Image Processing Reference
In-Depth Information
12.5 Translating Lines
Assuming that the normal of the tilting plane is k , how is the 2D normal flow ob-
tained from this 3D vector? In this case we have a linearly symmetric image in 2D,
g (cos( θ ) x 0 +sin( θ ) y 0 )
(12.48)
where g ( τ ) is a one-dimensional function, the vectors
s 0 =( x 0 ,y 0 ) T ,
a =(cos( θ ) , sin( θ )) T
(12.49)
represent the coordinates of an arbitrary point in the image plane, and the normal
of the line(s) defining the linearly symmetric image, respectively. The 2D image in
Eq. (12.48) is manufactured by replacing the argument of the 1D function with the
“equation” of a line:
cos( θ ) x 0 +sin( θ ) y 0 = τ (12.50)
Clearly, the gray value does not change as long as we are on a certain line, i.e.,
the x 0 ,y 0 pair that satisfies Eq. (12.50), and therefore the notion of line is justified
when speaking about linearly symmetric images. We take this CT reasoning one step
further and translate one of the lines in the pattern g .
We can assume that the position vector s 0 =( x 0 ,y 0 ) T represents a (spatial) point
in the image plane at the time instant t =0and we wish to move it with the velocity
v a , where a is the direction of the velocity (
=1) and v is the absolute speed.
A velocity in the direction orthogonal to a , i.e., the line moves “along itself”, will
not be observable. This problem is also known as the aperture problem . That is why
only in the direction a can a motion be observed in a linearly symmetric image, Eq.
(12.48). It may not be the true motion, but it is the only motion that we can observe.
Accordingly, after time t , a point on the line can be assumed to have moved to the
position
a
s ( t )=( x ( t ) ,y ( t )) T
= s 0 + vt
·
a
(12.51)
so that s 0 = s
a , although we should be aware that the point may have actually
moved to any place along the line. The vector v a is called the normal image velocity
or normal optical flow . Substituting, the gray-values expression in Eq. (12.48) yields
a spatio-temporal image (sequence) in which the lines of g move with the same
velocity:
vt
·
vt )= g ( k T r )
g ( a T s 0 )= g ( a T s ( t )
vt a T a )= g ( a T s
(12.52)
Here we have defined the new variables k and r as the spatial variables augmented
with the temporal variables
v and t , respectively.
k =[ a T
v ] T
r =[ s T
t ] T
|−
E 3 ,
|
E 3
(12.53)
However, Eq. (12.52) represents a linearly symmetric 3D image having paral-
lel planes as isosurfaces. The vector k is thus equal to the normal of the plane
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