7 Part B3. Istanbul 2004
ges through equation (1)
figure 1. The Euclidean
and the line segment | is
> xl and x2 are points
-
x X», -) 5
nd
vector equation with 3-D
with the second equation
es.
fies the equality d (xp, xo)
sultant corresponding point
ed using least squares
ves
ed from model/object space
tion parameters as exterior
d for the recovery of the
metric model and the plane
n observed to initial values
satisfying results and the
a possible high correlation
the bending angles of the
tion could lead to numerical
of X and Y are nearly
er. We therefore suggest a
>
Z4
IZ -20)50 0
axis
ormal
ane from origin
International Archives of the Photogrammetry, Remote Sensing and Spatial Information Sciences, Vol XXXV , Part B3. Istanbul 2004
The last representation changes the sought parameters to 0, Q
and D, but leads to another problem. If the plane is horizontal so
0 = 0, the derivative according to ¢ is infinite, because it does
not change the normal vector when multiplying with 0.
Therefore, when dealing with planes close to horizontal,
numeric problems are expected.
3. FINDING FUNDAMENTAL MATRIX
USING HOMOGRAPHY MAPPING
3.1 Homography mapping
Homography mapping transfer points from one image to the
second image as if they were on the plane in the object space
(Hartley 2000). As seen in figure 2. points on a plane are related
to correspondences point on the images of the photogrammetric
model. In fact this is a projective, having 8 free parameters. The
8 parameters can be obtained from the 5 relative orientation
parameters and the 3 planar parameters.
/ 7i. |
/ \
/ XT \
/ ~ \
f \
\
s > x =
= ;
= Ë
Bed
X
M 2
= \ „x
SS =
Lt T ~~ ; d
> a T te x
e S | >
OL H
Figure 2 . homography mapping
The homography induced by the plane is unique (see Tsai
(1982)), meaning that every planar curve can contribute one
homography. The homography transfer operator is linear for
homogenous coordinates and the mapping from one image to
the other is unique up to a scale factor.
The homography matrix:
D ^m 1H
H=|h h 4 (4)
h nh 3
and the mapping from right to left image vectors are readily
given by
U,-H-U, (5)
where Ur, and Ul are homogenous coordinates in right and left
images respectively.
One should notice that the mapping can be from the right image
to the left image and vice versa. In this paper we have chosen
the one from right image to the left image.
The homography matrix can be computed directly from the
relative orientation parameters and the plane parameters. The
common way to compute the homography matrix is by
determining the coordinate system of the model parallel to the
coordinate system of the left image. By choosing this coordinate
system the rotation matrix of the left image is the identity
matrix and the translation vector is zero. The rotation matrix
and the translation vector of the right image can be obtained
from the relative orientation parameters.
The homography matrix can be computed with rotation matrix
R and the displacement vector T according to equation (6).
Hz[Rx*T-m/D] (o
where : R — rotation matrix
T - displacement vector
n — unit vector of the plane normal
D — distance from the origin
The homogenous coordinate are obtained by dividing the image
coordinate by focal length as follows:
X u
ys vq
-f l
Using homogenous coordinate makes the homography mapping
correct up to scale. Multiplying the homogenous coordinate
with the homography matrix H is simply linear procedure, but
getting the right scale requires a determination of a scale factor.
Dividing the outcome vector by the 3" coordinate can answer
this question so the transformation from one image to the other
can be written as follows:
1
U za
mf
A: X, VE
© h,-x"+h, - y'+1
7 ay (8)
hix EI
1
= n n
hx hy"
where: x", y" — right image coordinate
X' y' - left image coordinate
hl .. h8 component of the homography matrix
Equations (8) remind the collinear equations, but one should
remember that the collinear equations transform from 3D object
space to 2D image space while those equations transform from
2D image space to another 2D image space.
3.2 Fundamental matrix
Yet another well-known relation between two images is the
epipolar geometry. One point from first image determines line
in the second image. The fundamental matrix defines this
relation with the constrain x" F x" = 0, obeying the coplanar
condition. Using the homography matrix we can write the
constrain (Hx")' F x" = x" H' F x = 0 for any point on the plane
that induced the specific homography. Hence, the matrix (Ht F)
must be skew-symmetric, namely
HF+FH=0 (9)
