‖2
‖Ck‖2
c3 c4
c3 c4
c3−2 c4
−3
c3 c4
c1 ≪ c3, c4
c3−2 c4
−3
c3−2 c4
−3
−3 −2
cx =g
h(cx − zx)−
j
mhθ
cx =g
h(cx − zx)−
j
mhθ ± x
pfFcast
xpfFcast
−2
nd
−−2
−2
π/2 π/3
lf ×Tprep
Treac
Tstep
f ′
−2
θ π/3τmax
−
−
c1 c6
Tstep
◦
⋆
⋆
W i = W g +W c
W i W g W c
1
c21‖Ck+1 − C
refk+1‖2 +
1
c22‖Ck‖2 +
1
c23‖Zk+1 − Fk+1‖2,
c1 c2 c3 u =
[
Ck
Fk+1
]
Ck Fk+1
mh cx + j θ = mg(cx − zx)
z = cx −h
gcx −
j
mgθ.
j θ
Ck+1
Θk+1
F ′
k+1
1
c21‖Ck+1‖2 +
1
c22‖Θk+1‖2 +
1
c23‖F ′
k+1‖2.
Ck Θk
Zk+1
1
c21‖Ck+1‖2 +
1
c22‖Θk+1‖2 +
1
c23‖F ′
k+1‖2
+1
c24‖Ck‖2 +
1
c25‖Θk‖2 +
1
c26‖Zk+1 − Fk+1‖2,
u =[
Ck Θk Fk+1
]T
c1 c6
‖F ′
k+1‖2
‖Ck‖2
−2
