expanding flakes: nut vs. cam?
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 May 21, 2011 The other day a buddy of mine placed a nut behind a somewhat sketchy-looking flake, and it made me wonder: does anyone know if the expansion or x-axis forces are different with a cam and a nut for a given load? In other words, was he any less likely to pop the flake off with a nut than with a cam? Most especially, is anyone aware of this having been actually studied and measured in a lab? (Anticipating other comments: 1) yes, he knew that this is a situation to be avoided whenever possible, and 2) my apologies if this has already been covered in another thread, but I couldn't locate it.) Cheers, David OptimisticFrom New PaltzJoined Aug 29, 2007247 points
 May 21, 2011 I don't know the numbers and I doubt there is lab results since there are so many variables to a thin flake scenario. The major thing to think about is nuts are more sheer forces while cams are expansion. You are more likely to pop a flake with a cam, because it will expand more than the nut. A tapered nut is only slightly larger from top to bottom, whereas a cam has a much larger difference between open and closed. On a side note, if there is an expanding flake that you are placing gear behind, it should be done with a cam not a nut. The nut will most likely pull in this scenario. With the cam you will just have to run the risk of blowing apart the rock. TomCaldwellFrom Clemson, S.C.Joined Jun 2, 20092,692 points
 May 21, 2011 TomCaldwell wrote:I don't know the numbers and I doubt there is lab results since there are so many variables to a thin flake scenario. The major thing to think about is nuts are more sheer forces while cams are expansion. You are more likely to pop a flake with a cam, because it will expand more than the nut. A tapered nut is only slightly larger from top to bottom, whereas a cam has a much larger difference between open and closed. On a side note, if there is an expanding flake that you are placing gear behind, it should be done with a cam not a nut. The nut will most likely pull in this scenario. With the cam you will just have to run the risk of blowing apart the rock. You logic is kind of backwards, having a smaller taper on a nut would create larger outward forces. It's all about force, not size or range. A cam has roughly the equivalent of a 14 degree taper. However, you can fit nuts in so that their weight is carried at the bottom so that there is almost no expansion force. That would be best for your flake if it is possible. DannyUncannyFrom VancouverJoined Aug 27, 201078 points
 May 22, 2011 DannyUncanny wrote: You logic is kind of backwards, having a smaller taper on a nut would create larger outward forces. It's all about force, not size or range. A cam has roughly the equivalent of a 14 degree taper. However, you can fit nuts in so that their weight is carried at the bottom so that there is almost no expansion force. That would be best for your flake if it is possible. What's the rationale for smaller taper creating larger outward forces? It's been a while since physics class, but it seems like if the wedge had absolutely no taper at all, it would develop no outward force. Any aid climbers reading this? Any anectdotal experience w/ cams vs. nuts behind grievous expando flakes? OptimisticFrom New PaltzJoined Aug 29, 2007247 points
 May 22, 2011 David Horgan wrote:if the wedge had absolutely no taper at all, it would develop no outward force. In that case, it would just slide through. Under appropriate simplifying assumptions, a chock behaves like an inclined plane. The most important such assumption is smooth surfaces (at the macroscopic level). Then the reaction of the rock on the wedge is perpendicular to the face of the rock. Its component in the direction opposite to the load has to equal half the load. Hence, the smaller the taper, the higher the outward force. In practice a chock well seated at a constriction may not work like an inclined plane, which seems to be DannyUncanny's point. On the other hand, nuts don't work in parallel-sided cracks. TomCaldwell's point seems to be that with smooth walls (but not parallel), it would take less force to expand the flake enough for the nut to pop out than for a cam to pop out. In summary: look at the flake and at the placement, then make up your mind. brentaFrom Boulder, COJoined Feb 2, 200672 points
 May 22, 2011 I'd believe that a nut may produce less outward force, but if the flake flexed by 1/8 inch then the cam is much more likely to be stuck in there. If I were leading, I'd use the cam and try so seat it as far back as possible to minimize levering. Jesse DavidsonFrom san diego, caJoined May 30, 200771 points
 May 22, 2011 Cams place a significant amount of outward force on a flake or block. Try it on the ground. Place a cam under a block and see how easy the block moves. Do some math. I won't but here are some links. bigwalls.net/climb/camf/index.... vainokodas.com/climbing/cams.h... en.wikipedia.org/wiki/Spring-l... justinreina.com/portfolio/ME47... BASE99999Joined Mar 10, 20111 points
 May 22, 2011 Jesse Davidson wrote:I'd believe that a nut may produce less outward force, but if the flake flexed by 1/8 inch then the cam is much more likely to be stuck in there. If I were leading, I'd use the cam and try so seat it as far back as possible to minimize levering. I do agree about seating the placement as deep as possible to minimize the lever arm of the flake. On the rest of it, I guess my thinking is that if the flake is unstable enough to move 1/8", more than likely it'll move 1/2" too. The question is, is a nut less likely to start the flake moving to begin with? I think if the plan is "when I take a leader fall the flake is going to move some, and then stop moving, still within the expansion range of the cam" that is a dubious plan. Again, what I'm wondering is whether anyone knows for a fact that one piece or the other develops LESS horizontal force for a given vertical load, in identical placements. One thought: most of you have probably heard of the phenomenon whereby as the angle between two closely horizontally spaced pieces (usually bolts) increases, the resultant load on the two pieces increases rapidly. The extreme case would be 2 bolts 6 inches apart, connected by two slings 3 inches long which results in doubling the vertical load or whatever it is. Conversely, if the runners are long, they have a very small angle between them, and the horizontal load approaches zero. So would the same increase in outward force apply as the cams rolled open behind the flake, so that it would take only a tiny downward load to open the cams from 170 degrees to the full 180? Is the physics the same as with the runner/bolt example above? OptimisticFrom New PaltzJoined Aug 29, 2007247 points
 May 22, 2011 BASE1361 wrote:Cams place a significant amount of outward force on a flake or block. Try it on the ground. Place a cam under a block and see how easy the block moves. Do some math. I won't but here are some links. bigwalls.net/climb/camf/index.... vainokodas.com/climbing/cams.h... en.wikipedia.org/wiki/Spring-l... justinreina.com/portfolio/ME47... Intuitively I think you're right that the cams actually develop a lot more horizontal force. But I'm not sure I can quite do the math. Or rather, the math isn't very hard (algebra/trig) but I'm not sure I can handle the physics. Here's the wiki link for a simple inclined plane, but what we're using is a wedge, two planes opposing each other. The Kodas paper you linked does offer a way to compute the horizontal force for a cam as equaling the vertical load/2*.25 (Kodas says .25 is the coefficient of the typical cam used in climbing, I'll take his word for it.) Maybe someone (who knows the coefficient of friction of aluminum on granite) knows how to calculate the horizontal force developed by a wedge? By the way, DannyUncanny was absolutely correct (at least according to Wikipedia) that a narrower cam offers a greater mechanical advantage (ie, more horizontal force) than a wider one. This is why an axe blade is tapered at a very narrow angle, not a very wide one. Never looked at it that way before... OptimisticFrom New PaltzJoined Aug 29, 2007247 points
 May 22, 2011 Reducing a cam (with 14 degree angle) to the simplest statics problem you get: Outward force = Downward force / tan(14 degrees). Tan(14 degrees) = ~.25, so the outward force is actually 4 times the downward force. A nut is way more complicated and depends entirely on the geometry of the placement, it could only be the downward force with no expanding force or it could be some complicated function of the taper, etc with a substantial outward force being generated. As people said earlier reduce the lever arm and make a call depending on the placement available. LauerFrom Duluth, MNJoined Apr 29, 20082 points
 May 22, 2011 If cam will ALWAYS create a large outward force unless it is completely open and functioning as a chock. A nut can be placed where the outward force is small or zero (a constriction at the nut's shoulders). If there's any questions as to the flake's integrity, then try a large nut at a small constriction. Someone mentioned smooth surface - yes, in this case the physics of nut and cam are basically the same. Eric KrantzFrom Black HillsJoined Feb 21, 2004469 points
 May 22, 2011 stack some pins at the bottom of the flake to expand it, then go with whatever afterwards. obviously only usefull on aid and depends on the size of the flake rock_fencerFrom Columbia, SCJoined Dec 20, 2009310 points
 May 22, 2011 David Horgan wrote:if the flake is unstable enough to move 1/8", more than likely it'll move 1/2" too. Some flakes, and I'd say granite flakes in particular, behave like springs, with force proportional to displacement (for small displacements). In that case, you need four times the force to get four times the displacement. David Horgan wrote:I think if the plan is "when I take a leader fall the flake is going to move some, and then stop moving, still within the expansion range of the cam" that is a dubious plan. No arguing with that, but the original question was whether the plan based on a nut is more dubious than the one based on a cam or vice versa. David Horgan wrote:what I'm wondering is whether anyone knows for a fact that one piece or the other develops LESS horizontal force for a given vertical load, in identical placements. For smooth surfaces, same angle implies same outward force. Note that if the walls are not parallel, as should be the case for a nut placement, then the effective cam angle increases. So, the details are... details, but it's the same equation that governs the outward force in the two cases. David Horgan wrote:most of you have probably heard of the phenomenon whereby as the angle between two closely horizontally spaced pieces (usually bolts) increases, the resultant load on the two pieces increases rapidly. The extreme case would be 2 bolts 6 inches apart, connected by two slings 3 inches long which results in doubling the vertical load or whatever it is. The vertical load does not change. If the vertical load is nonzero, the force on the two anchors tends to infinity as the angle between the two runners approaches 180 degrees. In practice, the runners stretch even a little bit, and eventually something breaks well before infinity is even in sight. David Horgan wrote: So would the same increase in outward force apply as the cams rolled open behind the flake, so that it would take only a tiny downward load to open the cams from 170 degrees to the full 180? Is the physics the same as with the runner/bolt example above? No. The outward force produced by a cam in a parallel-sided crack is only a function of the cam angle and the load. Until and unless the cam pulls out, that is. brentaFrom Boulder, COJoined Feb 2, 200672 points
 May 23, 2011 brenta wrote: So, the details are... details, but it's the same equation that governs the outward force in the two cases. Is that because both a cam and a nut are essentially two opposing inclined planes, with similar angles? OptimisticFrom New PaltzJoined Aug 29, 2007247 points
 May 23, 2011 Eric Krantz wrote:If cam will ALWAYS create a large outward force unless it is completely open and functioning as a chock. A nut can be placed where the outward force is small or zero (a constriction at the nut's shoulders). If there's any questions as to the flake's integrity, then try a large nut at a small constriction. Someone mentioned smooth surface - yes, in this case the physics of nut and cam are basically the same. This sounds like a key point to me: that in a tight constriction you may be able to use a nut more like a sort of hook, where the bottom of the nut is actally carrying the load rather than the sides. I guess modern cams work can work in that mode also, although that always looks sketchy to me even though the manufacturers say it isn't. In the case of the placement that prompted me to post this thread, by the way, the placement was a totally match-fit HB Offset (so I guess really TWO wedges working in tandem...awesome pieces, those Offsets) and most likely would've blasted the flake to kingdome come if he'd fallen on it. OptimisticFrom New PaltzJoined Aug 29, 2007247 points
 May 23, 2011 David Horgan wrote: Is that because both a cam and a nut are essentially two opposing inclined planes, with similar angles? Yes, essentially (that is, under the appropriate simplifications). brentaFrom Boulder, COJoined Feb 2, 200672 points
 May 23, 2011 David Horgan wrote:The other day a buddy of mine placed a nut behind a somewhat sketchy-looking flake, and it made me wonder: does anyone know if the expansion or x-axis forces are different with a cam and a nut for a given load? In other words, was he any less likely to pop the flake off with a nut than with a cam? Most especially, is anyone aware of this having been actually studied and measured in a lab? (Anticipating other comments: 1) yes, he knew that this is a situation to be avoided whenever possible, and 2) my apologies if this has already been covered in another thread, but I couldn't locate it.) Cheers, David A cam will ALWAYS create a higher expansion force than a nut. Why you may ask, the answer is friction. On a cam, the only friction keeping the cam from expanding is the friction between the lobes and the axel (smooth aluminum against smooth steel). This creates very little friction and alows the cam to expand very easily. With a nut on the other hand, there is friction between the faces of the nut and the rock keeping the nut from sliding further into the crack and expanding. Of course this friction can vary due to rock type, nut material, nut geometry and soforth, but this friction is always going to be much higher than the friction between a cam's lobe and axel. No lab test can be performed to determine the difference in expanding force because there are too many variables with a nut, but the friction keeping a nut from expanding will always be greater than the friction keeping a cam from expanding. kennoyceFrom Layton, UTJoined Aug 12, 20101,627 points
 May 23, 2011 kennoyce wrote: No lab test can be performed to determine the difference in expanding force because there are too many variables with a nut, but the friction keeping a nut from expanding will always be greater than the friction keeping a cam from expanding. Not to be argumentative, but it seems like your logic could be applied by extension to testing ANY piece of climbing gear, because the real world faced by ropes, cams, runners, pitons, bolts, nuts, biners, and harnesses is so much more complex than the lab. Couldn't you at least test both under ideal conditions, eg. have the test placement be two pieces of steel (or steel faced with stone, if you wanted to be aggro about it) tilted at a perfect angle to hold the nut, with strain gauges set up to measure lateral forces? Obviously this wouldn't capture every variable, but it would tell you roughly whether the forces were pretty much the same or wildly different. I do agree with you though that, intuitively, the axle seems like a major difference between the pieces. But on the other hand a nut doesn't really move in a placement unless it's failing, so maybe the friction isn't relevant in a non-parallel sided crack? Isn't the force holding the nut in a constriction the same one keeping me from falling through the sidewalk, i.e. covalent chemical bonds, as opposed to friction? In other words, if the constriction were frictionless, wouldn't the nut still be trapped in the constriction? OptimisticFrom New PaltzJoined Aug 29, 2007247 points
 May 23, 2011 I answered the question back at the start of the thread. If the taper is less than 14 degrees, then a nut exerts more force than a cam. If the taper is greater than 14 degrees, then the nut exerts less force. The outward force of most cams is approximately 4 times the downwards force (tan 14 ~ 0.25). Friction doesn't matter because the motion is constrained by the rock. There is no possible relative motion even if the surfaces were frictionless. The only real world effect to consider is that the aluminum might deform around crystals or such, so the actual local surface under load may have an actual contact angle greater than the overall taper of the nut. If you need help getting your head around the idea of taper and normal forces, imagine the extremes, a nut with a taper that is barely above 0, wedged perfectly between two parallel round bars. If you pull on it, the bars will flex open. Now imagine a the other extreme, a taper that is 90 or 100 degrees wedged between the same bars. When you pull on it, it will have much less force. To go to the far extreme, imagine a taper of 180 degres (flat surface). When you pull on it, there is no expansion. DannyUncannyFrom VancouverJoined Aug 27, 201078 points
 May 23, 2011 rock_fencer wrote:stack some pins at the bottom of the flake to expand it, then go with whatever afterwards. obviously only usefull on aid and depends on the size of the flake That works well in an aid climbing scenario, which is the context I think you were discussing. As has been said earlier, much of the discussion really depends on the flake in question. For aid, I typically prefer alternating nuts and cams to minimize the amount of expansion at play but, again, it depends on the amount of flex, the nature of crack behind the flake, the amount of weight you're putting on it, etc. To be honest, if I could just nut it (on aid), I'd probably do that. There's a big expando flake on the Nothing Atolls pitch of the PO Wall, and that approach worked well. It seemed pretty to work really well, though my partner said the flake groaned several times when he cleaned the pitch. Fat DadFrom Los Angeles, CAJoined Nov 9, 2007152 points
 May 23, 2011 David Horgan wrote: Not to be argumentative, but it seems like your logic could be applied by extension to testing ANY piece of climbing gear, because the real world faced by ropes, cams, runners, pitons, bolts, nuts, biners, and harnesses is so much more complex than the lab. Couldn't you at least test both under ideal conditions, eg. have the test placement be two pieces of steel (or steel faced with stone, if you wanted to be aggro about it) tilted at a perfect angle to hold the nut, with strain gauges set up to measure lateral forces? Obviously this wouldn't capture every variable, but it would tell you roughly whether the forces were pretty much the same or wildly different. I do agree with you though that, intuitively, the axle seems like a major difference between the pieces. But on the other hand a nut doesn't really move in a placement unless it's failing, so maybe the friction isn't relevant in a non-parallel sided crack? Isn't the force holding the nut in a constriction the same one keeping me from falling through the sidewalk, i.e. covalent chemical bonds, as opposed to friction? In other words, if the constriction were frictionless, wouldn't the nut still be trapped in the constriction? You are right that this logic could be applied to any climbing gear, but in this particular case we have many more variables than a typical gear scenario. Some of these extra variables include things like the spring force of the flexing flake, the roughness of the rock, the angle of the nut, things like that which don't matter in a crack that doesn't expand under a load. As for the second bolded part of your quote, this would be true if the nut weren't placed in an expanding flake. The OP asked about an expanding flake which means that friction does play a major role in this particular case. In an expanding flake, as more outward force is applied, the size of the crack grows which allows the nut to slide and fill up the extra space. Higher friction between the nut and the rock limits the amount of outward force applied by the nut because the frictional force (parallel to the surface of the nut) helps to offset some of the downward force which limits the expansion of the flake. kennoyceFrom Layton, UTJoined Aug 12, 20101,627 points
 May 23, 2011 DannyUncanny wrote:I answered the question back at the start of the thread. If the taper is less than 14 degrees, then a nut exerts more force than a cam. If the taper is greater than 14 degrees, then the nut exerts less force. The outward force of most cams is approximately 4 times the downwards force (tan 14 ~ 0.25). Friction doesn't matter because the motion is constrained by the rock. There is no possible relative motion even if the surfaces were frictionless. The only real world effect to consider is that the aluminum might deform around crystals or such, so the actual local surface under load may have an actual contact angle greater than the overall taper of the nut. If you need help getting your head around the idea of taper and normal forces, imagine the extremes, a nut with a taper that is barely above 0, wedged perfectly between two parallel round bars. If you pull on it, the bars will flex open. Now imagine a the other extreme, a taper that is 90 or 100 degrees wedged between the same bars. When you pull on it, it will have much less force. To go to the far extreme, imagine a taper of 180 degres (flat surface). When you pull on it, there is no expansion. This post is simply not true, see my previous post. In an expanding flake there most certainly is relative motion even if the surfaces are not frictionless. DannyUncanny is unfortunately still living in a physics 101 rigig frictionless world. kennoyceFrom Layton, UTJoined Aug 12, 20101,627 points
 May 23, 2011 kennoyce wrote: This post is simply not true, see my previous post. In an expanding flake there most certainly is relative motion even if the surfaces are not frictionless. DannyUncanny is unfortunately still living in a physics 101 rigig frictionless world. Fair enough, before the flake actually starts to move, friction will come into play. It is still a very solvable problem with a threshold taper angle and coefficient of friction. At least 14 degrees is a conservative limit, and I don't imagine many nuts have a taper less than that. DannyUncannyFrom VancouverJoined Aug 27, 201078 points
 May 23, 2011 DannyUncanny wrote: Fair enough, before the flake actually starts to move, friction will come into play. It is still a very solvable problem with a threshold taper angle and coefficient of friction. At least 14 degrees is a conservative limit, and I don't imagine many nuts have a taper less than that. Sorry if I came of as a jerk, but it should be known that friction does come into play before the flake starts to move, after it moves, durring movement, all the time (technically it even comes into play in the case of a non expanding flake due to the deformation of the nut and rock). Unfortunately, this still is not a "very solvable problem" since you don't know the spring constant of the flake (assuming that the flake even flexes linear elastically). You are right though that a nut will put less outward force on the flake. kennoyceFrom Layton, UTJoined Aug 12, 20101,627 points
 May 23, 2011 kennoyce wrote:Unfortunately, this still is not a "very solvable problem" since you don't know the spring constant of the flake (assuming that the flake even flexes linear elastically). You are right though that a nut will put less outward force on the flake. The spring constant is only relevant for an actual fall calculation and a given load. If you want to compare a nut and a cam, you can do so without knowing anything about the fall or how attached or flexible the flake is. DannyUncannyFrom VancouverJoined Aug 27, 201078 points
 May 23, 2011 DannyUncanny wrote: The spring constant is only relevant for an actual fall calculation and a given load. If you want to compare a nut and a cam, you can do so without knowing anything about the fall or how attached or flexible the flake is. Sorry but this is once again not true. The frictional force keeping the nut from expanding the flake further (like it would in a completely frictionless case) is directly related to the normal force of the flake on the nut which could only be determined if you know the spring constant of the flake. So the frictional force is dependent on the normal force which is dependent on the spring constant. Without the spring constant you would no longer be looking at an expanding flake scenario which was the OPs question. kennoyceFrom Layton, UTJoined Aug 12, 20101,627 points

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