How to figure working loads?

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NebClimber

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I've decided it is time I start knowing the working capacity of my equipment rather than guessing. But I don't know how to account for the loss of strength due to knots and hitches I use.

Please help.

I'll set up an example, first listing the rated capacity of my rigging gear:

Sampson bull rope: 8,500# tensile
block: 2 ton
whoopie: 3,380 WLL
loopie: 3,000 WLL

Example: I girth hitch the whoopie to the base of the tree and install a port-a-wrap. I girth hitch the loopie high in the tree and install the block. I run the bull rope through the block, and tie it to piece to be removed with a running bowline. Assuming a static load, what is the biggest piece of wood this system can lower (assuming I want to work at 10% of tensile strength)?

At first blush, the answer appears to be 850# (i.e., the WLL of the weakest link in the system - the bull rope).

But I have not accounted to strength lost in knots, hitches, etc. So how do I figure this out?

Steven
 
Good question..

I think the 10% SWL is designed to account for loss due to knots.. You can also through a tosemite tie off on that running bowline which should reduce loss of strength..
 
This is what I've gathered from my books:

Bowline decreases strength by 35%

A 5/8" rope bending over a sheave with a 2" "tread diameter" creates a bend ratio of 3:1. A 3:1 bend ratio decreases rope strength by 20%.

So, by running my 5/8" rope through a block with 2" tread diameter and tying off with a bowline, I've lost 55% of tensile strength.

So should I automatically reduce my WLL by 55%?

Also, how do you figure in dynamic loading? Blair's book says just add a unit of weight for each foot of fall. E.g., if 500# is dropped one foot, add one unit of weight to the original 500# and you get 1,000#. Drop it another foot and the load is 1,500#. Drop it four foot and the load is 2,500#.

But another souce says for each foot of drop, double the load. E.g., if 500# is dropped 1' then the load is 1000#. If dropped another foot, the load is 2000#. 3'=4,000#. 4'=8,000#.

Big difference between the two formulas.

Thoughts?
 
A little common sense is on order..
Rope bending over sheave has nothing to do with loss of knot..
So why add the two... I fyou're concerned about strength in a system look for the weakest link.. Which would be the rope at the knot...
 
Rescueman addressed this theoretical accumulation of strength losses in some other thread; the proper analysis is to look at the weakest point, not add them up...just as Daniel says.
 
How far and whether to double the load for ratings depends on your point of view. The sling that attaches the block "high in the tree" experiences a 2:1 loading b/c of the porty on one side, and the load on the other. However your setup involves a much larger working limit there so I wouldn't worry about it.

The real issue is how far you have to stop the load compared to how much of a free fall it undergoes. The "doubling" of a load assumes you're going to stop it in the same distance regardless of rigging. "Knowing" has to be coupled with experience to determine the limits of your rigging. If in doubt, take a smaller chunk.

I'm of the mind that the 55% is built into the WLL rating. You probably would be safe at the full 850 lb if the rope is pretty fresh and your groundie knows what they're doing on the porty.

TT
 
The higher load is on the block's position, perhaps the 3800# tensile Whoopie should go there? And the 3000# Loopie would go where the less loading of the Porty is? The doubling of the load also depends on how spread the legs of line to the pulley are.... at loading.

Alternatively, perhaps basket the Loopie at the Block position if the locking grip of choke could be forsaked for increased strength at this higher loaded position? Part of it i beleive is matching componenet positions, so that the weakest/ most loaded is fortified as high as possible. The chain of support positions will only be as strong as the weakest link; but their weaknesses will not be accumalative per position of weakness (as per Daniel and Burnham), only accumalative in the same position (doubling of load force on 80% positioned hitching)(?)

So with a chain strong as the weakest link, upgrade the whole system instantly by treating only the weakest position per loading. Especially made strategically with no more equipment or time, than sighting and running a different, sounder strategy.

i believe that running an 8/1 ratio of sheave to flexible braid diameter is ~100% rope efficiency (10/1 in 3 strand, ~22/1 in kevlar/aramids). Best with 10% wider sheave than rope diameter on the sides. Kevlar, wire etc, even take a specific shaped sheave base to support the 'belly' of the loaded line correctly too, or else overloading can result quite easier!

A 4/1 sheave to diameter of line for non-stiff braids is ~50% line strength and minimal recomended i beleive 3/1 would be less strength, certainly not maintain 80%?

Fixed terminations (not rotating pulley, just fixed hanging); 3/1 is recomended for ~only 40%(?) tensile loss. The tensile loss per specific situation is not the whole story, dropping below the 4/1 turning ratio can cause much less cycles to failure; giving out in the future(?).

The amount of braking force invoked by the support system determines the loading on the support system, so free falling or running loads don't load the system, or the doubling factor of the redirect of Porty's Brake Force.

Originally posted by murphy4trees
A little common sense is on order..
Rope bending over sheave has nothing to do with loss of knot..
So why add the two... I fyou're concerned about strength in a system look for the weakest link.. Which would be the rope at the knot...

So actually, 50% strength loss in 4/1 sheave; doesn't compound a 50% strength in the knot holding load on line running through sheave; but actually matches the componenets to a consistant chain strength of properly mathced elements.

Or something like that.........
:alien:
 
Last edited:
Thinking it through, I do agree that the strength reductions should not be added together.

Does anyone know the answer to the falling log question? Granted, the dynamic loads can be offset by letting the line run, rope stretch, etc. But still - in theory - what is the more accurate rule of thumb?

Note that I am not trying to figure out how to rig a specific piece. I just want to know the underlying rules so I can apply them to array of situations in the field.

Steven
 
Reply

In this figure, do you also add in the size of the tree and the size of the part of the tree you are riggigng off of? Also what type of tree your rigging off of. So to double check my self. What is the general rule of thumb of the diameter of your rigging spot compared to the diameter of your tree. In order to be 100% sure of where to set your rigging in the tree to get the maximum performance out of your total rigging system. I know there is a degree of comon sense in all this and you need a experianced groundman who can run the ropes.
 
You might be able to find the answers to the dynamic load questions in the ISA archives. I know the best discussions got deleted but there is probably enough other info there to thoroughly answer the question without having to muddle through the process again.

The Blair formula is close enough to actual calculations. There isn't a doubling effect, well, maybe in a vacuum, but not in the real world.

Tom
 
Brain,

What is your need to spray on anything connected with the ISA? Did the ISA do something so terribly rotten to you that you feel compelled to stink up any conversation about a professional organization. If you don't have anything to contribute, why not pipe down. Give it a rest will ya?

"Everyone" hasn't ingnored the forum, granted it isn't real busy though. When has popularity always meant quality?

I don't do homework for people but I will lend a little help. When I've needed info in the archives I've been able to find it. Maybe I have a less sophisticated knowledge level or computer system. Don't know.

Tom
 
Try reading a thread posted here titled 'oh, for pete's sake'. A guy showed how to figure loading of ropes. Brian's true sense of intelligence showed in that thread as well as this 1.

Joe
 
Originally posted by Joe
Try reading a thread posted here titled 'oh, for pete's sake'. A guy showed how to figure loading of ropes.Joe

But the equations in that thread are for energy, not force which is what NebClimber is asking about.

Force = mass x acceleration - or -
Force = mass x change of velocity/time

The key variable here is what is the time of deceleration?

The impact force is the weight of the falling object times the ratio of the time of falling to the time of stopping.

So if an object falls for one second and takes a quarter second to decelerate, then the impact force is 1/.25 or 4 times the weight.

That is why the force is so much less on a more dynamic system than on a more static system - because the deceleration time is larger.

And that's why the impact force of a dropped chainsaw on sod is much less than on concrete - because the deceleration time is much longer.

How's that for muddying the water?

- Robert
 
Reply

So in reality there is no real true answer. It all depends on the tree and what your doing to it and what type of rigging system you have set up. As long as you don't over load the tree or your rigging you should be just fine.
 
Re: Reply

Originally posted by Froggy
So in reality there is no real true answer. It all depends on the tree and what your doing to it and what type of rigging system you have set up. As long as you don't over load the tree or your rigging you should be just fine.

Thank you. :)
 
As long as you don't over load the tree or your rigging you should be just fine.

That kind of thinking is a red flag for me...
There are a lot of ways to make mistakes and get hurt or killed when rigging trees... Understanding safe working loads is just part of a much bigger performance...
 
Rescueman:

The deceleration time I am talking about is 0. I.e., the load comes to an abrupt halt.

I use zero b/c I know I can then decrease the impact by running the line, or using a rope with more stretch. But if I first start with the theory that the load will come to an abrubt halt at the end of its fall, at least I will have a constant baseline by which to measure the load on my rigging system.

So, Rescueman, which rule of thumb do you recommend? Blair's or the other one?
 
Originally posted by RescueMan
But the equations in that thread are for energy, not force which is what NebClimber is asking about.

Nope. You need to read near the end of the thread. The force was calculated, but the energy was used to help find the force.

Force = mass x acceleration - or -
Force = mass x change of velocity/time

The key variable here is what is the time of deceleration?

The impact force is the weight of the falling object times the ratio of the time of falling to the time of stopping.

So if an object falls for one second and takes a quarter second to decelerate, then the impact force is 1/.25 or 4 times the weight.

That is why the force is so much less on a more dynamic system than on a more static system - because the deceleration time is larger.

And that's why the impact force of a dropped chainsaw on sod is much less than on concrete - because the deceleration time is much longer.

How's that for muddying the water?

- Robert

http://www.glenbrook.k12.il.us/gbssci/phys/Class/momentum/u4l1c.html

Joe
 
Originally posted by NebClimber
Rescueman:

The deceleration time I am talking about is 0. I.e., the load comes to an abrupt halt.

So, Rescueman, which rule of thumb do you recommend? Blair's or the other one?

Sorry, but the deceleration time is never zero. Even bull rope stretches under load (about 5% at 10% of tensile strength), and that assumes an otherwise completely static system (no slip in the portawrap).

I don't see how either "rule of thumb" is valid, because as I said above the impact force has a lot less to do with the length of the fall than the length of the stop. The longer the fall, the more stretch in the rope to absorb the impact, so the impact doesn't change that much.

The impact is greater, though, if there is less rope to stretch. If you were lowering the chunk from a porty at the top of the spar, there would be a lot more impact force than if you were catching it by a porty at the base of the tree with a good amount of rope out.

- Robert
 


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