2 2 go HAD 132 2 
x ty ert (K^ t (D^ o, (64) 
from which we get: 
kyl = à 2 2 + f£? — 2 7 
y Vx? T f (k £) (65) 
! i f = . 
k y^ assumes the sign o (y. yy) 
Returning to the equations in (58) to solve for r r and Lays WE obtain: 
12° 22 
+ 
r : e D) ris (66) 
Ye 
= s RD as a 
22 kv! 
ce 
: : =f (k £) Tag ue 
32 k y! 
ce 
11° 21 and fap we first recognize the fact that the space angle between x 
and x' ís less than 90°, which means that r 
To solve for r 
11 is positive. Hence, again employing the 
property of orthogonality and making ri positive, we get: 
PO WM are 
Y1 r1» f (69) 
ru 7 13 
| FE Fit an 
E ^ TEC (70) 
11 
Teale Ys 
T31 - -r (71) 
11 
This completes the solution of [R]. Now the vanishing points in the fictitious photograph 
must be found. 
Equation (56) expresses the real photo image coordinates in terms of the fictitious 
photo coordinates. The inverse transformation is given by: 
  
t d T (72) 
1 
-f -f 
from which we obtain: 
1 + -— 
x: Aug t 5a a (73 
T Y 
8 Tate * roov. 7 fgof 
= 17 -