Conics

1
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
2
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?

3
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

4
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

5
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
6
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

7
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

8
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)

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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

10
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

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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

12
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

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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

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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

15
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

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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

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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

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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

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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