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my WW FRI 
  
  
ISPRS Commission III, Vol.34, Part 3A »Photogrammetric Computer Vision“, Graz, 2002 
  
  
  
  
  
  
  
  
rank(1;) Properties for I, Properties for Ij 
2 I. 3 P ~ dá bn qup E : e ev (s boas o: As 
Vay V2i 
T: : : 
5 0 la DV = 0 J y 
O» € ris I.7Y5-Vi Lh 2 3 Va mais 3 3 V32 
pz = pencil of lines in Vo; V32 Sa 
1: x à 
= 0 la DV = 0 d 
Quer ien 9 Lp TR ré re 
Ae = pencil of lines in V3; 31 Vas 
  
  
  
  
Table 3: Additional mapping properties of the correlation slices I;; c.f. Table 2 
Summing it up, the presented minimal set of constraints, 
ensures, that the three correlation slices of a TFT are sin- 
gular mappings of the lines from one image (V2) to the 
points of another image (3) via three concurrent 3D-lines, 
which are made up by the principal lines of image v. 
6 The minimal parameterization for the TFT 
In the previous section we did not only prove, that (14), 
(15) and (19) constitute a minimal set of constraints for the 
TFT, but we also found a minimal parameterization for it. 
If we adopt the equality of coefficients (v — v9 — v3 — v 
and w, = w2 = W3 = w) to the parameterization (17), we 
get this minimal parameterization (having 18 DOF): 
I; = là, b,ê] = (81, 7-81 +5 Wr + 5 Yu 
Isl ,ê,f] = [$2, v-82 + k - Vai, w - 82 + t - Van] 
Is = [& h, ] = [83, 0-83 +1 - Va1 , W - 83 + U - Va1] 
(20) 
A few remarks need to be given: 
e Obviously, this parameterization is not linear. Thus 
approximations are required, which can be obtained 
by an initial solution using the well-known eigen-value 
or linear solution for the TFT; e.g. [Hartley 1994]. 
e The scale of V31 needs to be fixed, e.g. by setting its 
length to 1. 
e Observe, that the vectors {81,82,83} in (20) param- 
eterize the same column (index c,) in the matrices 
I.. For numerical reasons, this index should be that 
one, for which the respective columns are farthest 
away from Vai. This index c, may be found by 
Il: Ze . ec, X Vai || — Max. 
e The overall scale in this parameterization needs also 
to be fixed, e.g. by setting the length of the concate- 
nated vectors (81,$2,$3] to 1. This yields in total 3 
possible mappings; i.e. the choice of c;. 
With this parameterization (20) (i.e. c, — 1) we get the 
homographic slices J, as follows: 
. Ji [à d, &] = [£1, 82, $a] 
Jo = [b, é, h = [v-81 + j-V31, v-82 + k-V31, v-83 + l-V31] 
Ja = T f. i| = [w-$1 + S-V31 ; Ww:$9 + t-V31 ; w-:$3 T u:Vai] 
(21) 
And so we can look at the general eigen-value problem 
Ja — p+ Ji. It is easy to see, that u — v yields a 2- 
dimensional general eigen-space, the line (j kl)'. Thus v 
is an eigen-value with multiplicity 2. This is in accordance 
with [Canterakis 2000]. And so we can summarize the ge- 
ometrical interpretation of this minimal parameterization 
in the following way: 
e V3; is the epipole of base O10 in image vs. 
e [$1, $2, $3] is a homography from image v1 to image 
13; i.e. the homographic slice J;. 
e (1vw)! is the epipole V2: (of base O10» in image 
V2) - its component at position c, is set to 1. 
e (jk D is the 2-dimensional general eigenspace of 
(Ja — u - J1). Since J, resp. J2 is a homography due 
to 731 resp. T22, the general eigenvector of (Ja—u-J1) 
must be the projection of the intersection of these two 
principal planes of image v»; i.e. the projection of the 
principal ray ro3 of image v» into image v1. 
e (stu)! is the 2-dimensional general eigenspace of 
(Ja — v. J1), ie. the projection of the principal ray 
roo Of image v»? into image v1. 
Of interest are the critical configurations for this minimal 
parameterization. Since it is part of the parameterization, 
that the length of V31 and one component in vo; are set 
to 1, problems surely arise if either of these epipoles is the 
zero-vector — O, = Os resp. O1 = O». This problem 
can be solved - as long as not all three projection centers 
coincide - by changing the role of the images in the way 
that the image with the unique projection center plays 
the role of image w%1. Note: The identity of two or all 
three projection centers might be of practical relevance 
during the work with a moving camera acquiring images 
in a constant frequency and which stops at a particular 
position for a moment. In case of O1 = O2 = Os the 
respective TF'T becomes the zero-tensor. 
Another problem with this parameterization could come 
from the fact, that the vectors ($1,92,$3) parameterize 
the same column (with index c,) in all three correlation 
slices. Still keep in mind that we choose the best column 
for this parameterization - the one that is farthest away 
from V31. If we take the minimal parameterization exactly 
as it is given in equation (20), we see, that the columns of 
I, are parameterized by vs1 and the vector s;. Thus it 
must be assured, that 8; is different from v3; and different 
from the zero-vector, because otherwise the column vectors 
b and/or ¢ (being different from V3; and 0) can not be 
parameterized by $1 and Vai. Of course, if b and 6 are 
similar to V31 or 0, than we would have no problem. So, 
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