Forums: Climbing Information: Technique & Training: Re: [ptlong] Best way to redirect a belay using a sliding X: Edit Log




dugl33


Apr 22, 2010, 8:36 PM

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Registered: Oct 6, 2009
Posts: 740

Re: [ptlong] Best way to redirect a belay using a sliding X
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ptlong wrote:
dugl33 wrote:
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.

The key is that in a toprope fall the climber doesn't carefully hang his weight on the rope, he falls onto it. If the climber gently eased onto the rope the tension would equal his weight, and the force on the top piece would be about 5/3 his weight. But when the climber falls onto a rope with no slack and no initial tension, he first falls past the equilibrium point. At the low point of this fall the tension in the rope would theoretically be twice his weight, and the force on the top piece 5/3 of that, or 3.3 times his weight.

The only thing Jay left out is that the simple analysis assumes a belay tied off statically a foot below the redirect, and a dead weight on the other end. The dynamics of a real belayer and the give in a real climber in a fall of just two feet would significantly reduce the predicted force. Unfortunately there's no easy formula to calculate this.

Oh man you guys are killing me. I started out merely trying to point out the downside of a redirect is the force multiplier of the pulley effect.

edit -- 5/3 does not equal 3.3 ok, on second glance you're not really saying this, but you are also not backing up your statement regarding tension theoretically being 2x weight. [Its as if you are calculating tension in a pendulum swing now (??)]

I imagine Jay meant to say 5/3 x weight, but he did not say this, he said 5/3 x 2 x weight.

When I say "carefully hang the weight" I'm just trying to isolate in simple terms static forces from dynamic forces, and point out that the force multiplier from just the pulley effect is 2.0 x or less. It will never be exactly 2.0 because even pulleys are not entirely without friction, much less biners.

(If you really wanted to figure out the force on the bolt the best way would be to empirically measure it with a strain gauge, otherwise you have to consider the seconds weight, fall distance, friction- slide or freefall properties of the second's fall, other friction of the rope running through gear and features below, how much rope is out, dynamic properties of the rope at that moment in time, knot style, air density, surface area of the second, pulley effect reduced by friction, the belayers weight and dynamic give when the belayer gets lifted up a bit...Crazy)


(This post was edited by dugl33 on Apr 22, 2010, 8:58 PM)



Edit Log:
Post edited by dugl33 () on Apr 22, 2010, 8:58 PM


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