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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 – PowerPoint PPT presentation

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Title: 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
  • 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
  • 2-minute exercise how many degrees of freedom
    does a conic have?

Conic representation 2
  • matrix C is projective (only defined up to a
  • 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 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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,
  • 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
  • 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

  • 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,
  • mxn A
  • mxm orthogonal U
  • mxn diagonal D (containing singular values
  • 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
  • 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
  • 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