BASE_689
September 19th, 2005, 07:07 AM
Due a disadventure (gone well, for God's sake) of a mate of mine, I am going to ask the folllowing to which ever bloke knows the answer.
1) Which is the break load of a bridle "properly" secured onto handrail (=bridle nicely fastened onto handrail WITHOUT sharp bends and without touching sharp edges and without touching anything "bad" (=laboratory conditions))?
2) Which is the break load of a bridle that touches small radius metallic pieces and/or sharp/cutting edges onto/around handrail?
Answer to point 1 could be 2500 lb - 1134 kg (according to data on ParaGear online catalog for type 17 nylon webbing 1" wide, if this is the webbing used for manufactiring bridle).
Answer to point 2 could be: very low load, I am afraid, depending on how sharp is the "razor" bridle leant onto, possibily 20 kg - 44 lb, possibly 10 kg - 22 lb or perhaps less....
What happened is the following.
SL solo jump, rigged everything (with a backup 46" AV ZP PC in "normal" position at the end of bridle within its own proper loop) with SL secured onto handrail (very bad, very old, very rusty, very "everything else") and break cord between SL and bridle loop in "shortened" configuration (bridle loop for tightening break cord has been created 1 m off the pins with an 8 knot), slack of bridle kept in place with very thin yellow elastic bands.
Exit, quick pull by "break cord" (definitely slightly higher that the pull in a "properly conducted" and "normally executed" SL jump), opening of parachute, flight, standup landing.
What my mate discovered once landed is that with its parachute he was ONLY trailing a 1.5 m (or so) of bridle, AND THAT'S IT!!!!!!!!!!!!
Looking at exit point he could see his PC and remaing of bridle nicely floating in the wind, still attached onto handrail.
What happened is, obviously, that his bridle became trapped/entangled somewhere between pins and "newly created" loop with break cord.
Luck on his side that whichever was the break load has been, was higher than the force required to open pins (on his Vertex 2) and extract parachute.
Now, what we are wondering is: in a condition of bridle "nicely" (=no sharp bends, no razor cutting edges) trapped onto handrail, can the dynamic (=applied after a 10 m or so "freefall") weight of jumper break the bridle, breaking it at "laboratory break load"?
Or a "nicely" trapped bridle CANNOT be broken by a jumper (pulling with his own weight 10 m below) and so, in order a bridle breaks with a jumper 10 m below it requires COMPULSORILY that same bridle must lean onto something sharp, in which case (sharp cutting edge), break load of bridle can be any number, from 40kg-88lb/20kg-44lb/10kg-22lb down to ZERO?
Whichever will be your answers/opinions, MY opinion is that THAT jump has been a close call.
We need to revise the conditions a PC/bridle(=slack of)/break cord/SL are with reference to "doubty" point near exit point. VERY CAREFULLY.
Just my 0.02€
1) Which is the break load of a bridle "properly" secured onto handrail (=bridle nicely fastened onto handrail WITHOUT sharp bends and without touching sharp edges and without touching anything "bad" (=laboratory conditions))?
2) Which is the break load of a bridle that touches small radius metallic pieces and/or sharp/cutting edges onto/around handrail?
Answer to point 1 could be 2500 lb - 1134 kg (according to data on ParaGear online catalog for type 17 nylon webbing 1" wide, if this is the webbing used for manufactiring bridle).
Answer to point 2 could be: very low load, I am afraid, depending on how sharp is the "razor" bridle leant onto, possibily 20 kg - 44 lb, possibly 10 kg - 22 lb or perhaps less....
What happened is the following.
SL solo jump, rigged everything (with a backup 46" AV ZP PC in "normal" position at the end of bridle within its own proper loop) with SL secured onto handrail (very bad, very old, very rusty, very "everything else") and break cord between SL and bridle loop in "shortened" configuration (bridle loop for tightening break cord has been created 1 m off the pins with an 8 knot), slack of bridle kept in place with very thin yellow elastic bands.
Exit, quick pull by "break cord" (definitely slightly higher that the pull in a "properly conducted" and "normally executed" SL jump), opening of parachute, flight, standup landing.
What my mate discovered once landed is that with its parachute he was ONLY trailing a 1.5 m (or so) of bridle, AND THAT'S IT!!!!!!!!!!!!
Looking at exit point he could see his PC and remaing of bridle nicely floating in the wind, still attached onto handrail.
What happened is, obviously, that his bridle became trapped/entangled somewhere between pins and "newly created" loop with break cord.
Luck on his side that whichever was the break load has been, was higher than the force required to open pins (on his Vertex 2) and extract parachute.
Now, what we are wondering is: in a condition of bridle "nicely" (=no sharp bends, no razor cutting edges) trapped onto handrail, can the dynamic (=applied after a 10 m or so "freefall") weight of jumper break the bridle, breaking it at "laboratory break load"?
Or a "nicely" trapped bridle CANNOT be broken by a jumper (pulling with his own weight 10 m below) and so, in order a bridle breaks with a jumper 10 m below it requires COMPULSORILY that same bridle must lean onto something sharp, in which case (sharp cutting edge), break load of bridle can be any number, from 40kg-88lb/20kg-44lb/10kg-22lb down to ZERO?
Whichever will be your answers/opinions, MY opinion is that THAT jump has been a close call.
We need to revise the conditions a PC/bridle(=slack of)/break cord/SL are with reference to "doubty" point near exit point. VERY CAREFULLY.
Just my 0.02€