# Pulling power ratio for a cable drum at different capacities

Prior to my life as a database administrator I was a hi lead logger in coastal British Columbia, which involves harvesting logs with machines capable of putting huge strain on cable and hydraulic systems. What got me to thinking this was a friend just let me know someone we know was recently killed in an accident where the machine was operating outside its intended purpose while doing a very hard pull with a cable drum and we were discussing how it may have happened.

In the discussion we were discussing how different strains were put on certain parts of a cable yarder, depending on angle of pull, lean of the spar, etc. In the discussion I was recollecting how a drum that is capable of holding 1000' of 1 3/8" cable has a lot more pulling power the less line is on it, as it drops the gear ratio. I used to run these machines and extra care had to be taken with the throttle if you were pulling with the drum on the bottom tiers of cable as things could break easier. The drums are powered by a torque converter, so it's more like an automatic transmission, if you will.

One thing I always wondered about was how would one figure out or what formula would you use trying to figure out the torque difference for the drum as each tier of cable was added (or removed) The drums are about 2' deep, (I'm just going from memory here) so the diameter of it would range from around 18" to 48" through the full capacity.

I'm sure there'd be a mathematical formula that would calculate the difference in pulling power decreasing as the drum got fuller from the bottom tier to the top tier, but I have no idea what it would be.

Mech_Engineer
Gold Member
It's a straightforward calculation actually, it's basically analyzing a torque with different length moment arms.

Take for example a drum that has (let's guess at a number) 1000 N-m of torque being applied to it by a drive system. The max torque applied by the drive systems stays the same regardless of how much cable is on the drum, so if we want to know how much force is applied to the cable from the drum, we only need to know the effective drum radius (1/2 the diameter).

Torque is force times radius, so to find force we can divide torque by radius. In our example we are applying 1000 N-m of torque, and you have stated the drum diameter ranges from 18" to 48" (0.457 m - 1.219 m). This means the drum radius ranges from 0.229 m - 0.610 m (half the diameter).
• When the drum is at its maximum radius (0.610 m), the force applied on the cable would be (1000 N-m) / (0.610 m) = 1.639 kN
• When the drum is at its minimum radius (0.229 m), the force applied on the cable would be (1000 N-m) / (0.229 m) = 4.367 kN
So you can see that the force on the cable scales with the radius of the drum; half the drum radius results in twice as much force on the cable (assuming it is pulling on a fixed object).

Usually, an automotive winch like used on a 4x4 truck will be rated based on its maximum load rating (which would correspond to its minimum drum radius, e.g. all of the cable is let out), this means it is necessary to take into account a lower pull strength if there are a few layers of cable on the drum. In the example below, you can see that 5 layers of winch cable reduce the effective pulling power of the winch to 53% of max:

http://www.pangaea-expeditions.com/resources/winchworksheet/index.html

WINCH CAPACITY

One of the common misconceptions regarding winches is that the maximum rated winch capacity is available any time the winch is hooked up. Unfortunately, this is false. Maximum winch capacity is in fact determined by the number of layers of cable wrapped on the winch drum. The best way to think of this is like the gears on a bicycle- the fewer the wraps, the smaller the "gear", the greater the pulling power.

Below is the rated winch capacity for typical 8,000lb winch.

# of layers Pulling capacity % of capacity
1st (Drum)
8,000 lbs 100%
2 6,800 lbs 85%
3 5,760 lbs 72%
4 4,960 lbs 62%
5 4,240 lbs 53%

billy_joule
Thanks - exactly what I was looking for!

jim hardy
Gold Member
Dearly Missed
Sorry to hear of that accident.

One thing I always wondered about was how would one figure out or what formula would you use trying to figure out the torque difference for the drum as each tier of cable was added (or removed) The drums are about 2' deep, (I'm just going from memory here) so the diameter of it would range from around 18" to 48" through the full capacity.

I'm sure there'd be a mathematical formula that would calculate the difference in pulling power decreasing as the drum got fuller from the bottom tier to the top tier, but I have no idea what it would be.

Maybe an easy way to remember the formula is to think about why you put a pipe on a breaker bar to get more leverage on a stubborn bolt.

Distance is radius to cable layer
and you know from everyday experience it takes more force to torque a bolt with a short wrench than with a long one.
So just reverse that - same torque pulls harder on a short radius arm than a long one.
Changing the radius from 48 inches to 18 inches almost triples the pull , 48/18 = 22/3
..........

With a torque wrench and a fish scale you could work up a demo for next safety meeting ?

In the discussion I was recollecting how a drum that is capable of holding 1000' of 1 3/8" cable has a lot more pulling power the less line is on it, as it drops the gear ratio.

You might mean ?, "as it increases the gear ratio" ... that is, the ratio becomes larger, or "higher".

Kinda like calling truck rear ends, "high geared" or "low geared"... low geared rear ends have a higher ratio, and high geared rear ends have a lower ratio...

A "lower' ratio is closer to 1:1... I think I have that right ? ...[COLOR=#black]..[/COLOR]lol

Although, I do know what you mean ...[COLOR=#black]..[/COLOR]

Prior to my life as a database administrator I was a hi lead logger in coastal British Columbia

Wow!... now that's a "career" change, in the extreme ...

Logged till the money ran out, did ya ? ...[COLOR=#black]..[/COLOR]

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... think about why you put a pipe on a breaker bar...

Lol, and here I always thought ... it was to break you're lower jaw...[COLOR=#black].[/COLOR]

jim hardy
Baluncore
In the discussion I was recollecting how a drum that is capable of holding 1000' of 1 3/8" cable has a lot more pulling power the less line is on it, as it drops the gear ratio. I used to run these machines and extra care had to be taken with the throttle if you were pulling with the drum on the bottom tiers of cable as things could break easier. The drums are powered by a torque converter, so it's more like an automatic transmission, if you will.
There should be some way of sensing the tension in the cable. That way when tension exceeds a pre-set limit, the power can be backed off to protect the equipment, and therefore the workers.

There are two obvious places to put a sensor. The cable could run through a slight deviation or 'S' bend around two idler pulleys, the force on those pulley shafts will be proportional to cable tension. Alternatively, a strain gauge on the winch mountings will indicate the tension in the cable, independent of the variable winch drum radius. The sensor could be as simple as a strong compression or leaf-spring, with a mechanical limit switch or hydraulic valve.

In transmission lingo high ratios are those that are closer to 1:1, low ratios are those that are farther from 1:1. It's counter-intuitive perhaps, but that's the way it is. It may be better to think of "output relative to input".

And pulling power is a misuse of the term power. Pulling force is the proper term. Power is a rate.

Wow!... now that's a "career" change, in the extreme ...

Logged till the money ran out, did ya ? ...[COLOR=#black]..[/COLOR]

Actually no - I eventually became a high lead contractor, and bought a $4500 computer in the late 80's thinking it would lower the cost of having my accountant do my company year end tax return. My naivete and knowledge of computers at the time was such that I thought I could type in something like "do my years end" then sit back with a beer.. I sat up many late nights with a borrowed MSDOS book and taught myself the Operating System, then a friend got me a disk with Lotus 123 on it and I took a night school course on Lotus and learned how to use spreadsheets - which were much better than the$400 accounting program I bought for doing what I wanted to do. I eventually got knowledgeable enough I designed an entire macro driven payroll module and one for forecasting cost & profit on timber sales.

Eventually I decided to leave contracting and took a job with a large mill as a truck and loader supervisor. Because I was reasonably proficient at computers all the computer stuff found its way to my lap, and I ended up designing overweight programs for the trucks, first in Excel then in MSAccess. The main woodlands program was on a UNIX AIX box and I became the defacto woodlands expert on it, working closely with the developers in its implementation. It ran on an Oracle 6 database I and I eventually became proficient enough at the SQL command language I could extract information not available in canned reports. After almost a decade in the Woodlands division the DBA position came up in the IT dept, so I applied for it and have been a DBA since then

OCR
Actually no...
...

Hello
new member , thank you for having me,

as this thread is old, I felt it wouldn't necessarily be pirating by continuing the topic -

I've got a winch - i'd like to determine It's speed in FT per SEC and lifting capacity in pounds.

the electric motor is a 3hp,208volt -9.26amp the motor plate says 39 rpm - SEW motor & gearbox combination

the gearbox has a 43.88:1ratio the wire rope drum is 6.25inch diameter. it's grooved for quarter inch wire rope.

would anyone be able to share the formula one would use to determine the machines mathematical capacities ?

thanks

marc

jim hardy
Gold Member
Dearly Missed
the electric motor is a 3hp,208volt -9.26amp the motor plate says 39 rpm - SEW motor & gearbox combination

the gearbox has a 43.88:1ratio the wire rope drum is 6.25inch diameter. it's grooved for quarter inch wire rope.

Back of the envelope calculation here, you should check my arithmetic and figure out why i did each step.

Basically it's work equals force X distance. But Torque has the same units as work and that confuses folks.
Think about a rope wrapped around a one foot radius shaft - a pound of pull makes a foot-pound of torque.
A foot pound of work is lifting a pound a foot against gravity, or dragging something a foot with a force of one pound.

A horsepower is a rate of doing work, like pulling or lifting something.
550 foot-pounds per second is one horsepower because it's how much they observed a good horse could lift in the 1800's.

...........................................................end basic physics lecture................................................................................

a 39 RPM AC motor ? Surely that's after the gearbox 39 X 43.88 =. 1711 which sounds a lot more reasonable.

Now HP = 2pi X Torque X RPM/33,000, torque in Foot-Pounds
so motor torque would be HP X 33,000 /(2pi X RPM)
3HP X 33,000 /(2pi X1711 RPM) = 99,000 /(10,750.5) = 9.209 ft-lbs of torque at the motor end of the gearbox.
Multiplied by your gear ratio of 43.88 = 404.08 ft-lbs of torque at the drum end but that neglects friction
which at the periphery of a 6.25 inch diameter (3.125 radius) drum equals 1551 pounds of pull. (Remember torque = Force X Radius....)

As the spool fills with cable your rated torque drops in accordance with increasing diameter and radius.

Subtract something for friction - say about 10% - and you're left with approximately 1400 pounds of pull.

Now a sanity check on my arithmetic:
Every turn of the 6.25 inch diameter spool will take up 6.25pi = 19.6 inches = 1.64 feet , at 39 RPM that's 1.066 feet per second.
Dragging or lifting something 1.066 feet per second against a force of 1400 pounds = 1.066 X 1400 = 1492.4 ft-lbs/second of work ,
and 1492 ft-lbs/sec divided by 550 ft-lbs/sec/hp gives 2.71 hp.
That's our 3hp from motor less the 10% i assumed for friction in the gearbox,
So what went in equals what came out plus losses... So arithmetic is probably okay.

Can you reproduce my thought process, or improve on it ?

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Hi Jim,
I feel you're correct that the listed rpm must be at the gearbox and not the motor (despite the information plate being on the motor)
1711-1800 rpm would be far more typical. - thank you - seeing the math makes perfect sense.
much appreciated sir.

marc

jim hardy
Baluncore
the electric motor is a 3hp, 208volt -9.26amp
208 V * 9.26 A = 1926 watt.
1926 W / 745.7 W / HP = 2.58 HP
Why is that different from the 3 HP on the plate ?

208 V looks OK, as it is 3 phases of 120V.
What current should we expect?
3 HP * 745.7 W/HP = 2.237 kW
2.237 W / 208 V = 10.75 amp.

jim hardy
Gold Member
Dearly Missed
3 phase ? dont forget √3 ?

208 v X 9.26 a X √3 = 3336 va,
2237 watts / 3336 va = 67% PF ?

old jim

Glenstr
I think I'll give this a shot as well (I'm using metric units to calculate, please bear with me, I'll convert to imperial at the end)
Motor output power = 3 hp = 2237 Watts
Gearbox efficiency = 90% (I did suck from my thumb)
Windage efficiency =90% (I once again consulted my thumb)
Overall Efficiency = 0.9x0.9 = 0.81
Power available at rope = 2237 x 0.81 = 1811 watts
Torque = 60P/{2(PI)N} = 60x1811/2/3.14/39 = 443 Nm
Force = Torque/(Torque Arm) = 443x2/(0.158+0.00635x1) = 5369 Newtons = 547.37 kg = 1206 Pounds assuming one wire thickness on spool.

Force = Torque/(Torque Arm) = 443x2/(0.158+0.00635x5) = 4650 Newtons =474 kg = 1045 pounds assuming 5 wire thicknesses on spool.
You lose quite a bit of your load capacity the more wire there is wound around the spool, interesting...
I used 2.205 pounds in a kilogram for the conversion calc.

For speed (Once again I will do it at 1 wire thickness and 5 wire thicknesses, it will illustrate a good point):
V = PI x D x N/60= 3.14 x 0.165x39/60 = 0.337 m/s (One wire thickness on spool) = 1.1 feet/second
V = V = PI x D x N/60= 3.14 x 0.19x39/60= 0.38 m/s (five wire thicknesses added to spool) = 1.24 feet/second

I hope I'm not making any mistakes. If there are, please feel free to point them out.

I dug out some of my old notes on wire ropes and decided to do some calcs with the loads that I got on a "typical" 6mm (1/4 inch) type wire rope. I ended up feeling bad feelings. Safety factors way low. I don't think your wire rope will last long at all if you push the winch this hard, but that's just my unprofessional opinion. I have to admit that I don't work with wire ropes much, but the safety factors that I got made me feel uncomfortable. I personally wouldn't stand anywhere near the unit if it's pulling this type of load. I've seen what wire ropes can to to 10mm plate when they suddenly decide to violently retire. (the rope in question was bigger than 6mm and the load much bigger, but the illustration stands) Go to a winch shop and get some advice from them before you actually attempt anything like this.

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this assumes of course that every wrap on the drum s perfectly round - which in real world terms is incorrect - a wire keeper on a drum build following standard rigging
practices will be bent into the drum itself and an internal keep will hold it in place - this causes a deformation for the second wrap of wire over the first precisely where the cable terminates in the drum. - not a problem in a helical grooved drum but something to keep in mind in a 'yoyo' style drum

Yeah, that's true. They are very rough calcs. To be a hundred percent honest with you I don't even know how you would account for those things in calculations...

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