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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

- 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

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

- 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

- 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)

- 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)

- 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)

- 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

- K = diag(mx,my,1) [f 0 px; 0 f py; 0 0 1]
- HZ156-7

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

- 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

- 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

- 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

- 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

- 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

- 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

- 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

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

- 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

- 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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