Forums: Climbing Information: The Lab: Re: [jt512] tension: Edit Log


Oct 5, 2008, 12:22 PM

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Registered: Nov 27, 2007
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Re: [jt512] tension
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jt512 wrote:
angry wrote:
So what is the question?

Are we talking about tension? From a toprope VS belaying on top there is more rope out and since it's dynamic rope, it's harder to get it tight.

Or are we talking about the weight the anchor sees? Assuming nothing dynamic, the anchor see's your weight and the belayers weight. Kindof. Actually this brings up another question. Lets assume we have a junk anchor that will fail at 300lbs. The climber weighs 100lbs and the belayer lowering her off is 200. The way I'm looking at it, the anchor feels the belayers opposing weight up to the weight of the climber (actually a bit less due to friction). I might accept being wrong about this but that seems to make the most sense to me.

The climber is dead the moment he weights the rope, regardless of whether the belay is directly off the anchor or redirected through the anchor. Weighting a top rope is equivalent to taking a fall-factor-0 fall, which produces peak tension in the climber's rope of two times his weight, assuming, of course, that the anchor hasn't already failed. If the belay has been redirected through the anchor, then the tension in the belayer's side of the rope is about 2/3 the tension in the climber's side; that makes the peak tension in the belayer's rope in the toprope scenario 4/3 times the climber's weight. The peak force on the anchor is the sum of these two tensions, or 10/3 times the climber's weight that is, a little more than triple the climber's weight.


where does the 2/3 come from? is this an empirical value derived from research somewhere (in which case i'd like to read the link if you remember where to find it)? or is the 2/3 a value that can be determined by math and physics? in that case would you mind posting the math or PMing me?

also, why is the climber's force double the static weight? clearly the climber must realistically undergo some acceleration and therefore an increase in force, but why is it double? is this value research based or math based?

edited to add:
if this is all explained in the above link to Rich's paper, then never mind.

(This post was edited by colatownkid on Oct 5, 2008, 12:26 PM)

Edit Log:
Post edited by colatownkid () on Oct 5, 2008, 12:26 PM

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