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 of the 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 represented on the cartesian plane by the equation below: x y
 h
 e h
 e
 x h
2
 h
 cosh
 x h
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 apex of the catenary curve, we have
H = q · h h being the parameter of the catenary curve
The calculation of the configuration of an unladen rope tightened between two ends A and B, is obtained by taking into consideration the diagram below:

Solution of 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
- Ta t b) Unknown values :
- S unit weight of the rope ension at the downstream end (=counterweight/2 for the 1° span)
- f
- Tb
-
 a
-
 b c) Symbols:
- Cos rope development between the ends A and B deflection at any point in the span tension of the rope at the upstream end rope mouth angle downstream from the span rope mouth angle upstream from the span hyperbolic cosine
- arctg
- cos
- tg arc cotangent trigonometric cosine trigonometric tangent
The formulae used for the calculus of the unladen spans are the following:
1) we calculate y
A
= T
A
/ q

A
 arctg



L


2

L y a



2) we calculate h
 y a
 cos
 
A

B
 arctg

 sen
L h arcsen
 tg

A


 y
B
 h cos

B
3) we check y
B
- y
A
-

< 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
T
B
= h · ( tg(
= y
B
· q

B
) - tg(

A
) )
6) in the end, we find the deflection of the generical abscissa "x" y f x x

 h x
 cos
 x h

Δ
L
 y
A arcsen
 tg

A

 y
A
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)”).

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 downstream end
Ordinate calculus and angle of
1 tangency of the rope downstream : y
A
= T
A
/q

A
= arctg(

/L - L/(2 · y
A
))
We find:
- h = y
A
· cos(
-

B

A
)
= arctgsen ( L/h + arcsen ( tg
2

A
))
- y
B
= h/cos(

B
)
3
Yb-Ya-

< 0.001 NO
YES
We correct angle
 a
4
then we repeat the cycle
In the end, we fin:
- S = h
·
(tg(

B
)-tg(

A
))
- T
B
= y
B
· q
5 and the deflection of the generic abscissa x as to the downstream end A
- y x
= h
· cos(x/h+arcsentg

A
)-y
A
- f x
= x · 
/L-y
A
END

Formulas used for the subsequent unladen spans
Regarding the subsequent spans, the calculation method remains unchanged, provided that the new value for the initial tension downstream from the new span to be tested (Ta) is given. 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 the pressure, and pressure depends on the rope tension and angle of deviation values, it is necessry to go on through subsequent iterations, according to the subsequent sequence:
1) as a first put and take technique, we put friction null ( Atp = 0 )
Ta = Tb' of the previous span
Ts = Ta
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)
P s

2

T s
 sen


B'

2

A
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
Ta
= Tb' + At /2
= Ts + At / 2
(tension at tower middle)
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:
1) 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).

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-tower friction coeff.
As a first approximation, we put the downstream tension equal to the tension Tb of the previous span:
At = 0
T a
= T b’
T s
= T a
(i-1)
1
We find: with the PCS procedure the concerned span, then, we find the downstream mouth angle imbocco
 a 2
We find the pressure on the tower downstream from the span:
P s
= 2 · T s
· sen((f b’
-f a
)/2) 3
We find the rope-tower friction value
At =

*Ps
4
5
At-Atp < 0.01 NO
YES
Now, we can go on calculating
7 the subsequent spans by following the same criterium.
The friction value is stored, then, the tension values are updated for the tower and downstream from the concerned
6 span:
- At p
- T s
- T a
= A
= T
= T t b’ s
+ At/2
+ At/2
END

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:
Diagram of forces at load application point
In the specific case of freight ropeways with only a carrying rope, it is necessary to annull the vectors of the actions induced by the counter rope, i.e. the values for t
2

2 etc…….

Ignoring the slide friction of the load on the carrying rope, the balance equation is the following:
T
T
2
2 cos
 sen

2
2


T
3
P cos

T
3

3 sen
 t
3

3 cos

 t
3
3 sen

3
From which (being T2=T3) we can obtain the components of the tensile force on the carrier truck : t
3 cos

3 t
3 sen

3


T 2 (cos
P

T
2

2

(sen

2 cos

3
)
 sen

3
) 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 haul 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: tg

3

P

T
2
T
2
(sen
(cos

2


2
 sen

3 cos

3
)
)

N
M t
3

N
2 
M
2
Also for the loaded span we must work by subsequent iterations according to the sequence below:
1) we find fc f c


 q
2 p 
P l





L
1

T
1
L
2

 (first put and take technique)

1

L
1


L
 f c
2) we find

2
=

1
3) we find the two unladen half-spans near the load:
- carrying half-span
- carrying half-span


1
2
,L
1
,T
1
,L
2
,T
3
=T
2
4) we find and store, with the balance equation:
-

'
3
= N / M storing T
2
, storing T
4
,


1
3
,


4
2
,S
1p
,S
2p
t
3

N
2 
M
2
5) we find the half-span of the carrying rope upstream from the load:
- carrying rope span D
2
,L
2
,t
3
storing t
4
,

3
,

4
,St
6) we check that y
3
- 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 f c
 t
P t
= L
1
·
= (


/ L -

2
+

= P · sen( 
1
(final) deflection on the load
3
) / 2 t
) course angle of the load carrying rope component

FLOW-CHART OF THE PROGRAM FOR THE AUTOMATIC CALCULATION OF THE LADEN SPAN
WITH THE EXACT FORMULAE OF THE CATENARY CURVE
- L
-

Known values input span horizontal length span difference in height
- L1 load distance from downstream
- qp unit weight of the carrying rope
- qz counter rope weight
- qt haul rope weight
- T1 carrying rope tension downsrt.
- t1 counter rope tension downstr.
- Pload weight on the carrying rope
-

truck-rope friction coeff.
Initially we find: f
L
2
= L - L
1 c
= (qp/2+P/L)
·
L
1
·
L
2
/T

1
=L
1
· 
/L - f c
1
We find
 
2

1
1
2
Through procedure PCS, we find:
1) carrying rope half-span downstream the load

1
,L
1
,T
1
=> T
2
,
2) carrying rope half-span upstream the load D
2
,L
2
,T
3
=T
2
=> T
4
,


1
,
3
,


2
,S
4
,S
1p
2p
We apply the balance equation of forces and we find:
ψ
3
*  arctg
P



1

P
 μ 
μ  cos

 sen


2

2

2

2

3

3




T
2
T
2

(sen

(cos

2

2

 sen cos

3

)
3
)

N
M
3
4 t
3

N
2 
M
2
Through the procedure PCS we find the carrying rope span with

2,L2,t3 obtaining => t
4
,

3
,

4
,S t
5
6
Check that

3
-

3
*

3

0.001
NO
We correct the difference in height
D1 then we repeat the cycle from item n° 2
7
YES

Now we can find the deflection value in coincidence with load, load course angle and the component of tensile force of the haul rope in coincidence with the point where the rope is fixed to the carrier truck.
- f c
= L
1
· 
/L-

1 (finale)
-
 t
= (

2
+
- P t
= P · sen


3
)/2 t
END
8
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 rope:
It is only considered the case of the carrying rope anchored to the two ends.
The rope tension at the downstream end is not known, but we know the laying tension of the unladen carrying ropes at a certain datum temperature (considered zero). The problem can be solved by put and take technique according to the stages below: a) it is assumed the the initial tension on the rope at the downstream end is alike the laying tension. b) line calculation is carried out with the initial tension set by determining the total length of the carrying rope on each one of the two branches. c) considering the average tension values of the single spans, obtained by line calculation and the possible temperature change in comparison with the rope laying temperature, it is possible to calculate again the length of the carrying rope, for every branch, considering that the temperature change induces elongation
 ’S= 
S o
 t , which is

' ' S

S o

T
EA being known that:
- S o
length of the unladen rope with laying tension and temperature
-

rope linear expansion coefficient
-
 t temperature change
-

T tension change
- E modulus of elasticity of the rope, A its metallic secion
Thus, the new length of the rope will be Sm = S o
+
 ’S -  ’’S to be compared with the value obtained by the calculus with pre-set downstream tension.
If the difference between the two lengths is greater than the required value (usually 0.001 m) the initial downstream pre-set tension value must be changed, then, the cycle is repeated starting from item b).
Haul rope:
From the tension value in coincidence with the point where the rope is fixed to the carrier truck, it is easy to find the tension value at the power unit (rope winding drum), applying the sequence of formulae of the unladen span, and keeping into an account (if necessary) the friction values on the rullers in the towers