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
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