BALANCE OF THE CHRONOMETER. 183 



3 



about 0' it is m^ Zm^p - 1 nearly. Let G be the centre 

 of gravity of AB' : then, in the triangle 00' Gr, we have 



OG*= oo f2 + (0' ay +200'. a a cos \e^ since * GO A = $0. 



00'* or (Ar) 2 may be neglected, then, approximately, 

 (OGy=(0'G?- 2Ar.<7 cos $0, 



-, ~, ~ sin 



and, as G = r l * 



we have (OG) Z - 



Now the moment of inertia round is equal to that round 0' 

 increased by m {(OG) 2 - (O'G)*} ; hence, finally, 



/ x being the moment of inertia of the arc AB. 



(In accordance with the rest of the approximation the ex- 

 pansion of A is neglected.) 



Again, we will suppose the weight at C to be a material 

 particle, and that the angle AOG is equal to <f>. Then, 7 2 being 

 the moment of inertia of this weight, whose mass we will de- 

 note by w 2 , we shall have 



-(l- cos<) ......... ..... (4). 



Consequently, as the inertia of the bar OA does not undergo 

 any sensible alteration, and as every thing which has been proved 

 of OAB is true of Oab, we have, finally, 



1 -cos <) ...... (5). 



It appears that the variation of e is exactly proportional to t : 

 so that e becomes e (1 - vt) 9 v being some constant. Consequently 



. i T ,-, e (1 vt] e 



we must have, in order that -j -^ = -j. , 



m, (1 - cos </>) ...... (6). 



In calculating the value of / we may take into account the 

 moment of inertia of a OA moreover, instead of the approxi- 



