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jonathanjcooke
Feb 16, 2007, 9:43 AM
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Okay, so a factor 2 fall always creates the same amount of force, right? That's what I've been led to believe by what I've read, heard. If this is true, then a 100' fall onto 50' of rope would create the same amount of force (on the climber, piece(s) of gear holding the climb, belayer, etc.) as a 40' fall onto 20' of rope. This is true because the rope stretches proportionally to the length of the rope fallen on at a fixed percentage (excepting of loss of elasticity over time or not allowing your rope to rebound, etc.)? My further questions about fall factors are as follow: 1. Does the measurement, "fall factor," actually measure force or does it simply measure the potential force for a set of circumstances? 2. I know this would be hard to replicate in a practical climbing situation, but the theory applies. A 2-inch fall onto 1-inch of rope is technically a factor 2 fall. Does this produce the same amount of force as a 2-foot fall onto 1-foot of rope? What about a 20-footer onto 10-feet? Does another variable (for example, speed) determines the severity of a factor 2 fall? It seems that there would not be enough time (and thus, speed) in a 2-inch fall to generate a force significantly greater than one's bodyweight. 3. I found this web page that can calculate force of a falling object. It seems to account for all of the factors needed to determine the force created by a climber in a climbing fall. Is it accurate to use for calculating such forces (in theory, of course) when: m=mass of climber in kilograms h=length of the fall (before rope stretch) in meters d=rope stretch in meters The page will calculate the rest for you. http://hyperphysics.phy-astr.gsu.edu/hbase/flobi.html#c1%22 Thanks for reading this and thanks even more for any answers. Sincerely, Jon P.S. Does anyone know the specific details of the anchor-failure situation (where the climber was clipped into the anchor with a daisy chain and then fell on it) mentioned in the new edition of "Climbing Anchors" (John Long, Bob Gaines)?
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jonathanjcooke
Feb 16, 2007, 10:22 AM
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I read the other threads on fall factors and they resolved some of my confusion. However, I still find it hard to believe that a 2-inch fall onto 1-inch of rope would create extreme forces comparable to those created by a much longer fall. I guess I could try it. Would it break my pelvis? I'm definitely not trying it.
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cantbuymefriends
Feb 16, 2007, 10:46 AM
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jonathanjcooke wrote: 1. Does the measurement, "fall factor," actually measure force or does it simply measure the potential force for a set of circumstances?)? It measures the potential force for a set of circumstances. The actual force measured in lbf or Newtons is determined by: a) The weight of the climber (including gear, etc). A heavier climber puts more force in the system, but (s)he will generate a slower decceleration. b) The properties of the rope (stretch elongation) c) Rope slip in the belay device. d) Friction in the protection chain. (Not applicable if you fall straight onto the belay.)
jonathanjcooke wrote: 2. I know this would be hard to replicate in a practical climbing situation, but the theory applies. A 2-inch fall onto 1-inch of rope is technically a factor 2 fall. Does this produce the same amount of force as a 2-foot fall onto 1-foot of rope? What about a 20-footer onto 10-feet? Does another variable (for example, speed) determines the severity of a factor 2 fall? It seems that there would not be enough time (and thus, speed) in a 2-inch fall to generate a force significantly greater than one's bodyweight. Yes, yes and yes. Stop thinking velocity, and start thinking potential energy (=height)! Velocity is just a middle step in the transformation from potential energy to braking energy (=rope stretch). And the velocity does nothing but confuse by introducing square roots in the equations!
(This post was edited by cantbuymefriends on Feb 16, 2007, 10:49 AM)
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hillbilly
Feb 16, 2007, 2:22 PM
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Howdy Jon, perhaps a better way to look at the rope fall factor is, performance ability of your rope. Ropes and gravity are a mismatch of properties. Gravity supplies energy in the square of the time spent in a fall. Ropes supply a linear absorption of energy dependent on length of rope involved. You are right, the time spent in a 2 inch fall, 0.10 sec creates very little energy but the rope is unable to perform its linear magic. Remember the first second of air time in a fall will double your weight on the anchor. The following second will quadruple your weight. In the next second your weight explodes to 9x resting energy and force. Although a three second fall is unlikely, 144 feet, You could run out a climb and fall a full rope length, 300 feet. I have heard of it happening a few times. I crunched the numbers many years ago before the term fall factor entered our vocabulary. You can grind through the numbers yourself. Distance covered in a fall d=16t2, velocity at end of fall v=32t. I have seen variety of table and graphs that include biner drag, rope lengths, rope types.... As I have treated fall factor zeros before, high velocity ground falls are disturbingly memorable. but that is another forum.
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rocknice2
Feb 16, 2007, 2:57 PM
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cantbuymefriends wrote: jonathanjcooke wrote: 2. I know this would be hard to replicate in a practical climbing situation, but the theory applies. A 2-inch fall onto 1-inch of rope is technically a factor 2 fall. Does this produce the same amount of force as a 2-foot fall onto 1-foot of rope? What about a 20-footer onto 10-feet? Does another variable (for example, speed) determines the severity of a factor 2 fall? It seems that there would not be enough time (and thus, speed) in a 2-inch fall to generate a force significantly greater than one's bodyweight. Yes, yes and yes. Stop thinking velocity, and start thinking potential energy (=height)! Velocity is just a middle step in the transformation from potential energy to braking energy (=rope stretch). And the velocity does nothing but confuse by introducing square roots in the equations! This would be true on the moon but on Earth with an atmosephere, we run into treminal velocity. So whether you fall 100 meters or 1000 meters you'll leave the same crater. I think T/V is reached around _(edited)_ meters but I'm not 100% on that. An F2 fall from 1 inch has far less force then from 10 feet, but after a certain it stays constant. The higher you are over the pro the more potential energy. Air resitance is definitly a factor b/c it limits velocity.
(This post was edited by rocknice2 on Feb 16, 2007, 3:01 PM)
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redpoint73
Feb 16, 2007, 3:06 PM
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The best way to think of fall factor is not as a measure of force (which it is not at all), but an index of the severity of the fall. The forces are going to vary greatly depending on weight of climber, friction in the system, rope properties, and many other things. It also depends on whether you are talking about force felt by the climber, or at the anchor point. The case of the short factor-2 fall versus a longer one is easily explained. The more rope you have out from the belayer, the more rope there is to stretch and absorb the force. This principle makes a lot of sense if you take a fall on the first bolt of a sport climb, versus a fall on the 7th. You will feel much more of a jolt on the lower fall, and the main reason for this is the amount of rope out. This is an example of a somewhat high fall factor versus very low fall factor situation, but helps explain how rope stretch has so much to do with fall factor.
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cracklover
Feb 16, 2007, 4:30 PM
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Fall Factor will tell you only one thing: the peak tension in the rope. That translates to peak force on the body and peak force on the top piece. If you removed all other factors, then yes, a 2" fall on 1" of rope would be approximately identical to a 20' fall on 10' of rope in terms of the peak tension in the rope. However, all other factors are not removed. For example, the knot can absorb a very small amount of energy. If there is only a very small amount of energy in the fall, the knot can absorb a large proportion of it. So a 2" fall on 1" of rope will mostly just tighten the knot. That means a significantly smaller proportion of energy that needs to be absorbed by the rest of the rope. However, take a 20' fall on 10' of rope, and the tiny amount of energy the knot can absorb will have no appreciable effect, because it's such a small proportion. Does that help you understand? GO
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cracklover
Feb 16, 2007, 4:45 PM
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Here's another way of looking at it: Think of the force absorbed by other elements as though they were more little sections of rope. Because functionally, that's what they are. So, for example, you could think of your fig-8 tie-in knot and the foot of rope in a loop behind it as though they acted as another 18 inches of rope. If you have a mammut sling at the top instead of just a locking carabiner, perhaps add a 1/2 inch of rope. If you have a nylon sling, perhaps add 2 inches of rope. Okay, now let's do the calculation: You fall 2 inches on 1 inch of rope. With the nylon sling at the top (2 extra inches) and the tie-in knot (18 extra inches), that's really like factor 0.1 fall. 2' on 1' of rope with those same extra factors: FF = 0.75 4' on 2' of rope: FF = 1.1 10' on 5' of rope: FF = 1.5 Etc. So you see, as the amount of rope increases, so the effect of the other energy absorbers decreases. So you're kind of right. But from a fall on around 5' of rope or more, you're effectively feeling the full force of the fall, as if those mitigating factors weren't there. Hope that helps. GO
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112
Feb 16, 2007, 4:50 PM
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jonathanjcooke wrote: Okay, so a factor 2 fall always creates the same amount of force, right? Not true. But what is true , ALWAYS, is that for any given 'set-up', increasing the fall factor increases the forces involved in the rope system. It is interesting that Fall Factor is a dimensionless number, like Mach, Reynolds, etc. I just wonder what 'other' dimensionless numbers we shhould/could be using. Because as others have pointed out in other threads, a FF = 2 is NOT constant over all 'set-ups'. RGold or someone really smart like that might know?
(This post was edited by 112 on Feb 16, 2007, 4:51 PM)
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cracklover
Feb 16, 2007, 4:51 PM
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Oh, and peak force (which is all the FF can tell you) doesn't tell you the whole story, either. Because how fast that peak force comes on, I suspect, makes a bit of a difference in how your body can respond. Take a massive force that comes on slowly, and I suspect your bones can deform a bit before they break. But that same exact force in a sharp burst, would probably cause the bones to shatter more easily. So, in my opinion, all other things being equal, a longer fall of the same FF is probably less damaging to bones than a shorter one. By the way, you know that in real life, the rope running over rock and around biners affects the amount of rope that can really be effective in absorbing the force of the fall, thus affecting the practical FF? GO
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trenchdigger
Feb 16, 2007, 5:14 PM
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Given a set of assumptions: - Rope is miraculously connected at each end directly and without a knot. - The object (anchor/climber) at each end does not "give" at all to absorb energy. - You perform the test in a vacuum. - The rope stretches "ideally" with an unchanging spring constant. - The climber's weight is a constant. A FF2 fall regardless of length - whether it's a 2in. fall on 1in. of rope or a 200ft fall on 100ft of rope - will theoretically generate the exact same peak force (tension) in the rope. The amount of energy absorbed by the rope, however, varies proportionally to the length of the fall. As others have pointed out, there are many other factors involved, including knots, anchor materials, your harness, belay device slip, an even your body's movement to help absorb impact. Those are the reasons an actual 2" fall with 1" of rope between you and the anchor won't feel anything like a longer FF2 fall.
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shockabuku
Feb 16, 2007, 5:25 PM
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I think the fall factor is an ideal generalization and can be thought of as modeling the rope as a spring. It's a generalization because different ropes will have different elastic properties, i.e. spring constant, that determines how much it will stretch and how much energy it absorbs per unit length of stretch. It's ideal (though in the FF2 case it's more accurate because some of the following factors don't apply) because it doesn't take into account real world factors such as friction over the biner, friction over the rock, knots tightening, belayer giving a dynamic catch, etc. The fall factor gives you an easy value to assess in terms of the danger involved in a fall but, as has been stated, there are so many other variables in a real world situation that it can be somewhat misleading.
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greenketch
Feb 16, 2007, 6:23 PM
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I agree that FF is the "ideal" number to be concerned about. As has been looked at from severeal directions now FF is part of a much larger equation. Of the component parts we have very little control. We can't adjust gravity, there is only partial control over the time over which the force is disapated(via charecteristics of the rope you choose), but, FF we have direct control of and it also has a huge affect on the force equation. Unless you choose to make a laptop part of your rack and strart calculating optimal placments it is way easier to just consider FF as you go.
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jonathanjcooke
Feb 16, 2007, 6:35 PM
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"Peak force" has been mentioned a couple of times. How does it differ from general force? -Jon
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112
Feb 16, 2007, 6:37 PM
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I think I meant non-dimensional. Buckyham's Pie Therom (sp) and stuff like that. Summary: Dimensions are man made; God doesn't use man's dimesnsions while controlling the Universe. Anyone know of a 'Moody' chart for climbing?
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jonathanjcooke
Feb 17, 2007, 4:23 PM
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Okay, so say that we have a 2-foot factor 2 fall and a 20-foot factor 2 fall, both directly onto the belay. The climber in each fall weighs exactly the same and there is absolutely no slippage through a belay device. Let's just say (for both falls) that the climber is tied directly into the anchor and that the knot is as tight as it can be so there would be no shock absorption from the knot (the same for the knot at his harness). Also, the properties of the rope are identical for both falls. So: 1. Weight of the climber is constant for both falls 2. The properties of the rope are constant 3. No slip in the belay 4. No friction in chain, because he falls directly onto the anchor. 5. No shock absorption by knots. Now, how would the forces differ between the 2-footer and the 20-footer. Because they are both factor 2, they would be considered equally "severe;" but how can falls with different forces be equally severe? I've heard a lot of people say that factor-2 measures how a climber (or/and anchor?) feels the force of a fall more than the force itself, but isn't a climber always feeling a percentage of the total force created by the fall (total force minus that which is absorbed by the rope in these instances)? So a fall that creates more force would put more strain on the climber and anchor, right? So if the 20-footer creates more force, it should be felt more severely by the climber and anchor. Of course, the rope stretches more in a longer fall, but it is still absorbing a fixed percentage of the force of the fall, right? The amount of force absorbed by the rope doesn't increase exponentially as the length of the fall increases, right? So why would the 20-footer be considered as severe as the 2-footer in terms of fall factor? Does fall factor just mean-most severe fall possible for the specific amount of rope out? Thanks, Jon
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shockabuku
Feb 17, 2007, 5:21 PM
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In reply to: Because they are both factor 2, they would be considered equally "severe;" but how can falls with different forces be equally severe? The point of the fall factor is that falls with the same fall factor have, ideally, the same maximum force. You're assuming that because the falls are of different length that they will require a different force to stop them. That's not true. The most simple way to consider this is that as the body falls it builds up energy. That energy is absorbed by the rope stretching. The more rope there is to stretch, the more energy the rope can absorb at the same maximum force and arrest the fall over a longer distance. The maximum force is determined by how quickly the body is stopped. In a short fall the body is stopped relatively quickly (and in a short distance)but moves at a slower speed; in a long fall the body is moving faster but takes more time to come to rest This isn't a perfect example, but consider dropping a baseball into the water. The higher you drop it from, the deeper it goes. The water provides a bouyant force to stop the ball. That bouyant force (except for when it breaks the surface) has the same value regardless of the depth to which the ball descends - it just takes longer to stop it when it has more energy from falling from a greater height.
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sactownclimber
Feb 17, 2007, 5:21 PM
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112 wrote: jonathanjcooke wrote: Okay, so a factor 2 fall always creates the same amount of force, right? Not true. But what is true , ALWAYS, is that for any given 'set-up', increasing the fall factor increases the forces involved in the rope system. Actually, it is true. Peak force is modeled by the following equation: T = mg + ((mg)^2+2kmgr)^1/2 where m is the mass of the climber, g is the gravitational constant (9.8 m/s^2), k is the rope's modulus, and r is the fall factor. Fixing m and k, then T is only dependent on r . . . meaning it doesn't matter if the fall is a 10' FF2 or a 10" FF2 fall, the peak force on the rope is the same.
cracklover wrote: Because how fast that peak force comes on, I suspect, makes a bit of a difference in how your body can respond. Your intuition is right, but your explanation isn't quite right. F = ma, which means that it doesn't matter if it's a 10' FF2 or a 10" FF2, the (negative) acceleration is the same. The time required to come to a complete stop is dependent on the initial velocity (a = delta(v)/t), but that doesn't change the peak force felt by the rope, because we've fixed the peak force by assuming that the fall is FF2. However, your intuition is correct in the sense that if there is more rope in the system, this increases the amount of time it takes the rope to absorb your kinetic energy . . . but this is the same thing as saying the peak force is only dependent on fall factor. I've attached Rgolds derivation of the peak force equation.
(This post was edited by sactownclimber on Feb 17, 2007, 6:01 PM)
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StandardEqn.pdf
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sactownclimber
Feb 17, 2007, 6:03 PM
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ahh, 112, went back and read your post and realized we were saying the same thing . . . that given a fixed mass and rope, peak force is only dependent on fall factor (not taking into account other factors that others have mentioned). -J-
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cantbuymefriends
Feb 17, 2007, 6:42 PM
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rocknice2 wrote: This would be true on the moon but on Earth with an atmosephere, we run into treminal velocity. So whether you fall 100 meters or 1000 meters you'll leave the same crater. I think T/V is reached around _(edited)_ meters but I'm not 100% on that. An F2 fall from 1 inch has far less force then from 10 feet, but after a certain it stays constant. The higher you are over the pro the more potential energy. Air resitance is definitly a factor b/c it limits velocity. Well, if you factor in the air resistance, then a 100 m FF2 fall (or 1000 m) would actually put LESS force on the anchor than a 1 inch fall... But I'm uncertain that you would come anywhere near "terminal velocity" even if you fall a full rope-length (2*60 or 2*70 m)?
(This post was edited by cantbuymefriends on Feb 17, 2007, 6:49 PM)
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rgold
Feb 17, 2007, 7:02 PM
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I don't know if I can add anything useful to this discussion. What follows is long enough that it would be sad if it wasn't helpful, but of course utility is determined by the reader, not the author. Part of the problem in this and most such discussions is that there are at least two different (correct) viewpoints floating around as well as the usual scattering of misunderstandings, most of which stem from viewing energy and force as equivalent. The concept of fall factor emerges from an ideal mathematical model of the rope as a spring (sactownclimber attached my derivation of the consequences of that model a few posts earlier). So when discussing questions about the fall factor, it is essential to specify whether we are speaking about consequences of the mathematical model, or about how that model deviates from reality. The fundamental condition in the mathematical model that makes fall factor emerge as the sole parameter determining the peak tension in the rope is Hooke's Law, in the form that the tension in the rope is proportional to its relative stretch. (This is the absolute amount of stretch divided by the original length of rope before stretching, typically expressed as a percentage.) In this ideal set-up, a piece of rope that stretches 20% will develop the same tension, regardless of whether it is 1 inch long or 100 feet long. An analysis, either by conservation of energy or by way of the differential equation for simple harmonic motion (I think my attachment has both approaches) shows that the maximum relative stretch in the rope depends only on the H/L ratio that defines fall factor, and in this way the fall factor determines the peak tension in the rope and so the maximum loads experienced by falling climber and the protection system. So: in the ideal world of the model, you get the same maximum tension in the rope if you fall 2" on a 1" piece of rope or 100 feet on a 50 foot piece of rope. End of story, as far as it goes. The model (any model) is only an approximation of reality. (We use models precisely because they ignore some of the details of reality.) A number of people have mentioned things that change the conclusions of the model, but I think almost all of them are items, like energy absorbing knots and protection slings and friction in the system, that were never part of the model to begin with. More sophisticated mathematical models have been proposed for including many of the proposed reality items. In the proposals I've seen, the fall factor is still a critical parameter, but it is no longer the only parameter. The fundamental reason for this is that other reality elements add energy-absorbing mechanisms that are not governed by Hooke's Law. A good example of this is the screamer. It is capable of removing a fixed amount of fall energy. The effect it can have in reducing the load on pro therefore depends on the total fall energy, which is in turn proportional to the height of the fall. So a screamer's ability to mediate fall impacts decreases as the height of the fall increases, even if the fall factor remains constant. In other words, the fall factor is no longer the sole determinant of maximum rope tension. So it is one thing to point out inaccuracies in the ideal model that the ideal model never tried to account for. It is another thing to point out that the assumptions of the ideal model are not satisfied in reality. Perhaps the most central issue is the extent to which real ropes actually satisfy Hooke's Law. Hooke's law says that the graph of rope tension vs. relative stretch is a straight line (whose slope is the proportionality constant, sometimes called the rope modulus.) In reality, rope (and other material) behavior can be divided into an elastic range and and a plastic range that occurs after a certain maximum level of stretch has been exceeded. Of course, Hooke's Law only applies to the elastic range of deformation. This isn't a serious problem for climbing falls, which are in the elastic range of our ropes. But even within this elastic range, the real graph of tension vs. relative stretch is not a straight line but rather more of an "S"-shaped curve with a relatively straight middle section and levelling curves at both ends. This means that models based on Hooke's law are most appropriate for the middle range of stretching. Using them for extreme extensions requires that a straight line be used to approximate the actual tension-extension curve, and how this straight line is chosen will determine the accuracy of the predictions. If, as is typical, the straight line is chosen from the UIAA impact data, then predictions based on it will be too large for the range of relative extensions one cares about. (How far off depends on the actual shape of the actual tension-extension curve.) This is a possibly substantial flaw in most of the fall calculators and indeed in practical formulas I included in the notes I wrote up. Now getting back to the original question, the postulate that rope tension depends on relative stretch may itself be inaccurate. For example, any given fixed length of rope could be governed by Hooke's Law, but different lengths might involve different constants of proportionality. Something like this is probably true for ropes. If you take a short enough piece, the same relative stretch may result in higher a higher tension than a longer piece would experience. I have seen some tests somewhere that suggest this is true, but I don't remember the differences being very large in percentage terms. Although I've never seen a precise analysis, most authors seem to assume that the relative stretch form of Hooke's law will operate without serious modification once you have a few feet of rope in the system. For very short falls on even shorter pieces of rope, the ideal model may be underestimating the impact levels. There are several other linearity assumptions in the standard mathematical model that might not prove to be true. But it does seem that the basic model has done well in estimating impact loads for a broad range of situations.
(This post was edited by rgold on Feb 18, 2007, 5:13 PM)
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cantbuymefriends
Feb 17, 2007, 7:18 PM
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jonathanjcooke wrote: Now, how would the forces differ between the 2-footer and the 20-footer. Because they are both factor 2, they would be considered equally "severe;" but how can falls with different forces be equally severe? I've heard a lot of people say that factor-2 measures how a climber (or/and anchor?) feels the force of a fall more than the force itself, but isn't a climber always feeling a percentage of the total force created by the fall (total force minus that which is absorbed by the rope in these instances)? So a fall that creates more force would put more strain on the climber and anchor, right? So if the 20-footer creates more force, it should be felt more severely by the climber and anchor. Of course, the rope stretches more in a longer fall, but it is still absorbing a fixed percentage of the force of the fall, right? The amount of force absorbed by the rope doesn't increase exponentially as the length of the fall increases, right? So why would the 20-footer be considered as severe as the 2-footer in terms of fall factor? Does fall factor just mean-most severe fall possible for the specific amount of rope out? Thanks, Jon The rope doesn't absorb force. It just transfer force between the belay device and the falling climber. The rope absorb energy. Fall factor is a measure of the severity of a fall. And fall factor 2 is in most practical situations the most severe fall possible. You are above your belayer with a somewhat tight rope, you fall off, there's no protection to catch you so you fall straight onto the belay, The rope will go tight and start to brake your fall at about the same distance below your belayer that you were above him when the fall started. Right? (Unless the ground or a ledge stops you first...) There are however some situations where you can get a fall factor >2, a belayer that is reeling in rope through the belay device during the fall is one example.
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greenketch
Feb 17, 2007, 9:42 PM
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jonathanjcooke wrote: So: 1. Weight of the climber is constant for both falls 2. The properties of the rope are constant 3. No slip in the belay 4. No friction in chain, because he falls directly onto the anchor. 5. No shock absorption by knots. Now, how would the forces differ between the 2-footer and the 20-footer. Because they are both factor 2, they would be considered equally "severe;" but how can falls with different forces be equally severe? Jon, The quick and dirty answer is you missed a factor. You are correct in assuming that one person will have accumulated more energy than the other in their fall. But just as one took more time to get there they will also disapate the energy over more time. The catch (or the stop) at the end of a 2 foot fall will be pretty quick and will feel harsh. The catch at the end of the 20footer will be much more sustained and will feel fairly soft. The instantaneous force is the same either way but it is the time over which it is disapaited that keeps them identicle.
(This post was edited by greenketch on Feb 18, 2007, 2:06 AM)
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112
Feb 21, 2007, 12:02 AM
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sactownclimber wrote: ahh, 112, went back and read your post and realized we were saying the same thing . . . that given a fixed mass and rope, peak force is only dependent on fall factor (not taking into account other factors that others have mentioned). -J- I was attempting to capture ALL factors, other than fall height, with the term 'set-up'.
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cracklover
Feb 21, 2007, 4:30 PM
Post #25 of 27
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Registered: Nov 14, 2002
Posts: 10162
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greenketch wrote: jonathanjcooke wrote: So: 1. Weight of the climber is constant for both falls 2. The properties of the rope are constant 3. No slip in the belay 4. No friction in chain, because he falls directly onto the anchor. 5. No shock absorption by knots. Now, how would the forces differ between the 2-footer and the 20-footer. Because they are both factor 2, they would be considered equally "severe;" but how can falls with different forces be equally severe? Jon, The quick and dirty answer is you missed a factor. You are correct in assuming that one person will have accumulated more energy than the other in their fall. But just as one took more time to get there they will also disapate the energy over more time. The catch (or the stop) at the end of a 2 foot fall will be pretty quick and will feel harsh. The catch at the end of the 20footer will be much more sustained and will feel fairly soft. The instantaneous force is the same either way but it is the time over which it is disapaited that keeps them identicle. I don't know what you meant by the sentence I italicized. There is no "instantaneous" force. The total energy in the short fall and the total energy in the long fall are quite different. What's (approximately) the same is the peak force in each fall. What determines the peak force is simply the maximum amount the rope stretches, measured by percent. What you need to understand is two things: 1 - A rope that's stretched 30% of its length will always have the same amount of tension at the ends, and that tension is what's putting force on the gear. 2 - More rope can absorb more energy. Specifically, 10 feet of rope stretched 30% can absorb 10 times as much energy as 1 foot of rope stretched 30%. So if the small amount of energy in the short fall is able to stretch 1 foot of rope 30% and the larger amount of energy in the larger fall is able to stretch 10 feet of rope 30%, the tension in either rope stretched 30% is, of course, identical. Thus the peak force on the gear and the climber is the same. Skip the next two paragraphs if you don't like math and physics. They'll just confuse you, and they're not central to the explanation. **** Okay, just as a dummy check, to see if what I'm saying holds true mathematically, I just ran some quick physics. If what I'm saying above is true, then the energy absorbed by the rope after a 20 foot fall must be exactly ten times the energy absorbed by the rope after a 2 foot fall. You follow? Because there's ten times as much rope out, and it's converting all the kinetic energy of the fall into heat, and stretching by the same percentage in each fall. Well it turns out that kinetic energy is proportional to the square of the speed the object is travelling. Okay, so if a body falls 2 feet, it will be going at about 3.5 meters per second. Square that, and you get 12.2. Now the body that falls 20 feet will be going around 11 meters per second. Square that, and you get 121. Woo hoo, the second one is ten times the first, the dummy check passes! End of math section. **** Hope that helps! GO
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