
jrathfon
May 13, 2009, 4:48 PM
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desertwanderer81 wrote: majid_sabet wrote: I was only saying in bold that I do get an enjoyment out of messing with other people who disrespect me and I was not referring to you at any shape or forml . I also wanted to ask you about the falling climber while loading the piece below and you kind of answered my question.that is all That's the thing though! The only way you EVER mess with people is by filling thread after thread with disinformation. Your understanding of math, physics, and even climbing is limitted to the point that you should never, ever comment on anything as almost without fail, it will be wrong! AMEN!!! But somehow, I don't think he'll get the message... edit: quotes and PTFTW!!!!111one!elevens!!!
(This post was edited by jrathfon on May 13, 2009, 4:49 PM)





ptlong
May 13, 2009, 4:50 PM
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shockabuku wrote: hafilax wrote: You will never measure the force calculated from a unmodified fall factor 2 calculation in a climbing situation. That's a rough sentence. What does it mean? Is halifax saying it's impossible to get 10kN tension in a rope in a climbing situation? Or what??





desertwanderer81
May 13, 2009, 4:53 PM
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hafilax wrote: Maybe it comes from years of marking 1st year physics labs, but I don't find Majid to be as 'wrong' as the haters make him out to be. You just have to filter out the trolling. What?? How? These are pretty basic concepts that he's butchering. If I had an engineer working for me who had his understanding of the world, I'd fire his ass in a heartbeat.





hafilax
May 13, 2009, 4:53 PM
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Try 2: Real world forces will never approach the value calculated with the fall factor formula.





majid_sabet
May 13, 2009, 4:54 PM
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I am not giving any wrong info. I am providing related info to the topic and whatever I have said so far, it was related to topic. If this information is wrong then people should bring it up so everyone can learn (that includes me) from it as true educational discussion. I will not tolerate against people who continue calling me names or take their personal problems and use these forums to become even with me over some unknown old business. If those sort of people want to use that sort of techniques or behavior to prove their points then I grantee you, I would the worse as*hole you guys ever wanted to deal with. peace and lets get back on topic
(This post was edited by majid_sabet on May 13, 2009, 5:11 PM)





ptlong
May 13, 2009, 4:55 PM
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hafilax wrote: Try 2: Real world forces will never approach the value calculated with the fall factor formula. Why not?





jt512
May 13, 2009, 4:56 PM
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colatownkid wrote: the significance of the fall factor becomes evident when one realizes that the stretch in dynamic rope serves to dissipate energy in the system and absorb some of the force. The stretch in the rope reduces the impact force (relative to a less stretchy rope). It does not "absorb" the force. This is not a pedantic distinction, as you'll see presently.
In reply to: if rope were totally static, the force of a fall would simply be a matter of distance fallen. the greater the distance fallen the greater the gain in velocity overtime, and the greater total force. That is false. If the rope were truly completely static, then the force of any fall would be infinite, and completely independent of the distance fallen. The longer the fall, the greater the (kinetic) energy, not the impact force. You seem to be implying that force builds up in a fall. It doesn't. The only force acting on a falling climber is gravity.
In reply to: what a fall factor DOES NOT do, is tell us anything specific about a definite peak load. No, but it's trivial to calculate the peak force. All we need to know is the UIAA impact force rating of the rope (which allows us to calculate the rope's modulus of elasticity), the weight of the climber, and the fall factor. If the rope has been clipped through an anchor, we can add in the "pulley effect" after adjusting for friction through the anchor. The result will be the maximum impact force assuming a static belay and no other frictional forces in the system.
In reply to: the one thing the fall factor does not tell us is any value in kilonewtons that describes the max force in the system. Right, but it's easy enough to calculate, given the assumptions above.
In reply to: majid is correct in stating that the friction over the carabiner does reduce some of the force. Majid's statement was confusing, and possibly outright wrong. Many climbers misunderstand the effect of friction in the belay system. All friction in the system is a braking force. It increases the impact force on the climber, and decreases the impact force on the belayer. Some people speak of friction increasing the "effective" fall factor, because friction reduces the stretch in the rope, much as if there were less rope out in the first place.
In reply to: (one final note: the fall factor is inherently an approximation since we all know from experience that rope stretch makes for a longer fall than 20 feet when 10 feet of rope is out.) No. The fall factor is defined as the distance the climber falls with no stretch in the rope divided by the amount of unstretched rope out. Jay





desertwanderer81
May 13, 2009, 5:03 PM
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majid_sabet wrote: That's the thing though! The only way you EVER mess with people is by filling thread after thread with disinformation. Your understanding of math, physics, and even climbing is limitted to the point that you should never, ever comment on anything as almost without fail, it will be wrong! I am not giving any wrong info. I am providing related info to the topic and whatever I have said so far, it was related to topic. If this information is wrong then people should bring it up so everyone can learn (that includes me) from it as true educational discussion. I will not tolerate against people who continue calling me names or take their personal problems and use these forums to become even with me over some unknown old business. If those sort of people want to use that sort of techniques or behavior to prove their points then I grantee you, I would the worse as*hole you guys ever wanted to deal with. peace and lets get back on topic That's the thing though! Your "related info" is not correct! You made the statement that 1kN is not equal to 224 lbs! 1kN IS however equal to 224lbs. Another instance of you providing misinformation is your complete lack of understanding of what a factor of safety is! You were saying that a person who weighs (say 224lbs) and falls on a rope that is rated for 10kN would have a factor of safety of 10! That statement is so far from reality that it is absurd!!





majid_sabet
May 13, 2009, 5:04 PM
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jt512 wrote: colatownkid wrote: the significance of the fall factor becomes evident when one realizes that the stretch in dynamic rope serves to dissipate energy in the system and absorb some of the force. The stretch in the rope reduces the impact force (relative to a less stretchy rope). It does not "absorb" the force. This is not a pedantic distinction, as you'll see presently. In reply to: if rope were totally static, the force of a fall would simply be a matter of distance fallen. the greater the distance fallen the greater the gain in velocity overtime, and the greater total force. That is false. If the rope were truly completely static, then the force of any fall would be infinite, and completely independent of the distance fallen. The longer the fall, the greater the (kinetic) energy, not the impact force. You seem to be implying that force builds up in a fall. It doesn't. The only force acting on a falling climber is gravity. In reply to: what a fall factor DOES NOT do, is tell us anything specific about a definite peak load. No, but it's trivial to calculate the peak force. All we need to know is the UIAA impact force rating of the rope (which allows us to calculate the rope's modulus of elasticity), the weight of the climber, and the fall factor. If the rope has been clipped through an anchor, we can add in the "pulley effect" after adjusting for friction through the anchor. The result will be the maximum impact force assuming a static belay and no other frictional forces in the system. In reply to: the one thing the fall factor does not tell us is any value in kilonewtons that describes the max force in the system. Right, but it's easy enough to calculate, given the assumptions above. In reply to: majid is correct in stating that the friction over the carabiner does reduce some of the force. Majid's statement was confusing, and possibly outright wrong. Many climbers misunderstand the effect of friction in the belay system. All friction in the system is a braking force. It increases the impact force on the climber, and decreases the impact force on the belayer. Some people speak of friction increasing the "effective" fall factor, because friction reduces the stretch in the rope, much as if there were less rope out in the first place. In reply to: (one final note: the fall factor is inherently an approximation since we all know from experience that rope stretch makes for a longer fall than 20 feet when 10 feet of rope is out.) No. The fall factor is defined as the distance the climber falls with no stretch in the rope divided by the amount of unstretched rope out. Jay Jay Lets say we use two cheap biner and both are rated to snap at 6 kn. we place one on the fig8 and attached to leader and another one to belayer's gri gri. based on your remark, the chances of leader's biner breaking is by greater than belayer's biner ? Right
(This post was edited by majid_sabet on May 13, 2009, 5:05 PM)





jt512
May 13, 2009, 5:06 PM
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hafilax wrote: Try 2: Real world forces will never approach the value calculated with the fall factor formula. I doubt that is true. With sufficient rope drag, the actual force should exceed the calculated force. Jay





jrathfon
May 13, 2009, 5:07 PM
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On that last point, and something I thought was a touch iffy, in the paper majidiot actually provided, they define fall factor as Dt, the total distance fallen by the climber after stretch, divided by the length of rope, prestretch. I've always seen it defined as the way you just stated, fall pre stretch/rope pre stretch.





desertwanderer81
May 13, 2009, 5:09 PM
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In reply to: Jay Lets say we use two cheap biner and both are rated to snap at 6 kn. we place one on the fig8 and attached to leader and another one to belayer's gri gri. based on your remark, the chances of leader's biner breaking is by greater than belayer's biner ? Right The force on leader's end is higher than the belayers end. The amount is based upon the amount of friction in the system.





hafilax
May 13, 2009, 5:10 PM
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jt512 wrote: hafilax wrote: Try 2: Real world forces will never approach the value calculated with the fall factor formula. I doubt that is true. With sufficient rope drag, the actual force should exceed the calculated force. Jay Damnit. I was thinking of the fall factor 2 case and got too general. I give up. My attempted point was that the fall factor force calculation is a poor model for climbing situations.





jt512
May 13, 2009, 5:10 PM
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majid_sabet wrote: jt512 wrote: colatownkid wrote: the significance of the fall factor becomes evident when one realizes that the stretch in dynamic rope serves to dissipate energy in the system and absorb some of the force. The stretch in the rope reduces the impact force (relative to a less stretchy rope). It does not "absorb" the force. This is not a pedantic distinction, as you'll see presently. In reply to: if rope were totally static, the force of a fall would simply be a matter of distance fallen. the greater the distance fallen the greater the gain in velocity overtime, and the greater total force. That is false. If the rope were truly completely static, then the force of any fall would be infinite, and completely independent of the distance fallen. The longer the fall, the greater the (kinetic) energy, not the impact force. You seem to be implying that force builds up in a fall. It doesn't. The only force acting on a falling climber is gravity. In reply to: what a fall factor DOES NOT do, is tell us anything specific about a definite peak load. No, but it's trivial to calculate the peak force. All we need to know is the UIAA impact force rating of the rope (which allows us to calculate the rope's modulus of elasticity), the weight of the climber, and the fall factor. If the rope has been clipped through an anchor, we can add in the "pulley effect" after adjusting for friction through the anchor. The result will be the maximum impact force assuming a static belay and no other frictional forces in the system. In reply to: the one thing the fall factor does not tell us is any value in kilonewtons that describes the max force in the system. Right, but it's easy enough to calculate, given the assumptions above. In reply to: majid is correct in stating that the friction over the carabiner does reduce some of the force. Majid's statement was confusing, and possibly outright wrong. Many climbers misunderstand the effect of friction in the belay system. All friction in the system is a braking force. It increases the impact force on the climber, and decreases the impact force on the belayer. Some people speak of friction increasing the "effective" fall factor, because friction reduces the stretch in the rope, much as if there were less rope out in the first place. In reply to: (one final note: the fall factor is inherently an approximation since we all know from experience that rope stretch makes for a longer fall than 20 feet when 10 feet of rope is out.) No. The fall factor is defined as the distance the climber falls with no stretch in the rope divided by the amount of unstretched rope out. Jay Jay Lets say we use two cheap biner and both are rated to snap at 6 kn. we place one on the fig8 and attached to leader and another one to belayer's gri gri. based on your remark, the chances of leader's biner breaking is by greater than belayer's biner ? Right Yes, assuming that the rope is clipped through one or more anchors. Jay





desertwanderer81
May 13, 2009, 5:13 PM
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jt512 wrote: hafilax wrote: Try 2: Real world forces will never approach the value calculated with the fall factor formula. I doubt that is true. With sufficient rope drag, the actual force should exceed the calculated force. Jay I was thinking the same thing when I saw that statement earlier. Ultimately the forces which are calculated will be fairly similiar to the actual forces in the field.





jt512
May 13, 2009, 5:15 PM
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jrathfon wrote: On that last point, and something I thought was a touch iffy, in the paper majidiot actually provided, they define fall factor as Dt, the total distance fallen by the climber after stretch, divided by the length of rope, prestretch. I've always seen it defined as the way you just stated, fall pre stretch/rope pre stretch. I haven't looked at the paper Majid referenced. I suppose that in principle you could define fall factor either way, but I would think that defining it in terms of maximum stretch would complicate the derivation of the equation for maximum impact force from Hooke's Law. Edit: Glancing at that paper, the author's expression for the maximum impact force, based on the total distance fallen (including maximum rope stretch) is actually simpler than the one based on the "free fall" distance. Jay
(This post was edited by jt512 on May 13, 2009, 5:24 PM)





ptlong
May 13, 2009, 5:20 PM
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hafilax wrote: jt512 wrote: hafilax wrote: Try 2: Real world forces will never approach the value calculated with the fall factor formula. I doubt that is true. With sufficient rope drag, the actual force should exceed the calculated force. Jay Damnit. I was thinking of the fall factor 2 case and got too general. I give up. My attempted point was that the fall factor force calculation is a poor model for climbing situations. Attaway came to the same conclusion as Jay: http://www.jrre.org/ropes_101.pdf But there are factor 2 falls that might approach the simple theoretical values as well. Think of a rope soloist run way out. Or someone belaying and the rope gets snagged in such a way as to provide a nearly static belay. The further out the leader the less important are mitigating factors like knot tightening and harness stretching.





ptlong
May 13, 2009, 5:24 PM
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hafilax wrote: I don't really know what happens when a piece pulls. I think it depends on how much the rope is able to recover before being stretched again since the damping in the rope is the primary energy dissipation mechanism (AFAIK).... Beverly and Attaway did drop tests to measure this. Their conclusion was that the rope recovered. http://www.mra.org/...equential_Falls2.pdf (edited to fix URL)
(This post was edited by ptlong on May 13, 2009, 5:35 PM)








colatownkid
May 13, 2009, 8:45 PM
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jt512 wrote: colatownkid wrote: the significance of the fall factor becomes evident when one realizes that the stretch in dynamic rope serves to dissipate energy in the system and absorb some of the force. The stretch in the rope reduces the impact force (relative to a less stretchy rope). It does not "absorb" the force. This is not a pedantic distinction, as you'll see presently. In reply to: if rope were totally static, the force of a fall would simply be a matter of distance fallen. the greater the distance fallen the greater the gain in velocity overtime, and the greater total force. That is false. If the rope were truly completely static, then the force of any fall would be infinite, and completely independent of the distance fallen. The longer the fall, the greater the (kinetic) energy, not the impact force. You seem to be implying that force builds up in a fall. It doesn't. The only force acting on a falling climber is gravity. In reply to: what a fall factor DOES NOT do, is tell us anything specific about a definite peak load. No, but it's trivial to calculate the peak force. All we need to know is the UIAA impact force rating of the rope (which allows us to calculate the rope's modulus of elasticity), the weight of the climber, and the fall factor. If the rope has been clipped through an anchor, we can add in the "pulley effect" after adjusting for friction through the anchor. The result will be the maximum impact force assuming a static belay and no other frictional forces in the system. In reply to: the one thing the fall factor does not tell us is any value in kilonewtons that describes the max force in the system. Right, but it's easy enough to calculate, given the assumptions above. In reply to: majid is correct in stating that the friction over the carabiner does reduce some of the force. Majid's statement was confusing, and possibly outright wrong. Many climbers misunderstand the effect of friction in the belay system. All friction in the system is a braking force. It increases the impact force on the climber, and decreases the impact force on the belayer. Some people speak of friction increasing the "effective" fall factor, because friction reduces the stretch in the rope, much as if there were less rope out in the first place. In reply to: (one final note: the fall factor is inherently an approximation since we all know from experience that rope stretch makes for a longer fall than 20 feet when 10 feet of rope is out.) No. The fall factor is defined as the distance the climber falls with no stretch in the rope divided by the amount of unstretched rope out. Jay i stand corrected. as for this part:
jt512 wrote: ] In reply to: the one thing the fall factor does not tell us is any value in kilonewtons that describes the max force in the system. Right, but it's easy enough to calculate, given the assumptions above. the only reason i mentioned this is that there seemed to be some implication that a specific fall factor could be equated with some specific impact force (across independent scenarios).





curt
May 13, 2009, 8:55 PM
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colatownkid wrote: jt512 wrote: colatownkid wrote: the significance of the fall factor becomes evident when one realizes that the stretch in dynamic rope serves to dissipate energy in the system and absorb some of the force. The stretch in the rope reduces the impact force (relative to a less stretchy rope). It does not "absorb" the force. This is not a pedantic distinction, as you'll see presently. In reply to: if rope were totally static, the force of a fall would simply be a matter of distance fallen. the greater the distance fallen the greater the gain in velocity overtime, and the greater total force. That is false. If the rope were truly completely static, then the force of any fall would be infinite, and completely independent of the distance fallen. The longer the fall, the greater the (kinetic) energy, not the impact force. You seem to be implying that force builds up in a fall. It doesn't. The only force acting on a falling climber is gravity. In reply to: what a fall factor DOES NOT do, is tell us anything specific about a definite peak load. No, but it's trivial to calculate the peak force. All we need to know is the UIAA impact force rating of the rope (which allows us to calculate the rope's modulus of elasticity), the weight of the climber, and the fall factor. If the rope has been clipped through an anchor, we can add in the "pulley effect" after adjusting for friction through the anchor. The result will be the maximum impact force assuming a static belay and no other frictional forces in the system. In reply to: the one thing the fall factor does not tell us is any value in kilonewtons that describes the max force in the system. Right, but it's easy enough to calculate, given the assumptions above. In reply to: majid is correct in stating that the friction over the carabiner does reduce some of the force. Majid's statement was confusing, and possibly outright wrong. Many climbers misunderstand the effect of friction in the belay system. All friction in the system is a braking force. It increases the impact force on the climber, and decreases the impact force on the belayer. Some people speak of friction increasing the "effective" fall factor, because friction reduces the stretch in the rope, much as if there were less rope out in the first place. In reply to: (one final note: the fall factor is inherently an approximation since we all know from experience that rope stretch makes for a longer fall than 20 feet when 10 feet of rope is out.) No. The fall factor is defined as the distance the climber falls with no stretch in the rope divided by the amount of unstretched rope out. Jay i stand corrected. as for this part: jt512 wrote: ] In reply to: the one thing the fall factor does not tell us is any value in kilonewtons that describes the max force in the system. Right, but it's easy enough to calculate, given the assumptions above. the only reason i mentioned this is that there seemed to be some implication that a specific fall factor could be equated with some specific impact force (across independent scenarios). It can if the only variable is the length of the fall. Curt





colatownkid
May 13, 2009, 8:58 PM
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curt wrote: colatownkid wrote: jt512 wrote: colatownkid wrote: the significance of the fall factor becomes evident when one realizes that the stretch in dynamic rope serves to dissipate energy in the system and absorb some of the force. The stretch in the rope reduces the impact force (relative to a less stretchy rope). It does not "absorb" the force. This is not a pedantic distinction, as you'll see presently. In reply to: if rope were totally static, the force of a fall would simply be a matter of distance fallen. the greater the distance fallen the greater the gain in velocity overtime, and the greater total force. That is false. If the rope were truly completely static, then the force of any fall would be infinite, and completely independent of the distance fallen. The longer the fall, the greater the (kinetic) energy, not the impact force. You seem to be implying that force builds up in a fall. It doesn't. The only force acting on a falling climber is gravity. In reply to: what a fall factor DOES NOT do, is tell us anything specific about a definite peak load. No, but it's trivial to calculate the peak force. All we need to know is the UIAA impact force rating of the rope (which allows us to calculate the rope's modulus of elasticity), the weight of the climber, and the fall factor. If the rope has been clipped through an anchor, we can add in the "pulley effect" after adjusting for friction through the anchor. The result will be the maximum impact force assuming a static belay and no other frictional forces in the system. In reply to: the one thing the fall factor does not tell us is any value in kilonewtons that describes the max force in the system. Right, but it's easy enough to calculate, given the assumptions above. In reply to: majid is correct in stating that the friction over the carabiner does reduce some of the force. Majid's statement was confusing, and possibly outright wrong. Many climbers misunderstand the effect of friction in the belay system. All friction in the system is a braking force. It increases the impact force on the climber, and decreases the impact force on the belayer. Some people speak of friction increasing the "effective" fall factor, because friction reduces the stretch in the rope, much as if there were less rope out in the first place. In reply to: (one final note: the fall factor is inherently an approximation since we all know from experience that rope stretch makes for a longer fall than 20 feet when 10 feet of rope is out.) No. The fall factor is defined as the distance the climber falls with no stretch in the rope divided by the amount of unstretched rope out. Jay i stand corrected. as for this part: jt512 wrote: ] In reply to: the one thing the fall factor does not tell us is any value in kilonewtons that describes the max force in the system. Right, but it's easy enough to calculate, given the assumptions above. the only reason i mentioned this is that there seemed to be some implication that a specific fall factor could be equated with some specific impact force (across independent scenarios). It can if the only variable is the length of the fall. Curt yes, but not on two different climbs at two different anchors, etc.





curt
May 13, 2009, 9:02 PM
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colatownkid wrote: curt wrote: colatownkid wrote: jt512 wrote: colatownkid wrote: the significance of the fall factor becomes evident when one realizes that the stretch in dynamic rope serves to dissipate energy in the system and absorb some of the force. The stretch in the rope reduces the impact force (relative to a less stretchy rope). It does not "absorb" the force. This is not a pedantic distinction, as you'll see presently. In reply to: if rope were totally static, the force of a fall would simply be a matter of distance fallen. the greater the distance fallen the greater the gain in velocity overtime, and the greater total force. That is false. If the rope were truly completely static, then the force of any fall would be infinite, and completely independent of the distance fallen. The longer the fall, the greater the (kinetic) energy, not the impact force. You seem to be implying that force builds up in a fall. It doesn't. The only force acting on a falling climber is gravity. In reply to: what a fall factor DOES NOT do, is tell us anything specific about a definite peak load. No, but it's trivial to calculate the peak force. All we need to know is the UIAA impact force rating of the rope (which allows us to calculate the rope's modulus of elasticity), the weight of the climber, and the fall factor. If the rope has been clipped through an anchor, we can add in the "pulley effect" after adjusting for friction through the anchor. The result will be the maximum impact force assuming a static belay and no other frictional forces in the system. In reply to: the one thing the fall factor does not tell us is any value in kilonewtons that describes the max force in the system. Right, but it's easy enough to calculate, given the assumptions above. In reply to: majid is correct in stating that the friction over the carabiner does reduce some of the force. Majid's statement was confusing, and possibly outright wrong. Many climbers misunderstand the effect of friction in the belay system. All friction in the system is a braking force. It increases the impact force on the climber, and decreases the impact force on the belayer. Some people speak of friction increasing the "effective" fall factor, because friction reduces the stretch in the rope, much as if there were less rope out in the first place. In reply to: (one final note: the fall factor is inherently an approximation since we all know from experience that rope stretch makes for a longer fall than 20 feet when 10 feet of rope is out.) No. The fall factor is defined as the distance the climber falls with no stretch in the rope divided by the amount of unstretched rope out. Jay i stand corrected. as for this part: jt512 wrote: ] In reply to: the one thing the fall factor does not tell us is any value in kilonewtons that describes the max force in the system. Right, but it's easy enough to calculate, given the assumptions above. the only reason i mentioned this is that there seemed to be some implication that a specific fall factor could be equated with some specific impact force (across independent scenarios). It can if the only variable is the length of the fall. Curt yes, but not on two different climbs at two different anchors, etc. Yes, then tooso long as the rope has the same maximum impact force, the climber weighs the same, etc. Curt





antiqued
May 13, 2009, 9:45 PM
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Registered: Aug 18, 2005
Posts: 243

jt512 wrote: hafilax wrote: Try 2: Real world forces will never approach the value calculated with the fall factor formula. I doubt that is true. With sufficient rope drag, the actual force should exceed the calculated force. Jay Both of these statements can be useful, but under different conditions. [Jay's is truer, never say never!] The fall factor data is measured in a high FF, with real friction over the top biner, belay end knotted to a steel fixture, and a steel block. Real falls typically have a belay device with slip, attached to a belayer of finite weight, not rigidly tied in on the one end, and a soft, floppy body on the other end. For low friction falls (zero or minimal rope drag up to the top biner), the impact force will be lower than that calculated from the UIAA test data because of auxiliary energy absorption mechanisms, usually quite a bit lower. In any serious fall the rope will slip and the belayer float, reducing impact force. For low to modest FF falls with a convoluted rope path, the effective FF can easily be much higher than the belayerclimber calculation, because a good portion of the rope doesn't see much force, and therefore can't be absorbing much of the fall energy. We know this because many of us weighing less than a kN catch decent size falls without leaving the ground. For example, a 20' fall on 50' of rope would be a 0.4FF, with a ~3.8kN force on the climber, and after friction, ~2.6kN (=0.7*3.8) on the belayer. That should yank people up and away, and indeed you can see it happen in gyms around the country, with their arrrow straight rope paths. But out on trad routes, many (not all, of course) such falls are trivially caught. So we know that the effective length of the rope has been reduced, and the impact force must be higher. What about hard falls? The UIAA test is so aggressive, it is almost impossible to exceed the measured impact force. Only 12% of the rope is between the tieoff and the protection in that test. In order to duplicate it, one might clip the 'Jesus nut' 4 feet from the belay, run it out 28 more feet and then take a 56 footer with the belay tied off. The only likely way is to misroute the rope, so that it is jammed in a crack, bends sharply under a roof, or gets pinned by the falling end of the rope (below). This can lead to real FF~2 falls if the jam or pin is very close to the top piece.





colatownkid
May 13, 2009, 9:51 PM
Post #100 of 211
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Registered: Nov 27, 2007
Posts: 512

curt wrote: colatownkid wrote: curt wrote: colatownkid wrote: jt512 wrote: colatownkid wrote: the significance of the fall factor becomes evident when one realizes that the stretch in dynamic rope serves to dissipate energy in the system and absorb some of the force. The stretch in the rope reduces the impact force (relative to a less stretchy rope). It does not "absorb" the force. This is not a pedantic distinction, as you'll see presently. In reply to: if rope were totally static, the force of a fall would simply be a matter of distance fallen. the greater the distance fallen the greater the gain in velocity overtime, and the greater total force. That is false. If the rope were truly completely static, then the force of any fall would be infinite, and completely independent of the distance fallen. The longer the fall, the greater the (kinetic) energy, not the impact force. You seem to be implying that force builds up in a fall. It doesn't. The only force acting on a falling climber is gravity. In reply to: what a fall factor DOES NOT do, is tell us anything specific about a definite peak load. No, but it's trivial to calculate the peak force. All we need to know is the UIAA impact force rating of the rope (which allows us to calculate the rope's modulus of elasticity), the weight of the climber, and the fall factor. If the rope has been clipped through an anchor, we can add in the "pulley effect" after adjusting for friction through the anchor. The result will be the maximum impact force assuming a static belay and no other frictional forces in the system. In reply to: the one thing the fall factor does not tell us is any value in kilonewtons that describes the max force in the system. Right, but it's easy enough to calculate, given the assumptions above. In reply to: majid is correct in stating that the friction over the carabiner does reduce some of the force. Majid's statement was confusing, and possibly outright wrong. Many climbers misunderstand the effect of friction in the belay system. All friction in the system is a braking force. It increases the impact force on the climber, and decreases the impact force on the belayer. Some people speak of friction increasing the "effective" fall factor, because friction reduces the stretch in the rope, much as if there were less rope out in the first place. In reply to: (one final note: the fall factor is inherently an approximation since we all know from experience that rope stretch makes for a longer fall than 20 feet when 10 feet of rope is out.) No. The fall factor is defined as the distance the climber falls with no stretch in the rope divided by the amount of unstretched rope out. Jay i stand corrected. as for this part: jt512 wrote: ] In reply to: the one thing the fall factor does not tell us is any value in kilonewtons that describes the max force in the system. Right, but it's easy enough to calculate, given the assumptions above. the only reason i mentioned this is that there seemed to be some implication that a specific fall factor could be equated with some specific impact force (across independent scenarios). It can if the only variable is the length of the fall. Curt yes, but not on two different climbs at two different anchors, etc. Yes, then tooso long as the rope has the same maximum impact force, the climber weighs the same, etc. Curt i give up.








