Loading in 5 sec....

Computation on Parametric CurvesPowerPoint Presentation

Computation on Parametric Curves

- 65 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' Computation on Parametric Curves' - sacha-sellers

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

Computation on Parametric Curves

Yan-Bin Jia

Department of Computer Science

Iowa State University

Ames, IA 50011-1040, USA

Dec 16, 2002

Why Curved Objects?

Frequent subjects of maneuver (pen, mouse, cup, etc.)

Actions and mechanics are inherently continuous /

differential and subject to local geometry of bodies

Contact kinematics

Dynamics

Control

Shape localization, recognition, reconstruction

Not well studied compared to polygonal and polyhedral

objects

Outline of the Talk

1. Constructing common tangents of two curves

2. Computing all pairs of antipodal points

antipodal

grasp

Primitive used by other algorithms

Hong et al. 1990; Chen & Burdick 1992; Ponce et al. 1993;

Blake & Taylor 1993

Grasping & Fixturing

Salisbury & Roth 1983; Mishra et al. 1987; Nguyen 1988;

Markenscoff et al. 1992; Trinkle 1992;Brost & Goldberg 1994;

Bicchi 1995, Kumar & Bicchi 2000

Curve Processing

Goodman 1991; Kriegman & Ponce 1991;

Manocha & Canny 1992; Sakai 1999; Jia 2001

Computational Geometry

Preparata & Hong 1977; Yao 1982; Chazelle et al. 1993;

Matousek & Schwarzkopf 1996; Ramos 1997; Bespamyatnikh 1998

Previous Worka

s

b

T(s )

T(s )

T(s )

T(s )

b

b

a

a

t

t

a

b

Common Tangent of Two SegmentsAssumptions for clarity (removable):

i) Curvature everywhere positive

or everywhere negative

T

ii) Opposite normals at corresponding

endpoints

S

iii) Total curvatures (–, )

(and with equal absolute value)

total curvature

How Many Common Tangents?

Theorem

Two segments satisfying conditions (i)—(iv)

- have at most twocommon tangents incident on
- points with opposite normals.

b) are always on the same side of every common

tangent or always on different sides.

0

s

0

s

t

1

1

t

t*

s

s*

2

2

Sequences { s } and { t } converge quadratically to s* and t*.

k

k

An Iterative Procedure( k 15, # low-level iterations 300)

To find common tangencies with the same normal,

reverse the normal vector field on one segment.

= 0 but 0

Preprocessing IStep 1Compute all simple inflections.

Step 2Split every segment whose total curvature [–, ].

> 0

> 0

< 0

< 0

Assumptions i)-iii)

satisfied.

Step 3 Shortening either or both segments

= 4 + 2.5 cos

elliptic lemniscate

2 2 2 2 1/2

= (6 cos + 3 sin )

Application II – Antipodal Pointsclosed cubic spline (10 pairs)

= 3/(1+0.5 cos(3))

s*

t*

Antipodal PointsTwo pointswhose inward normals are

i) collinear

ii) pointing at each other.

a simple close curve

Incomplete – not guaranteed to find one, not to mention

all of them.

Existing Methodsformulation nonlinear programming + heuristics

Inefficient – many initial estimates often needed.

Why not exploit global & local (differential) geometry together?

Locally, define

opposite points by

N(t )

N(s) + N(t) = 0

N(s)

: antipodal

angle

Assumption

iv) Normals at opposite points

(s, t) point towards each

other.

s

Antipodal Angles and t are antipodal if and only if (s) = 0.

t*

r(s)

s*

simple antipodal pointss* and t* if (s*) = 0 but (s*) 0.

uniquepair of antipodal points if (s )< 0 and (s )> 0.

a

b

s

s

b

a

(s )

(s )

a

b

s*

Bisection over [s , s ].

a b

t*

t

t

a

b

Opposite Angle Sign at Endpointsantipodal angleincreases monotonicallyas s increases.

no antipodal points otherwise.

Both segments are concave.

S

T

t

1

2

t =

0

0 1

t

t

b

a

monotone sequences

s ,s , …

t , t , …

0 1

s

s

a

a

s =

s

0

2

s

1

Same Angle Sign at EndpointsMarching I (convex-convex)

t*

If exist, linear convergence to the

closest pair.

s*

Otherwise, termination at the other

endpoints.

The ray of N(t ) or N(t ) must intersect segment S

a

b

t

t

b

a

s

s

b

a

monotone sequences:

s ,s , …

0 1

t , t , …

t

0 1

t =

t*

1

0

t

2

s*

s

2

s

1

s =

0

Cont’dMarching II (convex-concave)

Same convergence result

2

t*

s*

1

1

Recursion tree:

s* , t*

1 1

t*

Case 1 (march)

2

march

s* , t*

bisection

Case 2 (bisection)

2 2

Case 1

Case 1

Recursively Finding All pairss

b

a

t

b

t

a

Preprocessing IIa pair of normals pointing

away from each other

Steps 1, 2, 3

Step 4Compute common

tangent lines

2

Efficiency – O(n + m)two-level calls of numerical primitives

Conclusion Design of algorithms for curve computing

Dissecting the curve into monotone segments

(preprocessing of global geometry)

Interleaving marching with numerical bisection

(exploiting of local geometry)

Provable convergence rates

(depending on curvatures)

Completeness – up to numerical resolution

#inflections

#antipodal pairs

Download Presentation

Connecting to Server..