May 2012 1

Motion Planning

Shmuel WimerBar Ilan Univ., Eng. Faculty

Technion, EE Faculty

May 2012 2

Outline

• Problem definition

• Point robot

• Work space and configuration space

• Minkowski sums

• Translational motion planning

• Rotational motion planning

May 2012 3

Types of Robots and Motions

Articulated Robot

Translation Motion

start goal

May 2012 4

Rotational Motion

start goal

Some robots can move in any direction (e.g., ants)

Some robots cannot translate (e.g., cars)

We’ll study translational and rotational motions

Given a robot, is there a free paths (no collisions)

from start to goal?

May 2012 5

Work Space and Configuration Space

2

1

A robot is moving in ,called ,

consisting of a set , , of obstacles.tS

R

P P

work space

Reference Point

0,0R

9,7R

2 Degrees of freedom 3 Degrees of freedom

0,0,0R

9,7,45R

45

May 2012 6

The parameter space of a robot is called its

, denoted . A point

corresponds to a certain placement

of the robot in the work space.

p

p

R

C R

C R

R

configuration space

Configuration spaceWork space

May 2012 7

Free Space Computation

Divide into trapezoids.

It takes O(nlogn)

expected time.

Remove trapezoids of obstacle in O(n) time.

May 2012 8

Building a Road Map

Allocate node at center of vertical edges

Allocate node at center of every trapezoid

Connect center nodes to edge nodes

Done in O(n) with doubly-connected edge list

May 2012 9

Computing a Path

Get from start to center of trapezoid in O(logn) time

Get from goal to center of trapezoid in O(logn) time

Connect center of trapezoids by BFS in O(n) time

startp

goalp

May 2012 10

Let be a point robot moving among a

set of polygonal obstacles with edges in total.

can be processed in log expected time, such

that between any start and goal position a collision-

fr

S n S

O n n

: RTheorem

ee path for can be computed in time, if it

exists.

O nR

May 2012 11

Convert a problem with polygonal robot into point robot

by modifying the obstacles in the configuration space to

incorporate the geometry of the robot.

Minkowski Sums

obstacle

robot

May 2012 12

- convex obstacle, , - convex robot x yP R

, | ,

configuration space obstacle

x y x y P R P

2 21 2 1 2

1 2 1 2

, . -

| ,

S S S S

S S p q p S q S

R R Minkowski sum

, , , , ,x y x y x x y yp p p q q q p q p q p q

Reflection of a set about the origin

, , | x yp p p S p p S

May 2012 13

Let be a planar translating robot and

let be an obstacle. Then 0,0 .

:Theorem R

P P P R

We'll show:

, , 0,0x y x y

:

Proof

R P P R

Let , ,

, ,

, 0,0

, 0,0 .

x y

x y

x y

x y

q q q x y

q q q x y

q x q y

q x q y

R P

R

R

R

May 2012 14

But since , also,

, , ,

, 0,0 .

x y

x y x y

q q q

x y q q q x q y

x y

P

P R

Converesely, let , 0,0 . Then

there are points , and , 0,0

such that , ,

, , ,

x y x y

x x y y

x y x y

x y

p p r r

x y p r p r

x r y r p p x y

P R

P R

R P ■

May 2012 15

Extreme Points and Directions

p rP R

r

R

P

p

An extreme point of in direction

is the sum of extreme points of and in .d d

:����������������������������Observation P R

P R

May 2012 16

When a vertex of interacts (slides)

with an edge of , the direction of the resulting edge

in is that of the edge of .

:Observation R

P

P R P

How complex is Minkowski sum of two polygons?

Let and be two -vertex and

-vertex convex polygons, respectively.

is convex and has edges at most.

n

m

n m

:Theorem R P

P R

Convexity follows directly by definition

of Minkowski sum and convexity of and .

:Proof

R P

May 2012 17

is extreme in the direction of its outer

normal, hence must be defined by extreme points

in and in that direction.

e P R

P R

At least one of and must have an edge in that

direction. We charge that edge to and it will not

be charged again.

e

P R

■

May 2012 18

-R

a

b

cd

-R

a

b

cd

-R

a

b

cd

-R

a

b

cd

-R

a

b

cd

-R

a

b

cd

Construction of Minkowski Sum

1

2

3 4

5P

Edges of P and R are labeled counterclockwise

b

1

2

34

5

c

d

d

c

aa

b

May 2012 19

The self-intersecting polygonal path bounding

is labled, including edges near reflex vertices of .

The pattern of lables along can be derived from

a of the edge vectors of and .

P

P

P Rstar diagram

1

2

3

45

a

b

c

d

1

2

3 4

5P

-R

a

b

cd

May 2012 20

Starting at 1, circle around the star counterclockwise.

Between and 1 of , write the encountered lables

of .

i i

P

Rb

1

2

34

5

c

d

d

c

aa

b

1

2

3

45

a

b

c

d

May 2012 21

2 2

convex convex

convex nonconvex log

nonconvex nonconvex log

O n O n

O n O n n

O n O n n

Size of Time ComplexityR P P

Let have vertices and have fixed

(constant) number of vertices. Then

can be constructed in the following complexities:

n

:Theorem P R

P P R

May 2012 22

Rotation – Moving a Ladder

Minkowski sum for

ladder at 0º rotation.

Blockage exists.

May 2012 23

Minkowski sum for

ladder at 30º rotation.

Blockage exists.

Minkowski sum for

ladder at 60º rotation.

Blockage changed.

May 2012 24

Conversion to 3D Motion Problem

Bottom viewFront view. θ varies from 0º to 75º.

Ladder’s reference point can move in the 3D space!

May 2012 25

Cell Decomposition

Minkowski sums

A

B

C

Cell decomposition

108

9

2 1

3

4

56

7

Obstacles

∞

R

Ladder

May 2012 26

• A Cell is the collection of all free points labeled

with the same front/back edge label pairs.

– A: (3,2); B: (3,8); C: (1,9)

• Cell decomposition has discontinuities when

ladder is oriented similar to an edge.

• There are finite number of ladder rotation where

cell decomposition is changing.

– New cells can appear and old ones may disappear.

May 2012 27

56

7

108

9

2 1

3

4

∞

R

AB

C disappeared

Ladder is rotated

May 2012 28

In Connectivity Graph Gθ nodes are cells of decomposition

and edges are connecting nodes corresponding to adjacent

cells in free area (a kind of dual graph).

A: (3,2)B: (3,8)

(1,8)

C: (1,9)(1,∞)

(10,∞)

(5,∞)

(3,∞)

G0º

May 2012 29

(4,∞)

(1,∞)

(10,∞)

(5,∞)

(3,∞)

A: (3,2)B: (3,8)

(1,8)

(7,8)

Critical Orientations correspond to slants of edges.

May 2012 30

Connectivity Graph G is constructed by stacking the

connectivity graphs Gθ corresponding to the critical

orientations.

Vertices of two distinct Gθ are connected iff they are

labeled with the same edge pair. Starting from G0, Gθ

are added in increasing order of θ, thus creating a

layered 3D graph.

A paths from start to goal if exists can be found by a

BFS algorithm.

May 2012 31

5

2 *

2

2

2

2

Schwatz & Sharir 1983

O'Dunlaing et. al. 1987 log log

logLeven & Sharir 1987

logSifroni & Sharir 1987

Vegter 1990

O'Rourke 1985

O n

O n n n

O n n

O n n

O n

n

Time ComplexityAuthors

History