Dr Searle, A bifilar method of measuring the rigidity of wires 67 



e = YTC i<f>+V)+^ = D{c/>+r))+e (10) 



If Oq, Cq correspond to ^ = 0, we have, since €q = P, 



e, = Drj+e,^Drj + P. (11) 



Thus 



6 - 6^ = Dcf>+e- P = D<f) + P cos (l> + Qsincf) - P (12) 



Since this equation is linear in 6, we may take 6^ as corresponding 

 to any initial position of the bar which is yiear its ideal zero position. 

 Thus, if /S is the angle at any time between the bar and some 

 nearly ideal zero position, 



^ = Q-K (13) 



Since ^, though small — say less than 0*2 radian — is not infinitesimal, 

 some correction should be made. An exact solution cannot be 

 given, but accuracy is gained by writing sin ^ for ^, and then the 

 final formula becomes 



sin^ = Z)(/. + P cos </. + Q?,\\\(j> - P (14) 



To eliminate P and Q, we combine the observations. Let /3^ 

 correspond to (f> = imr/i. Then, putting cf) = tt and (f) = — tt, so that 

 m = i and m = — 4, we have 



ttD = i (sin ^4 - sin ^_4). (15) 



A second value for ttD is found by giving m the values 3, — 3, 

 1, - 1. Then 



7tD = sin ^3 — sin/3_3 — (sin^j — sin /3_i) (16) 



The two values of D are usually in good agreement, although, when 

 P is plotted against <f), the curve differs considerably from a straight 

 line. The mean value of ttD is used to find C. Thus 



ttD 



C' = -^. (17) 



Then n is found by (7). 



The actual values of P and Q are easily found. Thus 



P = - 1 (sin ^4 + sin ^_4), (18) 



Q = l (sin ^2 - sin ^.^ - ttD) (19) 



§ 6. Conversion table. A goniometer, such as those used at 

 the Cavendish Laboratory, gives the tangent of the angle iJj through 

 which the arm is turned from its zero. To find sin ip we subtract 

 from tan j/» the small quantity s given in the table. 



5—2 



