CS 395/495-26: Spring 2004

CS 395/495-26: Spring 2004CS 395/495-26: Spring 2004

IBMR: IBMR: 2-D Conics: Introduction2-D Conics: Introduction

Jack TumblinJack [email protected]@cs.northwestern.edu

Intuition for Conics: 2D EllipsesIntuition for Conics: 2D EllipsesStill a circle/ellipse after PStill a circle/ellipse after P22 projective transform projective transform

(Better!) every transformed conic is another conic!(Better!) every transformed conic is another conic!

HH11

HH22

HH33

Intuition for Conics: 2D EllipsesIntuition for Conics: 2D Ellipses• Cartesian axis Circle/Ellipse: Cartesian axis Circle/Ellipse: a a xx22 ++ b b yy22 – r– r22 = 0= 0• PP22 axis Circle/Ellipse: axis Circle/Ellipse: a a xx11

22 + b + b xx2222 – r – r2 2 xx33

22 = 0 = 0

• 5 parameters (deg. of freedom):5 parameters (deg. of freedom):– center point center point (cx, cy)(cx, cy)– major, minor axes major, minor axes (1/a, 1/b)(1/a, 1/b)– orientation orientation

1/a1/a1/b1/b

Conics: Key IdeasConics: Key Ideas

– ‘‘Conics’Conics’ == intersection of cone & plane:== intersection of cone & plane:– Many possible shapes:Many possible shapes:

circles, ellipses, parabola, hyperbola, circles, ellipses, parabola, hyperbola, degenerate lines & pointsdegenerate lines & points

In PIn P22, Conics have Dual Forms:, Conics have Dual Forms:– ‘‘Point Conics’Point Conics’ defined by points on the curve defined by points on the curve– ‘‘Line Conics’Line Conics’ defined by lines tangent to curve defined by lines tangent to curve

Conics written as a MatrixConics written as a Matrix– All points on any conic solve a 2D quadratic: All points on any conic solve a 2D quadratic:

a axx22 + b + bxyxy + c + cyy22 +d +dxx +e +eyy +f = 0 +f = 0– In homogeneous coordinates: (mpy. all by xIn homogeneous coordinates: (mpy. all by x33

22))

aaxx1122 + b + bxx11xx22 + c + cxx22

22 +d +dxx11xx33 +e +exx22xx33 + f + fxx3322 = 0 = 0

xxTTCx = 0Cx = 0– C is symmetric, 5DOF C is symmetric, 5DOF ((not 6;not 6; ignores ignores xx33 scaling) scaling)– Can find any Can find any CC from 5 homog. points... How? from 5 homog. points... How?

(SVD solves Null-space problems easily…)(SVD solves Null-space problems easily…)

aa b/2 d/2 b/2 d/2 b/2 c e/2 b/2 c e/2 d/2 e/2d/2 e/2 f f

xx11 xx22 xx33

= 0 = 0 xx11 x x22 x x33

(and how could conics help you find (and how could conics help you find HH?...)?...)

Familiar Conics: Circles, Ellipses...Familiar Conics: Circles, Ellipses...xxTT C x = 0 C x = 0 oror

• General form of point conic: General form of point conic: aaxx22 + b + bxyxy + c + cyy22 +d +dxx +e +eyy +f = 0 (or) +f = 0 (or)aaxx11

22 + b + bxx11xx22 + c + cxx2222 +d +dxx11xx33 +e +exx22xx33 + f + fxx33

22 = 0 = 0

• Circle at d with radius r:Circle at d with radius r:11xx11

22 + + bxbx11xx22 + + 11xx2222 + +dxdx11xx33 +ex +ex22xx33 –r–r22xx33

22 = 0; = 0;

CCr r ==• ‘‘Degenerate’ Cases Degenerate’ Cases ((CC rank<3) rank<3)::

lines, parabolas, hyperbolae... Example: lines, parabolas, hyperbolae... Example:

aaxx1122 + + bxbx11xx22 + + 00xx22

22 + +dxdx11 +ex +ex22 + f+ fxx3322 = 0; = 0;


xx11 xx22 xx33

= 0 = 0 xx11 x x22 x x33

11 0 0 0 0 0 1 0 0 1 0 0 0 -r0 0 -r22

aa 0 0 0 0 0 0 0 0 0 0 0 0 f0 0 f(Try it—plot points for x3 = 1…)

• Matrix Matrix CC makes conic curves from points makes conic curves from points xx: : xxTTCx = 0Cx = 0 CC is a is a ‘point conic’‘point conic’

• The tangent line The tangent line LL for point for point xx is: is: L = C xL = C x

• P2 point conic transformed by H:P2 point conic transformed by H:C’ = HC’ = H-T-T C H C H-1-1

(messy! requires an invertible H !)(messy! requires an invertible H !)

Point Conics: CPoint Conics: C

LLxx

Prove it! Use the two equations above…Prove it! Use the two equations above…

• Matrix Matrix C*C* makes conic curves from lines makes conic curves from lines LL: : LLT T C* L = 0C* L = 0 CC is a is a ‘line conic’‘line conic’

• The tangent point The tangent point xx for line for line LL is: is: x = C* Lx = C* L

• Projective transform by H is:Projective transform by H is:C*’ = H C* HC*’ = H C* HT T (better -- no H inversion required)(better -- no H inversion required)

• If If C C is non-singular (rank 3), then is non-singular (rank 3), then C* = CC* = C-1-1

‘‘Dual’ or Line Conics: CDual’ or Line Conics: C

LLxx

Prove it! Use two equations above…Prove it! Use two equations above…

Conic SummaryConic Summary– Matrix C identifies all the points on a conic: Matrix C identifies all the points on a conic:

xxTTCx = 0Cx = 0 CC is a is a ‘point conic’‘point conic’

– Given a point Given a point xx on a conic curve, the on a conic curve, the homog. tangent line homog. tangent line ll is given by is given by l = C xl = C x

– Matrix C* identifies all lines tangent to a conic:Matrix C* identifies all lines tangent to a conic:llTTC*l = 0C*l = 0 C*C* is a is a ‘Dual Conic’‘Dual Conic’

– Can you prove this? Try it! Can you prove this? Try it! C* = CC* = C-1 -1 (hint: use the boxed equations)(hint: use the boxed equations)

Degenerate ConicsDegenerate ConicsxxTT C x = 0 C x = 0 oror

• General form of point conic: General form of point conic: aaxx22 + b + bxyxy + c + cyy22 +d +dxx +e +eyy +f = 0 (or) +f = 0 (or)aaxx11

22 + b + bxx11xx22 + c + cxx2222 +d +dxx11xx33 +e +exx22xx33 + f + fxx33

22 = 0 = 0

• Very special ‘degenerate’ case:Very special ‘degenerate’ case:‘infinite radius circle’ as x‘infinite radius circle’ as x330, aka 0, aka CC

11xx1122 + + bxbx11xx22 + + 11xx22

22 + +dxdx11xx33 +ex +ex22xx33 + +00xx3322 = 0 = 0

CC == only solutions to only solutions to xxTT C Cx =0x =0

are 2 are 2 complexcomplex xx vectors: vectors: and and


xx11 xx22 xx33

= 0 = 0 xx11 x x22 x x33

11 0 0 0 0 0 1 0 0 1 0 0 0 00 0 0

1 1 i i 0 0

1 1 -i -i 0 0

Conics: Key IdeasConics: Key Ideas– Transformed conics tell you how Transformed conics tell you how HH will change angles: will change angles:

– one strange, degenerate conic one strange, degenerate conic C*C* can can measuremeasure angle angle

– Angles between line pairs yield transformed Angles between line pairs yield transformed C*C*

2D world space 2D world space 2D image 2D image space space

H = HH = HS S HHA A HHPP

C*C*==11 0 0 0 0 0 1 0 0 1 0 0 0 00 0 0

LL mm m’m’L’L’

C*C*’’==

00

Conic Weirdness IConic Weirdness I

‘‘Circular Points’ Circular Points’ I= , J=I= , J= ; and ; and CC ==

– ‘‘Ideal points’Ideal points’; 2 points at infinity / horizon, but…; 2 points at infinity / horizon, but…– Complex :Complex : real part on real part on xx11 axis, imag. on axis, imag. on xx22

– ““Where infinite circle hits horizon line LWhere infinite circle hits horizon line L [0,0,1]” [0,0,1]”– Where? imaginary! (infinite x,y axes, maybe?)Where? imaginary! (infinite x,y axes, maybe?)

?WHY BOTHER??WHY BOTHER? – To measure To measure anglesangles in projective space in projective space– To use angle correspondences to find To use angle correspondences to find HHAA,H,HPP

1 1 i i 0 0

1 1 -i -i 0 0

11 0 0 0 0 0 1 0 0 1 0 0 0 00 0 0

Conic Weirdness 2Conic Weirdness 2

‘‘Circular Points’ Circular Points’ I= , J=I= , J= ; and ; and CC ==

• CC is a is a pointpoint conic, affected by conic, affected by HHAAHHP P only.only.

• C*C* is is lineline conic, but is the conic, but is the samesame matrixmatrix (!?!)(!?!) – Is same as this outer product: Is same as this outer product: IJIJTT + JI + JITT – Intersects LIntersects L, the infinity line, the infinity line (proof: L(proof: LC*C*LL=0)=0)

– Affected by Affected by HHP P andand HHA A only, ( only, ( HHSS does nothing)does nothing)

ALL transforms of ALL transforms of C*C* have only 4DOF have only 4DOF (pg 33 )(pg 33 )

1 1 i i 0 0

1 1 -i -i 0 0

11 0 0 0 0 0 1 0 0 1 0 0 0 00 0 0

Conics: Angle MeasuringConics: Angle Measuring

• Matrix Matrix HH transforms transforms C*C* to another space: to another space:

C*C*’’ == H C*H C* H HT T

Angles:Angles:– define ‘world space’ define ‘world space’ C*C* as as

– (for any two world-space homogeneous lines) (for any two world-space homogeneous lines) LL and and mm are are perpendicularperpendicular iff iff LLTT C* C* m = 0 m = 0

– Angle Angle between lines between lines LL and and m m given bygiven by

11 0 0 0 0 0 1 0 0 1 0 0 0 00 0 0

(L(LTT C*C* m) m) (L(LT T C*C* L)(m L)(mT T C*C* m) m)

cos(cos() = ) =

(Remember, it is simpler (Remember, it is simpler to transform a line conic)to transform a line conic)

Conics: Angle MeasuringConics: Angle Measuring

• Matrix Matrix HH transforms transforms C*C* to another space: to another space:

C*C*’’ == H C*H C* H HT T

Angles:Angles:– ALSO true for ALSO true for transformedtransformed L, mL, m and and C* C*, , so…so…

– If you FIND If you FIND C*C*’’ in image space, you can in image space, you can measure 2 image-space lines measure 2 image-space lines L’, m’ L’, m’ and find and find the angle the angle between them in world space! between them in world space! Use the same expression:Use the same expression:

(L’(L’TT C* C*’ m’)’ m’) (L’(L’T T C*C*’ L’) (m’ L’) (mT T C*C*’ m’)’ m’)

cos(cos() = ) =

(Remember, it is simpler (Remember, it is simpler to transform a line conic)to transform a line conic)

? How can we find ? How can we find C*C*’ without H ?’ without H ?One Answer: measure perpendicular lines…One Answer: measure perpendicular lines…

– We know 2 world-space lines We know 2 world-space lines LL and and mm are are

perpendicularperpendicular iff iff LLTT C* C* m = 0 m = 0 – Also true for Also true for transformedtransformed L, mL, m and and C* C*: :

(e.g. in image space)(e.g. in image space)

L’L’TT C* C*’’ m’ = 0 m’ = 0

Hmmm. Hmmm. L’ and m’ are known, L’ and m’ are known, C*C*

’ ’ unknown.unknown.

Isn’t this a null-space problem?Isn’t this a null-space problem?Once we have Once we have C*C*

’’ then what? then what?

See any other uses for conics in P2?See any other uses for conics in P2?

11 0 0 0 0 0 1 0 0 1 0 0 0 00 0 0

Polar Lines and Pole PointsPolar Lines and Pole Points

• Line Conic Line Conic CC’s tangent line ’s tangent line LLtt at point at point xxtt by: by:C xC xtt = L = Ltt (given (given xxtt is on the conic: is on the conic: xxtt

T T C xC xtt=0=0))

xxtt

LLttCC

(You can skip this)(You can skip this)


• Line Conic Line Conic CC’s tangent line ’s tangent line LLtt at point x at point xtt by: by:C xC xtt = L = Ltt (given (given xxtt is on the conic: is on the conic: xxtt

T T C xC xtt=0=0))

• But if But if xx is is NOTNOT on the conic? try on the conic? try CxCxpp = L = Lpp

xxppLLppCC




T T C xC xtt=0=0))


• ‘‘Polar line’ Polar line’ LLpp = conic at = conic at pp11, p, p22 (find them?ugly!) (find them?ugly!)

xxppLLppCC

pp11

pp22




T T C xC xtt=0=0))


• ‘‘Polar line’ Polar line’ LLpp = conic at = conic at pp11, p, p22 (to find them?ugly!) (to find them?ugly!)

• pp11, p, p22 tangent lines meet at ‘Pole point’ tangent lines meet at ‘Pole point’ xxpp

xxppLLppCC

pp11

pp22




T T C xC xtt=0=0))


• ‘‘Polar line’ Polar line’ LLpp = conic at = conic at pp11, p, p22 (to find them?ugly!) (to find them?ugly!)

• pp11, p, p22 tangent lines meet at ‘Pole point’ tangent lines meet at ‘Pole point’ xxpp

xxppLLppCC

pp11

pp22

Interesting, But Interesting, But why does book why does book

show it?show it?

Question: Can you apply DLT?Question: Can you apply DLT?• Choose pairs of lines that are known to be Choose pairs of lines that are known to be

perpendicular in perpendicular in worldworld space; measure those space; measure those lines in lines in imageimage space. space.

• Solve for Solve for C*C*’ ’ using L’using L’TT C*C*’’ m’ = 0 m’ = 0

• We also know that We also know that HHC*C*HHTT== C*C*’’Can you solve for H?Can you solve for H? Zisserman tried, but book method is Zisserman tried, but book method is confused, contains errors, and may be incorrect…confused, contains errors, and may be incorrect…

END.END.STOP HERESTOP HERE

x’x’

Apply the Apply the 3x3 matrix 3x3 matrix HH

x’ = Hx x’ = Hx

Recall: Projective Transform Recall: Projective Transform HH

xx33

xx22

xx11

xx

yy

x’x’

y’y’

(x,y)(x,y) xx

(x’,y’)(x’,y’)

Homog. coordsHomog. coords x’ =x’ = [x’,y’,1][x’,y’,1]TT

2D image 2D image (x’,y’)(x’,y’)

2D image 2D image (x,y)(x,y)

Homog. coords Homog. coords

[x,y,1][x,y,1]TT = x = x

x’x’

Apply the Apply the 3x3 matrix 3x3 matrix HH

x’ = Hx x’ = Hx

Recall: Projective Transform Recall: Projective Transform HH

xx33

xx22

xx11

xx

yy

x’x’

y’y’

(x,y)(x,y) xx

(x’,y’)(x’,y’)

Homog. coordsHomog. coords x’ =x’ = [x’,y’,1][x’,y’,1]TT

2D image 2D image (x’,y’)(x’,y’)

2D image 2D image (x,y)(x,y)

Homog. coords Homog. coords

[x,y,1][x,y,1]TT = x = x

Goal: Find H forGoal: Find H for‘‘Image Rectification’Image Rectification’

‘‘World Plane’World Plane’ ‘‘View Plane’View Plane’

Recall: Rectification Undo parts of HRecall: Rectification Undo parts of H

x’ = H xx’ = H x where where H = HH = HS S HHA A HHPP

GOAL: GOAL: Put world plane x’ into view plane xPut world plane x’ into view plane x– Affine Rect.; Affine Rect.; (find only (find only HHPP (2DOF)); (2DOF)); – Metric Rect.;Metric Rect.; (find (find HHA A and and HHP P (6DOF));(6DOF));– Full Rect.;Full Rect.; (find all: (find all: HHSSHHAAHHPP (8DOF)); (8DOF));

METHODS:METHODS:1.1. Affine: ‘Vanishing Point’, Horizon line methodsAffine: ‘Vanishing Point’, Horizon line methods2.2. Metric: Conics & Circular PointsMetric: Conics & Circular Points3.3. Full: 4-point correspondence Full: 4-point correspondence

Recall: The bits and pieces of Recall: The bits and pieces of HHHH has 8 independent variables (DOF) has 8 independent variables (DOF)

• Computer Vision Jargon (2D projective):Computer Vision Jargon (2D projective):Isometry Isometry --3DOF(2D translate --3DOF(2D translate ttxx,t,tyy; 2D rotate ; 2D rotate zz; ); )

Similarity Similarity --4DOF (add uniform scale --4DOF (add uniform scale s;s;))AffineAffine --6DOF (add orientable scale --6DOF (add orientable scale ss,/s, s,/s, s/s/s))

ProjectiveProjective--8DOF (changes --8DOF (changes xx33; 3D-rotation-like); 3D-rotation-like)

Rectification “up to a Similarity” (finds Rectification “up to a Similarity” (finds HHPP, H, HAA))

Conics: Key IdeasConics: Key Ideas– Transformed conics tell you how Transformed conics tell you how HH changes angles changes angles

– one strange, degenerate conic one strange, degenerate conic C*C* measures angle measures angle

– Angles between line pairs yield transformed Angles between line pairs yield transformed C*C*

– We can compute We can compute HHAA and and HHP P from transformed from transformed C*C*


HHS S HHA A HHPP

C*C*== C*C*’’==11 0 0 0 0 0 1 0 0 1 0 0 0 00 0 0

LL MM mmLL

Undoing H: Metric Rect. 1Undoing H: Metric Rect. 1• C*C*’ =’ =

• First, ignore projective part: set v=0.First, ignore projective part: set v=0.• Choose two pairs of perpendicular lines L,mChoose two pairs of perpendicular lines L,m• Second, Use Second, Use LLTT’ ’ C*C*’ m’ = 0 ’ m’ = 0 to solve for to solve for KK

KKKKT T KvKv

vvTTKK ss


HHS S HHA A HHPPLL MM mmLL

KKKKTT 00

00 ssaka ‘Weak Perspective’aka ‘Weak Perspective’

Undoing H: Metric RectificationUndoing H: Metric Rectification• C*C*’’ has only 4DOFhas only 4DOF (before,after (before,after ANYANY H) H)

• All C*All C*’ ’ DOF are contained in (HDOF are contained in (HA A HHPP))

• SVD can convert C*SVD can convert C*’ to ’ to (H(HA A HHPP) matrix!) matrix!

• Task is then Task is then “find “find C*C*’’, then use it to find , then use it to find HH””


HHS S HHA A HHPP

C*C*== C*C*’’==

LL MM mmLL

00

11 0 0 0 0 0 1 0 0 1 0 0 0 00 0 0

Undoing H: Metric Rect. 1Undoing H: Metric Rect. 1• H = HH = HSS H HA A HHP P ==

• Book shows how to write transformed Book shows how to write transformed C*C*

as (simple, tedious):as (simple, tedious): C*C*’ =’ =

where where KK is 2x2 symmetric (affine part: 2DOF) is 2x2 symmetric (affine part: 2DOF) vv is 2x1 vector, (projective part: 2DOF) is 2x1 vector, (projective part: 2DOF)

ssRR tt

00TT 11

II 00

vvTT vv

KK 00

00TT 11

KKKKT T KvKv

vvTTKK 00

Undoing H: Metric Rect. 1Undoing H: Metric Rect. 1• H = HH = HSS H HA A HHP P ==

• Book shows how to write transformed Book shows how to write transformed C*C*

as (simple, tedious):as (simple, tedious): C*C*’ =’ =

where where KK is 2x2 symmetric (affine part: 2DOF) is 2x2 symmetric (affine part: 2DOF) vv is 2x1 vector, (projective part: 2DOF) is 2x1 vector, (projective part: 2DOF)

ssRR tt

00TT 11

II 00

vvTT vv

KK 00

00TT 11

KKKKT T KvKv

vvTTKK 00

!!!BUT DERIVATION & !!!BUT DERIVATION & RESULT IS WRONG!!! RESULT IS WRONG!!!

SKIP SKIP

‘‘Metric Rectification I’ pg 35,36!Metric Rectification I’ pg 35,36!

‘‘Let SVDs Explain it All for You’Let SVDs Explain it All for You’SVD finds H for you (?) SVD finds H for you (?) (pg 35) (pg 35)

Book’s Method for finding Book’s Method for finding HHAAHHPP from from C*C*’ :’ :

– C*C*’’ = H C*= H C* H HTT is symmetric; is symmetric;

– Assume SVD(Assume SVD(C*C*’) = USV’) = USVTT = ( = (U S ) I ( S VU S ) I ( S VTT ) ) = ( H ) C*= ( H ) C* ( H ( HTT) )

– BUT THIS DOESN’T WORK!BUT THIS DOESN’T WORK!

– ( Recall C*( Recall C* = ) = )

A = A = USVUSVT T ‘Find Input & Output Axes, Linked by Scale’‘Find Input & Output Axes, Linked by Scale’

11 0 0 0 0 0 1 0 0 1 0 0 0 00 0 0

Eigen -values,-vectors, Fixed pt & lineEigen -values,-vectors, Fixed pt & line

• Formalizes ‘invariant’ notion:Formalizes ‘invariant’ notion:– if x is ‘fixed’ for H, then Hx only scales xif x is ‘fixed’ for H, then Hx only scales x

H x =H x = xx (( is a constant scale is a constant scale factor)factor)

– xx is an ‘eigenvector’, is an ‘eigenvector’, is its ‘eigenvalue’is its ‘eigenvalue’– again, SVD helps you find them.again, SVD helps you find them.

• Elaborate topic (but not hard). Skip for now.Elaborate topic (but not hard). Skip for now.


Book Errata Website:Book Errata Website:http://www.robots.ox.ac.uk/~az/HZbook/HZerrata.htmlhttp://www.robots.ox.ac.uk/~az/HZbook/HZerrata.html

pg. 35, eqn. 1.22 is dubious / erroneous;pg. 35, eqn. 1.22 is dubious / erroneous; C*C*’ = HC*’ = HC*HHTT = (H = (HSSHHAAHHPP) C*) C* (H(HSSHHAAHHPP))TT

is the transformed line conic, but book uses is the transformed line conic, but book uses C* C*’ = (H’ = (HPPHHAAHHSS) C*) C* (H(HPPHHAAHHSS))TT instead, AND has error: instead, AND has error:

==But I get: But I get:

C*C*’ = ’ =

KKKKT T KvKv

vvTTKK ??

(book: (book: ?=0?=0; ; errata website: errata website: ??

=v=vTTKKKKTTvv))KKKKT T KKKKTTvv

vvTTKKKKTT vvTTKKKKTTvv

http://www.robots.ox.ac.uk/~az/HZbook/HZerrata.html

Dubious Book Aggravations:Dubious Book Aggravations:– Similarity transform Similarity transform HHSS leaves leaves C*C* unchanged:unchanged:

C*C* == HHSS C* C* H HSST T

– HH transforms transforms C*C* to image space to image space C*C*’’ by: by:

C*C*’’ == H C*H C* H HTT = ( = (HHSS H HA A HHPP) C*) C* ((HHSS H HA A HHPP))TT

– Book claims to derive Book claims to derive

C*C*’ = ’ = (H(HP P HHAA) C*) C* ((HHP P HHAA))T T (!?!?!)(!?!?!)

Then errs in simplifying to:Then errs in simplifying to:

11 0 0 0 0 0 1 0 0 1 0 0 0 00 0 0

KKKKT T KvKv

vvTTKK 00**

SVDs and ConicsSVDs and Conics• Conics (both C and C*) are symmetric;Conics (both C and C*) are symmetric;• SVD of any symmetric A is also symmetric:SVD of any symmetric A is also symmetric:

SVD(A) = USUSVD(A) = USUTT !Not Always! !Not Always!• Find conics’ singular values’ sign: Find conics’ singular values’ sign:

ssii =0,1, or –1, to classify conic type: =0,1, or –1, to classify conic type: (pg 40)(pg 40)

SSii values | Equation | Type . values | Equation | Type . (1, 1, 1)(1, 1, 1)

(1, 1,-1)(1, 1,-1)(1, 1, 0)(1, 1, 0)(1,-1, 0)(1,-1, 0)(1, 0, 0)(1, 0, 0)

xx22 + y + y22 + w + w22 = 0 = 0xx22 + y + y22 - w - w22 = 0 = 0 xx22 + y + y22 = 0 = 0xx22 - y - y22 = 0 = 0xx2 2 = 0 = 0

imaginary-only imaginary-only circle/ellipse circle/ellipse single real point (0,0,1)single real point (0,0,1)2 lines: x+/- y2 lines: x+/- y2 co-located lines: x=0 2 co-located lines: x=0

•(recall)(recall)H = H = HHSS H HA A HHP P ==

• Tedious algebraTedious algebra** (pg35)(pg35) shows symmetry: shows symmetry:

C*C*’ ’ == H C*H C* H HT T == where where KK is 2x2 symmetric ( is 2x2 symmetric (affine partaffine part: 2DOF): 2DOF)

vv is 2x1 vector, ( is 2x1 vector, (projective partprojective part: 2DOF): 2DOF)• ?But what do ?But what do KK and and VV really control? really control?

KKKKT T KvKv

vvTTKK 00**

K K 00

00T T 11

Undoing H: Metric RectificationUndoing H: Metric Rectification

ssR R tt

00T T 11

I I 00 vvT T vv

*==error?!?*==error?!?

•(recall)(recall)H = H = HHSS H HA A HHP P ==

• Tedious algebra Tedious algebra (pg35)(pg35) shows symmetry: shows symmetry:

C*C*’ ’ == H C*H C* H HT T == where where KK is 2x2 symmetric ( is 2x2 symmetric (affine partaffine part: 2DOF): 2DOF)

vv is 2x1 vector, ( is 2x1 vector, (projective partprojective part: 2DOF): 2DOF)• ?But what do ?But what do KK and and VV really control? really control?

KKKKT T KvKv

vvTTKK ??

K K 00

00T T 11

Undoing H: Metric RectificationUndoing H: Metric Rectification

ssR R tt

00T T 11

I I 00 vvT T vv

(book: (book: ?=0?=0; ; errata website: errata website: ??

=v=vTTKKKKTTvv))

Undoing H: Metric RectificationUndoing H: Metric Rectification• OK, then how do we find OK, then how do we find KK and and vv??

• (book: (book: ?=0?=0; ; errata website: errata website: ?=v?=vTTKKKKTTvv))• Choose known-perpendicular line pairsChoose known-perpendicular line pairs ((LLii, , mmii)),,

then compute by:then compute by:– Method 1a: Method 1a: (pg 36)(pg 36) Assume v=0, solve for KAssume v=0, solve for K– Method 1b: Method 1b: (NOT in book)(NOT in book)Assume K=I, solve for v Assume K=I, solve for v – Method 2:Method 2: Rearrange, solve for full Rearrange, solve for full C*C*’’

then get Hthen get HAAHHPP using SVD. using SVD.

KKKKTT KvKv

vvTTKK ??C*C*’ ’ == H C*H C* H HT T ==

Method 1a: Method 1a: (pg 36)(pg 36) Assume v=0, solve for KAssume v=0, solve for K

• Choose 2 Choose 2 line pairs ( line pairs (LLaa,m,maa) and () and (LLbb,m,mbb))

• Both pairs satisfy Both pairs satisfy LLT T C*C*’’ m = 0 m = 0 or: or:• (note x(note x33 term is ignored!) term is ignored!)

• ‘‘flatten’ to one equation:flatten’ to one equation:

Undoing H: Metric Rect. 1aUndoing H: Metric Rect. 1a

KKKKT T KvKv

vvTTKK 00C*C*’ ’ == H C*H C* H HT T = =

KKKKT T 00

00 00

s1 s2 0s1 s2 0s2 s3 0s2 s3 0

0 00 0 00

mm11 mm22 mm33

ll11 l l22 l l33 =0=0

ll11mm11 l l22mm11+l+l11mm22 l l22mm22

ss11 ss22 ss33

= 0= 0

Method 1a: Method 1a: (pg 36)(pg 36) Assume v=0, solve for KAssume v=0, solve for K

• ‘‘Stack’ to combine both line pairs;Stack’ to combine both line pairs;

• Solve for Solve for ss by SVD: find ‘input null space’ ( by SVD: find ‘input null space’ (Ax=0Ax=0) ) • From From ss make make C*C*’’ = = KKKKTT, then extract , then extract HHAAHHPP using using

SVD: SVD: recall recall C*C*’ ’ == H C*H C* H HTT , , it is symmetric…it is symmetric…

Undoing H: Metric Rect. 1aUndoing H: Metric Rect. 1a

KKKKT T KvKv


s1 s2 0s1 s2 0s2 s3 0s2 s3 0

0 00 0 00

lla1a1mma1a1 l la2a2mma1a1+l+la1a1mma2a2 l la2a2mma2a2ss11 ss22 ss33

= 0= 0llb1b1mmb1b1 l lb2b2mmb1b1+l+lb1b1mmb2b2 l lb2b2mmb2b2

Method 1b: Method 1b: (not in book)(not in book) Assume K=I, solve for vAssume K=I, solve for v

• Choose 2 Choose 2 line pairs ( line pairs (LLaa,m,maa) and () and (LLbb,m,mbb))

• These satisfy These satisfy LLT T C*C*’’ m = 0 m = 0 or: or:

• ‘‘flatten’ to one equation flatten’ to one equation (messy)(messy), stack, solve for , stack, solve for vv11,v,v22

• Extract Extract HHpp from from C*C*’ ’ using SVDusing SVD

Undoing H: Metric Rect. 1bUndoing H: Metric Rect. 1b

KKKKT T KvKv


I I vv

vvTT 00

1 0 v11 0 v10 1 v20 1 v2v1 v2v1 v2 ??

mm11 mm22 mm33

ll11 l l22 l l33 =0=0

Documents

CS 395/495-26: Spring 2004