Well, considering that no-one appears to be sure about this, I'll ramble on
a little more.

As I understand it, the orientation of the tangent plane can be specified by
the
inclination angle of the helix together with the inclination angle of plane
from the principle normal vector, not to be confused with the surface normal
vector orthogonal to the tangent plane on the pipe surface. The angle
between the normal force N and the gravitational force mg is thus

(1) chi = (phi + psi)/2
(Does this follow?)

Therefore

(2) N * Cos[chi] = mg

(3) N^2 * Cos^2[chi] = m^2 * g^2

is the vertical component of N.

(4) N^2 * Sin^2[chi] = m^2 * g^2 * Tan^2[psi] + m^2 * a^2

where mg*Tan[psi]= mv^2/(R + r*Sin[psi]) is the radial force, ma is the
tangential force.

Dividing (4) by (3) gives

Tan^2[chi] = Tan^2[psi] + a^2/g^2

0 = a^2/g^2
a = 0

Clearly this is incorrect. Any ideas?

"James Stokes" wrote:
[snip]
A particle inside a helical tube is subject to the force of gravity mg
acting vertically down and the normal force N exerted by the pipe wall.
The
normal force can be decomposed into three mutually orthogonal components;
a
vertical component balancing the graviational force, a radial component
providing the centripetal force, and a component providing the tangential
acceleration along the path. The question is, how do I derive these
forces
in terms of the inclination angles and the other forces? Any help would
be
greatly appreciated.

Thanks in advance.

James