the Times of Vibration of Watch Balances. 155 
angular distance of 90° from the quiescent point will be = 
1.00408926 (p. 129) : this force, in the present instance, is com- 
pounded of the force of the balance spring, and that of the auxi- 
liary spring, which are assumed to be in the proportion of 20 
to 1 ; consequently the force of the balance spring to accele- 
rate the circumference at the tension 90°, or f = 0.9562754 
and the force of the auxiliary spring at the same 
tension, or ~ of/ = - .0478138 
Joint force of both springs to accelerate the cir- 
cumference at the tension, 90° or/ -j- J^f = 1.0040892 
With* this force the balance vibrates five times in one se- 
* From the investigation in page 128 it appears, that if F is the accelerative force 
on the circumference of the balance at the angular distance from quiescence c°, p — 
3.14159, &c. r — the radius of the balance expressed in inches, l — 193 inches ; the 
time of a semivibration will be = . / P 3 r c ° . 
v 8 / F x 180 0 
In the present case c° =90°, r - 1 inch F -/+ 1.0040892; wherefore the 
time of a semivibration = — r ^ — r \ part of a second precisely. 
If the balance . should vibrate by the force of the auxiliary springs only, the force of 
acceleration on the circumference at the distance 90° from quiescence is / f~ 
0.0478138; wherefore ^/being substituted for F in the quantity*./ p 3 r c° 
^ 8/Fxi8o* 
the other values remaining the same as in the former case, the time of a semivibration 
... / p z r c J P ts - of a sec> 
will now be y/ — — - — — — 45825 ; and the time shewn by the watch in 
8 * X xoj X 180 
any portion of mean time t will — l - — — t x .21822. Thus if t should be 
4.5825 
taken = to one minute, or 6o seconds of time, the time shewn by the watch in 60 se- 
conds will be — 60 x .21822 — 13 seconds nearly. 
This would be the result if the forces of the balance and auxiliary springs were pre- 
cisely in the proportion of 20 to 1. Mr. Mudge mentioned this proportion by esti- 
X2 
