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Chockstone Forum - Gear Lust / Lost & Found

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Topic Date User

Impact factor calculations and microwire placement 26-Nov-2009 At 8:08:48 AM mikl law

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It's got to help, but
a) it's better to have a strategy (equalising, double ropes, slipepry belay) which makes it stay in.
b) the amount of energy absorbed (= speed reduction) is small if the gear fails.

A few assumptions follow here, particularly that ropes have a linear force stretch, that air rsistance is ignored, and I've ignored the extra energy you gain falling the meter or so whle your gear break, these affect the answer by less than 10%.

you fall energy is e =mhg
amount absorbed is z=fd, where f is the force the piece is putting on the rope and d is the distance over which the force is acting. typically f rises from zero up the breaking force.

The area under the curve is about z=1/2 fd (the area of the triangle).
Typical values for a fall are
m=80kg
g= 9.8 ms^2
h=5m
so e= 80*9.8*5=3920 J and your speed is v^2=u^2 + 2gh
your velocity v= 9.89m/s

Effect of gear
f=2kn
d=1m
z = 1000J
your new energy is a-e=2920J, which is 74% of what you had before the piece pulled

The decelaration while the gear is pulling is a=g-f/m=-15.2m/s,
which occurs over a time of t=d/v=1m/ 9.89 m/s= 0.101 seconds
you new velocity is v=u+at=9.898+-15.2*0.101 = 8.36 m/s

you velocity has only dropped 16%

if the rope impact force was 4kn and the gear held, e would equal a, the distance the gear loaded you for would be 1.96m