322 EXPLORATION GEOPHYSICS 



TABLE 8 (Continued) 



I = the ^2 length of the beam (horizontal portion) = 20 cm. 



h ■= distance between upper and lower masses: 21.8 cm. from beam to upper mass 

 and 17.6 cm. from beam to lower mass, "h" enters the formula as the sum 

 of these two values, or 39.4 cm. 

 Uvz = the E-W component of the gradient (4). 

 Uxt = the N-S component of the gradient (5). 



Numbers (1) to (5) denote the five iinknozvns. 



il 2-fmlh 

 The quantities -^ — = a and -^ ■ = b are set up as mstrument constants, m 



T T 



which form they apply to an optical system where the double angle (2 o) of deflection 

 was measured (see Equation 72). Where the angle is 4 a (see Equation 73), these 



constants would be -^ = a and JMJl— b^ which applies to the Z-beam instrument 



T T 



considered. 



The general equation is thus simplified to : 



n — Mo = a (sin 20f7A + cos 2<p 2Uxy) + b (cos (j) Uyz — sin Uxz) (76) 



The curvature components are functions of 20, and the gradient components of only. 



Solution of the Fundamental Equation. — In Equation 76 there are 5 unknowns ; 

 namely, the torsionless position of the balance and the 4 components of the gravity 

 quantities. These 5 unknowns may be obtained by setting the instrument in 5 azimuths 

 or with <p successively 0°, 72°, 144°, 216°, and 288°, allowing the beam to come to rest, 

 and taking readings. As it takes 40 minutes to take one reading, 3 hours and 20 minutes 

 are required for the minimum number of readings. 

 With = 0° or North, Equation 76 would be : 



Hi — iio = 2a.Uxy -{-hUyz {77) 



With = 72° 



n* — no = a sin 36° (7a — 2a cos 36° Uxy + b cos 72° Uyz — h sin 72° Uxz (78) 



and in like manner for the other azimuths equations of similar form result. Setting up 

 these 5 simultaneous equations and solving them for the gravity components, we have : 



^ _ «i + «2 + na + >t« + ns ("79) 



where the subscript for the scale reading n represents the azimuth of orientation start- 

 ing with <^ = 0° = 1, <^ = 72° = 2 etc. 



The result of this solution of these 5 equations gives : 



Usz =4- [A(ns-n2) +B(ni-iu)] 

 b 



[/», = —— [F(w6 + M2 — 2mi) — £(m4 + «3 — 2ni) ] 

 a 



Ux =— [B(nn-n2) -A(ni-ns)] 

 a 



2C/«v = — [E(ns + M2 — 2ni) —F(iu-\-tH — 2th) ] 

 a 



(80) 



