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International Archives of the Photogrammetry, Remote Sensing and Spatial Information Sciences, Vol XXXV, Part B3. Istanbul 2004 
  
  
  
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Figure 2. Ellipse characteristics. 
The coverage of the calculated ellipse is a very small area. 
This is expected considering that the final ambiguities along 
the two main directions are very small, at the magnitude of 
+0.1 pixels or even lower. Hence a scale factor must be 
applied in the two main axes. In fact the information for the 
best size of the square template is already available from a 
previous step of the algorithm, described in Skarlatos, 2000. 
An algorithm is applied prior to matching to decide about the 
best possible size of the square template. The decision is 
based on statistical values about information around the pixel 
in the left (template) image. It should be noted here that this 
algorithm is location invariant and investigates each patch 
size based on the square template concept, not on the ellipse 
itself. It is not repeated during iterations, instead it is applied 
once prior to matching in each point. 
It is possible though to use the existing self-adaptive template 
algorithm to recalculate the best size of the ellipse, based on 
its shape and orientation. The constraint is that the checking 
should be done for selected areas of 80,100,150...900 pixels 
(equivalent to 9x9-31x31 window size). Recalculations of the 
description of the ellipse and the pixels within cause 
unacceptable delay and therefore such modification was 
rejected. 
Find appropriate scale factor for the ellipse. 
The area of the ellipse is mab, where a and b are the main 
semi-axes, or in this case n- Omi, - Omax - 1herefore the scale 
factor for each semi-axis is a . If each semiaxis 
T1: min * O max 
is multiplied by this factor, the new ellipse has area equal to 
new area. Proportions and orientation of the ellipse are 
maintained, absolutely necessary to the concept of this 
algorithm. 
Find the pixels in the ellipse. 
Theoretically pixels belonging to the ellipse should have 
more than 5094 of their area in it. This method of pixel 
identification consumes a lot of computer power, therefore a 
simpler method was used. If the center of the pixel is inside 
the ellipse then the pixel belongs to the ellipse. 
Therefore the two focal points of the ellipse are calculated e; 
and e» (fig. 3). The focal points are located on the large axis 
at distance y »« Jo24. - o2,, from the centre of the ellipse, 
hence their coordinates on the local coordinate system of the 
ellipse are (+y,0). By applying a rotation angle 0 and two 
shifts X, and Y,, (X, Y,) being the centre of the ellipse, the 
coordinates are transformed in the image coordinate system 
X|  |cos(0) sin(8) |x " Xo (4) 
Y ei > -sin(8) cos(8)]y ei Yo ellipse _ centre 
For a point (center of pixel) to be inside the ellipse it is 
necessary: 
dist(e,,p) + dist(e,,p) < 2a (5) 
Where a is the big semi-axis and p the pixel under 
investigation. 
This check is being done on all pixels within a square with 
sides of 2a, ensuring all possibilities for the direction of the 
ellipse are included in the check. This check is simple and 
fast. The only drawback is that the number of finally selected 
pixels does not coincide exactly with the desired area of the 
ellipse, as calculated on step 2. Statistically this is less than 
2% for the 99% of the cases. For small ellipses this 
percentage may go up to 3%, but drops rapidly when size 
increase, and therefore returns the aforementioned results 
over a matched model. In any case such discrepancies do not 
affect the general idea of the proposed method. 
bs À + je mim od de mom oo ow pe wmf do 
  
   
TL A 1 ele JL had 0 Tt adn 
ne pr rs [17171 E 
Figure 3. Ellipse (green solid line) with check area (red 
dashed), finally selected pixels (green dashed). 
Expected area 221 (=11x11) and finally selected 
pixel 218, representing differentiation less than 
1.5%. 
Use the inverse affine transformation to locate these 
pixels on the left image 
In order to perform LSQM the same pixels should also be 
located on the left (template) image. In order to do so the 
inverse affine transformation from the previous iteration is 
used to find the co-ordinates of these pixels on the left image. 
It is expected that after the transformation the left pixels will 
be in random positions (not integer values) and therefore 
interpolation is necessary to find the grey level values for 
these positions. The values are used as floating point numbers 
for further calculations. 
Formulate the matrices, A and | for least squares and 
solve them for the 8 parameters