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venezuela
Aug 10, 2003, 3:55 AM
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I figure that this has ben posted before, since I have the vague idea I read about it once, but I can't find any thread that answers my question.
the other day, a couple of friends and me argued for a long while on this question: while belaying the second, what setup puts the most stress: belaying from the anchor (reverso, munter hitch, ABS...) or from the waist (figure 8, ATC, grigri, munter hitch...) re-directed through the anchor?
that question brought other topics to the discussion, like: where on the anchor do you have to re-direct when belaying the second from the waist?; when you only have 2 points to anchor in, what setup do you use?; when the placements are body-wheight only, or very marginal, what setup do you use?...etc...
I won't post my opinions about those questions because I wouldn't want to "contaminate" your answers...
I would very much appreciate your colaboration on the matter. i'm very eager to finally know the "facts and fiction" on the subject.
thank you.
Diego
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janiszewski11
Aug 10, 2003, 4:56 AM
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In most situations it is better to belay off of the powerpoint of the anchor. By belaying off the anchor only the bodyweight and minimal force of a toprope fall will be put on the anchor, and the belayer won't get a harness wedgie. However if the anchor is poor and you doubt it's ability to hold a toprope fall you should belay from the harness. The belayer however must have a solid stance that he/she can't get pulled out of or else the weight of both the belayer and climber plus their falling force will be put on the poor anchor(bad idea). Finally the setup will not make much difference unless the belayer allows too much slack to accumulate and a dynamic belay is required, thus a munster would be best as the others autolock. Oh yeah remember never use an ATC. Also when redirecting the anchor should be the strongest thing around so yes you often do redirect through the anchor's powerpoint.
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rgold
Aug 10, 2003, 4:57 AM
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Assume the belayer does not let any rope slip through the belay device and, if giving a redirected belay, does not move. Assume further that there is negligible rope friction between belayer and second.
When a climber weights the rope, the rope stretch gives rise to twice the climber's weight in rope tension. Consequently, the very least the anchor can be subjected to if the belay device is attached directly to the anchor is twice the second's weight. (A bit more, because the force exerted by the belayer must be added.) If there is any slack in the rope when the second falls, then the load on the anchor can be higher, in some cases much higher.
If the belay is from the waist and redirected through the anchor, then it is commonly accepted that friction over the carabiner will reduce the force felt by the belayer by 1/3. This gives double the climber's weight on one side of the biner and 2/3 of that on the other side, or about 2 2/3 the climber's weight on the anchor.
So a 150 lb second caught without slack with the belay device on the anchor will impose a bit more than 300 lbf on the anchor; the same situation but with a redirected belay will impose a 400 lbf load.
In general, for the redirected belay, the anchor will be subjected to about 1 2/3 the load that it would get if the device was directly on the anchor, so the redirected belay always imposes a larger load.
However, if your anchors are so bad that choosing between these small loads is a consideration, than either the belayer should set up a braced belay that does not load the "anchor" at all (if this is possible) or else the party should think seriously about unroping. If you have any kind of ordinarily decent anchor, whether or not you redirect the belay is of no consequence.
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muncher
Aug 11, 2003, 12:02 AM
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In reply to: When a climber weights the rope, the rope stretch gives rise to twice the climber's weight in rope tension.
Sorry mate but I think that maybe you should check you physics notes again. The stretch in the rope reduces the force on anchors etc in the event of a fall but has nothing to do with the tension in a static situation.
If you hang an 80kg object on a rope in a static situation (regardless of how much elongation) the tension in the rope is 80kg multiplied by the accelleration due to gravity (80kg*9.8m/s^-2=784N). The force is in no way doubled.
Didn't bother reading further than this so will not comment on the rest of your post.
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alpnclmbr1
Aug 11, 2003, 12:25 AM
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In reply to: In reply to: When a climber weights the rope, the rope stretch gives rise to twice the climber's weight in rope tension. Sorry mate but I think that maybe you should check you physics notes again. The stretch in the rope reduces the force on anchors etc in the event of a fall but has nothing to do with the tension in a static situation. If you hang an 80kg object on a rope in a static situation (regardless of how much elongation) the tension in the rope is 80kg multiplied by the accelleration due to gravity (80kg*9.8m/s^-2=784N). The force is in no way doubled. Didn't bother reading further than this so will not comment on the rest of your post.
try again muncher, the rule of thumb for a dynamic rope fall belayed directly off the anchor is two times the climbers weight. His wording may have been off a little, but his numbers are correct.
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muncher
Aug 11, 2003, 12:33 AM
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I'll take your word for it alpnclmbr1. I guess I missunderstood the situation in that the force in question was from a climber falling on a top rope rather than just hanging on it. Sorry for causing any confusion (my physics is still correct though, just for a different situation).
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muncher
Aug 11, 2003, 12:37 AM
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Oops, forgot to apoligise to rgold for misinterpreting and criticising their post. Sorry mate. :oops:
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alpnclmbr1
Aug 11, 2003, 2:27 AM
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This stuff gets confusing to say the least, and agreed that your numbers were correct too.
cheers,
d.
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muncher
Aug 11, 2003, 2:35 AM
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Thanks mate.
There you go, a little misunderstanding and no flames in sight.
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rgold
Aug 11, 2003, 4:46 AM
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Muncher,
No offense taken. I am using the spring model for elastic rope behavior, a model which, like all models, only approximates reality, although this one does rather well. When the climber weights the rope it stretches a certain amount, causing a short fall. (Not a free fall, but a loss of elevation nonetheless.) The energy of this fall must be equal to the work done in stretching the rope. This gives an equation that can be solved for the stretch, and the stretch can then be converted to rope tension. Twice the climber's weight is the result. The actual math is tenth grade algebra.
Twice the climbers weight is the peak load on the anchor. There is a small "bounce," returning the climber to a height at which the rope tension is equal to the climber's weight, as you say.
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muncher
Aug 11, 2003, 5:21 AM
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That is a pretty clear expanation.
Thanks rgold.
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pico23
Aug 11, 2003, 6:50 PM
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Unless the anchor is completely bogus, which should almost never ever be the case in normal cragging belay either directly off the anchor or redirected as per the situation. The only time I don't use the anchor at all is if I am belaying the top of a (shoot I don't remember what term we sort of agreed meant the same as 4th classing/simulclimbing/running belay) running belay and thats because I normally don't have an anchor, I just brace behind a boulder and belay off the harness.
If the anchor is bogus on a 5th class pitch you probably don't want to belay or redirect through it but I haven't been in this situation yet.
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joegoesup
Aug 11, 2003, 8:16 PM
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I would say belay from the harness to keep your belay as dynamic as possible.
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fear
Aug 11, 2003, 8:27 PM
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In reply to: I would say belay from the harness to keep your belay as dynamic as possible.
You wouldn't want a dynamic belay for the second unless he's a really, really big dude....
-Fear
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puma
Aug 11, 2003, 10:01 PM
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In reply to: You wouldn't want a dynamic belay for the second unless he's a really, really big dude.... In reply to: No, I would use a dynamic belay (off the waist, etc.) if my anchor is sketch, regardless if my follower is big or not. But if my anchor is bomber I usually belay straight off it. Lg
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venezuela
Aug 12, 2003, 7:33 PM
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well..thanks everyone for giving your advice on te matter...
now that that issue is resolved between my friends and I, another question rose: "when you have a 2 point anchor, tying in with the slider knot, and decide to belay from the waist you redirect from a). the strongest point alone, or b). from the slider knot point ??"
The situation is when the anchor points are NOT bomb proof...
should I tie in with independents slings?? and if so, where do I redirect from??
thanks...
Diego Martinez.
edited for spelling
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thegreytradster
Aug 12, 2003, 7:48 PM
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Throw in another piece. Redirect from that. Always a better option than redirecting from the anchor. You can usually put it where it will do the most good. Sometimes that's a ways from the anchor, (moved accross a ledge etc.)
The squawking about redirecting for the second mystifies me a little. It's not that big a deal. More important to make sure they won't take a big swing somewhere.
Making sure there's a bomber non directional first piece for the next lead is a big deal!
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climbrc
Aug 12, 2003, 7:56 PM
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Boy ego's seem to get in the way here. Let me see if I can keep mine out. A little more clarification: Let's see if I can explain this properly. Back to physics. Assume the re-directional carbiner is a frictionless pulley. Thus the force on one side must equal the force or "tension" on the other side (i.e. belayer's, for no movement to occur). Therefore the force on the anchor is 2wice the climbers. 2nd item: The stretch does not directly reduce the force. The force is the force. However, the stretch reduces the STRESS in the rope and the resulting SHOCK force onto the anchor (conservation of energy). That's why we use dynamic ropes, otherwise we should just use static ropes. I like to belay from my hip through a directional similar to TR. That way I keep my balance, I am comfortable, and can keep control of the situation much better.. Unless I use my REVERSO.. :-)
That's worst case.. no friction.. no stretch.. non-dynamic belay...but also excludes impact factor..
PS. Why is your second even falling?? The big sissy.
hehe
Cheers ALL..
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rgold
Aug 12, 2003, 9:56 PM
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In reply to: Assume the re-directional carbiner is a frictionless pulley.
It isn't.
In reply to: Thus the force on one side must equal the force or "tension" on the other side
In fact, the frictional force over the carabiner has been measured to be about 1/3 of the applied load. Thus, the holding force on one side is about 2/3 the tension on the other side.
In reply to: Therefore the force on the anchor is twice the climbers.
Not twice, but 5/3. Also, the climber's what? One suspects the climber's weight is what is intended, but this is wrong if you care (as you should) about the peak tension in the rope, which will be twice the climbers weight (calculation below for those who care).
In reply to: ...the stretch reduces the STRESS in the rope and the resulting SHOCK force onto the anchor (conservation of energy).
Does anyone know what these terms mean? For example, rope tension isn't reduced by stretch, it is increased by it. What then is the "stress" quantity that is reduced? And what is the distinction between a force and a shock force?
[physics]When the climber weights the rope, it stretches. As it stretches, tension builds up in the rope. According to the approximation afforded by the spring model, the tension in the rope is proportional to the percentage stretch of the rope. The maximum tension built up in the rope is the peak force the anchor has to hold. As indicated above, the anchor load will be 5/3 of the maximum rope tension, and that maximum tension will be twice the climber's weight and not simply the climbers's weight.
Conservation of energy comes in when one calculates the rope stretch that occurs when the climber's weight is applied. The rope stretch causes the climber to lose elevation and therefore potential energy. By conservation of energy, the loss of potential energy must be equal to the work done in stretching the rope. If s is the rope stretch, L is the length of rope involved, and and W is the weight (not the mass) of the climber, than the potential energy loss is Ws, the tension in the rope corresponding to a particular value of s is T=K(s/L) for some constant K, and the work done in stretching the rope is (1/2)(K/L)s^2. Solve the conservation of energy equation Ws=(1/2)(K/L)s^2. for s (discarding the solution s=0) and multiply the s-value by L/K to get T=2W.
If the belay is redirected through the anchor, the tension value 2W must be multiplied by 5/3 and the anchor load is thus 10W/3.[/physics]
Correction: In my previous post, I somehow converted 10/3 to 2 2/3 rather than 3 1/3 (sigh). The 150 lb second whose fall is caught, without slack, on a redirected anchor imposes a 500 lbf peak load on the anchor, not 400 lbf as I said.
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robmcc
Aug 12, 2003, 10:16 PM
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In reply to: In reply to: Assume the re-directional carbiner is a frictionless pulley. It isn't.
Can we skip to the part where we use spherical climbers of uniform density?
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mreardon
Aug 12, 2003, 10:19 PM
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Wow, who knew you could make belaying a physics lesson. It's pretty simple, you belay attached to you, and put a piece in above you with the rope clipped through (almost a mini-toprope) which gives no impact to the anchor at all. Anything other is getting ridiculous, and provides more ways to screw up.
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puma
Aug 13, 2003, 4:40 AM
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In reply to: now that that issue is resolved between my friends and I, another question rose: "when you have a 2 point anchor, tying in with the slider knot, and decide to belay from the waist you redirect from a). the strongest point alone, or b). from the slider knot point ??" The situation is when the anchor points are NOT bomb proof... should I tie in with independents slings?? and if so, where do I redirect from??
This is when you wouldn't re-direct, I'd belay from my waist if my 2-point anchor is sketch. I tie-in with the rope.
Lg
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alpnclmbr1
Aug 13, 2003, 5:01 AM
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In reply to: "when you have a 2 point anchor, tying in with the slider knot,
The situation is when the anchor points are NOT bomb proof...
In reply to: should I tie in with independents slings?? and if so, where do I redirect from??
Yes/ sling and the rope, equalized.
redirect/ I probably wouldn't if the anchor was bad
rgold: thanks for the physics review, I think it is good to know the why's involved
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rgold
Aug 13, 2003, 5:42 AM
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In reply to: You belay attached to you, and put a piece in above you with the rope clipped through (almost a mini-toprope) which gives no impact to the anchor at all. Anything other is getting ridiculous, and provides more ways to screw up.
When belaying the second, you usually have the choice to redirect the belay through a bomber multi-piece set-up, i. e. the belay anchor. Replacing this redundant distributed anchor with a single point above the belay anchor raises the screw-up potential rather than lowering it. The belay now depends on a single piece, whose failure (however unlikely) will subject the second to an honest-to god leader fall (equal to double the distance from belay device to redirection anchor) and load the belayer and belay anchor with a corresponding leader fall sized impact load. This outcome may be unlikely, but if we are counting ways to screw up, this method rates higher than redirecting through the belay anchor.
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lightandfast
Aug 13, 2003, 7:06 AM
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Essentialy with a re-direct you are creating a vector force. (pulley) With all factors constant, a direct anchor belay is best. With a re-direct you must make damn sure the piece you re-direct off of is (a) a bolt or (b) 110% BOMBPROOF PIECE.
You dont want a dynamic belay at all. All you are doing is increasing fall forces, on a single piece further stressing the anchor, the tighter belay the better.
Enjoy
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