Cam engagement
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Hey any engineers out there that can explain why cams must be cammed by a certain amount? This question is in response to the cams that Grog M placed when he tore his finger off. |
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The more you retract a cam lobe, the closer it gets to the center of a cam. This effectively reduces the torque applied if you were to fall. A cam placed completely undercammed can have almost double the torque, compared to a cam placed completely overcammed. If your torque is high enough, your cam will break. |
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Ray Lovestead wrote:Hey any engineers out there that can explain why cams must be cammed by a certain amount? This question is in response to the cams that Grog M placed when he tore his finger off. I'm looking for physics and math here. Not anecdotal evidence. It's not surface area, that doesn't change. Is it leverage? If so, what would the forces look like WRT to cam amount (%)? Thanks!Engineer here, Mathematically there is no reason why a cam must be cammed by a certain amount. The holding force will be exactly the same regardless of the amount it is cammed. Anecdotally, people suggest what Grog M stated about desert climbing, but there are no physics or mathematical reasons for doing it. The only real reasons for not overcamming or undercamming is that an undercammed cam can pull out due to the axle bending under load and allowing the cams to fully open up, and that an overcammed cam is much more likely to get stuck. |
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kennoyce wrote: Engineer here, Mathematically there is no reason why a cam must be cammed by a certain amount. The holding force will be exactly the same regardless of the amount it is cammed. Anecdotally, people suggest what Grog M stated about desert climbing, but there are no physics or mathematical reasons for doing it. The only real reasons for not overcamming or undercamming is that an undercammed cam can pull out due to the axle bending under load and allowing the cams to fully open up, and that an overcammed cam is much more likely to get stuck.Math aside I believe the reason the undercammed placement is not preferable is due to it's likelihood of walking when climbing past. A more "crammed" placement should exert more force outward, and thus prevent it from walking as much. |
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Greg G wrote: Math aside I believe the reason the undercammed placement is not preferable is due to it's likelihood of walking when climbing past. A more "crammed" placement should exert more force outward, and thus prevent it from walking as much.correct with one nitty picky change to say that a more "crammed" placement will exert more outward spring force, and thus prevent it from walking as much. I say this to point out that the outward force exerted in a fall will be the same either way, but the outward force of the cams on the rock due to the spring tension will be higher. |
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I think the torque idea might have something there.... Will have to go ask some physicists |
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It's not about the amount of force the cam exerts, but about the remaining range of the cam if the crack/rock yields at all. This blog post explains it well: andy-kirkpatrick.com/cragma… |
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several good reasons listed above. Vaughne's post lists one of the most applicable reasons for small cams. when you load a cam, the cams will rotate a small amount to press into the rock. if a small cam is tipped out, this small amount can basically max out the range of the cam and it will not produce the outward force needed to hold the load. |
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To add a little more: |
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Some interesting answers above. |
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Thanks everyone. I always wondered what was going on and my gut said that the force against the rock between under and overcammed was equal. |
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Greg D wrote:Some interesting answers above. The constant cam angle guarantees the same outward whether 5% retracted or 95% retracted. Of course if you only retracted a can 5% you are working with a very small margin of error and I would consider it very untrustworthy. With small cams the margin of error is very small. The tiniest bit of walking could make it useless. Generally speaking retracting a cam about 50 to 90% is ideal. That gives you some range on the open side if it walks a bit or the rock breaks and gives you 10% on the retracted side to make removal easy. In sandstone the rock is not super hard either, even in Indian Creek. I bet you you can see where this cam pulled out and the rock was damaged.Right, constant cam angle was greatly quoted by Wild Country BITD. Sometimes by trying to get more range, you can screw things up And as Ray Jardine used to say "when your camming a quarter inch, you gotta be precise" |
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James Sledd wrote:I believe there's a tradeoff between range and holding power but that math is beyond me.The tradeoff is between range and outward force that the lobes exert on the rock. More outward force doesn't always mean that a cam has more "holding power". But that is beside the point. As you know, cam lobes are logarithmic spirals. The "cam angle" determines the aggressiveness of the spiral. A smaller cam angle produces a smaller (more squished) spiral. The smaller the spiral, the less range the cam will have. At the same time, the smaller the cam angle, the more outward force a cam will produce. The math for how the cam angle relates to outward force is as follows... Theta : the cam angle F_out : the force perpendicular to the crack wall, which is exerted by the cam on the wall during a fall. F_down : the force exerted by the climber downward on the cam's axle F_out = (0.5 * F_down)/(tan(Theta)) The math for how range relates to cam angle... theta : rotation of spiral spiral_radius = e^(theta/tan(90 - cam_angle)) so if we set theta to say 90 (since cams have roughly 90 degrees of usable range) , and vary the cam_angle then spiral_radius increases as the cam_angle increases. |
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Here is a pic of my top cam that blew. It makes me feel better knowing that the fall destroyed the cam - indicating it was to some degree a decent placement. If you read the first comment on Fingers in a Lightsocket, I believe what he is saying is true, that that crack is wallowed out in the back. |
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Thanks Cyclestupor |
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Ray Lovestead wrote:Thanks Cyclestupor Is something wrong with the expression for spiral_radius? I'm reading that as infinity as cam angle approaches theta. RayNo. I just double checked to be sure, and "e^(theta/tan(90 - cam_angle))" definitely doesn't go to infinity when theta == cam_angle. That said, I threw it together pretty quickly so I could have screwed something else up. |
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So now that the cam angle and force on the rock statics problem has been covered, I would like to ask a related question. A lot of cam failures are attributed to rock failure. |
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Alex Zucca wrote:The more you retract a cam lobe, the closer it gets to the center of a cam. This effectively reduces the torque applied if you were to fall. A cam placed completely undercammed can have almost double the torque, compared to a cam placed completely overcammed. If your torque is high enough, your cam will break.So much rubbish here it is hilarious. Vaughne wrote:Compression and deformation can take place, especially with soft cams like Totems and AliensSorry to nitpick, but don't confuse Totem cams with Basic Cams made by the company Totem. Totem cams are NOT soft cams! grog m wrote:Here is a pic of my top cam that blew. It makes me feel better knowing that the fall destroyed the cam - indicating it was to some degree a decent placement.Sorry to disappoint you but it doesn't indicate that. You broken cam shows evidence of umbrella damage. This could be caused by being undercammed at the time of the fall and/or rock damage causing undercamming . It is quite unlikely that the fall as described, put 6kN force onto the cam. (Comments about this climb do mention the dangers of undercammed cams due to the crack opening up inside.) |
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cyclestupor wrote: The math for how range relates to cam angle... theta : rotation of spiral spiral_radius = e^(theta/tan(90 - cam_angle)) so if we set theta to say 90 (since cams have roughly 90 degrees of usable range) , and vary the cam_angle then spiral_radius increases as the cam_angle increases.A couple of caveats:
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brenta wrote: A couple of caveats: * The formula computes the ratio between the radius for a given value of theta and the radius when the cam is fully retracted (theta=0). * The value of theta must be given in radians. If the formula is rewritten as e^(theta * tan(cam_angle)) explicit references to degrees are eliminated.Correct. Nice catch. |
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George, I have noticed a similar phenomenon when placing them in horizontal cracks. I've had several placements where I found a perfect 0.1 or 0.2 crack, jammed it in, then realized that the trigger was lying right on the lip of the crack. In these situations, I was usually able to replace the piece with a tricam, which I felt much better about. I could possibly see the scenario you're describing in a vertical crack, but it seems like a pretty "perfect storm" type scenario, and shearing out of soft desert sandstone seems like the more likely cause for failure. |