the Times of Vibration of Watch Balances. 159 
Suppose the semiarc of vibration BO to be 135 0 
Separation of the points of quiescence = O Q i° 
Arc BQ in fig. 5, or B N in fig. 8. - 136° 
The time of describing the semiarc B O will be determined 
by referring to the theorem (page 146). 
making a — 1.5707963 = an arc of 90° 
b = 2.3561945 = an arc of 135 0 
c — 2.3736478 = an arc of 136° 
d— 0.0174533 == an arc of T 
l = 193. 
/— 0.9562754. 
» = 
The time of describing the semiarc B O = \ / — z ~ ■ 
& - //x» + 1 
x a circular arc of which the sine is 
y 
2 IS X 1 + 2 n 
2 b -f 2 n 1 
x a circular arc of which the sine 
Parts of a second. 
0.09950617 
is \/^- 
d* x 1 + 2 n 
6 1 -f n c 1 — 2 n d z 
Time of a semivibration in the arc B O 
24 * 
= 0.00047141 
= 0.09997758 
= 24 h o' 19".38, 
Time shewn by the watch in 24 h = 
J * • 999775 8 
giving a daily rate of i9/'38 fast. 
Every thing else remaining, let the semiarc of vibration be 
diminished from 135 0 to 125 0 , and make 
b = 2.1816616 = an arc of 125 0 
c = 2.1991149 = an arc of 126° 
The time of describing BO will now be y/ • 
l J X n + 1 
x a 
