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jt512
May 18, 2010, 10:16 PM
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JimTitt wrote: The UIAA decided way back in the 70´s that a 80kg solid weight represents the impact of a 100kg (in fact 101 and a little bit if I remember rightly) climber after tests by Troll who made some of the early harnesses.
Jim, I'd have to see how that result was derived and convince myself of its validity before I'd incorporate it into a model.
Jay
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theguy
May 18, 2010, 11:28 PM
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jt512 wrote: JimTitt wrote: The UIAA decided way back in the 70´s that a 80kg solid weight represents the impact of a 100kg (in fact 101 and a little bit if I remember rightly) climber after tests by Troll who made some of the early harnesses. Jim, I'd have to see how that result was derived and convince myself of its validity before I'd incorporate it into a model. Jay
While you are not explicitly stating that the weight in your calculator refers to the climber's weight rather than the test weight, many users may infer this; if they do, you are already (apparently unknowingly) incorporating the UIAA result in your calculator by basing your calculation in one scenario on impact force ratings provided by the UIAA.
Your options would appear to be:
a) Make it explicit that the weight refers to the drop test weight rather than the climber's weight when UIAA impact force ratings are used
b) Make it explicit that the weight refers to the climber's weight and correct for the results used in the UIAA impact force ratings
If, as you say, you need to better understand how the UIAA results were derived before pursuing the second option, the first option would appear to be your only viable choice for now.
For corroboration (though not explanation) of Jim's info, see the UIAA article "How strong does your climbing gear need to be" he has referenced elsewhere:
"The test uses an 80kg falling mass. Some time ago, tests were done which showed that the peak forces were approximately the same as would arise with a 100kg (15 1/2 stone) falling human body."
jt512 wrote: the impact force on the belayer is assumed to be a percentage of the impact force on the climber. Therefore, the impact force on the belayer is higher under the friction-adjusted model than under the standard model.
Jay, you're usually not a fan of assumptions, especially when they're demonstrably false; to maintain a consistent stance, you may wish to adjust this assumption.
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theguy
May 18, 2010, 11:33 PM
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jt512 wrote: I've put an online version of a simple impact force calculator I wrote on my website here.
majid_sabet wrote: If you could do this on windows, that would be very nice
No good deed goes unpunished 
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jt512
May 19, 2010, 12:01 AM
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theguy wrote: jt512 wrote: JimTitt wrote: The UIAA decided way back in the 70´s that a 80kg solid weight represents the impact of a 100kg (in fact 101 and a little bit if I remember rightly) climber after tests by Troll who made some of the early harnesses. Jim, I'd have to see how that result was derived and convince myself of its validity before I'd incorporate it into a model. Jay While you are not explicitly stating that the weight in your calculator refers to the climber's weight rather than the test weight, many users may infer this; if they do, you are already (apparently unknowingly) incorporating the UIAA result in your calculator by basing your calculation in one scenario on impact force ratings provided by the UIAA. Your options would appear to be: a) Make it explicit that the weight refers to the drop test weight rather than the climber's weight when UIAA impact force ratings are used b) Make it explicit that the weight refers to the climber's weight and correct for the results used in the UIAA impact force ratings
The weight you enter in the calculator is the climber's weight, regardless of whether you check the "rope modulus" button or the "impact force rating" button. If that results in the impact force being too high because the model assumes a rigid mass, then so be it—it's a model. There are many factors in a real-world fall that the model does not take into consideration, such as the dynamic nature of the belay, friction between the rope and the rock, etc, etc, etc.
In reply to: jt512 wrote: the impact force on the belayer is assumed to be a percentage of the impact force on the climber. Therefore, the impact force on the belayer is higher under the friction-adjusted model than under the standard model. Jay, you're usually not a fan of assumptions, especially when they're demonstrably false; to maintain a consistent stance, you may wish to adjust this assumption.
First of all, there are lots of assumptions in a model that are "demonstrably false," in terms of how things behave in real life. For instance, as mentioned above, there is an assumption in these impact force models that the belay is perfectly static. But that said, what are you talking about when you say that it is demonstrably false that impact force on the belayer is proportional to the impact force on the climber? That assumption is made in every model of impact force I've ever seen. I think you've misunderstood something.
Jay
(This post was edited by jt512 on May 19, 2010, 3:45 AM)
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jt512
May 19, 2010, 12:17 AM
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yokese wrote: jt512 wrote: yokese wrote: Mac OS X 10.6.3 Safari 4.0.5 [image]http://www.rockclimbing.com/cgi-bin/forum/gforum.cgi?do=post_attachment;postatt_id=4890;[/image] Based on your earlier post, it appears that you were able to access the page. Was that by using a different computer, or was this a temporary error? Jay Nope, I couldn't access it before. I just viewed the screen shot that you attached. I've also tried with firefox to no avail. Same error (the server is taking too long to respond). I'll check later on from a different location.
Has anyone else using a Mac either been able to access the page, or not?
Jay
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MS1
May 19, 2010, 12:48 AM
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The page loaded and ran fine for me, using MacOS 10.5.8 and firefox 3.6.3.
Thanks for providing a nice tool for the community.
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theguy
May 19, 2010, 1:30 AM
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jt512 wrote: What are you talking about when you say that it is demonstrably false that impact force on the belayer is proportional to the impact force on the climber?
Thanks for catch... what I should have said is that the proportion differs depending on the friction.
I see that on the website I'm now getting some funky results (e.g. no variance based on fall factor; standard and friction-adjusted numbers are the same), so presume you're now taking some of the feedback into account and tweaking the calculator.
Behaviour seems erratic, so may be a caching issue.
That said, what is intent of the standard vs. friction/adjusted columns? Would the standard column assume a friction of zero?
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jt512
May 19, 2010, 2:03 AM
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theguy wrote: jt512 wrote: What are you talking about when you say that it is demonstrably false that impact force on the belayer is proportional to the impact force on the climber? Thanks for catch... what I should have said is that the proportion differs depending on the friction.
I'm still not sure what you are trying to say. In both models the impact force on the belayer T2 is simply a fraction (1 – µ) of the impact force on the climber T1:
T2 = (1 – µ) * T1, where µ is the "friction factor" in the calculator.
In reply to: I see that on the website I'm now getting some funky results (e.g. no variance based on fall factor; standard and friction-adjusted numbers are the same), so presume you're now taking some of the feedback into account and tweaking the calculator.
I have not tweaked the calculator at all. If the fall factor is 0 or 2 (calculating using the rope modulus), or the friction factor is 0, the two models are equivalent, and hence produce the same results.
In reply to: Behaviour seems erratic, so may be a caching issue.
As far as I can tell the calculator is behaving quite well, and it is your comments that are erratic. According to my logs, none of your responses have been from cache. Can you provide some concrete examples of this erratic behavior you think you have seen? I suspect that you are mistaken.
In reply to: That said, what is intent of the standard vs. friction/adjusted columns? Would the standard column assume a friction of zero?
The standard model assumes, for the purpose of calculating the impact force on the climber, that there is no friction between the rope and the top anchor. The friction-adjusted model doesn't.
All of your questions are answered in the adjusted-impact-force.pdf document cited in the OP.
Jay
(This post was edited by jt512 on May 19, 2010, 2:48 AM)
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theguy
May 19, 2010, 2:14 AM
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majid_sabet wrote: If you could do this on windows, that would be very nice.
I've PM'd you an Excel spreadsheet based on the Wikipedia formulas, which are in turn of course based on other sources.
The spreadsheet takes into account Jim Titt's and Yokese's suggestions re. climber weight and friction. You may wish to have Jay review it.
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jt512
May 19, 2010, 2:34 AM
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[Disregard]
(This post was edited by jt512 on May 21, 2010, 4:33 AM)
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ptlong
May 19, 2010, 5:49 AM
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Use for comparison purposes only. Your actual mileage may vary.
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ptlong
May 19, 2010, 4:31 PM
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jt512 wrote: JimTitt wrote: The UIAA decided way back in the 70´s that a 80kg solid weight represents the impact of a 100kg (in fact 101 and a little bit if I remember rightly) climber after tests by Troll who made some of the early harnesses. Jim, I'd have to see how that result was derived and convince myself of its validity before I'd incorporate it into a model. Jay
I've read this before and wondered how they did those tests. Presumabely they did not use live volunteers!
But even if this is accurate for a UIAA drop, it seems intuitive that the fudge factor would not be constant for all falls. If more rope than 2.3m is out, the human "spring/dashpot" would not grow proportionally.
Did Petzl's calculator make such an adjustment? I don't remember but I thought it was also based on a pretty simplistic model.
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cracklover
May 19, 2010, 4:52 PM
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ptlong wrote: jt512 wrote: JimTitt wrote: The UIAA decided way back in the 70´s that a 80kg solid weight represents the impact of a 100kg (in fact 101 and a little bit if I remember rightly) climber after tests by Troll who made some of the early harnesses. Jim, I'd have to see how that result was derived and convince myself of its validity before I'd incorporate it into a model. Jay I've read this before and wondered how they did those tests. Presumabely they did not use live volunteers! But even if this is accurate for a UIAA drop, it seems intuitive that the fudge factor would not be constant for all falls. If more rope than 2.3m is out, the human "spring/dashpot" would not grow proportionally.
There are a number of factors that could be incorporated into models. For example, the tightening of a knot takes up a certain amount of energy. For small falls, with small amounts of energy, this can be a significant factor in lowering the max force. However, this cannot be factored into an equation like the one underlying Jay's model, because Jay's model does not take fall length into consideration.
I suspect that body elasticity (for lack of a better term) is similar, but even more difficult to model. It probably varies quite a bit depending on such factors as the position of the body when the rope begins to catch. But even if this weren't the case, it shares the same issue as the knot: unlike the rope involved in the catch, the affect it has on acceleration is not proportional to the fall distance. So there's no way it could be plugged into a calculation based on Fall Factor.
In reply to: Did Petzl's calculator make such an adjustment? I don't remember but I thought it was also based on a pretty simplistic model.
If they ever made public what the formulas behind their calculator were, I never heard of it.
GO
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ptlong
May 19, 2010, 7:00 PM
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cracklover wrote: There are a number of factors that could be incorporated into models. For example, the tightening of a knot takes up a certain amount of energy. For small falls, with small amounts of energy, this can be a significant factor in lowering the max force. However, this cannot be factored into an equation like the one underlying Jay's model, because Jay's model does not take fall length into consideration. I suspect that body elasticity (for lack of a better term) is similar, but even more difficult to model. It probably varies quite a bit depending on such factors as the position of the body when the rope begins to catch. But even if this weren't the case, it shares the same issue as the knot: unlike the rope involved in the catch, the affect it has on acceleration is not proportional to the fall distance. So there's no way it could be plugged into a calculation based on Fall Factor.
Jay's model doesn't include fall distance, but it could. The question is whether a correction that is also a function of fall distance or rope out could be determined.
Like you, I would also guess that the effect would vary depending on factors like body position. This makes the precision of the correction factor that Jim Titt cited intriguing. He didn't say it was just a rough 100 kg, but "101 and a little bit".
In reply to: In reply to: Did Petzl's calculator make such an adjustment? I don't remember but I thought it was also based on a pretty simplistic model. If they ever made public what the formulas behind their calculator were, I never heard of it.
I could be wrong but I thought they had the Wexler formula displayed on that web page. If so the rope was modeled simply as an ideal spring. The belay was fixed (no belayer mass), but the rope could slip at a fixed force depending on the selected belay device. They let you choose different rope diamaters corresponding to different moduli. They also allowed you to select the number of placements and whether or not the rope ran in a straight line. So they must have been doing some carabiner friction calculations. I seem to recall also reading that they included knot tightening in the calculation.
I always treated it as a toy.
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JimTitt
May 19, 2010, 8:31 PM
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I´ve forgotten over the years exactly the values used but the UIAA chose 80kg as it was a nice round number which was near the force from a 100kg climber, I´ll see if I can find anyone still alive from those days who knows more The fall protection guys have done a lot more work on this recently but there are two major differences in that they use full body harnesses and they have to account for the extra tools and clothing a worker normally would carry.
There are plenty of ifs and buts in the whole question of fall force calculators and with all respect to jt512 whether we need yet another calculator giving the wrong answers is debatable. As a tool for equipment developers a much higher level of precision is essential and for forum punters they just add another level of confusion.
The simplist form of fall calculator (and as it happens the most accurate) is:- I fell, the gear held therefore the forces were less than the breaking load of the gear. More is only confusing the issue.
In a real fall there are so many complicated things happening, some simultaneously and some with time lag that any simple model remains a crude tool and therefore has no real use. Attempts to define a better model have been made but founder on the failure to incorporate all the effects (many of which are little researched or understood) in a satisfactory way. The change of approach by Manin and Bedogni to modelling the results of tests seems a better approach than following Wexler as it is not nescessary to understand why the forces are what they are, only know what they are.
For gear manufacturers however it is useful to identify the seperate forces and influences more exactly and so we need both, a theory-driven formula of everything and a experience-driven model, hopefully both of which agree.
Currently we see the time-line of a fall (real life, not test drop) can be split it the following.
1. Free fall- of little interest!
2. Take up of slack between belayer and faller- depending on the number of bends in the rope this is of more of less interest.
3. Initial loading- at this stage life gets complicated. As the tension comes on the top karabiner (and subsequent ones and/or the rock) the coefficient of friction is about 0.5 and then drops to zero, it then rises again to nearly the same value and drops progressively depending on the velocity and load. So this friction coefficient factor varies with time and the length of the fall (which will define the velocity and load).
The initial tensioning in the rope is inside the "static stretch" dynamic area of the rope which stretches considerably more than a simple `one size fits all´ modulus would suggest.
5. The load then comes onto the belay device which also has the difficulty that the coefficient of friction varies throughout it braking cycle, the extent depending on exactly how the device works. the peaks in the braking effect are damped out up the rope giving a different load pattern at any point along the rope, and of course varying timewise as well.
This load is imposed (or transferred) by the climbing rope and its modulus will be different for any combination of fall distance and climber weight.
6. The braking load then comes onto the belayer through his harness and onto the climber through his, obviously how this force is dissapated (or transferred) through both bodies is a function of physical stature, body make-up, the length of the fall and the dynamics of the belaying, as you say it is going to be most marked in a small fall and the effect proportionally less in a larger fall.
7. Rebound- as all these forces (or loads) occur at different times they are balanced out by what could best be described as shock waves travelling along the rope, these play some funny tricks, for example the force on the belayer side of the rope may well exceed that on the other side of the top karabiner which suggests the rope must be travelling backwards.
There is obviously more to fall dynamics than just this since stuff like rope condition, humidity etc plays an important role but the above are at least the basics one needs to clarify to get started.
Fall factor is too simplistic to be useful, the difference between a F1 fall direct onto a fixed rope and a F1 over a karabiner is enormous, to say nothing of a F1 through 10 karabiners, over two rough edges and belayed by a nymphette!
We are tring to approach the problem from both directions, first by doing more basic research into mainly the friction and a bit on soft body impacts as this is part of another project, the other approach is to datalog a lot of climbing falls but the weather is piss at the moment!
Beer time!!!!
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CCSRacer
Aug 4, 2010, 8:51 PM
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Good write up. What are your thoughts on the force-on-belayer being reported as higher when considering friction? I have to agree with the earlier person that said it should be lower. The friction at the anchor point is doing negative work on the moving rope and dissipating that energy as heat (mostly); it never reaches the belayer.
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theguy
Aug 4, 2010, 9:23 PM
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jt512 wrote: [Disregard] Familiarity with rampant willful ignorance breeds contempt. —Anonymous
Preserved for posterity
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curt
Aug 5, 2010, 1:26 AM
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theguy wrote: jt512 wrote: the impact force on the belayer is assumed to be a percentage of the impact force on the climber. Therefore, the impact force on the belayer is higher under the friction-adjusted model than under the standard model. Jay, you're usually not a fan of assumptions, especially when they're demonstrably false; to maintain a consistent stance, you may wish to adjust this assumption.
Jay,
I think your assumption here is false--unless I'm misunderstanding something.
Curt
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jt512
Aug 5, 2010, 1:42 AM
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curt wrote: theguy wrote: jt512 wrote: the impact force on the belayer is assumed to be a percentage of the impact force on the climber. Therefore, the impact force on the belayer is higher under the friction-adjusted model than under the standard model. Jay, you're usually not a fan of assumptions, especially when they're demonstrably false; to maintain a consistent stance, you may wish to adjust this assumption. Jay, I think your assumption here is false--unless I'm misunderstanding something. Curt
Well, all assumptions are false; it's just a question of how false. ;)
In the standard model of impact force, the force on the belayer T2 = T1 * (1 – f), where T1 is the force on the climber and f is a constant, 0 <= f <= 1, representing the friction between the rope and the top anchor. In my friction-adjusted model f is the same as in the standard model; however, T1 is greater than in the standard model because my model takes into account the effect on T1 of friction between the rope and the top anchor. Hence T2 is greater in my model than in the standard model.
Jay
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theguy
Aug 5, 2010, 2:43 AM
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Attached is a spreadsheet version which applies a "conservation of force" model to the belayer. You can adjust as you see fit: seize the power! ;)
If you want to get really fancy, you could use Jay's Greasemonkey approach to tweak the results provided by his online calculator.
And to save you having to delete more posts, this is not directed at you Jay :)
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Impact Force rev3.xls
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jt512
Aug 5, 2010, 3:05 AM
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theguy wrote: And to save you having to delete more posts, this is not directed at you Jay :)
?
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curt
Aug 5, 2010, 4:00 AM
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jt512 wrote: curt wrote: theguy wrote: jt512 wrote: the impact force on the belayer is assumed to be a percentage of the impact force on the climber. Therefore, the impact force on the belayer is higher under the friction-adjusted model than under the standard model. Jay, you're usually not a fan of assumptions, especially when they're demonstrably false; to maintain a consistent stance, you may wish to adjust this assumption. Jay, I think your assumption here is false--unless I'm misunderstanding something. Curt Well, all assumptions are false; it's just a question of how false. ;) In the standard model of impact force, the force on the belayer T2 = T1 * (1 – f), where T1 is the force on the climber and f is a constant, 0 <= f <= 1, representing the friction between the rope and the top anchor. In my friction-adjusted model f is the same as in the standard model; however, T1 is greater than in the standard model because my model takes into account the effect on T1 of friction between the rope and the top anchor. Hence T2 is greater in my model than in the standard model. Jay
Well, I think T2 should be less instead of greater. Using an extreme example where the friction over the top piece is unity (i.e. just like the rope being jammed at the top piece) the force on the climber is much higher--because there is less rope available to absorb the energy of his fall, but the force on the belayer would be zero.
Curt
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jt512
Aug 5, 2010, 4:07 AM
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curt wrote: jt512 wrote: curt wrote: theguy wrote: jt512 wrote: the impact force on the belayer is assumed to be a percentage of the impact force on the climber. Therefore, the impact force on the belayer is higher under the friction-adjusted model than under the standard model. Jay, you're usually not a fan of assumptions, especially when they're demonstrably false; to maintain a consistent stance, you may wish to adjust this assumption. Jay, I think your assumption here is false--unless I'm misunderstanding something. Curt Well, all assumptions are false; it's just a question of how false. ;) In the standard model of impact force, the force on the belayer T2 = T1 * (1 – f), where T1 is the force on the climber and f is a constant, 0 <= f <= 1, representing the friction between the rope and the top anchor. In my friction-adjusted model f is the same as in the standard model; however, T1 is greater than in the standard model because my model takes into account the effect on T1 of friction between the rope and the top anchor. Hence T2 is greater in my model than in the standard model. Jay Well, I think T2 should be less instead of greater. Using an extreme example where the friction over the top piece is unity (i.e. just like the rope being jammed at the top piece) the force on the climber is much higher--because there is less rope available to absorb the energy of his fall, but the force on the belayer would be zero. Curt
Read the paper.
Jay
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curt
Aug 5, 2010, 4:30 AM
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jt512 wrote: curt wrote: jt512 wrote: curt wrote: theguy wrote: jt512 wrote: the impact force on the belayer is assumed to be a percentage of the impact force on the climber. Therefore, the impact force on the belayer is higher under the friction-adjusted model than under the standard model. Jay, you're usually not a fan of assumptions, especially when they're demonstrably false; to maintain a consistent stance, you may wish to adjust this assumption. Jay, I think your assumption here is false--unless I'm misunderstanding something. Curt Well, all assumptions are false; it's just a question of how false. ;) In the standard model of impact force, the force on the belayer T2 = T1 * (1 – f), where T1 is the force on the climber and f is a constant, 0 <= f <= 1, representing the friction between the rope and the top anchor. In my friction-adjusted model f is the same as in the standard model; however, T1 is greater than in the standard model because my model takes into account the effect on T1 of friction between the rope and the top anchor. Hence T2 is greater in my model than in the standard model. Jay Well, I think T2 should be less instead of greater. Using an extreme example where the friction over the top piece is unity (i.e. just like the rope being jammed at the top piece) the force on the climber is much higher--because there is less rope available to absorb the energy of his fall, but the force on the belayer would be zero. Curt Read the paper. Jay
OK, I did and see that you even mention the case of friction over the top piece equalling unity--and you agree that the force on the belayer in this case would be zero. Why isn't this also true, however for any other amount of friction 0 < F < 1? That is, as you increase friction over the top piece, the effective ff to the leader increases so the force felt by him increases--and the force felt by the belayer decreases? Unless I'm missing something you appear to be saying there is a direct, as opposed to inverse relationship.
Curt
(This post was edited by curt on Aug 5, 2010, 4:32 AM)
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jt512
Aug 5, 2010, 3:45 PM
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curt wrote: jt512 wrote: curt wrote: jt512 wrote: curt wrote: theguy wrote: jt512 wrote: the impact force on the belayer is assumed to be a percentage of the impact force on the climber. Therefore, the impact force on the belayer is higher under the friction-adjusted model than under the standard model. Jay, you're usually not a fan of assumptions, especially when they're demonstrably false; to maintain a consistent stance, you may wish to adjust this assumption. Jay, I think your assumption here is false--unless I'm misunderstanding something. Curt Well, all assumptions are false; it's just a question of how false. ;) In the standard model of impact force, the force on the belayer T2 = T1 * (1 – f), where T1 is the force on the climber and f is a constant, 0 <= f <= 1, representing the friction between the rope and the top anchor. In my friction-adjusted model f is the same as in the standard model; however, T1 is greater than in the standard model because my model takes into account the effect on T1 of friction between the rope and the top anchor. Hence T2 is greater in my model than in the standard model. Jay Well, I think T2 should be less instead of greater. Using an extreme example where the friction over the top piece is unity (i.e. just like the rope being jammed at the top piece) the force on the climber is much higher--because there is less rope available to absorb the energy of his fall, but the force on the belayer would be zero. Curt Read the paper. Jay [A]s you increase friction over the top piece, the effective ff to the leader increases so the force felt by him increases--and the force felt by the belayer decreases? Curt
That's correct. See Figure 1 and the last paragraph of Section 5 on page 7 of the paper.
In reply to: Unless I'm missing something you appear to be saying there is a direct, as opposed to inverse relationship.
No, what I said is that the impact force on the belayer is higher under the friction-adjusted model than under the standard model, as explained above, and here.
Jay
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