Geometry of perspective imaging ■ Coordinate transformations ■ Image formation ■ Vanishing points ■ Stereo imaging Image formation Images of the 3-D world ■ What is the geometry of the image of a three dimensional object? – Given a point in space, where will we see it in an image? – Given a line segment in space, what does its image look like? – Why do the images of lines that are parallel in space appear to converge to a single point in an image? ■ How can we recover information about the 3-D world from a 2-D image? – Given a point in an image, what can we say about the location of the 3-D point in space? – Are there advantages to having more than one image in recovering 3-D information? – If we know the geometry of a 3-D object, can we locate it in space (say for a robot to pick it up) from a 2-D image? Image formation Euclidean versus projective geometry ■ Euclidean geometry describes shapes “as they are” – properties of objects that are unchanged by rigid motions » lengths » angles » parallelism ■ Projective geometry describes objects “as they appear” – lengths, angles, parallelism become “distorted” when we look at objects – mathematical model for how images of the 3D world are formed Image formation Example 1 ■ Consider a set of railroad tracks – Their actual shape: » tracks are parallel » ties are perpendicular to the tracks » ties are evenly spaced along the tracks – Their appearance » tracks converge to a point on the horizon » tracks don’t meet ties at right angles » ties become closer and closer towards the horizon Image formation Example 2 ■ Corner of a room – Actual shape » three walls meeting at right angles. Total of 270o of angle. – Appearance » a point on which three lines segments are concurrent. Total angle is 360o Image formation Example 3 q ■ B appears between A and C from point p A ■ But from point 1, A appears between B C and C ■ Apparent displacement of objects due to B change in viewing position is called parallax shift p – Greeks knew that is the earth revolved around the sun there would be parallax shift of the stars C – Tycho Brahe looked for this shift, but A his instruments were not accurate B enough. Image formation Vanishing points and lines ■ To begin with, let’s assume we’re looking at a scene on the plane, σ, which is perpendicular to the image plane ρ. – To each point P on σ we associate the v C point p on ρ corresponding to the V intersection of the line CP with the plane ρ. – C is called the center of projection p ■ The point of intersection of the line P through the optical center, C, that is perpendicular to the image plane (and so Object plane, σ parallel to the object plane) is called the principal vanishing point, V. ■ The line, v, which is the intersection of the picture plane and the plane through P Image plane, ρ parallel to the object plane is called the vanishing line, or the horizon line. Image formation Vanishing points and lines ■ Our mapping between σ and ρ does not define a 1-1 correspondence. – Points on v are not the image of any point on σ – Points on m (intersection of plane ρ through C parallel to ) have no v C image on σ V ■ We’ll fix this later. ■ Add railroad tracks to σ. p P m Object plane, σ Image plane, ρ Image formation Vanishing points and lines ■ The images of all the points on L1 lie on the π plane 1 defined by L1 and C. – this is because the lines of sight for each point on v L1 lie on this plane. C ■ π V The image of L1 is then the intersection of 1 with the picture plane - l1. ■ L1 is parallel to CV since they are both perpendicular to the picture plane. l2 l1 π L2 – But, then CV must also lie on 1, since parallel lines are co-planar, and V lies on π 1. L1 ■ So, the principal vanishing point must lie on l1, Object plane, σ π since it lies on both 1 and the picture plane. ■ Similarly, V must lie on l2, the image of L2. ■ In fact, ALL lines in the object plane that are ρ perpendicular to the image plane image to lines Image plane, passing through V. – This is the image of the “point at infinity” for that set of parallel lines Image formation Vanishing points and lines ■ M1 and M2 are parallel, but not perpendicular to the image plane. ■ The image of M is the line m where the plane, i i v π , containing C and M intersects the image Λ C i i V plane. M V ■ There is a unique line, Λ, through C which is π parallel to Mi and so is this line is also on i. m2 ■ Since this line is horizontal, it also lies in the M m1 plane determined by C and v, the vanishing line. 2 M ■ Since v and Λ lie in the same plane, and are not 1 Object plane, σ parallel, they must intersect at some point Vm. Image plane, ρ Image formation Vanishing points and lines ■ π This places Vm on both i and the image plane, so it must lie on their intersection, mi. v C – So, any family of parallel lines on the Λ V V object plane will image to a set of lines M passing through a point (vanishing point) on the vanishing line. ■ As we rotate the set of lines, the vanishing m1 m2 point moves along the vanishing line. M2 ■ So even though parallel lines don’t “meet” we M1 can see where they meet in images! Object plane, σ Image plane, ρ Image formation General case ■ Generally, the vanishing line for any object plane is the line of intersection of the –image plane – plane parallel to the object plane through C (called the horizon plane). ■ The vanishing point for a line on the object plane is the intersection of the – vanishing line – plane containing the original line and the center of projection. V C v π ρ Image formation Vanishing points and lines Image formation Homogeneous coordinates ■ Classical Euclidean geometry: through any point not on a given line, there exists a unique line which is parallel to the given line. – For 2,000 years, mathematician tried to “prove” this from Euclid’s postulates. – In the early 20’th century, geometry was revolutionized when mathematicians asked: What if this were false? – That is, what if we assumed that EVERY pair of lines intersected? – To do this, we’ll have to add points and lines to the standard Euclidean plane. ■ If (x,y) are the rectangular coordinates of a point, P, and if (x1, x2, x3) are any three real numbers such that: –x1/x3 = x –x2/x3 = y then (x1, x2, x3) are a set of homogeneous coordinates for (x,y). ■ So, in particular, (x,y,1) are a set of homogeneous coordinates for (x,y) Image formation Homogeneous coordinates ■ Given the homogeneous coordinates, (x1, x2, x3), the rectangular coordinates can be recovered. ■ But (x,y) has an infinite number of homogeneous coordinate representations, because if (x1, x2, x3) are homogeneous coordinates of (x,y), then so are (kx1, kx2,k x3) for any k <>0. Image formation Points and lines ■ In rectangular coordinates, we can represent a line by: – l1x + l2y + l3 = 0 l1, l2 not both 0. ■ In homogeneous coordinates, the equation becomes – l1x1 + l2x2 + l3x3 = 0 ■ So a line is determined by the three coefficients (l1, l2, l3) ■ If l is the line l1x1 + l2x2 + l3x3 = 0, then we say that [l1, l2, l3] are the homogeneous coordinates of l. – So, in homogeneous coordinates, points and lines look the same - triples of numbers. – We’ll represent points as (…) and lines as […] to keep things straight. ■ A line is uniquely determined by a set of homogeneous coordinates, but a given line has an infinite number of homogeneous coordinate representations. –If [l1, l2, l3] represents a given line, then so does [kl1, kl2,k l3]. » I.e., the same set of points satisfy the equations l1x1 + l2x2 + l3x3 = 0 and kl1x1 +k l2x2 + kl3x3 = 0 Image formation Vectors: Inner Product ■ T Inner product of two vectors a=(ax ay az) and b=(b b b )T is a scalar b x y z α α a·b =|a|·|b|cos = axbx + ayby+ az bz |b|cosα a Note that we are projecting one vector onto the other. ■ Length: ≡ T 1/2 2 2 2 1/2 |a| ||a|| =(a ·a) = (ax + ay + az ) Image formation Vectors: Cross product ■ Vector product of two vectors a and b is a vector a×b i j k b × a b =det ax ay az = α bx by bz a = i(aybz –azby)+ j(azbx –axbz)+ k(axby –aybx) ■ This vector is orthogonal to the plane formed by a and b (in the sense of a right-handed screw) ■ Its magnitude is the area of the parallelogram whose sides are a and b |a×b| = |a|·|b| |sinα| ■ For two vectors in the xy-plane the vector product can be used to determine convexity! How? Image formation Vectors: Mixed vector product ■ Mixed vector product of two vectors a c b and b is a scalar α ax ay az × a (a b)·c =det bx by bz cx cy cz ■ Its magnitude is the volume of the parallelepiped whose sides are a, b, and c Image formation Points and lines ■ We usually think of the equation of a line as specifying the set of points that lie on that line.
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