Rigid Geometric Transformations COMPSCI 527 — Computer Vision COMPSCI 527 — Computer Vision Rigid Geometric Transformations 1 / 15
Outline 1 Motivation 2 Projection 3 Cross Product 4 Triple Product 5 Rotations 6 Rigid Transformations COMPSCI 527 — Computer Vision Rigid Geometric Transformations 2 / 15
Motivation Motivation • The relative motion between a camera and an otherwise static scene is a rigid transformation • It underlies 3D reconstruction • A rigid transformation is rotation + translation • Understanding rotation requires orthogonality and projection, which we have seen in general • All vectors in R 3 • Reconstruction techniques also require knowing about cross product and triple product • Read notes for details COMPSCI 527 — Computer Vision Rigid Geometric Transformations 3 / 15
Projection bib BI Projection a pi p b • Definition of projection of a onto b : the point p on the line through b that is closest to a c IR O • p is on the line through b : p = x b for some x • p is closest to a when ( a , p ) is orthogonal to b : b T ( a � x b ) = 0, which yields x = b T a b T b so that of F tEf MAGNITUDE p = x b = b x = bb T Ow b T b a COMPSCI 527 — Computer Vision Rigid Geometric Transformations 4 / 15
Projection The Projection Matrix • p = P b a where P b = bb T b T b a • P b is rank 1, symmetric, and idempotent: P 2 b = P b • Proof: I T EIS Ps Pj • Norm squared of p : IP k p k 2 = at I btp Pb a Pb sign Eb EIJI at bgbb a p latte b BT Pb • When k b k = 1, 1lb COMPSCI 527 — Computer Vision Rigid Geometric Transformations 5 / 15
Cross Product The Cross Product • Geometry: The cross product of two three-dimensional vectors a and b is a vector c orthogonal to both a and b , oriented so that the triple a , b , c is right-handed, and with tea magnitude k c k = k a ⇥ b k = k a k k b k sin θ where θ is the smaller angle between a and b • The magnitude of a ⇥ b is the area of a rectangle with sides a and b ETI UEI LIE � � O a x a y a z • Algebra: c = a ⇥ b = � � � � b x b y b z � � = ( a y b z � a z b y , a z b x � a x b z , a x b y � a y b x ) T • Easy to check that a ⇥ b = � b ⇥ a COMPSCI 527 — Computer Vision Rigid Geometric Transformations 6 / 15
Cross Product The Cross-Product Matrix • c = ( a y b z � a z b y , a z b x � a x b z , a x b y � a y b x ) T is linear in b • Therefore, there exists a 3 ⇥ 3 matrix [ a ] × such that c = a ⇥ b = [ a ] × b a 2 3 2 3 2 3 c x b x 5 = c = c y b y 4 4 5 4 5 O c z b z ay Qx • The matrix [ a ] × is skew-symmetric: [ a ] T × = � [ a ] × COMPSCI 527 — Computer Vision Rigid Geometric Transformations 7 / 15
Triple Product The Triple Product 5 ta E • Definition: det([ a , b , c ]) = a T ( b ⇥ c ) = a x ( b y c z � b z c y ) � a y ( b x c z � b z c x ) + a z ( b x c y � b y c x ) • Easy to check: a T ( b ⇥ c ) = b T ( c ⇥ a ) = c T ( a ⇥ b ) = � a T ( c ⇥ b ) = � c T ( b ⇥ a ) = � b T ( a ⇥ c ) • Signed volume of parallelepiped d oi BE θ a c g b COMPSCI 527 — Computer Vision Rigid Geometric Transformations 8 / 15
Rotations Multiple Reference Frames • If we associate a reference system to a camera and the camera moves, or we consider multiple cameras, or we consider one camera and the world, we have multiple reference systems • Point coordinates are x , y , z • Left superscript denotes which reference system coordinates are expressed in: 1 y • Subscripts denote which point or reference system we are talking about: x 2 0 • 2 y 3 is the y coordinate of point 3 in reference system 2 COMPSCI 527 — Computer Vision Rigid Geometric Transformations 9 / 15
Rotations Multiple Reference Frames • A zero left superscript can be omitted: 0 z = z if • The origin of a reference system is t (for “translation”) 2 3 0 • We always have i t i = 0 4 5 t 0 • If i , j , k are the unit points of a reference system, we always se ⇥ i i i O i j i i k i ⇤ have = I , the 3 ⇥ 3 identity matrix COMPSCI 527 — Computer Vision Rigid Geometric Transformations 10 / 15
Rotations z f 1 z Rotations 1 y j k 1 1 y i 1 x 1 x 2 3 0 • No translation: 0 t 1 = t 1 = 0 4 5 0 • Both systems right-handed O • i 1 , j 1 , k 1 are the unit vectors of reference system 1 expressed in reference system 0 • Given p = 0 p , what is 1 p ? COMPSCI 527 — Computer Vision Rigid Geometric Transformations 11 / 15
Rotations z Rotations p 1 z k 1 y 1 y j 1 x i 1 1 x p = 1 x i 1 + 1 y j 1 + 1 z k 1 1 x = i T 1 y = j T 1 z = k T 1 p , 1 p , 1 p 000 2 1 x 3 2 i T 3 1 p 5 = 5 = R 1 p 1 p = 1 y j T 1 p 4 4 1 z k T 1 p 2 i T 3 O 1 where R 1 = 0 R 1 = j T (unit vectors are the rows ) 4 5 1 k T 1 COMPSCI 527 — Computer Vision Rigid Geometric Transformations 12 / 15
Rotations Rotations in General 2 a i T 3 O b b p = a R ba p a R b = a j T • More generally, where 4 b 5 a k T b • Rotations are reversible, so there exists b R a = a R − 1 b • b R a = a R T b O 2 3 2 a i T 3 r 11 r 12 r 13 b 5 = • Intuition: a R b = a j T r 21 r 22 r 23 4 4 5 b a k T r 31 r 32 r 33 b • r mn is the signed magnitude of the projection of the m -th vector in S b onto the n -th vector in S a • Therefore, vice versa (direction cosines) • Therefore, what we want in the rows of R − 1 is in the columns of R COMPSCI 527 — Computer Vision Rigid Geometric Transformations 13 / 15
Rotations Properties of Rotation • det( R ) = i T ( j ⇥ k ) = i T i = 1 because i ⇥ j = k , j ⇥ k = i , k ⇥ i = j • Cross-product is covariant with rotations: e ( R a ) ⇥ ( R b ) = R ( a ⇥ b ) COMPSCI 527 — Computer Vision Rigid Geometric Transformations 14 / 15
Rigid Transformations lap bp Coordinate Transformation R Y p p t ftp.bR.bpbta bp p − t 1 bpatap btabp o p k 1 l2aTeptbta j 1 t 1 k 0 bR i 1 fptbReOD0ooDIIi.I i 0 j 0 • A.k.a. rigid transformation • First translate, then rotate: 1 p = R 1 ( p � t 1 ) • Inverse: p = R T 1 p + t 1 1 Owo • Generally, if b p = a R b ( a p � a t b ) then a p = b R a ( b p � b t a ) where b R a = a R T b and b t a = � a R ba t b COMPSCI 527 — Computer Vision Rigid Geometric Transformations 15 / 15
Recommend
More recommend
