264 Proceedings of the Royal Society of Edinburgh. [Sess. 
In point of fact, an aeroplane moving in a straight path at full speed is 
subject to accelerations and retardations in the direction of motion, but 
these never last for long. If in the above equation we make 0 very small, 
we obtain 
Id = Sin B = — 
2 ^ T' 
Since 
sin jS = 
T = 
mil 
and T = 2tt 
mgh 
It is obvious from the figure that when t is small the angle 9 is due 
to turning of the pivoted system about the pivots P 2 P 2 > which it is to 
be remembered are fore and aft. 
The above result may be obtained in 
a simpler manner as follows. Let the 
aeroplane be moving with uniform speed, 
and suppose the gyroscopic pendulum 
upright. If now the aeroplane is ac- 
celerated in the fore-and-aft direction 
the mass m experiences a force, in the 
fore-and-aft direction, of amount ma. 
The torque acting on the gyroscopic 
pendulum is mah, and this tends to turn 
the pivoted system about the pivots 
PtPi^ Such a torque would, in fact, 
turn the pendulum about p^Pi in the 
absence of the gyroscope. But in point 
of fact the pendulum turns about the pivots P 2 P 2 angular speed 9 given 
by mahjCn. Hence if 0 is the deviation of the pendulum from the vertical 
at time t (considered very small), we have 
mailt ‘lirat 
PPn ^ 7E ’ 
as before. It is to be observed that at is the increase of speed attained by 
the aeroplane in time t. 
We now consider the behaviour of the gyroscopic pendulum when 
it is mounted on an aeroplane which moves in a curved path. Let us 
suppose that the aeroplane moves in the path ABCDEFBG (fig. 5), 
where the parts AB and BG are straight, and the part BCDEFB is 
circular. Assuming that at the instant at which the plane is at B the 
pendulum is upright, it is required to find its inclination to the vertical 
after the plane has turned through an angle ijj, at which instant the plane 
is at We have the following sufficiently approximate solution. 
