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Chockstone Forum - General Discussion

General Climbing Discussion

Topic Date User
calculator for impact forces from a climbing fall 4-Nov-2004 At 2:57:48 PM M
Message
On 3/11/2004 bne wrote:
>the numbers are very scary. If you take a 1 meter factor 2 fall, you put
>nearly 20 KN force on the anchor (enough to break a bolt) and the climber
>faces 15.3 G-forces.
>
>these numbers seem very big.
>


This calculation is for a rope modulus of 40 which corresponds to an about 2% elongation in the rope. typically dynamic ropes have 7 or 8% elongation. To get more accurate results you need to input an appropriate rope modulus which you can calculate using the rope calculations function. i am not sure why the default value is 40. a 7 % elongation rope gives a force on the climber of 6.8 kN and 11.2 kN on the top anchor.

A wild climbing fall with the same amount of rope out as inputed into the equations will result in a lower impact force at all points due to other dynamic factors as outlined above. the actual forces involved would vary sigificantly under different circustances but i would estimate that the actual forces would be much lower than those given above.

on the subject of climbing gear, the strenght of most gear when new far excedes any likely impact forces it may sustain. repeated impacts, corrosion and other damage can significaly reduce the strength of gear, particaurly fixed placements. this is why such large saftey margins are neccasary. a CORRECTLY PLACED AND NEW 15 kN rated bolt is not really more likely to fail than a 30 kN bolt. With a bit of corrosion and repeat impacts the 15 kN bolt is likely to fail far sooner than the larger bolt because it is no longer capable of sustaining the original forces. Even when half a strong as new a 25 or 30 kn bolt will still sustain most conceivable impacts. Of course this all assumes that the rock is solid the placement is perfect and the bolts arent so loose or corroded that they can only sustain a small fraction of their original intended force.





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