Break testing 3 knots

Arborist Forum

Help Support Arborist Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.
It seems logical to me that used rope would have sections already stretched and thus any test performed would be moot. This would also explain why Dans rope broke in the middle rather than at the knot. I wonder if the middle of a length of rope actually stretches further than the ends which are tied to rigid objects or is that stretching uniform over length....?

Waiting patiently for more data... :popcorn:
 
Since I would NEVER nitpick, ....

....Conclusion: with the setup shown in the picture, the tension in the rope would change very little where it passes through the loop, and therefore that bent passage should be the weakest spot in the entire system.

I'll take those comments as mostly true, with pretty good analysis on the loaded legs. I don't think that at 120° angles, that all the legs must be equally loaded. It is only essential that the "bight end" have a force on it equal to the force applied by deflecting the loaded side of the rope, and that the sum of the vector forces of the "after bight" and the "bight end" equal the loaded leg. Some of the load is absorbed by friction going around the log, which would be converted to torque on the log. Not all logs are free floating. Some are rigidly mounted to the ground. The torque on this log might very well be hidden by the log not wanting turn; as in say, a tree being pulled over?

Sure, the main line MUST be weakened where bent, even if only a little. OK! Technically, the weakest point.

But how much does that little bend weaken the rope? Not much, I'll bet. And beyond that bend, the load is carried by two legs of rope. So any imperfections in the rope closer to the log than the bight become moot.

Then you must compare that bend-point against the other possibilities along the entire length of the rope. Given that most ropes are used when we load them to breaking (after all, when they were new, they were strong enough to not break!), then there must be other imperfections along their length. So the problem now becomes a statistical analysis: what is the probability that there is another imperfection along the length of loaded rope that exceeds the weakening that occurs at the loop where the loaded line is tied to the log?

That probability on a well-used rope is pretty good, hence The Dan's observation that they usually break in the middle.

My own personal observation is that they ALWAYS break where they bend most sharply around the strongest piece of rigging in the system. Bumper hitch, figure-8, or the half-hitch holding the timber hitch securing the giant log that really shouldn't have been cut off at that size. I don't think I have ever broken a rope "at the knot", since I never use a knot to carry a rope breaking load (they are such a b**** to untie afterward!). I always try to isolate the loads on the rope around structural items that are stronger than knots.

Really though, I can't remember breaking enough ropes to qualify as an expert on the topic. We usually end up cutting them with a chainsaw, or destroying them in some other fashion than breaking them.
 
Last edited:
When a load is put onto a rope, the rope would continually constrict to a point and then it would expand back out beyond that point. The apex of that constriction is where the rope has failed Dan. At that apex, the most energy is being forced onto the fibers causing heat and failure. Now, with that being said, general use of our ropes would cause degradation and could skew the location of the break. Also, pulling from only one end (dragging a tree) would shift that apex closer to the weighted end. If you have one of those exercise bands, you'll see how the middle constricts. Rope, technically and theoretically, stretches under the same principles.
 
If the teepee is flat; the line is generally leveraged hard; giving also tighter grip; but uses more of the potential tensile strength to do so.

Leveraged hard around a tight bight, that excludes all but the outer strands is more trouble than same around on larger bight. Stiffer line will show this more. It is the resistance to bend that gives the leveraging, that is why you can leverage a tensioned, but not a slack line with perpendicular force. Once again this (d)effect is much less compromising in flat lines, that give less dimension on this bent axis; so can't be leveraged against you as much(for they don't stand as tall at bend, and are more flexible generally.

Some of the tweaks yield less; but i due believe the art is in the practice of putting out the most maximum security/strength in an equitable amount of time, with as little wear as possible. And developing an eye that can immediately spot a problem; and 'polish out' any imperfections quickly and correctly. If these things take 'too long' perhaps more practice is needed; is what i'd always tell myself. Also, the more intently you try to maximize these things to their pure/unadulterated state, the more truer relationships and commonalities your eye can catch- in time. Anything else can go wrong, and the more chips ya got stacked on your side the better over all.
 
Slowly and in English please!

Hey Moray the results from the tests were way more interesting than the theory.:cheers:
 
I'll take those comments as mostly true, with pretty good analysis on the loaded legs. I don't think that at 120° angles, that all the legs must be equally loaded...

Yes, and your discussion is mostly right on, but I must complain about this 120° business, even though we are straying far into new territory and we have jointly nitpicked this thing to the bone. The 120° angle between 3 loaded legs leads to a mathematical truth that really has nothing to do with physics or mechanics per se. If three forces act on a central point, and no others, and they are respectively 120° apart, then they are equal, or they are not equal and the central point is in accelerated motion. 120° (how do you make that degree sign?) is not special. If you have 3 forces on a central point and no motion, then the size of one of the forces unconditionally and unambigously determines the sizes of the other two. This can be a very good reality check on your reasoning. There may be good reasons for believing something is very large or very small, but the angles tell the story.

But I agree; the loop does not apply much force to the main line, it is not bent much nor weakened much. The visible breaking point is surely due to some existing weakness or damage in the rope.
 
Slowly and in English please!

Hey Moray the results from the tests were way more interesting than the theory.:cheers:

I'm glad you liked at least some of it. Some serious tests are yet to come; I am ordering a bunch of new rope that should be nice and uniform and give meaningful results...
 
2 Vertical poles supporting a clothesline with a 120°(thanx°) spread, and load in the middle; will exert 1xLoad on each hitch point on pole. While the same with poles next to each other, and Zer0° deflection betwixt the support lines; will exert .5xLoad on each hitch point on pole.

To my weigh of thinking...(OOOOooooooooooooPs....) -

This is because; there are 2 legs of support, so if self equalizing (theoretical Zer0 % friction pulley on load point/ center bight of line); each leg would have to support half the load everytime, regardless of angle/ deflection from inline.

But, rope is not like wood, steel etc.; it will not support on any of the cross axises( i know bad grammar, but somehow less confusing than calling this plural "axes" on a tree site...), only on the inline axis.

Anyway; we can't get support from both the leveraged and inline axises like a non flexible device; only on the inline axis. So, the cosine of Zer0 (no spread angle of lines)is 1; so each leg can support it's half share of the load; with cos(1 or rather 1/1) x Load/2(supports)=line tension of .5xLoad.

Now, with a 120°spread; there are still 2 support lines sharing the load; so whar does the 'extra' tension come from?? Well, with a 120° spread; each leg is 60°from inline (for a 120° spread). The cosine of 60 is .5; so each half loaded support leg can only support that cosine (.5) x the line tension; or 1/cosine(.5) x load supported on that support leg (Load/2) gives 2 (1/.5) x 1/2 (half of load supported on each leg); for a total line tension = 1xLoad (2x 1/2 x Load); thus that pull on each supporting leg.

Now whippin'out (y)our windows calculator and placing it's view in scientific mode; we can see that 170°spread gives 85°deflection from inline on each leg. Place 85°on the calc and press cos key and we get ~0.0871; press the 1/x key; and we get ~11.4737. A 1000# Load would need to get 500# support from each of the 2 legs; but each load support leg can only give ~11.4737 inline support to each leg. 500(support needed per leg to support the 1000#Load ) X ~11.4737 = ~5736.8566 line tension. Which is a consistent pattern to other calculations(?).


Oooops gotta run, but i hope this is close.
 
Last edited:

Latest posts

Back
Top