CS 39549526: Spring 2003

1
CS 395/495-26 Spring 2003

• IBMR Week 3 A 2D Projective GeometryConics
  Measure Angles
• Jack Tumblin
• jet_at_cs.northwestern.edu

2
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
3
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
4
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 remove

5
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

6
Intuition for Conics 2D Ellipses

• Cartesian Circle/Ellipse a x2 b y2
  r2 0
• P2 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 ?

?
7
Intuition for Conics 2D Ellipses

• Still a circle/ellipse after P2 projective
  transform

H1
H2
H3
8
Conics Key Ideas

• Conic method Clever math, but NOT intuitive!
• (and book gives ugly, scattered description)
• P2 Conics are the family of ALL quadratics of
  (x1, x2, x3)
• Includes simpler degenerate forms point,
  line, line-pair, parabolas, hyperbolae.
• Closed ANY Conic transformed by H Another
  Conic

(imaginary)
9
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
10
Conic Review

• Review conics first (pg.8)
• Conics intersection of cone plane
• Many possible shapes circles, ellipses,
  parabola, hyperbola, degenerate lines points
• Dual Forms
• Point Conics defined by points on the curve
• Line Conics defined by lines tangent to curve

11
Point Conics C

• 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. point pairs...
  (SVD solves for
  null spacesee book pg 9)

x1 x2 x3
a b/2 d/2 b/2 c e/2 d/2 e/2
f
x1 x2 x3
0
(but not very useful for finding H...)
12
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 (ranklt3) lines, parabolas,
  hyperbolae... Example
• 1x12 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
13
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 conic transformed by H
• C H-T C H-1
• (messy! requires invertible H !)

x
L
14
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
15
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
16
Conic Weirdness I

• Circular Points I , J or C?
• Ideal points 2 points at infinity / horizon,
  but
• Complex real part on x axis, imag. on y
• Where infinite circle hits infinity line L?
  0,0,1
• Where? imaginary! (infinite x,y axes, maybe?)
• I,J are not affected by HS . ( only by HA, HP)
• ?WHY BOTHER?
• To measure angles in projective space
• To use angle correspondences to find HA,HP

17
Conic Weirdness 2

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

18
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
• Also true for transformed L, m and C?, so
• C? in image space?angle ? in world space

(simpler to transform a line conic)
(LT C? m) (LT C? L)(mT C? m)
cos(?)
19
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
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? But how do we find C? ?

• One Answer use perpendicular lines
• (Recall) we defined world space C? as
• Then 2 world-space lines L and m are
  perpendicular iff LT C? m 0
• Also true for transformed L, m and C?
• H transforms C? to image space C? by C? H
  C? HT
• Now split up H and try it H HSHAHP, so

1 0 0 0 1 0 0 0 0
21
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
22
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
23
END
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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
25
Matrix Math SVDs

• Matrix multiply is a change of coordinates
• Rows of A coordinate axis vectors
• Ax a dot product for each axis
• input space?output space
• SVD factors matrix A into 3 parts
• U orthonormal basis

svd(A) U S VT
26
Matrix Math SVDs

• Matrix multiply is a change of coordinates
• Rows of A coordinate axis vectors
• Ax a dot product for each axis
• input space?output space
• SVD factors matrix A into 3 parts
• U orthonormal basis

svd(A) U S V
27
Recall Projective Transform Hx x

Finding H from point pairs (correspondences)
x1 x2 x3
h11 h12 h13 h21 h22 h21 h31 h32 h33
x1 x2 x3
x y 1
Input (or output) image is on central plane
x1 x2 x3


x y
Write R2 expressions(for each x,x pair)
x (h31 x h32 y h33) (h11 x h12 y h13)
Rearrange, solve as a matrix problem
y (h31 x h32 y h33) (h11 x h12 y h13)
28
DLT Four-Point Correspondence

• 4 point correspondence
• Book shows
• Rearrange known vector (dot) unknown vector
• stack, solve for null space
• Hint Files
• Added max commands, examples
• Make, test your own H matrices!

x (h31 x h32 y h33) (h11 x h12 y h13)
y (h31 x h32 y h33) (h11 x h12 y h13)
x y 1 0 0 0 -xx -xy -x
0 0 0 0 x y 1 -yx -yy -y
h11 h12 h13 h21 h22 h23 h31 h32 h33
29
Direct Linear Tranform (DLT)

• Project 2 method Solve Hx x 0
• Requires constant scale (x31 in measurement)
• compare
• DLT method Solve Hx ? x 0
• Accepts any scale (x3w?1 OK, xx3)

h11 h12 h13 h21 h22 h23 h31 h32 h33
0 0 0 x y 1
-yx -yy -y x y
1 0 0 0 -xx -xy
-x
0 0 0 -wx -wy -ww yx
yy yw 0 wx wy ww
0 0 0 -xx -xy -xw
30
Direct Linear Tranform (DLT)

• Project 2 method Solve Hx x 0
• Requires constant scale (x31 always)
• compare
• DLT method Solve Hx ? x 0
• Accepts any scale (x3w?1, x3w?1)

h1 h2 h3
0 0 0 x y 1
-yx -yy -y x y
1 0 0 0 -xx -xy
-x
0 0 0 -wx -wy -ww yx
yy yw 0 wx wy ww
0 0 0 -xx -xy -xw
0T -wi xiT yI xiT wi
xiT 0T -xi xiT
31
Direct Linear Tranform (DLT)

• DLT method Solve H ? x 0
• Accepts any scale, any point (x3w?1 is OK)
• Pure, Compatible -- P2 terms only
• Much better suited to error measurements.
• Subtlety
• Wont divide-by-zero if w0 or h330 (it
  happens!)
• has a 3rd row it is not degenerate if w0
• OK to use it(Solve 8x12)

0T -wi xiT yI xiT wi
xiT 0T -xi xiT 0 -yi xiT
-xi xiT 0T
h1 h2 h3
32
Deceptive Robustness

• Suppose we have 4 pt-correspondences
• Use DLT to write 8x9 (or 12x9) matrix A Ah0
• Solve for h null space. ALWAYS gives H matrix
• But what if points are bad / fictional?
• 3 collinear input pts, crooked out IMPOSSIBLE!
• Yet we get an H solution! Why?
• A matrix rank is 7 or 6? rank 1 H result(s)
• Null Space of A may contain gt1 H solution!
• Degenerate H solution of form aH1 bH2
• Answer SVD ranks A reject bad point sets.

33
Actual Robustness

• Vectorizing (Flatten, Stack, Null Space method)
  works for almost ANY input! (Points, lines,
  planes, , ?, conics, quadrics, cross-ratios,
  vanishing points, twisted cubics
• Use DLT formulation (Hx ? x) 0
• Rearrange as dot product (known)(unknown) 0
• Be careful to have ENOUGH constraints
• tricky when you mix types points, lines, (pg
  75)

0
measured output
measured inputtransformed
?