Physics experts? Rope breaking strain question.

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coote

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Apr 3, 2006
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I use rope and cord alot. I am well aware that a strong rope that can hold a tremendous static load can sometimes easily break if that load is allowed to fall a short distance before the strain is taken by the rope.

I think in physics terms we probably have to consider things like mass of the weight, accelaration due to gravity, momentum, decelaration due to stretch etc etc.

Rock climbers and cavers who use climbing ropes have to be sure to consider the effects of this phenomenon when choosing a rope to climb with. A rope which could easily hold a man on a swing, might snap if that same man fell six feet before the rope takes up the load. Obviously stretchy ropes like nylon are better for this job because they stretch and the load doesn't stop with such a 'jerk'.

I have been testing some snare materials. I set pole snares. When the animals get caught in these snares on a high pole or branch they generally jump and hang themselves. So I have to select cord or cable which will take the strain of, say, a 10 pound animal dropping two or three feet. I have been testing my snares using a bit of firewood which weighs eight pounds and I have been amazed to see wire cable and sturdy nylon braid snap when the load comes to the end of its travel.

So.... here is my question. Can anybody provide me with some sort of a simple formula which can be a good rule of thumb to calculate the strength of rope required for a job where the weight the rope has to carry might be falling?

Many thanks in advance.... Coote.
 
I have a physics background and have looked into ropes and knots for mountaineering and the problem isn't very simple. Naively, you could use F=ma where a is the deceleration, say 10ft/sec in 1/10sec. But as you said, the force on the rope will be less if it stretches. But you then have to work out what is the length that can stretch, for instance, in a simple case the falling weight is tied to the rope and the other end to the anchor. But if you have a climber on a rope that is carabinered to protection, then the carabiners act as pulleys and the 'live length' is longer. This doesn't even take into account the sharpness of the corner you may have, the carabiner, cliff edge even the knot being kind of a corner. Then there is the issue of knots, larger radius knots being stronger etc...

I did some googling and here is a nice article describing the situation and does have a way to compute the breaking strength, but it needs a force/elongation curve for the particular rope you have and you need to integrate for the energy of the falling object.
http://www.bstorage.com/speleo/Pubs/rlenergy/Default.htm
 
Probably best just to find a rope that is designed and rated for the intended use. Rope science is like paint science - deceptively complex.
 
Another option might be to improvise a snubber like one uses to do the same thing when anchoring a boat or on some fishing rigs:

Snubber.jpg
 
Hiya coote,

I just spent about 20 minutes writing a bunch of stuff here, only to ditch it all. Long story short, I helped found the rope rescue team at our Town FD, and I work for a company that sells some rope rescue equipment. I have seen a vast variety of inquiries about rope applications, fall factors, safety margins, and so on. Please, if you are at all considering human applications for what you are asking, please seek professional instruction. Please.


Ok. No, I don't have a simple equation for you. I can tell you that in the fire service, we needed to insure a safety margin of 15:1 for rope applications. So, plugging in your numbers to that, a 10lb animal can hang from a 150lb-rated line. That is for a static load.

A fall factor is an estimate of the force of a potential freefall. you can figure a fall factor by dividing the distance the load attached to the end of the rope is expected to fall by the length of the rope between the load and an anchor point. For example, a 3-foot fall on a 10-foot rope would have a fall factor of 0.3. A fall of 10 feet on a 10-foot rope would have a fall factor of 1.0. In the fire service, fall factors less than 0.25 were considered acceptable.

550 should handle your requirements for 10lb animals. Now, if you're tying a 10lb animal to a 100' length of 550 cord and tossing the critter off a cliff, your fall factor is going to come into play heavily. Odds are that in this case, said critter would dislocate almost every joint, and maybe even loose an appendage or two before the 550 cord would break.
 
asfried2 said:
Probably best just to find a rope that is designed and rated for the intended use. Rope science is like paint science - deceptively complex.
Click to expand...

Here is your best answer.

Not only do I have a number of physics degrees, I've been a working physicist for many decades.

I've also climbed and been a caver in Colorado and New Mexico.

If you believe there's a chance your loads will fall any distance larger than a foot or so, you really should buy rope designed to take falls. The design begins with the fibers themselves, but continues in the weave and the way the fibers are melded into the rope. Climbing/caving rope is 'dynamic' rope -- it is specifically made to stretch under load.

If the rope is not specifically made to stretch under load, it will have to sustain a large 'impulse' (large force over a short period of time) whenever an on-rope mass is dropped.

I should also mention that dynamic ropes must be protected at all costs from abrasion and sharp edges. When under load, these ropes have a disturbing tendency to snap. When much younger, I participated in a demonstration of this. A dynamic rope of 30 feet in length was used to suspend 150 pounds of sand in a burlap sack. The instructor placed the sharp edge of a pocket knife against the rope where it ran over (90 degree bend) the edge of the cliff above. Snap! The rope parted instantly and the load struck the ground -- 90 feet below.

Mind you, the rope was protected from the cliff edge with a very thick cloth pad. And the instructor did not saw at the rope or attempt to sever it. He just pushed the edge into the rope...rather gently. Catastrophic, instant failure.

After each and every outing, I always cleaned and inspected every inch of my rope. If it didn't pass muster...it was immediately retired.

I hope this helps you.
 
Thanks very much for your thoughts.... and you have confirmed what I believed: it doesn't seem like there is a simple answer. There are just too many variables.

Yes...I most certainly would take expert advice if my rope problem involved suspending people above long drops. But right now I'm just curious. I've done a bit of caving etc, but nowadays my cordage use mainly involves boats, moorings, fishing and trapping.

I think as far as my pole snares are concerned, my experiments with dropping a dummy weight on the end of my cords to be tested gives me a good enough idea of what to expect. Like I said, the ease with which the tough cord breaks when the weight is dropped has been quite surprising.

I use a rule of thumb for bowstrings where I think the breaking strain of the string should be around five times the nominal draw weight of the bow.

For snares set at ground level (no drop) I've said to myself that the breaking strain of the cord should be maybe five to ten times the weight of the animal. I believe that an animal is less likely to break a snare if the snare is tethered up high...the critter runs out of traction as it rushes to the end of the rope. I think it also helps to keep tethers short, and where possible to tie them to a springy sapling.

Using my eight pound test weight, I've broken cord rated at over 100 pounds with only a short drop. So it seems that the 10:1 ratio would be the minimum that I should work with for pole snares that animals can jump from. 15:1 or 20:1 would be more secure...especially when the cable is getting a bit worn or kinked. I have to say that the hard bit of wood I've been using probably comes to a quicker stop than a more flexible animal...but the experiments are good food for thought.

Thanks for your thoughts....much appreciated. Isn't the internet a wonderful thing !

Regards....Stephen Coote.
 
For what your doing it seems like your using smaller cord? I really don't know of any "dynamic" cordage. The smallest dynamic rope I have ever used is 8mm.

Perhaps you could improvise a "load limiter" or "screamer" basically it is a bartacked piece of webbing that when loaded the stitches rip out thus reducing the force the rope is subjected to.

You could possibly do a lighter weight version.

http://www.yatesgear.com/climbing/screamer/index.htm

Check out the Yates site for info on how they work and you'll get the idea.

I dunno, maybe something like this could help.
 
Thanks for your thoughts. Perhaps some sort of shock absorber could be useful in some circumstances. My initial 'fix' is simply to use heavier cord.
 
another thought would be to purchase climbing rope (dynamic) and simple remove the exterior sheath. You will find various different manufacturers all use various diameters and number of interior strands. Perhaps you can simple use one (or some) of the disasssembled fibers as your dynamic snare?

If you have a local climbing/mountaineering club in your area I am certain that for the price of a couple of beers, you would have more than enough retired dynamic rope. I would suggest you try to get your hands on a rope of 10.5mm or 11mm diameter as they have the most interior fibers, and from what I recall, the largest diametered interior strands.

Cheers,
D
 
Now that is a great thought Diligence. Never thought of that, even though I've pulled apart some very strong Dynex rope to make snares. Dynex doesn't stretch though.

There must be a ton of old climbing rope retired into odd corners of clubhouses and garages all over the place.

In case anybody is interested, when I twisted up some two-ply cord from the fine fibers of the Dynex rope, I soaked it in acrylic house paint to help prevent the fine fibres from catching on everything. It also added a nice bit of stiffness to the cord.
 
I would have thought that you wanted the animal falling/jumping off the pole to come to a sudden, swift, stop. I know that squirrels have very strong necks that don't break easily, and assume that this would be common to other tree climbers, but even if you don't break the animal's neck, I would think that the sharp jolt would bed the noose bed in better. For strength and no stretch the lines used on sailing boats are very good. Many are designed to take some shock load, as when a sudden squall hits the sail. But that asside, they can carry some silly loads at really small diameters.
 
Diligence said:
another thought would be to purchase climbing rope (dynamic) and simple remove the exterior sheath.
Click to expand...

As far as I know dynamic ropes are meant to protect the climber from dislocation by deceleration, not really the cord itself. Actually when designing a trap, you'll want the deceleration to be pretty suddent to break the critters'neck so it doesn't suffer or try to escape.
Plus dynamic ropes get damaged when used so you have to discard it in the end anyway.

As for the difference between static load and shocks: a 200lbs man can stand on a brick holding a hammer without any damage to the brick, but a child with the same hammer will be capable of breaking it by striking.

As for choosing material as already said there are loads of factors to consider. For some material you'll probably need a ratio many time higher than static load, while it might not be necessary to some others. Problem is probably as complicated as those blade material/heat treatment discussions.
Actually, your "crash dummy" method is probably the best possible in that situation, at least with the tooling you got at hand anyway.
 
Here's an easy way to picture the strain spike of a falling object.

If you set a 10 lb ham on a scale it reads 10 lbs. If you drop a 10 lb ham on a scale from 10 feet up it will register much heavier for a second as it hits. Mac
 
The force acting on the rope is:
F = v*sqrt(k*m) [Newton]

The maximum velocity is:
v = sqrt(2*g*h) [m/s]

where

g = 9.81 [m/s^2]
h = height [m]
k = spring rate of the rope [N/m]
m = mass [kg]

So the answer is:

F = sqrt(2*g*h*k*m)


This applies if the rope behaves like a linear spring.
 
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