10 
Mathematically, the compensation in the control points means that 
the sum of the corresponding differentials (13)—(16) and (19)—(22) 
must be zero. Hence we have 
dx, + db — bdx + Sdx, — 0 
dy, + db + bdx + S (— dx, + dx, + dy,) = 0 
(23) 
dx, + bdx + Sda; zz 
dy, — db -- S (dx, — da; 4- dy;) — 0 
Solving the equation system (23) we obtain 
S , ” , ” , , 
das a (— dx, — dx, — 4 dx, + dx; + dy, — dy;) (24) 
S , ” , " , ” 
dy, = = (2 dx, — 3 dx, — 2 dz, 4- 3 dx, — 2dy, — 3dy,) (25) 
0 f s 
S 
db = — (2 da, — 3 dx, + 3 dx; — 2 dx; — 2dy, + 2dy.) (26) 
5 : 
S 
dx — —, (de, + dx, — da; — da; — dy, + dy;) (27) 
5b 
After substitution of the expressions (24)—(27) into (17)—(18) and 
after addition of the expressions (3)—(4) except the terms containing 
db we find after some rearrangement 
da: 1 , " , " , " 
"9 7 E. |- dx, — dx, — 4 dx; + dx, + dy, — dy; + 
x z , "H , " , " 
+ b (2 dx, — 3 dx, + 3 dx, — 2 dx, — 2 dy, + 2 dy.) 
y , " , " , " 
TO (dx, + dæ, — dx, — dx; — dy, -- dy) — 
x ; x ; æ [x ; ; 
— 1-7 — 11 da’ + — da" + — — ] hy — dy" 28 
b 6 Logo (dy "y ) (26)