
Hoisting and Cable Tensioning with Cylinders
FLUID POWER  Design Data Sheet 6
Hoisting with Cylinders

Fugure 1  Direct Pull
When used in a direct pull, with one cable supporting the load,
the ratio between load lift height and cylinder stroke is 1:1. In
this example, to raise a 1000 lb. weight a height of 6 inches, the
cylinder must have a 6inch stroke and must develop 1000 lbs. pull
in retraction. Since the load is supported on one cable, the cable
tension will be 1000 lbs.


Figure 2  One Cable Wrap
By using suitable idler pulleys, two cables support the 1000 lb.
load, with a tension of 500 lbs. in each cable. The
cylindertoload ratio is now changed, and a 12inch stroke is
required to lift the weight a height of 6 inches, but the cylinder
force is reduced to 500 lbs.


Figure 3  Multiple Cable Wrap
The use of additional idler pulleys and more wraps further
changes the cylindertoload ratio. For example, with four cables
supporting the load, the tension in each cable is reduced to 250
lbs. The cylinder stroke must be increased to 24 inches, but a pull
of only 250 lbs. is required.
Notice that cable tension, cylinder stroke, and cylinder pull are
all changed in the same ratio as the change in the number of cables
supporting the load.
An interesting point to notice on the three hoisting examples is
that whether a large bore, short stroke cylinder is used as in Fig.
1, or a smaller bore, longer stroke model as in Figs. 2 and 3, the
lift capacity, lifting speed, hydraulic system PSI and GPM is
identical in each case provided the cylinder is matched to the
application. The same hydraulic pumping unit could be used with any
of these arrangements.

CYLINDER ECONOMICS...
The table shows that at the same HP input, a substantial savings
can be realized by reducing the cylinder bore and increasing its
stroke. This is the important advantage of a multiple wrap system.
You can buy more HP per pound and per dollar with small bore,
longer stroke cylinders.
Cylinder Description 
Weight, Lbs. 
Price 
4" bore x 6" stroke 
55 
$176.00 
2" bore x 24" stroke to do same job 
31 
$127.00 
6" bore x 6" stroke 
155 
$385.00 
3 1/4" bore x 24" stroke to do same job 
73 
$190.00 
8" bore x 6" stroke 
310 
$675.00 
4" bore x 24" roke to do same job 
95 
$244.00 
On multiwrap systems, the savings on cylinder cost by using a
smaller bore cylinder may be far greater than the extra cost of
mechanical pulleys. Further advantages of a multiwrap system are:
lighter cylinder weight; and the cable, while longer, can be
smaller in diameter, lighter in weight, more flexible, and can be
less expensive.
One disadvantage of multiple wrap is the additional friction load
of pulleys and other mechanisms, and the internal friction of the
wire rope. The chart below, given by one authority, shows the extra
pull to be added to make up for friction of active pulleys. For
each idler pulley in the system, an additional 5 to 7% may also be
needed.
2part line (2 cables supporting the load)  add 18%
3part line (3 cables supporting the load)  add 25%
4part line (4 cables supporting the load)  add 33%
Figure 4  Suspended Load on a Cable.
This is a simplified method of finding cylinder force for
tensioning a cable, and was derived from trigonometry, although a
knowledge of trig is not necessary in order to use the method.
Ratio values in the left column of the chart are "cotangent" values
taken from a standard table of natural trigonometric functions.
Multipliers in the right column are "cosecant" values for the same
angle. For intermediate ratio values or those beyond the chart,
regular trig tables may be used.
Assume the weight of the cable is relatively small as compared
with the suspended weight. Tension in each leg of the cable will be
equal if the weight is hanging in the exact center of the cable.
Otherwise, the tension will be different in each leg. Solve for
each leg.
The first step, always, is to determine what percent of the load
is supported by each leg of the cable. The load on each leg will be
inversely proportional to distance A or A' from weight to support.
In Fig. 4, total span is 20 feet, so 9,000 lbs. (75% of the weight)
will be supported by the short leg and 3,000 lbs. (25% of the
weight) will be supported by the longer leg.
To solve for tension in a leg, after finding what per cent of the
load it carries, take distanceA and divide it by the distance B
(the sag) to find the ratio A ÷ B. Enter the left column of the
chart and find this ratio. Use the multiplier from the right column
to multiply times the portion of load carried by that leg. This
gives the cylinder force necessary to support the load in this
position.
If the exact value of your calculated ratio does not appear in the
chart, interpolate between the next higher and next lower ratio;
then interpolate in the same way to find the multiplier.
Solve for tension in short leg (Fig. 4): Ratio A ÷ B = 5 ÷ 2 =
2.50. Enter left column of chart and get multiplier, 2.69, on same
line as ratio 2.50. Tension = 2.69 x 9,000 lbs. = 24,210 lbs.
Solve for tension in longer leg: Ratio A ÷ B = 15 ÷ 2 = 7.50.
Enter left column of chart and find ratio 7.50. Use multiplier 7.57
shown on same line. Tension= 7.57 x 3,000 lbs. = 22,710 lbs.
In sizing a cylinder to tension the above cable, do not add
tension in both legs. One cylinder will tension both legs of the
cable. If pulling on the short leg, it must have 24,210 lbs. force;
if pulling on the longer leg it must have 22,710 lbs. force.
A small difference in elevation between the two end supports will
not seriously affect the accuracy of tension calculations. However,
if the difference in elevation results in an angle between cable
and horizontal of more than 10 to 15°, a more involved
trigonometric calculation will have to be made.
Hydraulic Motor Torque...
When pulling tension on a loaded cable with a winch drum driven
with a hydraulic motor, use procedure described above for finding
cable tension. Then solve for torque on winch drum by taking
tension times drum effective radius. Then calculate theoretical
motor torque by dividing drum torque by ratio between drum sheave
or sprocket and hydraulic motor sheave or sprocket.
Example: Cable tension = 15,000 lbs; winch drum
diameter 14 inches; winch drum drive sprocket diameter= 2 inches;
motor sprocket diameter = 6 inches. Find theoretical torque
required from the hydraulic motor.
Solution: Winch drum torque is 15,000 x 7 (radius
of drum)  105,000 inchlbs. Motor torque is found by dividing
winch torque by drive sprocket ratio. Torque= 105,000 / 2 = 52,500
inchlbs. This is theoretical torque required on the motor shaft.
Allow extra torque for mechanical losses in sprocket drive and in
winch drum bearings.

Ratio
A + B 
Multi
plier 
60.00 
60.01 
55.00 
55.01 
50.00 
50.01 
45.00 
45.01 
40.00 
40.01 
35.00 
35.01 
30.00 
30.02 
29.00 
29.02 
28.00 
28.02 
27.00 
27.02 
26.00 
26.02 
25.00 
25.02 
24.00 
24.02 
23.00 
23.02 
22.00 
22.02 
21.00 
21.02 
20.00 
20.03 
19.00 
19.03 
18.00 
18.03 
17.00 
17.03 
16.00 
16.03 
15.00 
15.03 
14.00 
14.03 
13.00 
13.03 
12.00 
12.04 
11.00 
11.04 
10.00 
10.04 
9.00 
9.05 
8.00 
8.06 
7.50 
7.57 
7.00 
7.07 
6.50 
6.58 
6.25 
6.33 
6.00 
6.08 
5.70 
5.79 
5.40 
5.49 
5.20 
5.30 
4.90 
5.00 
4.70 
4.81 
4.50 
4.62 
4.30 
4.42 
4.15 
4.27 
4.00 
4.12 
3.80 
3.93 
3.70 
3.83 
3.60 
3.74 
3.50 
3.64 
3.40 
3.54 
3.30 
3.45 
3.20 
3.35 
3.10 
3.26 
3.00 
3.16 
2.90 
3.07 
2.80 
2.97 
2.70 
2.88 
2.60 
2.79 
2.50 
2.69 
2.40 
2.60 
2.30 
2.51 
2.20 
2.42 
2.10 
2.33 
2.05 
2.28 
2.00 
2.24 
1.95 
2.19 
1.90 
2.15 
1.85 
2.10 
1.80 
2.06 
1.75 
2.02 
1.65 
1.93 
1.60 
1.89 
1.50 
1.80 
1.45 
1.76 
1.40 
1.72 
1.35 
1.68 
1.30 
1.64 

Download a PDF of Fluid
Power Design Data Sheet 6  Hoisting and Cable Tensioning with
Cylinders.
© 1988 by Womack Machine Supply Co. This company assumes no
liability for errors in data nor in safe and/or satisfactory
operation of equipment designed from this information.
