There's some chat on the How Not 2 forum about working out the outward force of cams and nuts. I found a reasonable paper discussing the industrial use of wedges and used the formula to write some code to work out the force of a typical 20 degree nut if you also include the friction between the nut and the rock.
[nut/wedge equations figure]
I found some values of friction coefficients for Aluminium and granite (somewhere in the range 0.38 to 0.5
The force multiplier (outward force / downward force) worked out to be somewhere in the region of 1.1x to 1.25x. Even if we used a slippy 0.3 for the friction coefficient, the outward force was only about 1.4x.
I also worked out a force multiplier for cams against camming angle. For the commonly used 13.75 the outward force was approximately 4x.
[cam multiplier figure]
So if you're looking at a flake attached on one edge, getting a small nut as close to the pivot point as possible is substantially better than a cam - possibly 8x less force!
Also, a bigger nut acting more like a piton (i.e. a sharp constriction, a bit like a cam acting as a nut, which greatly increases the equivalent friction coefficient) can reduce the outward force down to 0.5x or less.
Post edited at 11:07
