Optical fibre cables are made by placing optical fibres inside a loose tube packed with a water based gel, and then winding these loose tubes on to a central strength member in helically wound sections of alternating twist separated by reversing sections. The length of the loose tubes and their position on the strength member was modelled along with an analysis of where the optical fibres lie in the loose tubes

Fackerell, T.

Fraser, B.

English

Mathematics in Industry Information Service Eprints Archive

92 MM Cables
MODELLING OPTICAL FIBRE CABLE
Optical fibre cables are made by placing optical fibres inside a loose
tube packed with a water based gel, and then winding these loose
tubes on to a central strength member in helically wound sections of
alternating twist separated by reversing sections. The length of the
loose tubes and their position on the strength member was modelled
along with an analysis of where the optical fibres lie in the loose
tubes.
1. Introduction
The topic was introduced by Mr Allan Davies, Engineering Manager for MM
Cables Australia. A typical optical fibre cable is made in two stages: six optical
fibres are enclosed by a loose fitting plastic tube, then six to twelve of these loose
tubes are wrapped around a central strength member (CSM). These tubes are
wrapped helically for a number of turns and then the rotation is reversed, so
that reversal sections lie between helical sections of opposite rotation (see Figure
1 and Figure 2). The helical winding ensures that if the outer sheath of a loose
tube is put under strain, the actual optical fibres themselves are not put under
strain, provided the strain applied to the loose tube is not too great. Reversal of
the helical winding is required not only for reasons connected with construction
but also for reasons connected with installation. The strains that the cables
experience can be due to tensions when being laid, and also to strains due to
thermal expansion and contraction, since the cables have to be able to cope with
ambient temperatures which can vary between -40°C and 70°C. An important
concept is that of the strain free window, that is, the range of extension and
contraction of the cable for which the optical fibres remain strain free. It is
known that the strain free window is increased as the ratio of the length of
helical sections to the length of the reversal sections is increased. For a given
lay length, that is, the length along the CSM for one complete helical turn,
the way to increase this ratio is to increase the number of turns in each helical
section. However, this increases the amount of optical fibre and loose tube for
a given length of CSM. It is clearly important to be able to model the reversal
section accurately, since there is good evidence indicating that poor performance
is associated with the reversal section.
Important questions for which answers are desired are:
1. How much optical fibre and how much loose tube is required to make a
given length of cable?
Modelling optical fibre cable 93
2. Where is the loose tube laid on the CSM during manufacture?
3. When there is no tension on the cable, where do the fibres lie, especially
in the reversal section?
4. Where do the fibres go when the cable is elongated under tension or com-
pressed by low temperature?
5. Is there an optimal lay length and number of turns in the helical sections?
6. How tightly controlled does manufacturing have to be to comply with
specifications?
Figure 1: Schematic diagram of the loose tubes wrapping around the central
strength member. The wrapping starts offhelically in an anticlockwise direction,
then reverses to wrap clockwise.
2. Modelling the length of the loose tubes
The first problem to be investigated was the modelling of the length of a loose
tube through a full cycle of a helical section, a reversal section, a reversed helical
section and a further reversal section. A sequence of increasingly complicated
models were investigated and compared with one particular sample of cable.
Figure 2: Helically stranded loose tube, showing dotted centre-line of the loose
tube.
94 MM Cables
We denote by Dc the diameter of the CSM and by Do the outside diameter
of the loose tube. Then the centre-line of a loose tube is a curve on the surface
of a cylinder of diameter DCL = Dc + Do (see Figure 2). If (J and z are the
standard polar coordinates for the surface of this cylinder, the curve representing
the tube is defined once (J and z are given as piece-wise continuously differentiable
functions of a suitable parameter.
In helical sections (J and z obviously satisfy
(J _ (Jo = ± 21r z
I
where the + sign applies to right-handed (Z) twist, the - sign applies to left-
handed (S) twist sections, and I is the lay length (= pitch) of the helical windings
(see Figure 2). A major problem is to know how to model the reversal sections.
Two approaches were attempted:
1. Using geometrical models involving simple mathematical relations between
(J and z.
2. Using a kinematic model based on the motion of the stranding machinery,
that is, the actual machinery that winds the loose tubes on to the CSM.
2.1 Geometric modelling of the reversal section
Three simple curves were used by members of the group to model the reversal
section. In all three cases we have that if Np denotes the number of turns in
each helical section and P is the length (along the CSM) of each complete cycle,
then
P = 2(Npl + 2>'), (1)
where 2>' is the length of the reversal section.
1. The first curve considered was the one for which the ()-z relation in the
reversal section is modelled by a quadratic equation in a way that ensures
continuity of the tangent to the curve representing the loose tube. By a
suitable choice of origin we may take the ()-z relation to be
1r 2
()= >.lz . (2)
Some basic geometry leads to the expression for the length LT of tube per
cycle,
Modelling optical fibre cable 95
Straightforward integration yields
>'1 2>.1. 17rDcL
LT = 2(Np+ T)yJ2 +7r2DbL + 7rDCL sinh" (-1-)· (3)
2. In a second geometric model (G. Byrnes) the quadratic reversal section
discussed above was replaced by the appropriate arc of a circle with half-
angle </J= tan-1 (7rDCL/I) and radius p = >./ sin</J.The expression for LT
then becomes
3. In the third geometric model (P-F. Siew) the quadratic reversal section
was replaced by a cosine curve
O _ 2>'DcL [ (7rZ)]- 1- cos -
1 2>'
which results in
LT = 2NpJJ2 + 7r2DbL + 410>'
yielding
7r2D2 (7rZ)1+ CL sin2 - dz12 2>'
where E (~\ arctan(7r DCL/I)) is the complete elliptic integral of the sec-
ond kind (see Abramowitz and Stegun, p. 590).
In many cases the dimensionless parameter ( = 7rDCL/I is a small quantity
(e.g. for the sample cable ( ~ 0.225). It is interesting to note that, correct to
terms of order (4, equations (3), (4) and (5) respectively are
LT = 2NpJI2 + 7r2DbL + 4>' [1 + ~(2 - 410(4+ 0((6)] , (6)
LT = 2Np /12 + 7r2DbL + 4>' [1+ ~(2 - ~(4 + 0((6)] (7)y 6 120
and
LT = 2NpJJ2 + 7r2DbL + 4>' [1+ ~(2 - :4(4 + 0((6)] , (8)
showing the essential robustness of these simple geometrical models.
96 MM Cables
Now consider the sample cable. This has DeL = 5.65 mm, P/2 = 720 mm
and I = 79 mm. IT we take Np = 6, the nominal number of turns between
reversals, then from equation (1) we find that A = 123 mm. Entering these
values into equation (3) or equation (4), we obtain LT/2 = 734 mm, for a length
difference (LT - P)/2 per half cycle of 14 mm. Equation (5) similarly gives
LT /2 = 733 mm for a length difference per half cycle of 13 mm. Both of these
are slightly high when compared with the value 10.5 mm obtained by measuring
the actual cable.
The number Np allows some tuning of the model, since it is not clear quite
where the helical section can be said to end and the reversal section begin. IT
we set N p = 4 for the above cable, the first rotation at each end of the "helical"
section is taken to be part of the reversal. This is borne out somewhat by
the fact that the pitch of this winding is measurably longer than the others.
Corresponding to Np = 4 we have A = 202 mm. Entering these new values
into equation (3) or equation (4) we obtain LT/2 = 731 mm and we find that
equation (5) gives LT /2 = 730 mm, all of which are in in better agreement with
measurement.
However, the real problem that we would like to address is, given the draw
speed of the cable, the number of turns that are completed by the stranding
machine before a reversal is begun, and the angular velocity of the stranding
machine as a function of the time, to determine the length of loose tube used as
well as its location around the CSM.
2.2 Kinematic modelling of the reversal section
As a function of the time t, the stranding machinery has an angular velocity
w(t) of the form given in Figure 3. As a first approach to using this information,
the problem was modelled (P-F. Siew, S. Lord) by making the assumption that
dB = w(t)
dt
dz
dt
v,
(9)
(10)
where, as before, B and z are cylindrical polar coordinates and V is the draw
speed of the cable. It is convenient to use the notation
n
(11)
Modelling opticallibre cable 97
w(t)
o t
-0
Figure 3: Angular velocity of the stranding machine. Note that t5 = tt and
t7 = t3'
We find that the angular velocity w(t) is given by
0:::; t:::; Tt,
Tt :::;t :::;T2'
T2 s t :::;T3'
T3 :::;t :::;T4•
The arc length s is given as a function of the time t as
lot D2 w(t')2s(t) = V2 + CL dt'.
o 4
If we write kt = DCLOj(2Vtt}, k2 = V VI + (DcLOj(2V))2 and also k3 =
DCLOj(2Vt3), and if we introduce the function f(k, T, t) defined by
f(k, T, t) = ~ [(t - Th/1 + k2(t - T)2 +lsinh-t (k(t - T))] (12)
we find that s(t) is given by
0:::; t:::; Tt,
Tt :::;t :::;T2'
T2 :::;t :::;T3,
T3 :::;t:::; T4•
For the cable being considered which has a cable length of 1440 mm for a full
cycle, we obtain an excess length of 29.0 mm, corresponding to a fractional
increase of 2.01%.
98 MM Cables
One problem with this model is that it implies that the loose tubes are
straight in part of the reversal section, and this is not borne out by observation
of the manufactured cable. Furthermore, from the data presented, it was noted
that the number of complete helical turns actually obtained on the cable differed
from the number that the stranding machine is set to give, which suggests that
there is a lag or advance factor which is probably a function of the inertia of the
system, the stiffness of the loose tube and the friction between the loose tubes
and the CSM. The influence of these factors has not been modelled in the above
treatment.
A partial explanation for the difference between the the number of turns
actually laid down on the CSM and the number of turns preset on the stranding
machine was given by N. Zoubtchenko. He reasoned that in the process of
laying down the loose tubes on the CSM there is an important geometrical
consideration, which may be illustrated with reference to Figure 4.
Figure 4: The coordinate setup of the cable twisting.
If <p is the angular coordinate of the hole in the strander wheel through which
one of the loose tubes is being supplied, and if ()is the angular coordinate on the
point on the CSM at which this tube makes contact, then it is only true that
iJ = ~ so long as ~ is one-signed, which for convenience we will assume is positive.
As soon as ~ = 0, i.e. as soon as <p achieves a maximum, () remains constant
until <p has rotated back through an angle 2'}',where'}' = cos-1(Dc L / D), where
D is the diameter of the circle of holes in the strander wheel. Zoubtchenko is
currently modelling this process in more detail.
Modelling optical fibre cable 99
3. The strain free window (G. Byrnes)
Consider the situation of a bundle of fibres running inside a loose tube. In
the case where the fibre length is exactly equal to that of the tube, we can assume
the bundle of fibres and the tube to be concentric (see Figure 2). If the tube is
now strained, so that the fibres are shorter than the centre-line of the tube, the
fibres must either lie along a path inside the tube whose length is equal to that
of the unstrained fibres, or the fibres must strain. The straining of the fibres is
unavoidable if the tube is straight, and in this case the optical attenuation in
the fibres will increase to unacceptable levels.
In a cable where the tubes are wound around a central member, the curvature
of the tube makes it possible for the fibres to find a shorter path than the
centre-line of the loose tube and thereby avoid an increase in attenuation. We
will simplify the problem by making the approximation, common in the cable
industry, of treating a bundle of n fibres, each of diameter D f, as a single fibre
with diameter given by the approximate formula
J5n-2Db=DI -3-'
Then since this bundle runs in a tube with internal diameter Di, we replace this
with a line running inside a tube of diameter Del I = D, - Db.
In the case of a helix, the difference between the centre-line and shortest
lengths can be calculated exactly. For a helix of radius a and lay length I, the
arc-length of one revolution of the helix is given by
The shortest path inside the tube runs along the inside surface closest to the
central axis of the helix. This curve is itself a helix: the centre-line is a helix
of radius DCL/2, while the shortest path has radius (DcL - Dell )/2. Thus the
relative change is given by
JI2 + 7r2DbL - Jp + 7r2(DCL - Deff)2
~H= ~--------r========---------JZ2 + 7r2DbL
For a cable with n = 6, D I = 0.25 mm (including coating), DCL = 5.65 mm,
Z= 79mm and D; = 1.6 mm, we find Del I = 0.836 mm and hence ~H = 0.66%.
While there is no exact expression for the shortest path through the reversal
section, we can still obtain an expression for the relative change in path length
accurate to first order. Note that to shorten a curve as much as possible for
100 MM Cables
a given perturbation, it should be displaced in the direction of the principal
normal vector N to the curve, given by
d
N=-T,
ds
where T is the unit tangent to the curve and s is the arc-length. The modulus
of N is called the curvature, denoted by K" with K, = 1Nl = 1/ p, where p is
the radius of curvature. Note that in general p is non-constant. Note also that
the relative change in the arc-length of a curve which has been displaced in the
direction N by an amount £( z) is, to first order,
~ foL ;~:~ dz = ~ foL £(z)K,(z)dz,
where L is the arc-length of the curve.
Now when looking for the shortest path inside a tube, we have £(z) = Deff/2
constant, since we can only move from the centre to the inside wall of the tube.
Thus for a tube following an arbitrary curve C, the relative change in length
from centre to shortest path is approximately
1
t1c = "2KDef!>
where K is the average curvature along the path.
For the quadratic model of the tube path, it should be noted that K, is not
continuous across the joins between the helical and parabolic sections. However
this should not introduce significant error.
For a curve on the surface of a cylinder of radius a, with position vector in
cartesian coordinates, we have
r(z) = (acosO(z),asinO(z),z).
The corresponding unit tangent vector is given by
T _ (-asinO(Z) acosO(z) 1)
- (1 + a20/2)l/2' (1 + a2()12)l/2' (1 + a20/2)l/2 '
where 0/ = dO/dz. We then have K, = 1Nl, where
1 dT
N = (1+ a20/2)l/2 dz '
A tedious calculation then gives
K,= a 3 2 {0,,2 + (1+ a20/2)0/4 f/2
(1 + a20/2) /
Modelling optical fibre cable 101
For the helical sections this is simply
27\"2DeL
K-H(Z) = z2+7\"2DbL'
which for the cable parameters above yields PH = 1/ K-H = 58.8 mm. The
calculated relative path length change using this is ~ X 0.836/58.8 = 0.71%, not
too far from the exact value of 0.66%.
For the case of the reversal section of the quadratic model, we find that
a {47\"2 ( 7\"2D2 ) 167\"4z4}1/2
"Q( z) = (1+ ':~,~'z2)'/2 ~212 + 1+ ~21~L z2 ~414
The relative change for the complete quadratic model may be found from
a _ 6.6 x 1O-3Npl + Den loA K-Q(z)dz
Q - Npl + 2). .
For the values Np = 6, ). = 129 mm and I = 79 mm we find that
loA K-Q(z) dz = 0.789,
which results in aQ = 0.53%, showing a substantial reduction in the strain free
window. If we use the values Np = 4, ). = 202 mm and I = 79 mm we find that
loA K-Q(z) dz = 1.231,
giving an even worse strain free window of aQ = 0.43%. This clearly shows
the importance of having a long helical section compared with the length of the
reversing section.
4. Discussion
Two outstanding problems remain, namely, (1) to develop a detailed dynam-
ically based model for the way in which the stranding machine lays the loose
tubes on to the central strength member, and (2) to investigate the helical buck-
ling modes of the optical fibres inside the loose tubes. As well as including the
effect of the stiffness of the loose tubes, the first problem needs to take into
account the fact that the binding tapes, which ultimately fix the location of the
loose tubes on the CSM, are layed down in a preferential direction on the ca-
ble. Investigation of the second problem will need to be both experimental and
theoretical (see Thompson and Champneys, 1996), but will need to go beyond
this work in order to take into account the effect of the optical cable touching
the inside wall of the loose tube.
102 MM Cables
Acknowledgements
This problem was eo-moderated by Barrie Fraser and Ted Fackerell. We
are grateful to Allan Davies and Kate McKeand from MM Cables for their
readiness at all times to answer our questions, to Peg-Foo Siew for a Maple
program to calculate the length of the loose tube in Section (2.2), and to Jim
Clark for the diagrams. The problem was attended by Graham Byrnes, Jim
Clark, Allan Davies, Vincent Hart, Stephen Lord, Kate McKeand, Bernhard H.
Neumann, Peg-Foo Siew, Anthony Stace, Dave Stump, Phil Unterberger and
Nikolai Zoubtchenko.
References
M. Abramowitz and I.A. Stegun, (Eds) Handbook of Mathematical Functions
(Dover Publications, NY, 1970).
J .M.T. Thompson and A.R. Champneys, "From helix to localized writhing in the
torsional post-buckling of elastic rods", Proc. R. Soc. Lond. A452 (1996),
117-138.


Modelling optical fibre cable

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