182 



GRAVITATIONAL METHODS 



[Chap. 7 



Hence, for balance I: 



(1 + 2) = aI + A2 = 2b' (0.707) ^« 



(2 + 3) = A2 + A3 = 2b' {Q.10^)Uy^ 



(3 + 4) = A3 + aI = -2b' (0.707) f/„ 



(4 + 1) = aI + Ai' = -2b' (0.707) f/„., 

 and for balance II: 



(1 + 2) = a'/ + a;' = -2b" (0.707) t/« 



(2 + 3) = ^!^ + A'3' = -2b" (0.707) (7,,, 



(3 + 4) = a;' 4- aI' = 2b" (0.707) [/., 



(4 + 1) = a'/ + aI' = 2b" (0.707) t/„.. 



From beam I : 



and from beam II : 



In these equations 



f/« = 



f/„. = 



f/« = 



f/„. = 



a; + A2 



1.414b' 



Ag + A3 

 1.414b' 



a^+aT 



1.414b" 



A4 + Ai 

 1.414b" 



(Aa + Al) 

 1.414b' 



(a( + aI) 



1.414b' 



-(Ar + A;o 



1.414b" 



-(a;^ + a:/) 



1.414b" 



no = 



nx -\- 712 + nz -{• nt 



(7-53o) 



(7-536) 



an4 



Wo = 



ni + W2 + W3 + W4 



Therefore, f/„ = +(^4 - rig) /2b and C/i,^ = ±(713 - wi)/2b, where the 

 upper signs are for beam I and the lower for beam II. 



If only one balance beam is functioning and if both gradients and 

 curvatures are desired, observations in five azimuths are required. Then 

 for beam II: 



