1 / 37

CS 395/495-26: Spring 2004

CS 395/495-26: Spring 2004. IBMR: Intro to P 3 3 -D Projective Geometry Jack Tumblin jet@cs.northwestern.edu. 3D Homogeneous Coordinates. Extend Projective space from 2D to 3D: P 2 : map 3  2 dimensions (Easy to imagine) From 2D ‘world space’ (x,y) make

vea
Télécharger la présentation

CS 395/495-26: Spring 2004

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CS 395/495-26: Spring 2004 IBMR: Intro to P3 3-D Projective Geometry Jack Tumblin jet@cs.northwestern.edu

  2. 3D Homogeneous Coordinates Extend Projective space from 2D to 3D: • P2: map 3  2 dimensions (Easy to imagine) • From 2D ‘world space’ (x,y) make • 2D homogeneous coordinates (x1,x2,x3). • 2D projective ‘image space’ (x’,y’) = (x1/x3, x2/x3) • P3: map 4 3 dimensions (Harder to imagine) • From 3D ‘world space’ (x,y,z) make • 3D homogeneous coordinates (x1,x2,x3,x4). • 3D projective ‘image space’ (x’,y’,z’) = (x1/x4, x2/x4, x3/x4)

  3. 3D Homogeneous Coordinates • Unifies points and planes • (but lines are messy) • Puts perspective projection into matrix form • No divide-by-zero, points at infinity defined… in R3, write point x as But in P3, write same point x as: where: x1 x2 x3 x4 x y z x = x1 / x4, y = x2 / x4, z = x3 / x4, x4 = anything, even zero!(but usually defaults to 1) y (x,y,z) x z

  4. A Very Common Mistaek: • P2 homog. coords: 2D projective map (“one 2D pointone 3D ray” is correct) • P3 homog. coords. 3D projective map ( But “one 3D pointone 3D ray” is WRONG! It’s a 4D ray) (x,y) (x1, x2, x3) (x,y,z) (x1, x2, x3, x4) x2 x2 x3 y x x1 x1 (seen in some graphics books)

  5. A Very Common Mistaek: • P2 homog. coords: 2D projective map (“one 2D pointone 3D ray” is correct) • P3 homog. coords. 3D projective map ( But “one 3D pointone 3D ray” is WRONG!) (x,y) (x1, x2, x3) (x,y,z) (x1, x2, x3, x4) x2 NO! Don’t confuse 3D ‘z’ with projective x4! x2 (x,y,z) x3 y x x1 x1 (seen in some graphics books)

  6. 3D Homogeneous Coordinates • P2 homog. coords: 2D projective map (A 2D point in P2 maps to a 3D Cartesian ‘ray’ through the origin) • P3 homog. coords. 3D projective map (A 3D point in P3 maps to a 4D Cartesian ‘ray’ through the origin) (x,y) (x1, x2, x3) (x,y,z) (x1, x2, x3, x4) x2 x2 (x,y,z) x3 x3 y x x1 x1 P3: a scale-able 3D volume of points. Each point gets a 4D ray (constant x4 shown) P2: a scale-able 2D plane of points. Each point gets a 3D ray (constant x3 shown)

  7. 3D Homogeneous Coordinates P3 = homog. coordinates = 3D projective map 3D point4D ‘ray’: --Superset of P2 transformations: --Includes translation, and projection from any point H (x,y,z) (x1, x2, x3, x4) (x1’, x2’, x3’, x4’) (x,y,z) x2 x2 x3 x3 x1 x1 (Approx. as many 2D planes of constant z)

  8. P3: Point  Plane duality Recall Plane Equations in 3D: Normal vector: (a,b,c) = n Unaffected by scale k, with Min. Distance from plane to origin: d Write in 3D homog. coordinates: Point x and Plane  are duals (?Lines? wait….) ax + by + cz +d = 0 ^ kax + kby + kcz + kd= 0 ax + by + cz +d = 0 a b c d x1 x2 x3 x4 = 0 xT.= 0 T.x = 0

  9. Points Plane Conversions • Find plane  thru points P1,P2,P3? • Easy! Stack points: • Find null space (use SVD) • (Rank 2? collinear points!) • OR: use 3D cross products:  = 0 PT1 PT2 PT3 1 2 3 4 p11 p12 p13 p14 p21 p22 p23 p24 p31 p32 p33 p34 = 0 (p1 - p3)  (p2 - p3) • • -p3T · (p1 p2) (pg 47 in Zisserman book) Find ‘normal’ vector perpendicular to plane…  =

  10. Points Planes in P3 Plane can define a 3D coordinate system (find u,v coordinates within plane, w along plane normal) • Because  is equiv. to plane’s normal vector • Find 3  vectors in P3 (null space of [0 0 0] ) • assemble them as rows of 4x3 M vector • To get 3D coords of any P3 point p:M p = x’ • Finds a 2D coords in P2plane (of ): x’ = u v w

  11. Lines in P3: Awkward • Geometrically, Lines are: • Intersection of 2 (or more) planes, • An axis or ‘pencil’ of planes, • Linear combo of 2 points, p1+ A(p2-p1) • a 4 DOF entity in P3; a 4-vector won’t do! • Symbolically: Three forms of P3 lines: • ‘Span’ : Null Space of matrix: W • Plűcker Matrix • Plűcker Line Coordinates Skip this!

  12. P3 Lines 1a: (Point-Point) Span W • Recall that a point xis on a plane  iff • if 2 given points A, B intersectwith a ‘pencil’ of planes on a line, • Define a ‘P3 line’ as that intersection: stack AT, BTto make 2x4 matrix W: W = • If line W contains the plane , then W  = 0 xT.= 0 (or T.x=0 ) B A AT BT .= 0 a1 a2 a3 a4 b1 b2 b3 b4

  13. P3 Lines 1b: (Plane-Plane) Span W* • Recall that a point xis on a plane  if • If 2 given planes P,Q intersect ata ‘pencil’ of points on a line, • Define a ‘P3 line’ as that intersection: stack PT, QTto make 2x4 matrix W*: W* = • If line W* contains the point x,thenW* x = 0 xT.= 0 (or T.x=0 ) P Q P Q T. x = 0 p1 p2 p3 p4 q1 q2 q3 q4

  14. P3 Lines Summarized: Spans • (Point) Span W • (Found from points A,B) • Used to test plane : • (Plane) Span W* • (Found from planes P,Q) • Used to test point x: B W  = 0 A Useful property: WT W* = W* WT = 02x2 (the 2x2 null matrix) P Q W* x = 0

  15. P3 Lines 1: ‘Join’ reverses Spans... • Join Line W and Point p  Plane  W  = 0 iff plane  holds line W, andpT = 0 iff plane  holds point p; stack … Let M = ; solve for  in M  = 0 • Join Line W* and Plane  Point p W* p = 0 iff point p is on line W* Tp = 0 iff point p is on plane ; stack… Let M* = ; solve for p in M* p = 0 W pT W* T

  16. Projective Transformations • Use H for transforms in P3: • Has 15 DOF (4x4 -1) • Superset of the P2H matrix: Question: ? How could a 3D cube become a 2D image of a cube in P3? h11 h12 h13 h14 h21 h22 h23 h24 h31 h32 h33 h34h41 h42 h43 h44 H = h11 h12 0 h13 h21 h22 0 h23 0 0 0 0h31 h32 0 h33 h11 h12 h13 h21 h22 h21 h31 h32 h33 HP2 =

  17. P3 Transformations • Transform a point p or plane  with H: • Transform point span W or plane span W*: • (Optional): Transform a Plűcker Matrix: p’ = H.p’ = H-T.  W’ = H.W W*’ = H-T.W* L’ = H.L.HT L*’ = H-T.L*.H-1

  18. P3’s Familiar Weirdnesses • The plane at infinity:  = (0 0 0 1)T (this is the 2D set of all…) • ‘Ideal Points’ at infinity: d =p =(x1 x2 x3 0)T • (Also called ‘direction’ d in Zisserman book) • Parallel Planes intersect at a line within  • Parallel Lines intersect at a point within  • Any plane  intersects  at line l (put  in P2) l x2 d as x40, points (x1,x2,x3,x4) infinity 1/x4 x3 x1

  19. P3’s Familiar Weirdnesses • The plane at infinity:  = (0 0 0 1)T • Only Hp ignores  (stays  for HSHA) • Recall ’ = H-T. • Careful!HS, and HA will move points within  • Both HP and  have 3DOF • use one to find the other: Find ’in image space; use  in world-space to findHP • Find directions in  with known angles in P3

  20. What else can we DO in P3? View Interpolation! for non-planar objects! Find H (or its parts:HSHAHP): • By  Plane Pairs (1T Q* 2 = 0)… • Find Q* by flatten/stack/null space method • Find HAHP from Q*Q*’relation (symmetric, so use SVD) Simpler: • By Point (or Plane) Correspondence • Can find full H (15DOF) in P3 with 5 pts (planes) • Just extend the P3method (see Project 2)

  21. What else can we DO in P3? View Interpolation! for non-planar objects! Find H (or its parts:HSHAHP): • By Parallel Plane pairs (they intersect at ’) • Find ’ by flatten/stack/null space method, • Solve for Hp using HP-T = ’ • By  Direction Pairs ( d1 ’ d2 = 0 )… • Find ’ from flatten/stack/null space method, • Find HA from   ’ relation • (symmetric, so use SVD…)

  22. What else can we DO in P3? Best of all: Define Cameras as 3D2D transform… Define multiple cameras… Define transforms from one camera image to another camera image…

  23. What can we DO in P3? Questions to help you explore: • Can you do line-correspondence in P2? in P3? • 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 P2 ‘vanishing point’ methods to P3? • 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)?

  24. Optional: Describe P3 lines by --Plucker Matrices --Plucker Coords. Quadrics; angles and infinity

  25. P3 Lines 2: Plűcker Matrices Line == A 4x4 symmetric matrix, rank 2, 4DOF • Line L through known points A, B: L = A.BT - B.AT = • LineL*through known planesP,Q: L* = P.QT – QPT Two forms: B A a1a2a3a4 b1b2b3b4 - b1 b2 b3 b4 a1 a2 a3 a4 P Q

  26. P3 Lines 2: Plűcker Matrices Line == A 4x4 symmetric matrix, rank 2, 4DOF • Line L through A, B pts: L = A.BT - B.AT • LineL*through P,Q planes: L* = P.QT - QPT • L  L* convert?Note that: • Skew-symmetric; • L’s has 6 params: or • Note det|L|=0 is written l12l34+ l13l42 +l14l23= 0 • Simple! Replace lijwithlmn so that {i,j,m,n} = {1,2,3,4}Examples: l12 l34,or l42l13 etc. l12l13l14l23l24l34 l12l13l14l23l34l42 (to avoid minus signs)

  27. P3 Lines 2: Plűcker Matrices • Join Line L* and Point p  Plane  L*.p =(if point is on the line, then L*.p = 0) Join Line L and Plane  Point p L. =p (if line is in the plane, then L. =0)

  28. P3 Lines 3: Plűcker Line Coords ==Plűcker Matrix has 6 vital elements: • Off-diagonals or • Just make them a 6-vector: and require that det|L| = 0, or: l12l13l14l23l24l34 l12l13l14l23l34l42 (to avoid minus signs) l12l13l14l23l42l34 L = l12l34+ l13l42 +l14l23= 0

  29. Quadrics Summary Quadrics are the ‘x2 family’ in P3: • Point Quadric: xT Q x = 0 • Plane Quadric: T Q*  = 0 • Transformed Quadrics: • Point Quadric: Q’ = H-TQ H • Plane Quadric: Q*’ = HQ* HT • Symmetric Q, Q* matrices: • 10 parameters but 9 DOF; 9 points or planes • (or less if degenerate…) • 4x4 symmetric, so SVD(Q) = USUT Ellipsoid: 1 of 8 quadric types

  30. Quadrics Summary • SVD(Q) =USUT: • U columns are quadric’s axes • S diagonal elements: scale • On U axes, write any quadric as:au12 + bu22 +cu32 + d = 0 • Classify quadrics by • sign of a,b,c,d: (>0, 0, <0) • 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

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

  32. New Weirdness: Absolute Conic  • WHY learn ? Similar to C for P2… • Angles from directions (d1, d2) or planes (1, 2) •  has 3DOF for HP;  has 5DOF for HA •  Requires TWO equations:  : •  is complex 2D Point Conic on the  plane (?!?!?!) • Recall plane at infinity = [ 0, 0, 0, 1]T holds ‘directions’ d = [x1, x2, x3, 0]T x12 + x22 + x32 = 0, or ‘2D point conic where C = I’ x4 = 0, or ‘all points are on ’ .

  33. New Weirdness: Absolute Conic  •  is complex 2D Point Conic on the  plane • Only HAHP transforms  (stays  for HS) • All circles (in any ) intersect  circular pts. • (recall: circular pts. hold 2 axes: x iy) • All spheres (in P3) intersect  at all  pts. (not clear what this reveals to us) x12 + x22 + x32 = 0, or ‘2D point conic where C = I’ x4 = 0, or ‘all points are on ’ .

  34. New Weirdness: Absolute Conic   measures angles between Directions (d1,d2) • World-space  isI3x3 (ident. matrix) within  • Image-space ’ is transformed (How? as part of ?) • Euclidean world-space angle  is given by: • Directions d1,d2 are orthogonal iff d1T’ d2 = 0 • ?!?! What is  in P3? Perhaps ? but…. • JACK!!!CHECK THIS!! (d1T’ d2). (d1T’ d1) (d2T’ d2) cos() = 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 (do not change)

  35. 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 conic has no real points, all ‘outer limits’ of ) Q* is a Plane Quadric that matches  • Defined by tangent planes  (e.g. TQ*=0) • Conic  is on the ‘rim’ of quadric Q* • In world space, Q* = • Image space: 8DOF (?same as ?) (up to similarity) 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0

  36. Absolute Dual Quadric Q* 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 • In world space, Q* == • Q* always has infinity plane  as tangentQ*  = 0 and Q*’’= 0 • Find angles between planes 1,2 with Q*: . • Can test for  planes 1,2 : 1TQ*’ 2 = 0 0 0 0 1 (1TQ*’ 2) . (1TQ*1) (2TQ*2) cos() = End of Chapter 2. Now what?

  37. 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” meansChanges Q* to Q*’unless H is only Hp(Recall thatQ*’ =H Q* HT ) • THUS known Q*’ can solve for HAHP • Q*’ is symmetric, use SVD...

More Related