It follows that, in order that the spin-axle may remain in the meridian, the spin-axle must be caused to precess about a vertical axis with an angular velocity equal in magnitude to we„ but in the opposite direction. This precession requires a torque about a horizontal axis. The requirement of a torque about a horizontal axis, and the additional requirement that for damping there must at the same time be a torque about the vertical axis, are fulfilled in the Sperry Mark V compass and later models by connecting the mercury ballistic to a point of the gyro-casing slightly to the east of the point vertically below the center of the gyro-wheel, as we have seen in Arts. 107 and 121. By this scheme, the applied torque is about an axis making a small angle 0 to the horizontal, and the precessional velocity wp thereby produced is about an axis inclined at the angle 0 to the vertical.
The horizontal component of the torque produces a precessional velocity wp, about a vertical axis, Fig. 160. The vertical component of the torque produces a precession wph about a horizontal axis perpendicular to the gyro-axle of the value
wph = wp, tan 0
If the spin-axle of the compass is to remain in the meridian, the vertical component of the velocity of precession must equal the vertical component of the earth's rotation. That is,
Wp, = w,