CS 39549526: Spring 2004

1
CS 395/495-26 Spring 2004

• IBMR
• 2-D Conics Introduction
• Jack Tumblin
• jet_at_cs.northwestern.edu

2
Intuition for Conics 2D Ellipses

• Still a circle/ellipse after P2 projective
  transform
• (Better!) every transformed conic is another
  conic!

H1
H2
H3
3
Intuition for Conics 2D Ellipses

• Cartesian axis Circle/Ellipse a x2 b
  y2 r2 0
• P2 axis Circle/Ellipse a x12 b x22 r2
  x32 0
• 5 parameters (deg. of freedom)
• center point (cx, cy)
• major, minor axes (1/a, 1/b)
• orientation ?

?
4
Conics Key Ideas

• Conics intersection of cone plane
• Many possible shapes circles, ellipses,
  parabola, hyperbola, degenerate lines points
• In P2, Conics have Dual Forms
• Point Conics defined by points on the curve
• Line Conics defined by lines tangent to curve

5
Conics written as a Matrix

• All points on any conic solve a 2D quadratic
  ax2 bxy cy2 dx ey f
  0
• In homogeneous coordinates (mpy. all by x32)
• ax12 bx1x2 cx22 dx1x3 ex2x3 fx32
  0
• xTCx 0
• C is symmetric, 5DOF (not 6 ignores x3 scaling)
• Can find any C from 5 homog. points... How?
  (SVD solves
  Null-space problems easily)

x1 x2 x3
a b/2 d/2 b/2 c e/2 d/2 e/2
f
x1 x2 x3
0
(and how could conics help you find H?...)
6
Familiar Conics Circles, Ellipses...
x1 x2 x3
a b/2 d/2 b/2 c e/2 d/2 e/2
f

• xT C x 0 or
• General form of point conic
• ax2 bxy cy2 dx ey f
  0 (or)
• ax12 bx1x2 cx22 dx1x3 ex2x3 fx32 0
• Circle at d with radius r
• 1x12 bx1x2 1x22 dx1x3 ex2x3 r2x32 0
• Cr
• Degenerate Cases (C ranklt3) lines, parabolas,
  hyperbolae... Example
• ax12 bx1x2 0x22 dx1 ex2 fx32 0

x1 x2 x3
0
1 0 0 0 1 0 0 0 -r2
a 0 0 0 0 0 0 0 f
(Try itplot points for x3 1)
7
Point Conics C

• Matrix C makes conic curves from points x
  xTCx 0 C is a point conic
• The tangent line L for point x is L C x
• P2 point conic transformed by H
• C H-T C H-1
• (messy! requires an invertible H !)

x
L
Prove it! Use the two equations above
8
Dual or Line Conics C

• Matrix C makes conic curves from lines L
  LT C L 0 C is a line conic
• The tangent point x for line L is x C L
• Projective transform by H is
• C H C HT (better -- no H inversion
  required)
• If C is non-singular (rank 3), then C C-1

x
L
Prove it! Use two equations above
9
Conic Summary

• Matrix C identifies all the points on a conic
  xTCx 0 C is a point conic
• Given a point x on a conic curve, the homog.
  tangent line l is given by l C x
• Matrix C identifies all lines tangent to a
  conic
• lTCl 0 C is a Dual Conic
• Can you prove this? Try it!
• C C-1 (hint use the boxed equations)

10
Degenerate Conics
x1 x2 x3
a b/2 d/2 b/2 c e/2 d/2 e/2
f

• xT C x 0 or
• General form of point conic
• ax2 bxy cy2 dx ey f
  0 (or)
• ax12 bx1x2 cx22 dx1x3 ex2x3 fx32 0
• Very special degenerate case infinite
  radius circle as x3?0, aka C?
• 1x12 bx1x2 1x22 dx1x3 ex2x3 0x32 0
• C? only solutions to xT C?x 0
  are 2 complex x vectors and

x1 x2 x3
0
11
Conics Key Ideas

• Transformed conics tell you how H will change
  angles
• one strange, degenerate conic C? can measure
  angle
• Angles between line pairs yield transformed C?

?
?
?
C?
C?
?
?
?
?
?
0
L
m
H HS HA HP
2D image space
2D world space
12
Conic Weirdness I

• Circular Points I , J and C?
• Ideal points 2 points at infinity / horizon,
  but
• Complex real part on x1 axis, imag. on x2
• Where infinite circle hits horizon line L?
  0,0,1
• Where? imaginary! (infinite x,y axes, maybe?)
• ?WHY BOTHER?
• To measure angles in projective space
• To use angle correspondences to find HA,HP

13
Conic Weirdness 2

• Circular Points I , J and C?
• C? is a point conic, affected by HAHP only.
• C? is line conic, but is the same matrix (!?!)
• Is same as this outer product IJT JIT
• Intersects L?, the infinity line (proof
  L?C?L?0)
• Affected by HP and HA only, ( HS does nothing)
• ALL transforms of C? have only 4DOF (pg 33 )

14
Conics Angle Measuring

• Matrix H transforms C? to another space
• C? H C? HT
• Angles
• define world space C? as
• (for any two world-space homogeneous lines) L
  and m are perpendicular iff LT C? m 0
• Angle ? between lines L and m given by

(Remember, it is simpler to transform a line
conic)
(LT C? m) (LT C? L)(mT C? m)
cos(?)
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Conics Angle Measuring

• Matrix H transforms C? to another space
• C? H C? HT
• Angles
• ALSO true for transformed L, m and C?, so
• If you FIND C? in image space, you can measure
  2 image-space lines L, m and find the angle ?
  between them in world space! Use the same
  expression

(Remember, it is simpler to transform a line
conic)
(LT C? m) (LT C? L) (mT C? m)
cos(?)
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? How can we find C? without H ?

• One Answer measure perpendicular lines
• We know 2 world-space lines L and m are
  perpendicular iff LT C? m 0
• Also true for transformed L, m and C?
  (e.g. in image space) LT C? m 0
• Hmmm. L and m are known,
• C? unknown.
• Isnt this a null-space problem?
• Once we have C? then what?
• See any other uses for conics in P2?

1 0 0 0 1 0 0 0 0
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Polar Lines and Pole Points
(You can skip this)
xt
C
Lt

• Line Conic Cs tangent line Lt at point xt by
• C xt Lt (given xt is on the conic xtT C
  xt0)

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Polar Lines and Pole Points
(You can skip this)
xp
Lp
C

• Line Conic Cs tangent line Lt at point xt by
• C xt Lt (given xt is on the conic xtT C
  xt0)
• But if x is NOT on the conic? try Cxp Lp

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Polar Lines and Pole Points
(You can skip this)
p1
xp
Lp
C
p2

• Line Conic Cs tangent line Lt at point xt by
• C xt Lt (given xt is on the conic xtT C
  xt0)
• But if x is NOT on the conic? try Cxp Lp
• Polar line Lp conic at p1, p2 (find
  them?ugly!)

20
Polar Lines and Pole Points
(You can skip this)
p1
xp
Lp
C
p2

• Line Conic Cs tangent line Lt at point xt by
• C xt Lt (given xt is on the conic xtT C
  xt0)
• But if x is NOT on the conic? try Cxp Lp
• Polar line Lp conic at p1, p2 (to find
  them?ugly!)
• p1, p2 tangent lines meet at Pole point xp

21
Polar Lines and Pole Points
p1
Interesting, But why does book show it?
xp
Lp
C
p2

• Line Conic Cs tangent line Lt at point xt by
• C xt Lt (given xt is on the conic xtT C
  xt0)
• But if x is NOT on the conic? try Cxp Lp
• Polar line Lp conic at p1, p2 (to find
  them?ugly!)
• p1, p2 tangent lines meet at Pole point xp

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Question Can you apply DLT?

• Choose pairs of lines that are known to be
  perpendicular in world space measure those lines
  in image space.
• Solve for C? using LT C? m 0
• We also know that HC?HT C?
• Can you solve for H? Zisserman tried, but
  book method is confused, contains errors, and
  may be incorrect

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END. STOP HERE
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Recall Projective Transform H
2D image (x,y) ?? Homog. coords x,y,1T x

• Apply the 3x3 matrix H x Hx

Homog. coords x x,y,1T ?? 2D image
(x,y)
x
x
x2
(x,y)
y
x
y
y
y
x3
x
x
x
y
(x,y)
x1
x
25
Recall Projective Transform H
Goal Find H for Image Rectification
2D image (x,y) ?? Homog. coords x,y,1T x

• Apply the 3x3 matrix H x Hx

Homog. coords x x,y,1T ?? 2D image
(x,y)
x
x
x2
(x,y)
y
x
y
y
y
x3
x
x
x
y
(x,y)
x1
x
World Plane
View Plane
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Recall Rectification Undo parts of H

• x H x where H HS HA HP
• GOAL Put world plane x into view plane x
• Affine Rect. (find only HP (2DOF))
• Metric Rect. (find HA and HP (6DOF))
• Full Rect. (find all HSHAHP (8DOF))
• METHODS
• Affine Vanishing Point, Horizon line methods
• Metric Conics Circular Points
• Full 4-point correspondence

27
Recall The bits and pieces of H

• H has 8 independent variables (DOF)
• Computer Vision Jargon (2D projective)
• Isometry --3DOF(2D translate tx,ty 2D rotate
  ?z )
• Similarity --4DOF (add uniform scale s)
• Affine --6DOF (add orientable scale s?,/s,
  s??/s)
• Projective--8DOF (changes x3 3D-rotation-like)
• Rectification up to a Similarity (finds HP, HA)

28
Conics Key Ideas

• Transformed conics tell you how H changes angles
• one strange, degenerate conic C? measures angle
• Angles between line pairs yield transformed C?
• We can compute HA and HP from transformed C?

?
?
?
C?
C?
?
?
?
? ?
?
L
m
HS HA HP
2D image space
2D world space
29
Undoing H Metric Rect. 1

• C? ?
• First, ignore projective part set v0.
• Choose two pairs of perpendicular lines L,m
• Second, Use LT C? m 0 to solve for K

KKT ? Kv ? ? ? vTK ? s
KKT ? 0 ? ? ? 0 ? s
aka Weak Perspective
L
m
HS HA HP
2D image space
2D world space
30
Undoing H Metric Rectification

• C? has only 4DOF (before,after ANY H)
• All C? DOF are contained in (HA HP)
• SVD can convert C? to (HA HP) matrix!
• Task is then find C?, then use it to find H

?
?
?
C?
C?
?
?
?
?
?
0
L
m
HS HA HP
2D image space
2D world space
31
Undoing H Metric Rect. 1

• H HS HA HP
• Book shows how to write transformed C? as
  (simple, tedious)
• C?
• where K is 2x2 symmetric (affine part 2DOF)
• v is 2x1 vector, (projective part 2DOF)

?
?
?
sR t 0T 1
I 0 vT v
K 0 0T 1
?
?
?
?
?
?
?
?
?
?
?
KKT ? Kv ? ? ? vTK ? 0
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Undoing H Metric Rect. 1

• H HS HA HP
• Book shows how to write transformed C? as
  (simple, tedious)
• C?
• where K is 2x2 symmetric (affine part 2DOF)
• v is 2x1 vector, (projective part 2DOF)

?
?
?
sR t 0T 1
I 0 vT v
K 0 0T 1
?
?
?
?
?
?
?
?
?
?
?
!!!BUT DERIVATION RESULT IS WRONG!!! SKIP
Metric Rectification I pg 35,36!
KKT ? Kv ? ? ? vTK ? 0
33
Let SVDs Explain it All for You

• SVD finds H for you (?) (pg 35)
• Books Method for finding HAHP from C?
• C? H C? HT is symmetric
• Assume SVD(C?) USVT (U S ) I ( S VT )
• ( H ) C? ( HT)
• BUT THIS DOESNT WORK!
• ( Recall C? )

A USVT Find Input Output Axes, Linked
by Scale
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Eigen -values,-vectors, Fixed pt line

• Formalizes invariant notion
• if x is fixed for H, then Hx only scales x H
  x ? x (? is a constant scale factor)
• x is an eigenvector, ? is its eigenvalue
• again, SVD helps you find them.
• Elaborate topic (but not hard). Skip for now.

(You can skip this)
35
Book Errata Website

• http//www.robots.ox.ac.uk/az/HZbook/HZerrata.htm
  l
• pg. 35, eqn. 1.22 is dubious / erroneous
• C? HC?HT (HSHAHP) C? (HSHAHP)T
• is the transformed line conic, but book uses
  C? (HPHAHS) C?
  (HPHAHS)T instead, AND has error
• But I get
• C?

KKT ? Kv ? ? ? vTK ? ?
(book ?0 errata website ?vTKKTv)
KKT ? KKTv ? ?
? vTKKT ? vTKKTv
36
Dubious Book Aggravations

• Similarity transform HS leaves C?
  unchanged C? HS C? HST
• H transforms C? to image space C? by C? H
  C? HT (HS HA HP) C? (HS HA HP)T
• Book claims to derive C? (HP HA) C? (HP
  HA)T (!?!?!)
• Then errs in simplifying to

1 0 0 0 1 0 0 0 0
37
SVDs and Conics

• Conics (both C and C) are symmetric
• SVD of any symmetric A is also symmetric SVD(A)
  USUT !Not Always!
• Find conics singular values sign si 0,1, or
  1, to classify conic type (pg 40)

Si values Equation Type
.
imaginary-only circle/ellipse single real point
(0,0,1)2 lines x/- y2 co-located lines x0
x2 y2 w2 0x2 y2 - w2 0 x2 y2
0x2 - y2 0x2
0
(1, 1, 1) (1, 1,-1) (1, 1, 0) (1,-1, 0) (1, 0, 0)
38
Undoing H Metric Rectification

• (recall)H HS HA HP
• Tedious algebra (pg35) shows symmetry
• C? H C? HT
• where K is 2x2 symmetric (affine part 2DOF)
• v is 2x1 vector, (projective part 2DOF)
• ?But what do K and V really control?

K ? 0 ? ? ? 0T ? 1
sR ? t ? ? ? 0T ? 1
I ? 0 ? ? ? vT ? v
KKT ? Kv ? ? ? vTK ? 0
error?!?
39
Undoing H Metric Rectification

• (recall)H HS HA HP
• Tedious algebra (pg35) shows symmetry
• C? H C? HT
• where K is 2x2 symmetric (affine part 2DOF)
• v is 2x1 vector, (projective part 2DOF)
• ?But what do K and V really control?

K ? 0 ? ? ? 0T ? 1
sR ? t ? ? ? 0T ? 1
I ? 0 ? ? ? vT ? v
KKT ? Kv ? ? ? vTK ? ?
(book ?0 errata website ?vTKKTv)
40
Undoing H Metric Rectification

• OK, then how do we find K and v?
• (book ?0 errata website ?vTKKTv)
• Choose known-perpendicular line pairs (Li, mi),
  then compute by
• Method 1a (pg 36) Assume v0, solve for K
• Method 1b (NOT in book)Assume KI, solve for v
• Method 2 Rearrange, solve for full
  C? then get HAHP using SVD.

KKT? Kv ? ? ? vTK ? ?
C? H C? HT
41
Undoing H Metric Rect. 1a

• Method 1a (pg 36) Assume v0, solve for K
• Choose 2 ? line pairs (La,ma) and (Lb,mb)
• Both pairs satisfy LT C? m 0 or
• (note x3 term is ignored!)
• flatten to one equation

KKT ? 0 ? ? ? 0 ?
0
KKT ? Kv ? ? ? vTK ? 0
C? H C? HT ?
s1 s2 0 s2 s3 0 0 0 0
0
l1m1 l2m1l1m2 l2m2
0
42
Undoing H Metric Rect. 1a

• Method 1a (pg 36) Assume v0, solve for K
• Stack to combine both line pairs
• Solve for s by SVD find input null space
  (Ax0)
• From s make C? KKT, then extract HAHP using
  SVD recall C? H C? HT , it is symmetric

KKT ? Kv ? ? ? vTK ? 0
s1 s2 0 s2 s3 0 0 0 0
C? H C? HT ?
la1ma1 la2ma1la1ma2 la2ma2
0
lb1mb1 lb2mb1lb1mb2 lb2mb2
43
Undoing H Metric Rect. 1b

• Method 1b (not in book) Assume KI, solve for v
• Choose 2 ? line pairs (La,ma) and (Lb,mb)
• These satisfy LT C? m 0 or
• flatten to one equation (messy), stack, solve
  for v1,v2
• Extract Hp from C? using SVD

I ? v ? ? ? vT ?
0
KKT ? Kv ? ? ? vTK ? 0
C? H C? HT ?
1 0 v1 0 1 v2 v1 v2 ?
0