# 1911 Encyclopædia Britannica/Gyroscope and Gyrostat

GYROSCOPE AND GYROSTAT. These are scientific models or instruments designed to illustrate experimentally the dynamics of a rotating body such as the spinning-top, hoop and bicycle, and also the precession of the equinox and the rotation of the earth.

The gyroscope (Gr. γῦρος, ring, σκοπεῖν, to see) may be distinguished from the gyrostat (γῦρος, and στατικός, stationary) as an instrument in which the rotating wheel or disk is mounted in gimbals so that the principal axis of rotation always passes through a fixed point (fig. 1). It can be made to imitate the motion of a spinning-top of which the point is placed in a smooth agate cup as in Maxwell’s dynamical top (figs. 2, 3). (Collected Works, i. 248.) A bicycle wheel, with a prolongation of the axle placed in a cup, can also be made to serve (fig. 4).

 Fig. 1. Fig. 2.

The gyrostat is an instrument designed by Lord Kelvin (Natural Philosophy, § 345) to illustrate the more complicated state of motion of a spinning body when free to wander about on a horizontal plane, like a top spun on the pavement, or a hoop or bicycle on the road. It consists essentially of a massive fly-wheel concealed in a metal casing, and its behaviour on a table, or with various modes of suspension or support, described in Thomson and Tait, Natural Philosophy, serves to illustrate the curious reversal of the ordinary laws of statical equilibrium due to the gyrostatic domination of the interior invisible fly-wheel, when rotated rapidly (fig. 5).

The toy shown in figs. 6 and 7, which can be bought for a shilling, is acting as a gyroscope in fig. 6 and a gyrostat in fig. 7.

 Fig. 3. Fig. 4.

The gyroscope, as represented in figs. 2 and 3 by Maxwell’s dynamical top, is provided with screws by which the centre of gravity can be brought into coincidence with the point of support. It can then be used to illustrate Poinsot’s theory of the motion of a body under no force, the gyroscope being made kinetically unsymmetrical by a setting of the screws. The discussion of this movement is required for Jacobi’s theorems on the allied motion of a top and of a body under no force (Poinsot, Théorie nouvelle de la rotation des corps, Paris, 1857; Jacobi, Werke, ii. Note B, p. 476).

To imitate the movement of the top the centre of gravity is displaced from the point of support so as to give a preponderance. When the motion takes place in the neighbourhood of the downward vertical, the bicycle wheel can be made to serve again mounted as in fig. 8 by a stalk in the prolongation of the axle, suspended from a universal joint at O; it can then be spun by hand and projected in any manner.

 Fig. 5.
 Fig. 6. Fig. 7.
 Fig. 8. Fig. 9.

The first practical application of the gyroscopic principle was invented and carried out (1744) by Serson, with a spinning top with a polished upper plane surface for giving an artificial horizon at sea, undisturbed by the motion of the ship, when the real horizon was obscured. The instrument has been perfected by Admiral Georges Ernest Fleuriais (fig. 9), and is interesting theoretically as showing the correction required practically for the rotation of the earth. Gilbert’s barogyroscope is devised for the same purpose of showing the earth’s rotation; a description of it, and of the latest form employed by Föppl, is given in the Ency. d. math. Wiss., 1904, with bibliographical references in the article “Mechanics of Physical Apparatus.” The rotation of the fly-wheel is maintained here by an electric motor, as devised by G. M. Hopkins, and described in the Scientific American, 1878. To demonstrate the rotation of the earth by the constancy in direction of the axis of a gyroscope is a suggestion that has often been made; by E. Sang in 1836, and others. The experiment was first carried out with success by Foucault in 1851, by a simple pendulum swung in the dome of the Pantheon, Paris, and it has been repeated frequently (Mémoires sur le pendule, 1889).

A gyroscopic fly-wheel will preserve its original direction in space only when left absolutely free in all directions, as required in the experiments above. If employed in steering, as of a torpedo, the gyroscope must act through the intermediary of a light relay; but if direct-acting, the reaction will cause precession of the axis, and the original direction is lost.

The gyrostatic principle, in which one degree of freedom is suppressed in the axis, is useful for imparting steadiness and stability in a moving body; it is employed by Schlick to mitigate the rolling of a ship and to maintain the upright position of Brennan’s monorail car.

Lastly, as an application of gyroscopic theory, a stretched chain of fly-wheels in rotation was employed by Kelvin as a mechanical model of the rotary polarization of light in an electromagnetic field; the apparatus may be constructed of bicycle wheels connected by short links, and suspended vertically.

Theory of the Symmetrical Top.

1. The physical constants of a given symmetrical top, expressed in C.G.S. units, which are employed in the subsequent formulae, are denoted by M, h, C and A. M is the weight in grammes (g) as given by the number of gramme weights which equilibrate the top when weighed in a balance; h is the distance OG in centimetres (cm.) between G the centre of gravity and O the point of support, and Mh may be called the preponderance in g.-cm.; Mh and M can be measured by a spring balance holding up in a horizontal position the axis OC in fig. 8 suspended at O. Then gMh (dyne-cm. or ergs) is the moment of gravity about O when the axis OG is horizontal, gMh sin θ being the moment when the axis OG makes an angle θ with the vertical, and g = 981 (cm./s2) on the average; C is the moment of inertia of the top about OG, and A about any axis through O at right angles to OG, both measured in g-cm.2.

To measure A experimentally, swing the top freely about O in small plane oscillation, and determine the length, l cm., of the equivalent simple pendulum; then

(1)
l = A/Mh, A = Mhl.

Next make the top, or this simple pendulum, perform small conical revolutions, nearly coincident with the downward vertical position of equilibrium, and measure n, the mean angular velocity of the conical pendulum in radians / second; and T its period in seconds; then

(2)
4π2/T2 = n2 = g/l = gMh/A;

and f = n/2π is the number of revolutions per second, called the frequency, T = 2π/n is the period of a revolution, in seconds.

2. In the popular explanation of the steady movement of the top at a constant inclination to the vertical, depending on the composition of angular velocity, such as given in Perry’s Spinning Tops, or Worthington’s Dynamics of Rotation, it is asserted that the moment of gravity is always Steady motion of the top. generating an angular velocity about an axis OB perpendicular to the vertical plane COC′ through the axis of the top OC′; and this angular velocity, compounded with the resultant angular velocity about an axis OI, nearly coincident with OC′, causes the axes OI and OC′ to keep taking up a new position by moving at right angles to the plane COC′, at a constant precessional angular velocity, say μ rad./sec., round the vertical OC (fig. 4).

If, however, the axis OC′ is prevented from taking up this precessional velocity, the top at once falls down; thence all the ingenious attempts—for instance, in the swinging cabin of the Bessemer ship—to utilise the gyroscope as a mechanical directive agency have always resulted in failure (Engineer, October 1874), unless restricted to actuate a light relay, which guides the mechanism, as in steering a torpedo.

An experimental verification can be carried out with the gyroscope in fig. 1; so long as the vertical spindle is free to rotate in its socket, the rapidly rotating wheel will resist the impulse of tapping on the gimbal by moving to one side; but when the pinch screw prevents the rotation of the vertical spindle in the massive pedestal, this resistance to the tapping at once disappears, provided the friction of the table prevents the movement of the pedestal; and if the wheel has any preponderance, it falls down.

Familiar instances of the same principles are observable in the movement of a hoop, or in the steering of a bicycle; it is essential that the handle of the bicycle should be free to rotate to secure the stability of the movement.

The bicycle wheel, employed as a spinning top, in fig. 4, can also be held by the stalk, and will thus, when rotated rapidly, convey a distinct muscular impression of resistance to change of direction, if brandished.

3. A demonstration, depending on the elementary principles of dynamics, of the exact conditions required for the axis OC′ of a spinning top to spin steadily at a constant inclination θ to the vertical OC, is given here before proceeding to the more complicated question of the generalElementary demonstration of the condition of steady motion. motion, when θ, the inclination of the axis, is varying by nutation.

It is a fundamental principle in dynamics that if OH is a vector representing to scale the angular momentum of a system, and if Oh is the vector representing the axis of the impressed couple or torque, then OH will vary so that the velocity of H is represented to scale by the impressed couple Oh, and if the top is moving freely about O, Oh is at right angles to the vertical plane COC′, and

(1)
Oh = gMh sin θ.
In the case of the steady motion of the top, the vector OH lies

in the vertical plane COC′, in OK suppose (fig. 4), and has a component OC = G about the vertical and a component OC′ = G′, suppose, about the axis OC; and G′ = CR, if R denotes the angular velocity of the top with which it is spun about OC′.

If μ denotes the constant precessional angular velocity of the vertical plane COC′ the components of angular velocity and momentum about OA are μ sin θ and Aμ sin θ, OA being perpendicular to OC′ in the plane COC′; so that the vector OK has the components

OC′ = G′, and C′K = Aμ sin θ,
(2)

and the horizontal component

CK = OC′ sin θ − C′K cos θ
= G′ sin θ − Aμ sin θ cos θ.

(3)

The velocity of K being equal to the impressed couple Oh,

gMh sin θ = μ·CK = sin θ (G′μ − Aμ2 cos θ),
(4)

and dropping the factor sin θ,

Aμ2 cos θ − G′μ + gMh = 0, or Aμ2 cos θ − CRμ + An2 = 0,
(5)

the condition for steady motion.

Solving this as a quadratic in μ, the roots μ1, μ2 are given by

 μ1, μ2 = G′ sec θ [1 ± √ (1 − 4A2n2 cos θ) ]; 2A G′2
(6)

and the minimum value of G′ = CR for real values of μ is given by

 G′2 = cos θ, CR 2√(cos θ); 4A2n2 An
(7)

for a smaller value of R the top cannot spin steadily at the inclination θ to the upward vertical.

Interpreted geometrically in fig. 4

μ = gMh sin θ/CK = An2/KN, and μ = C′K/A sin θ = KM/A,
(8)
KM·KN = A2 n2,
(9)

so that K lies on a hyperbola with OC, OC′ as asymptotes.

4. Suppose the top or gyroscope, instead of moving freely about the point O, is held in a ring or frame which is compelled Constrained motion of the gyroscope. to rotate about the vertical axis OC with constant angular velocity μ; then if N denotes the couple of reaction of the frame keeping the top from falling, acting in the plane COC’, equation (4) § 3 becomes modified into

gMh sin θ − N = μ·CK = sin θ G′μ − Aμ2 cos θ,

(1)
N = sin θ (Aμ2 cos θ − G′μ + gMh)
= A sin θ cos θ (μμ1) (μμ2);
(2)

and hence, as μ increases through μ2 and μ1, the sign of N can be determined, positive or negative, according as the tendency of the axis is to fall or rise.

When G′ = CR is large, μ2 is large, and

μ1gMh/G′ = An2/CR,
(3)
 Fig. 10.

the same for all inclinations, and this is the precession observed in the spinning top and centrifugal machine of fig. 10 This is true accurately when the axis OC′ is horizontal, and then it agrees with the result of the popular explanation of § 2.

If the axis of the top OC′ is pointing upward, the precession is in the same direction as the rotation, and an increase of μ from μ1 makes N negative, and the top rises; conversely a decrease of the procession μ causes the axis to fall (Perry, Spinning Tops, p. 48).

If the axis points downward, as in the centrifugal machine with upper support, the precession is in the opposite direction to the rotation, and to make the axis approach the vertical position the precession must be reduced.

This is effected automatically in the Weston centrifugal machine (fig. 10) used for the separation of water and molasses, by the friction of the indiarubber cushions above the support; or else the spindle is produced downwards below the drum a short distance, and turns in a hole in a weightCentrifugal machine. resting on the bottom of the case, which weight is dragged round until the spindle is upright; this second arrangement is more effective when a liquid is treated in the drum, and wave action is set up (The Centrifugal Machine, C. A. Matthey).

Similar considerations apply to the stability of the whirling bowl in a cream-separating machine.

We can write equation (1)

N = An2 sin θμ·CK = (A2n2 − KM·KN) sin θ/A,
(4)

so that N is negative or positive, and the axis tends to rise or fall according as K moves to the inside or outside of the hyperbola of free motion. Thus a tap on the axis tending to hurry the precession is equivalent to an impulse couple giving an increase to C′K, and will make K move to the interior of the hyperbola and cause the axis to rise; the steering of a bicycle may be explained in this way; but K1 will move to the exterior of the hyperbola, and so the axis will fall in this second more violent motion.

Friction on the point of the top may be supposed to act like a tap in the direction opposite to the precession; and so the axis of a top spun violently rises at first and up to the vertical position, but falls away again as the motion dies out. Friction considered as acting in retarding the rotation may be compared to an impulse couple tending to reduce OC′, and so make K and K1 both move to the exterior of the hyperbola, and the axis falls in both cases. The axis may rise or fall according to the direction of the frictional couple, depending on the shape of the point; an analytical treatment of the varying motion is very intractable; a memoir by E. G. Gallop may be consulted in the Trans. Camb. Phil. Soc., 1903.

The earth behaves in precession like a large spinning top, of which the axis describes a circle round the pole of the ecliptic of mean angular radius θ, about 2312°, in a period of 26,000 years, so that R/μ = 26000 × 365; and the mean couple producing precession is

CRμ sin θ = CR2 sin 2312° /26000 × 365,
(5)

one 12 millionth part of 12CR2, the rotation energy of the earth.

5. If the preponderance is absent, by making the C·G coincide with O, and if Aμ is insensible compared with G′,

N = −G′μ sin θ,
(1)

the formula which suffices to explain most gyroscopic action.

Thus a carriage running round a curve experiences, in consequence Gyroscopic action of railway wheels. of the rotation of the wheels, an increase of pressure Z on the outer track, and a diminution Z on the inner, giving a couple, if a is the gauge,

Za = G′μ,

(2)

tending to help the centrifugal force to upset the train; and if c is the radius of the curve, b of the wheels, C their moment of inertia, and v the velocity of the train,

μ = v/c, G′ = Cv/b,
(3)
Z = Cv2/abc (dynes),
(4)

so that Z is the fraction C/Mab of the centrifugal force Mv2/c, or the fraction C/Mh of its transference of weight, with h the height of the centre of gravity of the carriage above the road. A Brennan carriage on a monorail would lean over to the inside of the curve at an angle α, given by

tan α = G′μ/gMh = G′v/gMhc.
(6)

The gyroscopic action of a dynamo, turbine, and other rotating machinery on a steamer, paddle or screw, due to its rolling and pitching, can be evaluated in a similar elementary manner (Worthington, Dynamics of Rotation), and Schlick’s gyroscopic apparatus is intended to mitigate the oscillation.

6. If the axis OC in fig. 4 is inclined at an angle α to the vertical, the equation (2) § 4 becomes

N = sin θ (Aμ2 cos θ − G′μ) + gMh sin (αθ).
(1)

Suppose, for instance, that OC is parallel to the earth’s axis, and that the frame is fixed in the meridian; then α is the co-latitude, and μ is the angular velocity of the earth, the square of which may be neglected; so that, putting N = 0, αθ = E,

gMh sin E − G′μ sin (α − E) = 0,
(2)
 tan E = G′μ sin α ≈ G′μ sin α. gMh + G′μ cos α gMh
(3)

This is the theory of Gilbert’s barogyroscope, described in Appell’s Mécanique rationnelle, ii. 387: it consists essentially of a rapidly rotated fly-wheel, mounted on knife-edges by an axis perpendicular to its axis of rotation and pointing east and west; spun with considerable angular momentum G′,The barogyroscope. and provided with a slight preponderance Mh, it should tilt to an angle E with the vertical, and thus demonstrate experimentally the rotation of the earth.

In Foucault’s gyroscope (Comptes rendus, 1852; Perry, p. 105) the preponderance is made zero, and the axis points to the pole, when free to move in the meridian.Foucault’s gyroscope.

Generally, if constrained to move in any other plane, the axis seeks the position nearest to the polar axis, like a dipping needle with respect to the magnetic pole. (A gyrostatic working model of the magnetic compass, by Sir W. Thomson. British Association Report, Montreal, 1884. A. S. Chessin, St Louis Academy of Science, January 1902.)

A spinning top with a polished upper plane surface will provide an artificial horizon at sea, when the real horizon is obscured. The first instrument of this kind was constructed by Serson, and is described in the Gentleman’s Magazine, vol. xxiv., 1754; also by Segner in his Specimen theoriaeGyroscopic horizon. turbinum (Halae, 1755). The inventor was sent to sea by the Admiralty to test his instrument, but he was lost in the wreck of the “Victory,” 1744. A copy of the Serson top, from the royal collection, is now in the Museum of King’s College, London. Troughton’s Nautical Top (1819) is intended for the same purpose.

The instrument is in favour with French navigators, perfected by 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 = 12π 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 = 12π, −12π, 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

 12A ( dφ ) 2 sin γ + 12Aμ2 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 (14π + 12φ) = 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 (14π + 12φ) = 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 = ±12π.

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 = 12π, cos θ = sin α sin β; and putting φ = 12πδ + 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 = h3h1θ. = 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(θα),
(28)
(MK2 + A)θ.. + (K2/A) (θα) − gMHθ = 0;
(29)

so that the upright position will be stable if K2 > gMHA, or the rotation energy of the wheel greater than 12A/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.

The velocity of H is in the direction KH perpendicular to the plane COC′, and equal to gMh sin θ or An2 sin θ, so that if a point in the axis OC′ at a distance An2 from O is projected on the horizontal plane through C in the point P on CK, the curve described by P, turned forwards through a right angle, will be the hodograph of H; this is expressed by

 An2 sin θ e(ψ + 1/2π)i = iAn2 sin θ eψi = d (ρeῶi) dt
(3)

where ρei is the vector CH; and so the curve described by P and the motion of the axis of the top is derived from the curve described by H by a differentiation.

Resolving the velocity of H in the direction CH,

d·CH/dt = An2 sin θ sin KCH = An2 sin θ KH/CH,
(4)
d·12CH2/dt = A2n2sin θ dθ/dt.
(5)

and integrating

12CH2 = A2n2 (E − cos θ),
(6)
12OH2 = A2n2 (F − cos θ),
(7)
12C′H2 = A2n2 (D − cos θ),
(8)

where D, E, F are constants, connected by

F = E + G2/2A2n2 = D + G′2/2A2n2.
(9)

Then

KH2 = OH2 − OK2,
(10)
OK2 sin2 θ = CC′2 = G2 − 2GG′ cos θ + G′2,
(11)
A2 sin2 θ (dθ/dt)2 = 2A2n2 (F − cos θ) sin2 θ − G2 + 2GG′ cos θ − G′2;
(12)

and putting cos θ = z,

 ( dz ) = 2n2 (F − z) (1 − z2) − (G2 − 2GG′z + G′2) /A2 dt
(13)
 = 2n2 (E − z) (1 − z2) − (G′ − Gz)2 /A2 = 2n2 (D − z) (1 − z2) − (G − G′z)2 /A2 = 2n2 Z suppose.

Denoting the roots of Z = 0 by z1, z2, z3, we shall have them arranged in the order

z1 > 1 > z2 > z > z3 > −1.
(14)
(dz/dt)2 = 2n2 (z1z) (z2z) (zz3).
(15)
nt = zz3 dz/ √(2Z),
(16)

an elliptic integral of the first kind, which with

 m = n √ z1 − z3 , κ2 = z2 − z3 , 2 z1 − z2
(17)

can be expressed, when normalized by the factor √(z1z3)/2, by the inverse elliptic function in the form

 mt = ∫ zz3 √ (z1 − z3) dz √ [4 (z1 − z) (z2 − z) (z − z3)]
 = sn−1 √ z − z3 = cn−1 √ z2 − z = dn−1 √ z1 − z . z2 − z3 z2 − z3 z1 − z3
(18)
zz3 = (z2z3) sn2mt, z2z = (z2z3) cn2mt, z1z = (z1z3) dn2mt.
(19)
z = z2sn2mt + z3cn2mt.
(20)

Interpreted dynamically, the axis of the top keeps time with the beats of a simple pendulum of length

L = l/12 (z1z3),
(21)

suspended from a point at a height 12 (z1 + z3)l above O, in such a manner that a point on the pendulum at a distance

12 (z1z3) l = l2/L
(22)

from the point of suspension moves so as to be always at the same level as the centre of oscillation of the top.

The polar co-ordinates of H are denoted by ρ, in the horizontal plane through C; and, resolving the velocity of H perpendicular to CH,

ρd/dt = An2 sin θ cos KCH.
(23)
ρ2d/dt = An2 sin θ·CK = An2 (G′ − G cos θ)
(24)
 ῶ = 12 ∫ G′ − Gz dt = ∫ z3 (G′ − Gz) / 2An dz , E − z A E − z √ (2Z)
(25)

an elliptic integral, of the third kind, with pole at z = E; and then

ψ = KCH = tan−1 KH/CH
 = tan−1 A sin θ dθ/dt = tan−1 √ (2Z) , G′ − G cos θ (G′ − Gz) / An
(26)

which determines ψ.

Otherwise, from the geometry of fig. 4,

C′K sin θ = OC − OC′ cos θ,
(27)
A sin2 θ dψ/dt = G − G′ cos θ,
(28)
 ψ = ∫ G − G′z dt = 12 ∫ G − G′ dt + 12 ∫ G + G′ dt , 1 − z2 A 1 − z A 1 + z A
(29)

the sum of two elliptic integrals of the third kind, with pole at z = ±1; and the relation in (25) (26) shows the addition of these two integrals into a single integral, with pole at z = E.

The motion of a sphere, rolling and spinning in the interior of a spherical bowl, or on the top of a sphere, is found to be of the same character as the motion of the axis of a spinning top about a fixed point.

The curve described by H can be identified as a Poinsot herpolhode, that is, the curve traced out by rolling a quadric surface with centre fixed at O on the horizontal plane through C; and Darboux has shown also that a deformable hyperboloid made of the generating lines, with O and H at opposite ends of a diameter and one generator fixed in OC, can be moved so as to describe the curve H; the tangent plane of the hyperboloid at H being normal to the curve of H; and then the other generator through O will coincide in the movement with OC′, the axis of the top; thus the Poinsot herpolhode curve H is also the trace made by rolling a line of curvature on an ellipsoid confocal to the hyperboloid of one sheet, on the plane through C.

Kirchhoff’s Kinetic Analogue asserts also that the curve of H is the projection of a tortuous elastica, and that the spherical curve of C′ is a hodograph of the elastica described with constant velocity.

Writing the equation of the focal ellipse of the Darboux hyperboloid through H, enlarged to double scale so that O is the centre,

x2/α2 + y2/β2 + z2/O = 1,
(30)

with α2 + λ, β2 + λ, λ denoting the squares of the semiaxes of a confocal ellipsoid, and λ changed into μ and ν for a confocal hyperboloid of one sheet and of two sheets.

λ > 0 > μ > −β2 > ν > −α2,
(31)

then in the deformation of the hyperboloid, λ and ν remain constant at H; and utilizing the theorems of solid geometry on confocal quadrics, the magnitudes may be chosen so that

α2 + λ + β2 + μ + ν = OH2 = 12k2 (F − z) = ρ2 + OC2.
(32)
α2 + μ = 12k2 (z1z) = ρ2ρ12,
(33)
β2 + μ = 12k2 (z2z) = ρ2ρ22,
(34)
μ = 12k2 (z3z) = ρ2ρ32,
(35)
ρ12 < 0 < ρ22 < ρ2 < ρ32,
(36)
F = z1 + z2 + z3,
(37)
λ − 2μ + ν = k2z, λν = k2,
(38)
 λ − μ = 1 + z , μ − ν = 1 − z λ − ν 2 λ − ν 2
(39)

with z = cos θ, θ denoting the angle between the generating lines through H; and with OC = δ, OC′ = δ′, the length k has been chosen so that in the preceding equations

δ/k = G/2An, δ′/k = G′/2An;
(40)

and δ, δ′, k may replace G, G′, 2An; then

 2Z = 1 ( dθ ) 2 = 4KH2 , 1 − z2 n2 dt k2
(41)

while from (33-39)

 2Z = 4 (α2 + μ) (β2 + μ) μ , 1 − z2 k2 (μ − λ) (μ − ν)
(42)

which verifies that KH is the perpendicular from O on the tangent plane of the hyperboloid at H, and so proves Darboux’s theorem.

Planes through O perpendicular to the generating lines cut off a constant length HQ = δ, HQ′ = δ′, so the line of curvature described by H in the deformation of the hyperboloid, the intersection of the fixed confocal ellipsoid λ and hyperboloid of two sheets ν, rolls on a horizontal plane through C and at the same time on a plane through C′ perpendicular to OC′.

Produce the generating line HQ to meet the principal planes of the confocal system in V, T, P; these will also be fixed points on the generator; and putting

(HV, HT, HP,)/HQ = D/(A, B, C,),
(43)

then

Ax2 + By2 + Cz2 = Dδ2
(44)

is a quadric surface with the squares of the semiaxes given by HV·HQ, HT·HQ, HP·HQ, and with HQ the normal line at H, and so touching the horizontal plane through C; and the direction cosines of the normal being

x/HV, y/HT, z/HP,
(45)
A2x2 + B2y2 + C2z2 = D2δ2,
(46)

the line of curvature, called the polhode curve by Poinsot, being the intersection of the quadric surface (44) with the ellipsoid (46).

There is a second surface associated with (44), which rolls on the plane through C′, corresponding to the other generating line HQ′ through H, so that the same line of curvature rolls on two planes at a constant distance from O, δ and δ′; and the motion of the top is made up of the combination. This completes the statement of Jacobi’s theorem (Werke, ii. 480) that the motion of a top can be resolved into two movements of a body under no force.

Conversely, starting with Poinsot’s polhode and herpolhode given in (44) (46), the normal plane is drawn at H, cutting the principal axes of the rolling quadric in X, Y, Z; and then

α2 + μ = x·OX, β2 + μ = y·OY, μ = z·OZ,
(47)

this determines the deformable hyperboloid of which one generator through H is a normal to the plane through C; and the other generator is inclined at an angle θ, the inclination of the axis of the top, while the normal plane or the parallel plane through O revolves with angular velocity dψ/dt.

The curvature is useful in drawing a curve of H; the diameter of curvature D is given by

 D = dp2 = 12k2 sin3 θ , 12D = 14k2 . dp δ − δ′ cosθ p KM·KN
(48)

The curvature is zero and H passes through a point of inflexion when C′ comes into the horizontal plane through C; ψ will then be stationary and the curve described by C′ will be looped.

In a state of steady motion, z oscillates between two limits z2 and z3 which are close together; so putting z2 = z3 the coefficient of z in Z is

 2z1z3 + z23 = −1 + GG′ = −1 + (OM cosθ + ON) (OM + ON cosθ) , A2n2 OM·ON
(49)
 2z1z3 = OM2 + ON2 cos θ, z1 = OM2 + ON2 , OM·ON 2OM·ON
(50)
 2 (z1 − z3) = OM2 − 2OM·ON cos θ + ON2 = MN2 . OM·ON OM·ON
(51)

With z2 = z3, κ = 0, K = 12π; and the number of beats per second of the axis is

 m = n √ z1 − z3 = MN n , π π 2 √ (OM·ON) 2π
(52)

beating time with a pendulum of length

 L = l = 4OM·ON l. 12 (z1 − z3) MN2
(53)

The wheel making R/2π revolutions per second,

 beats/second = MN n = C · MN , revolutions/second √ (OM·ON) R A OC′
(54)

from (8) (9) § 3; and the apsidal angle is

 μ 12π = Aμ · n ·12π = ON · 2√ (OM·ON) ·12π = ON π, m An m √ (OM·ON) MN MN
(55)

and the height of the equivalent conical pendulum λ is given by

 λ = g = n2 = OM = KC = OL , l lμ2 μ2 ON KC′ OC′
(56)

if OR drawn at right angles to OK cuts KC′ in R, and RL is drawn horizontal to cut the vertical CO in L; thus if OC2 represents l to scale, then OL will represent λ.

9. The gyroscope motion in fig. 4 comes to a stop when the rim of the wheel touches the ground; and to realize the motion when the axis is inclined at a greater angle with the upward vertical, the stalk is pivoted in fig. 8 in a lug screwed to the axle of a bicycle hub, fastened vertically in a bracket bolted to a beam. The wheel can now be spun by hand, and projected in any manner so as to produce a desired gyroscopic motion, undulating, looped, or with cusps if the stalk of the wheel is dropped from rest.

As the principal part of the motion takes place now in the neighbourhood of the lowest position, it is convenient to measure the angle θ from the downward vertical, and to change the sign of z and G.

Equation (18) § 8 must be changed to

 mt = nt √ z3 − z1 = ${\displaystyle \int _{z}^{z_{3}}}$ √ (z3 − z1) dz , 2 √ (4Z)
(1)
 Z = (z − F) (1 − z2) − (G2 − 2GG′z + G′2) / 2A2n2  = (z − D) (1 − z2) − (G − G′z)2 / 2A2n2  = (z − E) (1 − z2) − (G′ − Gz)2 / 2A2n2  = (z3 − z) (z − z2) (z − z1),
(2)

1 > z3 > z > z2 > −1, D, E > z1,
(3)
z1 + z2 + z3 = F = D − G′2 / 2A2n2 = E − G2 / 2A2n2,
(4)

and expressed by the inverse elliptic function

 mt = sn−1 √ z3 − z = cn−1 √ z − z2 = dn−1 √ z − z1 , z3 − z2 z3 − z2 z3 − z1
(5)
z = z2sn2mt + z3cn2mt, κ2 = (z3z2) / (z3z1).
(6)

Equation (25) and (29) § 8 is changed to

 ῶ = 12 ${\displaystyle \int }$ G′ − Gz dt = 12 ${\displaystyle \int }$ G′ − GE dt − Gt , z − E A z − E A 2A
(7)
 ψ = ${\displaystyle \int }$ G′z − G dt = 12 ${\displaystyle \int }$ G′ + G dt − 12 ${\displaystyle \int }$ G′ − G dt , 1 − z2 A 1 − z A 1 + z A
(8)

while ψ and change places in (26).

The Jacobian elliptic parameter of the third elliptic integral in (7) can be given by ν, where

 v = ${\displaystyle \int _{E}^{z_{3}}}$ √ (z3 − z1) dz = ${\displaystyle \int _{z_{2}}^{z_{3}}+\int _{E}^{z_{2}}}$ = K + (1 − ƒ) Ki′, √ (4Z)
(9)

where ƒ is a real fraction,

 (1 − ƒ) K′ = ${\displaystyle \int _{E}^{z_{2}}}$ √ (z3 − z1) dz, √ (−4Z)
(10)
 ƒK′ = ${\displaystyle \int _{z_{3}}^{E}}$ √ (z3 − z1) dz, √ (−4Z)
 = sn−1 √ E − z1 = cn−1 √ z2 − E = dn−1 √ z3 − E , z2 − z1 z2 − z1 z3 − z1
(11)

with respect to the comodulus κ′.

Then, with z = E, and

2ZE = −{ (G′ − GE) / An}2,
(12)

if II denotes the apsidal angle of , and T the time of a single beat of the axle, up or down,

 II + GT = ∫ z3z2 √ (−2ZE) dz ,= 12πƒ + Kznƒ K′, 2A z − E √ (2Z)
(13)

in accordance with the theory of the complete elliptic integral of the third kind.

Interpreted geometrically on the deformable hyperboloid, flattened in the plane of the focal ellipse, if OQ is the perpendicular from the centre on the tangent HP, AOQ = amƒK′, and the eccentric angle of P, measured from the minor axis, is am(1 − ƒ) K′, the eccentricity of the focal ellipse being the comodulus κ′.

A point L is taken in QP such that

QL/OA = znƒK′,
(14)
QV, QT, QP = OA (zs, zc, zd) ƒK′;
(15)

and with

mT = K, m/n = √ (z3z1) /2 = OA/k,
(16)
 GT = G · k K = QH K, 2A 2An OA OA
(17)
 II = 12πƒ + QL + QH K = 12πƒ + HL K. OA OA
(18)

By choosing for ƒ a simple rational fraction, such as 12, 13, 14, 15, ... an algebraical case of motion can be constructed (Annals of Mathematics, 1904).

Thus with G′ − GE = 0, we have E = z1 or z2, never z3; ƒ = 0 or 1; and P is at A or B on the focal ellipse; and then

= −pt, p = G/2A,
(19)
 ψ + pt = tan−1 n√ (2Z) , 2p (z−E)
(20)
sin θ exp (ψ + pt) i = i√ [(−z2z3) (zz1)] + √ [(z3z) (zz2)],
 z1 = 1 + z2 z3 , √ −z2 − z3 = G = p = G′ , or z2 + z3 2 2An n 2Anz1
(21)
sin θ exp (ψ + pt)i = i√ [(−z1z3) (zz2)] + √ [(z3z) (zz1)],
 z2 = 1 + z1 z3 , √ −z1 − z3 = G = p = G′ . z1 + z3 2 2An n 2Anz1
(22)

Thus z2 = 0 in (22) makes G′ = 0; so that if the stalk is held out horizontally and projected with angular velocity 2p about the vertical axis OC without giving any spin to the wheel, the resulting motion of the stalk is like that of a spherical pendulum, and given by

 sin θ exp (ψ + pt)i = i √ ( 2p2 cos θ ) + √ ( sin2 θ − 2 p2 cos θ ), n2 n2
= i sin α √ (sec α cos θ) + √ [(sec α + cos θ) (cos α − cos θ)],
(23)

if the axis falls in the lowest position to an angle α with the downward vertical.

With z3 = 0 in (21) and z2 = −cos β, and changing to the upward vertical measurement, the motion is given by

sin θ eψi = eint12 cos β [√ (1 − cos β cos θ) + i√ (cos β cos θ − cos2 θ)],
(24)

and the axis rises from the horizontal position to a series of cusps; and the mean precessional motion is the same as in steady motion with the same rotation and the axis horizontal.

The special case of ƒ = 12 may be stated here; it is found that

 p exp (ῶ − pt) i = √ (1 + x) (κ − x) + i √ (1 − x) (κ + x) , a 2 2
(25)
ρ2 = a2 (κx2),
(26)
 12λ2 sin θ exp (ψ − pt) i = (L − 1 + κ − x) √ (1 − x) (κ + x) 2
 ⁠+ i(L − 1 + κ + x) √ (1 + x) (κ − x) , 2
(27)

L = 12 (1 − κ) + λp/n,
(28)

so that p = 0 and the motion is made algebraical by taking L = 12 (1 − κ).

The stereoscopic diagram of fig. 12 drawn by T. I. Dewar shows these curves for κ = 1517, 35, and 13 (cusps).

10. So far the motion of the axis OC′ of the top has alone been considered; for the specification of any point of the body, Euler’s third angle φ must be introduced, representing the angular displacement of the wheel with respect to the stalk. This is given by

 dφ + cos θ dψ = R, dt dt
(1)
 d(φ + ψ) = ( 1 − C ) R + G′ + G , dt A A (1 + cos θ)
(2)
 d(φ − ψ) = ( 1 − C ) R + G′ − G . dt A A (1 − cos θ)

It will simplify the formulas by cancelling a secular term if we make C = A, and the top is then called a spherical top; OH becomes the axis of instantaneous angular velocity, as well as of resultant angular momentum.

When this secular term is restored in the general case, the axis OI of angular velocity is obtained by producing Q′H to I, making

 HI = A − C , HI = A − C , Q′H C Q′I A
(3)

and then the four vector components OC′, C′K, KH, HI give a resultant vector OI, representing the angular velocity ω, such that

OI/Q′I = ω/R.
(4)
 Fig. 12.

The point I is then fixed on the generating line Q′H of the deformable hyperboloid, and the other generator through I will cut the fixed generator OC of the opposite system in a fixed point O′, such that IO′ is of constant length, and may be joined up by a link, which constrains I to move on a sphere.

In the spherical top then,

 12 (φ + ψ)= ${\displaystyle \int }$ G′ + G dt ,   12 (φ − ψ)= ${\displaystyle \int }$ G′ − G dt 1 + z 2A 1 − z 2A
(5)

depending on the two elliptic integrals of the third kind, with pole at z = ±1; and measuring θ from the downward vertical, their elliptic parameters are:—

 v1 = ${\displaystyle \int _{1}^{\infty }}$ ∞1 √ (z3 − z1) dz = ƒ1K′i, √ (4Z)
(6)
 v2 = ${\displaystyle \int _{-\infty }^{-1}}$ √ (z3 − z1) dz K + (1 − ƒ2) K′i, √ (4Z)
(7)
 f1K′ = ${\displaystyle \int _{1}^{\infty }}$ √ (z3 − z1) dz √ ( −4Z)
 = sn−1 √ z3 − z1 = cn−1 √ 1 − z3 = dn−1 √ 1 − z2 , 1 − z1 1 − z1 1 − z1
(8)
 (1 − ƒ2) K′ = ${\displaystyle \int _{z_{1}}^{-}1}$ √ (z3 − z1) dz √ ( −4Z)
 = sn−1 √ −1 − z1 = cn−1 √ 1 + z2 = dn−1 √ 1 + z3 . z2 − z1 z2 − z1 z3 − z1
(9)

Then if v′ = K + (1 − ƒ′)K′i is the parameter corresponding to z = D, we find

f = ƒ2 − ƒ1, ƒ′ = ƒ2 + ƒ1,
(10)
v = v1 + v2, v′ = v1v2.
(11)

The most symmetrical treatment of the motion of any point fixed in the top will be found in Klein and Sommerfeld, Theorie des Kreisels, to which the reader is referred for details; four new functions, α, β, γ, δ, are introduced, defined in terms of Euler’s angles, θ, ψ, φ, by

α = cos 12θ exp 12 (φ + ψ)i,
(12)
β = i sin 12θ exp 12 (−φ + ψ)i,
(13)
γ = i sin 12θ exp 12 (φψ)i,
(14)
δ = cos 12θ exp 12 (−φψ)i.
(15)

Next Klein takes two functions or co-ordinates λ and Λ, defined by

 λ = x + yi = r + z , r − z x − yi
(16)

and Λ the same function of X, Y, Z, so that λ, Λ play the part of stereographic representations of the same point (x, y, z) or (X, Y, Z) on a sphere of radius r, with respect to poles in which the sphere is intersected by Oz and OZ.

These new functions are shown to be connected by the bilinear relation

 λ = αΛ + β ,   αδ − βγ = 1, γΛ + δ
(17)

in accordance with the annexed scheme of transformation of co-ordinates—

 Ξ Η Ζ ξ α2 β2 2αβ η γ2 δ2 2γδ ζ αγ βδ αδ + βγ

where

ξ = x + yi,   η = −x + yi,   ζ = −z,
Ξ = X + Yi,   Η = −X + Yi,   Ζ = −Z;
(18)

and thus the motion in space of any point fixed in the body defined by Λ is determined completely by means of α, β, γ, δ; and in the case of the symmetrical top these functions are elliptic transcendants, to which Klein has given the name of multiplicative elliptic functions; and

αδ = cos2 12θ,   βγ = −sin2 12θ,

αδβγ = 1,   αδ + βγ = cos θ,

√ ( −4αβγδ) = sin θ;
(19)

while, for the motion of a point on the axis, putting Λ = 0, or ∞,

λ = β/δ = i tan 12θeψi, or λ = α/γ = −i cot 12θeψi,
(20)

and

αβ = 12i sin θeψi, αγ = 12i sin θeψi,
(21)

giving orthogonal projections on the planes GKH, CHK; and

 α dβ − dα β = n ρ eῶi, dt dt k
(22)

the vectorial equation in the plane GKH of the herpolhode of H for a spherical top.

When ƒ1 and ƒ2 in (9) are rational fractions, these multiplicative elliptic functions can be replaced by algebraical functions, qualified by factors which are exponential functions of the time t; a series of quasi-algebraical cases of motion can thus be constructed, which become purely algebraical when the exponential factors are cancelled by a suitable arrangement of the constants.

Thus, for example, with ƒ = 0, ƒ′ = 1, ƒ1 = 12, ƒ2 = 12, as in (24) § 9, where P and P′ are at A and B on the focal ellipse, we have for the spherical top

(1 + cos θ) exp (φ + ψqt)i
= √ (sec β − cos θ) √ (cos β − cos θ) + i(√ sec β + √ cos β) √ cos θ,
(23)
(1 − cos θ) exp (φψqt)i
= √ (sec β − cos θ) √ (cos β − cos θ) + i(√sec β − √ cos β) √ cos θ,
(24)
q, q′ = n√ (2 sec β) ± n√ (2 cos β);
(25)

and thence α, β, γ, δ can be inferred.

The physical constants of a given symmetrical top have been denoted in § 1 by M, h, A, C, and l, n, T; to specify a given state of general motion we have G, G′ or CR, D, E, or F, which may be called the dynamical constants; or κ, v, w, v1, v2, or ƒ, ƒ′, ƒ1, ƒ2, the analytical constants; or the geometrical constants, such as α, β, δ, δ′, k of a given articulated hyperboloid.

There is thus a triply infinite series of a state of motion; the choice of a typical state can be made geometrically on the hyperboloid, flattened in the plane of the local ellipse, of which κ is the ratio of the semiaxes α and β, and am(1 − ƒ) K′ is the eccentric angle from the minor axis of the point of contact P of the generator HQ, so that two analytical constants are settled thereby; and the point H may be taken arbitrarily on the tangent line PQ, and HQ′ is then the other tangent of the focal ellipse; in which case θ3 and θ2 are the angles between the tangents HQ, HQ′, and between the focal distances HS, HS′, and k2 will be HS·HS′, while HQ, HQ′ are δ, δ′. As H is moved along the tangent line HQ, a series of states of motion can be determined, and drawn with accuracy.

 Fig. 13.

11. Equation (5) § 3 with slight modification will serve with the same notation for the steady rolling motion at a constant inclination α to the vertical of a body of revolution, such as a disk, hoop, wheel, cask, wine-glass, plate, dish, bowl, spinning top, gyrostat, or bicycle, on a horizontal plane, or a surface of revolution, as a coin in a conical lamp-shade.

The point O is now the intersection of the axis GC′ with the vertical through the centre B of the horizontal circle described by the centre of gravity, and through the centre M of the horizontal circle described by P, the point of contact (fig. 13). Collected into a particle at G, the body swings round the vertical OB as a conical pendulum, of height AB or GL equal to g/μ2 = λ, and GA would be the direction of the thread, of tension gM(GA/GL) dynes. The reaction with the plane at P will be an equal parallel force; and its moment round G will provide the couple which causes the velocity of the vector of angular momentum appropriate to the steady motion; and this moment will be gM·Gm dyne-cm. or ergs, if the reaction at P cuts GB in m.

Draw GR perpendicular to GK to meet the horizontal AL in R, and draw RQC′K perpendicular to the axis Gz, and KC perpendicular to LG.

The velocity of the vector GK of angular momentum is μ times the horizontal component, and

horizontal component /Aμ sin α = KC/KC′,
(1)

so that

gM·Gm = Aμ2 sin α(KC/KC′),
(2)
 A = KC′ g Gm = GQ·Gm. M KC μ2 sin α
(3)

The instantaneous axis of rotation of the case of a gyrostat would be OP; drawing GI parallel to OP, and KK′ parallel to OG, making tan K′GC′ = (A/C) tan IGC’1; then if GK represents the resultant angular momentum, K′K will represent the part of it due to the rotation of the fly-wheel. Thus in the figure for the body rolling as a solid, with the fly-wheel clamped, the points m and Q move to the other side of G. The gyrostat may be supposed swung round the vertical at the end of a thread PA′ fastened at A′ where Pm produced cuts the vertical AB, and again at the point where it crosses the axis GO. The discussion of the small oscillation superposed on the state of steady motion requisite for stability is given in the next paragraph.

12. In the theoretical discussion of the general motion General motion of
a gyrostat rolling on
a plane.
of a gyrostat rolling on a horizontal plane the safe and shortest plan apparently is to write down the most general equations of motion, and afterwards to introduce any special condition.

Drawing through G the centre of gravity any three rectangular axes Gx, Gy, Gz, the notation employed is

 u, v, w, the components of linear velocity of G; p, q, r, the components of angular velocity about the axes; h1, h2, h3, the components of angular momentum; θ1, θ2, θ3, the components of angular velocity of the coordinate axes; x, y, z, the co-ordinates of the point of contact with the horizontal plane; X, Y, Z, the components of the reaction of the plane; α, β, γ, the direction cosines of the downward vertical.

The geometrical equations, expressing that the point of contact is at rest on the plane, are

ury + qz = 0,
(1)
vpz + rx = 0,
(2)
wqx + py = 0.
(3)

The dynamical equations are

du/dtθ3v + θ2w = gα + X/M,
(4)
dv/dtθ1w + θ2u = gβ + Y/M,
(5)
dw/dtθ2u + θ1v = gγ + Z/M,
(6)

and

dh1/dtθ3h2 + θ2h3 = yZ − zY,
(7)
dh2/dtθ1h3 + θ3h1 = zX − xZ,
(8)
dh3/dtθ2h1 + θ1h2 = xY − yX.
(9)

In the special case of the gyrostat where the surface is of revolution round Gz, and the body is kinetically symmetrical about Gz, we take Gy horizontal and Gzx through the point of contact so that y = 0; and denoting the angle between Gz and the downward vertical by θ (fig. 13)

α = sin θ,   β = 0,   γ = cos θ.
(10)

The components of angular momentum are

h1 = Ap,   h2 = Aq,   h3 = Cr + K,
(11)

where A, C denote the moment of inertia about Gx, Gz, and K is the angular momentum of a fly-wheel fixed in the interior with its axis parallel to Gz; K is taken as constant during the motion.

The axis Gz being fixed in the body,

θ1 = p,   θ2 = q = −dθ/dt,   θ3 = p cot θ.
(12)

With y = 0, (1), (2), (3) reduce to

u = −qz,   v = pzrx,   w = qx;
(13)

and, denoting the radius of curvature of the meridian curve of the rolling surface by ρ,

 dx = ρ cos θ dθ = −qρ cos θ, dz = −ρ sin θ dθ = qρ sin θ; dt dt dt dt
(14)

so that

 du = − dq z − q2ρ sin θ, dt dt
(15)
 dv = dp z − dr x + pqρ sin θ + qrρ sin θ, dt dt dt
(16)
 dw = dq x − q2ρ cos θ. dt dt
(17)

The dynamical equations (4) . . . (9) can now be reduced to

 X = − dq z − p2z cotθ + q2 (x − ρ sin θ) + prx cot θ − g sin θ, M dt
(18)
 Y = dp z − dr x − pq (x + z cot θ − ρ sin θ) + qrp cos θ, M dt dt
(19)
 Z = dq x + q2 (z − ρ cos θ) + p2z − prx − g cos θ, M dt
(20)
 −zY = A dp − Apq cot θ + qh3, dt
(21)
 −zX − xZ = A dq + Ap2 cot θ − ph3, dt
 xY = dh3 = C dr = −Cq d . dt dt dθ
(23)

Eliminating Y between (19) and (23),

 ( C + x2 ) dr − xz dp + pqx (x + z cot θ − ρ sin θ) − qrxρ cos θ = 0, M dt dt
(24)
 ( C + x2 ) dr − xz dp − px (x + z cot θ − ρ sin θ) + rxρ cos θ = 0. M dθ dθ
(A)

Eliminating Y between (19) and (21)

 ( A + z2 ) dp − xz dr − A pq cot θ + q h3 M dt dt M M
pqz (x + z cot θρ sin θ) + qrzρ cos θ = 0,
(25)
 −xz dr + ( A + z2 ) dp + A p cot θ − h3 dθ M dθ M M
+ pz (x + z cot θρ sin θ) + rzρ cos θ = 0.
(B)

In the special case of a gyrostat rolling on the sharp edge of a circle passing through G, z = 0, ρ = 0, (A) and (B) reduce to

 p = ( C + 1 ) dr = ( 1 + 1 ) dh3 , Mx2 dθ Mx2 C dθ
(26)
 dp + p cot θ = h3 , d·p sin θ = h3 sin θ ; dθ A dθ A
(27)
 d2h3 + dh3 cot θ = CMx2 h3, dθ2 dθ A (Mx2 + C)
(28)

a differential equation of a hypergeometric series, of the form of Legendre’s zonal harmonic of fractional order n, given by

n (n + 1) = CMx2 / A (Mx2 + C).
(29)

For a sharp point, x = 0, ρ = 0, and the previous equations are obtained of a spinning top.

The elimination of X and Z between (18) (20) (22), expressed symbolically as

(22) − z(18) + x(20) = 0,
(30)

gives

 ( A + x2 + z2 ) dq − p h3 + ( A + z2 ) p2 cot θ + p2xz M dt M M
+ q2ρ (x cos θz sin θ) − prx (x + z cot θ) − g (x cos θ − z sin θ) = 0,
(C)

and this combined with (A) and (B) will lead to an equation the integral of which is the equation of energy.

13. The equations (A) (B) (C) are intractable in this general form; but the restricted case may be considered when the axis moves in steady motion at a constant inclination α to the vertical; and the stability is secured if a small nutation of the axis can be superposed.

It is convenient to put p = Ω sin θ, so that Ω is the angular velocity of the plane Gzx about the vertical; (A) (B) (C) become

 ( C + x2 ) dr − xz sinθdΩdθ − Ωx (x sin θ − 2z cos θ − ρ sin2 θ) + rxρ cos θ = 0, M dθ
(A*)
 −xz dr + ( A + z2 ) sin θ dΩ − h3 + 2Ω ( A + z2 ) cos θ dθ M dθ M M
+ Ωz sin θ (xρ sin θ) − rzρ cos θ = 0,
(B*)

 ( A + x2 + z2 ) dq + q2p (x cos θ − z sin θ) − Ω h3 sin θ + Ω2 (AM + z2 ) M dt M
sin θ cos θ + Ω2xz sin2 θΩrx (x sin θ + z cos θ)−g (x cos θz sin θ) = 0.
(C*)

The steady motion and nutation superposed may be expressed by

θ = α + L, sin θ = sin α + L cos α, cos θ = cos α − L sin α, Ω = μ + N, r = R + Q,
(1)

where L, N, Q are small terms, involving a factor enti, to express the periodic nature of the nutation; and then if a, c denote the mean value of x, z, at the point of contact

x = a + Lρ cos α, z = c − Lρ sin α,
(2)
x sin θ + z cos θ = a sin α + c cos α + L (a cos αc sin α),
(3)
x cos θz sin θ = a cos αc sin α − L (a sin α + c cos αρ).
(4)

Substituting these values in (C*) with dq/dt = −d2θ/dt2 = n2L, and ignoring products of the small terms, such as L2, LN, ...

 ( A + a2 + c2 ) Ln2 − (μ + N) ( CR + K + CQ ) (sin α + L cos α) M M M
 + (μ2 + 2μN) (A/M + c2 − 2Lρc sin α) (sin α cos α + L cos α) + (μ2 + 2μN) [ac − Lρ (a sin α − c sin α)] (sin2 α + L sin 2α) − (μ + N) (R + Q) (a + Lρcos α) [a sin α + c cos α + L (a cos α − c sin α)] − g (a cos α − c sin α) + gL (a sin α + c cos α − ρ) = 0,
(C**)

which is equivalent to

 −μ CR + K sin α + μ2 ( A + c2 ) sin αcos α M M
+ μ2 ac sin2 αμRa (a sin α + c cos α) − g (a cos αc sin α) = 0,
(5)

the condition of steady motion; and

DL + EQ + FN = 0,
(6)

where

 D = ( A + a2 + c2 ) n2 − μ CK + K cos α − 2μ2ρc sin2 α cos α M M
 + μ2 (A/M + c2) cos α − μ2ρ (a sin α − c cos α) sin2 α + μ2ac sin 2α − μRρ cos α (a sin α + c cos α) − μRa (a cos α − c sin α) + g (a sin α + c cos α − ρ),
(7)
 E = −μ C sin α − μa (a sin α + c cos α), M
(8)
 F = − CR + K sin α + 2μ ( A + c2 ) sin α cos α M M
+ 2μac sin2 α − Ra (a sin α + c cos α).
(9)

With the same approximation (A*) and (B*) are equivalent to

 ( C + a2 ) Q − ac sin α N − μa (a sin α + 2c cos α − ρ sin2 α) + Raρ cos α = 0, M L L
(A**)
 −ac Q + ( A + c2 ) sin α N − CR + K + 2μ ( A + c2 ) cos α L M L M M
+ μc sin α (aρ sin α) − Rcρ cos α = 0.
(B**)

The elimination of L, Q, N will lead to an equation for the determination of n2, and n2 must be positive for the motion to be stable.

If b is the radius of the horizontal circle described by G in steady motion round the centre B,

b = v/μ = (cP − aR) / μ = c sin αaR / μ,
(10)

and drawing GL vertically upward of length λ = g/μ2, the height of the equivalent conical pendulum, the steady motion condition may be written

(CR + K) μ sin αμ2 sin α cos α = −gM (a cos αc sin α)
+ M (μ2c sin αμRa) (a sin α + c cos α)
 = gM [bλ−1 (a sin α + c cos α) − a cos α + c sin α] = gM·PT,
(11)

LG produced cuts the plane in T.

Interpreted dynamically, the left-hand side of this equation represents the velocity of the vector of angular momentum about G, so that the right-hand side represents the moment of the applied force about G, in this case the reaction of the plane, which is parallel to GA, and equal to gM·GA/GL; and so the angle AGL must be less than the angle of friction, or slipping will take place.

Spinning upright, with α = 0, a = 0, we find F = 0, Q = 0, and

 − CR + K + 2μ ( A + c2 ) − Rcp = 0, M M
(12)
 ( A + c2 ) n2 = μ CR + K − μ2 ( A + c2 ) + μRρc − g (c − ρ), M M M
(13)
 ( A + c2 ) 2 n2 = 14 ( CK + R + Rcρ ) 2 − g ( A + c2 ) (c − ρ). M M M
(14)

Thus for a top spinning upright on a rounded point, with K = 0, the stability requires that

R > 2k′√ {g (cρ)} / (k2 + cρ),
(15)

where k, k′ are the radii of gyration about the axis Gz, and a perpendicular axis at a distance c from G; this reduces to the preceding case of § 3 (7) when ρ = 0.

Generally, with α = 0, but a ± 0, the condition (A) and (B) becomes

 ( C + a2 ) Q = 2μac − Raρ, −ac Q = CR + K + Rcρ − 2μ ( A + a2 ), M L L M M
(16)

so that, eliminating Q/L,

 2 [ A + c2 ) ( C + a2 ) − a2c2 ] μ = ( C + a2 ) ( CR + K ) + C Rcρ, M M M M M
(17)

the condition when a coin or platter is rolling nearly flat on the table.

Rolling along in a straight path, with α = 12π, c = 0, μ = 0, E = 0; and

N/L = (CR + K)/A,
(18)
 D = ( A + a2 ) n2 + g (a − ρ), M
 F = − CR + K − Ra2, M
(19)
N = − D =
 ( A + a2 ) n2 + g (a − ρ) M
,
L F
 ( C + a2 ) R + K M M
(20)
 ( A + a2 ) n2 = (CR + K) [ C + a2 ) R + K ] − g (a − ρ). M A M M
(21)

Thus with K = 0, and rolling with velocity V = Ra, stability requires

 V2 > a − ρ > 12 A a − ρ , 2g 2CA (CMa2 + 1) C CMa2 + 1
(22)

or the body must have acquired velocity greater than attained by rolling down a plane through a vertical height 12 (aρ) A/C.

On a sharp edge, with ρ = 0, a thin uniform disk or a thin ring requires

V2/2g > a/6 or a/8.
(23)

The gyrostat can hold itself upright on the plane without advance when R = 0, provided

K2/AM − g (aρ) is positive.
(24)

For the stability of the monorail carriage of § 5 (6), ignoring the rotary inertia of the wheels by putting C = 0, and replacing K by G′ the theory above would require

 G′ ( aV + G′ ) > gh. A A

For further theory and experiments consult Routh, Advanced Rigid Dynamics, chap. v., and Thomson and Tait, Natural Philosophy, § 345; also Bourlet, Traité des bicycles (analysed in Appell, Mécanique rationnelle, ii. 297, and Carvallo, Journal de l’école polytechnique, 1900); Whipple, Quarterly Journal of Mathematics, vol. xxx., for mathematical theories of the bicycle, and other bodies.

14. Lord Kelvin has studied theoretically and experimentally the vibration of a chain of stretched gyrostats (Proc. London Math. Soc., 1875; J. Perry, Spinning Tops, Gyrostatic chain. for a diagram). Suppose each gyrostat to be equivalent dynamically to a fly-wheel of axial length 2a, and that each connecting link is a light cord or steel wire of length 2l, stretched to a tension T.

Denote by x, y the components of the slight displacement from the central straight line of the centre of a fly-wheel; and let p, q, 1 denote the direction cosines of the axis of a fly-wheel, and r, s, 1 the direction cosines of a link, distinguishing the different bodies by a suffix.

Then with the previous notation and to the order of approximation required,

θ1 = −dq/dt, θ2 = dp/dt,
(1)
h1 = Aθ1, h2 = Aθ2, h3 = K,
(2)

to be employed in the dynamical equations

 dh1 − θ3h2 + θ2h3 = L, ... dt
(3)

in which θ3h1 and θ3h2 can be omitted.

For the kth fly-wheel

−Aq..k + Kp.k = Ta (qk − sk) + Ta (qk − sk+1),
(4)
Ap..k + Kq.k = −Ta (pkrk) − Ta (pkrk+1);
(5)

and for the motion of translation

Mx..k = T (rk+1rk), My..k = T (sk+1sk);
(6)

while the geometrical relations are

xk+1xk = a (pk+1 + pk) + 2lrk+1,
(7)
yk+1yk = a (qk+1 + qk) + 2lsk+1.
(8)

Putting

x + yi = w, p + qi = ω, r + si = σ,
(9)

these three pairs of equations may be replaced by the three equations

A..k − K..ki + 2Tak − Ta (σk+1 + σk) = 0,
(10)
M..k − T (σk+1σk) = 0,
(11)
ωk+1ωk − a(k+1 + k − 2lσk+1) = 0.
(12)

For a vibration of circular polarization assume a solution

ωk, k, σk = (L, P, Q) exp (nt + kc) i,
(13)

so that c/n is the time-lag between the vibration of one fly-wheel and the next; and the wave velocity is

U = 2 (a + l) n/c.
(14)

Then

P (−An2 + Kn + 2Ta) − QTa (eci + 1) = 0,
(15)
−LMn2 − QT (eci − 1) = 0,
(16)
L (eci − 1) − Pa (eci + 1) − 2Qleci = 0,
(17)

leading, on elimination of L, P, Q, to

 cos c = (2 Ta + Kn − An2) (1 − Mn2l/T) − Mna2 , 2Ta + Kn − An2 + Mna2
(18)
 2 sin2 12c = Mn2 2Ta (a + l) + KNl − An2l . T 2Ta + Kn − An2 + Mn2a2
(19)

With K = 0, A = 0, this reduces to Lagrange’s condition in the vibration of a string of beads.

Putting

ρ = M/2 (a + l),   the mass per unit length of the chain,
(20)
κ = K/2 (a + l),   the gyrostatic angular momentum per unit length,
(21)
α = A/2 (a + l),   the transverse moment of inertia per unit length,
(22)
1/2c = (a + l) n/U,
(23)

equation (19) can be written

{sin (a + l) n/U}2
 = (a + l)2n2 ρ · Ta + κnl − αn2l , T Ta + κn (a + l) − αn2 (a + l) + ρn2a2 (a + l)
(24)
 { (a + l) n } 2 sin (a + l) n/U
 = T · T + (κn − αn2) (1 + l/a) + ρn2a (a + l) . ρ T + (κn − an2) l/a
(25)

In a continuous chain of such gyrostatic links, with a and l infinitesimal,

 U2 = T { 1 + κn − αn2 } ρ T + (κn − αn2 l/a)
(26)

for the vibration of helical nature like circular polarization.

Changing the sign of n for circular polarization in the opposite direction

 U′2 = T { 1 − κn + αn2 } ρ T − (κn + αn2 l/a)
(27)

In this way a mechanical model is obtained of the action of a magnetized medium on polarized light, κ representing the equivalent of the magnetic field, while α may be ignored as insensible (J. Larmor, Proc. Lond. Math. Soc., 1890; Aether and Matter, Appendix E).

We notice that U2 in (26) can be positive, and the gyrostatic chain stable, even when T is negative, and the chain is supporting a thrust, provided κn is large enough, and the thrust does not exceed

(κnan2) (1 + l/a);
(28)

while U′2 in (27) will not be positive and the straight chain will be unstable unless the tension exceeds

(κn + αn2) (1 + l/a).
(29)

15. Gyrostat suspended by a Thread.—In the discussion of the small vibration of a single gyrostat fly-wheel about the vertical position when suspended by a single thread of length 2l = b, the suffix k can be omitted in the preceding equations of § 14, and we can write

A.. − K.i + Ta − Taσ = 0,
(1)
Mw.. + Tσ = 0, with T = gM,
(2)
w − abσ = 0.
(3)

Assuming a periodic solution of these equations

w, , σ, = (L, P, Q) exp nti,
(4)

and eliminating L, P, Q, we obtain

(−An2 + Kn + gMa) (gn2b) − gMn2a2 = 0,
(5)

and the frequency of a vibration in double beats per second is n/2π, where n is a root of this quartic equation.

For upright spinning on a smooth horizontal plane, take b = ∞ and change the sign of a, then

An2 − Kn + gMa = 0,
(6)

so that the stability requires

K2 > 4gAMa.
(7)

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

(A + Ma2) n2 − Kn + gMa = 0,
(8)

requiring for stability, as before in § 3,

K2 > 4g (A + M2) Ma.
(9)

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 ξ, η, then

br = ξ, bs = η, bσ = ξ + ηI = λ (suppose),
(10)
ω = α + λ.
(11)

If the point is forced to take the motion (ξ, η, ζ) by components of force X, Y, Z, the equations of motion become

−Aq.. + Kp. =    Ya − Zaq,
(12)
Ap.. + Kq. =    −Xa + Zap,
(13)
Mw.. = X + Yi, M (ζ..g) = Z;
(14)

so that

A.. − K.i + gMa + Maw.. = Maῶζ..,
(15)

or

(A + Ma2).. − K.i + gMa + Maλ = Maῶζ..
(16)

Thus if the point of the gyrostat is made to take the periodic motion given by λ = R exp nti, ζ = 0, the forced vibration of the axis is given by = P exp nti, where

P { −(A + Ma2) n2 + Kn + gMa} − RMn2a = 0;
(17)

and so the effect may be investigated on the Fleuriais gyroscopic horizon of the motion of the ship.

Suppose the motion λ 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 λ = b1, 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

(A1 + M1h12)..1 − K1. 1i
 = −g (M1h1 + Mb) ῶ1 − b (X + Yi) = −g (M1h1 + Mb) ῶ1 − Mb(aῶ.. + λ..);
(18)

so that, in the small vibration,

Rb${\displaystyle \scriptstyle {\left\{{\begin{matrix}\ \\\ \end{matrix}}\right.}}$−(A1 + M1h12) n2 + K1n + g (M1h1 + Mb)${\displaystyle \scriptstyle {\left.{\begin{matrix}\ \\\ \end{matrix}}\right\}\,}}$= Mn2b (aP + R),
(19)
R { −(A1 + M1h12 + Mb2) n2 + K1n + g (M1h1 + Mb)} − PMn2ab2 = 0.
(20)

Eliminating the ratio of P to R, we obtain

 { −(A + Ma2) n2 + Kn + gMa} × { −(A1 + M1h12 + Mb2) n2 + K1n + g (M1h1 + Mb)} − M2n4a2b2 = 0,
(21)

a quartic for n, giving the frequency n/2π of a fundamental vibration.

Change the sign of g for the case of the gyrostats spinning upright, one on the top of the other, and so realize the gyrostat on the top of a gyrostat described by Maxwell.

In the gyrostatic chain of § 14, the tension T may change to a limited pressure, and U2 may still be positive, and the motion stable; and so a motion is realized of a number of spinning tops, superposed in a column.

16. The Flexure Joint.—In Lord Kelvin’s experiment the gyrostats are joined up by equal light rods and short lengths of elastic wire with rigid attachment to the rod and case of a gyrostat, so as to keep the system still, and free from entanglement and twisting due to pivot friction of the fly-wheels.

When this gyrostatic chain is made to revolve with angular velocity n in relative equilibrium as a plane polygon passing through Oz the axis of rotation, each gyrostatic case moves as if its axis produced was attached to Oz by a flexure joint. The instantaneous axis of resultant angular velocity bisects the angle πθ, if the axis of the case makes an angle θ with Oz, and, the components of angular velocity being n about Oz, and −n about the axis, the resultant angular velocity is 2n cos12 (πθ) =2n sin12θ; and the components of this angular velocity are

(1) −2n sin 12θ sin 12θ = −n (1 − cos θ), along the axis, and

(2) −2n sin 12θ cos 12θ = −n sin θ, perpendicular to the axis of the case. The flexure joint behaves like a pair of equal bevel wheels engaging.

The component angular momentum in the direction Ox is therefore

L = −An sin θ cos θ − Cn (1 − cos θ) sin θ + K sin θ,
(3)

and Ln is therefore the couple acting on the gyrostat.

If α denotes the angle which a connecting link makes with Oz, and T denotes the constant component of the tension of a link parallel to Oz, the couple acting is

Ta cos θk (tan αk+1 + tan αk) − 2Tα sin θk,
(4)

which is to be equated to Ln, so that

−An2 sin θk cos θk − Cn (1 − cos θk) sin θk + Kn sin θk −Ta cos θk (tan αk+1 + tan αk) + 2Tα sin θk = 0.
(5)

Mn2xk + T (tan αk+1 − tan αk) = 0,
(6)

with the geometrical relation

xk+1xka (sin θk+1 + sin θk) − 2l sin αk+1 = 0.
(7)

When the polygon is nearly coincident with Oz, these equations

can be replaced by
(8)
(−An2 + Kn + 2Ta) θk − Ta (αk+1 + αk) = 0,
(9)
Mn2xk + T (αk+1αk) = 0,
(10)
xk+1xka (θk+1 + θk) − 2lak = 0,

and the rest of the solution proceeds as before in § 14, putting

(11)
xk, θk, αk = (L, P, Q) exp cki.

A half wave length of the curve of gyrostats is covered when ck = π, so that π/c is the number of gyrostats in a half wave, which is therefore of wave length 2π (a + l)/c.

A plane polarized wave is given when exp cki is replaced by exp (nt + ck)i, and a wave circularly polarized when w, ῶ, σ of § 14 replace this x, θ, α.

Gyroscopic Pendulum.—The elastic flexure joint is useful for supporting a rod, carrying a fly-wheel, like a gyroscopic pendulum.

Expressed by Euler’s angles, θ, φ, ψ, the kinetic energy is

(12)
T = 12A (θ. 2 + sin2 θψ. 2) + 12C′ (1 − cos θ)2ψ. 2 + 12C (φ. + ψ. cos θ)2,

where A refers to rod and gyroscope about the transverse axis at the point of support, C′ refers to rod about its axis of length, and C refers to the revolving fly-wheel.

The elimination of ψ. between the equation of conservation of angular momentum about the vertical, viz.

(13) A sin2 θψ. − C′ (1 − cos θ) cos θψ. + C(φ. + ψ. cos θ) cos θ = G, a constant, and the equation of energy, viz.

(14) T − gMh cos θ = H, a constant, with θ measured from the downward vertical, and

(15) φ. + ψ. cos θ = R, a constant, will lead to an equation for dθ/dt, or dz/dt, in terms of cos θ or z, the integral of which is of hyperelliptic character, except when A = C′.

In the suspension of fig. 8, the motion given by φ. is suppressed in the stalk, and for the fly-wheel φ. gives the rubbing angular velocity of the wheel on the stalk; the equations are now

(16)
T = 12A (θ. 2 + sin2 θψ. 2) + 12C′ cos2 θψ. 2 + 12CR2 = H + gMh cos θ,
(17)
A sin2 θψ. + C′ cos2 θψ. + CR cos θ = G,

and the motion is again of hyperelliptic character, except when A = C′, or C′ = 0. To realize a motion given completely by the elliptic function, the suspension of the stalk must be made by a smooth ball and socket, or else a Hooke universal joint.

Finally, there is the case of the general motion of a top with a spherical rounded point on a smooth plane, in which the centre of gravity may be supposed to rise and fall in a vertical line. Here

(18)
T = 12 (A + Mh2 sin2 θ) θ. 2 + 12A sin2 θψ. 2 + 12CR2 = H − gMh cos θ,

with θ measured from the upward vertical, and

(19)
A sin2 θψ. + CR cos θ = G,

where A now refers to a transverse axis through the centre of gravity. The elimination of ψ. leads to an equation for z, = cos θ, of the form

 (20) ⁠ ( dz ) 2 = 2 g Z = 2 g (z1 − z) (z2 − z) (z3 − z) , dt h 1 − z2 + A/Mh2 h (z4 − z) (z − z5)

with the arrangement

(21)
z1, z4 > / > z2 > z > z3 > − / > z5;

so that the motion is hyperelliptic.

Authorities.—In addition to the references in the text the following will be found useful:—Ast. Notices, vol. i.; Comptes rendus, Sept. 1852; Paper by Professor Magnus translated in Taylor’s Foreign Scientific Memoirs, n.s., pt. 3, p. 210; Ast. Notices, xiii. 221–248; Theory of Foucault’s Gyroscope Experiments, by the Rev. Baden Powell, F.R.S.; Ast. Notices, vol. xv.; articles by Major J. G. Barnard in Silliman’s Journal, 2nd ser., vols. xxiv. and xxv.; E. Hunt on “Rotatory Motion,” Proc. Phil. Soc. Glasgow, vol. iv.; J. Clerk Maxwell, “On a Dynamical Top,” Trans. R.S.E. vol. xxi.; Phil. Mag. 4th ser. vols. 7, 13, 14; Proc. Royal Irish Academy, vol. viii.; Sir William Thomson on “Gyrostat,” Nature, xv. 297; G. T. Walker, “The Motion of a Celt,” Quar. Jour. Math., 1896; G. T. Walker, Math. Ency. iv. 1, xi. 1; Gallop, Proc. Camb. Phil. Soc. xii. 82, pt. 2, 1903, “Rise of a Top”; Price’s Infinitesimal Calculus, vol. iv.; Worms, The Earth and its Mechanism; Routh, Rigid Dynamics; A. G. Webster, Dynamics (1904); H. Crabtree, Spinning Tops and Gyroscopic Motion (1909). For a complete list of the mathematical works on the subject of the Gyroscope and Gyrostat from the outset, Professor Cayley’s Report to the British Association (1862) on the Progress of Dynamics should be consulted. Modern authors will be found cited in Klein and Sommerfeld, Theorie des Kreisels (1897), and in the Encyclopädie der mathematischen Wissenschaften.  (G. G.)