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Two piece anchors are plenty strong! Poll!
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Poll: Two piece anchors are plenty strong!
I always use at least 3 pieces, more if needed. 80 / 37%
I always use exactly three pieces, quoth the raven. 6 / 3%
I use two sometimes if they are good. 115 / 53%
I never use more than two. 2 / 1%
I pancake. 13 / 6%
216 total votes
 

jeremy11


May 25, 2009, 11:30 PM
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I've used zero pieces before, with a totally bomber stance (top of a squeeze chimney) and used 5 too. just depends on what is available on the rock and on the rack and how freaked out I am.


altelis


May 26, 2009, 12:16 AM
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jmeizis wrote:
There's still a small chance that all the pieces will fail so why not place 4 or 27. If you have a 12 piece anchor and 11 of the pieces fail then you'll be on one. So by that trend of thinking we can't possibly be safe...ever. We're all gonna die!

1-2-3 What are we fighting for?
Don’t ask me, I don’t give a damn. The next stop is Vietnam.
5-6-7 Open up the pearly gates.
It ain’t no time to wonder why. Yippee! We’re all going to die.


bill413


May 26, 2009, 1:53 AM
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jmeizis wrote:
There's still a small chance that all the pieces will fail so why not place 4 or 27. If you have a 12 piece anchor and 11 of the pieces fail then you'll be on one. So by that trend of thinking we can't possibly be safe...ever. We're all gonna die!
Because 3 is a magic number. 4 or 27 are not. Therefore, 3 is better.

pendereki - really liked your answer.

altelis - nice choice of music.


jmeizis


May 26, 2009, 1:56 AM
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27 is divisible by 3, 9 times, that's gotta be better than just three. Thus far we have nothing.


altelis


May 26, 2009, 2:04 AM
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bill413 wrote:
jmeizis wrote:
There's still a small chance that all the pieces will fail so why not place 4 or 27. If you have a 12 piece anchor and 11 of the pieces fail then you'll be on one. So by that trend of thinking we can't possibly be safe...ever. We're all gonna die!
Because 3 is a magic number. 4 or 27 are not. Therefore, 3 is better.

pendereki - really liked your answer.

altelis - nice choice of music.

haha. not to date myself (in the opposite way that most people use it) but i used to LOVE sitting around and listening to my dad's woodstock lps.....!


jt512


May 26, 2009, 2:20 AM
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jmeizis wrote:
There's still a small chance that all the pieces will fail so why not place 4 or 27.

Because 3 is the fewest pieces of removable protection that, by visual inspection, appear individually infallible under the highest load believed possible, for which the probability of total anchor failure is infinitesimal.

Mods, lock 'er up.

Jay


bill413


May 26, 2009, 2:24 AM
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jt512 wrote:
jmeizis wrote:
There's still a small chance that all the pieces will fail so why not place 4 or 27.

Because 3 is the fewest pieces of removable protection that, by visual inspection, appear individually infallible under the highest load believed possible, for which the probability of total anchor failure is infinitesimal.

Mods, lock 'er up.

Jay
I told you 3 was a magic number.


jt512


May 26, 2009, 2:26 AM
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bill413 wrote:
jt512 wrote:
jmeizis wrote:
There's still a small chance that all the pieces will fail so why not place 4 or 27.

Because 3 is the fewest pieces of removable protection that, by visual inspection, appear individually infallible under the highest load believed possible, for which the probability of total anchor failure is infinitesimal.

Mods, lock 'er up.

Jay
I told you 3 was a magic number.

I know, and you were right.

Jay


reno


May 26, 2009, 12:19 PM
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jmeizis wrote:
We're all gonna die!

Now you're catching on. Wink


mikeo


May 26, 2009, 8:28 PM
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tomcat wrote:
Seemingly overlooked in this discussion is actually getting the climb done.I lead 95% of the pitches I climb.What use is a terrific three piece cam anchor that sucks up your red,yellow and blue C4 if there is a hand crack looming above?Belay anchors can't always,or even frequently look like Majid's truckstops,because some of us want to climb above that point,and get some gear in.

So,in reality,you may belay from four iffy stoppers equalised however you prefer,despite the existance of a great cam crack,because the person leading the next pitch is going to need the goods.

I'll take a solid Jesus piece over a nine piece POS like Majid posted every time.

tricams make great anchors, usually don't want to fiddle with them on lead if it will take a cam, and they are bomber for building anchors. the number 4 in the pic got replaced with a tricam slightly above when my 2nd arrived. guess the climb for bonus points, it's a classic in eldo.
link: http://img29.imageshack.us/img29/8560/p1030638.jpg


(This post was edited by mikeo on May 26, 2009, 8:33 PM)


jsj7051


May 26, 2009, 8:51 PM
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See if you can get MalDay to give you his opinion on this one. I liked it.


kennoyce


May 26, 2009, 9:27 PM
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In reply to:
In reply to:
Dynamic ropes are tested for a maximum impact force which is far below 30kN. As long as they can stretch far enough to dissipate the energy of the fall they will have a tension that is not much above that impact force.

That is false. The tension is proportional to the relative elongation of the rope. Hooke's Law.

Jay

Sorry Jay, I usually agree with things you say, but hooke's law only applies to the linear elastic region, and with dynamic ropes, once you are about at the maximum impact force you leave this region (if you assume that a rope behaves in a linear elastic manner to begin with).


jt512


May 26, 2009, 10:53 PM
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kennoyce wrote:
In reply to:
In reply to:
Dynamic ropes are tested for a maximum impact force which is far below 30kN. As long as they can stretch far enough to dissipate the energy of the fall they will have a tension that is not much above that impact force.

That is false. The tension is proportional to the relative elongation of the rope. Hooke's Law.

Jay

Sorry Jay, I usually agree with things you say, but hooke's law only applies to the linear elastic region, and with dynamic ropes, once you are about at the maximum impact force you leave this region...

Huh? The rope does not have an absolute maximum impact force. The maximum impact force for a particular rope is a function of the fall factor and the weight of the climber.

Jay


majid_sabet


May 26, 2009, 11:33 PM
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kennoyce wrote:
I chose sometimes 2 pieces, not because as majidiot wants everyone to believe, 2 piece anchors are the norm (they most certainly are NOT the norm), but because that was all that was available to me. I have also rapped off of a single nut on occasion, but my typical anchor consists of either 3 or 4 pieces (more if necessary).

As has been said it is all situation dependent.

Rapping off a single piece is different than belaying off the same one. In rapping, the load is static so whatever is on the anchor is not going to affect the anchor unless you do monkey move or a sudden stop while rapping.


On belaying, you could shock load your anchor from 5kn to 30 kn and that is where the two piece anchor may not be sufficient to handle the shock load.


hafilax


May 26, 2009, 11:39 PM
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The maximum impact force figure for a rope is the maximum force measure in the first drop of the UIAA test. The requirement is that it be less than 12kN for single ropes. The force quickly goes up in the subsequent drops in the test; I've read up to 15kN by the 4th drop. I accidentally started this whole thing when I was thinking of the rope like a screamer which it isn't at all.

I just took a quick look at the literature and it seems that ropes are highly non linear and that the force-extension curve has significant upward concavity. The springiness increases as it stretches.

The following statement that I made on page 3 is incorrect as well:
I wrote:
Let me restate that as long as you don't exceed the energy of the test by much you won't get a tension in the rope much higher than the rated impact force.
It's not the energy but the fall factor that is important. Even then, for high fall factors the experimental results will high by up to 30% for a Hook's law model and that a second order fit gives better results. This is also ignoring knots but that's a whole other kettle of fish.

My main point is that there is no reason to think that ropes will hold a 30kN load. The UIAA tests must have a peak force number for the ropes when they snap but that is under highly unusual circumstances. Otherwise even in the extreme case of the 4th drop of the UIAA test the maximum tension in the rope is limited to around 15kN.


majid_sabet


May 26, 2009, 11:45 PM
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USnavy wrote:
summerprophet wrote:
majid_sabet wrote:
and not by some imaginary numbers out of your as* like how some of the climbers are using these days.

Ropes are assumed to have a breaking strength of 30kN.
.
Not even close. Not even the most extreme duty 11 mm dynamic rope will hold that. A brand new 10.5 mm dynamic rope will hold about 9 - 10 kN with a figure eight. The used 10.5 mm rope I tested held about 6.1 kN with a munter hitch.

Check this link out: http://www.blackdiamondequipment.com/...p_archive.php#120308

You will notice that the obviously new ropes tested on that page broke around 2200 lbs. Although I am not sure what diameter that rope is, it would be a safe to bet they are 10.5 mm or close.

[img]http://img255.imageshack.us/img255/8503/img2056eg5.jpg[/img]

you test what with munter ?

A box apples or 100 KG of iron block ?


jmeizis


May 26, 2009, 11:55 PM
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In reply to:
Because 3 is the fewest pieces of removable protection that, by visual inspection, appear individually infallible under the highest load believed possible, for which the probability of total anchor failure is infinitesimal.

How much less probable is it than a removeable two piece anchor in which each piece is bomber by visual inspection? How much more probable is three than four? I sucked at calculus and statistics but wouldn't it increase logarithmically for each piece placed (many things being equal, which might as well be ignored).

In which case if the probability of anchor failure for a two piece anchor is 50%, a three piece anchor is 5%, and a four piece anchor is only 2.5% then it makes sense to have more than two pieces but less so to have four pieces.

On the other hand if the probability for failure of a two piece anchor is 10%, a three piece anchor is 5%, and a four piece anchor is 2.5%, then it seems that a three piece anchor is not significantly better than a two piece anchor.

Obviously I'm pulling statistics out of my ass but I didn't see anyone putting forth any actual evidence besides three is better than two.


jt512


May 27, 2009, 12:45 AM
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hafilax wrote:
The maximum impact force figure for a rope is the maximum force measure in the first drop of the UIAA test. The requirement is that it be less than 12kN for single ropes. The force quickly goes up in the subsequent drops in the test; I've read up to 15kN by the 4th drop. I accidentally started this whole thing when I was thinking of the rope like a screamer which it isn't at all.

I just took a quick look at the literature and it seems that ropes are highly non linear and that the force-extension curve has significant upward concavity. The springiness increases as it stretches.

Could you cite the specific literature you are referring to, because it seems to contradict what rgold, who seems very familiar with the literature, has said; namely, that Hooke's Law is a very reasonable model of the max. impact force of a dynamic rope, except when the fall is very short (and very long?). Note, that these exceptions are for extremes of fall length, not fall factor.

In reply to:
The following statement that I made on page 3 is incorrect as well:
I wrote:
Let me restate that as long as you don't exceed the energy of the test by much you won't get a tension in the rope much higher than the rated impact force.

I know. I already pointed out that error, and you already agreed, IIRC.

In reply to:
It's not the energy but the fall factor that is important.

Both the energy and the fall factor are important determinants of the maximum impact force. Under Hooke's Law, the maximum impact force on the climber

T = w + sqrt(w^2 + 2krw),

where T is the maximum impact force on the climber, w is the climber's weight, k is the rope modulus (assumed constant here), and r is the fall factor.

The kinetic energy of the fall is determined by the weight of the climber and the height of the fall. The fall factor takes care of the height of the fall, but not the weight of the climber, which has an independent contribution to T, as the above equation shows.

In reply to:
My main point is that there is no reason to think that ropes will hold a 30kN load. The UIAA tests must have a peak force number for the ropes when they snap but that is under highly unusual circumstances. Otherwise even in the extreme case of the 4th drop of the UIAA test the maximum tension in the rope is limited to around 15kN.

There may be no reason to think that ropes will hold a 30 kN load, but based on the analysis you presented, there is no reason to think they won't. However, USNavy did post a link to several static pull tests in which ropes broke at loads less than 30 kN. That's the only thing in this thread, as far as I can see, that provides any insight into the question.

Jay


jt512


May 27, 2009, 12:57 AM
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jmeizis wrote:
In reply to:
Because 3 is the fewest pieces of removable protection that, by visual inspection, appear individually infallible under the highest load believed possible, for which the probability of total anchor failure is infinitesimal.

How much less probable is it than a removeable two piece anchor in which each piece is bomber by visual inspection?

How much more probable is three than four? I sucked at calculus and statistics but wouldn't it increase logarithmically for each piece placed (many things being equal, which might as well be ignored).

Under two mild assumptions, if p is the probability of one piece pulling, n is the number of pieces in the anchor, and P is the probability of all the pieces pulling, then P = p^n. Each piece added to the anchor, therefore, decreases the probability of total anchor failure by a factor of p.

In reply to:
In which case if the probability of anchor failure for a two piece anchor is 50%, a three piece anchor is 5%, and a four piece anchor is only 2.5% then it makes sense to have more than two pieces but less so to have four pieces.

On the other hand if the probability for failure of a two piece anchor is 10%, a three piece anchor is 5%, and a four piece anchor is 2.5%, then it seems that a three piece anchor is not significantly better than a two piece anchor.

None of those numbers is consistent with the equation for P above. Plug in some realistic numbers, see what you come up with, and if you still disagree that 3 is the "magic" number, post up.

Jay


altelis


May 27, 2009, 1:23 AM
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jt512 wrote:
jmeizis wrote:
In reply to:
Because 3 is the fewest pieces of removable protection that, by visual inspection, appear individually infallible under the highest load believed possible, for which the probability of total anchor failure is infinitesimal.

How much less probable is it than a removeable two piece anchor in which each piece is bomber by visual inspection?

How much more probable is three than four? I sucked at calculus and statistics but wouldn't it increase logarithmically for each piece placed (many things being equal, which might as well be ignored).

Under two mild assumptions, if p is the probability of one piece pulling, n is the number of pieces in the anchor, and P is the probability of all the pieces pulling, then P = p^n. Each piece added to the anchor, therefore, decreases the probability of total anchor failure by a factor of p.

In reply to:
In which case if the probability of anchor failure for a two piece anchor is 50%, a three piece anchor is 5%, and a four piece anchor is only 2.5% then it makes sense to have more than two pieces but less so to have four pieces.

On the other hand if the probability for failure of a two piece anchor is 10%, a three piece anchor is 5%, and a four piece anchor is 2.5%, then it seems that a three piece anchor is not significantly better than a two piece anchor.

None of those numbers is consistent with the equation for P above. Plug in some realistic numbers, see what you come up with, and if you still disagree that 3 is the "magic" number, post up.

Jay

is there a relatively and equally simple formula for approximating P when p is different for the different pieces in the anchor? or does that fuck the simple-factor up nice and good?


bill413


May 27, 2009, 1:53 AM
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altelis wrote:
jt512 wrote:
Under two mild assumptions, if p is the probability of one piece pulling, n is the number of pieces in the anchor, and P is the probability of all the pieces pulling, then P = p^n. Each piece added to the anchor, therefore, decreases the probability of total anchor failure by a factor of p.

is there a relatively and equally simple formula for approximating P when p is different for the different pieces in the anchor? or does that fuck the simple-factor up nice and good?
You multiply the individual probabilities of failure. So, if the probability of piece one failing is p1, and piece 2 is p2, and piece 3 is p3...
P = p1 * p2 * p3....
(This gives the formula that Jay stated if p1 = p2 = p3...).


jt512


May 27, 2009, 1:57 AM
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altelis wrote:
jt512 wrote:
jmeizis wrote:
In reply to:
Because 3 is the fewest pieces of removable protection that, by visual inspection, appear individually infallible under the highest load believed possible, for which the probability of total anchor failure is infinitesimal.

How much less probable is it than a removeable two piece anchor in which each piece is bomber by visual inspection?

How much more probable is three than four? I sucked at calculus and statistics but wouldn't it increase logarithmically for each piece placed (many things being equal, which might as well be ignored).

Under two mild assumptions, if p is the probability of one piece pulling, n is the number of pieces in the anchor, and P is the probability of all the pieces pulling, then P = p^n. Each piece added to the anchor, therefore, decreases the probability of total anchor failure by a factor of p.

In reply to:
In which case if the probability of anchor failure for a two piece anchor is 50%, a three piece anchor is 5%, and a four piece anchor is only 2.5% then it makes sense to have more than two pieces but less so to have four pieces.

On the other hand if the probability for failure of a two piece anchor is 10%, a three piece anchor is 5%, and a four piece anchor is 2.5%, then it seems that a three piece anchor is not significantly better than a two piece anchor.

None of those numbers is consistent with the equation for P above. Plug in some realistic numbers, see what you come up with, and if you still disagree that 3 is the "magic" number, post up.

Jay

is there a relatively and equally simple formula for approximating P when p is different for the different pieces in the anchor? or does that fuck the simple-factor up nice and good?

The two assumptions in my formula are (1) that the p's for each piece are equal and (2) that the failure of any piece does not affect probability of failure of any other piece (ie, independence). If you throw out equality of the p's but retain independence then the formula for n pieces becomes

P = p[1]*p[2]*p[3]*...*p[n],

in other words, P is just the product of the probabilities of the individual pieces failing. And, if the anchor contains k pieces, adding a (k+1)th piece reduces P by a factor of p[k+1].

If you throw out the independence assumption, you'd have to have, for each piece k in the anchor, an estimate of the conditional probability of the failure for each other piece in the anchor, should the kth piece fail, which, except for simple situations, like "the whole anchor will fail if a single piece blows," would be a mess.

Jay


(This post was edited by jt512 on May 27, 2009, 1:57 AM)


jmeizis


May 27, 2009, 2:48 AM
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So if we just peg the probability of one piece pulling at say 50% for simplification then the probability for total failure of:

2 piece anchor: 25%
3 piece anchor: 12.5%
4 piece anchor: 6.25%
5 piece anchor: 3.12%
6 piece anchor: .02%

If we go with a more likely probability of individual gear failure like say 10% then the difference is even smaller:

2 piece anchor: 1%
3 piece anchor: .1%
4 piece anchor: .01%
5 piece anchor: .001%
6 piece anchor: .0001%

So I don't disagree that three is a magic number. It's got the highest difference in probability besides a two piece anchor over a one piece anchor, but that goes along with what I was saying before each piece makes for a better anchor. I think it makes the point quite obvious that the quality of the gear is more important than the number of pieces. If the gear has a 1% chance of pulling then it seems that a two piece anchor is highly likely to not fail (1 time out of 10,000). If their is a 75% chance of gear pulling then I personally wouldn't feel comfortable till I had more than five pieces and even then I would be afraid to weight the anchor.

Once you add in factors like probability of falling, dynamics of the belay, and all the many factors that play into climbing it seems that with bomber gear (x<25% of individual anchor pieces failing?) that two would be adequate, albeit not perfect. Already I'm injecting my own risk tolerance into things by accepting at maximum a 6.25% probability of total failure. With more probability of individual pieces failing it seems that three or more is necessary. This all goes along with what others were saying about the quality of rock, availability of gear, and what the ground ahead looks like.

Either way it makes for an interesting thing to think about when building anchors. It also makes me ever more scared to climb on sandstone. Maybe we aren't all gonna die. Just some of us.


(This post was edited by jmeizis on May 27, 2009, 2:49 AM)


jt512


May 27, 2009, 3:19 AM
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jmeizis wrote:
So if we just peg the probability of one piece pulling at say 50% for simplification then the probability for total failure of:

2 piece anchor: 25%
3 piece anchor: 12.5%
4 piece anchor: 6.25%
5 piece anchor: 3.12%
6 piece anchor: .02%

I said give it some realistic numbers. If the probability of failure of a piece that you place and deem to be bombproof is actually 50%, then the optimal number of pieces your anchors should contain is 0, because you have no business leading trad in the first place.

In reply to:
If we go with a more likely probability of individual gear failure like say 10% then the difference is even smaller:

2 piece anchor: 1%
3 piece anchor: .1%
4 piece anchor: .01%
5 piece anchor: .001%
6 piece anchor: .0001%

Those are interesting numbers, but I don't know why you posted them, given the following:

In reply to:
If the gear has a 1% chance of pulling then it seems that a two piece anchor is highly likely to not fail (1 time out of 10,000).

Actually, you've just made my point. If a single piece has a 1% chance of pulling, then, yes, a 2-piece anchor has a probability of failing of 1/10,000. That is not an infinitesimal probability (which was my criterion). But, if you add a third piece, now you're talking about a 1-in-a-million probability. Now that represents practically no risk. I'm willing to call an anchor that has a 1-in-a-million chance of failing "bombproof." But one that has a 1-in-10,000 chance of failing? No way.

In reply to:
Already I'm injecting my own risk tolerance into things by accepting at maximum a 6.25% probability of total failure.

Then you're a fucking lunatic. I hope you have disclosed your tolerance for risk to your partners. There is absolutely no fucking way I'd ever climb with anyone who was as casual about risk as you are.

Jay


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May 27, 2009, 3:50 AM
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Posts: 558

Re: [bill413] Two piece anchors are plenty strong! Poll! [In reply to]
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bill413 wrote:
I told you 3 was a magic number.

that was the best of the Schoolhouse Rock vids---

---funny, but i used 0 pieces on a fourth class walk-off because the second was sketchy, rapped off 1 piece, had a hanging belay on 2 pieces [cuz' there was nothing else], but built most anchors on 3 pieces out of 18 pitches just this weekend. 4 is a rarity [unless you count opposing nuts as individual pieces]. i guess---

---i'm just gonna die

---like Jay said-- 'lock 'er up'

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