Mῶk − T (σk+1 − σk) = 0,
(11)
ωk+1 − ωk − a(ῶk+1 + ῶk − 2lσk+1) = 0.
(12)
For a vibration of circular polarization assume a solution
ωk, ῶk, σk = (L, P, Q) exp (nt + kc) i,
(13)
so that c/n is the time-lag between the vibration of one fly-wheel and the next; and the wave velocity is
U = 2 (a + l) n/c.
(14)