AUTOMATIC LINE CALCULATION PROGRAM
THEORY FOR PASSENGER ROPEWAYS TESTING
Introduction
The calculation routines for the configuration of the ropes in the spans according to the
different load conditions are mainly based on the theoretical principles described in the
publication "Elementi di Progetto per Impianti a Fune ediz.1981" and in the report "Calcolo
di una linea funiviaria bifune a più campate, con l'ausilio del calcolatore elettronico", by Prof.
Pietro D'Armini.
Calculation theory
As we know, the balance configuration for a rope only subject to its own weight and to the
tensions at the ends, is represented by a curve, called catenary curve, which may be shown
on the Cartesian plane by the equation below:
x
eh  e
y  h
2

x
h
 x
 h  cosh 
h
h being defined as a catenary "parameter" .
The fundamental properties of the catenary curve make it possible to obtain the relation
expressing the tension value at any point along the rope, only as a function of the ordinate "y"
and the unit weight "q" of the rope:
T = q·y
Particularly, at the vertex V in the catenary curve (cf. Figure 1) :
H = q·h
Wherein h is the ordinate for V, vertex of the catenary curve.
Figure 1. Graphical representation of a rope length only subject to its own weight.
Unladen rope between two ends (catenary).
The calculation of the configuration of an unladen rope laid between two ends A and B, is
obtained by taking into consideration the diagram in Figure 1.
1
Solution for the unladen span, considering:
a) Known values :
-L
horizontal length of the span between the ends A and B
-
span difference in height between the ends A and B
-q
unit weight of the rope
- Ta
tension at the downstream end (=counterweight/2 for the 1° span)
b) Unknown values :
-S
rope development between the ends A and B
-f
deflection at any point in the span
- Tb
tension of the rope at the upstream end
- a
rope mouth angle downstream from the span
- b
rope mouth angle upstream from the span
c) Symbols:
- Cos
hyperbolic cosine
- arctg
arc cotangent
- cos
trigonometric cosine
- tg
trigonometric tangent
The formulae used for the calculus of the unladen spans are the following:
1) we calculate
yA
= TA / q
Δ
L 


 L 2  y A  
 A  arctg 
2) we calculate
h  y A  cos A 

L

 arcsen tg A 
h

 B  arctg sen

yB 
3) we check
h
cos  B 
yB - yA -  < 0.001 (convergence)
4) if convergence has not been checked out, we must correct angle a, then, we can
continue the iterative calculus from stage NR.2.
5) if convergence has been checked out, we find
S
= h · ( tg(B) - tg(A) )
TB
= yB · q
6) in the end, we find the deflection of the generic abscissa "x"
x

y x  h  cos  arcsen tg A   y A
h

fx  x 

 yA
L
The application software makes use of the Excel functions for the solution of the above
formulae and for the automatic seek of the calculus routine convergence (specific function
“expression.GoalSeek(Goal, ChangingCell)”).
2
FLOW-CHART OF THE PROGRAM FOR THE AUTOMATIC CALCULATION OF THE SPAN WITH
UNLADEN ROPE, WITH THE EXACT CATENARY CURVE FORMULAE
Known Values Input
- L Span horizontal length
-  span difference in height
- q rope unit weight
- Ta tension at the downstream end
Ordinate calculus and angle of 1
tangency downstream from the
rope
yA = TA/q
A = arctg(/L - L/(2·yA))
We calculate:
2
- h = yA·cos(A)
- B = arctgsen(L/h+arcsen(tg A))
- yB = h/cos(B)
3
NO
YB - yA - < 0.001
YES
We calculate:
- S = h· (tg(B)-tg(A))
- TB = yB·q
5
and the deflection of the generic
abscissa x as to the downstream end
A
- yx = h·cos(x/h+arcsentg A)- yA
- fx = x·/L- yA
END
3
4
We correct angle a then we repeat
the cycle
Formulae used for the subsequent unladen spans
Regarding the subsequent spans, the calculation method remains unchanged, provided that
it is given the new value for the initial tension downstream from the new span to be tested
(Ta). It is the result of the sum of the value Tb' of the previous span plus the rope-tower
friction value. But, as the rope-tower friction value depends on pressure, and pressure
depends on the rope tension and angle of deviation values, it is necessary to go on through
subsequent iterations, according to the subsequent sequence:
1) as a first put and take technique, we put friction null
Ta
= Tb' of the previous span
Ts
= Ta
( Atp = 0 )
2) we find, through the previous procedure, the downstream mouth angle a
3) we find the pressure value on the tower (b' and a being the rope mouth angles
downstream and upstream from the tower)
Ps
= 2 * Ts * sen ( b' - a ) / 2
4) we find the rope-tower friction value
At
=  * Ps
 being the friction coefficient
5) we compare the friction value At with the previous one Atp, then, we check that
At - Atp < 0.01
6) it the put and take technique has not been checked out, we put
Atp
= At
Ts
= Tb' + At /
(tension at tower middle)
Ta
= Ts + At / 2
And we repeat the sequence starting from item NR.2.
7) if the put and take technique has been checked out, we go on calculating the subsequent
span.
NB:
The rope-tower friction coefficient is positive or negative depending on the direction of
the rope, thus, for the carrying ropes the result is that:
) it is positive for the towers downstream from the carrier truck
2) it is negative for the towers upstream from the carrier truck
3) it is null in the case of end spans less than 20 m in length
(relating to which it is assumed that the rope is not subject to such remarkable spring
shifts to induce friction pressures).
4
FLOW-CHART OF THE PROGRAM FOR THE AUTOMATIC CALCULATION OF SPANS, WITH
UNLADEN ROPE, SUBSEQUENT TO THE FIRST ONE
Known values input
- i span number
- L span horizontal length
-  span difference in height
- q rope unit weight
- Ta tension at downstream end
- Ts tension on the tower
-  rope-shoe friction coeff.
1
As a first approximation, we
put the downstream tension equal to
the tension Tb of the previous span:
At = 0
TA = TB’(i-1)
Ts = TA
We find: with the PCS procedure
2
the concerned span, then, we
find the downstream mouth angle
a
We find the pressure on the tower
downstream from the span:
3
Ps = 2∙Ts∙sen((fB’-fA)/2)
We find the friction value on the
tower:
At = ∙Ps
5
At-Atp < 0.01
4
NO
YES
Now, we can go on calculating 7
the subsequent spans by following
the same criterium
END
5
The friction value is stored,
then, the tension values are
6
updated for the tower and
downstream from the concerned
span:
- Atp
= At
- Ts
= TB’+ At/2
- TA
= Ts + At/2
Calculation of the laden span
Referring to the typical ropeway span with load kept by the carrying rope and the counter
rope, we take into consideration the balance condition for the forces at the load application
point C, which is the point of convergence for load weight, tensions towards downstream
and upstream from the carrying rope, tension of the carrying rope and tension of the
counter rope. The test diagram is the following:
cf. Figure 2 and Figure 3):
Figure 2 Balance of forces in the single span with load P
Where:
P is the load weight;
T1 is the carrying rope tension at the downstream tower;
T2 is the downstream tension of the carrying rope;
T3 is the upstream tension of the carrying rope;
T4 is the tension of the carrying rope at the upstream tower;
t1 is the tension of the hauling rope at the downstream tower;
t2 is the tension of the counter rope at position C;
t3 is the tension of the hauling rope at position C;
t4 is the tension of the hauling rope at the downstream tower;
1 is the angle between the horizontal and tension t1 of the hauling rope;
2 is the angle between the horizontal and tension t2 of the counter rope;
3 is the angle between the horizontal and tension t3 of the hauling rope;
4 is the angle between the horizontal and tension t4 of the hauling rope.
Here below, the graphical representation of the forces at the load application position is shown.
6
Figure 3 Polygon of the forces applied to position C
Ignoring the slide friction of the load on the carrying rope, the balance equation is the following:
T2cos 2  t2cosψ2  T3cos 3  t3cosψ3
T2 sen 2  P  t2 senψ2  T3 sen 3  t3 senψ3
From which we can obtain the components of the tensile force on the carrier truck :
t3cosψ3  T2 (cos 2  cos 3 )  t2 cosψ2
t3 senψ3  P  T2 (sen  2  sen 3 )  t2 senψ2
dividing the second expression by the first one and summing up the squares make it possible
to find the angle of tensile force of the hauling rope on the carrier truck and the value of the
tension module in coincidence with the point where the carrying rope is fixed to the carrier
truck:
tgψ3 
P  T2 (sen  2  sen 3 )  t2 senψ2 N

T2 (cos 2  cos 3 )  t2cosψ2
M
t3  N 2  M 2
Also for the loaded span we must work by subsequent iterations according to the sequence
below:
1) we find (by first put and take technique)
q
P L L 
f c   p     1 2 
l   T1 
 2
Δ1  L1 
2) we find 2
Δ
 fc
L
= 1
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3) we find the two unladen half-spans near the load:
- carrying rope half-span 1,L1,T1
- carrying rope half-span 2,L2,T3=T2
- counter rope span D1,L1,t1
we store T2,1 2,S1p
we store T4,3,4,S2p
we store t2,1,2,Sz
4) we find and store, with the balance equation:
- ' 3 = N / M
- t3 
N2  M2
5) we find the half-span of the carrying rope upstream from the load:
- carrying rope span D2,L2,t3 storing t4,3,4,St
6) we check that
y3 - y'3= | 3 | < 0.001
7) in case of no check, we correct the value 1, then we restart from stage NR.2
8) if the check has been carried out, we find
fc
= L1 *  / L - 1 (final) deflection on the load
t
= ( 2 + 3 ) / 2
course angle of the load
Pt
= P * sen(t)
carrying rope component
8
FLOW-CHART OF THE PROGRAM FOR THE AUTOMATIC CALCULATION OF THE LADEN SPAN
WITH THE EXACT FORMULAE OF THE CATENARY CURVE
-L
-
- L1
- qp
- qz
- qt
- T1
- t1
-P
-
Known values input
span horizontal length
span difference in height
load distance from downstream
unit weight of the carrying rope
counter rope weight
haul rope weight
carrying rope tension downsrt.
counter rope tension downstr.
load weight on the carrying rope
truck-rope friction coeff.
Initially we find:
L2 = L - L1
fc
= (qp/2+P/L) ·L1·L2/T1
1 =L1·/L - fc
1
We find 21
2
Through the automatic calculation procedure, we find:
3
1) carrying rope half-span downstream from load 1,L1,T1
=> T2,1,2,S1p
2) carrying rope half-span upstream from load D2,L2,T3=T2 => T4,3,4,S2p
3) counter rope half-span with 1,L1,t1 ==> t2,1,2,Sz
We apply the balance equation of forces and we find:
ψ3  arctg
*
P  (1  μ  sen(
P  μ  cos
2  3
2
2  3
2
)  T2  (sen  2  sen 3 )  t2  senψ 2
 T2  (cos 2  cos 3 )  t2  cosψ 2
t3  N 2  M 2

N
M
4
Through the procedure PCS we find the carrying rope span with 2,L2,t3
And we get => t4,3,4,St
5
6
We check that
NO
3-3*30.001
We correct the difference in height
D1 then we repeat the cycle from
item NR. 2
7
YES
9
Now we can find the deflection
value in coincidence with load,
load course angle and the component
of tensile force of the hauling rope in
coincidence with the point where
the rope is fixed to the carrier truck.
- fc = L1 . /L-1(final)
- t = (2+3) / 2
- Pt = P . sent
8
END
Line global test mode
The line test is carried out by applying the above calculation principles to each single span.
Calculation always starts from downstream and goes on towards upstream, considering the
load at a certain position on the ascent branch, thus determining the corresponding position
of the load on the descent branch.
Carrying ropes:
It is provided the case of carrying ropes anchored upstream and with counter weight
downstream, or anchored at the two ends.
1) Ropes with counter weight downstream.
In this case, we know the initial tension at the downstream end, therefore, all the
unknown values may be found univocally.
2) Ropes anchored at the ends.
In this case, the rope tension at the downstream end is not known, but we know the
laying tension of the unladen ropes at a certain datum temperature. The problem may
be solved by put and take technique, as follows:
a) we assume that the initial tension in the rope at the downstream end is equal to the
laying tension.
b) we carry out the line calculation with the initial tension pre-set by defining the total
development of the carrying rope regarding each one of the two branches.
c) considering the mean tensions of the single spans obtained by line calculation and
the temperature difference from the laying temperature, we find the new length of the
carrying rope per each single branch, also considering that the temperature difference
induces an elongation ’S=Sot and that the tension difference induces and
elongation
' ' S 
S o T
being:
EA
- So rope length at the laying tension and temperature
-  rope linear expansion coefficient
- t thermal head
- T tension difference
- E the rope modulus of elasticity, A rope metallic section
Thus, the new rope length will be Sm = So + ’S - ’’S to be compared with the result of
the calculation obtained with pre-set downstream tension.
d) If the difference between the two lengths is higher than the desired value (usually
0.001 m) we must suitably change the value for the initial downstream tension,
then we repeat the cycle from B).
Hauling ropes and counter ropes:
The ropes are provided with counter weight, therefore, we can know the initial downstream
tension value. In the case of counter weight upstream and power unit downstream, we go on
by subsequent iterations so as to make converge the tension values to the counterweight
known value. In the case of counterweight downstream and power unit upstream, the values
for the initial tensions are immediately known.
10