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rocknice2
Sep 5, 2012, 12:09 AM
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I've been trying to find the answer to a question but the amount of anchor topics is overwhelming. So I bring up one more.
Consider 3 protection points @ 10kn each
[edit] The rope has a impact force of 11kn [/edit]
A fall directly on the anchor with no belayer
The sling/cord breaking strength is not a factor.
Lastly for the sake of argument the self equalizing anchor is perfect.
1 A self equalizing anchor where the outer 2 legs are 90 deg apart & equal length.
2 An anchor with a tied power point, 90 deg outer legs and center leg is somewhat loose
3 Imagine pro orientated horizontally and a PAS type sling is used to connect the pro in series and the rope is clipped to only one end. Forcing one piece to completely blow before the other even sees any load at all.
What is the ultimate load energy each anchor can take absorb before it collapses completely?
Edited to clarify more what I want to know.
Basically which anchor will hold the most in a fall before completely breaking
.
(This post was edited by rocknice2 on Sep 6, 2012, 5:17 PM)
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acorneau
Sep 5, 2012, 2:16 AM
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rocknice2 wrote: I've been trying to find the answer to a question but the amount of anchor topics is overwhelming. So I bring up one more. Consider 3 protection points @ 10kn each A fall directly on the anchor with no belayer and capable of ripping any anchor The sling/cord breaking strength is not a factor. Lastly for the sake of argument the self equalizing anchor is perfect. 1 A self equalizing anchor where the outer 2 legs are 90 deg apart & equal length. 2 An anchor with a tied power point, 90 deg outer legs and center leg is somewhat loose 3 Imagine pro orientated horizontally and a PAS type sling is used to connect the pro in series and the rope is clipped to only one end. Forcing one piece to completely blow before the other even sees any load at all. What is the ultimate load each anchor can take before it collapses completely?
Theoretically speaking...
Scenario #1: +24kN
Scenario #2: +14kN
Scenario #3: +10kN
... or so.
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shockabuku
Sep 5, 2012, 3:50 AM
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rocknice2 wrote: I've been trying to find the answer to a question but the amount of anchor topics is overwhelming. So I bring up one more. Consider 3 protection points @ 10kn each A fall directly on the anchor with no belayer and capable of ripping any anchor The sling/cord breaking strength is not a factor. Lastly for the sake of argument the self equalizing anchor is perfect. 1 A self equalizing anchor where the outer 2 legs are 90 deg apart & equal length. 2 An anchor with a tied power point, 90 deg outer legs and center leg is somewhat loose 3 Imagine pro orientated horizontally and a PAS type sling is used to connect the pro in series and the rope is clipped to only one end. Forcing one piece to completely blow before the other even sees any load at all. What is the ultimate load each anchor can take before it collapses completely?
I think:
1. ~14 kN, assuming symmetric loading.
2. Somewhat loose is somewhat vague so hard to say, but at least ~14 kN assuming symmetric loading.
3. 10 kN.
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JimTitt
Sep 5, 2012, 7:27 AM
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acorneau wrote: rocknice2 wrote: I've been trying to find the answer to a question but the amount of anchor topics is overwhelming. So I bring up one more. Consider 3 protection points @ 10kn each A fall directly on the anchor with no belayer and capable of ripping any anchor The sling/cord breaking strength is not a factor. Lastly for the sake of argument the self equalizing anchor is perfect. 1 A self equalizing anchor where the outer 2 legs are 90 deg apart & equal length. 2 An anchor with a tied power point, 90 deg outer legs and center leg is somewhat loose 3 Imagine pro orientated horizontally and a PAS type sling is used to connect the pro in series and the rope is clipped to only one end. Forcing one piece to completely blow before the other even sees any load at all. What is the ultimate load each anchor can take before it collapses completely? Theoretically speaking... Scenario #1: +24kN Scenario #2: +14kN Scenario #3: +10kN ... or so.
+1
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jkd159
Sep 5, 2012, 3:22 PM
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Do these scenarios ignore energy dissipated by the failure of each piece of protection? For example, in the third scenario, blowing the first piece of protection takes 10kN out of the system. That energy is mostly used to stretch the rope (assuming this factor 2 fall is on a rope). So is it the case that a rope tensioned to 10kN and given no time to recover will break the next piece of protection no matter the load at this point?
The numbers make sense to me if the anchor is stressed by a pull tester which can maintain a constant force, but they do not make sense in the case of the anchor arresting a fall.
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madrasrock
Sep 5, 2012, 4:05 PM
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A few years ago I was at a Peak Rescue course at J-tree, and we pulled on three cams in a horizontal crack, at 29k one of the cams broke at the cable. One thing I noted is slack in the 7mm cord did not matter. Conclusion if the pro is 10 kn, that is what you will get. If you place four pieces, you will most likely break your carabiner first. This was a static pull.
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dagibbs
Sep 5, 2012, 4:11 PM
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jkd159 wrote: Do these scenarios ignore energy dissipated by the failure of each piece of protection? For example, in the third scenario, blowing the first piece of protection takes 10kN out of the system. That energy is mostly used to stretch the rope (assuming this factor 2 fall is on a rope). So is it the case that a rope tensioned to 10kN and given no time to recover will break the next piece of protection no matter the load at this point? The numbers make sense to me if the anchor is stressed by a pull tester which can maintain a constant force, but they do not make sense in the case of the anchor arresting a fall.
You can't "take 10kN out of the system".
As you state, you need energy dissipation -- which means you need to decelerate the falling mass (climber), which means you need a force applied over time.
So, to know how much the blowing of the first piece affects what the further pieces need to do, is very complicated.
1.) Is there time for rope recovery (to absorb more force)?
2.) Is there extra fall distance, so more energy/velocity into the system?
3.) How does the piece fail -- how quickly? Does it snap instantly? Does it drag out, crumbling rock as it goes, taking a bit of time, but with a reduced force load over that time? Some combination of the above?
That's why the answer for case three was 10+ kN.
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jkd159
Sep 5, 2012, 4:46 PM
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dagibbs wrote: 1.) Is there time for rope recovery (to absorb more force)? 2.) Is there extra fall distance, so more energy/velocity into the system? 3.) How does the piece fail -- how quickly? Does it snap instantly? Does it drag out, crumbling rock as it goes, taking a bit of time, but with a reduced force load over that time? Some combination of the above?
So assuming the pieces are close together (but not equalized), there is negligible time for rope recovery and a negligible amount of extra fall distance. Then the amount of energy dissipated from the system is simply a function of how long each piece resists the fall before breaking? I do get that stopping the falling climber requires an acceleration (deceleration). But my guess is that the time it takes for gear to fail is 1) short relative to the duration of the fall; and 2) similar in length for most types of gear failure. Gear that pulls through rock does so in about the same amount of time that gear stretches and snaps, and that time is an order of magnitude faster than the total time a climber falls. What happens when gear fails is that rope stretches until the tension in the rope exceeds the failure point of the gear, right? Stretching the rope slows the climber. So once the first piece fails the rope comes tight against the second piece. The climber is moving slower, but the rope is pre-stretched. So a climber falling with any velocity at all on a rope which is already stretched to 10kN will cause the failure of the next piece of gear. Is that true?
I'll be the first to admit that I don't understand the mechanics of this process. I'd love to have it explained.
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JimTitt
Sep 5, 2012, 7:18 PM
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jkd159 wrote: Do these scenarios ignore energy dissipated by the failure of each piece of protection? For example, in the third scenario, blowing the first piece of protection takes 10kN out of the system. That energy is mostly used to stretch the rope (assuming this factor 2 fall is on a rope). So is it the case that a rope tensioned to 10kN and given no time to recover will break the next piece of protection no matter the load at this point? The numbers make sense to me if the anchor is stressed by a pull tester which can maintain a constant force, but they do not make sense in the case of the anchor arresting a fall.
Yes, we ignored all that because the original scenario specifically stated the fall was capable of breaking any anchor, thus you can dissipate as much enrgy as you like in any way you like and it won´t effect the inevitable failure of the three proposed anchors.
I was a bit hasty giving acorneau a +1 as he shouldn´t really have put the + in front of 10kN for scenario 3, maybe a +0.9 was in order but at least he knows it is written kN so can have the other 0.1 back!
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patto
Sep 5, 2012, 7:41 PM
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jkd159 wrote: Do these scenarios ignore energy dissipated by the failure of each piece of protection?
You set up the problem not us. You wanted the ultimate breaking force and it was given. Energy doesn't come into it.
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acorneau
Sep 5, 2012, 8:40 PM
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JimTitt wrote: Yes, we ignored all that because the original scenario specifically stated the fall was capable of breaking any anchor, thus you can dissipate as much enrgy as you like in any way you like and it won´t effect the inevitable failure of the three proposed anchors. I was a bit hasty giving acorneau a +1 as he shouldn´t really have put the + in front of 10kN for scenario 3, maybe a +0.9 was in order but at least he knows it is written kN so can have the other 0.1 back!
I wrote "+10kN" because each placement can hold 10kN then you need more than 10kN to break it.
10.0000000001kN would work... theoretically, of course.
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JimTitt
Sep 5, 2012, 10:34 PM
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acorneau wrote: JimTitt wrote: Yes, we ignored all that because the original scenario specifically stated the fall was capable of breaking any anchor, thus you can dissipate as much enrgy as you like in any way you like and it won´t effect the inevitable failure of the three proposed anchors. I was a bit hasty giving acorneau a +1 as he shouldn´t really have put the + in front of 10kN for scenario 3, maybe a +0.9 was in order but at least he knows it is written kN so can have the other 0.1 back! I wrote "+10kN" because each placement can hold 10kN then you need more than 10kN to break it. 10.0000000001kN would work... theoretically, of course.
True enough, +1.1 then, need some stars as well I guess!
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rocknice2
Sep 6, 2012, 5:14 PM
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JimTitt wrote: jkd159 wrote: Do these scenarios ignore energy dissipated by the failure of each piece of protection? For example, in the third scenario, blowing the first piece of protection takes 10kN out of the system. That energy is mostly used to stretch the rope (assuming this factor 2 fall is on a rope). So is it the case that a rope tensioned to 10kN and given no time to recover will break the next piece of protection no matter the load at this point? The numbers make sense to me if the anchor is stressed by a pull tester which can maintain a constant force, but they do not make sense in the case of the anchor arresting a fall. Yes, we ignored all that because the original scenario specifically stated the fall was capable of breaking any anchor, thus you can dissipate as much enrgy as you like in any way you like and it won´t effect the inevitable failure of the three proposed anchors. I was a bit hasty giving acorneau a +1 as he shouldn´t really have put the + in front of 10kN for scenario 3, maybe a +0.9 was in order but at least he knows it is written kN so can have the other 0.1 back!
Sorry guys I did mean to take into account energy absorbed by a rope. I should have picked a lower kn for the anchors seeing as any roped fall will be held by 1 pro.
Assume a rope is in the system and it's rate to 11kn impact
My theory is the anchor 3 would fail @ +30kn and the other two slightly less because of the angle between the legs.
Again sorry for not being clear from the beginning.
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majid_sabet
Sep 6, 2012, 5:48 PM
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Your last carabiner or last piece clipped in to master point sets the strength of your anchor so even if each lag is rated to 30kn and your last biner is rated to 15kn
your anchor is 15 kn rated
(This post was edited by majid_sabet on Sep 6, 2012, 5:49 PM)
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cracklover
Sep 6, 2012, 6:27 PM
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rocknice2 wrote: JimTitt wrote: jkd159 wrote: Do these scenarios ignore energy dissipated by the failure of each piece of protection? For example, in the third scenario, blowing the first piece of protection takes 10kN out of the system. That energy is mostly used to stretch the rope (assuming this factor 2 fall is on a rope). So is it the case that a rope tensioned to 10kN and given no time to recover will break the next piece of protection no matter the load at this point? The numbers make sense to me if the anchor is stressed by a pull tester which can maintain a constant force, but they do not make sense in the case of the anchor arresting a fall. Yes, we ignored all that because the original scenario specifically stated the fall was capable of breaking any anchor, thus you can dissipate as much enrgy as you like in any way you like and it won´t effect the inevitable failure of the three proposed anchors. I was a bit hasty giving acorneau a +1 as he shouldn´t really have put the + in front of 10kN for scenario 3, maybe a +0.9 was in order but at least he knows it is written kN so can have the other 0.1 back! Sorry guys I did mean to take into account energy absorbed by a rope. I should have picked a lower kn for the anchors seeing as any roped fall will be held by 1 pro. Assume a rope is in the system and it's rate to 11kn impact My theory is the anchor 3 would fail @ +30kn and the other two slightly less because of the angle between the legs. Again sorry for not being clear from the beginning.
You seem to be a bit confused about how force is transmitted to the protection and how energy is diffused.
Let's think about what's happening in your scenario 3:
The rope stretches until it gets to the breaking point of piece number 1. Let's call this 10kN. What is causing that 10kN of force? There is roughly 4kN of tension in the rope between the belayer and the top piece, and 6kN of tension in the rope between the climber and the top piece.
Now piece #1 rips out. Tests have shown that gear failure absorbs very little energy, so the rope at the moment of failure is pretty much unchanged by that.
Now the rope "unstretches" an inch or so in the split second before it comes into contact with the next piece. Assuming the rope has already stretched a good deal (10kN of force is a significant amount when you're talking about a number of meters of rope) that inch will only reduce the tension on either side of the rope a negligible amount. So let's say for the sake of argument that that rope "unstretch" drops the force on the next piece to only 9.9kN.
If the climber still has any downward movement (which he must, since you said that the fall was enough to generate more than 10kN of force on the rope) then the tension in the rope will immediately start to rise again. Almost instantly, the force will reach 10kN (with minimal time to effect on the speed of the falling climber), and piece #2 will rip.
The same thing happens with piece #3.
So you can see - pieces in close sequence that do not share load do *not* act significantly better than a single piece.
An interesting thing happens, though, when those pieces are far enough apart, not like you'd find in an anchor, but as you might find in running protection. After piece #1 fails, let's say piece #2 is two meters below. Some preliminary research has found that even in the very short amount of time it takes for the rope to "ping" from broken piece #1 down to piece #2, it can regain some of its elasticity. So the energy in the system actually drops a bit, and piece #2 might actually hold a little better than it would have if piece #1 had never been there. Cool, huh!?
GO
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JimTitt
Sep 6, 2012, 7:27 PM
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rocknice2 wrote: JimTitt wrote: jkd159 wrote: Do these scenarios ignore energy dissipated by the failure of each piece of protection? For example, in the third scenario, blowing the first piece of protection takes 10kN out of the system. That energy is mostly used to stretch the rope (assuming this factor 2 fall is on a rope). So is it the case that a rope tensioned to 10kN and given no time to recover will break the next piece of protection no matter the load at this point? The numbers make sense to me if the anchor is stressed by a pull tester which can maintain a constant force, but they do not make sense in the case of the anchor arresting a fall. Yes, we ignored all that because the original scenario specifically stated the fall was capable of breaking any anchor, thus you can dissipate as much enrgy as you like in any way you like and it won´t effect the inevitable failure of the three proposed anchors. I was a bit hasty giving acorneau a +1 as he shouldn´t really have put the + in front of 10kN for scenario 3, maybe a +0.9 was in order but at least he knows it is written kN so can have the other 0.1 back! Sorry guys I did mean to take into account energy absorbed by a rope. I should have picked a lower kn for the anchors seeing as any roped fall will be held by 1 pro. Assume a rope is in the system and it's rate to 11kn impact My theory is the anchor 3 would fail @ +30kn and the other two slightly less because of the angle between the legs. Again sorry for not being clear from the beginning.
What you have to understand is that while a piece may hold 10kN the energy involved to break it can be minimal. Substances are either brittle or tough and climbing gear no exception and rock in particular. For a cam to ping off a limstone flake is easy and takes virtually no energy out of the faller whereas a cam ripping at 10kN load down a perfect crack from top to bottom of El Cap will require an enormous amount of energy, more than can ever be supplied by a falling climber.
So you approach the problem by calculating the energy available, remove whatever energy is used in the failure (google `energy of fracture´ as a starting point as this is technically what it is called) and then recalculate the force the climber will impose on the next piece. Best of luck!
For sequential failure of protection then you whizz through http://amga.com/resources/various/Sequential_Failure_Paper.pdf
Best of luck there as well!
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rocknice2
Sep 6, 2012, 7:29 PM
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Thanks! I was forgetting the time it takes the rope to spring back. No wonder it sounded weird.
So acorneau was right
acorneau wrote: Theoretically speaking... Scenario #1: +24kN Scenario #2: +14kN Scenario #3: +10kN ... or so.
Scenario #1:
14kn for the outer legs + 10kn for center leg
Anchor fails @ 24kn
Scenario #2:
outer legs fails @ 14kn [ lets say a rope could generate this ]
rope regains 0kn [ there is exactly enough slack to allow failure before next pro engages ]
center legs fails [ even if the rope recovered 4kn the leg would still fail ]
Anchor fails @ 14kn
Scenario #3:
1st piece fails @ 10kN
rope regains .1kn [ let's just say ]
2nd piece fails @ 10kN
rope regains .1kn
3rd piece fails @ 10kN
Anchor fails @ 10.2kn
Am I understanding this better??
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cracklover
Sep 6, 2012, 7:51 PM
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rocknice2 wrote: Thanks! I was forgetting the time it takes the rope to spring back. No wonder it sounded weird.
It has little to do with time, in your case. The fact is, there is 10kN of tension in the rope at the moment that first piece fails. What do you think can possibly happen to change that? If there is five feet of stretch in the rope causing that tension, in order for the next piece, an inch below the first one, to be at square one again, you'd have to have some kind of miraculous force literally lift the climber five feet upwards into the air to unstretch the rope! How is that going to happen in any amount of time?
In reply to: Scenario #3: 1st piece fails @ 10kN rope regains .1kn [ let's just say ] 2nd piece fails @ 10kN rope regains .1kn 3rd piece fails @ 10kN Anchor fails @ 10.2kn Am I understanding this better??
You still don't seem to quite have it. I think you're still confusing energy and force?
If each of the pieces fails at 10 kN, then they each fail at 10 kN! Yes, a small amount of energy will be taken out of the system between when piece 1 rips and piece 3 rips (although almost certainly not enough to keep piece 3 from ripping), but still, each of the three will rip when the peak force gets to 10kN.
GO
(This post was edited by cracklover on Sep 6, 2012, 7:52 PM)
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rocknice2
Sep 6, 2012, 8:06 PM
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JimTitt wrote: What you have to understand is that while a piece may hold 10kN the energy involved to break it can be minimal. Substances are either brittle or tough and climbing gear no exception and rock in particular. For a cam to ping off a limstone flake is easy and takes virtually no energy out of the faller whereas a cam ripping at 10kN load down a perfect crack from top to bottom of El Cap will require an enormous amount of energy, more than can ever be supplied by a falling climber. So you approach the problem by calculating the energy available, remove whatever energy is used in the failure (google `energy of fracture´ as a starting point as this is technically what it is called) and then recalculate the force the climber will impose on the next piece. Best of luck! For sequential failure of protection then you whizz through http://amga.com/...al_Failure_Paper.pdf Best of luck there as well!
Thanks Jim
Yes I made somewhat of a mess of this thread and that link gave me a headache.
If the limestone flake would take 10kn to snap off. The rope would absorb those 10kn out of the fall at the point it breaks, no?
The cam tracking 3000ft down El Cap. Would the energy dissipate into the rock via friction.
The question was geared towards which system holds more, all things being equal.
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sittingduck
Sep 6, 2012, 8:20 PM
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Who would have thought that a 11kn fall would rip out 30kn worth of protection like that 
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rocknice2
Sep 6, 2012, 8:22 PM
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So if the pro rips when the rope has 5ft of stretch [10kn].
I regain the energy it took to stretch the rope from 4.5ft to 5ft assuming I had 6 inches of slack between pro.
I would need 2 meters of slack between the pro to go back to square 1. Assuming the rope could recover it's properties back instantly.
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cracklover
Sep 6, 2012, 8:37 PM
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rocknice2 wrote: So if the pro rips when the rope has 5ft of stretch [10kn]. I regain the energy it took to stretch the rope from 4.5ft to 5ft assuming I had 6 inches of slack between pro. I would need 2 meters of slack between the pro to go back to square 1. Assuming the rope could recover it's properties back instantly.
Precisely. Although the rope absolutely cannot regain its properties back instantly. It can, however, regain some of it's properties.
Have you ever belayed someone toproping (or seconding a slingshot belay) where it's a fairly long pitch, and there's a hard move right off the ground? You're concerned about them decking with all the rope stretch available with all that rope out, so you take in, hard, repeatedly, sitting on the rope, and/or having the climber climb up and jump off. If you have, you know that you can remove some of the elasticity from the rope temporarily. You can make it such that the climber feels very little "lift" from the rope, but should they fall, they will fall much less far, and/or hit the ground less hard.
The reason this works is precisely because the rope does not regains all its elasticity right away. When you stretch it and then release, even though you remove most of the tension from the rope with each move up the climber takes, the rope does not fully "relax".
GO
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JimTitt
Sep 6, 2012, 8:47 PM
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EEk, back to step one!
The rope applies a force onto the cam, the cam applies a force onto the flake, if there is enough force the stress (force per unit area of the material) exceeds the strength of the material and it breaks. The energy required to do this breaking depends on the material, a good steel takes several hundred times more energy to break as concrete for example. A steel karabiner with the same breaking strength as an aluminium one would take roughly twice as much energy to break, this is why some materials are known as tough and others as brittle. How much force is taken out of the fall depends on how much energy was removed by the flake breaking, not the force required to do so.
The energy from our tracking cam would either be friction or the work done to gouge the rock out or rip the aluminium off the lobes, probably a bit of each at a guess!
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cracklover
Sep 6, 2012, 9:05 PM
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sittingduck wrote: Who would have thought that a 11kn fall would rip out 30kn worth of protection like that
This type of cascade failure is, among other reasons, why many of us have objected for years to the standard cordelette as a universal method of anchoring.
Let's say I create a standard cordelette with three pieces well spaced apart in a horizontal crack. On the middle and left side, I have two good pieces, each capable of holding up to 10 kN, but on the right side I place a tiny cam that, as it turns out, will rip out at 2 kN. I mean, I can see that the third piece isn't so good, but three pieces are better than two, right?
Now let's say the leader falls on the anchor with a fall sufficient to generate 12 kN of force. What happens?
As the force builds, it's more or less evenly split across the three pieces*, but before it can get to that 12kN, the force on the right-side piece exceeds the 2kN it's good for, and the piece rips.
Now, suddenly, we're in a situation more like the OP's case #3. The loads are no longer anything resembling evenly shared. Remember, we're still going to see something like a 12kN peak force on the anchor, and the middle piece will take almost all the load until it fails, at which time the right-side piece will be subjected to the same force.
Now, to be fair, (and before Jim Titt corrects me, better to do so myself) the left side *will* take a little bit of the force, and as the cord stretches and the strands in the knot slide a little as the force continues to build, the pieces will equalize a little bit more.
But will it be enough to keep the middle piece from ripping? That depends on things like the angle between the pieces, the type of material, etc. If that middle piece fails, the left one is a for-sure goner. And the whole party of belayer and climber is riding on the outcome. Not a good situation, IMO.
GO
*Actually, the center piece will see more, but not enough so to change the outcome in this situation.
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cracklover
Sep 6, 2012, 9:20 PM
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And because a picture is worth a thousand words:
GO
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