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atanarjuat
Aug 1, 2004, 3:46 AM
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So I just took a course on top-rope anchors and rappelling, and it was mostly straightforward, sensible-sounding stuff, but my instructor introduced me to a rule of thumb that confused me.
Suppose you're building an equalized anchor for a pair of sturdy bolts. I'm told that I should place my focal point or equalizing knot low enough that the slings or cordelette make a narrow "V" reaching up to the bolts. The reasoning here is that "shallow angles multiply force" and focal point or an equalizing knot that pulls on the bolts in a wide "V" exert a higher effective force on these anchor points.
This, I don't get. I can see how this "angle" advice may be important for directional protection, like a cam, but no matter what happens, each bolt must support half my weight, and I don't see why it would care about the direction of pull.
Is anyone familiar with this? Physically speaking, what is the background reasoning behind this "multiplication of force?"
Thanks.
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atanarjuat
Aug 1, 2004, 4:31 AM
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Ah, brilliant. Thanks.
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pico23
Aug 1, 2004, 4:52 AM
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two well placed bolts in good rock won't fail regardless of the angle, but hey i don't trust bolts anyway, just use gear
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tedc
Aug 2, 2004, 4:04 PM
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In reply to: .... but no matter what happens, each bolt must support half my weight, and I don't see why it would care about the direction of pull.
Non-Physics Explaination:
In a simple (one dimensional) sense you are correct. Each bolt will "support" 1/2 your weight IN THE VERTICAL DIRECTION.
However, the greater the angle between the slings the more they pull AGAINST each other and the more force they exert on the bolts in the HORIZONTAL direction. This force is additive with the vertical force.
This type of understanding of force direction and force multiplication is critical to building protection systems. Make it a point to understand it before you decide to go building anchors.
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overlord
Aug 2, 2004, 4:12 PM
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you had simple physics in school, right??? remeber where you had to draw forces as vectors???.
now draw the anchor and see ho the forces work on bolts, pay special attention to the direction in wich the anchoring equipment works on the bolt. now take the two vectors (one for each bolt) and divide them into their horizontal and vertical components and youll see why the angle of the V is important.
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ben87
Aug 2, 2004, 5:57 PM
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If you clip the bolts at a very wide angle, you are creating a simple machine that generates additional force. Does the diagram below make sense?
>1kn X----------------------X >1kn
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V
1kn
This is a CRITICAL concept. If you want to test this concept, tie a rope to a tree. Now have two or three friends hold the rope taut. Grap the middle of the rope and pull at a right angle.
Two or three friends X----------------------X Tree
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V
You pulling
You should easily be able to win this tug of war, because you've created a mechanical advantage for your self. In just the same way, clipping two anchors at a very wide angle creates a mechanical advantage, multiplying the force created by a fall.
-Ben
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ben87
Aug 2, 2004, 5:58 PM
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damn, those diagrams didn't come out right...
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the_pirate
Aug 2, 2004, 6:00 PM
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Someone was dozing off in trigonometry class.
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ben87
Aug 2, 2004, 6:00 PM
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OK, this works. (always remember to preview!)
Two or three friends X----------------------X Tree
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----------------------------------V
-----------------------------You pulling
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ben87
Aug 2, 2004, 6:03 PM
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in this tug of war, you win, for the same reason that a wide angle clipping pro multiplies force on the pro.
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atanarjuat
Aug 3, 2004, 8:56 PM
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Thanks, guys.
So after reading your stuff, I sat down with the one sheet of paper I could find in my room and worked it out myself. It's one of those things that seemed intuitively appealing but had my suspicion going. I'm glad to know my instructor was right.
In the end, I was satisfied with a general expression to find the tension pulling on any given anchor bolt in any roughly-2D anchor geometry (very similar to what was in those websites, but I think there were some mistakes).
For those of you who're interested but lazy, the essential equation is
Ta*cos(alpha) = L/n
where Ta is the tension on the rope attached to a given anchor bolt "a" and "alpha" is the angle it makes with the vertical. "L" is the load force, which actually is roughly twice the climber's body weight in a free-hanging situation, and probably much more in a fall. "n" is the number of anchor points you have.
What I found interesting here is that it means that "equalizing" an anchor is a bit of a misnomer. Because the angles to the anchor points are always different, the tensions are still always different (imagine a four-bolt anchor, for example). Really, you're "minimizing" the distribution of tension.
Another interesting realization that comes from this is that you can remember this formula if, for whatever reason, you want to favour one anchor-point over another. Suppose you'd like to put less tension on a sketchy-looking piton, or a weaker-looking sling (God forbid). You can also make a quick mental-math check to decide if, given how much rope you have, it is actually safer to go with just two anchor points instead of three. Very nice.
Thanks again.
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fallenfreesoloist
Aug 3, 2004, 9:22 PM
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a recent rock and ice had a good article on anchors. cheack it out if u can find a copy.
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musicman
Aug 4, 2004, 5:12 AM
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wow, great post, after taking a basic physics class this last semester in school i pretty much get what your'e saying, in basic terms though you're saying that if the angle is really wide, there will be a lot of horizontal force on the bolts, where as if the angle is very acute then there will be little to no horizontal force on them and almost all downward/vertical. So if i'm safety-ed into the bolts at the top of a route and i have a very long sling connecting me to the two bolts there will be less horizontal force (which is bad?) than with a short sling?
sorry if i somewhat sabotaged the thread, been meaning to ask this but this is basically the same thing, so figured it'd fit here nicely.
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atanarjuat
Aug 4, 2004, 6:00 AM
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That's right-- it's as though the bolts are now pulling against each other. There's a larger horizontal component on the bolt when the rope pulls more to the side, which results in a larger tension than if the rope could hang straight down. So in your question, the short sling would be worse than the long sling.
But the good news is that you don't need to start extending your anchors to absurd lengths. If you look at a graph of y=cos(x), you'll see that there isn't much change in y for x= 0 to 30 degrees. In fact, cos(30deg)= ~0.866, which is pretty darn good when you look at the above equation. And cos(45deg)=0.707.
So shoot for an angle of less than 45 degrees (relative to the vertical) for each anchor-point, and you'll have a bomber rig.
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musicman
Aug 4, 2004, 6:11 AM
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oh, yeah, i'm not planning on all of a sudden getting super long slings or anything, i was mostly curious of how the horizontal force effects the anchor in general, thanks for the answer and keeping it simple enough for me to follow!
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rgold
Aug 4, 2004, 4:56 PM
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In reply to: the essential equation is Ta*cos(alpha) = L/n, where Ta is the tension on the rope attached to a given anchor bolt "a" and alpha is the angle it makes with the vertical. L is the load force, which actually is roughly twice the climber's body weight in a free-hanging situation, and probably much more in a fall. n is the number of anchor points you have.
This is approximately right when n=2 and the two anchor bolts are at the same level. For n>2 it is not, in general, correct. The result relies on the assumption that each anchor experiences the same vertical component of the total load, an assumption that is not true in general and may not even be approximately true in very simple special cases.
Most of the analyses of anchor forces I've seen do not account for the mechanism by which the cordalette transmits forces to the anchor points. What happens when the power point is loaded is that the cordalette arms stretch a little, lowering the power point. The tension in each cordalette arm is proportional to its relative elongation, and the relative elongations of different arms will, in general be different.
To see just how far from equalization this can be (and so how far off the above formula can be), consider the simple and quite prevalent case of three anchors in a vertical line. For the sake of an example, suppose that there is one foot from power point to lowest anchor and that each of the two higher anchors is one foot above its predecessor. This means that the arms from the power point to the anchors are 1, 2, and 3 feet respectively. When loaded, the power point experiences a downward displacement of e feet (e will typically be a small fraction). The relative displacements are thus e, e/2, and e/3, which means the tension in the 2 foot arm is half the tension in the 1 foot arm and the tension in the 3 foot arm is 1/3 the tension in the one foot arm. Put another way, the lowest piece takes about 55% of the load, the middle piece takes about 27% of the load, and the top piece takes about 18% of the load. The formula given above---and conventional "wisdom" about so-called anchor equalization---says that each anchor takes 1/3 of the load.
(When there are 3 or more anchors not in a vertical line, the calculations are complicated enough to make a computer algebra system desirable. The congnescenti will recognize that the three-anchor situation is what the civil engineers call statically indeterminate, which means that the force equilibrium data by itself is not enough to determine the tensions in each arm.)
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atanarjuat
Aug 4, 2004, 5:19 PM
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Oh! I see what you're saying. This difference is that I was treating the rope as a "perfect" rope that wouldn't stretch, yes? If the power point is far enough away from the bolts (probably too far away for rock climbing purposes), my approximation would agree with you, though.
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