In reply to duchessofmalfi:
A quick calculation on the basis of my email above suggests that for a near parallel sided crack (where frictional force dominate) the outward pull that can be resisted is equal to the downward pull which gets you a max angle of 45 degrees regardless of the coefficient of friction.
The reason is the nut is stopped from sliding down the crack purely by friction and, assuming the frictional force that needs to be overcome is isotopic, then the force required to move it in any other direction is the same.
For tapering cracks this drops quickly to about 1/2 this by the time you get to the average nut tapering angle (~15 degrees) but depends on the heavily on the coefficient of friction (slipperier falls out easier). and this gives about 25 degrees.
However, please don't try this! the frictional forces that can be resisted require the nut to be set with the forces exerted down the crack before you can resist the force outwards - there is no guarantee in the event of a fall this will occur and the outwards pull is could be as low as the tug applied to seat the nut.
[plus it use idealised physics where infinite forces can be invoked without damaging the rock or nut]
Still it goes to explain why nut can be quite such a bugger to recover from parallel sided cracks.