expanding flakes: nut vs. cam?

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!

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/ind… 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

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/ind… 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

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.

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.

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?

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.

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.

could be a typo in the table. i've got skunked a couple times by those :(

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. :-(

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).

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.

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.

slim wrote: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.

This is true, but you wouldn't place a cam in a crack with that shape because it would walk up and either umbrella (making it useless) or stop in a more parallel section of the crack (once again decreasing the cam angle). If you were choosing between placing a nut and a cam in the same crack, you most likely are going to pick a different place for each one. Of course there are always exceptions to the above (pods or pin scars for a couple) but generally speaking the above is true.

brenta wrote: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.

I get this, what I'm thinking though is that since in this case we're talking about an expanding flake, the flake will expand and limit the normal force to the minimun required for equilibirum (i.e where ff = u * f). I may still be wrong (if you can see a flaw in my thinking let me know) but reguardless, the frictional force of a nut will help limit the outward expansion of the flake when compared to a cam.

The argument about the flake expanding just as much as required to achieve equilibrium is rather convincing, and the expectation that friction will limit a wedge's outward force relative to that of a cam is very reasonable. However, the transitions from static to dynamic friction and irregularities in the crack may cause the minimum outward force to be exceeded. I've no idea by how much. Mechanical structures with friction are far removed from what I pretend to know something about.

brenta wrote:The argument about the flake expanding just as much as required to achieve equilibrium is rather convincing, and the expectation that friction will limit a wedge's outward force relative to that of a cam is very reasonable. However, the transitions from static to dynamic friction and irregularities in the crack may cause the minimum outward force to be exceeded. I've no idea by how much. Mechanical structures with friction are far removed from what I pretend to know something about.

This is true, but I'd guess that the numbers should still be fairly close (within 5% or so of what I calculated). I'd say that they are probably close enough for what we are trying to do here;)

Agreed. Close enough for Internet work.

Did you guys come up with an actual answer yet or are you not even attempting any longer to come up with a coherent response?

jcurl wrote:The expanding force from a cam in a constriction is 1/(2*tan(alpha+beta)), where alpha is the cam angle and beta is half the taper angle (Ken's theta/2).

I found a ratio of 1/(2*(tan(alpha)+tan(beta))). For the wedge, the ratio is lower bounded by 1/(2*(u+tan(beta))).

EDIT: This is wrong, because I left out the horizontal component of the friction force. The correct result for the cam is the one given by jcurl, while for the wedge, it's 1/(2*tan(alpha'+beta)), where alpha' is the angle between the reaction and the normal to the rock face. Since the system is statically indeterminate, this angle depends on its history.


This has a nice interpretation: the advantage of the wedge comes from being able to use all the friction available, whereas the cam must have a safety margin (tan(alpha) < u).

EDIT: This conclusion is still correct, because it is possible to have tan(alpha)) < tan(alpha') < u. Conversely, a hard-wedged nut can exert a significant outward force even without load, whereas a cam that is not stuck does not.