and putting tan (¼π + ½φ) = 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 (¼π + ½φ) = 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 = ±½π.
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 = ½π, cos θ = sin α sin β; and putting φ = ½π − δ + 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)