
brenta
Dec 18, 2008, 10:48 AM
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rgold wrote: Presumably, the manufacturer has some ability to "tune" this curve, I'd be surprised to learn they have the level of control suggested by JT. There may have been a bit of didactic oversimplification there. That comment, though, and your response, got me thinking. What is the lowest achievable impact force? That is, if a truly flat force curve were possible, what would the value be? If I'm not mistaken, letting f be the fall factor, e the fractional elongation, m the mass of the climber and g the usual acceleration due to gravity, this limit force is mg(f/e+1). For f=1.78, e=0.4, m=80 kg, and g=9.81 m/s^2, we get 4.28 kN. So, that's what we would see on a single rope tag if its designers had full control on the force it produces and chose to allow maximum elongation.





ptlong
Dec 18, 2008, 11:05 AM
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brenta wrote: What is the lowest achievable impact force? That is, if a truly flat force curve were possible, what would the value be? If I'm not mistaken, letting f be the fall factor, e the fractional elongation, m the mass of the climber and g the usual acceleration due to gravity, this limit force is mg(f/e+1). For f=1.78, e=0.4, m=80 kg, and g=9.81 m/s^2, we get 4.28 kN. So, that's what we would see on a single rope tag if its designers had full control on the force it produces and chose to allow maximum elongation. Wouldn't that force be applied over the distance of rope elongation? I think that means mgf(1+e)/e, or 4.9 kN. Either way I wouldn't want to even toprope on this rope.





JimTitt
Dec 18, 2008, 11:09 AM
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Since both a) and b) have the same impact force then ths catch will be equally as hard, but don´t forget that the greater the elongation the further the climber has fallen so he has gained more energy. So to stop him you will have to apply your 8kN for longer. In real life it doesn´t matter!
In reply to: So then is the ultimate goal to choose a rope that has both the lowest impact force and the lowest dynamic elongation? Because after all there is no reason to buy a rope the stretches more and has the same impact force because you just increase the chance of falling on an object. Right? Like always, there is no free ride. Like you say the goal is the lowest impact and lowest stretch (less chance of hitting anything like the ground). But we also want light, easy handling, durable and cheap! You can get most of these in one rope but probably not all at once, especially the cheap. Maybe a rope company techy can come and tell us all about it, they are incredibly clever guys and get paid fantastic moneyjavascript:%20addTag(':)')





brenta
Dec 18, 2008, 11:54 AM
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ptlong wrote: Wouldn't that force be applied over the distance of rope elongation? I think that means mgf(1+e)/e, or 4.9 kN. Here's my reasoning. Let h be the height of the fall. Then the rope that arrests the fall is l=h/f. The rope stretch is s=le =he/f; hence, h=sf/e. The energy balance equation reads: mg(h+s)=Fs, where F is the constant force whose magnitude we seek. Plug in the value of h and get mgs(f/e+1)=Fs, whence the result I gave.





cracklover
Dec 18, 2008, 12:39 PM
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USnavy wrote: JimTitt wrote: Rope a) could be constructed so that the modulus rises very rapidly to give say a force of 5kN, the rope is also constructed so that the modulus than stays constant so the force then continues to remain at 5kN. Rope b) could be constructed to have a constant modulus giving a force of say 0,1kN and then rises very rapidly. Rope a) never produces a higher force (impact) than 5kN but has worked at stopping the fall all the time so the elongation will be relatively low. Rope b) has not done any work to stop the fall until the last moment when it then has to apply a much higher force. A good way to think about this is to consider driving a car. A smooth driver applies the brakes consistently until the car come to a halt, a poor driver brakes too gently at first so at the last moment they have to stand on the pedal, throwing you into the windscreen. Both have braked from the same speed and for the same distance but the peak forces are wildly different. I would just look at the areas under the force/time graphs which represent the work done by the rope stopping the fall and then see where the elongation or max force lies but these graphs are not normally available. Personally I just buy a strong, cheap rope and don´t fall off too often! Ok, that makes sense. I now understand where your coming from and finally understand the answer to my question. But that leads to another question. Say you have two ropes: Rope A: Impact force: 8 kN Dynamic Elongation: 35% Rope B: Impact force: 8 kN Dynamic Elongation: 30% Which rope will provide a softer catch? Will rope “a” provide a softer catch because it slows the load down over more time or will it provide a harder catch because the first few percent of the rope stretch will not be doing any load dispersion work ultimately leavening a lower percentile of rope stretch then rope “b” in which to disperse the load? So then is the ultimate goal to choose a rope that has both the lowest impact force and the lowest dynamic elongation? Because after all there is no reason to buy a rope the stretches more and has the same impact force because you just increase the chance of falling on an object. Right? Rope a and b will provide equally soft catches. The softness of the catch is exactly what the impact force number on the rope tells you. This is the peak load on you, the climber. And it has a direct relationship to the peak load on the top piece of gear. The elongation will tell you how far it stretches. Why is this difficult for you to understand? GO





rgold
Dec 18, 2008, 12:40 PM
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Not sure what you mean by a "truly flat" force curve, but it seems from your calculation that you are assuming the rope provides constant tension regardless of the amount of elongation, a concept that is slightly problematical, for example, when the elongation is zero, and I don't think (without however doing any checking) that the concept of the fall factor emerges from such a model. If you mean that the force curve is really a straight line but not necessarily a horizontal line, then the work done in stretching the rope (i.e. the righthand side of the energy balance equation) is not Fs as in your equation, but rather (k/2L)s^2, where K is the rope "modulus." I have to run now, so can't compute, but it seems to me that by letting K approach zero, maximum tension in the rope would approach 2mg, of course at the expense of an unbounded amount of extension.





ptlong
Dec 18, 2008, 1:02 PM
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rgold wrote: Not sure what you mean by a "truly flat" force curve, but it seems from your calculation that you are assuming the rope provides constant tension regardless of the amount of elongation, a concept that is slightly problematical, for example, when the elongation is zero, and I don't think (without however doing any checking) that the concept of the fall factor emerges from such a model. I'm pretty sure that's what brenta meant. I was worried that a toprope fall would result in a 1000 lb impact. What hadn't occurred to me was that just hanging on the rope would propel you upward! Something about thremodynamics probably prevents this scenario. [brenta: my bad with the math]
(This post was edited by ptlong on Dec 18, 2008, 1:02 PM)





brenta
Dec 18, 2008, 2:28 PM
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I should have said: constant force during the arrest. Before the rope comes taut the force would be zero. After the climber has come to a stop, the force would equal the weight of the climber. My concern was not the practical realizability of a rope with such behavior. I was interested in a lower bound of the impact force that would tell us how far current ropes are from the limit dictated by fact that work is the integral of force over distance.





ptlong
Dec 18, 2008, 5:35 PM
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No rebound? Sounds like a screamer with a 4.28kN activation force.





brenta
Dec 19, 2008, 5:38 AM
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We can add a little rebound if you wish. We are working with an imaginary rope after all. The whole point was to see how far the impact forces of real ropes are from the lower bound. For instance, the Tendon Master 9.2, with its impressive 6.8 kN impact force and 35% elongation, is only 42% above the theoretical limit. The Sterling Evolution Velocity is 45% above the limit thanks to low elongation (26.4%), and so on. If, on the other hand, someone came to you with talk of graphene, smart materials, nanotransducers, and told you they've developed an ultralight rope with an impact force of 3.8 kN, you'd know your leg was being pulled. Along the same lines, one can compute the impact force one would get if the rope behaved like an ideal spring. It would be exactly twice that of the ideal rope. (The height of a triangle with the same base and area as a rectangle is twice the height of that rectangle.) The rope modulus would be 2mg(f/e+1)/e. So, for 40% elongation we obtain an impact force of 8.55 kN, for 30% we obtain 10.88 kN, and we see in general that most ropes sold today comfortably outperform the ideal spring.
(This post was edited by brenta on Dec 19, 2008, 7:37 AM)





ptlong
Dec 19, 2008, 11:35 AM
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I understood your intent. I wasn't convinced that it made physical sense, and hence might not be a valid lower limit, but now I see that it does. Two long parallel chains of offtheshelf screamers would just about do it (minus the obvious disadvantages). But if instead of passive materials you have a rope that reacts intelligently then that opens up all sorts of possibilities. Certainly it could be smart enough to give you a softer catch than 4.28 kN for shorter falls. And a claim of 3.8 kN max impact force could no longer be dismissed as impossible.





brenta
Dec 19, 2008, 5:16 PM
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ptlong wrote: But if instead of passive materials you have a rope that reacts intelligently then that opens up all sorts of possibilities. We have perhaps reached the point where the "rules of the game" need to be made reasonably precise. I assume that the fall is a "UIAA fall." Specifically, that the length of the unstretched rope available to arrest the fall is 1/f = 1/1.78 times the height of the free fall and that the rope does not stretch more than 40% during the arrest. The rope exerts no force on the falling mass before coming taut and does not change length until then. This assumption fixes the amount of energy to be converted per unit length to mg(f+e). Change the rules of the game to allow, for example, a rope to shrink during the freefall portion of the fall, and the lower bound does not apply any more. However, any rope, no matter how hightech it may be, must perform the same work on the falling mass to stop it in the same distance from the same initial speed. A constant force results in minimum peak force. Hence, when the above rules are followed, the lower bound mg(f/e+1) applies to smart and dumb ropes alike.





USnavy
Dec 21, 2008, 8:37 AM
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cracklover wrote: USnavy wrote: JimTitt wrote: Rope a) could be constructed so that the modulus rises very rapidly to give say a force of 5kN, the rope is also constructed so that the modulus than stays constant so the force then continues to remain at 5kN. Rope b) could be constructed to have a constant modulus giving a force of say 0,1kN and then rises very rapidly. Rope a) never produces a higher force (impact) than 5kN but has worked at stopping the fall all the time so the elongation will be relatively low. Rope b) has not done any work to stop the fall until the last moment when it then has to apply a much higher force. A good way to think about this is to consider driving a car. A smooth driver applies the brakes consistently until the car come to a halt, a poor driver brakes too gently at first so at the last moment they have to stand on the pedal, throwing you into the windscreen. Both have braked from the same speed and for the same distance but the peak forces are wildly different. I would just look at the areas under the force/time graphs which represent the work done by the rope stopping the fall and then see where the elongation or max force lies but these graphs are not normally available. Personally I just buy a strong, cheap rope and don´t fall off too often! Ok, that makes sense. I now understand where your coming from and finally understand the answer to my question. But that leads to another question. Say you have two ropes: Rope A: Impact force: 8 kN Dynamic Elongation: 35% Rope B: Impact force: 8 kN Dynamic Elongation: 30% Which rope will provide a softer catch? Will rope “a” provide a softer catch because it slows the load down over more time or will it provide a harder catch because the first few percent of the rope stretch will not be doing any load dispersion work ultimately leavening a lower percentile of rope stretch then rope “b” in which to disperse the load? So then is the ultimate goal to choose a rope that has both the lowest impact force and the lowest dynamic elongation? Because after all there is no reason to buy a rope the stretches more and has the same impact force because you just increase the chance of falling on an object. Right? Rope a and b will provide equally soft catches. The softness of the catch is exactly what the impact force number on the rope tells you. This is the peak load on you, the climber. And it has a direct relationship to the peak load on the top piece of gear. The elongation will tell you how far it stretches. Why is this difficult for you to understand? GO lmao, you are the one that is not understanding . I am not asking what impact force or elongation is. Anyway Jim answered my question.





cracklover
Dec 22, 2008, 10:10 AM
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How am I the one who doesn't understand. You wrote:
In reply to: But that leads to another question. Say you have two ropes: Rope A: Impact force: 8 kN Dynamic Elongation: 35% Rope B: Impact force: 8 kN Dynamic Elongation: 30% Which rope will provide a softer catch? Will rope “a” provide a softer catch because it slows the load down over more time or will it provide a harder catch because the first few percent of the rope stretch will not be doing any load dispersion work ultimately leavening a lower percentile of rope stretch then rope “b” in which to disperse the load? Your question itself showed a lack of understanding. Forgive me for trying to help you understand. GO





