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Re: [jt512] Best way to redirect a belay using a sliding X:
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dugl33
Apr 22, 2010, 6:44 PM
Views: 13459
Registered: Oct 6, 2009
Posts: 740
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jt512 wrote: dugl33 wrote: jt512 wrote: dugl33 wrote: absinthe52 wrote: I want to use a sliding X since I am without a cordellet for right now. As there is no Top-Shelf for the redirect, whats the best method to set this up so I can belay off my harness? Also, I'm usuing a B-52. If I want a direct belay, can I just clip another locker in to the belay anchor? Thanks Oh yes, question refers to a two bolt achor system, not trad. Instead of messing with the anchor system for a redirect, can I just clip a quickdraw into either bolt and use that? This may not be the textbook answer, but if the bolts look bomber, I'll set up the sliding x off a locker to each bolt hanger, and clip myself to it with another sling girth hitched to the harness proper and a locker to the sliding x. I will then back myself up with the rope to one of the bolts, say the right one, and tie a clove hitch directly to the locker. I leave the rope a bit slack. Next I'll clip a draw to the left hand locker and redirect the belay through this. Ok. So, its not perfectly equalized. If the second falls, twice his or her weight is on that one bolt. This common misconception is false. A clean toprope fall will weight the anchor by at least 3 times the climber's body weight. That assumes there is absolutely no slack in the rope. With slack, the impact force can be much higher. Consider what might be the worst case scenario. Say the climber fell making the last move to the anchor, and impacted the rope before the belayer could take up any of the slack, producing, say, a 2-foot fall onto three feet of rope—a factor-0.67 fall. Then the impact force on the anchor would be on the order of 12 times the climber's body weight. With a 200-lb climber, the anchor would feel about 2600 lb of force (11.5 kN). The bolt you think is bomber better had be. Jay I think what you are stating is more true if the climber is belayed in guide mode off an atc or grigri. No. For a factor-0 fall (toprope fall with no slack), the force on the anchor would still be three times the climber's weight, and for higher factor falls, the force on the anchor would be greater, but less than that caused by a static redirected belay. In reply to: But...with a redirected belay, the falling second will typically lift up the belayer somewhat if the forces are substantial. True, but the minimum force on the anchor will still be triple, not double, the belayer's weight. In reply to: Pulley effect will generally at least double the seconds weight as far as forces on the bolt..No. Friction reduces the pulley effect to around 5/3. With no pulley effect at all, a toprope fall with no slack in the system will weight the anchor two times the climber's body weight. With pulley effect, the anchor feels (5/3)*2 time the climber's weight. Slack in the rope increases the force. In reply to: What's your typical set-up with regards to the OPs query? I don't think I have such a standard setup. I have redirected the belay off of a single bolt, but I have also redirected through a dedicated sliding X not a part of the anchor. Jay
1.) I agree that friction reduces pulley effect. No argument there.
2.) How does a reduction in pulley effect result in *67% more force?
*5/3 = 1.6666...
Are you sure you don't mean 3/5, i.e. reducing for friction, factor 0 hang, (weight)(3/5)(2)= 1.2 x weight?
If you put a weight on a string, run it through a pulley, tie off the other side, carefully hang the weight, the force on the pulleys connection will equal 2 x weight, not 3 x weight. Same experiment with friction, the force must be less than 2x weight.
Edit to add: my apologies here. What I really meant to say was the sum of modest dynamic forces and the pulley effect will generally at least double the climbers weight with regards to forces on the bolt.
(This post was edited by dugl33 on Apr 22, 2010, 7:06 PM)
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Edit Log:
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Post edited by dugl33
() on Apr 22, 2010, 6:55 PM
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Post edited by dugl33
() on Apr 22, 2010, 7:06 PM
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