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Multiple View Geometry

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Multiple View Geometry

Marc Pollefeys

University of North Carolina at Chapel Hill

Modified by Philippos Mordohai

- 3-D Reconstruction
- Fundamental matrix estimation
- Chapters 9 and 10 of “Multiple View Geometry in Computer Vision” by Hartley and Zisserman

- Correspondence geometry: Given an image point x in the first image, how does this constrain the position of the corresponding point x’ in the second image?

- (ii)Camera geometry (motion): Given a set of corresponding image points {xi ↔x’i}, i=1,…,n, what are the cameras P and P’ for the two views?

- (iii)Scene geometry (structure): Given corresponding image points xi ↔x’i and cameras P, P’, what is the position of (their pre-image) X in space?

p

p

L2

L2

m1

m1

m1

C1

C1

C1

M

M

L1

L1

l1

l1

e1

e1

lT1

l2

e2

e2

Canonical representation:

l2

m2

m2

m2

l2

l2

Fundamental matrix (3x3 rank 2 matrix)

C2

C2

C2

Underlying structure in set of matches for rigid scenes

- Computable from corresponding points
- Simplifies matching
- Allows to detect wrong matches
- Related to calibration

reconstruction problem:

given xi↔x‘i , compute P,P‘ and Xi

for all i

without additional information possible up to projective ambiguity

- Compute F from correspondences
- Compute camera matrices from F
- Compute 3D point for each pair of corresponding points

computation of F

use x‘iFxi=0 equations, linear in coeff. F

8 points (linear), 7 points (non-linear), 8+ (least-squares)

(more on this next class)

computation of camera matrices

use

triangulation

compute intersection of two backprojected rays

Reconstruction ambiguity: projective

xi↔x‘i

Original scene Xi

Projective, affine, similarity reconstruction

= reconstruction that is identical to original up to

projective, affine, similarity transformation

Literature: Metric and Euclidean reconstruction

= similarity reconstruction

If a set of point correspondences in two views determine thefundamental matrix uniquely, then the scene and cameras may be reconstructed from these correspondences alone, and any two such reconstructions from these correspondences are projectively equivalent

- along same ray ofP2, idem for P‘2

two possibilities: X2i=HX1i, or points along baseline

key result:

allows reconstruction from pair of uncalibrated images

- Projective reconstruction
- Affine reconstruction
- Metric reconstruction

Projective to affine

(if D≠0)

theorem says up to projective transformation,

but projective with fixed p∞ is affine transformation

can be sufficient depending on application,

e.g. mid-point, centroid, parallellism

points at infinity (not necessarily visible) are fixed for a pure translation

reconstruction of xi↔xi is on p∞

Parallel lines

parallel lines intersect at infinity

reconstruction of corresponding vanishing point yields

point on plane at infinity

3 sets of parallel lines allow to uniquely determine p∞

remark: in presence of noise determining the intersection of parallel lines is a delicate problem

remark: obtaining vanishing point in one image can be sufficient

Scene constraints

Scene constraints

*

*

projection

constraints

identify absolute conic

transform so that

then projective transformation relating original and reconstruction is a similarity transformation

in practice, find image of W∞

image w∞back-projects to cone that intersects p∞ in W∞

note that image is independent of particular reconstruction

Affine to metric

given

possible transformation from affine to metric is

proof:

(Cholesky factorization to obtain A)

vanishing points corresponding to orthogonal directions

vanishing line and vanishing point corresponding

to plane and normal direction

rectangular pixels

square pixels

same intrinsics same image of the absolute conic

e.g. moving camera

given sufficient images there is in general only one

conic that projects to the same image in all images,

i.e. the absolute conic

This approach is called self-calibration

transfer of IAC:

provides 4 constraints, one more needed

(in general two solutions)

approach 1

calibrated reconstruction

approach 2

compute projective reconstruction

back-project w from both images

intersection defines W∞ and its support plane p∞

(2 lin. eq. in H-1per view,

3 for two views)

use control points XEi with know coordinates

to go from projective to metric

(3 lin. eq. in H per point,

H has 15 d.o.f.)

- Given two uncalibrated images compute (PM,P‘M,{XMi})
- (i.e. within similarity of original scene and cameras)
- Algorithm
- Compute projective reconstruction (P,P‘,{Xi})
- Compute F from xi↔x‘i
- Compute P,P‘ from F
- Triangulate Xi from xi↔x‘i

- Rectify reconstruction from projective to metric
- Direct method: compute H from control points
- Stratified method:
- Affine reconstruction: compute p∞
- Metric reconstruction: compute IAC w

- 3-D Reconstruction
- Fundamental matrix estimation
- Chapters 9 and 10 of “Multiple View Geometry in Computer Vision” by Hartley and Zisserman

separate known from unknown

(data)

(unknowns)

(linear)

SVD from linearly computed F matrix (rank 3)

Compute closest rank-2 approximation

one parameter family of solutions

but F1+lF2 not automatically rank 2

3

F7pts

F

F2

F1

(obtain 1 or 3 solutions)

(cubic equation)

Compute possible l as eigenvalues of

(only real solutions are potential solutions)

~10000

~100

~10000

~100

~10000

~10000

~100

~100

1

Orders of magnitude difference

between column of data matrix

least-squares yields poor results

!

Transform image to ~[-1,1]x[-1,1]

(-1,1)

(1,1)

(0,0)

(-1,-1)

(1,-1)

normalized least squares yields good results(Hartley, PAMI´97)

(0,500)

(700,500)

(0,0)

(700,0)

Gold standard

Sampson error

Symmetric epipolar distance

Maximum Likelihood Estimation

(= least-squares for Gaussian noise)

Initialize: normalized 8-point, (P,P‘) from F, reconstruct Xi

Parameterize:

(overparametrized F=[t]xM)

Minimize cost using Levenberg-Marquardt

(preferably sparse LM, see book)

Gold standard

Alternative, minimal parametrization (with a=1)

(note (x,y,1) and (x‘,y‘,1) are epipoles)

- problems:
- a=0

pick largest of a,b,c,d to fix to 1

- epipole at infinity

pick largest of x,y,w and of x’,y’,w’

4x3x3=36 parametrizations!

reparametrize at every iteration, to be sure