L.Jaulin,B.Desrochers Caractérisationdelazoneexploréeparunrobotsous-marin

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Caractérisation de la zone explorée par un robot sous-marin

L. Jaulin, B. Desrochers

Arras, 8 Nov 2018

L. Jaulin, B. Desrochers Caractérisation de la zone explorée par un robot sous-marin

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Polynesian navigation

Given a set of point islands m ₁ , m ₂ , . . . Boats have a visibility area.

Boats have to move in this environment without being lost, nd new islands, etc.

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Prove that geo-localized islands will be reached by one boat

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Prove that geo-localized islands will be reached by the n boats

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Find a control to reach the geo-localized islands

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Explore a given area entirely to nd new islands

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Visible area

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The robot has a state x. The visible area is V ( x ) Example. The robot is able to see all up to 3 meters

V ( x ) = n

(z ₁ , z ₂ ) | (z ₁ − x ₁ ) ² + (z ₂ − x ₂ ) ² ≤ 9 o .

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Blind observer

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Given 





x ˙ = f (x, u) , u (t) ∈ [u] (t) y = g ( x ) , y (t) ∈ [ y ] (t) x ( 0 ) = x ₀

An observer is blind if dimy= 0.

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Predict is a blind estimation problem

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





x(t) ˙ = f(x(t), u(t))

y (t) = g ( x (t)) + e y (t) e y (t) ∈ [ e y ] u (t) = r ( y (t)) + e u (t) e u (t) ∈ [ e u ]

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x(t) ˙ = f (x(t), r(g(x(t)) + e y (t))+e u (t)) e y (t ) ∈ [ e y ], e u (t) ∈ [ e u ]

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x(t) ˙ = h(x(t),v(t)) v (t) ∈ [ v ] with v = ( e y , e u )

h ( x , v ) = f ( x , r ( g ( x ) + e y ) + e u ) [v] = [e y ] × [e u ]

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Blind exploration

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The blind explored zone Z is dened by [2]







x ˙ = f ( x , u ) , u (t) ∈ [ u ] (t) Z = ^S _t _≥0 V (x (t))

x ( 0 ) = x 0

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We have

\

x ( · ) ∈X ( · )

[

t ≥0

V ( x (t))

| {z }

Z

⁻

⊂ Z ⊂ ^[

x ( · ) ∈X ( · )

[

t ≥0

V( x (t))

| {z }

Z

⁺

.

Z ⁻ is the certainly explored zone.

Z ⁺ is the maybe explored zone.

Z ⁺ \Z ⁻ is the penumbra.

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Spiral scan

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We have [4]

[

t ≥0

\

x∈X (t)

V ( x )

| {z }

{z | ∃ t ∀x∈X (t), z∈V ( x ) }

⊂ Z ⁻ = ^\

x ( · ) ∈X ( · )

[

t ≥0

V ( x (t ))

| {z }

{z | ∀x ( · ) ∈X ( · ),∃ t, z∈V ( x ) }

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Boustrophedon

Which pattern is the best for exploration?

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Thick sets

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y

x

y

∈

x

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Interval analysis

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Problem. Given f : R ⁿ → R and a box [ x ] ⊂ R ⁿ , prove that

∀x ∈ [ x ], f ( x ) ≥ 0 .

Interval arithmetic can solve eciently this problem.

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Example. Is the function

f ( x ) = x ₁ x ₂ − (x ₁ + x ₂ ) cos x ₂ + sin x ₁ · sin x ₂ + 2 always positive for x ₁ , x ₂ ∈ [ −1,1 ] ?

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Interval arithmetic

[ −1 , 3 ] + [ 2 , 5 ] =?, [ −1 , 3 ] · [ 2 , 5 ] =?, abs ([ −7,1]) =?

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Interval arithmetic

[ −1 , 3 ] + [ 2 , 5 ] = [ 1 , 8 ], [ −1 , 3 ] · [ 2 , 5 ] = [ −5 , 15 ],

abs ([ −7, 1]) = [0,7]

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The interval extension of

f (x ₁ , x ₂ ) = x ₁ · x ₂ − (x ₁ + x ₂ ) · cos x ₂ + sin x ₁ · sin x ₂ + 2 is

[f ] ([x ₁ ] ,[x ₂ ]) = [x ₁ ] · [x ₂ ] − ([x ₁ ] + [x ₂ ]) · cos [x ₂ ] + sin [x ₁ ] · sin [x ₂ ] + 2 .

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Theorem (Moore, 1970)

[f ] ([ x ]) ⊂ R ⁺ ⇒ ∀x ∈ [ x ] ,f ( x ) ≥ 0 .

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Set Inversion

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X = { (x ₁ ,x ₂ )

x ₁ ² + x ₂ ² ∈ [ 1 , 2 ] }

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Thick sets

The boat at a ∈ [ a ] sees all at distance in [y ^⊂ ] = [ 0,10 ] and nothing outside [y ^⊃ ] = [ 0 , 20 ].

What is the visible zone Z ?

We apply interval arithmetic to characterize Z.

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(i ) ([z ₁ ] − [a ₁ ]) ² + ([z ₂ ] − [a ₂ ]) ² ⊂ [ 0 , 10 ] ⇒ [ z ] ⊂ Z (ii ) ([z ₁ ] − [a ₁ ]) ² + ([z ₂ ] − [a ₂ ]) ² ∩ [ 0 , 20 ] = /0 ⇒ [ z ] ∩ Z = /0.

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Thick set inversion problem [3] can be written as

X = f ⁻¹ (Y), f ∈ [ f ] and Y ∈ [[Y]].

The set X is a solution if

∃f ∈ [ f ] ,∃Y ∈ [[ Y ]], X = f ⁻¹ ( Y ).

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Thick intervals

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A thick interval [[x]] is a set of intervals [1]

[[x]] = [[[x ⁻ ] ,[x ⁺ ]]]

=

[x ⁻ ,x ⁺ ] ∈ IR | x ⁻ ∈ [x ⁻ ] and x ⁺ ∈ [x ⁺ ] .

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[2,2]

[1,7]

[3,6]

[3,8] [7,8]

lb ub

0 1 2 3 4 5 6 7 8 9 10

JaK

JbK

In the endpoints diagram, an interval is a point

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Operations

Multiplication [5]. Since

[a,b] · [c ,d ] = [ min (ac ,ad ,bc ,bd ),max (ac ,ad ,bc ,bd )]

we have

[[[a],[b]]] · [[[c],[d ]]]

= [[ min ([a] · [c ],[a] · [d ], [b] · [c], [b] · [d ]), max ([a] · [c ], [a] · [d ],[b] · [c ],[b] · [d ]) ]]

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Thick boxes

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[x⁻]

[x⁺]

The four boxes belong to the thick box [[ x ]] = [[[ x ⁻ ],[ x ⁺ ]]]

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[x]

x

1

x

2

y

1

y

2

[f ] a

[f ](a)

[f⁻]([x]) [f⁺]([x])

f⁻(a)

f⁺(a)

[ f ] ( a ) ∈ [[

f ⁻ ([ x ]) ,

f ⁺ ([ x ])]]

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0 5 10 15 20 25

-18 -16 -14 -12 -10 -8 -6 -4 -2 0 2 4 6 8

-0.5 0 0.5 1 1.5 10

1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6

We conclude that [ x ] is inside the penumbra

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Applications

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Daurade DGA-TN

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During its boustrophedon Daurade explored Z ∈ [[ Z ]]

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A la recherche de la Cordelière

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Reconstitution de la bataille youtu.be/yP4cM1UGrqY

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Expériences à Guerlédan youtu.be/KXHyYeWYGOI

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Submeeting 2018

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youtu.be/VqXG9zO_q1A submeeting2018.html

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Boatbot

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Objectif: remorquer le SeaSpy du Shom avec un zodiac autonome.

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Boatbot tracte un magnétomètre youtu.be/cxVs1fDdm1s

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L.Jaulin,B.Desrochers Caractérisationdelazoneexploréeparunrobotsous-marin

Caractérisation de la zone explorée par un robot sous-marin