262 
Proceedings of the Royal Society of Edinburgh. [Sess. 
Consequently the rate of growth of angular momentum, in a horizontal 
direction perpendicular to the instantaneous direction of the horizontal 
component of spin, is Cn sin 0 ( 27 t/T) ; and since this must be equal to the 
resultant couple acting on the gyroscope at the instant, we have, neglecting 
the couple which results from the centrewards acceleration of the mass m 
(which couple is very small when T is great). 
Cn sin = mgh sin 0 , 
or 
Cn 
m.gh ’ 
where h is the distance of the C.G. of m from the intersection of the 
pivot axes. 
In connection with what follows, the nature of the steady precessional 
motion should be carefully noted by readers who have not handled gyro- 
scopes. If at a particular instant the pendulum is inclined wholly about 
the pivots P 2 P 2 » then at a later instant it is inclined about ’ ^^d also 
about Piy>j . After time T/4 it is inclined wholp^ about 
Further, if the pivots are good and Cn large, 0 remains sensibly constant 
in amount over a considerable interval of time. 
Let now the gyroscopic pendulum be supposed mounted on an aeroplane 
with the pivots P 2 P 2 lyi^g ^^^d aft. We suppose that initially the 
aeroplane is moving in a straight line with uniform speed and that the 
pendulum is upright. If now the aeroplane moves with constant accelera- 
tion a in the direction of motion, which we suppose remains unchanged, it 
is required to find the inclination of the axis of spin to the true vertical at 
time t, time being measured from the instant at which the state of uniform 
velocity is departed from. 
Consider a simple pendulum o'm (fig. 4) carried by the aeroplane. 
When the velocity is uniform, the thread of the pendulum is vertical. 
When the aeroplane is accelerated in a constant direction as stated, the 
thread takes up the direction om \ where om" lies in a vertical fore-and- 
aft plane. If F is the stretching force in the thread, and /3 the inclination 
of the thread to the true vertical, we have 
F cos p = mg ; F sin 6 = in a ; 
tan ~ ~ } F = \/| 7 ^ -f a^. 
The force of gravity is replaced by a force whose amount per unit mass 
is and whose direction makes with the true vertical an angle 
given by tan j3 = ajg. From what has been said already it will be clear, on 
