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 = 2π²θ²l/360², where θ is half the angle of swing expressed in degrees; hence h/8l = θ²/52520, 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 of inertia of the body I = Σ(ma²). If k be defined by the relation Σ(ma²) = Σ(m) X k², then k is called the radius of gyration. If k be the radius of gyration of a bob round a horizontal axis through its centre of gravity, h the distance of its centre of gravity below its point of suspension, and k’ the radius of gyration of the bob round the centre of suspension, then k′² = h² + k². If l be the length of a simple pendulum that oscillates in the same time, then lh = k′² = h² + k². Now k can be calculated if we know the form of the bob, and l is the length of the simple pendulum = 39.14 in.; hence h, the distance of the centre of gravity of the bob below the point of suspension, can be found.