expanding flakes: nut vs. cam?
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 May 23, 2011 Assuming the appropriate idealizations, if you put a tapered nut into a constricting crack, which expands with an associated spring constant, the nut would expand the crack until the (vertical) components of normal force and proportional frictional force combined were sufficient to offset the load. But this is also true for a cam in a constricting crack. The only difference would be that a nut would need to slide deeper into the crack to achieve this equalization whereas a cam could stay in place and expand with the crack, provided the cam's range had not been exceeded. jcurlJoined Apr 22, 20101 points
 May 23, 2011 A little nit picking. The outward force in a vertical, parallel-sided crack, when the cam angle is arctan(1/4) is twice the load, not four times. There are two frictional forces, one on each side, that together must balance the load. Each frictional force is related to the normal force by the tangent of the cam angle. Mix, shake well, and the result follows. brentaFrom Boulder, COJoined Feb 2, 200672 points
 May 23, 2011 Just to add a little clarity or confusion--pick one--when a nut slides down the constriction, part of the work done by the load goes into bending the flake and part is lost to friction. So, if one wants to analyze an actual fall, one has to take into account that the fixed quantity (OK, almost fixed) is the energy to be dissipated, while the load may peak at different values depending on the details of the anchor. brentaFrom Boulder, COJoined Feb 2, 200672 points
 May 23, 2011 brenta wrote:A little nit picking. The outward force in a vertical, parallel-sided crack, when the cam angle is arctan(1/4) is twice the load, not four times. There are two frictional forces, one on each side, that together must balance the load. Each frictional force is related to the normal force by the tangent of the cam angle. Mix, shake well, and the result follows. Your nit picking is incorrect. Friction force adds up to load force, normal force adds up to 4 x friction force, you can add it any way you want, it's still 4 times. DannyUncannyFrom VancouverJoined Aug 27, 201078 points
 May 24, 2011 DannyUncanny wrote: Your nit picking is incorrect. Wrong. Let's see why. DannyUncanny wrote: Friction force adds up to load force, True. DannyUncanny wrote: normal force adds up to 4 x friction force, Normal force equals four times the friction force on each side. The normal forces are equal in magnitude and opposite in direction and cancel each other. Draw a free body diagram if you are still incredulous. DannyUncanny wrote: you can add it any way you want, it's still 4 times. Wrong. brentaFrom Boulder, COJoined Feb 2, 200672 points
 May 24, 2011 Oh my. Anyone have a single end beam load cell handy? That'd be the easy way to settle this. All of mine are S-beam, which would take an annoying amount of fixturing to get to work for this. Aric DatesmanJoined Sep 16, 2008145 points
 May 24, 2011 Aric Datesman wrote:Oh my. Anyone have a single end beam load cell handy? That'd be the easy way to settle this. All of mine are S-beam, which would take an annoying amount of fixturing to get to work for this. I (the OP) was hoping that Aric would show up on this thread eventually! The reason that quite a lot more science got done in the last 350-ish years than in the preceding 300,000 was that we decided to go into the lab and MEASURE WHAT IS ACTUALLY HAPPENING as opposed to sitting around in our togas talking about what "makes sense". But I digress... I took a shot at measuring the angle between the wide faces of my HB Offset. It appears to be 7.5 degrees. So if it is in fact true (I don't know if it is) that a) both a cam and a nut can be viewed as governed by the equation describing an inclined plane and b) outward force increases in inverse proportion to the cam/nut angle Then: nuts, or at least my HB Offset, will put more outward force on a given placement, for a given load, than cams will, making the cam less likely to move the flake. But I still think I won't really know until someone measures it. OptimisticFrom New PaltzJoined Aug 29, 2007247 points
 AdministratorMay 24, 2011 not sure if brenta and dannyuncanny are maybe arguing the same thing, exept brenta is talking as a function of the friction force at one side and danny is talking as a function of total downward force at the stem(?), but each side (pair of lobes) will horizontally exert twice the vertical load applied at the stem. so, for a 4 lobe unit, each individual lobve would exert the equivalent of the vertical load at the stem. slimJoined Dec 1, 20042,097 points
 May 24, 2011 slim wrote:not sure if brenta and dannyuncanny are maybe arguing the same thing, exept brenta is talking as a function of the friction force at one side and danny is talking as a function of total downward force at the stem(?), but each side (pair of lobes) will horizontally exert twice the vertical load applied at the stem. so, for a 4 lobe unit, each individual lobve would exert the equivalent of the vertical load at the stem. Again, I don't really know, but that's not where the 4x comes from in the Kodas paper cited at the beginning of the thread. It comes from the load being divided by the cam angle coefficient, which is 0.25. 1/0.25=4. But you do raise an interesting point: is the whole cam unit equal to one inclined plane, as some are asserting here, or two (two sides to the crack) or four? As far as the nut goes, should it be treated as one inclined plane, or two planes opposing the midline? Does that make a difference? I don't know. Lots of answers might appear sensible. One answer would trump all of them: measuring it in the lab. OptimisticFrom New PaltzJoined Aug 29, 2007247 points
 AdministratorMay 24, 2011 i think that you are forgetting that the load on each side is F/2 though, which is the basis of my point. if you are talking relative to the load on one side (ie saying that F/2 is your F), or as a function of the total vertical load on the cam. slimJoined Dec 1, 20042,097 points
 May 24, 2011 slim wrote:so, for a 4 lobe unit, each individual lobe would exert the equivalent of the vertical load at the stem. That's exactly it. The point is that the outward force is not the sum of the forces at the two sides. And there's no more reason to go to the lab to test this than there is to write a proposal to get some time on the supercomputers at Oak Ridge National Labs to settle the dispute as to whether three plus three equals six or twelve. brentaFrom Boulder, COJoined Feb 2, 200672 points
 May 24, 2011 Thinking on this a bit more it's not to hard to measure indirectly.... simply bolt the ends of a pair of plates together (in a U shape) a fixed distance apart (with a bit of a constriction cut into them so the nut can hold), insert the cam/nut, load to a specific force and then measure the amount the plates spread. From there you could back into the outward force provided you know the modulus of elasticity of the material, but since you're only really looking for a comparison between the two it doesn't really matter. I really don't have time to spend on this at the moment, unfortunately. Aric DatesmanJoined Sep 16, 2008145 points
 AdministratorMay 24, 2011 even after 15 to 20 years of doing this, i always have an intuitive hangup on that one brenta (ie not summing the 2 forces such that the outward force on the flake is 4 times). i always pretend that the force is reacted by a vertical beam reaching down to a floor (instead of reacted by the other cam lobe). then it feels like it makes more sense. not sure if you know what i am talking/babbling about? slimJoined Dec 1, 20042,097 points
 May 24, 2011 brenta wrote: That's exactly it. The point is that the outward force is not the sum of the forces at the two sides. And there's no more reason to go to the lab to test this than there is to write a proposal to get some time on the supercomputers at Oak Ridge National Labs to settle the dispute as to whether three plus three equals six or twelve. That's what the Greeks said about all matter being composed of four elements. But anyway, my original question was about making a comparison between nuts and cams, and it doesn't seem like much has been said here to make an actual, numerical comparison between the two types of gear. Some folks have said very authoritatively that cams exert more force, some have said nuts. Which is it? OptimisticFrom New PaltzJoined Aug 29, 2007247 points
 May 24, 2011 David Horgan wrote: That's what the Greeks said about all matter being composed of four elements. But anyway, my original question was about making a comparison between nuts and cams, and it doesn't seem like much has been said here to make an actual, numerical comparison between the two types of gear. Some folks have said very authoritatively that cams exert more force, some have said nuts [Edit: because nuts have a narrower angle than cams]. Which is it? OptimisticFrom New PaltzJoined Aug 29, 2007247 points
 May 24, 2011 dumb question - i feel like the best gear for expanding flakes are sliding nuts, am I wrong? doligoJoined Sep 26, 2008393 points