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Pulling power ratio for cable drum at different capacities

  1. Apr 20, 2016 #1
    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.
  2. jcsd
  3. Apr 20, 2016 #2


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    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:


  4. Apr 20, 2016 #3
    Thanks - exactly what I was looking for!
  5. Apr 21, 2016 #4

    jim hardy

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    Sorry to hear of that accident.

    Your intuition is exactly right.
    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 ?
  6. Apr 21, 2016 #5


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    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]:oldbiggrin:

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

    Logged till the money ran out, did ya ? ...[COLOR=#black]..[/COLOR] :oldwink:
    Last edited: Apr 21, 2016
  7. Apr 21, 2016 #6


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    Lol, and here I always thought ... it was to break you're lower jaw...[COLOR=#black].[/COLOR] :blushing:
  8. Apr 25, 2016 #7


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    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.
  9. May 1, 2016 #8
    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.
  10. Jun 27, 2017 #9
    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.. :cool:

    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 :oldbiggrin:
  11. Jun 28, 2017 #10


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    ... :thumbup:
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