GRAVITATIONAL METHODS 



323 



where 



A = 

 B = 



E = 



sin 72° 



2 — cos 72° — cos 36° 



sin 36° 



2 — cos 72° — cos 36° 



1 — cos 36° 

 5(cos72°-cos36°) 



= 0.38042 

 ■--0.23511 

 = 0.32361 



(81) 



TORSION HEAD U 



TORSION HEAD 1 



BALANCE I 



F = , , ^ ~^^^ ^^° = 0.12361 



5 (cos 72 — cos 36 ) 



Double Beam Torsion Balance. — In order to shorten the time 

 required to take all the readings necessary at a station, the double beam 

 torsion balance was developed. In it are 

 two similar but separate and independent 

 torsion balance systems side by side, re- 

 versed 180° as regards their hanging 

 weight ends. (See Figure 182.) 



With this instrument, readings are 

 taken in three symmetrically-spaced azi- 

 muths, or orientations, 0°, 120° and 240° 

 from the north. At one setting of the 

 instrument, data for the solution of two 

 of the set of simultaneous equations are 

 obtained. Two dots representing deflec- 

 tions of the beams are laid down on the 

 photographic plate. 



A sixth unknown, namely, the torsionless position of the second bal- 

 ance, has been introduced which requires the solution of 6 equations. The 

 necessary information can be obtained with 3 readings or orientations, 

 requiring only 2 hours. It is the usual practice to take two check readings 

 on the first two orientations, making the time per station 3 hours and 20 

 minutes. 3 to 4 stations can be taken in 24 hours, more if the instrument 

 is moved at night. 



With the single prime ( ' ) representing balance I and the double prime ( " ) 

 balance II, the formulae for a double beam instrument may be set up using Equation 

 76. Two equations are obtained at each orientation. Balance II is read in azimuths 0°, 

 120°, and 240° successively for orientation positions 1, 2 and 3, while balance I is read 

 in azimuths 180°, 300° and 60°. The following equations are thus obtained: 



Position I 



Fig. 182. — -Diagram of double beams 

 in a torsion balance. Beams are reversed 

 in respect to hanging weight ends. 



0' = 180° 



(Balance I) n'l — n'o = 2a'Uxv — b't/y^ 



0' = O° 



(Balance II) n"i — h"o = 2a"Uxy -{- h"Uyz 



(82) 

 (83) 



