# Computing winch maximum line pull (torque vs lift capacity)

I am trying to ascertain the maximum line pull for an old winch I have in my workshop and I am arriving at values that are considerably higher than what is in the original manufacturer's manual. I am not sure if I am doing this right and want to validate my calculations.

The winch is rotated hydraulically with a motor. The motor shaft is connected to a wet brake which in turn is connected to a planetary gear hub. This hub is then connected directly to the drum.

Some specs:

If the motor builds up 3000 PSI and has 38 LPM of flow it produces a torque of 295Nm with 316RPM. (or 30.08163KGm)

The wet brake disengages and produces negligible friction.

The ratio in the gear hub is +53.58:1

The diameter of the drum on the winch is 0.81m

The diameter of the cable used to pull is 0.0202m

From what I understand, to find linear force for a winch I must use this formula: Torque = Force X Arm

Where Arm = radius of drum + radius of cable radius drum = 0.81m/2 = 0.405m radium cable = 0.0202m/2 = 0.0101m Arm = 0.405m + 0.0101m = 0.4151m

Where Torque = Torque of motor X ratio of gear hub Torque = 30.08163KGm X 53.58:1 = 1611.7737354KGm

Therefore Force = Torque / Arm Force = 1611.7737354KGm / 0.4151m = 3882.856505420381KG

Am I correct to therefore say that the maximum line pull before the winch stalls is 3882.85KG? Therefore We can say it can lift anything bellow this weight?

• You haven't accounted for losses. Efficiency of typical hydraulic motors would be about 75%. The gear box, which is probably a two stage, might be 85% efficient. And that rating is only bare drum direct pull. Pulleys and fairleads can reduce that by a lot. Additional layers of cable on the drum further reduce the pull. – Phil Sweet Aug 4 '18 at 21:43

The main constraint would be the cable braking strength. Per Engineering Toolbox, this size cable has a maximum safe load of 42.3 kN. Your device produces 38 kN at stall point, so the cable is under it's safe load. This implies that the device will indeed handle this level of load without a problem. The system may have derated the pump as approaching the stall point of most motors, including hydraulic motors, can wear excessively and cause early breaking, so it is best to stay within nameplate parameters for day-to-day operations.

Of course, due to angles, we could have additional loading beyond the weight. See this handy diagram and table from Engineering Toolbox:

α   β   Increased Force
0   90  1.00
5   85  1.00
10  80  1.02
15  75  1.04
20  70  1.07
25  65  1.10
30  60  1.16
35  55  1.22
40  50  1.31
45  45  1.41
50  40  1.56
55  35  1.74
60  30  2.00
65  25  2.37
70  20  2.92
75  15  3.86
80  10  5.76
85  5   11.5

• Thank you very much for this informative response. The breaking strength of the cable used is 180kN according to manufacturer. So the SWL should be 5:1 ratio for lifting equipment. 36kN. The maximum weight I want to lift is 2200KG. The manufacturer says 2350KG maximum line pull which is far from my calculations. Its an old manual with very little information and the manufacturer is closed. I wanted to be sure this was not the stall value and I got my answer.Also, the cable goes up 45 deg through a sheave and down straight. wouldn't this "cancel" the increased force? – E Demers Oct 17 '17 at 1:06