How do you add up an anchors KN?
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Basically, your being told here that there is no hard and fast rule for figuring anchor strength. The books you refer to basically assume all pieces to be thoroughly bomber and well-equalized, in which case the author is simply adding the strength of the individual pieces. |
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For a 25kN anchor.... add up the three weakest isolated points. Suppose you have a nut, and two finger sized cams. See each arm of the anchor as independent and look for the weakest point... the cam or nut should be. Add those together, and you can get in the general region of what the sort of fall it could withstand. Where the math gets complicated. The angles between each piece and the master point can act as multipliers for impact force. Mostly that aspect is ignored for this calculation. Most of us guesstimate.\ |
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ChrisCase wrote: I have read the books, but still don't see anything on how to add up the KN. It does say that for safety standards we make our anchors 4 to 5 times stronger than needed, and that a 3 piece anchor could hold more than 40kn. So how are they adding that all up? Is there any guides out there that can shed some light for me?heres why a "40 KN" anchor ... or even a "25 KN" one is simply irrelevant in the real recreational climbing world BC QC lab blackdiamondequipment.com/e… build your anchors as strong as reasonably feasible under the circumstances years ago at an alpine course someone insisted that the only acceptable anchor was a 3 piece with an 4th upward pull .... the instructor replied "so lets see youre going to waste time building a full anchor at every belay adding an extra hour to all the belays and carrying up an extra 500g of gear ... while yr exposed to changing weather and other objective risks .... safety is more than gear" no one in the field can give you an exact KN distribution of a multipoint ... those who say they can are making stuff out of their azz ... but there are some general rules of thumb you can use .... ;) |
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ChrisCase, I think this equation will help: |
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ChrisCase wrote:...When it comes to adding up the KN of an anchor, how do you add it? ... ChrisChrisCase, Perhaps this will answer your question: web.mit.edu/sp255/www/refer… |
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r m wrote:ChrisCase, I think this equation will help: Climbing forum physics = 5% reality + 95% bullshitKinda like dating |
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r m wrote:ChrisCase, I think this equation will help: Climbing forum physics = 5% reality + 95% bullshitFor once the math discussed here had made sense to me! |
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tradvlad wrote: Here's a cool article: fallpro.com/fall-protection… Looks like you can break your spine at around 4 kn.That is a great resource. Anyone know of a resource on actual forces generated during falls? |
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dave custer wrote: ChrisCase, Perhaps this will answer your question: web.mit.edu/sp255/www/refer…Dave, I'm not sure that the linked exercises do answer the question. Most of them begin with the assumption that the loads to each anchor point are equal, and the ones that don't make that assumption assign anchor loads a priori. As you know, the question of how much load each anchor point gets when the only information is the load vector to the power point is much harder, and this is essentially the OP's question---or should be, because the preassigned load questions have no practical relevance. Even forgetting about the practical issues of unequal strand tensions arising from imperfect rigging and the behavior of the power point knot under load, the classical three-point anchor is statically indeterminate---the equilibrium equations involving the strand tensions do not have a unique solution. In order to get a solution, the elongation behavior of the strands has to be added to the data supplying equations, eg via Hooke's Law, and this does complicate the math considerably. Since the strand elongations are not very small (as is the case if the anchor was made of steel beams), linear approximations are likely to be poor and so the whole mess probably has to be given to a computer algebra system for solution. In other words, bearbreeder's TI 83 isn't up to the task even if available. All this is at the theoretical level before the various practical realities cloud the issue further. Now that I've gone this far, perhaps it is worth mentioning that the classic three-point anchor has a cascade failure mode: if one of the lateral arms blows, the entire load is transferred to the middle strand, which gets no help from the strand on the other side, and if the middle strand blows, everything ends up on the remaining point. It seems unlikely that the system benefits much from energy absorbed by extracting pieces, and so there will be very little load reduction as the cascade proceeds. What this means is that such a three-piece anchor, say with all pieces able to withstand the same load L, could turn out to be no stronger than L all together, which is about as far as you can get from "adding up the kN" for each piece. |
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FrankPS wrote:My anchors have one of two ratings: 1. Good 2. Good enoughmine have only one: good. |
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As a useful heuristic, which is what we're after in this case - since we all admit a real calculation can't be done, I suggest the following to "calculate" anchor strength: |
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Now that I think about it, an even better heuristic (although more complicated) based on a somewhat similar logic is multiplying each piece's KN by inverse of the square root of one plus its rank. This yields the following results for the systems above: |
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Pick up John Long's latest book on anchors. He has re-evaluated his anchor building methodology based on more realistic physics models (mainly that the 7mm cord that was assumed to be static does not act in that manner whatsoever.) Using the traditional figure eight masterpoint cordalette anchor, all anchors would have to be evaluated on a case by case basis depending upon gear used in the anchor, length of cord from the gear to the masterpoint and angle of force on the masterpoint. And this would still assume perfect gear and rock. You would be left with a matrix of forces applied to the anchor over a range of angles. His equalette anchor is a decent solution to provide more evenly distributed forces and still retain redundancy. |
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Dylan B. wrote: This is something I've wondered about, and in particular the portion in bold. Why does it seem unlikely to you that a significant amount of the energy has already been dissipated by the time piece one blows, and still more by the time piece two blows? My totally uninformed intuition is the opposite.I say that because I don't think much work is done when a piece blows, and since it is energy absorbtion that reduces the peak load, I don't think much energy absorbtion has taken place. The situation is different when gear blows, because there is a moment after a piece blows when the rope is unweighted, and in that instant the rope springs back, (there are high-speed pictures of this), which means it can be "reset" to absorb more energy. But there is no "spring back" mechanism going on when an anchor leg blows, no part of the system is unweighted and free to contract, so I doubt there is any significant energy-absorbing effect. |
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I think it is probably best to assume most likely you will never have perfectly balanced points (even bolted anchors with 2 of the same draws alot of times will be slightly off in how the load really pulls) so it would vary how much weight truly goes to each point. I would guess in at least 50% of the anchors out there 100% of the weight will go to one spot until it breaks. |
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Dylan B. wrote: This is something I've wondered about, and in particular the portion in bold. Why does it seem unlikely to you that a significant amount of the energy has already been dissipated by the time piece one blows, and still more by the time piece two blows? My totally uninformed intuition is the opposite.All depends on what fails and why. The energy absorption of materials as they fail is called the work of failure or rupture and depends on the material itself, some stuff is brittle like rock, others like aluminium are ductile and as a generalisation brittle materials need very little energy input to fail whereas ductile ones need more. Glass has a tensile strength ca 6x that of steel but requires almost no energy to break and rock is usually very similar. Something like a cam would usually fail catastrophically and take very litle energy to do so, pulling an alloy nut down a crack on the other hand might take a considerable amount. To break the average modern karabiner requires suprisingly little energy, just a lot of force. |
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rgold wrote:In fact, almost no climbers have ever done any kind of feedback testing to see if judgements like "good" correspond to any level of load tolerance and, indeed, if there is any consistency in what the "good" judgement represents.For me, the best thing to come out of Aliengate was that I tested all my Aliens myself, and learned a hell of a lot about what a "good" placement really is and isn't. When I first started testing, a lot of the gear pulled from its placement before the fuse popped, showing that that placement was not sufficient to hold the rating of my fuse (which I believe was 3-4 KN). Soon enough I got a better eye, and in the second round of testing, very few cams popped. And in all the rounds of testing I did after that, none did. From what I learned then, and from what I've seen climbers do, I suspect that when placing small cams, most climbers are totally mistaken about when their gear will hold. Only the fact that they rarely fall on them keeps them all from getting dead. This would be a really valuable exercise that I think all new trad leaders (with supervision) should do. That, and practice holding a greater than 1 factor fall, like your generation used to do, and mine did not. GO |
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Thanks for the remark Gabe. I should modify the "almost no" part of my statement, since many climbers have put in some ground-school and/or mock aid leading time bounce-testing gear and so do have a sense of what is "good" and "not good." |
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I don't think so. In any case, I don't see the connection to the presence or absence of energy absorbtion from individual failures. |
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rgold wrote:Thanks for the remark Gabe. I should modify the "almost no" part of my statement, since many climbers have put in some ground-school and/or mock aid leading time bounce-testing gear and so do have a sense of what is "good" and "not good."I think aiding is a valuable learning tool for trad leaders for a number of reasons, among them the ability to get themselves past a tough section quickly and efficiently when needed, and yes, seeing how gear shifts when weighted. But one thing that is commonly claimed as a benefit, it doesn't really accomplish. It does not actually give you an eye for which gear is "good" and "not good". It will show you which pieces are "total shit", and it will show you how pieces might move in reaction to a fall, but it does not show which gear would rip out in even a small fall. A fairly small fall will generate 2 or more kN of peak force. However bounce testing does not generate this kind of force. So you're really not learning what will hold a small fall and what will not. Here's a simple test to prove what I'm saying: Take a screamer, attach it to a piece of gear (or a bolt, or an I-beam, it doesn't matter), and then put some static slings or aiders on it and bounce test it as hard as you can. If you can get to 2 kN, you will rip a stitch. I tried, I couldn't. GO |