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ptlong
Apr 22, 2010, 9:39 PM
Post #26 of 30
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Registered: Oct 4, 2007
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Okay, have it your way.
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dugl33
Apr 22, 2010, 9:55 PM
Post #27 of 30
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Registered: Oct 6, 2009
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ptlong wrote: dugl33 wrote: 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 (??)] You want to see the math? Richard Goldstone has already done this quite nicely for us: http://www.rockclimbing.com/...ent;postatt_id=2957; Here's a quote from page 3: "There is a consequence of Equation (10) that seems little appreciated in the climbing world. Simply weighting a rope, rather than having it catch a fall, corresponds to a fall factor r = 0. For that fall factor, we find that T = 2w; the maximum tension in the climbing rope is double the climber’s weight. What happens when a rope is weighted is that it stretches until the maximum tension is twice the climber’s weight, and then recovers to the point that the tension in the rope is just the climber’s weight. But the anchor will be momentarily subjected to double the climber’s weight."
Well, if I pull down with the force of 1*w, and the anchor pulls up with the force of 1*w, the tension in the rope may be the sum of these equal and opposite forces, i.e., 2*w, but isn't the force on the anchor still 1*w (if tied off directly), and the force at the climbers harness 1*w?
Now take your cord over your "pulley" and you have 1*w on one side, 1*w on the other, which add up to 2*w on the anchor, thus, the pulley effect.
But its not a pulley, its a biner, so we have 1*w on one side + (1*w - friction) on the other, or a little less than 2*w.
Now if you have slack in the system, and the second lobs onto the rope, thats different, obviously. You are adding dynamic forces to the mix. Empirically though, I've caught seconds this way and the forces on my side never seem like that much. If the climber is close you'll get pulled toward the redirect. Rarely will a falling second actually lift you up.
And coming full circle, I stated initially I do not redirect unless the anchor is bomber. And yes, you can redirect through an equalized point if you like.
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dugl33
Apr 22, 2010, 11:39 PM
Post #29 of 30
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Registered: Oct 6, 2009
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PT that is a compelling document and I see where you and Jay are coming from now. I don't know that it accounts for the tendency of the belayer to move and the reduction in force that would (probably) result.
However, I do see how you are coming up with 2*wt*5/3 = 3.33*wt for short term peak loading of a factor 0 hang.
Thanks. Forgive the stubborness. Just need to see things directly sometimes. (I'd be happy as a clam with a graphing strain gauge.)
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davewalks
May 5, 2010, 2:19 PM
Post #30 of 30
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Registered: Jun 24, 2008
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So--well--assuming a sliding x is what I'm thinking, the engineering stuff is cool, thing is if you're not sure there's a biner actually through the sling in case one anchor fails and the other one doesn't, you might have to do some fst math
Oh, well, probably I just missed something there
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