Assuming a periodic solution of these equations

(4) w, [~omega], [sigma], = (L, P, Q) exp nti,

and eliminating L, P, Q, we obtain

(5) (-An^2 + Kn + gMa)(g - n^2b) - gMn^2a^2 = 0,

and the frequency of a vibration in double beats per second is

n/2[pi], where n is a root of this quartic equation.

For upright spinning on a smooth horizontal plane, take b = [oo] and

change the sign of a, then

(6) An^2 - Kn + gMa = 0,

so that the stability requires

(7) K^2 > 4gAMa.

Here A denotes the moment of inertia about a diametral axis through

the centre of gravity; when the point of the fly-wheel is held in a

small smooth cup, b = 0, and the condition becomes

(8) (A + Ma^2)n^2 - Kn + gMa = 0,

requiring for stability, as before in S 3,

(9) K^2 > 4g(A + M^2)Ma.

For upright spinning inside a spherical surface of radius b, the sign

of a must be changed to obtain the condition at the lowest point, as

in the gyroscopic horizon of Fleuriais.

For a gyrostat spinning upright on the summit of a sphere of radius b,

the signs of a and b must be changed in (5), or else the sign of g,

which amounts to the same thing.

Denoting the components of horizontal displacement of the point of the

fly-wheel by [xi], [eta], then

(10) br = [xi], bs = [eta], b[sigma] = [xi] + [eta]i = [lambda]

(suppose),

(11) [omega] = [alpha][~omega] + [lambda].

If the point is forced to take the motion ([xi], [eta], [zeta]) by

components of force X, Y, Z, the equations of motion become

(12) -A :q + K .p = Ya - Zaq,

(13) A :p + K .q = -Xa + Zap,

(14) M :w = X + Yi, M( :[zeta] - g) = Z;

so that

(15) A :[~omega] - K .[~omega]i + gMa[~omega] + Ma :w

= Ma[~omega] :[zeta],

or

(16) (A + Ma^2) :[~omega] - K .[~omega]i + gMa[~omega] + Ma[lambda] =

Ma[~omega] :[zeta].

Thus if the point of the gyrostat is made to take the periodic motion

given by [lambda] = R exp nti, [zeta] = 0, the forced vibration of the

axis is given by [~omega] = P exp nti, where

(17) P{-(A + Ma^2)n^2 + Kn + gMa} - RMn^2a = 0;

and so the effect may be investigated on the Fleuriais gyroscopic

horizon of the motion of the ship.

Suppose the motion [lambda] is due to the suspension of the gyrostat

from a point on the axis of a second gyrostat suspended from a fixed

point.

Distinguishing the second gyrostat by a suffix, then [lambda] =

b[~omega]1, if b denotes the distance between the points of suspension

of the two gyrostats; and the motion of the second gyrostat influenced

by the reaction of the first, is given by

(18) (A1 + M1h1^2) :[~omega]1 - K1 .[~omega]1 i

= -g(M1h1 + Mb)[~omega]1 - b(X + Yi)

= -g(M1h1 + Mb)[~omega]1 - Mb(a :[~omega] + :[lambda]);

so that, in the small vibration,

R / \

(19) -- ( -(A1 + M1h1^2)n^2 + K1n + g(M1h1 + Mb) ) = Mn^2b(aP + R),

b \ /

(20) R { -(A1 + M1h1^2 + Mb^2)n^2 + K1n + g(M1h1 + Mb)} - PMn^2ab^2 = 0.

Encyclopaedia Britannica, 11th Edition, "Gyantse" to "Hallel" Volume 12, Slice 7 · 第 35 / 53 题