102 Proceedings of the Royal Society of Edinburgh. [Sess. 
If we have the development (15) of p. 96, 
R = A 0 + A x cos r + Bj sin r + . . . + A r cos rr + B r sin rr + . . . (15) 
t = n't + e 0 , 
where 
x" + kx + u 2 % = R, 
after the decaying terms of the free oscillation have disappeared the 
solution is 
where 
x = a 0 + a 1 cos r + b 1 sin r + . . . + a r cos rr + b r sin rr + . . 
( n 2 — r 2 n 2 )A r — Krn'B r ^ _ ( n 2 - r 2 // 2 )B r + kth A r _ 
(?i 2 — r 2 /i ' 2 ) 2 + K 2 r 2 ?i 2 ’ ' (^ 2 — r 2 n 2 ) 2 + K 2 r 2 n 2 ; 
in particular 
provided 
= 0 , b 1 = A 1 /K?i, 
n 2 + kAB^A-!^ - /i 2 = 0, 
as adopted in (165), p. 97. All the terms except are of inconsiderable 
value, and we may ignore k/'U, in connection with them, writing simply 
a r = - A r /(r 2 - \)n 2 , 5,.= - B r /(r 2 - \)n 2 ; [r=^=l. 
it remains then merely to obtain the development of R in the form (15) 
for each of the separate cases. 
1. Synchronome Clock. 
The changing value of the maintenance at any instant is given 
approximately as Q sin 0 on p. 81. It vanishes for 0 = 0 and at 0=y, 
attaining a maximum between. The angle y is connected with r by 
the equations a sin y — c sin \[s, \!s = a sin t, where \Js is the inclination of 
the crutch to the vertical and a cos y + c cos \fr — l is the distance of the 
centre of the roller from the pivot of the crutch ; but instead of following 
out this somewhat intricate relation, I shall merely keep its general effect 
and simplify the case by supposing that the graph of the maintenance 
is a half-loop of a sinusoid, vanishing at B ( T — 0) and D (t = S), where o 
is a given angle, so that we may put 
R = A sin 7rr/S from r = 0to8; =0 from r = S to 7r, 
= — A sin 7t(t — 7r)/8 ,, 7T ,, 7T + 8 ; =0 ,, 7T + S ,, 2-7T. 
Fig. 6, p. 104, illustrates this. The left hand shows the incidence of 
the maintenance while r progresses continuously ; the right shows, in the 
circle, the progression of t in relation to the period and phase, and within 
it the corresponding positions of the pendulum. 
