Cam Physics Explanation Please!

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?!

http://www.vainokodas.com/climbing/cams.html

Magnets

Ray Pinpillage wrote:Magnets

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.

I just stick it in and hope for the best.

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.

nicelegs wrote:I just stick it in and hope for the best.

"That's what she said"......

Another good source: totemcams.com/files/galeria….

It contains a full description of the physics of ordinary cams before doing the math for the Totem configuration.

Climb with them on progressively harder routes. You'll see soon enough.

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?

wouldn't that be similar as to passive or active pro?

a piton can be active, a bolt is active, also.

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.

A nut is actually a wedge so it excerts an outward force too.

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).

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...

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.

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.

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.

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...

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?