Loc: Holiday Inn Express
Quote: ...On the run here, but I think the chock exerts a normal force on each wall of the crack of F/(2sin a), where F is the load and a is the chock angle, measured from the vertical. (I'm assuming a straight-sided chock and crack perfectly matched to it, of course.)
This will, in general, be less than what you get with a cam. For example, in order for the normal force on one wall to be double the load, you'd need a chock angle around 14 degrees. I don't know what the actual angles are, but I think they're less than 14 degrees for most sizes.
I think I got tangent instead of sine, not that it matters that much for the low angles. Also the angle that is important is the angle of the crack, which might be the same as the angle of the stopper, but sometimes is not.
But the fact that it is inversely proportional to the tangent (or sin?) means that as the angle decreases the normal force on the wall increases. So if the angles are less than 14 degrees, the normal force exceeds the 4 times normal forces quoted previously for cams (which BTW was for parallel crack. It is, of course lower for less than parallel and higher for flaring).
For those whose eyes glaze over thinking about math and physics, there is a simple example to understand this principle. If you want to split very hard wood, you use a sharp narrow angle wedge, because it tranfers the sharp downward force into a higher outward force. (It is also why they sometimes get so solidly stuck - fixed weges )
[Sorry - minor hijack - continue with PAS discussion]