Page:EB1911 - Volume 12.djvu/797

Admiral Fleuriais (fig. 9); but it must be noticed that the horizon given by the top is inclined to the true horizon at the angle E given by equation (3) above; and if μ₁ is the precessional angular velocity as given by (3) § 4, and T = 2π/μ, its period in seconds,

tan E =	μ	cos lat =	T cos lat	, or E =	T cos lat	,
	μ1		86400		8π

(4)

if E is expressed in minutes, taking μ = 2π/86400; thus making the true latitude E nautical miles to the south of that given by the top (Revue maritime, 1890; Comptes rendus, 1896).

This can be seen by elementary consideration of the theory above, for the velocity of the vector OC′ of the top due to the rotation of the earth is

μ·OC′ cos lat = gMh sin E = μ₁·OC′ sin E,

sin E =	μ	cos lat, E =	T cos lat	,
	μ1		8π

(5)

Fig. 11.

in which 8π can be replaced by 25, in practice; so that the Fleuriais gyroscopic horizon is an illustration of the influence of the rotation of the earth and of the need for its allowance.

7. In the ordinary treatment of the general theory of the gyroscope, the motion is referred to two sets of rectangular axes; the Euler’s coordinate angles. one Ox, Oy, Oz fixed in space, with Oz vertically upward and the other OX, OY, OZ fixed in the rotating wheel with OZ in the axis of figure OC.

The relative position of the two sets of axes is given by means of Euler’s unsymmetrical angles θ, φ, ψ, such that the successive turning of the axes Ox, Oy, Oz through the angles (i.) ψ about Oz, (ii.) θ about OE, (iii.) φ about OZ, brings them into coincidence with OX, OY, OZ, as shown in fig. 11, representing the concave side of a spherical surface.

The component angular velocities about OD, OE, OZ are

ψ. sin θ, θ., φ. + ψ. cos θ;

(1)

so that, denoting the components about OX, OY, OZ by P, Q, R,

P =	θ. cos φ	+ ψ. sin θ sin φ,
Q =	−θ. sin φ	+ ψ. sin θ cos φ,
R =	φ.	+ ψ. cos θ.

(2)

Consider, for instance, the motion of a fly-wheel of preponderance Mh, and equatoreal moment of inertia A, of which the axis OC is held in a light ring ZCX at a constant angle γ with OZ, while OZ is held by another ring zZ, which constrains it to move round the vertical Oz at a constant inclination θ with constant angular velocity μ, so that

θ. = o, ψ. = μ;

(3)

P = μ sin θ sin φ, Q = μ sin θ cos φ, R = φ. + μ cos θ.

(4)

With CXF a quadrant, the components of angular velocity and momentum about OF, OY, are

P cos γ − R sin γ, Q, and A (P cos γ − R sin γ), AQ,

(5)

so that, denoting the components of angular momentum of the fly-wheel about OC, OX, OY, OZ by K or G′, h₁, h₂, h₃,

h₁ = A (P cos γ − R sin γ) cos γ + K sin γ,

(6)

h₂ = AQ,

(7)

h₃ = −A (P cos γ − R sin γ) sin γ + K cos γ;

(8)

and the dynamical equation

dh3	− h1Q + h2P = N,
dt

(9)

with K constant, and with preponderance downward

N = gMh cos zY sin γ = gMh sin γ sin θ cos φ,

(10)

reduces to

A	d2φ	sin γ + Aμ2 sin γ sin2 θ sin φ cos φ
	dt2

+ Aμ² cos γ sin θ cos θ cos φ − (Kμ + gMh) sin θ cos φ = 0.

(11)

The position of relative equilibrium is given by

cos φ = 0, and sin φ =	Kμ + gMh − Aμ2 cos γ cos θ	.
	Aμ2 sin γ sin θ

(12)

For small values of μ the equation becomes

A	d2φ	sin γ − (Kμ + gMh) sin θ cos φ = 0,
	dt2

(13)

so that φ = 1/2π gives the position of stable equilibrium, and the period of a small oscillation is 2π √{A sin γ/(Kμ + gMh) sin θ}.

In the general case, denoting the periods of vibration about φ = 1/2π, −1/2π, and the sidelong position of equilibrium by 2π/(n₁, n₂, or n₃), we shall find

n12 =	sin θ	{ gMh + Kμ − Aμ2 cos (γ − θ) },
	A sin γ

(14)

n22 =	sin θ	{ −gMh − Kμ + Aμ2 cos (γ + θ) },
	A sin γ

(15)

n₃ = n₁ n₂/μ sin θ.

(16)

The first integral of (11) gives

1/2A (	dφ	)	2	sin γ + 1/2Aμ2 sin γ sin2 θ sin2 φ
	dt

− Aμ² cos γ sin θ cos θ sin φ + (Kμ + gMh) sin θ sin φ − H = 0,

(17)

and putting tan (1/4π + 1/2φ) = z, this reduces to

dz	n √Z
dt

(18)

where Z is a quadratic in z², so that z is a Jacobian elliptic function of t, and we have

tan (1/4π + 1/2φ) = C (tn, dn, nc, or cn) nt,

(19)

according as the ring ZC performs complete revolutions, or oscillates about a sidelong position of equilibrium, or oscillates about the stable position of equilibrium φ = ±1/2π.

Suppose Oz is parallel to the earth’s axis, and μ is the diurnal rotation, the square of which may be neglected, then if Gilbert’s barogyroscope of § 6 has the knife-edges turned in azimuth to make an angle β with E. and W., so that OZ lies in the horizon at an angle E·β·N., we must put γ = 1/2π, cos θ = sin α sin β; and putting φ = 1/2π − δ + E, where δ denotes the angle between Zz and the vertical plane Zζ through the zenith ζ,

sin θ cos δ = cos α, sin θ sin δ = sin α cos β;

(20)

so that equations (9) and (10) for relative equilibrium reduce to

gMh sin E = KQ = Kμ sin θ cos φ = Kμ sin θ sin (δ − E),

(21)

and will change (3) § 6 into

tan E =	Kμ sin α cos β	,
	gMh + Kμ cos α

(22)

a multiplication of (3) § 6 by cos β (Gilbert, Comptes rendus, 1882).

Changing the sign of K or h and E and denoting the revolutions/second of the gyroscope wheel by F, then in the preceding notation, T denoting the period of vibration as a simple pendulum,

tan E =	Kμ sin α cos β	=	F sin α cos β	,
	gMh − Kμ cos α		86400 A/T2C − F cos α

(23)

so that the gyroscope would reverse if it were possible to make F cos α > 86400 A/T²C (Föppl, Münch. Ber, 1904).

A gyroscopic pendulum is made by the addition to it of a fly-wheel, balanced and mounted, as in Gilbert’s barogyroscope, in a ring movable about an axis fixed in the pendulum, in the vertical plane of motion.

As the pendulum falls away to an angle θ with the upward vertical, and the axis of the fly-wheel makes an angle φ with the vertical plane of motion, the three components of angular momentum are

h₁ = K cos φ, h₂ = Aθ. + K sin φ, h₃ = Aφ.,

(24)

where h₃ is the component about the axis of the ring and K of the fly-wheel about its axis; and if L, M′, N denote the components of the couple of reaction of the ring, L may be ignored, while N is zero, with P = o, Q = θ., R = o, so that

M′ = h₂ = Aθ.. + Kφ. cos φ,

(25)

0 = h₃ − h₁θ. = Aφ.. − Kθ. cos φ.

(26)

For the motion of the pendulum, including the fly-wheel,

MK²θ.. = gMH sin θ − M′
= gMH sin θ − Aθ.. − Kφ. cos φ.

(27)

If θ and φ remain small,

Aφ.. = Kθ., Aφ. = K(θ − α),

(28)

(MK² + A)θ.. + (K²/A) (θ − α) − gMHθ = 0;

(29)

so that the upright position will be stable if K² > gMHA, or the rotation energy of the wheel greater than 1/2A/C times the energy acquired by the pendulum in falling between the vertical and horizontal position; and the vibration will synchronize with a simple pendulum of length

(MK² + A) / [(K²/gA) − MH].

(30)

This gyroscopic pendulum may be supposed to represent a ship among waves, or a carriage on a monorail, and so affords an explanation of the gyroscopic action essential in the apparatus of Schlick and Brennan.

8. Careful scrutiny shows that the steady motion of a top is not steady absolutely; it reveals a small nutation General motion of the top. superposed, so that a complete investigation requires a return to the equations of unsteady motion, and for the small oscillation to consider them in a penultimate form.

In the general motion of the top the vector OH of resultant angular momentum is no longer compelled to lie in the vertical plane COC′ (fig. 4), but since the axis Oh of the gravity couple is always horizontal, H will describe a curve in a fixed horizontal plane through C. The vector OC′ of angular momentum about the axis will be constant in length, but vary in direction; and OK will be the component angular momentum in the vertical plane COC′, if the planes through C and C′ perpendicular to the lines OC and OC′ intersect in the line KH; and if KH is the component angular momentum perpendicular to the plane COC′, the resultant angular momentum OH has the three components OC′, C′K, KH, represented in Euler’s angles by

KH = A dθ/dt, C′K = A sin θd ψ/dt, OC′ = G′.

(1)

Drawing KM vertical and KN parallel to OC′, then

KM = A dψ/dt, KN = CR − A cos θ dψ/dt = (C − A) R + A dφ/dt

(2)

so that in the spherical top, with C = A, KN = A dφ/dt.