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CS 395/495-26: Spring 2002 IBMR: Week 5B Finish Chapter 2: 3D Projective Geometry + Applications Jack Tumblin

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Presentation on theme: "CS 395/495-26: Spring 2002 IBMR: Week 5B Finish Chapter 2: 3D Projective Geometry + Applications Jack Tumblin"— Presentation transcript:

1 CS 395/495-26: Spring 2002 IBMR: Week 5B Finish Chapter 2: 3D Projective Geometry + Applications Jack Tumblin

2 IBMR-Related Seminars 3D Scanning for Cultural Heritage Applications 3D Scanning for Cultural Heritage Applications Holly Rushmeier, IBM TJ Watson 3D Scanning for Cultural Heritage Applications Friday May 16 3:00pm, Rm 381, CS Dept. no title yet... no title yet... Steve Marschner, Cornell University Friday May 23 3:00pm, Rm 381, CS Dept.

3 Quadrics Summary Quadrics are the ‘x 2 family’ in P 3 : –Point Quadric:x T Q x = 0 –Plane Quadric:  T Q*  = 0 Transformed Quadrics:Transformed Quadrics: –Point Quadric:Q’ = H -T Q H –Plane Quadric:Q*’ = H Q* H T Symmetric Q, Q* matrices:Symmetric Q, Q* matrices: –10 parameters but 9 DOF; 9 points or planes –(or less if degenerate…) –4x4 symmetric, so SVD(Q) = USU T Ellipsoid: 1 of 8 quadric types

4 Quadrics Summary SVD(Q) =USU T :SVD(Q) =USU T : – U columns are quadric’s axes –S diagonal elements: scale On U axes, write any quadric as: au bu 2 2 +cu d = 0On U axes, write any quadric as: au bu 2 2 +cu d = 0 Classify quadrics byClassify quadrics by –sign of a,b,c,d: (>0, 0, 0, 0, <0) Book’s method:Book’s method: –scale a,b,c,d to (+1, 0, –1) –classify by Q’s rank and (a+b+c+d) Ellipsoid: 1 of 8 quadric types

5 Quadrics Summary All Unruled Quadrics are Rank 4: (See page 55) BUT Some Rank 4 Jack!check this!!! quadrics are Ruled: and All degenerate quadrics (Rank<4) are Ruled (or Conic)

6 P 3 ’s Familiar Weirdnesses The plane at infinity:   = ( ) TThe plane at infinity:   = ( ) T (this is the 2D set of all…) ‘Ideal Points’ at infinity: d = p  = (x 1 x 2 x 3 0) T‘Ideal Points’ at infinity: d = p  = (x 1 x 2 x 3 0) T –(Also called ‘direction’ d in book) Parallel Planes intersect at a line within  Parallel Planes intersect at a line within   Parallel Lines intersect at a point within  Parallel Lines intersect at a point within   Any plane  intersects   at line l  (put  in P 2 )Any plane  intersects   at line l  (put  in P 2 ) x2x2x2x2 x1x1x1x1 x3x3x3x3 as x4  0, points (x 1,x 2,x 3,x 4 )  infinity 1/x 4 llll d

7 P 3 ’s Familiar Weirdnesses The plane at infinity:   = ( ) TThe plane at infinity:   = ( ) T Only H p ignores   (stays   for H S H A )Only H p ignores   (stays   for H S H A ) –Recall  ’ = H -T.  –Careful! H S, and H A will move points within   Both H P and   have 3DOFBoth H P and   have 3DOF –use one to find the other: Find   ’ in image space; use   in world-space to find H P Find directions in   with known angles in P 3Find directions in   with known angles in P 3

8 New Weirdness: Absolute Conic   WHY learn   ? Similar to C  for P 2 …WHY learn   ? Similar to C  for P 2 … –Angles from directions (d 1, d 2 ) or planes (  1,  2 ) –   has 3DOF for H P ;   has 5DOF for H A   Requires TWO equations:   Requires TWO equations:   :   is complex 2D Point Conic on the   plane (?!?!?!)   is complex 2D Point Conic on the   plane (?!?!?!) Recall plane at infinity   = [ 0, 0, 0, 1] T holds ‘directions’ d = [x 1, x 2, x 3, 0] TRecall plane at infinity   = [ 0, 0, 0, 1] T holds ‘directions’ d = [x 1, x 2, x 3, 0] T x x x 3 2 = 0, or ‘2D point conic where C = I’ x 4 = 0, or ‘all points are on   ’ x 4 = 0, or ‘all points are on   ’.

9 New Weirdness: Absolute Conic     is complex 2D Point Conic on the   plane   is complex 2D Point Conic on the   plane Only H A H P transforms   (stays   for H S )Only H A H P transforms   (stays   for H S ) All circles (in any  ) intersect   circular pts.All circles (in any  ) intersect   circular pts. –(recall: circular pts. hold 2 axes: x  i y) All spheres (in P 3 ) intersect   at all   pts.All spheres (in P 3 ) intersect   at all   pts. (not clear what this reveals to us) (not clear what this reveals to us) x x x 3 2 = 0, or ‘2D point conic where C = I’ x 4 = 0, or ‘all points are on   ’ x 4 = 0, or ‘all points are on   ’.

10 New Weirdness: Absolute Conic     measures angles between Directions (d 1,d 2 ) –World-space   is I 3x3 (ident. matrix) within   –Image-space   ’ is transformed (How? as part of   ?) –Euclidean world-space angle  is given by: –Directions d 1,d 2 are orthogonal iff d 1 T   ’ d 2 = 0 –?!?! What is   in P 3 ? Perhaps ? but…. –JACK!!!CHECK THIS!! (d 1 T   ’ d 2 ). (d 1 T   ’ d 1 ) (d 2 T   ’ d 2 ) cos(  ) = (do not change)

11 Absolute Dual Quadric Q*  Exact Dual to Absolute Conic   in plane   –Any conic in a plane = a degenerate quadric (even though plane   consists of all points at infinity)(even though plane   consists of all points at infinity) (even though conic   has no real points, all ‘outer limits’ of   )(even though conic   has no real points, all ‘outer limits’ of   ) Q*  is a Plane Quadric that matches   –Defined by tangent planes  (e.g.  T Q*   =0 ) –Conic   is on the ‘rim’ of quadric Q*  –In world space, Q*  = –Image space: 8DOF (?same as   ?) (up to similarity) (up to similarity)

12 Absolute Dual Quadric Q*  –In world space, Q*  =   = –Q*  always has infinity plane   as tangent Q*    = 0 and Q*  ’   ’ = 0 –Find angles between planes  1,  2 with Q*  :. –Can test for  planes  1,  2 :  1 T Q*  ’  2 = (  1 T Q*  ’  2 ). (  1 T Q*   1 ) (  2 T Q*   2 ) cos(  ) = 0001 End of Chapter 2. Now what?

13 Absolute Dual Quadric Q*  How can we find Q*  ’ ? –From  plane pairs (flatten, stack, null space…) How can we use it? –Transforms: “ Q*  fixed iff H is a similarity” means Changes Q*  to Q*  ’ unless H is only H p (Recall that Q*  ’ = H Q*  H T ) –THUS known Q*  ’ can solve for H A H P Q*  ’ is symmetric, use SVD... Q*  ’ is symmetric, use SVD...

14 What else can we DO in P 3 ? View Interpolation! for non-planar objects! Find H (or its parts:H S H A H P ): By  Plane Pairs (  1 T Q*   2 = 0)…By  Plane Pairs (  1 T Q*   2 = 0)… –Find Q*  by flatten/stack/null space method –Find H A H P from Q*   Q*’  relation (symmetric, so use SVD) Simpler: By Point (or Plane) CorrespondenceBy Point (or Plane) Correspondence –Can find full H (15DOF) in P 3 with 5 pts (planes) –Just extend the P 3 method (see Project 2)

15 But what can we DO in P 3 ? View Interpolation! for non-planar objects! Find H (or its parts:H S H A H P ): By Parallel Plane pairs (they intersect at  ’  )By Parallel Plane pairs (they intersect at  ’  ) –Find  ’  by flatten/stack/null space method, –Solve for H p using H P -T   =  ’  By  Direction Pairs ( d 1  ’  d 2 = 0 )…By  Direction Pairs ( d 1  ’  d 2 = 0 )… –Find  ’  from flatten/stack/null space method, –Find H A from     ’  relation –(symmetric, so use SVD…)

16 What can we DO in P 3 ? Open questions to ponder: –Can you do line-correspondence in P 2 ? in P 3 ? –How would you find angle between two lines whose intersection is NOT the origin? –Can you find H from known angles,   90º ? –How can we adapt P 2 ‘vanishing point’ methods to P 3 ? –How might you find H using twisted cubics? using the Screw Decomposition? –Given full 3D world-space positions for pixels (‘image+depth), what H matrix would you use to ‘move the camera to a new position’? –What happens to the image when you change the projective transformations H (bottom row)?

17 END


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