3CG 2006 1 / 1

B4 Computational Geometry

David Murray

david.murray@eng.ox.ac.ukwww.robots.ox.ac.uk/∼dwm/Courses/3CG

Michaelmas 2006

3CG 2006 2 / 1

Vanishing points and horizons. Applications ofprojective transformations

Lecture 1: Euclidean, similarity, affine and projectivetransformations. Homogeneous coordinates and matrices.Coordinate frames.

Lecture 2Perspective projection and its matrix representation.Vanishing points.Applications of projective transformations.

Lecture 3: Convexity of point-sets, convex hull and algorithms.Conics and quadrics, implicit and parametric forms, computationof intersections.Lecture 4: Bezier curves, B-splines. Tensor-product surfaces.

3CG 2006 3 / 1

Recall: Perspective (central) projection — 3D to 2D

The camera modelMathematical idealized camera3D→ 2D• Image coordinates xy• Camera frame XYZ (origin atoptical centre)• Focal length f , image plane isat Z = f .Similar triangles

xf=

XZ

yf=

YZ

or x = fXZ

where x and X are 3-vectors, withx = (x , y , f )>, X = (X ,Y ,Z )>.

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Vanishing Points

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Vanishing Points in 1D and 2D

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All parallel lines meet at the same vanishing point

A line of 3D points isrepresentedas

X(λ) = A + λD

Using x = fX/Z the vanishingpoint of its image is

v = limλ→±∞

x(λ) = fA + λD

AZ + λDZ

= fD

DZ= f

DX/DZDY/DZ

1

• v depends only on thedirection D, not on A.• Parallel lines have the samevanishing point.

3CG 2006 7 / 1

The Homogeneous 3× 4 Projection Matrix

x = fXZ

Choose f = 1 from now on.

Homogeneous image coordinates (x1, x2, x3)> correctly represent

x = X/Z if

x1x2x3

P=

1 0 0 00 1 0 00 0 1 0

XYZ1

= [I | 0](

X1

)

because thenx =

x1

x3=

XZ

y =x2

x3=

YZ

Then perspective projection is a linear map, represented by a 3× 4projection matrix, from 3D to 2D.

3CG 2006 8 / 1

Vanishing points using homogeneous notation

A line of points in 3D through the point A with direction D is

X(µ) P= A + µD

Writing this in homogeneous notationX1(µ)X2(µ)X3(µ)X4(µ)

P=

(A1

)+ µ

(D0

)P=

1µ

(A1

)+

(D0

)

In the limit µ→∞ the point on the line is

(D0

)So, homogeneous vectors with X4 = 0 represent points “atinfinity”.Points at infinity are equivalent to directions

3CG 2006 9 / 1

Vanishing points using homogeneous notation: EgThe vanishing point of a line with direction D is the image of thepoint at∞

v =

1 0 0 00 1 0 00 0 1 0

( D0

)=

DXDYDZ

Exercise: Compute the vanishing points of lines on an XZplane:(1) parallel to the Z axis; (2) at 45◦ to the Z axix; (3) parallel tothe X axis.

1:

0010

→ 0

01

2:

1010

→ 1

01

3:

1000

→ 1

00

3CG 2006 10 / 1

The advantages of homogeneous notation ...

There are two advantages of using homogeneous notation torepresent perspective projection:

1 Non-linear projections equations are turned into linear equations.The tools of linear algebra can then be used.

2 Vanishing points are treated naturally, and awkward limitingprocedures are then avoided.

3CG 2006 11 / 1

Plane to Plane Projective Transformations

x1x2x3

= P

X1X2X3X4

Choose the world coordinate system such that the world plane haszero Z coordinate. Then the 3× 4 matrix P reduces to a 3× 3 planeto plane projective transformation.

x1x2x3

=

p11 p12 p13 p14p21 p22 p23 p24p31 p32 p33 p34

XY01

=

p11 p12 p14p21 p22 p24p31 p32 p34

XY1

• This is the most general transformation between the world planeand image plane under imaging by a perspective camera.• A plane-plane projective transformation is also called anhomography or a collineation.

3CG 2006 12 / 1

Computing planar transformations

How many pointcorrespondences are required tocompute a Euclideantransformation of the plane?

How many pointcorrespondences are required tocompute a similaritytransformation of the plane?

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Computing a projective transformation x ′1x ′2x ′3

P=

h11 h12 h13h21 h22 h23h31 h32 h33

x1x2x3

P= Hx

where H is a 3× 3 non-singularhomogeneous matrix withEIGHT dof.

Each point correspondence gives two constraints

x ′ =x ′1x ′3

=h11x + h12y + h13

h31x + h32y + h33, y ′ =

x ′2x ′3

=h21x + h22y + h23

h31x + h32y + h33

Multiplying out give two equations linear in the elements of H

x ′ (h31x + h32y + h33) = h11x + h12y + h13

y ′ (h31x + h32y + h33) = h21x + h22y + h23

3CG 2006 14 / 1

Simple ExampleSuppose the correspondences (x , y)↔ (x ′, y ′) are known for fourpoints (no three collinear), then H is determined uniquely.

1st correspondence(0,0)→ (0,0)

λ1

001

=

h11 h12 h13h21 h22 h23h31 h32 h33

001

=

h13h23h33

Whence h13 = h23 = 0.

2nd correspondence(1,0)→ (1,0)

λ2

101

=

h11 h12 0h21 h22 0h31 h32 h33

101

=

h11h21

h31 + h33

Whence

h21 = 0 and h11 = h31 + h33

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Simple example Continued

3rd correspondence (0,1)→ (0,1) gives h12 = 0 andh22 = h32 + h33.4th correspondence (1,1)→ (2,1)

λ4

211

=

h31 + h33 0 00 h32 + h33 0

h31 h32 h33

111

=

h31 + h33h32 + h33

h31 + h32 + h33

Take ratios⇒ 2 equations in 3 unknowns⇒ solve for ratio of matrixelements only.

H = λ

2 0 00 1 00 −1 2

P=

2 0 00 1 00 −1 2

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Computational Algorithm

The equations,

x ′ (h31x + h32y + h33) = h11x + h12y + h13

y ′ (h31x + h32y + h33) = h21x + h22y + h23

can be rearranged as[x y 1 0 0 0 −x ′x −x ′y −x ′

0 0 0 x y 1 −y ′x −y ′y −y ′

]h = 0

where h = (h11,h12,h13,h21,h22,h23,h31,h32,h33)> is the matrix H

written as a 9-vector.

3CG 2006 17 / 1

continued ...

For 4 points,

x1 y1 1 0 0 0 −x ′1x1 −x ′1y1 −x ′10 0 0 x1 y1 1 −y ′1x1 −y ′1y1 −y ′1x2 y2 1 0 0 0 −x ′2x2 −x ′2y2 −x ′20 0 0 x2 y2 1 −y ′2x2 −y ′2y2 −y ′2x3 y3 1 0 0 0 −x ′3x3 −x ′3y3 −x ′30 0 0 x3 y3 1 −y ′3x3 −y ′3y3 −y ′3x4 y4 1 0 0 0 −x ′4x4 −x ′4y4 −x ′40 0 0 x4 y4 1 −y ′4x4 −y ′4y4 −y ′4

h = 0

which has the form Ah = 0, with A a 8× 9 matrix. The solution h isthe (one dimensional) null space of A.If using many points, one can use least squares. Solution best foundthen using SVD of A — ie USV> ← A Then h is the column of Vcorresponding to smallest singular value. (The smallest singularvalue would be zero of all the data were exact ...)

3CG 2006 18 / 1

Some Matlabnpoints = 4 (or 5 later -- 5th point is noisy)x = [0,1,0,1, 1.01]; y = [0,0,1,1, 0.99];xd = [0,1,0,2, 2.01]; yd = [0,0,1,1, 1.01];A = zeros(2*npoints,9);for i=1:npoints,A(2*i-1,:)= [x(i),y(i),1,0,0,0, -x(i)*xd(i),-xd(i)*y(i),-xd(i)];A(2*i, :)= [0,0,0,x(i),y(i),1, -x(i)*yd(i),-yd(i)*y(i),-yd(i)];

end;if npoints==4

h = null(A);else

[U,S,V] = svd(A);h=V(:,9);

end;H=[h(1),h(2),h(3);h(4),h(5),h(6);h(7),h(8),h(9);]

With the 4 exact points ...H = 0.6325 −0.0000 0.0000

0.0000 0.3162 −0.00000.0000 −0.3162 0.6325

Adding the noisy point ...H = 0.6295 −0.0000 −0.0000−0.0001 0.3188 0.0001−0.0050 −0.3155 0.6344

3CG 2006 19 / 1

♣ 1: Removing Perspective DistortionObjective: Back project to world plane1. Find Euclidean coordinates of 4 points on the flat object plane(xi , yi)

>.2. Measure the corresponding image coordinates of these 4 points(x ′i , y

′i )>.

3. Compute H from the 4 (xi , yi)> ↔ (x ′i , y

′i )>.

4. Euclidean coords of any image point are(x , y ,1)> = H−1(x ′, y ′,1)>.

The image can be warped onto the world plane using H. How?

3CG 2006 20 / 1

♣ 2: Mosaicing from rotated image planes

An image is the intersection of a plane with the cone of raysbetween points in 3-space and the optical centre. Any two such“images” (with the same optical centre) are related by a planarprojective transformation.

X

x

x

1

2

As the camera is rotated the points of intersection of the rays withthe image plane are related by a planar projective transformation.Image points x1 and x2 correspond to the same scene point X.

3CG 2006 21 / 1

Relationship in terms of rotationsFor corresponding points x1 and x2 in two views 1 and 2,

x1P=

1 0 0 00 1 0 00 0 1 0

[ R1 00> 1

]XYZ1

= R1

XYZ

x2P= R2

XYZ

Hence

x2 = R2R−11 x1

The cameras could have different focal lengths — so one can do all ofthis while rotating and zooming. Then

x1P= K 1R1

(XYZ

)x2

P= K 2R2

(XYZ

)⇒ x2 = K 2R2R−1

1 K−11 x1

where in the simplest case

K 1 =

f1 0 00 f1 00 0 1

3CG 2006 22 / 1

♣ 2: MosaicsInput sequence: 30 frames, hand held camcorder

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♣ 2: Mosaics

Eight images (out of 30) acquired by rotating a camcorder about itsoptical centre.

Register all the images to one reference image by projectivetransformations.

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♣ 2: Keble Panoramic Mosaic

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♣ 3: Synthetic Rotations

Original image Warped: floor tile square Warped: door squareThe synthetic images are produced by projectively warping theoriginal image so that four corners of an imaged rectangle mapto the corners of a rectangle.Both warpings correspond to a synthetic rotation of the cameraabout the (fixed) camera centre.

3CG 2006 26 / 1

♣ 4: Projective pointing devices: modern DürerHere the rotation joint of apan-tilt camera become theprojection centre, and trackingpeople in the ground planeproduces a track on a notionalfrontal plane — like Durer’smarks. These frontal planetracks are then converted inCartesian tracks viewed fromabove.

3CG 2006 27 / 1

Video Example

Camera tracking by motion nulling Recovered ground plane map

3CG 2006 28 / 1

Summary

We have looked at four classes of transformation (in 2D):

Euclidean: 3 DOF

r11 r12 txr21 r22 ty0 0 1

Similarity: 4 DOF

sr11 sr12 txsr21 sr22 ty0 0 1

Affine: 6 DOF

a11 a12 txa21 a22 ty0 0 1

Projective: 8 DOF

h11 h12 h13h21 h22 h23h31 h32 h33

and have developed their 3D counterparts.