1917-18.] 
so that if 
we have 
Studies in Clocks and Time-keeping. 
^ = c o + c 1 cos <h + d 1 sin <£ + . . . 
107 
2 7 rC Q = C 
7+5 
V+y+S 
tan <fi dcf> + C 
Jir + 7 
= 2 C logjcos y/cOS (y + 8) } 
tan 9!) dcf> 
7TCv = C 
r 7 + 5 
tan <f> cos ?’</> d<f> + 
r 7+5 
7 rd T = C / tan 0 sin r</> d<f> + . . . ; 
J *V 
it is unnecessary to write out the forms further; they are not simple, but 
they involve no difficulty, as the range of integration does not include a zero 
of cos </>. As regards the steady state, we have from p. 101, equations (19), 
E = aJ K = CS, 
7TK 
ri — n= — J KC 0 la 0 = - Jk 8 1 log e { cos y/cos (y + S ) } 
= — kS 
-1 ( ■ 
O 8 
|tan-tan(^y + -j + Jtan 3 -tan 3 ^y + -j+ . . .| 
( 23 ) 
The escapement error has the same sign as for the other two clocks, viz. 
opposite to the circular error. 
Any other case of maintenance, even very complicated ones, can be 
treated in the manner applied above for the three typical clocks ; and 
allowances can readily be made for various slight departures from simple 
laws. I do not propose to follow these out, but shall take a few examples 
of some of the more important constructions. 
It will have been gathered that almost all that matters is the deter- 
mination of the coefficients a v b 1 in the development 
E , = « 0 + cos t + b l sin r + . . . ; 
that is to say, we have to estimate the phase and amplitude of the first 
harmonic of the development of R ; this can usually be done very fairly by 
graph, especially for symmetrical cases. When there are several distinct 
parts in the maintenance, unlocking, etc., it is best to treat them separately 
and afterwards unite the various sine curves arrived at. 
(1) Graham Escapement. — Omitting at first the pallet friction, let the 
maintenance be represented by a constant pressure A, exerted from t — y 
to r — y + £, and —A from T = ir-\-y to r — 7 r-fy + $. Generally y will be 
negative, the maintenance coming on before the lowest point is reached. 
It is evident from fig. 8 that the first harmonic will have its zero at 
i7r + y + £<$. By calculation it appears 
R = 2A/?r . sin ^8 cos (r — y — ^8) + . . . 
x = 2 A sin ^-8(7rtt 2 ) 1 . ( 11/ k ) cos (y -f -|8) sin t + . . . 
and 
