Part 10
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 = 2[pi]/µ, its period in seconds,
µ T cos lat T cos lat (4) tan E = -- cos lat = ---------, or E = ---------, µ1 86400 8[pi]
if E is expressed in minutes, taking µ = 2[pi]/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
(5) µ·OC´ cos lat = gMh sin E = µ1·OC´ sin E,
µ T cos lat sin E = -- cos lat, E = ---------, µ1 8[pi]
in which 8[pi] 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.
[Illustration: FIG. 11.]
Euler's coordinate angles.
7. In the ordinary treatment of the general theory of the gyroscope, the motion is referred to two sets of rectangular axes; the 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 [theta], [phi], [psi], such that the successive turning of the axes Ox, Oy, Oz through the angles (i.) [psi] about Oz, (ii.) [theta] about OE, (iii.) [phi] 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
(1) [.[psi]] sin [theta], .[theta], .[phi] + .[psi] cos [theta];
so that, denoting the components about OX, OY, OZ by P, Q, R,
(2) P = .[theta] cos [phi] + .[psi] sin [theta] sin [phi], Q = -.[theta] sin [phi] + .[psi] sin [theta] cos [phi], R = .[phi] + .[psi] cos [theta].
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 [gamma] with OZ, while OZ is held by another ring zZ, which constrains it to move round the vertical Oz at a constant inclination [theta] with constant angular velocity µ, so that
(3) .[theta] = 0, .[psi] = µ;
(4) P = µ sin [theta] sin [phi], Q = µ sin [theta] cos [phi], R = .[phi] + µ cos [theta].
With CXF a quadrant, the components of angular velocity and momentum about OF, OY, are
(5) P cos [gamma] - R sin [gamma], Q, and A(P cos [gamma] - R sin [gamma]), AQ,
so that, denoting the components of angular momentum of the fly-wheel about OC, OX, OY, OZ by K or G´, h1, h2, h3,
(6) h1 = A(P cos [gamma] - R sin [gamma]) cos [gamma] + K sin [gamma],
(7) h2 = AQ,
(8) h3 = -A(P cos [gamma] - R sin [gamma]) sin [gamma] + K cos [gamma];
and the dynamical equation
dh3 (9) --- - h1Q + h2P = N, dt
with K constant, and with preponderance downward
(10) N = gMh cos zY sin [gamma] = gMh sin [gamma] sin [theta] cos [phi],
reduces to
d²[phi] (11) A ------- sin [gamma] dt²
+ Aµ² sin [gamma] sin² [theta] sin [phi] cos [phi] + Aµ² cos [gamma] sin [theta] cos [theta] cos [phi] - (Kµ + gMh) sin [theta] cos [phi] = 0.
The position of relative equilibrium is given by
(12) cos [phi] = 0, and sin [phi]
Kµ + gMh - Aµ² cos [gamma] cos [theta] = --------------------------------------. Aµ² sin [gamma] sin [theta]
For small values of µ the equation becomes
d²[phi] (13) A ------- sin [gamma] - (Kµ + gMh) sin [theta] cos [phi] = 0, dt²
so that [phi] = ½[pi] gives the position of stable equilibrium, and the period of a small oscillation is 2[pi] [root]{A sin [gamma]/(Kµ + gMh) sin [theta]}.
In the general case, denoting the periods of vibration about [phi] = ½[pi], -½[pi], and the sidelong position of equilibrium by 2[pi]/(n1, n2, or n3), we shall find
sin [theta] (14) n1² = ------------- {gMh + Kµ - Aµ² cos ([gamma] - [theta])}, A sin [gamma]
sin [theta] (15) n2² = ------------- {-gMh - Kµ + Aµ² cos ([gamma] + [theta])}, A sin [gamma]
(16) n3 = n1 n2/µ sin [theta].
The first integral of (11) gives
/d[phi]\² (17) ½A ( ------ ) sin [gamma] \ dt /
+ ½Aµ² sin [gamma] sin² [theta] sin² [phi] - Aµ² cos [gamma] sin [theta] cos [theta] sin [phi] + (Kµ + gMh) sin [theta] sin [phi] - H = 0,
and putting tan (¼[pi] + ½[phi]) = z, this reduces to
dz (18) -- = n [root]Z dt
where Z is a quadratic in z², so that z is a Jacobian elliptic function of t, and we have
(19) tan (¼[pi] + ½[phi]) = C(tn, dn, nc, or cn)nt,
according as the ring ZC performs complete revolutions, or oscillates about a sidelong position of equilibrium, or oscillates about the stable position of equilibrium [phi] = ±½[pi].
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 ß with E. and W., so that OZ lies in the horizon at an angle E.ß.N., we must put [gamma] = ½[pi], cos [theta] = sin [alpha] sin ß; and putting [phi] = ½[pi] - [delta] + E, where [delta] denotes the angle between Zz and the vertical plane Z[zeta] through the zenith [zeta],
(20) sin [theta] cos [delta] = cos [alpha], sin [theta] sin [delta] = sin [alpha] cos ß;
so that equations (9) and (10) for relative equilibrium reduce to
(21) gMh sin E = KQ = Kµ sin [theta] cos [phi] = Kµ sin [theta] sin ([delta] - E),
and will change (3) § 6 into
Kµ sin [alpha] cos ß (22) tan E = --------------------, gMh + Kµ cos [alpha]
a multiplication of (3) § 6 by cos ß (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,
Kµ sin [alpha] cos ß F sin [alpha] cos ß (23) tan E = -------------------- = ---------------------------, gMh - Kµ cos [alpha] 86400 A/T²C - F cos [alpha]
so that the gyroscope would reverse if it were possible to make F cos [alpha] > 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 [theta] with the upward vertical, and the axis of the fly-wheel makes an angle [phi] with the vertical plane of motion, the three components of angular momentum are
(24) h1 = K cos [phi], h2 = A .[theta] + K sin [phi], h3 = A .[phi],
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 = 0, Q = [.[theta]], R = 0, so that
(25) M´ = h2 = A :[theta] + K .[phi] cos [phi],
(26) 0 = h3 - h1 .[theta] = A :[phi] - K .[theta] cos [phi].
For the motion of the pendulum, including the fly-wheel,
(27) MK² :[theta] = gMH sin [theta] - M´ = gMH sin [theta] - A :[theta] - K .[phi] cos [phi].
If [theta] and [phi] remain small,
(28) A :[phi] = K .[theta], A .[phi] = K([theta] - [alpha]),
(29) (MK² + A) :[theta] + (K²/A) ([theta] - [alpha]) - gMH[theta] = 0;
so that the upright position will be stable if K² > gMHA, or the rotation energy of the wheel greater than ½A/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
(30) (MK² + A)/[(K²/gA) - MH].
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.
General motion of the top.
8. Careful scrutiny shows that the steady motion of a top is not steady absolutely; it reveals a small nutation 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
(1) KH = A d[theta]/dt, C´K = A sin [theta] d[psi]/dt, OC´ = G´.
Drawing KM vertical and KN parallel to OC´, then
(2) KM = A d[psi]/dt, KN = CR - A cos [theta] d[psi]/dt = (C - A)R + A d[phi]/dt
so that in the spherical top, with C = A, KN = A d[phi]/dt.
The velocity of H is in the direction KH perpendicular to the plane COC´, and equal to gMh sin [theta] or An² sin [theta], 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
(3) An² sin [theta] e^{([psi] + ½[pi])i} = iAn² sin [theta] e^{[psi]i} = d/dt ([rho]e^{[pi]i})
where [rho]e[varpi]i 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,
(4) d·CH/dt = An² sin [theta] sin KCH = An² sin [theta] KH/CH,
(5) d·½CH²/dt = A²n²sin [theta] d[theta]/dt.
and integrating
(6) ½CH² = A²n²(E - cos [theta]),
(7) ½OH² = A²n²(F - cos [theta]),
(8) ½C´H² = A²n²(D - cos [theta]),
where D, E, F are constants, connected by
(9) F = E + G²/2A²n² = D + G´²/2A²n².
Then
(10) KH² = OH² - OK²,
(11) OK² sin² [theta] = CC´² = G² - 2GG´ cos [theta] + G´²,
(12) A² sin² [theta] (d[theta]/dt)² = 2A²n²(F - cos [theta]) sin² [theta] - G² + 2GG´ cos [theta] - G´²;
and putting cos [theta] = z,
(13) (dz/dt)² = 2n²(F - z) (1 - z²) - (G² - 2GG´z + G´²)/A² = 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
(14) z1 > 1 > z2 > z > z3 > -1.
(15) (dz/dt)² = 2n²(z1 - z)(z2 - z)(z - z3).
_z / / (16) nt = | dz/ \/(2Z), _/z3
an elliptic integral of the first kind, which with
/z1 - z3 z2 - z3 (17) m = n \ / -------, [kappa]² = -------, \/ 2 z1 - z2
can be expressed, when normalized by the factor [root](z1 - z3)/2, by the inverse elliptic function in the form
_z / [root](z1 - z3)dz (18) mt = | --------------------------------- _/z3 [root][4(z1 - z)(z2 - z)(z - z3)]
/ z - z3 /z2 - z /z1 - z = sn^(-1)\ / ------- = cn^(-1)\ / ------- = dn^(-1)\ / -------. \/ z2 - z3 \/ z2 - z3 \/ z1 - z3
(19) z - z3 = (z2 - z3)sn²mt, z2 - z = (z2 - z3)cn²mt, z1 - z = (z1 - z3)dn²mt.
(20) z = z2sn²mt + z3cn²mt.
Interpreted dynamically, the axis of the top keeps time with the beats of a simple pendulum of length
(21) L = l/½(z1 - z3),
suspended from a point at a height ½(z1 + z3)l above 0, in such a manner that a point on the pendulum at a distance
(22) ½(z1 - z3)l = l²/L
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 [rho], [varpi] in the horizontal plane through C; and, resolving the velocity of H perpendicular to CH,
(23) [rho] d[~omega]/dt = An² sin [theta] cos KCH.
(24) [rho]² d[~omega]/dt = An² sin [theta] CK
= An² (G´ - G cos [theta]) _ _ / G´ - Gz dt / (G´ - Gz)/2An dz (25) [~omega] = ½ | ------- -- = | ------------- ----------, _/ E - z A _/z3 E - z [root](2Z)
an elliptic integral, of the third kind, with pole at z = E; and then
(26) [~omega] - [psi] = KCH = tan^(-1) KH/CH
A sin [theta] d[theta]/dt [root](2Z) = tan^(-1) ------------------------- = tan^(-1) ------------, G´ - G cos [theta] (G´ - Gz)/An
which determines [psi].
Otherwise, from the geometry of fig. 4,
(27) C´K sin [theta] = OC - OC´ cos [theta],
(28) A sin² [theta] d[psi]/dt = G - G´ cos [theta], _ _ _ / G - G´z dt / G - G´ dt / G + G´ dt (29) [psi] = | ------- -- = ½ | ------ -- + ½ | ------ --, _/ 1 - z² A _/ 1 - z A _/ 1 + z A
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,
(30) x²/[alpha]² + y²/ß² + z²/O = 1,
with [alpha]² + [lambda], ß² + [lambda], [lambda] denoting the squares of the semiaxes of a confocal ellipsoid, and [lambda] changed into µ and [nu] for a confocal hyperboloid of one sheet and of two sheets.
(31) [lambda] > O > µ > -ß² > [nu] > -[alpha]²,
then in the deformation of the hyperboloid, [lambda] and [nu] remain constant at H; and utilizing the theorems of solid geometry on confocal quadrics, the magnitudes may be chosen so that
(32) [alpha]² + [lambda] + ß² + µ + [nu] = OH² = ½k²(F - z) = [rho]² + OC².
(33) [alpha]² + µ = ½k²(z1 - z) = [rho]² - [rho]1²,
(34) ß² + µ = ½k²(z2 - z) = [rho]² - [rho]2²,
(35) µ = ½k²(z3 - z) = [rho]² - [rho]3²,
(36) [rho]1² < 0 < [rho]2² < [rho]² < [rho]3²,
(37) F = z1 + z2 + z3,
(38) [lambda] - 2µ + [nu] = k²z, [lambda] - [nu] = k²,
[lambda] - µ 1 + z µ - [nu] 1 - z (39) --------------- = -----, --------------- = ----- [lambda] - [nu] 2 [lambda] - [nu] 2
with z = cos [theta], [theta] denoting the angle between the generating lines through H; and with OC = [delta], OC´ = [delta]´, the length k has been chosen so that in the preceding equations
(40) [delta]/k = G/2An, [delta]´/k = G´/2An;
and [delta], [delta]´, k may replace G, G´, 2An; then
2Z 1 /d[theta]\² 4KH² (41) ------ = -- ( -------- ) = ----, 1 - z² n² \ dt / k²
while from (33-39)
2Z 4([alpha]² + µ)(ß² + µ)µ (42) ------ = --------------------------, 1 - z² k²(µ - [lambda])(µ - [nu])
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 = [delta], HQ´ = [delta]´, so the line of curvature described by H in the deformation of the hyperboloid, the intersection of the fixed confocal ellipsoid [lambda] and hyperboloid of two sheets [nu], 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
(43) (HV, HT, HP,)/HQ = D/(A, B, C,),
then
(44) Ax² + By² + Cz² = D[delta]²
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
(45) x/HV, y/HT, z/HP,
(46) A²x² + B²y² + C²z² = D²[delta]²,
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, [delta] and [delta]´; 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
(47) [alpha]² + µ = x·OX, ß² + µ = y·OY, µ = z·OZ,
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 [theta], the inclination of the axis of the top, while the normal plane or the parallel plane through O revolves with angular velocity d[psi]/dt.
The curvature is useful in drawing a curve of H; the diameter of curvature D is given by
dp² ½k²sin³ [theta] ½D ¼k² (48) D = --- = -----------------------------, -- = -----. dp [delta] - [delta]´ cos[theta] p KM·KN
The curvature is zero and H passes through a point of inflexion when C´ comes into the horizontal plane through C; [psi] 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
GG´ (OM cos[theta] + ON)(OM + ON cos [theta]) (49) 2Z1z3 + z²3 = -1 + ---- = -1 + -----------------------------------------, A²n² OM·ON
OM² + ON² OM² + ON² (50) 2z1z3 = --------- cos [theta], z1 = ---------, OM·ON 2OM·ON
OM² - 2OM·ON cos [theta] + ON² MN² (51) 2(z1 - z3) = ------------------------------ = -----. OM·ON OM·ON
With z2 = z3, [kappa] = [omicron], K = ½[pi]; and the number of beats per second of the axis is
m n /z1 - z3 MN n (52) ---- = ---- \ / ------- = ------------- -----, [pi] [pi] \/ 2 [root](OM·ON) 2[pi]
beating time with a pendulum of length
l 4OM·ON (53) L = ---------- = ------ l. ½(z1 - z3) MN²
The wheel making R/2[pi] revolutions per second,
beats/second MN n C MN (54) ------------------ = ------------- -- = -- . ---, revolutions/second [root](OM·ON) R A OC´
from (8) (9) § 3; and the apsidal angle is
½[pi] Aµ n ON 2[root](OM·ON) ON (55) µ ----- = --·--·½[pi] = ------------- · -------------- · ½[pi] = -- [pi], m An m [root](OM·ON) MN MN
and the height of the equivalent conical pendulum [lambda] is given by
[lambda] g n² OM KC OL (56) -------- = --- = -- = -- = --- = ---, l lµ² µ² ON KC´ OC´
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 [lambda].
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 [theta] from the downward vertical, and to change the sign of z and G.
Equation (18) § 8 must be changed to
_z3 /z3 - z1 / [root](z3 - z1)dz (1) mt = nt / ------- = | -----------------, \/ 2 _/z [root](4Z)