Full text: XVIIth ISPRS Congress (Part B3)

1.1) of 
by a 
ise of 
3.1) 
on the 
| case. 
! used, 
nd the 
). YN 7C, 
3.2) 
AN ap- 
irallel 
nts in 
'ormula 
3.3) 
1. The 
rom P’ 
d from 
je ex- 
3.4) 
om the 
photo- 
at the 
|, the 
ng K’, 
ng K'. 
by the 
2.1.1) 
[.2) 
ecause 
i1s*i2"-i2*i3" ^3" ja "712! ja" - ia" ka" - do" Kg" 
Cz | js?i2"7-J2?da" ^ 3s? ja "-ja" ja"  ja'ke"-je" Ks" 
ks'i2"-k2'ia" ks'ja"-ke'ja" Ks'ka"-k2'ka" 
(2.13.3) 
and contains only the second and third components 
of the 1, j, k. 
2.2 Computation of the parameters 
First of all it is to be assumed that the coor- 
dinates xo’, yo’ and xo", yo" of the epipoles are 
already calculated from 
xg’ = 0 and ZX" =. 0. 
They are the images of the base given by 
Ao’xo’ = R’Tb and Ao"Xo" z-R'Tb, 
or 
xo’ cos?! cos Xo" cose"cosK" 
^o'|yo' ||-cose'sinK' | and Ao" |yo" |-- |-cose" sin" |, 
-C sine -C sine" 
from which independently from Q" follow 
yo -C 
and tan¢ = 
tank = - 
Xo 4 Xo? *yo? 
  
(2.3.1) 
for both images. By means of these parameters R’= 
z[i',j',k"] is known. 
The still missing parameter Q' of R" may be cal- 
culated now from any component of (2.1.3). The best 
way is to use the third column 
(iz'cosQ" * is'sinQ")cose" - cs2z13a 
(j2'cosQ" + js'sinQ")cose" - cs2zes 
- ks'sinQ" cos?" z ca2233a 
and to eliminate cos" by 
C32233 
cose” = 
- ks'sin2" 
Therefrom the two symmetric possibilities 
233 i2’ 
tan@'= ———————— — 
Z13 ks! - za3 is? 
233 j2' 
= re eme (2.3.2) 
Z23 Ks’ - z3s js” 
orientation has been linearized by more obser- 
vations than necessary. Moreover, C is calculated 
irrespective of the conditions of rectangularity 
and normalization of the unit vectors 1, j. k, so 
that an iterative post-processing must take place 
in order to get an algebraically and stochastically 
consistent set of parameters. 
2.3 Adjustment 
The rotation matrices of section 2.2 undoubtedly 
will be very close approximations (R) to the most 
probable solutions R. Hence small additional ro- 
tations dR will give the final position of the 
images according to 
0  -dK d4 
R = dR(R) = (E+dA)(R), dA = ax 0 -do 
-dé "dO 0 
By means of a vector v'z(vx,vy,0) of the residuals 
of coordinate measurement and by neglecting quanti- 
ties of second order, (1.2.3) turns to 
(x’+v’)T{(E+dA’ ) (R’)}TB{(E+dA") (R") } (X"+v")= 
zX'T(C)x" *v'T(C)x" *x'T(C)v" * (p' )T dA'TB(p")4 
+(p’)TBdA"(p")=o, 
wherein (C)=(R’)TB(R") and (p)=(R)x. Because of 
0. -dé -dk 
dA'TB - 0 0 0 » 
0 0 0 
0 0 0 
BdA" - dé)" -doO" 0 
dK” 0 -00" 
and using the substitutions dp=x’T(C)x"(=parallax), 
v'T(CO)x'zv'T(h'), x''(C)v'z(h")'v" (according to 
equ. (1.2.6)), one linearized coplanarity equation 
(without round brackets at h and pi) reads 
op + h'.v' * h'.v" = p1’p2"de®’ + p1’ps"dK’ - 
- (pa'"pe'*ps"ps')dQ" - p1"p2’de’ - p1"ps’dK" 
and represents formally the general case of least 
squares adjustment, i.e. conditions with unknowns. 
But as the residuals of one equation do not appear 
in any other equation (Tab. 2.3), the procedure can 
be simplified by introduction of the fictitious 
residuals (Wolf 1968, p.105, Rinner 1972, p.402) 
Ww = h1’vx’+h2’vy’+h1 "vx "+h2 "vy" 
and the related weights 
1 h1’2 hz’? hi"? h2"? 
  
  
  
  
arise for the determination of ©", which result me m + + + ; 
from the fact that the transcendental problem of g gx"? Oy’ ax" dy" 
Residuals Unknowns Par. 
i|vx'vy'vx" vy" vx'vy? vx" vy" ...... vx'vy'vx" vy" ]de'dC de" de"dK"| Sp 
1/59 8 B H B B B B B Sp1 
B H B BR BH 
2 8B HB NH HN 
8 
  
  
ôp2 
  
B B B H|H EH BR HB B | Ops 
  
  
  
Tab. 2.3: Scheme of the linearized equations of coplanarity 
703 
 
	        
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