Mountain Project Logo

Cam engagement


Original Post
Ray Lovestead · · Boulder, CO · Joined Jan 2008 · Points: 117

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!

Alex Zucca · · University Heights · Joined Oct 2013 · Points: 285

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.

Ken Noyce · · Layton, UT · Joined Aug 2010 · Points: 2,122
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.
Greg Gavin · · SLC, UT · Joined Oct 2008 · Points: 718
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.
Ken Noyce · · Layton, UT · Joined Aug 2010 · Points: 2,122
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.
Medic741 · · Red Hook, New York · Joined Apr 2012 · Points: 265

I think the torque idea might have something there.... Will have to go ask some physicists

Vaughn · · Colorado · Joined Mar 2011 · Points: 50

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…

From the link:
"Undercamming (AKA ‘tipped out’)
In very soft rock, loose rock, or expanding flakes (most flakes have some elasticity in them) the force applied to the rock will cause it to yield, either opening it up or compressing it (sandstone). If your cam is set at 50% and is medium sized (2 inch), then the loss of some small degree is OK, say the cam diggs into the crud by a few millimetres. A well placed cam has range to waste and will hold. On the other hand if your unit is tipped out (almost totally open), and you only have 3% range left, and the flake expands even just a millimetre the cam with open up and rip. By the way don’t put your hopes in cam stops (these stop the ‘umbrellaring’ - were they invert), because in the small cams the stops just snap off as the cam inverts, and in the bigger units, or the ones with double axles, nine times out of ten the open cam loses its stability and rips anyway.

Compression and deformation can take place, especially with soft cams like Totems and Aliens (the soft material also give it some ‘stick’ making the trade of worthwhile). In hard granite a medium cam made from aluminium (Camalot for example) you’re not going to get any real deformation in the lobes so in extremis you can have a little more confidence. If on the other hand it was a tiny black Alien with a range as big as a beaked bean, then you’re not going to get away with a tipped out unit. The danger comes with small and micro cams where the total contact surface area against the rock may be measured in square millimetres. Add rock that has a irregular shape, perhaps some lichen, and such a cam requires a safe cam margin in order to stay in place."

slim · · Unknown Hometown · Joined Dec 2004 · Points: 1,045

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.

Jimmy Downhillinthesnow · · Bozeman, Montana · Joined Mar 2013 · Points: 10

To add a little more:

The only thing holding your cam in place is friction, whether you've just placed it or you're mid-whip. Friction is a function of the normal force and the coefficient of friction for a given combination of objects. The coefficient of friction is constant for a given placement but varies between rock types and metals, the "stickier" metals have a higher coefficient of friction. The springs generate a small normal force and enough friction to keep it in place, a lead fall generates a huge normal force and hopefully enough to stop you from cratering. That huge normal force can, as the linked article describes, crush the rock, expand a flake, or deform the lobes. If that happens, there's just less room for expansion with a smaller cam.

The cam angle is constant throughout the range of the cam, so the normal force generated is the same with a given pull. I believe there's a tradeoff between range and holding power but that math is beyond me.

Cam math: valleygiant.com/cam_math.html

Greg D · · Here · Joined Apr 2006 · Points: 871

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.

Ray Lovestead · · Boulder, CO · Joined Jan 2008 · Points: 117

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.

Having said that, BD, metolius and others always show images of cams like >50% or more 'cammed'. Metolius goes a step further with green/yellow/red dots. I have never had much heartburn placing a Metolius in the 'yellow' range.

I have friends that insist that the cam be basically overcammed. I always figured they were over doing it.

Ray

john strand · · southern colo · Joined May 2008 · Points: 1,640
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"
cyclestupor · · Woodland Park, Colorado · Joined Mar 2015 · Points: 93
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.
grog m aka Greg McKee · · Unknown Hometown · Joined Aug 2012 · Points: 70

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.

broken cam
Ray Lovestead · · Boulder, CO · Joined Jan 2008 · Points: 117

Thanks Cyclestupor

Is something wrong with the expression for spiral_radius? I'm reading that as infinity as cam angle approaches theta.

Ray

cyclestupor · · Woodland Park, Colorado · Joined Mar 2015 · Points: 93
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. Ray
No. 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.
Pat A · · Unknown Hometown · Joined Oct 2013 · Points: 20

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.

Has any of you engineers out there done an analysis on cam failure's relationship to rock properties. Specifically what failure types will occur at what compressive strengths and coefficients of friction? Also what the failures might look like? Rotational shear plane maybe?

I realized this information is pretty useless unless you're doing to take rock samples to a lab from each climb, but it'd be interesting to see anyway.

patto · · Unknown Hometown · Joined Jul 2012 · Points: 25
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 Aliens
Sorry 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.)
brenta · · Boulder, CO · Joined Feb 2006 · Points: 75
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:

  • 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.
cyclestupor · · Woodland Park, Colorado · Joined Mar 2015 · Points: 93
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.
Ted Pinson · · Chicago, IL · Joined Jul 2014 · Points: 190

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.

Guideline #1: Don't be a jerk.

Post a Reply

Log In to Reply