‘est Hungary 
ce resection and in 
t'y complex and the 
inimally necessary 
he space resection, 
rientation elements 
alue following the 
-error and it can be 
thod or the robust 
tions, which needs 
space resection in 
1 initial values and 
t a unique solution 
ind the differences 
of the gross error 
1ethod. 
ve solution we can 
give approximate 
accuracy which is 
iterative process 
netry) 
rs is not so easy if 
s errors. The least 
ss errors to other 
> robust estimators 
ber of points with 
| avoid the above 
  
International Archives of the Photogrammetry, Remote Sensing 
1.2 The procedure 
First I give a direct solution for the space resection based on 
three control points (chapter 2.1). Than I explain how to get the 
adjusted values of unknowns using the Jacobian Mean Theorem 
(Gleinsvik, 1967), if the number of control points is more than 
three (chapter 2.2). And finally I present the basic formulas 
necessary to detect the points with gross error (chapter 2.3). 
2. DIRECT ANALYTICAL METHODS 
2.1 Space resection without adjustment 
As it is seen on Figure 1. we have a tetrahedron with a,b,c 
sides. The ABC triangle is known and formed by the control 
points, The 4'B'C'P tetrahedron including the ¢, B,y is also 
known after the measurement of image points. The goal is to 
determine the outer orientation elements (@,@,K and 
Xo»Yo>Zo). It is wise to first calculate only the projection 
center coordinates and after this the rotation angles can be 
calculated with well known direct equations. 
  
  
M D 
i 
A 
V 
e Y« 
T 
& nt s 
pe 
\ > 
AC 
7 \ 
A: 
uU 
n 
^ 
-— — — n9 HÀ— Lor 
f T mee f 
| v 
o HE 
ctt 
E p 
  
  
  
Figure 1. Space resection based on three control points 
Let’s derive equations for the a,b,c sides, since using these 
values we can calculate the P projection center coordinates 
using the well known distance equations from the coordinate 
geometry. 
If we take the a side as a basic distance the sides of b. and C 
will differ only with an n scale factor, so in this case we have 
only two unknowns ( a and 7 ). 
We can setup three independent equations for the triangles 
AABC , ABCP and AACP using the cosine-theorem: 
  
and Spatial Information Sciences, Vol XXXV, Part B3. Istanbul 2004 
2 2 5 
d'-a -(a-ny -2a’ncosa 
(2) 
e? z (a- ny. *« (a- my. — 2a? nmcos f 
f =(a-m} +a —2a^mcosy 
We can eliminate the side a from the equations reducing the 
three equations into one forth-degree equation (Jancso, 1994): 
Win' + Won’ - Wa -WnW,-0 (3) 
where W, - f(d.e,f,a, fy) 
After solving the equation (3) we can calculate the unknown 
sides: 
d 
a= + FA SATT (4) 
l+n° —2ncosa 
b=a-n 
e -f ea hl 
C-mm- 
2(acosy —bcos B) 
Now we can calculate the projection center coordinates using 
the distance equations: 
Xp XV ANY) HZ. Zu 
FEN HN ET 6 
c =(X, made)! *(Y, 7 Val *(, -Zoy 
The solution of (5) is : 
I? T [3 ; [2 > 
7. = -Y,E4Xv,-4u.w, SE EVEN MW LEN EYE Aw, 
ep -— 
  
= :Z,>0 
Qu, 2, 2u, , (6) 
AX mk -kT, 
Yo mk RE, 
where Hnc obe Wade k, , parameters are functions of 
coordinates of the A, B,C control points and the sides of 
a,b,c (Jancso, 1994). 
Finally we can calculate the rotation angles of @,w,x from the 
rotation matrix with the well-known direct equations (Hirvonen, 
1964): 
he a (Hs 
Rely, 1, fs (7) 
In fo Tay 
and 
pz -arctg(r,/r..) 
Q = arcsin(r,, ) (8) 
x = —arctg(n,/r,,)