
mtnrock
Oct 7, 2008, 9:14 PM
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im trying to find what the maximum height you can climb above your protection before it will pop and i need to know the spring constant of a rope does any one know what it is





rocknice2
Oct 7, 2008, 9:26 PM
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mtnrock wrote: im trying to find what the maximum height you can climb above your protection before it will pop and i need to know the spring constant of a rope does any one know what it is It shouldn't pop as your climbing above it. That's poor placement! If you mean maximum fall, then double your rope length divide by Pi sutract the number of passengers on the first bus you see and multiply itt all by your IQ. A good piece in good rock will not break. That's the way the system works. I STRESS WELL PLACED IN SOLID ROCK.





sungam
Oct 7, 2008, 9:49 PM
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mtnrock wrote: im trying to find what the maximum height you can climb above your protection before it will pop and i need to know the spring constant of a rope does any one know what it is Different ropes, different friction and different conditions of the rope (dirty core, wet, frozen)  these variables change the spring constant of the rope greatly. I guess I'm saying no no one can tell you the spring constant of all ropes.





armsrforclimbing
Oct 7, 2008, 9:55 PM
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Get to work li'l shaver: "The spring constant is a measure of the "stiffness" or "strength" of a spring. To determine this spring constant, you could do the following: hang the spring from some sort of support and note its unstretched length. Now add some mass to one end of the spring. The spring will stretch and come into equilibrium at a length x beyond the spring's unstretched length. The masses are pulled down by force of gravity and pulled up by the spring force. The two forces balance, and this means that the force of the spring on the masses is equal to mg, the weight of the masses. Adding more mass will further stretch the spring. By measuring and plotting the spring force, F, againt the stretching of the spring, x, you should get a straightline graph with slope k"





jt512
Oct 7, 2008, 10:01 PM
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mtnrock wrote: im trying to find what the maximum height you can climb above your protection before it will pop and i need to know the spring constant of a rope does any one know what it is You can calculate the spring constant of a rope from the manufacturer's published impact force rating. See equation (11) in rgold's paper. The spring constant is k, which rgold refers to as "rope modulus." I also know what punctuation is. Jay
(This post was edited by jt512 on Oct 7, 2008, 10:04 PM)





sungam
Oct 7, 2008, 10:12 PM
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jt512 wrote: mtnrock wrote: im trying to find what the maximum height you can climb above your protection before it will pop and i need to know the spring constant of a rope does any one know what it is You can calculate the spring constant of a rope from the manufacturer's published impact force rating. See equation (11) in rgold's paper. The spring constant is k, which rgold refers to as "rope modulus." I also know what punctuation is. Jay Jay, would you say that forces shown by the calculated spring constant would be similar to the actual forces involved in a situation with no drag on the rock and little or no drag from gear (lets say a steep sport route at a fairly steady just overhanging) or would it still be far off?





petsfed
Oct 7, 2008, 10:48 PM
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mtnrock wrote: im trying to find what the maximum height you can climb above your protection before it will pop and i need to know the spring constant of a rope does any one know what it is Welcome to the wonderful world of real world physics. You can find the idealized spring constant via rgold's paper in JT512's link, but that's based on the assumption of a massless (or at least trivially low mass) rope/spring. You have to take into account the mass of the rope, and that will throw off your spring constant calculation. Second, you have to take into account rope drag, which will yield different tensions for different parts of the rope, and finally nuances of the fall (sliding, rope getting caught, hitting ledges and the like). Its actually a very complicated dynamical system, one that does not yield to a "formulafishing" approach. One thing that I can tell you is that under normal circumstances (fall factor less than or equal to 2) with solid pro in good rock (so not micronuts/microcams/micropitons and likewise not rotten or otherwise soft rock), the protection will not fail.
(This post was edited by petsfed on Oct 7, 2008, 10:49 PM)





sungam
Oct 7, 2008, 10:51 PM
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Also, don't forget the reduced (i'm tired) "springyness" the rope has if it's already taken a fall (or falls).





armsrforclimbing
Oct 7, 2008, 11:07 PM
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Jay and/or Petsfed, has this paper been discussed on the page before? I am commiting a cardinal sin by admitting that maybe I don't understand the term "relative stretch" in equation #1, in that the y/L term causes the units of length to cancel each other creating a conflict in the conclusion T=k (i.e. units of force are equated to a units of force divided by unit length) It may be the engineer in me, but I am going back to the units here, and they are ultimately what give the equation meaning.
(This post was edited by armsrforclimbing on Oct 7, 2008, 11:08 PM)





jt512
Oct 7, 2008, 11:18 PM
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armsrforclimbing wrote: Jay and/or Petsfed, has this paper been discussed on the page before? Yeah, but it's been a while since it was discussed at any length. If you search rgold's posts, you'll eventually find the original discussion, but the reason I put the paper on my Web server is that I got tired of search for it every time.
In reply to: I am commiting a cardinal sin by admitting that maybe I don't understand the term "relative stretch" in equation #1, in that the y/L term causes the units of length to cancel each other creating a conflict in the conclusion T=k (i.e. units of force are equated to a units of force divided by unit length) It may be the engineer in me, but I am going back to the units here, and they are ultimately what give the equation meaning. You are right that y/L is unitless, but the units of k are force units, so the equation T = k is consistent. Jay





armsrforclimbing
Oct 7, 2008, 11:52 PM
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It's late, I'm fresh off the debates and a glass or two of wine, so bear with me. The equation of a line is y=mx+b set b=0 and the line passes through the origin, you also end up with the equation y=mx (similar to that of the Hooke's Law without direction of force) Hooke's law describes the linear relationship of force and deformation in an idealized spring. So in concept, the slope of the line obtained from plotting force on the Y axis and the resulting deformation on the X axis results in k's (read: slope) units of force/unit length. "k" is referred to as the spring constant, if it were a force value, as used in rgolds paper, it would by nature of a slope (change in y/change in x) be required to change and cannot be constant. This is the Hooke's law I am familiar with. The "k" value derived in the paper does not represent a physical constant unique to a particular spring, rather a variable force, and will not work with other equations when used as such.
(This post was edited by armsrforclimbing on Oct 8, 2008, 12:24 AM)





rgold
Oct 8, 2008, 12:36 AM
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Arms, the report you and others are referring to is an account I wrote up after explaining and reexplaining the various issues over and over. However, other than being reread by me, it has not been refereed and, as far as I know, has never had its substance, as opposed to its conclusions, discussed online. I am honored to have it a called a "paper," but considering the absence of expert review, I think we should find a less lofty term, and as you can see I've settled on "report." I have tried to free said report of typos and mathematical errors, and certainly any that remain are my responsibility, but the conclusions have been obtained over and over again by many different people and are, in my estimation, reliable. Relative stretch is, of course, defined by the equation y/L, there's nothing more to understand. The values taken on by y/L are, again by definition, what I called units of relative stretch. So when y=L, the equation becomes T (in Kn) = k (in kN per unit relative stretch) X 1 (unit of relative stretch) and there is no problem with any mismatch of units, and no problem with the meaning of the equation. But your point is welltaken, it would have been clearer for me to say that T=k kilonewtons, rather than saying that T=k. Petsfed, there isn't a mathematical model in the world that incorporates every detail of the reality being modeled. If one did, it would in some sense be the reality and not a model of it. Perhaps the essential ingredient in every model is the fact that it ignores details, so your criticism is in fact a foregone conclusion. I'd love to see your model incorporating the mass of the rope, and I would be interested to see, after the mass is properly taken into account, how much effect it actually has on the results. Please do post up. By the way, the fact that the real dynamical system is much more complicated than the model does not automatically disqualify the model from providing useful data. One has to demonstrate that the additional complexities, if they are incorporated, will significantly change the predictions. Assertions to this effect, without supporting arguments, are not evidence. Not to be outdone in knocking what I wrote (which never pretends to be any more than it is, by the way), I should mention that the source of all deductions, Hooke's Law, is most likely just the linear part of a more accurate higherorder formula for rope tension, so defects are built into the model from the very beginning. And believe me, I could go on about other defects. Nonetheless, it is a deep misunderstanding of the pervasive role of mathematics in science and engineering to equate the fact that the model has ignored details of reality with an automatic claim that the model is invalid. All this windy justification notwithstanding, I do think Petsfed is right to cite the complexities of reality in regard to the use of a simple model to calculate how far above your gear you can climb. Mtnrock, you can calculate the spring constant of the rope as I indicated in my report, but it will be of no use in your practical quest to know how far you can climb above your gear. Indeed, even within the realm of the theoretical model, that question has a conditional answer. Edit: In the meantime, Arms has posted another comment. It is true that the usual statement of Hooke's Law assumes a spring of fixed length, but that is not, and could not be the version I used in the report, where relative stretch (you could think percentage elongation if that makes it any clearer) has to replace the absolute change in length of a fixedlength spring. Look, there can be varying amounts of rope out in the system, so knowing that there has been two feet of stretch tells you nothing. The version of Hooke's Law that I (and everyone else) used makes k the same constant regardless of how long the rope is. The xaxis in Arms' comment should be marked in percentages, not in linear units. Otherwise, it is Arms' formulation that subverts the intrinsic nature of k.
(This post was edited by rgold on Oct 8, 2008, 12:52 AM)





jt512
Oct 8, 2008, 12:50 AM
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armsrforclimbing wrote: It's late, I'm fresh off the debates and a glass or two of wine, so bear with me. Then we are in exactly the same position.
In reply to: The equation of a line is y=mx+b set b=0 and the line passes through the origin, you also end up with the equation y=mx (similar to that of the Hooke's Law without direction of force) Hooke's law describes the linear relationship of force and deformation in an idealized spring. So in concept, the slope of the line obtained from plotting force on the Y axis and the resulting deformation on the X axis results in k's (read: slope) units of force/unit length. "k" is referred to as the spring constant, if it were a force value, as used in rgolds paper, it would by nature of a slope (change in y/change in x) be required to change and cannot be constant. This is the Hooke's law I am familiar with. The "k" value derived in the paper does not represent a physical constant unique to a particular spring, rather a variable force, and will not work with other equations when used as such. Your model states that force is proportional to the absolute stretch in the spring, whereas rgold's states that force is proportional to the relative stretch, ie, the absolute stretch divided by the length of the spring. I don't know which one, if either, is the official Hooke's Law, but it is unimportant. What is important is that your model and his are different, if the models are considered to apply to springs of arbitrary length. If, regardless of the length of the spring, the force is proportional to the absolute stretch; then the force cannot be, in general, proportional to the relative stretch, and vice versa. On the other hand, given a fixed spring length, the two models are equivalent, differing only in whether the length of the spring is taken into account in k or x. If the relative stretch model holds for climbing ropes, it is the more convenient model, because it holds for any length rope. I don't know if that clears anything up, or not. If not, it's the wine's fault. Jay
(This post was edited by jt512 on Oct 8, 2008, 1:09 AM)





armsrforclimbing
Oct 8, 2008, 6:40 AM
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So, it's early know and I haven't had any coffee yet, so bear with me. Having read (read: scanned for keywords) your posts I can draw several conclusions. 1. You are gentlemen and scholars both. 2. I think we are on the same page. 3. Given the "mathemagic" that normally goes on on this page your explanations offered are certaintly reasonable. 4. Good rock will always hold good pro (that we always agreed on.) 5. And in short, I NEVER (safe use of the word here) bring a calculator climbing.





rightarmbad
Oct 8, 2008, 9:38 AM
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http://jemez.org/lam/xRopes.pdf This is pretty good





petsfed
Oct 8, 2008, 2:16 PM
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rgold wrote: Petsfed, there isn't a mathematical model in the world that incorporates every detail of the reality being modeled. If one did, it would in some sense be the reality and not a model of it. Perhaps the essential ingredient in every model is the fact that it ignores details, so your criticism is in fact a foregone conclusion. I'd love to see your model incorporating the mass of the rope, and I would be interested to see, after the mass is properly taken into account, how much effect it actually has on the results. Please do post up. Its supposed to be too cold to climb this weekend, so I'll work on it. I'd expect that for a UIAA rope test, its pretty minor, but if you've got 100 feet of rope out, it might be pretty big (on the order of 5 or 10 percent difference). I'll try to work up some general equations and explanatory assumptions.





jt512
Oct 8, 2008, 3:27 PM
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petsfed wrote: rgold wrote: Petsfed, there isn't a mathematical model in the world that incorporates every detail of the reality being modeled. If one did, it would in some sense be the reality and not a model of it. Perhaps the essential ingredient in every model is the fact that it ignores details, so your criticism is in fact a foregone conclusion. I'd love to see your model incorporating the mass of the rope, and I would be interested to see, after the mass is properly taken into account, how much effect it actually has on the results. Please do post up. Its supposed to be too cold to climb this weekend, so I'll work on it. I'd expect that for a UIAA rope test, its pretty minor, but if you've got 100 feet of rope out, it might be pretty big (on the order of 5 or 10 percent difference). I'll try to work up some general equations and explanatory assumptions. 100 feet of rope weighs about 3 lb. I'm going to go out on a limb and say that rope mass can be ignored. Jay





mtnrock
Oct 8, 2008, 4:48 PM
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im trying to get a very rough idea but yea i for got that i could test for it on my own its just going to be the average k(spring constant) of my rope and to the one who said a good placed peace of gear will hold no matter what you might wana read whats on that paper you get with your gear it says what force it will fall at no matter how good the placement is





petsfed
Oct 8, 2008, 7:04 PM
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jt512 wrote: petsfed wrote: rgold wrote: Petsfed, there isn't a mathematical model in the world that incorporates every detail of the reality being modeled. If one did, it would in some sense be the reality and not a model of it. Perhaps the essential ingredient in every model is the fact that it ignores details, so your criticism is in fact a foregone conclusion. I'd love to see your model incorporating the mass of the rope, and I would be interested to see, after the mass is properly taken into account, how much effect it actually has on the results. Please do post up. Its supposed to be too cold to climb this weekend, so I'll work on it. I'd expect that for a UIAA rope test, its pretty minor, but if you've got 100 feet of rope out, it might be pretty big (on the order of 5 or 10 percent difference). I'll try to work up some general equations and explanatory assumptions. 100 feet of rope weighs about 3 lb. I'm going to go out on a limb and say that rope mass can be ignored. Jay You're right, it is pretty negligible, but the more rope you have out, the more measurable an affect it will have on the overall system. If you have all 200 feet out, that's nearly 10% of the UIAA test mass. And it definitely means that the spring constant is not constant per se, but rather a function of how much rope is out. I'm gonna try to develop it more than that and maybe produce some general equations so we can at least figure out limiting cases.





petsfed
Oct 8, 2008, 7:16 PM
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mtnrock wrote: im trying to get a very rough idea but yea i for got that i could test for it on my own its just going to be the average k(spring constant) of my rope and to the one who said a good placed peace of gear will hold no matter what you might wana read whats on that paper you get with your gear it says what force it will fall at no matter how good the placement is That's the rated breaking strength. The maximum allowable impact force on a UIAA rope is 12kN, and most ropes have a peak force of significantly less than that. In the limiting case of a factor 2 fall, then the force on the anchor system is 24 kN, enough to break a carabiner. But if your belay anchor consists of only 1 nut with only one carabiner and your leader falls clean past you with out sliding at all, then there's actually two nuts at the belay, if you follow my meaning. But that limiting case also presupposes absolutely no give in the system, which is simply unrealistic. It should explain why its a good idea to double up on pro before a long runout though.





jt512
Oct 8, 2008, 7:31 PM
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mtnrock wrote: im trying to get a very rough idea but yea i for got that i could test for it on my own its just going to be the average k(spring constant) of my rope and to the one who said a good placed peace of gear will hold no matter what you might wana read whats on that paper you get with your gear it says what force it will fall at no matter how good the placement is If you expect to get answers from educated adults, then I would suggest that you learn to use punctuation. You're not text messaging with adolescents. Jay





jt512
Oct 8, 2008, 7:41 PM
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petsfed wrote: jt512 wrote: petsfed wrote: rgold wrote: Petsfed, there isn't a mathematical model in the world that incorporates every detail of the reality being modeled. If one did, it would in some sense be the reality and not a model of it. Perhaps the essential ingredient in every model is the fact that it ignores details, so your criticism is in fact a foregone conclusion. I'd love to see your model incorporating the mass of the rope, and I would be interested to see, after the mass is properly taken into account, how much effect it actually has on the results. Please do post up. Its supposed to be too cold to climb this weekend, so I'll work on it. I'd expect that for a UIAA rope test, its pretty minor, but if you've got 100 feet of rope out, it might be pretty big (on the order of 5 or 10 percent difference). I'll try to work up some general equations and explanatory assumptions. 100 feet of rope weighs about 3 lb. I'm going to go out on a limb and say that rope mass can be ignored. Jay You're right, it is pretty negligible, but the more rope you have out, the more measurable an affect it will have on the overall system. If you have all 200 feet out, that's nearly 10% of the UIAA test mass. Less than 5%, unless I made an arithmetic error. Jay





mtnrock
Oct 8, 2008, 9:18 PM
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what im trying to figure out is the peak force that makes your protection fail not the rope. but are you saying that doesn't matter because the rope will break first. like it says the strength of a cam is 8kN does that mean if your fall generates a force higher then that it will fail?
(This post was edited by mtnrock on Oct 8, 2008, 9:20 PM)





hafilax
Oct 8, 2008, 9:26 PM
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If your cam feels an impact less than 8kN it won't fail. If the force is greater it might fail. The impact force of the rope is a rating that tells you how stretchy the rope is. The lower the impact force the stretchier it is. It has nothing to do with the rope breaking. If something is slowed down by a small force it comes to a stop further along than if a stronger force is applied. Similarly if the rope is stretchy and you fall further once it's under tension the force will be smaller.





mtnrock
Oct 8, 2008, 9:31 PM
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o got it. yea it was when i was finding an equation to find the force generated on the peace of equipment the one unknown i was missing was the spring constant but im just going to test for that on my rope unless theres a way buy using the impact force of the rope to find the spring constant(also shows how streachy the rope is)





