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Pole/polar consider projective geometry again we may want to compute the tangents of a curve that pass through a point (e.g., visibility) let C be a conic.

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Presentation on theme: "Pole/polar consider projective geometry again we may want to compute the tangents of a curve that pass through a point (e.g., visibility) let C be a conic."— Presentation transcript:

1 Pole/polar consider projective geometry again we may want to compute the tangents of a curve that pass through a point (e.g., visibility) let C be a conic and x a point if x lies on the conic, we know that Cx is the tangent at x if x does not lie on the conic, Cx is a line called the polar of x with respect to C x is the pole of the polar the polar of x passes through the 2 points of tangency of the 2 tangents of C through x –if tangent line at y, Cy, contains x, then x^t Cy=0 or (Cx)^t y = 0 so y lies on the polar the polar is the generalization of the tangent; it degenerates to the tangent when x lies on C HZ58

2 Conjugacy points x and y are conjugate if x^t C y = y^t C x = 0 if y lies on the polar Cx, then x and y are conjugate conjugacy is symmetric(!): if y lies on polar of x, then x lies on polar of y HZ59

3 Pole/polar for quadrics quadric Q polar plane of X = QX if X lies on Q, QX = tangent plane if X lies outside Q, QX passes through the points of tangency of the tangent cone through X HZ73-74

4 Conic fitting fitting point data is an important task in shape design –we shall explore in future, but let’s see what projective space has to say first fitting a conic is equivalent to finding the null space of a matrix each point (x,y) on a conic imposes a constraint on the conic coefficients –ax^2 + bxy +... + f = 0 –or, separating the unknowns from the knowns, –(x^2, xy, y^2, x, y, 1). C = 0, where C = (a,b,c,d,e,f) 5 points define a 5 x 6 matrix M: MC = 0 Q: how do you solve for conic C? A: 1d null space of M defines the coefficients C of the conic null space computation is also used to find fundamental matrix, epipoles (from F) and camera center (from camera matrix) HZ31

5 Projective 3-space 2D projective geometry was introduced through the duality of lines and points; 3D is motivated through the duality of planes and points; lines are self-dual plane = 4-vector point X lies on plane π iff π t X = 0 projective transformation is represented by a 4x4 nonsingular matrix H: X’ = HX homographies preserve lines, incidence and order of contact HZ65-66

6 3 points  plane consider 3 points X i that lie on the plane π build the 3x4 matrix M with ith row = X i t Mπ = 0 for the plane π if X i are in general position, M is rank 3 and its null vector is the plane if the three points are collinear, M is rank 2 and its 2D null space defines a pencil of planes i.e., plane = null space of point matrix the plane can also be expressed through determinants π = (D 234, - D 134, D 124, - D 123 ) t, where –D ijk = determinant of rows ijk of the 4x3 matrix with ith column = X i (note that this is transpose of above matrix) –proof: build 4x4 matrix [X,X1,X2,X3], whose determinant = 0 if X is on plane –3D analogue of line = cross product of two points equivalently, π = ((X 1 ’ – X 3 ’) x (X 2 ’ – X 3 ’), -X 3 ’ t (X 1 ’ x X 2 ’)) t where X i = (X i ’ 1) 3 planes  point is a dual result: null vector of the plane matrix HZ66-67

7 Lines how do you represent a line in 3-space? line = join of n-1 points or intersection of n-1 planes –shows duality line has 4dof –can define line by intersection with 2 orthogonal planes (2 parameters per plane) awkward to represent lines in 3-space since the natural representation by homogeneous 5-vector does not play well with the point and plane’s 4-vectors see HZ68-70 for representation of lines as null spaces (nice implementation consistency with other results)

8 Lines as Plücker matrices the most famous representation of a line is by Plücker represent the line through points A and B as: –L = AB^t – BA^t –4x4 skew-symmetric matrix called the Plücker matrix L is independent of points! L is rank 2 L has 4 dof (6 in skew-symm – 1 homogeneous – 1 det 0) L generalizes 2d line = point x point valency-2 tensor: if point transforms as X  HX, then Plücker matrix transforms as L  HLH^t dual Plücker matrix from two planes: L* = PQ^t – QP^t point X and line L  plane L* X –and L*X = 0 reflects X lying on L plane X and line L  point L X –and LX=0 reflects line L lying in the plane P HZ70-71

9 Plücker coordinates and Klein quadric Plücker coordinates are simply the 6 nonzero elements extracted from the Plücker matrix: {12,13,14,23,42 !,34} –from upper triangle with fifth element reversed since Plücker matrix is rank 2, det=0 imposes a quadratic constraint on these coordinates: –12*34 + 13*42 + 14*23 = 0 –dot product of 1 st half of L with reversed second half of L –this codimension-1 surface in projective 5-space is called the Klein quadric two lines L and L’ intersect if and only if L. reversed(L’) = 0 –Klein quadric encodes the fact that a line must intersect itself: L. reversed L = 0 HZ71-72


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