SWL questions

Newbieone

Ok i'v just started doing a bit of climbing after being on the ground, And i'v been told alot of stuff.

(1) If Breaking load of a rope is 5000kg is the SWL 1/6th or 1/10th in australia?

Does SWL include the reduction in strength for knots?

Does SWL take to consideration possible shock load? or if i have a 250 SWL rope can i chuck on a 250kg log and drop it 1m?

Great site lots for god discussion

Thanks guys

kiwi_tree_steve

in NZ we work to a 10% SWL eg. 22kn / 2200kg has a SWL of 220kg

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"You have to feel and touch a tree" Shigo

Garry Brockley

The equation for the force of gravity for all objects relatively close to the Earth is:

F = mg

where

•F is the force pulling objects toward the Earth; it is also the weight of the object
•m is the mass of the object
•g is the acceleration due to gravity; this number is a constant for all masses of matter
•mg is the product of m times g
This acceleration due to the force of gravity on Earth g equals 9.8 m/s² in the metric system and 32 ft/s² in the English system.

Note: g is often called the acceleration of gravity. That is incorrect and misleading, since gravity does not accelerate. The expression should be the acceleration due to the force of gravity, which is a more accurate definition for g.

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kiwi_tree_steve

as far as im aware:

SWL does take into account for shock loading, thats what its all about aint it? knowing and working to a limit that is SUBSTANTIALLY lower than its rating,

in tree work we can only make an educated and calculated guesses on weight of wood, this is why the SWL is only 10% on rating, this allows for the ups and downs in weight due to many many factors; seasons, species, country, ect


im prity sure that Knots do effect, but others out there can tell ya for sure, JimNZ1 if hes around





steve
/

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"You have to feel and touch a tree" Shigo

kiwi_tree_steve

ok can ya explan a bit more regarding Shockload, drops ect galbee?
'

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"You have to feel and touch a tree" Shigo

Garry Brockley

apart from avoiding it as often as you possibly can, its caused by not allowing the rope to ease the load to stationary by tieing it off too tight or not allowing it to run on the portawrap.
the forces caused can easily exceed the swl and can either snap the rope through over stretching , cause failure at the rigging point, throw the tree into a major rocking or even bounce the cut piece into the climber , but you know that one, i think we've all had that groundy that wraps a piece tighter than a ducks ar** and nearly takes you out. iight

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kiwi_tree_steve

yeah, i understand what shock load is, and your right avoid it all all times. but are the Calculations to work it what the shock load will be? say dropin 1m, 5m, 10m, for interests and learning's sake?

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"You have to feel and touch a tree" Shigo

MTS247

ForceOn rope =

weight of load (L) x Gravity x Distance of fall (F)
Distance of deceleration(R)




= L x 9.8 x F
R

Im no physicist but i found that somewhere

kiwi_tree_steve

thanks!

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"You have to feel and touch a tree" Shigo

MTS247

I'm not saying thats correct.

And if it was you would need to plug the righ values into it. and im not sure if the gravity changes above sea level.

And the distance of deceleration (strech) is a big modifier, more rope in the system the more the stretch.

MTS247

I got that fomula to work once. if anyone knows what units to use let me know

Garry Brockley

Sorry the wife nicked the laptop for the last hour
thats the calculation i have but its not absoloute the piece will increase its velocity therefore kinetic energy every meter until it reaches terminal velocity (air resistance) then you have to take into consideration the amount of elasticity in each rope which gives you your R (distance of deceleration) ie an 11mm climbing rope will give you approx 1/5 more stretch than a 16mm bull rope. hope that helps

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MTS247

If your game see if you can use the formula for a 250kg log, iv been trying and keep stuffing it up, i just dont know what "units" to put in
eg; grams,kg,tons mm,cm,m

Garry Brockley

A piece of wood weighs 25kgs before being released,
theoretically after falling for one meter (without being held back by the rope or air) it has increased its weight E = (9.8 m/s2)m x h = 1/2 mv2
Energy(force emitted by the log)= 9.8 meters persecond persecond (or however many meters it falls) x original weight x height = 1/2 x mass x velocity squared
If i remember it right then velocity is worked out as v=squareroot of 2GH or 2 x gravity x height V = 4.428

9.8 x 1 x 25 x 1 = 1/2 x 25 x 4.428 squared and if you can do that in your head in a tree your a better man than I

ok my heads hurting again now i havent done these in a while

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Garry Brockley

so a piece at 250kgs x 9.8 gives a force of 2450 kgs when it falls for 1 second you can also use this formula
a = W / m = (m * g) / m = g

Acceleration 9.8= (weight 2450 divided by mass 250) divided by mass 250= gravity 9.8 or acceleration equalls gravity thats the easy way.
you then add the pressure of the troposphere and it gets really complicated as this is what holds back the object or T = 59 - .00356 x h is the height in feet from sea level. that goes up to 36000 ish feet. and then you allow for the lapse rate p = 2116 x [(T + 459.7)/ 518.6]^5.256 p is pressure.

see pilot training does pay off (eventually)

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MTS247

*eyes glaze over* you lost me at "if"

So did we get a answer? 250kg log


Pilot training
Plane weighs 200ton going milion miles an hour heading straight toward the ground! = BAD

FlashD

How about a really easy way to estimate force.
Take the approximate weight of the load, multiply that by the distance of fall, and add an additional unit.

Say, I have a 100lb piece, that is allowed to drop three feet, you're going to get an end force of 400lbs give or take. (ie 100x3+100=400)

That's the easiest way I've ever heard of to approximate force. It's not the true physics formula, but it'll get ya in the neighborhood.

Am I mistaken in my understanding that SWL suggests the weight of load that should be placed on a line, block, or friction management device, not necessarily the total force that should be applied? The tensil strength discribes the total amound of force and load that can be handled by the equipment and the SWL discribes the size of object that can be placed upon that equipment under any normal, reasonable rigging situation, right?

Eric Frei

I was taught 1/6th

No, you need to calc that out.

No, you need to calc that out.

No, your log would be much heavier now. You would easily exceed the 250kg SWL.

.........

I use a 6tonne breaking strength lowering rope, it's SWL is 1 tonne. It has a spliced eye so retains 90% of rope strength so in theory I can lower 900kg pieces with it.

Harry the hero chooses to tie knots in his rigging though, he ties mainly bowlines and has effectively reduced the SWL by 45% because a bowline loses 45% of rope strength. So Harry the clever hero now has a lowering system only good for 550kg, and he started out with a 6000kg rope.

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Eric Frei

S afe
W orking
L oad

With ropes manufacturers often provide breaking strength due to variances of SWL around the place. Never ever use the breaking strength as the SWL.

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FlashD

Thank you.

Steve Tipton

I was taught ropes, synthetic slings etc SWL (Safe working load) = 1/6th of breaking strain.

SWL of metal items, wire slings, shackles etc = 1/4th of breaking strain.

SWL is the same as WLL (weight load limit)

Steve

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Heightmaster

Eric Frei

I was taught 1/5 or 20% for hardware. But most hardware comes stamped with SWL

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Steve Tipton

That it does Ekka.

I could be mistaken!

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Heightmaster

MTS247

Why would it be less for hardware?

Steve Tipton

Hardware is less likely to be weakened by exposure to damage from chemicals, UV rays etc.

Also, mechanical damage is much easier to detect on metal items.

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Heightmaster

woodmiller

Quintrex, sometimes damage to metal items is very difficult to see, as many failures begin as microscopic cracks unable to be seen without the aid of an x-ray machine.

In general, the SWL question boils down to knowing your loads, and having a basic appreciation of what increases the loading on your ropes. i.e. avoid sudden jolts and allow slow down branches gradually when lowering. The quicker something comes to rest, the higher the force acting on the rope. It comes down to Newtons second law of Force = mass x acceleration, so if we can decrease the acceleration component by gradually slowing branches as they descend toward the ground, we are also decreasing the force acting on the rope to very close to the weight of the branch.

Steve Tipton

Excellent point.

Thanks!

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Heightmaster

second-gen_monkey

i'm a new climber too and as such (mostly because i dont want to kill a family member) i am always carefull to never make any cuts i don't absolutely have to that result in a load more than i can lift (about 220) i use a 1000lb rated lowering line now. i started with a 600lb and have noticed that my new rope seems to have more stretch to it. the highest load it's taken is (an estimated) 500 or so pounds at the point of deceleration. i was actually able to watch the rope stretch... um, is this normal? the load (a good peice of live black olive) was suspended momentarily while the swing was brought under control and this is when i saw the stretch occur. my other rope was fairly well used as it was a hand me down, as is most of my equipment and i don't recall ever seeing this happen to it.

kel

Sounds like you've got a dynamic (designed to strech) rope as opposed to a static (no strech) rope. The streching helps take the shock out of a falling load ie. trunk I use static 12mm for everything.

second-gen_monkey

that would explain the lecture i got on not using that as a climbing line.... thanks.