Zip Wire Load Requirements: 50 - 80 Metres

[rob200]

Hi there, hopefully somebody can assist me with the following

I am looking to set up a zip wire of between 50 and 80 metres in length, can you provide me with details of the loadings required at each end of the wire to keep the tension whilst under weight of the user.

Thanks for your help

ROB

 

[madmike159]

That depends on the weight of the user. If you have some one of mass 100kg on it, their weight if 100*9.81 = 981N. So you would need 490.5N at each end to keep it tension in it. I'm not sure how much the thinkness etc of the wire will effect this. I'm guessing their thick enough to support more than the avrage person's weight.

 

[rob200]

the wire would be approx 12mm thick and presuming the average weight would be 80Kg, yes maxium 100kg What weight would i need to creat 490.5n at each end

 

[madmike159]

What are you suspending it between and what will it be used for? If it was between two trees like a high ropes course you could tie it to the tree. 490.5N of force won't break a large tree. The 490.5N is just a amount of force you will need. If oyu had some one with mass 80Kg and 2 people of mass 45Kg holding it at each end, they could hold it but that depends more on strength.

 

[rob200]

I am looking at a manmade structure so i need to calculate what will hold the weight of the cable, to keep in tensioned and take the weight of the individual.

 

[madmike159]

So some sort of scaffolding type thing. Really all you need to do is tie the wire round something and get some clamps to hold it in place. There shouldn't be a problem with it supporting someone, there will always be a little bit of slack though.

 

[gmax137]

madmike159 said "I'm not sure how much the thinkness etc of the wire will effect this. I'm guessing their thick enough to support more than the avrage person's weight."

hmmm, you guys better be more careful. 80 meters of the 1/2 inch cable (if steel) will weigh in at around 175 lb.

google around "catenary" - you can figure out the real length of the cable and the tension involved. Then you can figure out how to modify the calcs to include the point weight of the passenger.

 

[rob200]

we would be looking at 1/2 inch - not sure what the total weigh would be, rob

 

[rob200]

the weight of a 12mm 100 cable is 57KG

 

[stewartcs]

Hi there, hopefully somebody can assist me with the following

I am looking to set up a zip wire of between 50 and 80 metres in length, can you provide me with details of the loadings required at each end of the wire to keep the tension whilst under weight of the user.

Thanks for your help

ROB

Here is a similar thread that should answer your question.

https://www.physicsforums.com/showthread.php?t=207321

CS

 

[rcgldr]

The tension in a zip line could be over 10 times the weight of any user on the line, and that's ignoring the weight of the line itself. Say an object is hanging from the middle of the zip line, and the angle at the support point is 2.5 degrees. To support the objects weight the vertical component of the tension in the zip line to either side of the object has to support 1/2 of the objects weight, so T sin(2.5) = 1/2 weight, or T ~= 11.5 weight.

 

[madmike159]

What will this be used for, normaly I wouldn't thyink about something in this much detail. I would just test the wire at a safe hight before putting it up.

 

[rob200]

The zip line will come in at approx 90 to 100kg, and the maximum weight of any person will be 130kg. Can I calculate it from this,

 

[rcgldr]

The zip line will come in at approx 90 to 100kg, and the maximum weight of any person will be 130kg. Can I calculate it from this,

As posted before, the vertical component of tension at the point of support for the person has to be at least 1/2 the weight of person and support, and can easily be over 10 times the weight of person and support.

The cable is going to have a natrual component of droop depending on it's tension and weight. Supporting the weight of a person in the middle of the cable is the situation where the total tension requirement will be the highest. Plus you'd want some extra tension to handle any vertical movement of the person, which creates acceleration, and increases the load on the cable beyond the weight of the person.

 

[Toymkr]

rob200-

I did a search for "sling tension calculator" since what you are doing is an inverted case for a crane lifting sling.
Givens: 200 pound person in the middle of a cable stretched between two anchors 100 feet apart.
Results: for a cable sag of 3 feet loaded, the tension on the wire is 1667 pounds. Safe working load of a cable is a minimum of 3x the working load so you need a cable with that working load, nooooot ultimate breaking strength. If your rider bounces, it is easy to get 3 or 4 g's load which adds to the requirement for a stout cable...and anchorages !

For a 1 foot cable sag, loaded, the tension on the cable increases to 5000 pounds so you're going to need something on the order of 5/8" wire rope to be safe.

 

[rbj]

If you have some one of mass 100kg on it, their weight if 100*9.81 = 981N. So you would need 490.5N at each end to keep it tension in it.

that is both incorrect and dangerously mistaken since it grossly underestimates the tension in the line and what would be required of the anchors on the two ends. someone (maybe madmike) needs to draw a 2-dimensional force diagram (the 981 N is straight down and the force the zip line pulls on the two anchors are nearly horizontal). setting this problem up is not hard. what will take a little more research is determining what 1/2 inch steel cable can do as far as tensile strength (and what it weighs per meter).