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*Objects at infinity used in calibration Computing K from 1 image HZ8.8 IAC and K HZ8.5 Camera matrix...*

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Objects at infinity used in calibrationComputing K from 1 image

HZ8.8IAC and K

HZ8.5Cameramatrixfrom F

HZ9.5Objects atinfinity(w, Q*)

HZ3.5-3.7, 8.5Extractingcameraparameters

HZ6.2Cameramatrix

HZ6.1Calibrationusing Q*

HZ19.3Hartley 92

Thoughts about calibrationnow that we have motivated the necessity for moving from projective information about the camera to metric information, we consider calibration of the camera (Chapters 8 and 19)this requires an understanding of the absolute conic (Ch. 3.6-3.7) and the image of the absolute conic or IAC (Ch 8.5), as well as the absolute dual quadric Q* we will start with IAC and develop a method for finding the calibration matrix K using a single image (but also some calibration objects)later we will show how to develop the entire homography that corrects our camera matrix using IAC and Q* , using two views but no calibration objectsin short:stage 1: calibration from a single view, using IAC and calibration objectsstage 2: calibration from two views, using IAC and Q* without calibration objects

More thoughts about calibrationthe absolute conic -- and its dual, the absolute dual quadric Q* -- are structures that characterize a 3d similarity, like the dual conic did for the 2d similaritywe get interested in images of objects at infinity, since they depend only on camera rotation and calibration matrix KIAC, a special object at infinity, depends only on K, so it contains the key to solving for K in the calibration stageIAC and Q* allow angle to be measured in 3-spacegiven the relationship of IAC and Q* to angle, we can solve for them using angle constraints

Plane at infinity once again, we shall be fascinated by objects at infinityplane at infinity characterizes the affinityin projective 3-space, points at infinity (x,y,z,0) lie on the plane at infinity = (0,0,0,1)two lines (resp., two planes, a line and a plane) are parallel iff intersection lies on a projective transformation is an affinity iff is fixed (as a plane, not pointwise)i.e., this is a defining invariant of the affinityidentifies the excess: the difference between an affinity (12dof) and a projectivity (15dof) is exactly the 3dof of the plane at infinity intuition: once is known, parallelism is also known, which is a defining invariant of the affinitycan restore parallel lines (think of HW1)just as angle is a defining invariant of the similarityi.e., affine rectification involves restoring image of to its rightful positionHZ80-81

The image of we are interested in discovering the internal parameters of the camera such as focal length (or computationally in finding K)to isolate the internal camera parameters, we need to factor out the effects of camera translation and rotationwe get interested in the image of objects at infinity, since these images are independent of the position of the cameramakes sense, since points at infinity are directions, independent of positionthe homography between and the image plane is a planar homography H = KRproof: consider the image of an ideal point under the camera KR[I C]:image(d 0) = KR[I C] (d 0) = KRdhave eliminated the last columns translation vector

HZ209-210

Absolute conic absolute conic characterizes the similarityabsolute conic is 3D analogue to circular points (just as is 3D analogue of L )circular points are 0D subset of 1D line at infinityabsolute conic is 1D subset of 2D plane at infinityconic = plane/quadric intersectionabsolute conic = intersection of quadric x2 + y2 + z2 = 0 with plane at infinityequivalently, absolute conic is represented by the conic with identity matrix I on the plane at infinitya projective transformation is a similarity iff is fixed (as a set)proof: if A-t I A-1 = kI (so AAt = kI), then mapping A is scaled orthogonal, so represents scaled rotation (perhaps along with reflection)identifies the excess: difference between a similarity (7dof) and an affinity (12dof) is exactly the 5dof of HZ81-82

More on absolute conicmetric rectification involves restoring to its rightful position (Charles II)recall that every circle contains the circular points of its planeevery sphere contains any circle intersects in 2 points = circular points of the plane that contains the circleonce again, we can measure angle using (just as we could with dual of circular points)and it is invariant to projectivities; but angle measure with next construct Q* is better

The image of (aka or IAC)the image of was good, but the image of (a certain object on the plane at infinity) is even better: it is independent of both position and orientation of the camerathe image of the absolute conic is known by its acronym IAC, and since it is so valued for calibration, it is given its own symbol lowercase = uppercase well call it Williamcalibration is our end goal and often our last step (so omega) = (KKt)-1proof: we saw that the plane at infinity transforms according to the homography KR, so the conic I on the plane at infinity transforms to (KR)-t I (KR)-1 = K-t RR-1 K-1 = (KKt)-1since conics map contravariantly (HZ37) and R is orthogonal (Rt = R-1)conclusion: knowledge of implies knowledge of KHZ210

IAC and circular pointswe have seen that circular points lie on the absolute conic, so their image lies on Williamthis will be used to find William: if we can find enough points on William, we have William

every plane contains 2 circular pointsin the planes coordinate system, these are (1,i,0) and (1,-i,0)the plane intersects the plane at infinity in a line at infinity, which intersects the absolute dual conic in 2 points: these are the planes circular pointssince the planes circular points lie on the absolute conic , their images lie on the IAC how to find image of circular point?we shall find the image of a circular point on plane as H(1,+-i, 0) where H is the homography between and the image planeHZ82, 211

Calibrating with calibration objects: 3 squares yield 6 circular pointswe shall solve for William (the IAC) by finding 6 circular points that lie on Williamyou can find points on the vanishing line by intersecting parallel linesyou can find (imaged) circular points by finding a homography (to the image plane)each planar homography yields 2 pointsconsider 3 calibration squares in the image on 3 non-parallel planesassume wlog that the four vertices of each square are (0,0), (1,0), (0,1), (1,1)aligning the coordinate frame in this way is a similaritysimilarity does not affect circular pointsHZ211

3-square algorithmFor each calibration square [find 2 points on ]Find the corners of the calibration square using user input or automated tracking (see Davids excellent demo).Use the DLT algorithm to compute the homography H that maps each square vertex to its image [only 4 pts needed].Apply this homography to the circular points I and J to find 2 points on .they are h1 +- i h2, where h1 and h2 are first 2 columns of HFit to 6 points [using numerical computing]real and imaginary components of x^t x = 0 (applied to circular points) yield 2 eqns:h1t h2 = 0h1t h1 = h2t h2interesting: the pair of circular points encode the same information, so only one circular point gives useful information; but we still get 2dof because of real and imaginary components (so in effect we are not fitting to 6 points, but getting 6 constraints from 3 pts)encode these linear equations in Hw = 0 where w is the row-major encoded (see development of DLT algorithm)solve for w (and hence ) as null vectorExtract calibration matrix K from using Cholesky.-1 = KKt

HZ211

Overview of 3-square algorithmhomography of squares to imaged squares 6 imaged circular points (2 per square) image of absolute conic (by fitting) K (by Cholesky factorization) metric properties (see Chapter 19 for how to use K to get metric)