1921-22.] Gyroscope and “Vertical” Problem on Aircraft. 261 
pivots PiPj . The pivoted system hangs from the uprights as shown, and 
the base h is supposed supported from above in a horizontal position. 
We suppose the mass m to lie vertically below the intersection of the 
pivot axes. 
The principal properties possessed by this arrangement may be described 
briefly. In the first place, if the frame / is neutral with respect to the 
pivots P 2 P 2 is, if the C.G. of the frame lies in the line P 2 P 2 ) ^-^^d the 
axis of spin is upright, the apparatus may be trained (that is, turned in 
azimuth) without the introduction of errors. If the apparatus is inclined 
to the vertical on the pivots P 2 P 2 y ^be effect of turning the base through a 
right angle in azimuth is to transfer the inclination to the pivots p-^p ^ . 
If at a given instant the inclina- 
tion is wholly about the axis 
P 2 P 2 ^^d the base is turned 
quickly through 90°, the in- 
clination is transferred to the 
axis p-^p-^ (neglecting the small 
change which occurs in the in- 
terval as a consequence of the 
existing precessional motion, to 
be described directly). When 
the arrangement as described is 
mounted on an aeroplane with 
the pivots P 2 ^p 2 
turning of the plane does not, in itself, cause the inclination of the axis of 
spin to turn with the machine. 
Let us suppose the apparatus illustrated in fig. 3 set up in a room. If 
the axis of spin is tilted from the vertical and the apparatus is left to 
itself, it assumes after a short time a state of steady precession. The 
rod r traces out a cone, the vertex of which is at the intersection of the 
pivot axes, and the semi-vertical angle 0 of which is the inclination of the 
rod to the vertical. If the pivots were frictionless the angle 6 would 
remain constant in amount; in practice 0 diminishes very slowly (if the 
pivots are good) as a consequence of the retardation of the precessional 
motion by pivot friction. At any instant during steady precessional 
motion the horizontal component of angular momentum is Cn sin 0, where 
C is the moment of inertia of the flywheel of the gyroscope, n is its 
angular speed, and 0 is the inclination of the axis of spin to the vertical. 
If T is the periodic time of the motion, this horizontal component of 
angular momentum is turning in azimuth with angular speed 27 t/T. 
-P 
•a 
t-\ ^ 
i 6 
P 
% 
Fig. 3. 
