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Topic Date User

impulse force on cams 12-Sep-2005 At 1:02:21 PM shiltz

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This is a difficult problem to supply more than a rough answer.

5m above the cam (I'll assume no slack in the system - very tight rope).
Fall distance 10m until rope begins to catch fall.
Velocity after falling 10m would be approximately 14m/s.
Impulse = mass by change in velocity = 1120kg.m/s
Lets assume the rope stretches by 10% because the maths is easy.
Further assume that the cam is 15m off the ground.
Total rope from belay to climber 20m = stretch 2m
Climber goes from 14m/s to 0m/s across a distance of 2m
Deceleration takes about 0.3s (linear deceleration)
Impulse = force by deceleration time
therefore average force = 3.7kN (once again assuming the deceleration is linear which it won't be)

The force acting on the cam will actually increase as the rope begins to catch the climber and reach a peak before finally settling at 0.8kN when everything stops moving. I've left the weight of the rope out of the calculation because its simpler that way. I've also assumed the belayer doesn't move and the rope doesn't slip through the belay device (both unrealistic). I've also assumed the cam doesn't move or rotate during the fall.