(i) Optimisation using rays 
Referring to Fig.6, the four,rays R,(íi-z1,2,3,4). could fail to 
intersect at one point, thus a mathematical model must be devel- 
oped to obtain an optimum point.  Fig.7 shows only one of these 
rays. 
If p is the optimum point then: Tip. S1T, = 0 (13) 
  
and T,P = W ji = S;P - S,T, i1-z1,2,.:.,N 
therefore 
E 2 
(Sip S;T.).S,;T, = 40", Tor S,b.S,T, = (S; Ti) 
—— 
  
  
assuming Lom mn, are direction cosines of SiTi then 
  
3 T 2 
-— — -— 2 = 
(X X;,Y Yı,Z Z:)^( Lou) Ai (14) 
= - 2 - - 
Hence Ai (X Xi) i + (Y Yi)m; + (Z 2,)n; 
pia 2512 
also Mie = (Sip) Ai (15) 
or 
2 5 2 E 2 A 2 = 2 2... = 2. 2 
Hom (X Xi) - (X Yi) + (Z zZ.) (X Xi) 2; (Y Yi) m. 
2 2 
- (2721) ni - 2(X-X,) (Y-Yi) 21m; ~~ 2 (X-X1) (2-24) tini (16) 
assuming 4 
g= X M4 (17) 
i=l 
and differentiating Oo with respect to X,Y and Z, we get 
ein 45* 9843 X b11 
94257778491. 593 Y = {by (18) 
2135.33 233 Z b31 
where 4 ; ; 4 4 
8,7 AG 3, a2 7 5 NN or 0334 7 X AM, 
i=l i=l i=l 
4 4 4 
$ wn a L (1-m 2) X «1 2 
a = . . , £z LES ; à == -n. 
23 ra i 22:7 5, i gs Hee i 
4 2 
bii = E t (1-2, )X,-2,m,Y,-2,n,3,} 
i=l 
4 2 
b = I {-2.m,X, -* (1-m.^)Y,-m.n.Z.) 
21 isl lii i i "lii 
134