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Conics

Behaviour at infinity

Primitives pt/line/conic HZ 2.2

DLT alg HZ 4.1

Hierarchy of maps Invariants HZ 2.4

Projective transform HZ 2.3

Rectification HZ 2.7

Conic representation

- conic plane curve described by quadratic

polynomial section of cone - what are the conics?
- important primitives for vision and graphics
- conic in Euclidean space ax2 bxy cy2 dx

ey f 0 - homogenize to translate to projective space
- 2-minute exercise how do you homogenize this

polynomial and thereby translate to projective

space? - conic in projective space ax2 bxy cy2

dxw eyw fw2 0 - observation homogenize a polynomial by the

replacement x ? x/w and y ? y/w (ring a bell?) - this conic in projective space is encoded by a

symmetric 3x3 matrix - xt C x 0
- C a, b/2, d/2 b/2, c, e/2 d/2, e/2, f
- ijth entry encodes ijth coefficient (where

x1,y2,w3) - 2-minute exercise how many degrees of freedom

does a conic have?

Conic representation 2

- matrix C is projective (only defined up to a

multiple) - conic has 5dof ratios abcdef or 6 matrix

entries of C minus scale - kC represents the same conic as C (equivalent to

multiplying equation by a constant) - in P2, all conics are equivalent
- that is, can transform from a conic to any other

conic using projective transforms - not true in Euclidean space cannot transform

from ellipse to parabola using linear

transformation - transformation rule
- if point x ? Hx, then conic C ? (H-1)t C H-1
- HZ30-31, 37

Conic tangents

- another calculation is made simple in projective

space (and reduces to matrix computation) - Result The tangent of the conic C at the point x

is Cx. - Proof
- this line passes through x since xt Cx 0 (x

lies on the conic) - Cx does not contain any other point y of C,

otherwise the entire line between x and y would

lie on the conic - if yt C y 0 (y lies on conic) and xt C y 0

(Cx contains y), then x\alpha y (the entire line

between x and y) also lies on C - thus, Cx is the tangent through x (a tangent is a

line with only one point of contact with conic) - HZ31

Metric rectification with circular points

Behaviour at infinity

Primitives pt/line/conic HZ 2.2

DLT alg HZ 4.1

Hierarchy of maps Invariants HZ 2.4

Projective transform HZ 2.3

Rectification HZ 2.7

Material for metric rectification

- circular points (34)
- similarity iff fixed circular points (52)
- conic (55)
- dual conic
- degenerate conic
- C8 conic dual to circular points (or just dual

conic) - angle from dual conic
- image of dual conic under homography
- then were ready for the algorithm of Example

2.26 HZ

Line conic

- we have considered point conic
- points are dual to lines lets dualize
- suppose conic is nondegenerate (not 2 lines or

repeated line), so C is nonsingular - 3x3 matrix C encodes the points of a conic
- xt C x 0 if x lies on conic
- matrix C-1 encodes the tangent lines of that

conic - L is a tangent line of the conic iff
- Lt C-1 L 0
- proof tangent L Cx for some point x on conic

xt C x 0 so substituting x C-1L yields

(C-1L)t C (C-1L) Lt C-1L 0 (recall that C

is symmetric) - called a dual conic or line conic conic envelope
- transformation rule if point x ? Hx, then dual

conic C ? H C Ht - HZ31,37

Degenerate conics

- there are degenerate point conics and degenerate

line conics - recall the outer product
- if L and M are lines, LMt MLt is the

degenerate point conic consisting of these two

lines - notice that the matrix is singular (rank 2)
- if p and q are points, pqt qpt is the

degenerate line conic consisting of all lines

through p or q - 2 pencils of lines, drawing
- notice that this is a line conic (or dual conic)

The dual conic C8

- two circular points I and J
- make a degenerate line conic out of them
- C8 IJt JIt (1,0,0 0,1,0 0,0,0)
- all lines through the circular points
- C8 will be called the dual conic
- although it is the line conic dual to the

circular points - like circular points, C8 is fixed under a

projective transform iff it is a similarity - interesting fact null vector of C8 line at

infinity - HZ 52-54
- explore relationship to angle

Computing angle from C8

- how can we measure angle in a photograph?
- angle is typically measured using dot product
- but dot product is not invariant under homography
- consider two lines L and M in projective space

(viewed as conventional columns, not correct

rows) - Lt M is replaced by a normalized Lt C8 M
- (Lt C8 M) / \sqrt( (Lt C8 L) (Mt C8 M))
- this is invariant to homography
- note that Lt C8 M is equivalent to dot product

in the original space - (a1,a2,1) C8 (b1,b2,1) a1b1 a2b2
- Result (Lt C8 M) / \sqrt( (Lt C8 L) (Mt C8

M)) is the appropriate measure of angle in

projective space. - since angle arccos (A.B), this is of course a

measure of cos(angle), not angle - corollary angle can be measured once C8 is

known - HZ54-55

How C8 transforms

- we want to understand how C8 transforms under a

homography (since we dont want it to transform!) - a homography matrix can be decomposed into a

projective, affine, and similarity component - H I 0 K 0 sR t
- vt 1 0 1 0 1
- Hp Ha Hs
- note that K is the affine component
- recall that line conics transform by C ? H C

Ht, so - C8 ? (Hp Ha Hs) C8 (Hp Ha Hs)t
- but the dual conic is fixed under a similarity,

so - C8 ? (Hp Ha) C8 (Hp Ha)t
- which reduces as follows using C8 diag(1,1,0)
- image of C8 KKt KKt v
- vt KKt vt KKt v
- when line at infinity is not moving, v 0
- image of C8 in an affinely rectified image

KKt 0 - 0 0
- HZ43 for this decomposition chain of a

homography, HZ55

Metric rectification algorithm

- We have the technology. We can rebuild him!
- input 2 orthogonal line pairs
- assume that the image has already been affinely

rectified (i.e., line at infinity is in correct

position) - we are within an affinity of a similarity!
- what is this affinity K? use the known

orthogonality (known angle) to solve for K - let S KKt we actually solve for S, then

retrieve K using Cholesky decomposition - note S has 2 dof (symmetric 2x2)
- use the two line angle constraints to solve for

these dof

Metric rectification continued

- (L,M) image of orthogonal line pair
- Figure 2.17a
- choose two points P1,P2 on L L P1xP2
- L (L1,L2,L3) and M (M1,M2,M3)
- orthogonal pair (L,M) satisfies Lt C8 M 0
- Lt S 0 0 0 M (L1 L2) S (M1 M2)
- this imposes a linear constraint on S
- (L1 M1, L1M2 L2M1, L2M2) . (s11, s12, s22) 0
- first row of matrix . S 0
- a second orthogonal pair defines the 2nd row of

the system As 0 A for angle - A is 2x3 matrix, s is the vector (s11,s12,s22)
- S is the null vector of this matrix M
- Cholesky decompose S to retrieve K
- the true image (up to metric structure) must

have been mapped by the affinity K 0 0,1 - see how C8 transforms above
- so rectify by mapping image by K-1
- Moral use orthogonal constraint to solve for

affinity K, using the dual conic to help you

measure angle - note never actually compute dual conic, just

rely on its properties

Synopsis

- want affinity K
- want S KKt
- S is available from the image of C8
- get at image of C8 using angle relationship
- use orthogonal pairs to solve for image of C8
- back out from image of C8 to S to K

Numerical aside solving for null space

- to solve Ax 0
- compute SVD of A U D Vt
- if null space is 1d, x last column of V (called

null vector) - in general, last d columns of V span the

d-dimensional null space

Numerical aside SVD

- singular value decomposition of A is A U D Vt,

where - mxn A
- mxm orthogonal U
- mxn diagonal D (containing singular values

sorted) - nxn orthogonal V
- do not compute yourself!
- compute using CLAPACK or OpenCV

Other methods for metric rectification

- the 2-ortho-pair method is called stratified

rectification - stratification 2 step approach to removal of

distortion, first projective, then affine - Note 2 orthogonal lines are conjugate with

respect to the dual conic C8 - can also rectify (in a stratified fashion) using

an imaged circle - image of circle is ellipse intersect it with

vanishing line to find imaged circular points - can also rectify using 5 orthogonal line pairs
- why would you want to? because it doesnt assume

image has already been affinely rectified - method (your responsibility to read)
- find the dual conic as the null vector of a 5x6

matrix built up from 5 linear equations - see DLT algorithm below for a similar approach

(this is a very popular approach!) - note solving for entire 5dof conic, not just the

2dof affine component - HZ56-57

On OpenCV

- review openCVInstallationNotes
- OpenCV demo programs at my website
- cvdemo, cvmousedemo, cvresizedemo, cvpixeldemo
- opencvlibrary.sourceforge.net
- invaluable resource for full documentation, FAQ,

examples, Wiki - IPL Image Processing Library
- depth of bits in each data value of image
- 8 uchar, 32 float
- channels of data values for each image pixel
- 1 grayscale 3 RGB

Capturing images as test data

- I would like you to personalize your test data
- I would also like to gather an image database
- assignment find images replete with parallel

lines (and optimally, vanishing lines inside the

image) and orthogonal lines - also want different directions of parallel and

orthogonal lines e.g., brick wall is not as

useful since all parallel lines yield only two

distinct ideal points - challenge rectification without gaps
- challenge rectification to within a translation

using external cues - challenge rectification from other constraints

(e.g., in a natural scene with no visible lines

or circles) - challenge image contexts with circles and other

known nonlinear curves

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