example: least square line fitting data scatterdata as 2d vectors x y a a x i +b y i = c a. x = c =...
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Example: Least Square Line Fitting
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-2 -1.5 -1 -0.5 0 0.5 1 1.5 20
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Data scatter Data as 2D vectors
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y
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a xi +b yi = c
a.x = c = ||x||| cos ø
ø
Introducing Homogenous Coordinates
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Data as 3D homogenous vectors pi = [xi yi 1]’
In 3D, the set of points liesClose to a common plane
a xi +b yi = c
a.x = c = ||x||| cos ø
Becomesa xi +b yi + c 1 = 0a.pi = 0
Geometry of solution
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-0.5 -0.5 0.5 0.50.25 -0.25 -0.25 0.25 2 2 2 2
A =
[U,S,V] = svd(A)
U = 0 1 0 0 0 1 1 0 0
S =4 0 0 00 1 0 00 0 0.5 0
Fundamental Matrix, why are 8 point matches enough?
€
u'
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⎥ ⎥ ⎥=
f11 f12 f13
f21 f22 f23
f31 f32 f33
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⎥ ⎥ ⎥⇒ f33 =1
Thus only 8 free parameters => Need 8 or more constraints.
Stereo Reconstruction Ambiguity
Extrinsic Parameter ambiguity
Projective structure
Metric Structure
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B = 1 -0.5 -0.5 0 0.86603 -0.86603 2 2 2
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U = 0 -1 0 0 0 -1 1 0 0
[V,D]=eig(B)Yields z-axis andComplex eigenvaluesRepresenting the ambiguity
[U,S,V] = svd(B)
Matlab examples
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Homework #2 solns
Problem #2 Projected image of a cube
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Problem #3
% 3) Show that a straight line in 3D projects to a straight line % in an image, using the simplest projection model, u = f x/z, v = f y/z.% Hint: A straight line can be defined by two points L = a*p1+(1-a)*p2),% where a is a scalar. % p1 = [x1,y1,z1]';% Show that a third point on the line p3 = a*p1+(1-a)*p2% projects to an image point (u3,v3) that can be written in the form% u3 = b u1 + (1-b) u2% v3 = b v1 + (1-b) v2% where b = a*z1/(a z1 + (1-a) z2)
p1 = 1/z1 M x1 p2 = 1/z2 M x2
x3 = a x1 +(1-a) x2
p3 = 1/z3 M x3 = a 1/z3 M x1 + (1-a) 1/z3 M x2 = a 1/z3 (z1/z1) M x1 + (1-a) 1/z3 (z2/z2) M x2 = a z1/z3 p1 + (1-a) z2/z3 p2
But now x3 = a x1 +(1-a) x2And z3 = a z1 +(1-a) z2Thusp3 = a z1/(a z1 +(1-a) z2) p1 + (1-a) z2/(a z1 +(1-a) z2) p2
Set b = a z1/(a z1 +(1-a) z2)Then 1-b = (1-a) z2/(a z1 +(1-a) z2)
And we have:
p3 = b p1 + (1-b) p2 {QED}