optimisation of running strategies according to ...fiorini.perso.math.cnrs.fr/semjeuneslmv.pdf ·...

Optimisation of running strategies according to physiological parameters

5/2/2016 - Séminaire des Doctorants

Camilla Fiorini

Camilla Fiorini 5/2/2016 /29

Aim

2


Mathematical framework

Ordinary differential equation:

+ n initial (or final) conditions

f(y(n)(t), y(n�1)(t), . . . , y(t), t;p) = 0state

p vector of parametersy

Direct problem:

Inverse problem (parameter identification):

p

p, f+ n i.c. SOLVE y

y, f+ n i.c.

IDENTIFY

3


Mathematical framework

Optimal control problems

u

f(y(n)(t), y(n�1)(t), . . . , y(t), u(t), t;p) = 0 u controlState equation:

Cost functional: J(u) = g(y(u)) quantity to be minimised

There can be n+1 initial (or final) condition.

The aim is to find u J(u) = minu2U

J(u)such that

subject to the state equation and the i.c.

yp, f, J

n+1 cond.OPTIMISE

4


Introduction

States:position velocity energy

x(t)

v(t)

e(t)

Control: propulsive force f(t)

Cost functional: final time T

Equations:

8><

>:

x(t) = ?

v(t) = ?

e(t) = ?

5


Introduction

States:position velocity energy

x(t)

v(t)

e(t)

Control: propulsive force f(t)

Cost functional: final time T

Equations:

8><

>:

x(t) = ?

v(t) = ?

e(t) = ?

v(t)

5


Introduction

Energy

Propulsive forceFriction

6


Keller’s model (1974)Equations:

Physiological constraints:

Optimisation problem:

subject to (1) and (2).

(1)

(2)

Matches the world records times

Unreasonable velocity profile No competition with other runners

Straight line

e(t) � 0, 0 f(t) fM 8t � 0.

minf2F

T (f)

8><

>:

x(t) = v(t) x(0) = 0, x(T ) = D

v(t) = f(t)� v(t)⌧ v(0) = 0

e(t) = � � f(t)v(t) e(0) = e

0.

7


Oxygen uptake (VO2)

� 6= const.

.

e(t) = � � f(t)v(t)

�(e)

8


Two-runners model

Slipstream

Psychological factor

8>>>>>><

>>>>>>:

x1 = v1 x1(0) = 0

x

D

= v2 � v1 x

D

(0) = 0

v1 = f1 � v1⌧1

� cv

21(1� �(e�↵(xD��)2)) v1(0) = 0

v2 = f2 � v2⌧2

� cv

22(1� �(e�↵(xD+�)2)) v2(0) = 0

e

i

= �

i

(ei

)� f

i

v

i

e

i

(0) = e

0i

.

1

0.2

11

1-γ

β0 2β

9


Two-runners model

Physiological constraints:

ei(t) � 0, 0 fi(t) fM,i 8t � 0

Final condition:

(x1(T )�D)(x2(T )�D) = 0

Optimisation problem:

minfi2Fi

(T + cw|xD(T )|)

10


Curved path

Rcg

Rf

↵

(i) Leaning towards the centre

(ii) Increase of internal friction

(iii) Decrease of maximal force

11



Curved path

x(t)xcg(t)xf (t)feet

centre of gravityv(t)vf (t)vcg(t)

xf (0) = 0, xf (T ) = DInitial and final conditions:

Newton’s second law: vcg(t) = f(t)� vcg(t)

⌧

xf (t) = vf (t) = h(vcg)

12



Curved path

vfvcg

=Rf

Rcg

N cos↵ = mg N sin↵ = mv2cgRcg

) tan↵ =

v2cggRcg

sin2 ↵ =tan2 ↵

1 + tan2 ↵) sin↵ =

v2cg

gRcg

q1 +

v4cg

(gRcg)2

xf = vcg

0

@1 +Lv

2cg

Rcg

q(gRcg)2 + v

4cg

1

A

13

= 1 +L

Rcgsin↵

Rf �Rcg = L sin↵ , Rf

Rcg= 1 +

L

Rcgsin↵


(ii) Increase of internal friction

Curved path

⌧ = ⌧0(1� ⌧c tan↵) = ⌧0

✓1� ⌧c

v2

gRcg

◆

(iii) Decrease of maximal force

fM = fM,0(1� fc tan↵) = fM,0

✓1� fc

v2

gRcg

◆

14


Curved path

One runner model

tr(x) =

(0 if x 2 S1 if x 2 B

8>>>><

>>>>:

x

f

= v

✓1 + tr(x

f

) Lv

2

Rcg

p(gRcg)2+v

4

◆x

f

(0) = 0, x

f

(T ) = D

v = f � v

⌧0

⇣1�tr(xf )⌧c

v2gRcg

⌘ � cv

2v(0) = 0

e = �(e)� fv e(0) = e

0,

15

e(t) � 0, 0 f(t) fM,0

✓1� fc

v2

gRcg

◆


Curved path for two runners

Problem: definition of xD(t)

16


x

APPi,L (t) =

8>>>><

>>>>:

R1#0,L + R1RL

xi,L(t) if xi,L(t) 2 Ba,L

xi,L(t) + #0,LRL � ⇡(RL �R1) if xi,L(t) 2 Sa,L

R1RL

xi,L(t) +R1#0,L + L

⇣1� R1

RL

⌘if xi,L(t) 2 Bb,L

xi,L(t) if xi,L(t) 2 Sb,L


Idea: definition of an apparent position

For the 400-m:

x

APPD (t) = x

APP2,L2

(t)� x

APP1,L1

(t)

x

APPi,L (t) =

(R1RL

vi(t) if xi,L(t) 2 Bk,L, k = a, b

vi(t) if xi,L(t) 2 Sk,L, k = a, b,

17



Final model:

8>>>>>>>>><

>>>>>>>>>:

x1 = v1

✓1 + tr(x1)

Lv

21

Rcg

p(gRcg)2+v

41

◆x1(0) = 0

x

APP

D

= x

APP

2,L2� x

APP

1,L1x

APP

D

(0) = R1 (#0,L1 � #0,L2)

v1 = f1 � v1

⌧1

⇣1�tr(xf )⌧c

v2gRcg

⌘ � cv

21(1� �(e�↵(xAPP

D ��)2)) v1(0) = 0

v2 = f2 � v2

⌧2

⇣1�tr(xf )⌧c

v2gRcg

⌘ � cv

22(1� �(e�↵(xAPP

D +�)2)) v2(0) = 0

e

i

= �

i

(ei

)� f

i

v

i

e

i

(0) = e

0i

,

18

ei(t) � 0, 0 fi(t) fM,0,i

✓1� fc

v2

gRcg

◆


Numerical results

t[s]0 50 100 150 200 250

-1

0

1

2

3

4

5

6

7v[m/s]

fM

=5 N/kg

fM

=4.75 N/kg

fM

=5.25 N/kg

t[s]0 50 100 150 200 250

3.5

4

4.5

5

5.5f[N/kg]

fM

=5 N/kg

fM

=4.75 N/kg

fM

=5.25 N/kg

t[s]0 50 100 150 200 250

0

200

400

600

800

1000

1200

1400e[J/kg]

fM

=5 N/kg

fM

=4.75 N/kg

fM

=5.25 N/kg

t[s]0 50 100 150 200 250

5

10

15

20

25sigma[m2/s3]

fM

=5 N/kg

fM

=4.75 N/kg

fM

=5.25 N/kg

Optimal strategy for a single runner - 1500 m

19


Numerical results

[J/kg]

[N/kg]

Two-runners model - 1500 m

Two runners with the same parameters

normalized time0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-2

0

2

4

6

8

10

12

14

16

xD

[m]

20


Numerical results

[J/kg]

[N/kg]


The weaker can win starting behind

normalized time0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-1.5

-1

-0.5

0

0.5

1

1.5

2

xD

[m]

21


Numerical results


The stronger starts behind

x[m]0 500 1000 15000

1

2

3

4

5

6

7v[m/s]

Runner-1Runner-2

x[m]0 500 1000 15003

3.5

4

4.5

5f

Runner-1Runner-2

x[m]0 500 1000 15000

200

400

600

800

1000

1200

1400e

Runner-1Runner-2

x[m]0 500 1000 15005

10

15

20

25sigma[m2/s3]

Runner-1Runner-2

[J/kg]

[N/kg]normalized time

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-5

0

5

10

15

20

25

30

35

40

45

xD

[m]

22


Numerical results

Running alone vs running behind someone else

[J/kg]

[N/kg]

1500 m

23


Numerical results

0 500 1000 1500x[m]

0

1

2

3

4

5

6

7

8

v[m

/s]

bending partsmodel real trackmodel straight

Real track model vs straight line model

1500 m

24


Numerical results

Parameter identification

lane 1

lane 3

lane 6

40m

32.69m

21.66m

Six runners

Three paths

D = 180m

Time splits every 10m

25



directly measurable

Parameters to be identified:

�

⌧, fM , e0

0 = v(t⇤) = f(t⇤)� v(t⇤)

⌧� cv2(t⇤) = fM � vp

⌧� cv2p

(a) identification of and inputs: straight line model + lane 6 data

(b) Use parameters identified in step (a) and find the that minimises the least-square distance with respect to the experimental data.

Procedure:

e0 ⌧

⌧c

fM =vp⌧

+ cv2p

26


Numerical results


0 20 40 60 80 100 120 140 160 180x[m]

0

1

2

3

4

5

6

7

8

9

10

v[m

/s]

Gatien C1, tau1 = 0.2

Real trackStraight pathExperimental valuesBending parts

27


Numerical results


0 20 40 60 80 100 120 140 160 180x[m]

0

1

2

3

4

5

6

7

8

9

10

v[m

/s]



28


Numerical results


0 20 40 60 80 100 120 140 160 180x[m]

0

1

2

3

4

5

6

7

8

9

10

v[m

/s]



29

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optimisation of running strategies according to ...fiorini.perso.math.cnrs.fr/semjeuneslmv.pdf ·...

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