(1)
and transverse velocity
| u sin θ − v cos θ = ( l − | c1 | ) u sin θ ≈ (β − α) u sin θ; |
| c2 |
(2)
and the time of completing a turn of the spiral is 2π/μ.
When μ has the critical value in (7),
| 2π | = | 4π | C2 | cos θ = | 2π | (x2 + 1) cos θ, | |
| μ | p | C1 | p |
(3)
which makes the circumference of the cylinder on which the helix is wrapped
| 2π | (u sin θ − v cos θ = | 2πu | (β − α) (x2 + 1) sin2 θ cos θ |
| μ | p |