ISPRS Commission III, Vol.34, Part 3A „Photogrammetric Computer Vision“, Graz, 2002 
  
  
3. The method of quadratic weighted average, 
which is weighted too but the weight function is not 
linear but quadratic function of dependence from 
detection object pixels colors. 
V, = 2*P+1 - 2(255-F,) (5) 
2 
v, 2 2*P«41- AS Sd ; (6) 
P 4 
where 
v;- Weight of i-th pixel of the detection object 
P - threshold 
F;- color of i-th pixel of the detection object 
The tests showed that the best results were 
obtained with the second method. In some cases the 
first (less often the third) method was more accurate. 
The accuracy of the methods was dependent on the 
shape, size and color of the detection objects. 
To be able to determine relative position of 
the copys it is needed to know two detection objects at 
least. It means that in every copy I must be able to find 
the same detection objects identical to the objects in 
the scene. I suppose, that such pair of detection objects 
exists in every record. Then by the selection of the 
most suitable method I come out from following 
assumption: The smaller is the spread of the distances 
of identical pairs of detection objects centers 
(calculated by certain method) in the records the more 
accurate is that method. The spread means the size of 
the scale of distances of given pairs if centers calculed 
for all the records. 
R= max {| Vadial — Vzdial | | (7) 
R — the scale of the distances of given pair of detection 
objects 
Vzdial; — distance of given pair of detection objects in 
the i-th record 
index i (j) represents i-th (j-th) record (if there are N 
records then i 7 1, ..., N) 
Vij means that with indexes ij go through all the 
copys 
If I have more than two detection objects 
I perform described procedure for every pair. Then the 
best pair of the detection objects centers is the one, 
which scale if the distances was minimal. These 
centers I label by symbols T ; T! , Where i is the index 
of the copy. 
Transformation of records to equalize the detection 
objects coordinates 
At first the image with doubled resolution is 
created in the way that every pixel in first record is 
divided into four pixels containing the same color as 
the pixel from the first record. Le. by the resampling 
of the first record I get the base for image with 
enhanced resolution. 
When I have the centers of the detection 
objects pair for each record I can move the records on 
the first record that the corresponding centers are 
identical. T! zT, T — T, for each i. 
2. datantion 
abject of 
reoord 
   
1.dete 
object of 
avary 
racord 
  
Figure 4 The translation and rotation of the records 
on the first record 
To compute the pixels in the enhanced image 
it is needed to know the equations of the lines 
dividing two neighbouring rows (X-lines) or columns 
(Y-lines) in the records. It is necessary to determine 
these equations in such position of the record in which 
the detection objects T}, T, center coordinates are 
equal to detection objects centers coordinates T! Ti 
of the first snapped copy. The example for one of the 
copys (red pixel grid) which is translated and rotated 
to validate T! = T ; T. zT! is illustrated in the 
Figure 4. The gray pixel grid represents the enhanced 
image.All the records are translated in the way that the 
T coordinates for every record and T! coordinates 
for the first record have equal integer parts. Eg. if T! 
— (23.4,13.5) and in the second record is T^ = 
(38.1,24.3), then the second record pixels are 
translated about 15=38-23 in x-direction and about 
11=24-13 in y-direction. 
R-lina 
S-Line 
  
Figure 5 The translation and rotation of the record to 
the resulting image 
A - 260