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Camera diagram. Camera matrix HZ6.1. Extracting camera parameters HZ6.2. Camera matrix from F HZ9.5. IAC HZ3.5-3.7, 8.5. IAC and K HZ8.5. Computing K from 1 image HZ8.8. Calibration using Q* HZ19.3 Hartley 92. Camera terminology.

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camera diagram
Camera diagram

Camera

matrix

HZ6.1

Extracting

camera

parameters

HZ6.2

Camera

matrix

from F

HZ9.5

IAC

HZ3.5-3.7, 8.5

IAC and K

HZ8.5

Computing K

from 1 image

HZ8.8

Calibration

using Q*

HZ19.3

Hartley 92

camera terminology
Camera terminology
  • a camera is defined by an optical center c and an image plane
  • image plane (or focal plane) is at distance f from the camera center
    • f is called the focal length
  • camera center = optical center
  • principal axis = the line through camera center orthogonal to image plane
  • principal point = intersection of principal axis with image plane
    • an indication of the camera center in the image
  • principal plane = the plane through camera center parallel to image plane
camera matrix
Camera matrix

Camera

matrix

HZ6.1

Extracting

camera

parameters

HZ6.2

Camera

matrix

from F

HZ9.5

IAC

HZ3.5-3.7, 8.5

IAC and K

HZ8.5

Computing K

from 1 image

HZ8.8

Calibration

using Q*

HZ19.3

Hartley 92

encoding in a camera matrix
Encoding in a camera matrix
  • the act of imaging is encoded by the camera matrix P, which is 3x4:
    • x = point in image = 3-vector
    • X = point in 3-space being imaged = 4-vector
    • PX = x
  • in an early lecture, we saw that perspective projection is a linear operation in projective space, which allows this encoding
  • we will develop the camera matrix in stages, generalizing as we go
camera matrix 1
Camera matrix 1
  • assumption set:
    • square pixels
    • origin of 3D world frame = camera center
    • z-axis of 3D world frame = principal axis
    • origin of 2D image space = principal point
  • P = diag(f,f,1)[I 0]
    • (X,Y,Z,1)  (fX, fY, Z)
camera matrix 2 general image space
Camera matrix 2 (general image space)
  • remove assumption #4: origin of 2D image space is arbitrary, so principal point is (px,py)
  • Euclidean space: (X,Y,Z)  (fX/Z + px, fY/Z + py)
  • projective space: (fX + Zpx, fY + Zpy, Z)
  • matrix: P = K[I 0] where K = (f 0 px; 0 f py; 0 0 1)
camera matrix 3 general world frame
Camera matrix 3 (general world frame)
  • remove assumptions #2 and #3: origin and z-axis of the world frame are arbitrary, or equivalently, the world frame has no explicit connection to the camera
  • freeing the origin from the camera center involves a translation C
  • freeing the z-axis from the principal axis involves a rotation R
  • let X be a point in world frame coordinates and Xcam be the point in camera frame coordinates (with origin and z-axis aligned with camera):
    • Euclidean: Xcam = R(X-C)
    • projective: Xcam = [R –RC; 0 1] X
  • so imaging process is x = PXcam = K[I 0] Xcam

= [K 0][R –RC; 0 1] X = [KR -KRC] X = KR[I –C] X

  • x = KR[I –C]X or P = KR[I –C]
  • HZ154-156 for last 3 slides
camera matrix 4 rectangular pixels
Camera matrix 4 (rectangular pixels)
  • remove assumption #1: pixels may no longer be square
  • cameras typically have non-square pixels
    • let mx = # of pixels / x-unit in image coordinates
    • let my = # of pixels / y-unit in image coordinates
    • e.g., mx = 1.333 and my = 1 if pixels are wider
  • XcamRect = diag (mx, my, 1) XcamSquare = diag(mx,my,1) [R –RC; 0 1] X
  • still write P = KR [I –C], but now the calibration matrix is
    • K = diag(mx,my,1) [f 0 px; 0 f py; 0 0 1]

= [fmx 0 mxpx; 0 fmymypy; 0 0 1]

    • K is called the calibration matrix because it contains all of the internal camera parameters
    • we will be solving for K in the last stages as we solve for metric structure
    • K now has 10dof (encodes a CCD camera)
    • can also add a skew parameter s in k12 entry, yielding a finite projective camera
  • HZ156-7
camera parameters
Camera parameters

Camera

matrix

HZ6.1

Extracting

camera

parameters

HZ6.2

Camera

matrix

from F

HZ9.5

IAC

HZ3.5-3.7, 8.5

IAC and K

HZ8.5

Computing K

from 1 image

HZ8.8

Calibration

using Q*

HZ19.3

Hartley 92

extracting camera parameters from p
Extracting camera parameters from P
  • what do we want to know about a camera?
  • a camera is defined by its center (position) and image plane (part of orientation)
  • can compute image plane from center, principal plane, and focal length
    • can also compute image plane from center, principal axis, and focal length
  • all of these can be extracted from the camera matrix P
  • also important to know orientation of the camera
    • that is, the rotation from the world frame to the camera frame
    • gives orientation of image rectangle
  • therefore, want rotation matrix R (as in P = KR [I –C])
  • the principal point is extractable from the camera center and image plane, but we can also find it directly from the camera matrix, if so desired
camera center from p
Camera center from P
  • let’s characterize finite cameras (with finite camera centers)
  • Fact: camera matrices of finite cameras = 3x4 matrices with nonsingular 3x3 left submatrix (nonsingular KR)
    • easy to see that KR is nonsingular for finite focal lengths
  • the camera center C (of a finite camera) is the null vector of camera matrix P
    • P is 3x4 with nonsingular 3x3 left submatrix, so has a 1D null space
    • proof: let C be null vector; all points on a line through C will project to the same point since P((1-t)C + tA) = PA; the only lines that map to a point are those through the camera center
    • another proof: the projection of C, PC, is undefined, which characterizes the camera center
  • C = (-M-1 p4 1) where p4 = last column of P and M = left 3x3 submatrix
    • recall that C in KR[I –C] was the camera center (in Euclidean space)
    • we want to extract –C from P
    • let P = KR [I –C’]; if C = (C’ 1), then PC = 0
    • p4 = -KRC’ = -MC’
  • exercise: where do you think camera center of infinite camera is?
  • infinite camera: center is defined by the null vector of M: C = (d, 0) where Md = 0
    • M is singular in this case, so it does have a null vector
    • note that this is a point at infinity
  • HZ157-9
exercise
Exercise
  • before we talk about principal planes, we need a projective representation for the plane
  • What do you think it is?
  • Hint: it is the natural generalization of the projective representation of a line
  • Take 2 minutes.
principal plane from p
Principal plane from P
  • a plane may be represented by ax+by+cz+d = 0; which is represented in projective space by the 4-vector (a,b,c,d)
  • points on the principal plane are distinguished as the only ones that are mapped to infinity by the camera
    • that is, PX is an ideal point with x3=0
    • that is, P3T X = 0, where P3T is the 3rd row of P
    • that is, the principal plane is P3T
  • note that this plane does contain the camera center, as it should
    • PC = 0  P3T C = 0
  • HZ160
focal length and r from p
Focal length and R from P
  • finding focal length reduces to finding the calibration matrix K
    • focal length is encoded on the diagonal of K: (fmx fmy 1)
    • where (mx, my) is the pixel’s aspect ratio
  • recall that K and R are embedded in the front of the camera matrix P
    • P = KR [I –C]
  • we simply need to factor
  • let M = KR = left 3x3 submatrix of P
    • K is upper triangular, R is orthogonal
  • recall QR decomposition from numerical computing
    • A = QR where Q is orthogonal (e.g., product of Householders) and R is upper triangular
    • perfect, except wrong order
  • RQ decomposition is analogous
  • HZ157
rq decomposition
RQ decomposition
  • A = RQ, where R = upper triangular, Q = orthogonal
  • we want to reduce A to upper triangular: A  R
  • need to zero a21, a31 and a32
  • use 3 Givens rotations
    • A G1 G2 G3 = R
    • so A = R (G1 G2 G3)^t
    • Q = (G1 G2 G3)^t
  • zero carefully to avoid contamination
    • zero a32 with x-rotation G1, then a31 with y-rotation G2, then a21 with z-rotation G3
    • i.e., reverse order of A entries
    • Qy leaves 2nd column alone (so a32 remains zero)
    • Qz creates new 1st and 2nd columns from linear combinations of old 1st and 2nd columns, so 3rd row remains zeroed (cute!)
  • Q represents a rotation, which can be encoded by roll/pitch/yaw
  • angles associated with Gi are these 3 angles of roll/pitch/yaw (also called Euler angles)
  • HZ579
solving for euler angle
Solving for Euler angle
  • the first Givens rotation zeroes a32
  • let G1 = (1 0 0; 0 c –s; 0 s c) rotate about x
  • a32 element of AG1 = c*a32 + s*a33
  • a32’ = 0  c:s = -a33 : a32

c = k(-a33); s = k(a32); c2 + s2 = 1

k2(a332 + a322) = 1  k = sqrt(a322 + a332)

  • we now know G1
  • solve for G2 and G3 analogously
exercise17
Exercise
  • before we talk about principal axis, we need to recall a basic fact about the normal of a plane
  • what is the normal of the plane ax+by+cz+d=0? why?
  • take 2 minutes as a group
principal axis from p
Principal axis from P
  • the normal of a plane ax+by+cz+d = 0 is (a,b,c)
    • why? (X-P).N = 0 becomes X.N – P.N = 0
  • the principal axis is a normal of the principal plane
  • recall: the principal plane is the third row of P (we called it P3T)
  • Exercise: what is the normal of the principal plane?
  • let M3T = third row of M
  • normal of the principal plane = M3
  • so principal axis direction = M3
  • which direction, M3 or –M3, points towards front of camera?
  • answer: det(M) M3
    • det(M) protects it against sign changes
  • principal axis = det(M) M3
principal point from p
Principal point from P
  • the principal point is less important, but let’s look at it anyway
  • we could find the principal point by intersecting the line normal to the principal plane and through the camera center with the image plane
    • but this does not relate back directly to the camera matrix
    • let’s relate directly to P
  • principal point = image of the point at infinity in the direction of the principal plane’s normal
  • the point at infinity in this direction is (M3 0)
  • P (M3 0) = MM3
  • principal point = MM3, where M3T is the third row of M
  • HZ160-161
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