Rieter

Conditions at the traveler in the plane of the ring

Index

Fig. 91 – The forces acting at the traveler

The following forces act on the traveler (1) in the plane of the ring (2):

• A tensile force FF, which arises from the winding tension of the yarn and always acts at a tangent to the circumference of the cop (3).
• A frictional force FH between the ring and the traveler. In the stationary state, i.e. with constant traveler speed, this braking force FH is in equilibrium with the forward component FT of the yarn tension FF. Hence we have:

$F_H = F_T$  or  $F_H = F_F \times sin\alpha$

A force FN normal to the surface of the ring (pulling the traveler in the direction of the cop, diminishing the friction of the traveler at the ring created by the centrifugal force FZ ). The frictional force FH arises from this normal force in accordance with the relation:

$F_H = \mu \times F_N$

Where μ

• A centrifugal force FZ, which is the largest force acting on the traveler. This force can be calculated in accordance with the relations [20]:

$F_Z = m_L \times {\omega^2}_L \times d_R/2$

$\omega_L = n_{spindle} \times \pi/30$

where mL is the mass of the traveler, ωL is the angular velocity of the traveler, and dR is the diameter of the ring.

Professor Krause (ETH, Zurich) identifies the following relationships between these forces, solved for the tensile force:

$F_F = \frac { \mu \times F_Z}{sin \alpha + \mu \times cos \alpha}$

$F_F = \frac { \mu \times m_L \times { \omega^2}_L \times d_R}{2 \times (sin \alpha + \mu \times cos \alpha)}$

For a rough estimate, the term $\mu \times cos \alpha$ can be ignored. Approximately, therefore, we have:

$F_F = \frac { \mu \times m_L \times { \omega^2}_L \times d_R}{2 \times sin \alpha}$