In the case of the steady motion of the top, the vector OH lies in the vertical plane COC′, in OK suppose (fig. 4), and has a component OC = G about the vertical and a component OC′ = G′, suppose, about the axis OC; and G′ = CR, if R denotes the angular velocity of the top with which it is spun about OC′.
If μ denotes the constant precessional angular velocity of the vertical plane COC′ the components of angular velocity and momentum about OA are μ sin θ and Aμ sin θ, OA being perpendicular to OC′ in the plane COC′; so that the vector OK has the components
OC′ = G′, and C′K = Aμ sin θ,
(2)
and the horizontal component
CK = OC′ sin θ − C′K cos θ
= G′ sin θ − Aμ sin θ cos θ.
(3)
The velocity of K being equal to the impressed couple Oh,
gMh sin θ = μ·CK = sin θ (G′μ − Aμ2 cos θ),
(4)