1921-22.] Gyroscope and ‘‘Vertical” Problem on Aircraft. 271 
that it is proportional to the speed of the resisted body ; and if we assume 
that the pendulum is provided with a dashpot, which results in the 
resisting couple being proportional to the angular speed of the rod r, we 
have for the couple sin d, wdiere k is a constant. The rate at which 
energy is being dissipated in heat within the dashpot, being the product 
47t2 
of the couple and the angular speed, is /vTj^sin^ 0. The potential energy of 
the system at the instant is ?u^/i(l— cos d), and the rate of diminution of 
potential energy is — m^Asin dd. Hence 
k^^ sin^ (9 ^ - mqh sin 6*^^ . 
■ (it 
Cn 
Since T = ^bis equation may be written 
clt= - 
TC?z dO 
sin ^ 
which gives 
t = - log tan ^6-\- const. ; 
and denoting the value of d at time ^ = 0 by do, we have finally 
27tA* tan 2 0 
As an example we take the following. A gyroscopic pendulum is mounted 
on an aeroplane, the pendulum being damped by means of a dashpot. The 
aeroplane executes a half turn and thereafter proceeds in a straight line. 
It is required to investigate the behaviour of the pendulum. 
Let the speed of the aeroplane be 100 feet per second, the precessional 
period of the pendulum be 6 minutes, and the angular momentum of 
the gyroscope be 250, in foot, pound, second units. If the pendulum is 
initially upright, then after the half turn its inclination to the vertical 
is degrees, as shown in a previous example. Straight-line motion on 
the part of the aeroplane is now assumed. Precession takes place, and the 
value of d gradually diminishes. We assume a value of k such that when 
d = SO'" the value of d amounts to 1 degree in 10 seconds of time, which 
corresponds to a very powerful dashpot. This gives k — 50, and 
^=280 log 
tan 
tan 
) 
with the values for T and Cn which have been specified. 
