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expanding flakes: nut vs. cam?
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By Crag Dweller
From New York, NY
May 24, 2011
My navigator keeps me from getting lost
i don't want to distract from the original question but i was wondering...

would a thin dyneema sling slung around a dead, rotting tree exert more force than a fat nylon sling? and, please, no speculation. we need to be discussing in terms of quantitative analysis. this is important stuff.

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By kennoyce
From Layton, UT
May 24, 2011
Climbing at the Gallery in Red Rocks
Aric Datesman wrote:
. . Actually, he's not; see post above re: Empedocles. As for the cam/nut thing, I'd rather see someone put it to rest by measuring something rather than continue watching this turn into RC.


Don't you miss RC even just a little Aric? ;)

I'd also love to see someone measure something, unfortunately, you're the only person I know of who has that ability currently. If you do get time to do some measurements, make sure the plates are textured where the nut contacts them because I'm sure that the frictional forces will keep the nut from expanding the flake (or plates in this case) nearly as much as the cam would.

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By DannyUncanny
From Vancouver
May 24, 2011
Ok, I concede now, I was wrong.

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By brenta
From Boulder, CO
May 24, 2011
Cima Margherita and Cima Tosa in the Dolomiti di Brenta.  October 1977.
Aric Datesman wrote:
Actually, he's not; see post above re: Empedocles.

Aric, no one disputes that Empedocles put forth the four-element theory and that it held sway for a couple of millennia. What is incorrect and naive is to believe that the Greek had no interest in measurements. Look no further than at Eratosthenes' measurement of the circumference of the Earth. It is also incorrect to conclude from the uselessness of one experiment the uselessness of the experimental method.

Aric Datesman wrote:
As for the cam/nut thing, I'd rather see someone put it to rest by measuring something rather than continue watching this turn into RC.

It's much easier to question one's experimental setup than a mathematical derivation. The kind of person who is not convinced by the mathematical derivation may, for instance, argue that you only measured half the force. Another may believe that your load cell is defective, that rock flakes do not behave like your metal plates, and so on.

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By Aric Datesman
May 24, 2011
kennoyce wrote:
Don't you miss RC even just a little Aric? ;)


Not really. Too much of a time suck.

kennoyce wrote:
I'd also love to see someone measure something, unfortunately, you're the only person I know of who has that ability currently. If you do get time to do some measurements, make sure the plates are textured where the nut contacts them because I'm sure that the frictional forces will keep the nut from expanding the flake (or plates in this case) nearly as much as the cam would.


That's the beauty of fixturing it like that.... No special equipment needed. All you need are some nuts, bolts and some metal stock from Home Depot, along with something heavy and a way to measure accurately. Were it not for needing to finish the project I'm working on (and 2weeks late with) I'd do it, but alas, I can not.

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By Aric Datesman
May 24, 2011
brenta wrote:
Aric, no one disputes that Empedocles put forth the four-element theory and that it held sway for a couple of millennia.


Not to argue with you Brenta, but you did just a couple posts ago. As for the rest, as an objective observer of this discussion it's clear that there's several people putting forth theories for how it works based on their understanding of the subject. The easiest way to sort the wheat from the chaff is to get some empirical evidence to test the theories rather than each standing there proclaiming themselves to be the only one who really understands the problem.

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By Boissal
From Small Lake, UT
May 24, 2011
Sooooooooo... Are we any closer to getting any form of answer to the OP's question (even if it's intuitive/based on anecdotal evidence) or are you guys going to continue displacing large amounts of air by tooting your own horns?




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By Optimistic
From New Paltz
May 24, 2011
brenta wrote:
Aric, no one disputes that Empedocles put forth the four-element theory and that it held sway for a couple of millennia. What is incorrect and naive is to believe that the Greek had no interest in measurements. Look no further than at Eratosthenes' measurement of the circumference of the Earth. It is also incorrect to conclude from the uselessness of one experiment the uselessness of the experimental method. It's much easier to question one's experimental setup than a mathematical derivation. The kind of person who is not convinced by the mathematical derivation may, for instance, argue that you only measured half the force. Another may believe that your load cell is defective, that rock flakes do not behave like your metal plates, and so on.


Actually, I'd accept the lab data over the math as at least a starting point toward an actual comparison between nuts and cams. Then I'd accept objections about the lab setup as a way toward making a better setup.

But for a START I'd even accept some math which actually answered my question, which was nuts VERSUS cams, not cams cams cams cams cams, which is where this discussion is heading.

So are people willing to accept the logic that cams and nuts work on the same mechanical principle, and since nuts have the smaller angle, they have the greater expansion force?

Again, to be clear, I'm not asserting this myself, because I don't understand mechanics well enough. But among those that do understand it well enough, is there agreement on this point?

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By Optimistic
From New Paltz
May 24, 2011
Boissal wrote:
Sooooooooo... Are we any closer to getting any form of answer to the OP's question (even if it's intuitive/based on anecdotal evidence) or are you guys going to continue displacing large amounts of air by tooting your own horns?


Wow, that's great! Way better than any of that Icanhazcheezburger stuff. (the pic, I mean, although the comment is good too.)

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By brenta
From Boulder, CO
May 24, 2011
Cima Margherita and Cima Tosa in the Dolomiti di Brenta.  October 1977.
Aric Datesman wrote:
Not to argue with you Brenta, but you did just a couple posts ago.

No, I did not. Please, read more carefully. If you still think so, please, explain how you came to that conclusion.

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By berl
From Oregon
May 24, 2011
well, this is fun,isn't it? the internet has pictures of lions!

I think there are two cases that should be treated separately:

1. an flake that you're worried about breaking with outward force

and


2. an (big) flake that isn't going to break from being flexed a little bit.

in the first case, you might focus on trying passive pro in hopes of getting some protection without as much outward force- maybe a hex that really hangs up on a feature in the crack and would pull down on the flake instead of outward in the case of a fall.

in the second case, use a cam, because it will still provide an outward force (and friction) as the flake flexes beyond the size of the chock you're using.

If you have a situation where your nut isn't going to pull through even with the flex, then get back on MP and push this one to page 4.

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By Aric Datesman
May 24, 2011
brenta wrote:
No, I did not. Please, read more carefully. If you still think so, please, explain how you came to that conclusion.


Ah, I see. My apologies. Upon a quick reading it appeared you were responding to David's comment re: the Greeks and their 4 elements. I will say in that case I don't quite follow your response to how David's point that the Greeks were incorrect on this belief, but frankly I don't have a horse in this race and have more important things to do at the moment.

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By Optimistic
From New Paltz
May 24, 2011
berl wrote:
well, this is fun,isn't it? the internet has pictures of lions! I think there are two cases that should be treated separately: 1. an flake that you're worried about breaking with outward force and in the first case, you might focus on trying passive pro in hopes of getting some protection without as much outward force- maybe a hex that really hangs up on a feature in the crack and would pull down on the flake instead of outward in the case of a fall.

[I shortened your post a bit to reply to just part of it, hope that's ok]
I would have agreed with what you're saying about case 1 prior to reading all this stuff above, but it does indeed look (according to as exalted a source as wikipedia) as though a narrower wedge has more expanding power than a wider one, and so while I like the "hex hanging up on a feature" plan, I'm not sure now that "passive pro" in general is going to be gentler on a flake.

I do agree with you that this is getting a little long...I thought it was a pretty simple question!

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By kennoyce
From Layton, UT
May 24, 2011
Climbing at the Gallery in Red Rocks
David Horgan wrote:
Actually, I'd accept the lab data over the math as at least a starting point toward an actual comparison between nuts and cams. Then I'd accept objections about the lab setup as a way toward making a better setup. But for a START I'd even accept some math which actually answered my question, which was nuts VERSUS cams, not cams cams cams cams cams, which is where this discussion is heading. So are people willing to accept the logic that cams and nuts work on the same mechanical principle, and since nuts have the smaller angle, they have the greater expansion force? Again, to be clear, I'm not asserting this myself, because I don't understand mechanics well enough. But among those that do understand it well enough, is there agreement on this point?



Okay, here is the math for the nut (you can see the math for the cam in my previous post). We'll use the same "F" to be the load on the nut, we'll use the same "f" to be the normal force of the rock on the nut, and now we need to add a "ff" for the frictional force of the rock on the nut (again, see my totally awesome ms paint FBD below). So this looks about the same except for the frictional force which will actually help prevent the flake from expanding. Again, summing the forces in the vertical direction yields the following equation:

0 = 2f sin(theta/2) + 2ff cos(theta/2) - L

Now we can substitute in that the frictional force "ff" is equal to the coeficient of friction "u" times the normal force "f" and we get:

0 = 2f (sin(theta/2) + u cos(theta/2)) - L

Now we can't go any farther until we pick a coeficient of friction and a taper angle of the nut (theta). From this site bigwalls.net/climb/camf/index.... I got that a typical coeficient of friction between aluminum and rock is .3 and an earlier post had 8 degrees for a nuts taper angle so we'll use that. Plug thos numbers in and we get:

f = 1.3549F

now we need just the expansion force which is found by multiplying f by the cos(theta/2) and we get that the expansion force "fx" is equal to:

fx = 1.3516F

Now we compare that number to the previously calculated "fx" for a cam (with a 15 degree caming angle) which was equal to 1.866F and we can clearly see that a nut with almost half the taper angle of the cam's camming angle still puts much less outward force on the flake then the cam would.
FBD of a Nut
FBD of a Nut

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By Aric Datesman
May 24, 2011
kennoyce wrote:
Now we can't go any farther until we pick a coeficient of friction and a taper angle of the nut (theta). From this site bigwalls.net/climb/camf/index.... I got that a typical coeficient of friction between aluminum and rock is .3 and an earlier post had 8 degrees for a nuts taper angle so we'll use that.


Just FYI, a couple friction values for 6061 aluminum on rock can be found on US Patent Application #20060231708 (which deals with high friction inserts on cam lobes):

Nevermind.... somehow managed to read AL6106 as AL6061 a hundred times over the past couple years, so this chart likely has little bearing on the discussion. I assume AL6106 refers to a friction coating of some sort, but Google isn't finding it (as it does for the others). Sorry.
frictional coefficients between aluminum and rock
frictional coefficients between aluminum and rock

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By kennoyce
From Layton, UT
May 24, 2011
Climbing at the Gallery in Red Rocks
Aric Datesman wrote:
Just FYI, a couple friction values for 6061 aluminum on rock can be found on US Patent Application #20060231708 (which deals with high friction inserts on cam lobes):


Thanks Aric, so it looks like .3 is a pretty conservative number to use. Anything higher than .3 will just increase the force of friction which will in turn decrease the outward force of the nut on the rock.

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By Aric Datesman
May 24, 2011
No prob. Can't vouch for the numbers as they're not mine, but I have to assume that if they included them in a patent app that they're in the ballpark. I don't recall offhand if they describe how they obtained the numbers, but a quick Google should turn that info up.

Side note- this patent app was the result of a undergrad / master's thesis (I forget which) that was floated to Trango a long while back. According to Mal everyone involved lost interest once they graduated, and the whole idea was dropped.

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By brenta
From Boulder, CO
May 24, 2011
Cima Margherita and Cima Tosa in the Dolomiti di Brenta.  October 1977.
kennoyce wrote:
Now we can substitute in that the frictional force "ff" is equal to the coeficient of friction "u" times the normal force "f"

Isn't the relation between ff, f, and u an inequality and consequently the system statically indeterminate?

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By slim
Administrator
May 24, 2011
tomato, tomotto, kill mike amato.
i guess a simplified answer would be that the placement that has the greater sum of the 2 tangential contact angles (ie on the left wall of the rock and on the right wall of the crack) will exert less force. this isn't all that practical of an answer as most nuts have non-symmetric curved faces. also, measuring and comparing the contacgt angles would take some work.

my guess is that the sum of the tangent angles of the cam will be greater than the sum of the tangent angles of the nut faces, so the cam will actually exert a lower force. the nut will be more like a wedge. also, the cam can more easily deal with the flake expanding. for most cases, GENERALLY speaking, i would probably go with the cam.

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By Aric Datesman
May 24, 2011
Sorry everyone... I have a (more than a) slight dyslexia problem and after looking at that friction coefficient chart for literally the hundredth time over several years I now see that it says AL6106, not AL6061. All of the others in the chart are commercial friction coatings (think brakes or clutch for your car), and most are easy enough to find the manufacturer with some Google-fu. How I managed to mis-read this chart so many times over so long a period is beyond me, but there you go. There's a small chance AL6106 refers to an aluminum alloy, but I've never heard of it so it's most likely a manufacturer's name for a friction coating.

Again, my apologies.

-a.

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By slim
Administrator
May 24, 2011
tomato, tomotto, kill mike amato.
could be a typo in the table. i've got skunked a couple times by those :(

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By Aric Datesman
May 24, 2011
Thanks for the vote of confidence Slim, but my issues with dyslexia are fairly well established and likely genetic, given the problems several cousins have with it (one of which is borderline illiterate in spite of being quite smart). Seriously, I've read that patent app a hundred times and this is the _first_ time I've not seen it say AL6061. :-(

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By kennoyce
From Layton, UT
May 25, 2011
Climbing at the Gallery in Red Rocks
brenta wrote:
Isn't the relation between ff, f, and u an inequality and consequently the system statically indeterminate?


That was what I was thinking earlier in this thread, but for some reason now my mind doesn't want to believe that it's statically indeterminate now. If you can show me in more detail why it is I'm all ears (or eyes as the case may be).

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By brenta
From Boulder, CO
May 25, 2011
Cima Margherita and Cima Tosa in the Dolomiti di Brenta.  October 1977.
I'll try. Since ff < u * f, we only get a lower bound for the normal force f required to balance the load. This means that different combinations of normal and frictional forces result in equilibrium.

At the intuitive level, I find it useful to think of a piton, which is a wedge with a small taper angle. The position of a piton in a crack depends on its history, and the compression forces (usually) increase with the depth to which it was hammered in. Friction forces vary in both magnitude and direction, depending on whether one is driving the pin further in, extracting it, or just leaving it alone. Let's suppose the load is null. If the piton is deeper in the crack, the component of f that tries to "spit it out" is larger, and a larger ff develops to counter it.

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By slim
Administrator
May 25, 2011
tomato, tomotto, kill mike amato.
kennoyce, one interesting thing to note is that if you put a cam in your nut FBD, the cam angle would be quite a bit higher, thus reducing the horizontal load. basically your new cam angle to work with would be approx 14 degrees + theta/2 at each lobe.

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