What is the formula used when determining shock load? For the sake of discussion, a 500# log falling 3' vs 6' or does no such rule of thumb for the industry exist? I know wwl and let lines run when applicable, however, knowing the fomula and calculating prior to the cut would reduce some of the sweat in the tree I expieriance over property.
Making one think.. and not sure if I can get this 100% correct or not, digging out my old notes so will try. But am missing some info.. to get accurate need specs on the rope (amount of stretch, etc)..
But lets plug in some #s and see how it goes.. feel free to point out where I make errors
Now my formulas all are in metric, not sure if what the other formulas are .. so will play it by ear.
500 lbs is around 227 kg.. and we will call 3 ft as 1 m.
227 kg dropped 1 m is traveling at 4.4 m/s at place of stop. This equates to around 2225 J of energy (rounding here).
Now if the rope stretches .1 m or 10 cm (not sure how much it would realistically be).. then this means that there is an impact of around 22,000 N or 4,400 lbs of force.
That same 227 kg dropped 6 ft or 2 m would be traveling at around 6 m/s and have around 4450 J of energy.
If we have the same .1 m stretch (again not sure how realistic this is), then we have 44,500 J of energy or 8,900 lbs of force. With more J of energy, I expect the stretch would really be slightly more but not sure how much.
BUT if I recall correctly the rope stretch changes based upon mass and energy.. so the amount changes here.
Also dynamic rope, again if I remember correctly, looses it stretch rather quickly when used in heavy dynamic loading situations and each drop will be different loading due to the change in stretch of rope.
Now don't use the #s above for anything serious.. as too much guess in the #s ..
Now as it falls farther, it will be dropping faster.. 4 m would be dropping almost 9 m/s and would be around 17,000 lbs of force... and 10 m would be up to 14 m/s and 45,000 lbs of force.
Keep in mind the block at top, if the rope is going straight up and back down (at 180 degrees) is handling twice the stated load .. so at 1 m the block would have around 8,800 lbs of force.. As that angle decreases so does the impact on the block. I think 120 degrees would half that load as I recall..
Think this is right but been a long time since I did this stuff.
This is also taking some very simple concepts, and ruling out any friction, etc.. if using porty there is also some slippage as rope tightens (no matter how hard you try there will be some with this amount of impact).. and depending on wraps on porty and how much you letting it run.. will dramatically impact that force actually experienced.
Bottom line, you get into big #s very fast.. and we were taught to apply the following:
10:1 safety factor on all ropes
7:1 on slings
5:1 on steel hardware
Some apply the 10:1 for everything.. so this could be a point of discussion and may be controlled by WSIB or similar rules in your area (so be aware).