(3) L = -An sin [theta] cos [theta] - Cn (1 - cos [theta]) sin [theta]
+ K sin [theta],
and Ln is therefore the couple acting on the gyrostat.
If [alpha] denotes the angle which a connecting link makes with Oz,
and T denotes the constant component of the tension of a link parallel
to Oz, the couple acting is
(4) Ta cos [theta]_k(tan [alpha]_(k+1) + tan [alpha]_k)
- 2T[alpha]sin[theta]_k,
which is to be equated to Ln, so that
(5) -An^2 sin [theta]_k cos [theta]_k
- Cn(1 - cos [theta]_k) sin [theta]_k
+ Kn sin [theta]_k
- T[alpha] cos [theta]_k(tan [alpha]_(k+1) + tan [alpha]_k)
+ 2T[alpha] sin [theta]_k = 0.