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

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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.

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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

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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 1 2 + bu 2 2 +cu 3 2 + d = 0On U axes, write any quadric as: au 1 2 + bu 2 2 +cu 3 2 + 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

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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)

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P 3 ’s Familiar Weirdnesses The plane at infinity: = (0 0 0 1) TThe plane at infinity: = (0 0 0 1) 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 llll d

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P 3 ’s Familiar Weirdnesses The plane at infinity: = (0 0 0 1) TThe plane at infinity: = (0 0 0 1) 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

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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 1 2 + x 2 2 + 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 ’.

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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 1 2 + x 2 2 + 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 ’.

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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( ) = 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 (do not change)

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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) 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0

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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 = 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 ( 1 T Q* ’ 2 ). ( 1 T Q* 1 ) ( 2 T Q* 2 ) cos( ) = 0001 End of Chapter 2. Now what?

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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...

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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)

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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…)

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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)?

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