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 By From New England USAOct 26, 2013 Hey, so I just finished Rock Climbing Anchors by Craig Luebben (great book btw) and he mentions that an SLCD applies twice the downward force (the fall) on each side of the crack its placed in. Can anyone explain why this is the case (twice the force on each side)? My best guess was that due to equal and opposite reactions the walls of the crack end up pushing against each other in some nice round number way. (It's amazing how little you can learn and still ace all your physics classes haha) So, who's smarter than me?! FLAG
 By Oct 26, 2013 vainokodas.com/climbing/cams.h... FLAG
 By From West EggOct 26, 2013 Magnets FLAG
 By From Denver, COOct 26, 2013 Ray Pinpillage wrote:Magnets how do they work?mountainproject.com&modestbran... frameborder='0'> FLAG
 By Oct 26, 2013 It's more a log relationship. The camming angle is a critical part of this, somewhere around 14 degrees have been written up as results, with some manufacturers using about 13.75 degrees. You can model the outward with trig using tan(theta) and a coef of friction somewhere between 0.34 to 0.38. FLAG
 By From DenverOct 26, 2013 I just stick it in and hope for the best. FLAG
 By Oct 26, 2013 The whole number is a result of camming angle. I suppose it could be a coincidence, or maybe the common 14 degree angle was chosen from a rounded off force multiplication. Either way, tan(14 degrees) = .24931. That's going to be the ratio of horizontal and vertical forces in each cam, but since each cam takes half the applied load, the horizontal component of each cam is going to be just 2x the applied load. FLAG
 By Oct 26, 2013 nicelegs wrote:I just stick it in and hope for the best. "That's what she said"...... FLAG

 By From Poughkeepsie, NYOct 26, 2013 Another good source: totemcams.com/files/galeria/fi.... It contains a full description of the physics of ordinary cams before doing the math for the Totem configuration. FLAG
 By Oct 27, 2013 Climb with them on progressively harder routes. You'll see soon enough. FLAG
 By From New PaltzOct 27, 2013 Ok, so if, as in the Kodas article posted, the outward force is twice the downward force for cams, does anyone know how the outward force with a nut compares? Same? Less? More? Looked at in another way, if I have to put a placement behind a creaky flake, should it be a nut or a cam? FLAG
 By Oct 27, 2013 wouldn't that be similar as to passive or active pro? a piton can be active, a bolt is active, also. FLAG
 By From New England USAOct 27, 2013 Aha! Thanks all for your responses. Aric, that's a great resource. Can't say I understand it 100% but I get the point. I had no idea the force would be so much higher in flared cracks. Good question Optimistic, my guess would be that a nut would cause much less outward force. As most placements would not result in camming action (and thus no camming angle). Would love to see some info on that though. FLAG
 By From Montreal, QuebecOct 27, 2013 A nut is actually a wedge so it excerts an outward force too. FLAG
 By From New PaltzOct 28, 2013 rocknice2 wrote:A nut is actually a wedge so it excerts an outward force too. Most definitely there is outward force. Not sure if it's correct, but I think I remember reading somewhere that from the perspective of a physicist, a cam is essentially an inclined plane (ie, a wedge), translating force in one direction (down, in the case of a cam in a vertical crack) into force perpendicular to it (horizontally into the walls of the crack). This is basically the same as the classic inclined plane example of a box on a ramp: horizontal force (you pushing on the side of the box) translates into vertical force (the box rising upwards as you push it sideways). FLAG
 By From Los Angeles, CAOct 28, 2013 The takeaway is that there is 4x force outwards against rock due to the cam angle. So if you take a 5KN fall on a piece, then it's about 2x due to the pulley effect and 4x due to the cam so it's 40KN (4.5 tons) outwards on that flake you stuck that cam behind... FLAG

 By Oct 28, 2013 hikingdrew wrote:The takeaway is that there is 4x force outwards against rock due to the cam angle. So if you take a 5KN fall on a piece, then it's about 2x due to the pulley effect and 4x due to the cam so it's 40KN (4.5 tons) outwards on that flake you stuck that cam behind... In this case, the outward force does not add the way you may expect. It may be 2x the outward force on each side of the cam, but that means the force outward on the flake is 2x, and the force inward on the mountain is also 2x. The total force acting to seperate the flake is still only 2x the downward force on the cam. FLAG
 By Oct 28, 2013 rocknice2 wrote:A nut is actually a wedge so it excerts an outward force too. Except in cases where it's keyholed, in which case it's square peg/round hole and there's no outward force. Long story short, there's a whole lot of "it depends" going on with passive pro, and it's much harder to quantify than with active gear. FLAG
 By From HereOct 28, 2013 hikingdrew wrote: So if you take a 5KN fall on a piece, then it's about 2x due to the pulley effect... No, not quite right. FLAG
 By From Los Angeles, CAOct 28, 2013 ARowland wrote: In this case, the outward force does not add the way you may expect. It may be 2x the outward force on each side of the cam, but that means the force outward on the flake is 2x, and the force inward on the mountain is also 2x. The total force acting to seperate the flake is still only 2x the downward force on the cam. If the mountain is fixed and the cam and flake are free, then it should be 4x on the flake... FLAG
 By Oct 28, 2013 hikingdrew wrote: If the mountain is fixed and the cam and flake are free, then it should be 4x on the flake... If you do a free body diagram of the flake, you'll see what I mean. There's 2x at the cam, and the reacting force at the attachment of the flake. An analogous situation is two people playing tug-o-war with a rope. Person A pulls with 500N to the left, and person B pulls with 500N to the right. What is the tension in the rope? It's 500N. The forces should not be considered seperate, independant forces, but as a pair of balancing forces in a static system. It may be clearer if you look at the situating in which the tension in the rope is more intuitive. If you hold a 100N weight suspunded by a rope, the tension in the rope is going to be 100N. But there are two forces acting on the rope, the one exerted by the weight, as well as the one exerted by your hand holding the rope. Greg D wrote: No, not quite right. I thing a carabiners efficiency has been meansured to be in neighborhood of .7 IIRC. 1.7x isn't so far off. And hey, maybe the top piece is a DMM revolver? FLAG
 By From Los Angeles, CAOct 28, 2013 ARowland wrote: If you do a free body diagram of the flake, you'll see what I mean. There's 2x at the cam, and the reacting force at the attachment of the flake. Okay, if you include the flake attachment then it can be modeled as a wedge with a mechanical advantage ~ 1/tan(wedge angle) so for 1/tan(14x2 deg) is about 2. So maybe the oft quoted 4x factor must include the pulley effect? From outdooradventureclub.com/newsl... : Brand Cam Angle Outward Force multiplier Metolius 13.25º 4.36 WIld Country 13.75º 4.2 Black Diamond approx 15º 3.88 Aliens 16º 3.64 "So what does it mean? First off it means that it's true, cams do generate roughly four times the downward force as outward force." ARowland wrote: I thing a carabiners efficiency has been meansured to be in neighborhood of .7 IIRC. 1.7x isn't so far off. And hey, maybe the top piece is a DMM revolver? From Luebben's anchor book: books.google.com/books?id=IXqk... "The pulley effect. The top protection gets loaded by the falling climber and the belayer, increasing the load by 60 to 70 percent above the force on the climber." FLAG
 By From Montreal, QuebecOct 28, 2013 Aric Datesman wrote: Except in cases where it's keyholed, in which case it's square peg/round hole and there's no outward force. Long story short, there's a whole lot of "it depends" going on with passive pro, and it's much harder to quantify than with active gear. Same applies for a cam when placed above a constriction. The cam angle changes and a different outward force is applied. What about a #1 Camelot placed as a passive device. There's a shitload of depends but I think for the sake of arguing let's keep it to the basics. Wouldn't the outward force = tangent(cam angle) * downward force ? Not taking friction into account. FLAG
 By From Montreal, QuebecOct 28, 2013 Downforce / tangent(cam angle) = outward force. FLAG

 By Oct 28, 2013 hikingdrew wrote: Okay, if you include the flake attachment then it can be modeled as a wedge with a mechanical advantage ~ 1/tan(wedge angle) so for 1/tan(14x2 deg) is about 2. So maybe the oft quoted 4x factor must include the pulley effect? I think what's going on is that the 4x figure is just being misenterpreted. It's not that is wrong really, but applied in the wrong way. The reason why you would want to use the 4x figure is in computing the friction neccesary to keep a cam from slipping. On a given cam lobe, or side of the cam, or even the cam as a whole, the ratio of the outward forces to the vertical forces is going to be 4, which means that the coefficient of friction must be at least .25 in order to prevent slipping. Because this is an important figure for the relative security of a cam, I can understand why it's listed for various models. FLAG
 By From Los Angeles, CAOct 28, 2013 ARowland wrote: ...the coefficient of friction must be at least .25 in order to prevent slipping. Because this is an important figure for the relative security of a cam, I can understand why it's listed for various models. Sure, but when climbing or talking about cam placement, what I'm interested in is what is the force multiplier from the fall to the flake/crack I've stuck the cam in. It has to be burly enough to take the several x times fall force.. We used to say that it should be able to hold a truck (few tons ~ tens kilonewtons) FLAG

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