describe the dial or “motion-work.” The minute hand fits on
to a squared end of a brass socket, which is fixed to the wheel
M, and fits close, but not tight, on the prolonged arbor of the
centre wheel. Behind this wheel is a bent spring which is (or
ought to be) set on the same arbor with a square hole (not a
round one as it sometimes is) in the middle, so that it must
turn with the arbor; the wheel is pressed up against this spring,
and kept there, by a cap and a small pin through the end of the
arbor. The consequence is, that there is friction enough between
the spring and the wheel to carry the hand round, but not
enough to resist a moderate push with the finger for the purpose
of altering the time indicated. This wheel M, which is sometimes
called the minute-wheel, but is better called the *hour-wheel* as
it turns in an hour, drives another wheel N, of the same number
of teeth, which has a pinion attached to it; and that pinion
drives the *twelve-hour wheel* H, which is also attached to a large
socket or pipe carrying the hour hand, and riding on the former
socket, or rather (in order to relieve the centre arbor of that
extra weight) on an intermediate socket fixed to the *bridge* L,
which is screwed to the front plate over the hour-wheel M. The
weight W, which drives the train and gives the impulse to the
pendulum through the escapement, is generally hung by a
catgut line passing through a pulley attached to the weight,
the other end of the cord being tied to some convenient place
in the clock frame or *seat-board*, to which it is fixed by screws
through the lower pillars.

Fig. 4. |

*Pendulum.*—Suppose that we have a body P (fig. 4) at rest,
and that it is material, that is to say, has “mass.” And for
simplicity let us consider it a ball of
some heavy matter. Let it be free
to move horizontally, but attached
to a fixed point A by means of a
spring. As it can only move horizontally and not fall, the
earth’s gravity will be unable to impart any motion to it.
Now it is a law first discovered by Robert Hooke (1635–1703)
that if any elastic spring be pulled by a force, then, within its
elastic limits, the amount by which it will be extended is proportional
to the force. Hence then, if a body is pulled out against
a spring, the restitutional force is proportional to the displacement.
If the body be released it will tend to move back to its
initial position with an acceleration proportioned to its mass and
to its distance from rest. A body thus circumstanced moves with
harmonic motion, vibrating like a stretched piano string, and the
peculiarity of its motion is that it is isochronous. That is to say,
the time of returning to its initial position is the same, whether
it makes a large movement at a high velocity under a strong
restitutional force, or a small movement at a lower velocity under
a smaller restitutional force (see Mechanics). In consequence of
this fact the balance wheel of a watch is isochronous or nearly
so, notwithstanding variations in the amplitude of its vibrations.
It is like a piano string which sounds the same note, although the
sound dies away as the amplitude of its vibrations diminishes.

Fig. 5. |

A pendulum is isochronous for similar reasons. If the bob be
drawn aside from D to C (fig. 5), then the restitutional force
tending to bring it back to rest is approximately
the force which gravitation
would exert along the tangent CA, *i.e.*

g cos ACW = g |
BC | = g | displacement BC | . |

OC | length of pendulum |

Since *g* is constant, and the length of the
pendulum does not vary, it follows that
when a pendulum is drawn aside through
a small arc the force tending to bring it
back to rest is proportional to the displacement
(approximately). Thus the
pendulum bob under the influence of
gravity, if the arc of swing is small, acts
as though instead of being acted on by gravity it was acted on
by a spring tending to drag it towards D, and therefore is
isochronous. The qualification “If the arc of swing is small” is
introduced because, as was discovered by Christiaan Huygens,
the arc of vibration of a truly isochronous pendulum should
not be a circle with centre O, but a cycloid DM, generated by
the rolling of a circle with diameter DQ = ½OD, upon a straight
line QM. However, for a short distance near the bottom, the
circle so nearly coincides with the cycloid that a pendulum
swinging in the usual circular path is,
for small arcs, isochronous for practical
purposes.

Fig. 6. |

The formula representing the time of
oscillation of a pendulum, in a circular arc,
is thus found:—Let OB (fig. 6) be the
pendulum, B be the position from which
the bob is let go, and P be its position at
some period during its swing. Put FC = *h*,
and MC = *x*, and OB = *l*. Now when a
body is allowed to move under the force
of gravity in any path from a height *h*,
the velocity it attains is the same as a body would attain falling
freely vertically through the distance *h*. Whence if *v* be the
velocity of the bob at P, *v* = √2gFM = √2*g*(*h* − *x*). Let P*p* = *ds*, and
the vertical distance of *p* below P = *dx*, then P*p* = velocity at P × *dt*;
that is, *dt* = *ds*/*v*. Also

ds | = | 1 | = | 1 | , |

dx | MP | √x(2l − x) |

whence

dt = | ds | = | ldx |
. | 1 | = | 1 | √ | l | . | dx | . | 1 | . |

v | √x(2l − x) |
√2g(h − x) | 2 | g | √x(p − x) |
√1 − (x/2l) |

Expanding the second part we have

dt = | 1 | √ | l | . | dx |
. ( 1 + | x |
+ ... ) . |

2 | g | √x(h − x) |
4l |

If this is integrated between the limits of 0 and *h*, we have

t = π √ |
l | . ( 1 + | h | + ... ) , |

g | 8l |

where *t* is the time of swing from B to A. The terms after the second
may be neglected. The first term, π √*l*/*g*, is the time of swing in a
cycloid. The second part represents the addition necessary if the
swing is circular and not cycloidal, and therefore expresses the
“circular error.” Now *h* = BC²/*l* = 2π²θ²*l*/360², where θ is half the
angle of swing expressed in degrees; hence *h*/8*l* = θ;²/52520, and the formula becomes

t = π √ |
l | ( 1 + | θ² | ) . |

g | 52520 |

Hence the ratio of the time of swing of an ordinary pendulum of any length, with a semiarc of swing = θ degrees is to the time of swing of a corresponding cycloidal pendulum as 1 + θ²/52520 : 1. Also the difference of time of swing caused by a small increase θ′ in the semiarc of swing = 2θθ′ / 52520 second per second, or 3.3θθ′ seconds per day. Hence in the case of a seconds pendulum whose semiarc of swing is 2° an increase of .1° in this semiarc of 2° would cause the clock to lose 3.3 × 2 × 0.1 = .66 second a day.

Huygens proposed to apply his discovery to clocks, and since the evolute of a cycloid is an equal cycloid, he suggested the use of a flexible pendulum swinging between cycloidal cheeks. But this was only an example of theory pushed too far, because the friction on the cycloidal cheeks involves more error than they correct, and other disturbances of a higher degree of importance are left uncorrected. In fact the application of pendulums to clocks, though governed in the abstract by theory, has to be modified by experiment.

Neglecting the circular error, if L be the length of a pendulum and
*g* the acceleration of gravity at the place where the pendulum
is, then T, the time of a single vibration = π√(L/*g*). From this
formula it follows that the times of vibration of pendulums are
directly proportional to the square root of their lengths, and inversely
proportional to the square root of the acceleration of gravity
at the place where the pendulum is swinging. The value of *g* for
London is 32.2 ft. per second per second, whence it results that the
length of a pendulum for London to beat seconds of mean solar
time = 39.14 in. nearly, the length of an astronomical pendulum to
beat seconds of sidereal time being 38.87 in.

This length is calculated on the supposition that the arc of swing is cycloidal and that the whole mass of the pendulum is concentrated at a point whose distance, called the radius of oscillation, from the point of suspension of the pendulum is 39.14 in. From this it might be imagined that if a sphere, say of iron, were suspended from a light rod, so that its centre were 39.14 in. below its point of support, it would vibrate once per second. This, however, is not the case. For as the pendulum swings, the ball also tends to turn in space to and fro round a horizontal axis perpendicular to the direction of its motion. Hence the force stored up in the pendulum is expended, not only in making it swing, but also in causing the ball to oscillate to and fro through a small angle about a horizontal axis. We have therefore to consider not merely the vibrations of the rod, but the oscillations of the bob. The moment of the momentum of the system round the point of suspension, called its moment of inertia, is composed of the sum of the mass of each particle multiplied into the square of its distance from the axis of rotation. Hence the moment