Page:EB1911 - Volume 12.djvu/799

D =	dp2	=	1/2k2 sin3 θ	,	1/2D	=	1/4k2	.
	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 = 1/2π; 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.
	1/2 (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

1/2π

=

Aμ

·

n

·1/2π =

ON

·

2√ (OM·ON)

·1/2π =

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	= $\int _{z}^{z_{3}}$	√ (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 = (z₃ − z₂) / (z₃ − z₁).

(6)

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

ῶ = 1/2 $\int$	G′ − Gz		dt	= 1/2 $\int$	G′ − GE		dt	−	Gt	,
	z − E		A		z − E		A		2A

(7)

ψ = $\int$

G′z − G

dt

= 1/2 $\int$

G′ + G

dt

− 1/2 $\int$

G′ − G

dt

,

1 − z2

A

1 − z

A

1 + z

A

(8)

while ψ and ῶ change places in (26).

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

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

(9)

where ƒ is a real fraction,

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

(10)

ƒK′ = $\int _{z_{3}}^{E}$	√ (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 ῶ, and T the time of a single beat of the axle, up or down,

II +	GT	= ∫ z3z2	√ (−2ZE)		dz	,= 1/2πƒ + 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 = √ (z₃ − z₁) /2 = OA/k,

(16)

GT	=	G	·	k	K =	QH	K,
2A		2An		OA		OA

(17)

II = 1/2πƒ +	QL + QH	K = 1/2πƒ +	HL	K.
	OA		OA

(18)

By choosing for ƒ a simple rational fraction, such as 1/2, 1/3, 1/4, 1/5, ... 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

ῶ = −pt, p = G/2A,

(19)

ψ + pt = tan−1	n√ (2Z)	,
	2p (z−E)

(20)

sin θ exp (ψ + pt) i = i√ [(−z₂ − z₃) (z − z₁)] + √ [(z₃ − z) (z − z₂)],

z1 =	1 + z2 z3	, √	−z2 − z3	=	G	=	p	=	G′	, or
	z2 + z3		2		2An		n		2Anz1

(21)

sin θ exp (ψ + pt)i = i√ [(−z₁ − z₃) (z − z₂)] + √ [(z₃ − z) (z − z₁)],

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 = eint √ 1/2 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 = 1/2 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 (κ − x²),

(26)

1/2λ2 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 = 1/2 (1 − κ) + λp/n,

(28)

so that p = 0 and the motion is made algebraical by taking L = 1/2 (1 − κ).

The stereoscopic diagram of fig. 12 drawn by T. I. Dewar shows these curves for κ = 15/17, 3/5, and 1/3 (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)