Simulation dynamique de perte d'equilibre · HAL Id: tel-00982026 Submitted on 23 Apr 2014...

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HAL Id: tel-00982026https://tel.archives-ouvertes.fr/tel-00982026

Submitted on 23 Apr 2014

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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

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

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