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uckerian
line representation. The Pluckerian representation is the
canonical line representation in projective geometry.
The Pluckerian coordinates are defined as follow: Let
P be the cartesian coordinates of an arbitrary point on a
line D in a 3D space and 1 be the unit direction vector
of the line D. We introduce and often use the vector H
which is the orthogonal from the origin O to the line D. It
is easily seen that
PAI= HA
or
PAI=N=hn
where N = HA], and n = TN is the normal to the plane
defined by the 3D line D and the origin O, and finally
h = || H || = || N || represents the distance of the line to
the origin (see fig 2). Therefore, the line equation will be
PAI=N
The two vectors (1, N) define the Pluckerian coordinates of
the line D. Note that H, 1, and N form a right handed
coordinate system. n
Fig.2. The vectors n, 1, and h.
Using this line representation we need four parameters
to represent a 3D line. Two parameters for the unit line
direction l, and two parameters to define the vector IN or
H which are orthogonal to the line direction 1.
Image lines : A line d, the projection of a 3D line D
on the image plane is called a 2D line. In the camera
coordinate system, this line may be considered as a 3D
line which lies on the plane z — 1. Therefore its equation
is simply:
min =0
where m — (z, y, 1)? is an arbitrary point on d.
The vector n is the normal to the plane containing the
3D line, its image and the camera optical center. Therefore,
this vector n is the same as the vector n — INT introduced
in the previous paragraph. We may even use the vector
N to represent the image line when the 3D line is given in
camera coordinate system. Usually we have only access to
the image lines. Therefore, we prefer in general represent
the image lines by a unit vector n.
If we take n = (a, 8,7)" the line equation is written
as: :
az +pfy+y=0
The vector n is a unit vector and therefore, our 2D line
representation depend only on two parameters.
4 Ego-motion estimation
Much work has been done on the motion estimation from
straight lines. In the case of discrete motion, we can par-
ticularly mention the works of Liu, Huang, Spetsakis, Ag-
garwal, Chellapa, and Vieville [13, 12, 20, 1, 4, 25] on the
monocular sequences and that of Zhang [26] on the stereo
sequences. And in the case of continuous line motion analy-
sis approach we may refer to the works of Faugeras, Navab,
and Henriksen [8, 9, 19, 11]. One may easily verify that for
each of the fundamental formulae obtained in one case (dis-
crete or continue) one can find a similar and related one
in the other case [15]. In this paper, we use the continu-
ous approach. After the first steps, when we can use more
frames this approach shows its real advantages.
Let us take a 3D line represented by the vectors (N, 1).
We now describe its motion (N,1). In order to gain more
insight into the problem, we assume that the 3D line under
consideration is attached to a rigid body whose motion
is described by its instantaneous angular velocity, £2, and
linear velocity V, its kinematic screw at the origin O.We
can also suppose that the object is static and the camera
system has such a motion description.
We know that the velocity P of any point P attached
to the rigid body is given by
The normalized direction 1 satisfies a simpler differential
equation:
i=QAI (2)
The vector H can be expressed as
H = P — (PT) (3)
therefore,
H = P — (PTIi - (P71 + PTi)l
Replacing P andi by their values from equations 1 and 2,
H — V.CQOAIP- (P7 -(v?y
we obtain:
H-QAH-V -(V?DI (4)
Then it’s easy to obtain N, the time derivative of N = HAI:
HAL+HAI
(QAH-tV-(VTID)AI-HA(QA1)
QA(HAD)+VAI
N
and we obtain:
N=QAN+VAI (5)
Therefore, the motion of a 3D line can be defined as follows:
Asli] 6
where the matrix D is defined as follows:
Qo
»-[i 2]
Line Motion Field Equation: What we measure from
the images are the 2D lines represented by the unit vectors
n and their motion fields n. Therefore, we give here the
line motion field equation. Line motion field equation was
first given in [19]. We used two points on the line to obtain
that result. Here we draw the same equation from the
above equations. we have n = TNT: therefore