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camera terminology - PowerPoint PPT Presentation

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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 l.jpg
Camera diagram










from F



HZ3.5-3.7, 8.5

IAC and K


Computing K

from 1 image



using Q*


Hartley 92

Camera terminology l.jpg
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 l.jpg
Camera matrix










from F



HZ3.5-3.7, 8.5

IAC and K


Computing K

from 1 image



using Q*


Hartley 92

Encoding in a camera matrix l.jpg
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 l.jpg
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 l.jpg
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 l.jpg
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 l.jpg
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 l.jpg
Camera parameters










from F



HZ3.5-3.7, 8.5

IAC and K


Computing K

from 1 image



using Q*


Hartley 92

Extracting camera parameters from p l.jpg
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 l.jpg
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 l.jpg

  • 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 l.jpg
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 l.jpg
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 l.jpg
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 l.jpg
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 l.jpg

  • 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 l.jpg
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 l.jpg
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