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More on single-view geometry class 10

Multiple View GeometryComp 290-089Marc Pollefeys

Multiple View Geometry course schedule

(subject to change)

Jan. 7, 9 Intro & motivation Projective 2D Geometry

Jan. 14, 16 (no class) Projective 2D Geometry

Jan. 21, 23 Projective 3D Geometry (no class)

Jan. 28, 30 Parameter Estimation Parameter Estimation

Feb. 4, 6 Algorithm Evaluation Camera Models

Feb. 11, 13 Camera Calibration Single View Geometry

Feb. 18, 20 Epipolar Geometry 3D reconstruction

Feb. 25, 27 Fund. Matrix Comp. Structure Comp.

Mar. 4, 6 Planes & Homographies Trifocal Tensor

Mar. 18, 20 Three View Reconstruction Multiple View Geometry

Mar. 25, 27 MultipleView Reconstruction Bundle adjustment

Apr. 1, 3 Auto-Calibration Papers

Apr. 8, 10 Dynamic SfM Papers

Apr. 15, 17 Cheirality Papers

Apr. 22, 24 Duality Project Demos

Single view geometry

Camera model

Camera calibration

Single view geom.

Gold Standard algorithmObjective

Given n≥6 2D to 2D point correspondences {Xi

↔xi

’}, determine the Maximum Likelyhood Estimation of P

Algorithm(i)

Linear solution:(a)

Normalization: (b)

DLT

(ii)

Minimization of geometric error: using the linear estimate as a starting point minimize the geometric error:

(iii)

Denormalization:

ii UXX~ = ii Txx~ =

UP~TP -1=

~ ~~

More Single-View Geometry

•

Projective cameras and planes, lines, conics and quadrics.

•

Camera calibration and vanishing points, calibrating conic and the IAC

** CPPQ =T

coneQCPP =T

Action of projective camera on planes

[ ] [ ]⎥⎥⎦

⎤

⎢⎢⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡==

1ppp

10ppppPXx 4214321 Y

XYX

The most general transformation that can occur between a scene plane and an image plane under perspective imaging is a plane projective transformation

(affine camera-affine transformation)

Action of projective camera on lines

forward projection

( ) μbaμPBPAμB)P(AμX +=+=+=

back-projection

lPT=Π

PXlX TT =Π

Action of projective camera on conics

back-projection to cone

CPPQ Tco =

[ ] ⎥⎦⎤

⎢⎣⎡=⎥⎦

⎤⎢⎣⎡= 00

0CKK0|KC0KQ

TT

T

co

example:

Images of smooth surfaces

The contour generator Γ

is the set of points X on S at which rays are tangent to the surface. The corresponding apparent contour γ

is the set of points x which are the image of X, i.e. γ

is the image of Γ

The contour generator Γ

depends only on position of projection center, γ

depends also on rest of P

Action of projective camera on quadrics

back-projection to cone

TPPQC ** = 0lPPQlQ T*T*T ==ΠΠ

The plane of Γ

for a quadric Q is camera center C is given by Π=QC (follows from pole-polar relation)

The cone with vertex V and tangent to the quadric Q is the degenerate Quadric: TT

CO (QV)(QV)-QV)QV(Q = 0VQCO =

The importance of the camera center

]C~|[IR'K'P'],C~|KR[IP −=−=

( ) PKRR'K'P' -1=

( ) ( ) xKRR'K'PXKRR'K'XP'x' -1-1 ===

( )-1KRR'K'HHx with x' ==

Moving the image plane (zooming)

( ) xKK'0]X|[IK'x'0]X|K[Ix

1-===

( ) ⎥⎦⎤

⎢⎣⎡ −

==10

x~k)(1kIKK'H T01-

⎥⎦⎤

⎢⎣⎡=⎥⎦

⎤⎢⎣⎡=

⎥⎦⎤

⎢⎣⎡⎥⎦⎤

⎢⎣⎡ −

=⎥⎦⎤

⎢⎣⎡ −

=

100kIK

10x~kA

10x~A

10x~k)(1kIK

10x~k)(1kIK'

TT0

T0

T0

T0

'/ ffk =

Camera rotation

xKRK0]X|K[Rx'0]X|K[Ix

1-===

-1KRKH =

conjugate rotation

{ }θθ ii ee −μ,μμ,

Synthetic view

(i)

Compute the homography that warps some a rectangle to the correct aspect ratio

(ii)

warp the image

Planar homography mosaicing

close-up: interlacingcan be important problem!

Planar homography mosaicingmore examples

Projective (reduced) notation

T4

T3

T2

T1 )1,0,0,0(X,)0,1,0,0(X,)0,0,1,0(X,)0,0,0,1(X ====

T4

T3

T2

T1 )1,1,1(x,)1,0,0(x,)0,1,0(x,)0,0,1(x ====

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−−

=dcdbda

000000

P

Tdcba ),,,(C 1111 −−−−=

Moving the camera center

motion parallax

epipolar line

What does calibration give?

xKd 1−=

⎥⎦⎤

⎢⎣⎡= 0d0]|K[Ix

( )( ) ( )( )21-T-T

211-T-T

1

2-1-TT

1

2T

21T

1

2T

1

)xK(Kx)xK(Kx

)xK(Kx

dddd

ddcos ==θ

An image l defines a plane through the camera center with normal n=KTl measured in the camera’s Euclidean frame

The image of the absolute conic

KRd0d]C~|KR[IPXx =⎟⎠⎞⎜

⎝⎛−== ∞

mapping between π∞

to an image is given by the planar homogaphy x=Hd, with H=KR

image of the absolute conic (IAC)

( ) 1-T-1T KKKKω ==− ( )1TCHHC −−a

(i)

IAC depends only on intrinsics(ii)

angle between two rays(iii)

DIAC=ω*=KKT

(iv)

ω ⇔ K (cholesky factorisation)(v)

image of circular points

( )( )2T

21T

1

2T

1

ωxxωxx

ωxxcos =θ

A simple calibration device

(i)

compute H for each square (corners (0,0),(1,0),(0,1),(1,1))

(ii)

compute the imaged circular points H(1,±i,0)T

(iii)

fit a conic to 6 circular points(iv)

compute K from ω

through cholesky factorization

(= Zhang’s calibration method)

Orthogonality = pole-polar w.r.t. IAC

The calibrating conic

1T K1

11

KC −−

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

−=

Vanishing points( ) ( ) λKdaλPDPAλPXλx +=+==

( ) ( ) KdλKda limλ xlimvλλ

=+==∞→∞→

KdPXv == ∞

ML estimate of a vanishing point from imaged parallel scene lines

Vanishing lines

Orthogonality relation

( )( )2T

21T

1

2T

1

ωvvωvv

ωvvcos =θ

Five constraints gives us five equations and can determine w

Calibration from vanishing points and lines

Assumes zero skew, square pixels and 3 orthogonal

vanishing points

Principal point is the orthocenter of the trinagle made of 3 orthogonol vanishing lines

Assume zero skew, square pixels, calibrating conic is a circle;How to find it, so that we can get K?

Assume zero skew , square pixels, and principal point is at the image centerThen IAC is diagonal{1/f^2, 1/f^2,1) i.e. one degree of freedom need one more Constraint to determine f, the focal length two vanishing points corresponding To orthogonal directions.

Next class: Two-view geometry Epipolar geometry