Sean, Actually I think that Nick is more correct than you....but what you say MAY occur. The real meaning of the fall ratings is almost univerally misunderstood, The ratings are based upon a standard test in which a dummy load is dropped into a specific length of slack line routed over a specific diameter rod. The fall rating is for the number of repeated drops until failure. It is a standard test that has almost nothing to do with real world rope saftey save the assurance that the rope will hold a climber should he fall. Many many climbers mistakenly think that the fall rating is the maximum times a rope could be subjected to falls safely. In fact it is the minimum number of very extreme loadings that the rope can be expected to endure. In the real world the loads don't get concentrated on the same extreme bend location and do not occur repeatedly without recovery time for the fibers (high loads on Nylon may cause both permenent, semipermanent and transitory elastic deformation. Time between loadings allows much of the elasticity to be recovered). All of which means that climbing ropes are safe far beyond taking 8 or 9 big swings on them.
M
moray
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And, as Nick said, ropes that have endured a number of falls get stiffer. This is a form of degradation that would be very hard to tie to broken fibers.
The standard UIAA drop test is this: 80kg weight dropped 5 meters, caught by 2.8 meters of rope draped over a 10mm-radius curve. This is a fall factor of 1.78. My current jumps are down around .04 fall factor. My rope is rated for 10 standard falls. If I keep future jumps below .1 fall factor, and the support radius at 25mm, how many jumps should the rope survive? I would guess many thousands. I suspect that a .17 fall factor fall does not do 1/10 the damage of a 1.7 FF fall. I'll bet it is more like 1/100 or even less. Anyone know more about this?
The standard UIAA drop test is this: 80kg weight dropped 5 meters, caught by 2.8 meters of rope draped over a 10mm-radius curve. This is a fall factor of 1.78. My current jumps are down around .04 fall factor. My rope is rated for 10 standard falls. If I keep future jumps below .1 fall factor, and the support radius at 25mm, how many jumps should the rope survive? I would guess many thousands. I suspect that a .17 fall factor fall does not do 1/10 the damage of a 1.7 FF fall. I'll bet it is more like 1/100 or even less. Anyone know more about this?
No_Bivy
Treehouser
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Dan Osman was wild and a bad ass climber...made a bad choice. RIP Danno'. He would still be cranking today if he had not got into rope jumping stuff. He pushed the envelop to far..........so do many adventurers, but I dont judge them. Hell, my job is dangerous, does that me selfish?
B
Bounce
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Stumper - so if the line isn't getting damaged by repeated shock loading, why does it eventually fail due to repeated shock loading then? Regardless of the differences between the test and real life conditions, it seems fairly obvious that there must be something happening to make the rope fail eventually. If it isn't broken fibers, then why do you think it eventually breaks?
M
moray
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I think ANY loading, dynamic or not, must cause damage.
It is hard to find really good info on this on the Web, but in several places I have seen nylon (and polyester) rope fibers described as partly amorphous and party crystalline. Manufacturers can and do control this to achieve desired qualities like stretchiness and strength. Apparently one method of controlling the degree of crystallinity is by applying heat and strain. It seems plausible that loading the rope causes local areas, perhaps much smaller than a fiber, to heat up enough to allow changes in the crystalline structure. The result (perhaps) is the rope is now stiffer and weaker, yet no fibers have broken.
UV degradation certainly follows this model in that it acts at the molecular level, not the fiber level. It breaks molecules. The rope will be weaker as a result even if all the fibers are still intact.
Sean. I am NOT saying that the rope is not damaged by repeated shock loading. I was and am saying that the test is designed to standardize an extreme stress situation so it does not represent maximum useful life of a rope, rather a standard of comparison to other ropes and a minimum guaranteeable safe life for the rope. Reread my post about permanent, semipermant and transitory elastic deformation. Moray is hitting the same subject from a different angle with different specific language. The test pounds one location on the rope with cumulative effects of all types of elastic deformation. In normal use the transitory and semipermanent effects are allowed time to recover. Permanent deformation is there forever. Thus stresses on a rope ARE cumulative and lead to eventual failure but the accumulation of damage is a much slower process than the test suggests. This means that Nick's rope course allowing 250 falls on a rope was probably a well calculated thing based upon the fall factor and anticipated recovery times for the rope fibers.
Returning to the original question.........Fibers need not be broken for a rope to be weakened.
Weakening a rope without breaking fibers via cumulative stress and elastic deformation of fibers is a longer process when recovery time is available between stresses compared to rapidly repeated stresses.
Returning to the original question.........Fibers need not be broken for a rope to be weakened.
Weakening a rope without breaking fibers via cumulative stress and elastic deformation of fibers is a longer process when recovery time is available between stresses compared to rapidly repeated stresses.
M
moray
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Interesting Factoid
I discovered this on the Web yesterday. It applies directly to my jumping experiments and any other situation in which a rope catches a falling load. Say you are happily suspended with an arborist rope from a questionable TIP. The TIP gives way, and you fall, say, 2 feet before the rope catches on a new branch. If you had 30 feet of rope above you, there would probably be enough stretch in the rope to give you a fairly mild landing. What if you only fell 1 inch? What if you fell 1 inch and the rope above you was 150 feet of stretchy dynamic rope? In other words, what is the minimum shock (tension in the rope) from the softest conceivable fall?
The answer is: twice your weight. No matter how much rope there is, or how stretchy, or how short the free-fall distance, you (the load) will experience at least a minimum of 2g deceleration, or equivalently, a force equal to twice your weight.
When I drew some diagrams and checked the math, I realized there is a cool corollary to this, one that is directly useful to me when adjusting rope length for jumps. It is also very simple. Say you have just jumped. There was a small amount of free fall before the rope went taut and then caught you. At maximum stretch your feet are still 6 feet above the ground. How much more slack can you give yourself and still be sure you won't hit the ground? Answer: 4 feet. This means an extra 4 feet of free fall, of course, but the extra 4 feet will never cause even half that much in extra stretch! Altogether then, the 4 extra feet of rope and under 2 feet of extra stretch keeps you safely above the ground. More generally, if a given jump leaves you with X feet of ground clearance at maximum stretch, you can safely add 2/3 that amount in extra slack on the next jump.
These are not your grandfather's rigging problems...
I discovered this on the Web yesterday. It applies directly to my jumping experiments and any other situation in which a rope catches a falling load. Say you are happily suspended with an arborist rope from a questionable TIP. The TIP gives way, and you fall, say, 2 feet before the rope catches on a new branch. If you had 30 feet of rope above you, there would probably be enough stretch in the rope to give you a fairly mild landing. What if you only fell 1 inch? What if you fell 1 inch and the rope above you was 150 feet of stretchy dynamic rope? In other words, what is the minimum shock (tension in the rope) from the softest conceivable fall?
The answer is: twice your weight. No matter how much rope there is, or how stretchy, or how short the free-fall distance, you (the load) will experience at least a minimum of 2g deceleration, or equivalently, a force equal to twice your weight.
When I drew some diagrams and checked the math, I realized there is a cool corollary to this, one that is directly useful to me when adjusting rope length for jumps. It is also very simple. Say you have just jumped. There was a small amount of free fall before the rope went taut and then caught you. At maximum stretch your feet are still 6 feet above the ground. How much more slack can you give yourself and still be sure you won't hit the ground? Answer: 4 feet. This means an extra 4 feet of free fall, of course, but the extra 4 feet will never cause even half that much in extra stretch! Altogether then, the 4 extra feet of rope and under 2 feet of extra stretch keeps you safely above the ground. More generally, if a given jump leaves you with X feet of ground clearance at maximum stretch, you can safely add 2/3 that amount in extra slack on the next jump.
These are not your grandfather's rigging problems...
M
moray
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Stay tuned.
No_Bivy
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have fun :roll:
have fun :roll:
M
moray
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Nice set of vids, No_Bivy. I had seen most of it before, but some of it was new. I always wonder what sort of G forces he was experiencing at the bottom of his jumps/falls. I'm sure Dan and his engineers had that all figured out ahead of time. If you're jumping out of trees, and have a bunch of pulleys available, you can control the G forces and keep them very low.
