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• # Computer controlled winch project - Lookinng for some confirmation of calculations

Discussion in 'Calculations' started by Vanbot, Mar 18, 2015.

1. ### VanbotNew Member

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Hello everyone. I'm just getting started on a project that involves developing a computer controlled winch that will vertically raise and lower a payload. The winch drive will be located on top of a structure or cliff and will be lowering (and raising) the payload vertically from that position. I have very basic mechanical knowledge but am no engineer. Since I'm developing this project from scratch I'm hoping that I can run some ideas past the experienced and knowledgeable folks here. I'm sure some of my questions will be very elementary for you so hopefully its not going to bore everyone to tears! So...I think I have the basic concept worked out for how it will function and am now working on determining what I need for a drive motor. This requires knowing the torque requirement and that will depend on the mass being moved. The winch will be using steel cable of some kind to raise and lower a payload that I estimate will weigh about 15 kg up to (theoretically) 100 m of height. I need to be sure that the cable I am using meets the required safety factor so I have done the following calculations for three different thicknesses of cable (Bright wire, uncoated, fiber core (FC) wire rope, improved plow steel (IPS)). I have estimated that I'll want to be able to accelerate the payload straight up at a maximum 5 m/s/s which is where I get the 15 m/s/s I use in the equations: 3/8â€ Cable: Weight of 100 m = 36 kg Assume weight of inspection head = 15 kg So total weight = 51 kg Safe weight = F/a = 10.9 kN/15 m/s2 = 10900/15 = 726 kg Safety factor = 14 5/16â€ Cable: Weight of 100 m = 24 kg Assume weight of inspection head = 15 kg So total weight = 39 kg Safe weight = F/a = 7.56 kN/ 15 m/s2 = 7560 / 15 = 504 kg Safety factor = 12.9 1/4â€ Cable: Weight of 100 m = 16 kg Assume weight of inspection head = 15 kg So total weight = 31 kg Safe weight = F/a = 4.89 kN/15 m/s2 = 4890/15 = 326 kg Safety factor = 10.5 So, assuming my calculations are correct, what kind of safety factor would an engineer be looking for when designing something like this? I should also ask whether the type of cable I have specced is suitable for an outdoor application like the one I have described.

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3. ### LochnagarWell-Known Member

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Hi Vanbot,

Before you go into detailed calculations - it might be worth thinking outside the box for a minute. I don't know if you have been on a cable car - or a ski gondola - but the way these systems are set up - they have a continuous loop of rope which is fed round a pulley at the bottom of the mountain, and a pulley at the top of the mountain - and the "weight of gondolas" are equal on the ascending and descending sides of the rope. This means you are only doing mechanical work to raise the people inside the gondola from the bottom of the mountain to the top of the mountain. From an efficiency perspective this is the best way of doing things. Obviously, I don't know if this would do for your application.

See the link below from Bridon ropes on factors of safety (page 41).

If you are going down the winch route - then you need to think about the inertia of the winch too. It might also be worth thinking about how you are going to stop the load - something that is often given little thought Breaking forces can be just as significant as accelerating forces.

Hope this helps.

4. ### VanbotNew Member

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Hi Lochnagar.Thanks for your reply. The link you provided was very useful and has now been added to our technical library. Much of the information there was in the rigging handbook I have been reviewing but some of the calculations are new and useful.We were considering incorporating a wormgear into the drive to make braking less of a problem. Do you agree that this is a good way to go? This device has to work at a variety of locations where the elevation won't always be the same. On some days it will be working from a height of 100 m, on other a height of 20 m. Thus the winch. According to the current concept, the payload (30 kg) in this project will be steadied by two vertical static guide cables strung along either side of it. For example, the winch could be placed at the top of a 100 m cliff and the job is to lower the payload down the face of the cliff. Once the payload reaches the bottom, the whole rig is moved laterally a few meters and the process is repeated.In order to keep the payload from twisting and swaying, it is guided by two static cables, each about 18" to each side of the payload, that are fixed at top and at ground level. The payload has two outriggers that reach out to the guide cables and can run up and down the static cables as the payload is raised and lowered.So, I'm planning to use 1/4" wire rope for the winch cable. I'd like to use the same for the two static guide cables. I've been wondering about keeping enough tension in the guide cables that they will resist being blown around in windy conditions. How/where would I find out what kind of tension is required in a 100 m cable?

5. ### LochnagarWell-Known Member

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Hi Vanbot,
Every winch I have come across has had a fail safe brake - in the case of the Brevini winches below - it is spring applied and hydraulically released.
I wonder if you have seen forestry skylines in operation - see the video below where you have a catenary cable which supports a carriage - which is winched up the catenary cable. The carriage has another "winch" which lifts the timber up to the carriage - before it is brought up the catenary cable.

http://www.brevini.co.uk//I,O&M/Manuale ARGANI 2011 Winches.pdf

http://www.brevini.co.uk//pdf/winches/Hoisting-Recovery-Winches-2014.pdf

Hope this helps.

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