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# Full text

Title
The 3rd ISPRS Workshop on Dynamic and Multi-Dimensional GIS & the 10th Annual Conference of CPGIS on Geoinformatics
Author
Chen, Jun

196
ISPRS, Vol.34, Part 2W2, “Dynamic and Multi-Dimensional GIS", Bangkok, May 23-25, 2001
Figure3-1 parallelogram l,j,k,l and volume of a tetrahedron
The volumes of the 4 tetrahedrons(Vc-ijk,Vc-ilk,vc-ijl,Vc-jkl) can
be computed from the 3 vectors defined by the image
coordinates, the focal length and the distances from the
projection center to the 3 corner points. Volume of the
tetrahedron (Vc-ijk)
didjdk I , j, k jj
4№l
where
(20)
Xi = i = (Xi,yi-f)
|/j, |y|, |A:| express the size of the vectori.j.k.
di.dj.dk express the distance between c and l.j.k.
Because of the parallelity, a diagonal of the parallelogram always
splits its area in 2 equal parts. And thus, the areas of the 4
triangles^*,Aiik.Ajki,Aim) in the plane of the parallelogram are all
equal. As the same holds for the distance h of the projection
center to the plane of the parallelogram, ilt follows that the
volumes of the 4 tetrahedrons are identical.
dc0[i,j,k\ jkk^j,kj\ jldct[l,i, j\
~m~ = isr HIT TT
The distance ratios Cij can be written as follows
[kj,j]
[jM
(21)
Cik =
4 model coordinates and 3D coordinates
Model coordinates (X m i) can be written as a function of one
distance di that defines the scale of the model
tiXl\
(22)
x image coordinates.
X the available coordinates in the object system
X° the coordinates of the projection center.
5 experimental result
One simulated test and one real test have been conducted to
check the feasibility of the new method.
simulated test The first set of data are a simulated cube, whose
coordinates of 3D and corresponding to 2-D are designed by
back-projection. Three couples of parallel straight lines in cube
are parallel to X.Y.Z axes in object coordinate system. The
simulated orientation parameters of the camera are listed in table
5-1.The simulated control points are listed in table 5-2. The
results are listed in table 5-3 and use OpenGL function in visual
C++ to display result as figure 5-1.
Table 5-1 the simulated orientation parameters
Interior
orientation
parameters
xO(pixel)
yO(pixel)
F(pixel)
-274.50
-236.98
2604. 76
Exterior
orientation
parameters

w
K
-0.6910
0.5642
0.9172
JtM
Xs(mm)
Ys(mm)
Zs(mm)
57910.90
-47492.94
54088.57
Table 5-2 the simulated control points
number
X(m
m)
Y(mm)
Z(mm)
u(pixel)
v(pixel)
1
0
0
0
-492.26
-40.85
2
100
0
0
-490.25
-41.89
3
100
0
100
-490.57
-39.60
4
0
0
100
-492.54
-38.56
5
0
100
0
-490.19
-39.51
6
0
100
100
-490.51
-37.22
7
100
100
100
-488.54
-38.25
(23)
X m j = XiCtjXj
From the first step, interior and exterior orientation parameters
are known. So, object coordinates can be computed in following
form
ÁRx¡ = X,-X°
XRCijxj = Xj - X°
(24)
where X scale factor
R rotation matrix
Cij distance ratio projection center-points i and j from step 2.