# Force applied to a cam follower

Hello I am a simple rock climber, and would like to figure out how much force is applied to The lobe of a cam/the rock when the cam is loaded. So the cam lobe is circle the ramp or cam goes from 7/16”-11/16” I’m not sure that matters or if just the point of contact matters. If I ignore friction and those kind of nussance variables what is my formula if I 180lbs simply hang on the cam. (Is it a simple lever calculation)!

The curve of a spring loaded cam is certain curve called exponential spiral with the unique property that the angle it touches the wall of the crack and a horizon remains the same (13-14 degrees) no matter haw much the cam expands.

This small angle means the force applied by the cam to the crack is always $$F_{horiz}= W_{weight}*cotan(13)= W*4.3$$

Which after many tests has proven to be the safest and most practical.

here is the detail calculations. cam math.

This is a hugely complicated question that's not going to be answered both accurately and briefly, because there's too many variables presented by the real world. So I'm going to make some simplifying assumptions, to give you an idea.

What I'm going to present is a simple lever calculation, but it'll almost never apply to the real world.

Let's say that the movable and fixed part of the cam get an equal amount of force on them when you shove it into a crack, and that the crack faces are perfectly parallel. Then the free-body diagram of the cam in the crack would look something like this:

You can immediately simplify that down to something you'd find in a first Statics class:

Because the beam (the diagonal line) is free to pivot, in order for it to not rotate the torques on it must be balanced. This means that the condition $$\frac{F}{2} a = F_I b$$ must be satisfied.

That's pretty much it, for this ridiculously simplified case. But crack angles will vary, cam designs will vary, you may have differing grips on one side of the crack than the other, etc., etc., etc..

I think that for a given design of cam (i.e., how much does it spread as it opens, which determines the angle of the contact point with respect to the pivot) and for a given inward or outward taper of crack, you can get an idea of how much outward push you'd get. But even there, a crack that's got a rough enough surface with large enough irregularities would tend to catch the cam at a point where the effective taper is more favorable, leading to less outward force.

A good pessimistic estimate would probably be to assume that only one side or the other actually grips (so assume $$F$$ instead of $$F/2$$ in the calculations above), and that the taper of the crack is what it seems to be on a large scale (so assume no hidden advantages from catching on a high spot).

To do this really right you'd build a bunch of cams that are instrumented to measure the force on the rope and on the pivot, and then you'd go climbing, recording the forces on a datalogger and taking notes. Be sure to do it on faces that are famous for having cracks that cams slip out of, and on ones that are famous for having cracks that are secure even if someone with five thumbs on each hand is putting in the cams.

• That’s good enough for the math I’m trying to find I just want prove to my friend that my adjustable home built crack is strong enough! Thanks – Najn Oct 9 '19 at 18:48
• Test the @#\$% out of it! I've worked on areas of engineering where the unknowns are largely unknown -- what ends up happening is that people build stuff that breaks, and then they just beef it up a bit. The process ends up being more engineering by evolution than by induction. – TimWescott Oct 9 '19 at 19:23
• Haha thanks in my experience millwrighting we build everything far stronger than it needs to be! But yea often at work they make some it fails then they just keep fixing it the same over and over. Come on guys let’s figure out what we need to do to make it work properly. – Najn Oct 9 '19 at 20:41