ISPRS, Vol.34, Part 2W2, “Dynamic and Multi-Dimensional GIS”, Bangkok, May 23-25, 2001
195
The plane defined by perspective center of image S and straight
line I is called interpretation plane. According to coplanar
coordination of perspective geometry, the straight line L in the
object space is on the plane defined by the straight line I
(projection of L) and Sp. So, L is perpendicular to the normal
vector of plane defined by S and I in S-xyz .the dot-product of L
and n is equal to zero
-vyvX + v*-vY+ d - Vx .--^- + ^ y ~. JC ~. vZ = 0 (5)
/
Equation (5) may be expressed by
■Vy • a + Vx • ß +
di - Vx • JA) + Vy XO
/
= 0 (6)
vA"
where OC =
vZ
(7)
When L is parallel to the X-axis in the object coordinate system,
we can transform vector L from O-XYZ to S-xyz with the rotation
matrix R and scale X..
vX~
c
¿21
bl
C\
c
¿21c
vY
= Â-'R-'
0
= r'
¿22
bi
Cl
0
aie
(8)
vZ
0
¿23
bi
C 3
0
aie
where c,0,0 are three coordinates of L on the X-,Y- and Z- axes
in the object coordinate system respectively,
substituting equation (8) into equation (7)
a 1
a 1
A3
¿23
(9)
<22 = k-(h2 + xo)
ßi^kßgi + yo) (15)
<23 = k • (/23 + xo)
ßi = k-(gi + yo)
where h2,g2,h 3 ,g 3 have meaning similar to h 3 ,gi,respectively.
Considering orthogonal constrains of the nine elements in the
rotation matrix R, we can get the following equations
h-h+gi-gi-dh+h)-xo-d^+g2)-y)+X) 2 +y) 2 -\-f 2 =0
h-h+gI • &Mh +/»)*AD+(gi +^)->o W +>o 2 +f= 0 (16)
k-h+g2-&Mfa+h)-M-){gi+&)-y)+X) 2 +y) 2 +f 2 =0
Obviously, in the linear equation above, the parameter x 0 ,yoand f
can be solved.
If we know the interior orientation parameters(x 0 ,yo,f)of camera,
considering the equation (9),(12),(14) ,(15),we can get
1
<23 = —============
yj<21 2 + ß\ 2 — 1
1
C3 = . (17)
yjcn 2 +/?3 2 -l
y]oC2 2 + ¡3 2 2 —1
Substituting equation (17) into equation (9) and (14), we can
obtain nine elements of R
where R = (cii,bi,Ci) 1 ,i=1,2,3 is the rotation matrix defining
the camera orientation. ai,bi,ci are the direction cosines or the
elements of the rotation matrix R.
Suppose Hand I2 in the image plane are the projections of two
straight lines parallel to the X axis in the object coordinate
system. Equation (6) can be given by another expression of the
following form respectively
—Vy\ • <21+Vx\ • f3\+k • d\ - k ■ Vxi • yo+k ■ Vyi • xo = 0
-Vy2-OC\+Vx2-/3\+k-d2~k-Vc2'yo+k-Vy2-XO=0 (10)
7 1
where K = — (11)
/
According to equation (10) , we can obtain
a\ = k-(h\ + xo)
fi\ = k-(g\ + yo) (12)
Vx2 • di — Vx\ • di
where Vyl ’ Vx2 v >' 2 ' Vjrl (13)
Vy2 ■ d\ - Vxl ■ d2
gl =
Vyl ■ Vx2 - Vyl ■ Vxl
Similarly, when Li is parallel to the Y-axis in the object coordinate
system or Li is parallel to the Z-axis in the object coordinate
system we can get
Oil - —
bi
to
II
Cl
II
CO
a 3 = —
C3
b2
bi
C2
C 3
(14)
According to collinear equations
a\(X-Xs)+b\(Y-Ys)+c\(Z-Zs)
X X °~ J a3(X-Xs)+b3-(Y-Ys)+C3(Z-Zs)
a2(X-3S)+fe(y-K)+C2(Z-Z^)
y yn ~ 3 cn(X-Xs)+bi{Y-Y)+a(Z-2s)
If knowing the rotation matrix R, the camera interior orientation
parameter(x 0 ,yo,f)and two control points, according to Taylor’s
theorem, Eqs(18) may be expressed in linear form as
fix fix fix
X = (X) +—dXs + ——dYs + ——dZs
dXs dYs dZs (19)
y - 00+-¥-dXs + -^dYs + -^-dZs
dXs dYs dZs
where (x),(y) are approximates of function. dX s ,dY s ,dZs are
corrections of X s ,Y s ,Zs respectively. Xs,Y s ,Z s are unknown
parameters. x,y are observations. So, we can get the adjustment
model and the least-square solution to equation (19).
3 Computation of the distance ratios
These lines in the image are assumed to correspond to lines
in object space that are coplanar and parallel(see figure3-1). The
image coordinates of the 4 corner points(i,j,k,l) of the
parallelogram are arrived by measure manually or line
intersection after edge detection. In the camera system, the
image point can be expressed in vector built from the image
coordinates^,y) and the focal length(f).This is the vector(x,y,-f)
