Multi-view geometry. Multi-view geometry problems Structure: Given projections of the same 3D point in two or more images, compute the 3D coordinates.

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Presentation transcript:

Multi-view geometry

Multi-view geometry problems Structure: Given projections of the same 3D point in two or more images, compute the 3D coordinates of that point Camera 3 R 3,t 3 Slide credit: Noah Snavely ? Camera 1 Camera 2 R 1,t 1 R 2,t 2

Multi-view geometry problems Stereo correspondence: Given a point in one of the images, where could its corresponding points be in the other images? Camera 3 R 3,t 3 Camera 1 Camera 2 R 1,t 1 R 2,t 2 Slide credit: Noah Snavely

Multi-view geometry problems Motion: Given a set of corresponding points in two or more images, compute the camera parameters Camera 1 Camera 2 Camera 3 R 1,t 1 R 2,t 2 R 3,t 3 ? ? ? Slide credit: Noah Snavely

Structure: Triangulation Given projections of a 3D point in two or more images (with known camera matrices), find the coordinates of the point O1O1 O2O2 x1x1 x2x2 X?

Structure: Triangulation We want to intersect the two visual rays corresponding to x 1 and x 2, but because of noise and numerical errors, they don’t meet exactly O1O1 O2O2 x1x1 x2x2 X?

Triangulation: Geometric approach Find shortest segment connecting the two viewing rays and let X be the midpoint of that segment O1O1 O2O2 x1x1 x2x2 X

Triangulation: Linear approach Cross product as matrix multiplication:

Triangulation: Linear approach Two independent equations each in terms of three unknown entries of X

Triangulation: Nonlinear approach Find X that minimizes O1O1 O2O2 x1x1 x2x2 X?X? P1XP1X P2XP2X

Two-view geometry

Epipolar Plane – plane containing baseline (1D family) Epipoles = intersections of baseline with image planes = projections of the other camera center = vanishing points of the baseline (motion direction) Baseline – line connecting the two camera centers Epipolar geometry X xx’

The Epipole Photo by Frank Dellaert

Epipolar Plane – plane containing baseline (1D family) Epipoles = intersections of baseline with image planes = projections of the other camera center = vanishing points of the baseline (motion direction) Epipolar Lines - intersections of epipolar plane with image planes (always come in corresponding pairs) Baseline – line connecting the two camera centers Epipolar geometry X xx’

Example: Converging cameras

Example: Motion parallel to image plane

Example: Motion perpendicular to image plane

Points move along lines radiating from the epipole: “focus of expansion” Epipole is the principal point

Epipolar constraint If we observe a point x in one image, where can the corresponding point x’ be in the other image? x x’ X

Potential matches for x have to lie on the corresponding epipolar line l’. Potential matches for x’ have to lie on the corresponding epipolar line l. Epipolar constraint x x’ X X X

Epipolar constraint example

X xx’ Epipolar constraint: Calibrated case Assume that the intrinsic and extrinsic parameters of the cameras are known We can multiply the projection matrix of each camera (and the image points) by the inverse of the calibration matrix to get normalized image coordinates We can also set the global coordinate system to the coordinate system of the first camera. Then the projection matrices of the two cameras can be written as [I | 0] and [R | t]

X xx’ = Rx+t Epipolar constraint: Calibrated case R t The vectors Rx, t, and x’ are coplanar = (x,1) T

Essential Matrix (Longuet-Higgins, 1981) Epipolar constraint: Calibrated case X xx’ The vectors Rx, t, and x’ are coplanar

X xx’ Epipolar constraint: Calibrated case E x is the epipolar line associated with x (l' = E x) E T x' is the epipolar line associated with x' (l = E T x') E e = 0 and E T e' = 0 E is singular (rank two) E has five degrees of freedom

Epipolar constraint: Uncalibrated case The calibration matrices K and K’ of the two cameras are unknown We can write the epipolar constraint in terms of unknown normalized coordinates: X xx’

Epipolar constraint: Uncalibrated case X xx’ Fundamental Matrix (Faugeras and Luong, 1992)

Epipolar constraint: Uncalibrated case F x is the epipolar line associated with x (l' = F x) F T x' is the epipolar line associated with x' (l' = F T x') F e = 0 and F T e' = 0 F is singular (rank two) F has seven degrees of freedom X xx’

The eight-point algorithm Minimize: under the constraint ||F|| 2 =1

The eight-point algorithm Meaning of error sum of squared algebraic distances between points x’ i and epipolar lines F x i (or points x i and epipolar lines F T x’ i ) Nonlinear approach: minimize sum of squared geometric distances

Problem with eight-point algorithm

Poor numerical conditioning Can be fixed by rescaling the data

The normalized eight-point algorithm Center the image data at the origin, and scale it so the mean squared distance between the origin and the data points is 2 pixels Use the eight-point algorithm to compute F from the normalized points Enforce the rank-2 constraint (for example, take SVD of F and throw out the smallest singular value) Transform fundamental matrix back to original units: if T and T’ are the normalizing transformations in the two images, than the fundamental matrix in original coordinates is T’ T F T (Hartley, 1995)

Comparison of estimation algorithms 8-pointNormalized 8-pointNonlinear least squares Av. Dist pixels0.92 pixel0.86 pixel Av. Dist pixels0.85 pixel0.80 pixel

From epipolar geometry to camera calibration Estimating the fundamental matrix is known as “weak calibration” If we know the calibration matrices of the two cameras, we can estimate the essential matrix: E = K’ T FK The essential matrix gives us the relative rotation and translation between the cameras, or their extrinsic parameters