L1) 
I.3) 
lous 
for 
ase 
ela- 
for 
hich 
n is 
jua- 
tain 
are 
jum 
1 in 
ym- 
| of 
  
RELATIVE ORIENTATION IN MOUNTAINOUS TERRAIN, VAN DER WEELE 145 
  
—p, = —Z,.4B T AE 
—P = —Z,.4B qTb.4D-- AE 
Ep. — 23 Gt k2) AB + kAC + AE (11.4) 
—p, = —k.bAA — Z, (1 + k2) AB + kAC + b.AD + AE 
—p; = — Z; (1 + k2) AB — kAC + AE 
—pa = +k.bAA — Ze (1 + k2) AB — kAC + b.4AD + AE 
It will be obvious, from this group of formulae, that the only difference between the 
orientation problems for flat and mountainous terrain lies in the determination of the 
element AB (or Aw), and hence that, as soon as a solution has been obtained for 
Aw, there will be no differences between flat and mountainous terrain as far as the 
solution of the orientation problem is concerned. 
As a consequence of this, the remaining part of this paper will be chiefly concerned 
with the determination of omega. 
II.4. General consideration. 
In view of the fact that the coefficient of Aw, in the parallax formula, contains only 
Y- and Z-coordinates, it is evident that observations of parallax will be required in 
points, with different Y- and Z-coordinates in order to find a solution, although the X- 
coordinate may be constant. 
Following a suggestion of Poivilliers (Photogrammetria 1949/50, nr. 3) a further 
simplification of the parallax formula is obtained if, in equation II.2, X is considered as 
being constant: 
, 72 + Y? 
—P, = 7 (X.44—4€0) — 7 —AB t (X. AD t AE) 
Y Z2 + y? 
or: py mg dF gr AB AG iN RL (ILE) 
where: AF = XAA — AC — XAg, — (X—) Ag; — Ab, + Ab 
AG = XAD + AE = X An, — (X—b) Any - Ab, — Ab 
and B = Ao, — Joy, 
(IT.6) 
VII 
Taking a constant X dictates that we should use points lying in a cross-section 
through the model in a direction perpendicular to the base line, and it is necessary to 
observe p, in at least three such points in order to solve for the three unknowns of equa- 
tion (IL5). The fictitious elements AF, and AG, act, in the cross-section, as movements 
of the projection centre of the bundle of rays (or its orthogonal projection in the plane 
of the cross-section), in the Z- and Y-directions respectively, while 4B is a rotation of 
the bundle about that projection centre. 
Thus the solution of the orientation problem in one cross-section, is comparable with 
the solution of the resection problem in plane surveying. 
Il.5. Geometrical explanation. 
Il.5.3. Representation of parallaxes. 
In Fig. II.c, there is a sketch of an arbitrary cross-section through a model containing 
the points a, b and c, and showing the projection, in the plane of that cross-section, of 
the rays to these points, and the projection centre from which they emanate. The exist- 
ing parallaxes at the points a, b and c, are represented both in magnitude (at an enlarged 
scale) and direction, by arrows at those points. 
The orientation problem in this figure may be stated as follows: “Find the rotation 
Archives 4