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What method would I used to calculate the likely force placed on a piece of protection if I were to fall onto it. Using knowledge of my weight, distance between myself and the gear and the amount of rope between myself and the belay.

Scenario, I'm above a small but well placed nut. I'd like to know how far I can climb before the force of my fall will overcome its stated strength.

Perhaps ignore potential for a soft catch from the belayer for simplicity. No imperial measurements please, i.e. meters, kilograms, kilonewtons.

  • It still won't provide a direct answer to this question but for anyone fascinated by such calculations this software appears to be the best on the market: vrigger.com – Mr.Wizard Feb 3 '16 at 8:52
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Wikipedia has a simple treatment of this problem, as well as some notes on at least one of the reasons why the simple treatment is only a very rough approximation. Let be the impact force quoted by the manufacturer, which is normally 12 kN for a single rope. Let be the 80 kg mass used for lab tests, let , let and let f=h/L be the fall factor (ratio of length of fall h to the length of the rope L). Then in their simplified model, the maximum tension in the rope is

.

The simplified model ignores friction, including both friction in the rope and friction where the rope goes around the carabiner attached to your top piece of pro. In this approximation, the force on your hypothetical micronut is 2T. This is because there are two strands pulling down on the pro, and due to the lack of friction the tensions is the same in these two strands. In some online discussions I've seen, it's not really clear whether they have in mind the stress on the pro or the stress on your body, leading to an ambiguity by this factor of 2.

There is another surprising factor of 2, which is that in the special case where f=0, we get T=2mg, which is a little surprising. You would think that in a factor-zero fall, the rope would just support body weight. But the peak tension is actually more than body weight, because the rope stretches and the person does fall down through some height. In this model the peak tension is actually twice body weight for a factor-zero fall.

If you put in f = 2 as a worst-case scenario, then we have 2 T = 26 kN. This is more than 10 times the 2 kN rated strength of the smallest micronuts, and even exceeds the ~8 kN rating of much burlier pieces of pro. Uh oh!

OK, so it's not quite as bad as that. Internal friction in the rope seems to cut the maximum force by about 3 kN. If you're using an ATC rather than an autolocking belay device, the dynamic properties of the device will cut the force by maybe another 30%. If the belayer is giving an indirect belay, then they're going to get picked up by your fall, and their body becomes a dynamic element in the system, further reducing the maximum stress.

In a factor-2 fall, the combination of these factors might reduce the force on your nut by quite a but, but it's still very likely to fail.

So the basic answer is, don't take any kind of serious lead fall on the smallest 2 kN-rated micronuts. Just as a simple reality check, I climb pretty often with a guy who's 100 kg. His body weight is 1 kN, which would put 2 kN of force on a piece of pro if he simply hung his body weight on it.

  • wow, this is quite intense. I was hoping there would be a simpler 'rule of thumb' that could be used while climbing, I suppose one option would be to do the math on the ground/at home to give yourself the knowledge to make more informed judgements while on the sharp end. – aaaaargZombies Feb 23 '15 at 23:55
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    @aaaaargZombies: I don't think anybody actually does calculations like this for practical climbing. There are too many factors that are hard to quantify, such as rock quality, the quality of the placement, and how far beyond its rated strength the piece will really go before it's torn apart. Realistically, your time is better spent reading John Long's book on climbing anchors, and reading Accidents in North American Mountaineering to find out what really goes wrong and gets people hurt or killed. The biggest issues are: (1) minimizing fall factor, and (2) getting that first "Jesus nut" in. – Ben Crowell Feb 24 '15 at 20:37
  • You could do the calculations- probably completing them while accelerating at 10ms2 and finding out the definitive answer one way of the other, or concentrate on the climb - to many variables to do it in the class room. – user5330 Mar 10 '15 at 4:38
  • "There is another surprising factor of 2, which is that in the special case where f=0, we get T=2mg, which is a little surprising. You would think that in a factor-zero fall, the rope would just support body weight. But the peak tension is actually more than body weight, because the rope stretches and the person does fall down through some height." Isn't the reason in this equation simply the pulley effect mentioned in the paragraph right before this? "This is because there are two strands pulling down on the pro ..." -- you made it sound like two separate effects. +1 however. – Mr.Wizard Feb 3 '16 at 8:42

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