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CS 395/495-25: Spring 2003

CS 395/495-25: Spring 2003. IBMR: Week 7A The Camera Matrix Continued... Jack Tumblin jet@cs.northwestern.edu. IBMR-Related Seminars. 3D Scanning for Cultural Heritage Applications Holly Rushmeier, IBM TJ Watson Friday May 16 3:00pm, Rm 381, CS Dept.

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CS 395/495-25: Spring 2003

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  1. CS 395/495-25: Spring 2003 IBMR: Week 7A The Camera MatrixContinued... Jack Tumblin jet@cs.northwestern.edu

  2. IBMR-Related Seminars 3D Scanning for Cultural Heritage Applications Holly Rushmeier, IBM TJ Watson Friday May 16 3:00pm, Rm 381, CS Dept. no title yet ...<BRDF, BRSSDF capture? Optics of Hair? Inverse Rendering?> Steve Marschner, Cornell University Friday May 23 3:00pm, Rm 381, CS Dept.

  3. Reminders • ProjA graded: Good Job! 90,95, 110 • ProjB graded: Good! minor H confusions... • MidTerm graded: • ProjC posted tonight; Due next Friday, May 15 • Start Watson’s Late Policy: Grade -(3n) points; n=# of class meetings late • ProjD coming Thurs May 15, due Thurs May 29 • Take-Home Final Exam: Thurs June 5, due June 11

  4. Basic Cameras Revisited Plenty of Terminology: • Image Plane or Focal Plane • Focal Distance f • Camera Center C • Principal Point p • Principal Axis zc • Principal Plane (?!?! DOESN’T touch principle point!?!?) • Camera Coords (xc,yc,zc) • Image Coords (x’,y’) yc C p f zc y' x' xc

  5. Recall Basic Camera P0: • Basic Camera as a 3x4 matrix: • To translate image origin (x’,y’) away from zc axis:Shift principal point p from (0,0) to (px,py) • does NOT use ‘homogeneous’ 1 term in X • is NOT obvious: scales zc (think in P2 zc=x3 = 1,...) yc xc yc zc 1 f 0 px0 0 fpy00 0 1 0 x’ y’ z’ = y y’ p C z zc f y' P0 X = x x' xc

  6. Basic Camera P0: P3P2(or camera R3) • Basic Camera P0 is a 3x4 matrix: • Non-square pixels? change scaling (x, y) • Parallelogram pixels? set nonzero skew s K matrix: “(internal) camera calib. matrix” yc xc yc zc 1 xfspx0 0 yfpy00 0 1 0 x y z = y y’ p C z K (3x3 submatrix) zc f y' P0 X = x x' xc [K 0] = P0

  7. Complete Camera Matrix P y x X (world space) z yc C zc xc y x • K matrix: “internal camera calib. matrix” • R·T matrix: “external camera calib. matrix” • T matrix: Translate world origin to cam. origin, • R matrix: then Rotate world to fit cam. axes Combine: write (P0·R·T)·X = x as P·X = x Input:X P3 World Space Output: x P2 Camera Image x(camera space)

  8. The Pieces of Camera Matrix P yc p C f zc xc • • ••• • ••• • • • xw yw zw tw Columns of P matrix = image of world-space axes: • p1,p2,p3 == image of x,y,z axis vanishing points • Direction D = [1 0 0 0]T = point on P3’s x1 axis, at inifinity • PD = 1st column of P = P1. Repeat for y and z axes. • p4 == image of the world-space origin pt. • Proof: let D = [0 0 0 1]T = world origin • PD = 4th column of P = image of origin pt. • • • •• • • •• • • • xc yc zc p1 p3 p4 P p2 P = P·X = x, or =

  9. The Pieces of Camera Matrix P yc p C f zc xc xw yw zw tw Rows of P matrix: camera planes in world space • row 1 = P1T = image x-axis plane • row 2 = P2T = image y-axis plane • row 1 = P3T = camera’s principal plane P1T • • • •• • • •• • • • • • ••• • ••• • • • xc yc zc P P·X = x, or = P = P2T P3T

  10. The Pieces of Camera Matrix P xw yw zw tw Rows of P matrix: planes in world space • row 1 = P1 = image x-axis plane • row 2 = P2 = image y-axis plane • row 1 = P3 = camera’s principal plane P1T • • • •• • • •• • • • • • ••• • ••• • • • xc yc zc P P·X = x, or = P = P2T P3T yc p C f zc xc

  11. The Pieces of Camera Matrix P xw yw zw tw Rows of P matrix: planes in world space • row 1 = P1 = image x-axis plane • row 2 = P2 = image y-axis plane • row 1 = P3 = camera’s principal plane P1T • • • •• • • •• • • • • • ••• • ••• • • • xc yc zc P P·X = x, or = P = P2T P3T yc p C f zc xc

  12. The Pieces of Camera Matrix P yc zc p C f y' x' xc xw yw zw tw Rows of P matrix: planes in world space • row 1 = P1 = image x-axis plane • row 2 = P2 = image y-axis plane • Careful! Shifting theimage origin by px, pyshifts the x,y axis planes! P1T • • • •• • • •• • • • • • ••• • ••• • • • xc yc zc P P·X = x, or = P = P2T P3T

  13. The Pieces of Camera Matrix P xw yw zw tw Rows of P matrix: planes in world space • row 1 = P1 = image x-axis plane • row 2 = P2 = image y-axis plane • row 3 = P3 = camera’s principal plane P1T • • • •• • • •• • • • • • ••• • ••• • • • xc yc zc P P·X = x, or = P = P2T P3T yc p C f zc xc

  14. The Pieces of Camera Matrix P xw yw zw tw Rows of P matrix: planes in world space • row 1 = P1 = image x-axis plane • row 2 = P2 = image y-axis plane • row 3 = P3 = camera’s principal plane • princip. plane P3= [p31 p32 p33 p34]T • its normal direction: [p31 p32 p33 0 ]T • Why is it normal? It’s the world-space P3 direction of the zc axisBOOK has yet another inconsistent notation for this..... P1T • • • •• • • •• • • • • • ••• • ••• • • • xc yc zc P P·X = x, or = P = P2T P3T Principal plane p3 yc zc p C f xc

  15. The Pieces of Camera Matrix P xw yw zw tw • Principal Axis Vector (zc) in world space: • Normal of principal plane: = m3 = [p31 p32 p33 ]T • P3 Scaling  Ambiguous direction!! +/- m3 ? • Solution: use det(M)·m3as front of camera • Principal Point p in image space: • image of (infinity point on zc axis = m3)M·m3 = p = x0 (book renames p as x0) • • ••• • ••• • • • • • • •• • • •• • • • m1T xc yc zc P p4 P·X = x, or P = = m2T m3T M yc p x0 C f zc xc

  16. The Pieces of Camera Matrix P xw yw zw tw Where is the Camera? • Camera center C is at camera origin(xc,yc,zc)=0=C • What point C will the camera P transform to C=0 ?Answer: C is the Null Space of P :P C = C = 0 (solve for C. SVD works, but here’s an easier way:) Finite Camera: C = Affine Camera: C = • • ••• • ••• • • • • • • •• • • •• • • • m1T xc yc zc P p4 P·X = x, or P = = m2T m3T M ~ ~ yc ~ • • • 1 ~ • • • 0 d -M-1·p4 C p f zc (where Md=0) xc

  17. Uses for Camera Matrix P y x z xw yw zw tw Each world space R3 point d = (x,y,z)defines a world-space P3directionD = (x,y,z,0). What is the image point xd from direction D? xd= PD = [M | p4] D = Md (P4 column has no effect, because of D’s zero): xd = Md or M-1xd = d • • ••• • ••• • • • • • • •• • • •• • • • xc yc zc P M p4 P·X = x, or P = = yc xd d ~ p C f zc X (world space) xc

  18. Uses for Camera Matrix P xw yw zw tw Given image pointx0 and camera matrix P,Find ray X() in world space through both: Slow, Obvious way: ‘Pseudo-invert’ P, apply to x0: • Define pseudo-inverse P+ as = PT(PPT)-1 (note P·P+ = I) • Find a world-space point on ray: X0 = P+ x0 • LIRP with world-space camera point C: X() = C + (X0-C) • • ••• • ••• • • • • • • •• • • •• • • • xc yc zc P M p4 P·X = x, or P = = X() yc x0 ~ ~ ~ p ~ C f zc ~ xc

  19. Uses for Camera Matrix P xw yw zw tw Given image pointx0 and camera matrix P,Find ray X() in world space through both: Better way: ray from C to point D0 at infinity: • Point x0 is the image of (unknown) world-space directionD0 = (x,y,z,0)T. Define a point d0 = (x,y,z)T. • Recall: we can find d0 from the image: M-1x0 = d0 • Recall: world-space camera C = -M-1p4 X() =  M-1x0 + C = M-1(x0 – p4) 0 1 1 • • ••• • ••• • • • • • • •• • • •• • • • xc yc zc P M p4 P·X = x, or P = = ~ X() yc ~ x0 ~ ~ p C f zc xc

  20. Uses for Camera Matrix P xw yw zw tw Given world-space pointX0, camera matrix P,Find camera depth z0: • X0 = [xw, yw, zw, tw]Tseen thru camera P is X0·P = x0 = [xc, yc, 1]T·wc • Then signed depth z0 is: z0 = wc sign(det M)tw || m3 || • • ••• • ••• • • • • • • •• • • •• • • • m1T xc yc zc P p4 P·X = x, or P = = m2T m3T M X0 yc x0 zc z0 p C f xc

  21. Skipped: • P = [K|0]·R·T How can we separate K, R, T? • Answer: K is triangular; use QR decomposition • Cameras at Infinity: • Orthographic or ‘Parallel Projection’ Cameras • Transition to Orthographic: • Weak Perspective projection cameras • the ‘zoom’ lens (variable f) • Moving line-scan or ‘pushbroom’ cameras • Translation Scan: aerial/sattelite cameras • Cylindrical Scan: panoramic cameras • UNC ‘HiBall Tracker’: 6 tiny self-locating line-scan cameras

  22. Chapter 6 In Just One Slide: Given point correspondence sets (xi Xi), How do you find camera matrix P ? (full 11 DOF) Surprise! You already know how ! • DLT method: -rewrite H x = x’ as Hx  x’ = 0 -rewrite P X = x as PX  x = 0 -vectorize, stack, solve Ah = 0 for h vector -vectorize, stack, solve Ap = 0 for p vector -Normalizing step removes origin dependence • More data  better results (at least 28 point pairs) • Algebraic & Geometric Error, Sampson Error…

  23. END

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