188 
where x B 1s the vector of the contravarisnt 
coordinates related to B and the br are the columns 
of B®. The Inverse form of (1.2.3), namely 
x" = By = (B*)- 1 y, (1.2.4) 
gives the transformation of y Into the system B. 
The rows of B contain, 1n analogy to (1.2.2), the 
reciprocal vectors b 1 of the system bi . As the 
vectors myi establish the contravariant coordinate 
system, their transformation into 1t will yield the 
unit vectors ei 8 = By* and the transformation of 
yE must produce the unit point (e* B ) T = (1,1,1,1). 
Fig.1 illustrates, that the image y’= py of the 
vector y may be represented by affine coordinates 
of the image space B as 
y’“U 1 bi +u 2 b2+u 3 b3 + ( 1-u 1 -u 2 -u 3 )bo = B®Uu (1.2.5) 
with u T =(1,u 1 ,u 2 ,u 3 ), 
1 -1 -1 -1 
1111 
0 10 0 
and U' 1 = 
0 10 0 
0 0 10 
0 0 10 
0 0 0 1 
0 0 0 1 
■ 
. 
therefore results from B* = (UP) -1 1f, as usual in 
photogrammetry, the vector u contains a constant 
homogenizing component and two or more Inhomo 
geneous affine cordmates of the image space. 
l»-3—Ir.s<! 1 ^ format ion .between two different projective 
spaces 
A mam problem of photogrammetry is to find the 
transformation 
u" - Pu’ 
from a projective space P’ (e.g. model) into a 
reference space P" (system of object), by means of 
control points. As it was shown, generally n+2 
points (=5 in three dimensions) are required. 
From (1.2.6) and (1.2.4) 
u" = p"U _1 B"y and u’ = M ? U' 1 B’y . 
may be derived. The second equation yields 
1 
y = - (B’)- 1 Uu\ (1.3.1) 
p’ 
and thence by means of the first one 
M" 
u" = - U* 1 B"(B’)- 1 Uu’ = Mu’. (1.3.2) 
p’ 
Using four of the position vectors of the control 
points as base vectors bi", the base of P’ (compare 
section 1.2) will be 
bi’ = ui’bi" 
and therefor 
B'B” = 
b° " T 
b 1 ” T 
b 2 " T 
b 3 " T 
^ Mo * bo " M1 ’ bi ” M2 ’ b2 ” M3 ’ b3 " j . 
Fig. 1: Unit tetrahedron of a base B(bi). The Gi 
are the so-called basic points of the 
projective space P 3 
Because of (1.2.3) equation 
B*Uu = pB*x B , Uu = mx b , u = pU* 1 x B (1.2.6) 
follows and thus 
1 
Zx 1 
u 1 
= p 
X 1 
u 2 
x 2 
u 3 
x 3 
. 
In order to get Inhomogeneous coordinates of the 
image space the vector w of (1.1.1) must be divided 
by 
Its own first 
(homogenizing) component 
wo 
PO T y 
W1 
Pi T y 
— 
= 1 = — , 
— 
= u 1 = , 
Wo 
P0 T y 
wo 
po T y 
W2 
P2 T y 
W3 
P3 T y 
— 
= U 2= , 
— 
= u 3 = , 
Wo 
po T y 
wo 
PO T y 
so that at the left of (1.1.1) u may substitute w 
and the projective matrix P depends on the base B 
through the relation P=U _1 B. 
The base of any regular projective transformation 
According to (1.2.3) and (1.2.4) the product b’ T bi 
of a base vector and its reciprocal results 1n 1, 
the other products yield 0 and because of this fact 
the product of the matrices must be 
M" 
- B' B” 
p’ 
po 
Pi 
P2 
P3 
(1.3.3) 
where pi=p"pi’/p’. These four unknowns can be 
determined from the coordinates of the fifth point 
using the expanded form 
1 
po pi-po P2-pO P3~P0 
1 
u 1 " 
z 
0 Ml o 0 
u 1 ’ 
u 2 " 
0 0 M2 0 
u 2 ’ 
u 3 " 
0 0 0 M3 
u 3 ’ 
. 
» 
. 
of (1.3.2) giving the simple relations 
) 
1 
1 
c 
r 
c 
r 
c 
GJ 
u 1 " 
cu 
3 J 
1 
{ 
u 3 " 
i M1 - 
» H4. • 1 
, M3- 
• 
1-u 1 ’-u 2 ’-u 3 ’ 
u 1 ’ 
u 2 ’ 
u 3 ’ 
(1.3.4) 
From the first term 1t is seen, that the fifth 
point does not dare to coincide with the plane 
u 1 ‘+u 2 ’ + u 3 ’ = 1 ( = plane through G1,G2,G3 in fig. 1) 
nor with one of the affine coordinate axis as show 
the other terms.