projective geometry
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Projective Geometry. Lecture slides by Steve Seitz (mostly) Lecture presented by Rick Szeliski. Final project ideas. Discussion by Steve Seitz and Rick Szeliski …. Projective geometry. Readings - PowerPoint PPT PresentationTRANSCRIPT
CSE 576, Spring 2008 Projective Geometry 1
Projective Geometry
Lecture slides by Steve Seitz (mostly)Lecture presented by Rick Szeliski
CSE 576, Spring 2008 Projective Geometry 2
Final project ideasDiscussion by Steve Seitz and Rick Szeliski
…
CSE 576, Spring 2008 Projective Geometry 3
Projective geometry
Readings• Mundy, J.L. and Zisserman, A., Geometric Invariance in Computer Vision, Appendix:
Projective Geometry for Machine Vision, MIT Press, Cambridge, MA, 1992, (read 23.1 - 23.5, 23.10)
– available online: http://www.cs.cmu.edu/~ph/869/papers/zisser-mundy.pdf
Ames Room
CSE 576, Spring 2008 Projective Geometry 4
Projective geometry—what’s it good for?
Uses of projective geometry• Drawing• Measurements• Mathematics for projection• Undistorting images• Focus of expansion• Camera pose estimation, match move• Object recognition
CSE 576, Spring 2008 Projective Geometry 5
Applications of projective geometry
Vermeer’s Music Lesson
Reconstructions by Criminisi et al.
CSE 576, Spring 2008 Projective Geometry 6
1 2 3 4
1
2
3
4
Measurements on planes
Approach: unwarp then measure
What kind of warp is this?
CSE 576, Spring 2008 Projective Geometry 7
Image rectification
To unwarp (rectify) an image• solve for homography H given p and p’• solve equations of the form: wp’ = Hp
– linear in unknowns: w and coefficients of H
– H is defined up to an arbitrary scale factor
– how many points are necessary to solve for H?
pp’
work out on board
CSE 576, Spring 2008 Projective Geometry 8
Solving for homographies
CSE 576, Spring 2008 Projective Geometry 9
Solving for homographies
A h 0
Defines a least squares problem:
2n × 9 9 2n
• Since h is only defined up to scale, solve for unit vector ĥ• Solution: ĥ = eigenvector of ATA with smallest eigenvalue• Works with 4 or more points
CSE 576, Spring 2008 Projective Geometry 10
(0,0,0)
The projective planeWhy do we need homogeneous coordinates?
• represent points at infinity, homographies, perspective projection, multi-view relationships
What is the geometric intuition?• a point in the image is a ray in projective space
(sx,sy,s)
• Each point (x,y) on the plane is represented by a ray (sx,sy,s)– all points on the ray are equivalent: (x, y, 1) (sx, sy, s)
image plane
(x,y,1)-y
x-z
CSE 576, Spring 2008 Projective Geometry 11
Projective linesWhat does a line in the image correspond to in
projective space?
• A line is a plane of rays through origin– all rays (x,y,z) satisfying: ax + by + cz = 0
z
y
x
cba0 :notationvectorin
• A line is also represented as a homogeneous 3-vector ll p
CSE 576, Spring 2008 Projective Geometry 12
l
Point and line duality• A line l is a homogeneous 3-vector• It is to every point (ray) p on the line: l p=0
p1p2
What is the intersection of two lines l1 and l2 ?
• p is to l1 and l2 p = l1 l2
Points and lines are dual in projective space• given any formula, can switch the meanings of points and
lines to get another formula
l1
l2
p
What is the line l spanned by rays p1 and p2 ?
• l is to p1 and p2 l = p1 p2
• l is the plane normal
CSE 576, Spring 2008 Projective Geometry 13
Ideal points and lines
Ideal point (“point at infinity”)• p (x, y, 0) – parallel to image plane• It has infinite image coordinates
(sx,sy,0)-y
x-z image plane
Ideal line• l (a, b, 0) – parallel to image plane
(a,b,0)-y
x-z image plane
• Corresponds to a line in the image (finite coordinates)– goes through image origin (principle point)
CSE 576, Spring 2008 Projective Geometry 14
Homographies of points and linesComputed by 3x3 matrix multiplication
• To transform a point: p’ = Hp• To transform a line: lp=0 l’p’=0
– 0 = lp = lH-1Hp = lH-1p’ l’ = lH-1
– lines are transformed by postmultiplication of H-1
CSE 576, Spring 2008 Projective Geometry 15
3D projective geometryThese concepts generalize naturally to 3D
• Homogeneous coordinates– Projective 3D points have four coords: P = (X,Y,Z,W)
• Duality– A plane N is also represented by a 4-vector
– Points and planes are dual in 3D: N P=0
• Projective transformations– Represented by 4x4 matrices T: P’ = TP, N’ = N T-1
CSE 576, Spring 2008 Projective Geometry 16
3D to 2D: “perspective” projection
Matrix Projection: ΠPp
1************
ZYX
wwywx
What is not preserved under perspective projection?
What IS preserved?
CSE 576, Spring 2008 Projective Geometry 17
Vanishing points
image plane
line on ground plane
vanishing point v
Vanishing point• projection of a point at infinity
cameracenter
C
CSE 576, Spring 2008 Projective Geometry 18
Vanishing points
Properties• Any two parallel lines have the same vanishing point v• The ray from C through v is parallel to the lines• An image may have more than one vanishing point
– in fact every pixel is a potential vanishing point
image plane
cameracenter
C
line on ground plane
vanishing point v
line on ground plane
CSE 576, Spring 2008 Projective Geometry 19
Vanishing lines
Multiple Vanishing Points• Any set of parallel lines on the plane define a vanishing point• The union of all of vanishing points from lines on the same
plane is the vanishing line– For the ground plane, this is called the horizon
v1 v2
CSE 576, Spring 2008 Projective Geometry 20
Vanishing lines
Multiple Vanishing Points• Different planes define different vanishing lines
CSE 576, Spring 2008 Projective Geometry 21
Computing vanishing points
V
DPP t 0
1ZZ
YY
XX
t tDP
tDP
tDP
P
P0
D
CSE 576, Spring 2008 Projective Geometry 22
0/1
/
/
/
1Z
Y
X
ZZ
YY
XX
ZZ
YY
XX
t D
D
D
t
t
DtP
DtP
DtP
tDP
tDP
tDP
PP
Computing vanishing points
Properties• P is a point at infinity, v is its projection
• They depend only on line direction
• Parallel lines P0 + tD, P1 + tD intersect at P
V
DPP t 0
ΠPv
P0
D
CSE 576, Spring 2008 Projective Geometry 23
Computing the horizon
Properties• l is intersection of horizontal plane through C with image plane
• Compute l from two sets of parallel lines on ground plane
• All points at same height as C project to l– points higher than C project above l
• Provides way of comparing height of objects in the scene
ground plane
lC
CSE 576, Spring 2008 Projective Geometry 24
CSE 576, Spring 2008 Projective Geometry 25
Fun with vanishing points
CSE 576, Spring 2008 Projective Geometry 26
Perspective cues
CSE 576, Spring 2008 Projective Geometry 27
Perspective cues
CSE 576, Spring 2008 Projective Geometry 28
Perspective cues
CSE 576, Spring 2008 Projective Geometry 29
Comparing heights
VanishingVanishingPointPoint
CSE 576, Spring 2008 Projective Geometry 30
Measuring height
1
2
3
4
55.4
2.8
3.3
Camera height
What is the height of the camera?
CSE 576, Spring 2008 Projective Geometry 31
q1
Computing vanishing points (from lines)
Intersect p1q1 with p2q2
v
p1
p2
q2
Least squares version• Better to use more than two lines and compute the “closest” point of
intersection
• See notes by Bob Collins for one good way of doing this:– http://www-2.cs.cmu.edu/~ph/869/www/notes/vanishing.txt
CSE 576, Spring 2008 Projective Geometry 32
C
Measuring height without a ruler
ground plane
Compute Z from image measurements• Need more than vanishing points to do this
Z
CSE 576, Spring 2008 Projective Geometry 33
The cross ratio
A Projective Invariant• Something that does not change under projective transformations
(including perspective projection)
P1
P2
P3
P4
1423
2413
PPPP
PPPP
The cross-ratio of 4 collinear points
Can permute the point ordering• 4! = 24 different orders (but only 6 distinct values)
This is the fundamental invariant of projective geometry
1i
i
i
i Z
Y
X
P
3421
2431
PPPP
PPPP
CSE 576, Spring 2008 Projective Geometry 34
vZ
r
t
b
tvbr
rvbt
Z
Z
image cross ratio
Measuring height
B (bottom of object)
T (top of object)
R (reference point)
ground plane
HC
TBR
RBT
scene cross ratio
1
Z
Y
X
P
1
y
x
pscene points represented as image points as
R
H
R
H
R
CSE 576, Spring 2008 Projective Geometry 35
Measuring height
RH
vz
r
b
t
R
H
Z
Z
tvbr
rvbt
image cross ratio
H
b0
t0
vvx vy
vanishing line (horizon)
CSE 576, Spring 2008 Projective Geometry 36
Measuring height vz
r
b
t0
vx vy
vanishing line (horizon)
v
t0
m0
What if the point on the ground plane b0 is not known?
• Here the guy is standing on the box, height of box is known
• Use one side of the box to help find b0 as shown above
b0
t1
b1
CSE 576, Spring 2008 Projective Geometry 37
Computing (X,Y,Z) coordinates
CSE 576, Spring 2008 Projective Geometry 38
3D Modeling from a photograph
CSE 576, Spring 2008 Projective Geometry 39
Camera calibrationGoal: estimate the camera parameters
• Version 1: solve for projection matrix
ΠXx
1************
ZYX
wwywx
• Version 2: solve for camera parameters separately– intrinsics (focal length, principle point, pixel size)
– extrinsics (rotation angles, translation)
– radial distortion
CSE 576, Spring 2008 Projective Geometry 40
Vanishing points and projection matrix
************
Π 4321 ππππ
1π 2π 3π 4π
T00011 Ππ = vx (X vanishing point)
Z3Y2 , similarly, vπvπ
origin worldof projection10004 TΠπ
ovvvΠ ZYXNot So Fast! We only know v’s and o up to a scale factor
ovvvΠ dcba ZYX• Need a bit more work to get these scale factors…
CSE 576, Spring 2008 Projective Geometry 41
Finding the scale factors…Let’s assume that the camera is reasonable
• Square pixels• Image plane parallel to sensor plane• Principal point in the center of the image
1000
100
010
001
1000
0
0
0
0100
0010
0001
/100
010
001
3
2
1
333231
232221
131211
t
t
t
rrr
rrr
rrr
f
Π
ftfrfrfr
trrr
trrr
/'///
'
'
3333231
2232221
1131211
ovvv dcba ZYX
3
2
1
333231
232221
131211
3
2
1
'
'
'
t
t
t
rrr
rrr
rrr
t
t
t
CSE 576, Spring 2008 Projective Geometry 42
Solving for f
dtcrbrar
dtcrbrar
dtcrbrar
fofvfvfv
ovvv
ovvv
zyx
zyx
zyx
/'///
/'///
/'///
3333231
2232221
1131211
3333
2222
1111
Orthogonal vectors
Orthogonal vectors
dftcfrbfrafr
dtcrbrar
dtcrbrar
ovvv
ovvv
ovvv
zyx
zyx
zyx
/'///
/'///
/'///
3333231
2232221
1131211
3333
2222
1111
ovvv ZYX
0332
2211 yxyxyx vvfvvvv33
2211
yx
yxyx
vv
vvvvf
CSE 576, Spring 2008 Projective Geometry 43
Solving for a, b, and c
dtcrbrar
dtcrbrar
dtcrbrar
ovvv
fofvfvfv
fofvfvfv
zyx
zyx
zyx
/'///
/'///
/'///
////
////
3333231
2232221
1131211
3333
2222
1111
Norm = 1/a
Norm = 1/a
Solve for a, b, c• Divide the first two rows by f, now that it is known• Now just find the norms of the first three columns• Once we know a, b, and c, that also determines R
How about d?• Need a reference point in the scene
CSE 576, Spring 2008 Projective Geometry 44
Solving for d
Suppose we have one reference height H• E.g., we known that (0, 0, H) gets mapped to (u, v)
1
0
0
/'
/'
/'
3333231
2232221
1131211
Hdtrrr
dtrrr
dtrrr
w
wv
wu
dtHr
dtHru
/'
/'
333
113
HrHur
uttd
1333
31 ''
Finally, we can solve for t
3
2
1
333231
232221
131211
3
2
1
'
'
'T
t
t
t
rrr
rrr
rrr
t
t
t
CSE 576, Spring 2008 Projective Geometry 45
Calibration using a reference object
Place a known object in the scene• identify correspondence between image and scene• compute mapping from scene to image
Issues• must know geometry very accurately• must know 3D->2D correspondence
CSE 576, Spring 2008 Projective Geometry 46
Chromaglyphs
Courtesy of Bruce Culbertson, HP Labshttp://www.hpl.hp.com/personal/Bruce_Culbertson/ibr98/chromagl.htm
CSE 576, Spring 2008 Projective Geometry 47
Estimating the projection matrix
Place a known object in the scene• identify correspondence between image and scene• compute mapping from scene to image
CSE 576, Spring 2008 Projective Geometry 48
Direct linear calibration
CSE 576, Spring 2008 Projective Geometry 49
Direct linear calibration
Can solve for mij by linear least squares• use eigenvector trick that we used for homographies
CSE 576, Spring 2008 Projective Geometry 50
Direct linear calibration
Advantage:• Very simple to formulate and solve
Disadvantages:• Doesn’t tell you the camera parameters• Doesn’t model radial distortion• Hard to impose constraints (e.g., known focal length)• Doesn’t minimize the right error function
For these reasons, nonlinear methods are preferred• Define error function E between projected 3D points and image positions
– E is nonlinear function of intrinsics, extrinsics, radial distortion
• Minimize E using nonlinear optimization techniques
– e.g., variants of Newton’s method (e.g., Levenberg Marquart)
CSE 576, Spring 2008 Projective Geometry 51
Alternative: multi-plane calibration
Images courtesy Jean-Yves Bouguet, Intel Corp.
Advantage• Only requires a plane• Don’t have to know positions/orientations• Good code available online!
– Intel’s OpenCV library: http://www.intel.com/research/mrl/research/opencv/
– Matlab version by Jean-Yves Bouget: http://www.vision.caltech.edu/bouguetj/calib_doc/index.html
– Zhengyou Zhang’s web site: http://research.microsoft.com/~zhang/Calib/
CSE 576, Spring 2008 Projective Geometry 52
Some Related TechniquesImage-Based Modeling and Photo Editing
• Mok et al., SIGGRAPH 2001• http://graphics.csail.mit.edu/ibedit/
Single View Modeling of Free-Form Scenes• Zhang et al., CVPR 2001• http://grail.cs.washington.edu/projects/svm/
Tour Into The Picture• Anjyo et al., SIGGRAPH 1997• http://koigakubo.hitachi.co.jp/little/DL_TipE.html