# Splice Squeeze Force



## moray (Jun 10, 2008)

The rope manufacturers publish nice data tables for their ropes, including such things as tensile strength, stretch under load, density, and so on. I imagine most people who use such ropes, aware of the rated strength, have no worries about the rope failing under far smaller loads, such as a climber's weight. 

Unfortunately there are no such nice tables when it comes to splices, even though the manufacturers publish detailed instructions on how to make a safe splice. The obvious concern about splices is that they will simply pull apart under load, and I am certainly among those who have worried about this. If the manufacturers would publish numerical data for the various rope and splice types, the engineered safety factor for spliced eyes would be apparent and a lot of people would sleep better at night.

For a spliced eye in a 12-strand rope like Tenex, it is common knowledge that the splice works because the outer cover squeezes the inner core when the rope is under tension, and the squeeze force creates friction that holds the core in place. Just how hard is the core squeezed by the cover? Since the manufacturers won't tell us, I decided to find out.

But first a short detour. Even though it will be interesting to know how much squeeze force is generated by a given amount of tension, we _really_ care about the resulting friction that holds the splice together. This is simply the total squeeze force times the coefficient of friction between the two surfaces. To find the coefficient of friction of 3/8 in Tenex rubbing on itself, I used the setup shown below. 







The one-inch copper pipe is wound with a very tight spiral of Tenex which is clamped at each end. Then a separate Tenex line with a 20-lb weight on one end is used to take measurements, as detailed in the thread on Friction Saver Friction:  http://www.arboristsite.com/showthread.php?t=64807

The results showed an efficiency of 48%, meaning it would take 100 lbs pulling straight down on one leg to lift 48 lbs on the other. Mathematically this translates to a coefficient of friction of .234, a very middle-of-the-road figure for coefficients of friction. With this number in hand, it was time to return to the main problem, measuring the squeeze force.


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## ClimbinArbor (Jun 10, 2008)

*keep it up moray*

you have some great numbers moray, and i wish i could spend as much time as you do experimenting with ropes and splices. most of my experimenting is done tied in and standing on the ground to 5' in the air.

also on the current subject os splices. sherril had a couple of figures in one of their last catalogs(i think 07) on hand splicing and grizzly splicing. i think the WLL on the grizzly was almost twice that of the hand spliced. this drew ALOT of water with me and i have been working the grizzlies since, when they have them on what i need that its.


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## moray (Jun 10, 2008)

ClimbinArbor said:


> ...also on the current subject os splices. sherril had a couple of figures in one of their last catalogs(i think 07) on hand splicing and grizzly splicing. i think the WLL on the grizzly was almost twice that of the hand spliced. this drew ALOT of water with me and i have been working the grizzlies since, when they have them on what i need that its.



This is interesting. There are a lot of different hand splices, depending on the type of rope. Usually the manufacturer will specify that the splice preserves 90% to 100% of rope strength. I don't remember ever seeing a number less than 85%. If that is the case, then it is hardly possible for a grizzly to have twice the strength.

On the other hand, the grizzly can make a really strong eye in a rope where hand splicing is not possible or really hard to do. I haven't actually seen one yet, but I think they're a great idea.


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## moray (Jun 11, 2008)

*Experimental approach*

How can one measure the squeeze force of a splice under tension? Two ways occurred to me, one experimental and the other purely mathematical. 

For both the experimental and the computational methods, we need to know the tension on the rope when the splice slips. Unfortunately, we don't know that. But we can still get where we want to go if we know the _ratio_ between the tension in the core and the tension in the cover. It is the cover tension that is doing the squeezing and producing the holding friction; it is the core tension that this friction must resist if the splice is not to fail. We actually know this ratio from an earlier experiment.

In the thread on splicing 3/8 inch Tenex, here: http://www.arboristsite.com/showthread.php?t=68791, I described some pull tests on very short untapered splices. A splice with a 2-inch bury could not be pulled apart with a car, and a splice with 1.5-inch bury held for a number of seconds before finally coming apart. 


During the pull tests, the eye was held by a 12mm steel screw link. As the splice pulls apart, the material of the eye will be sliding over the screw link like a rope moving over a pulley. A lot of friction is generated by the rope sliding over the steel, and this friction resists the motion. The result, just as with a pulley or false crotch, is that the rope leg moving _away_ from the steel has more tension than the leg moving _toward_ the steel. I had already measured the efficiency of Tenex on a steel screw link at 48%, meaning it takes 100 lbs pull on one leg to lift a 48-lb weight on the other. Therefore, as the splice pulls apart, the ratio of core tension (the load) to cover tension (the pulling leg) is .48.

To proceed with the experimental method, let's assume that the car was pulling with a force of 1000 lbs when the splice failed. This assumption has no significance, but it makes it easy to describe the reasoning with real numbers instead of algebraic variables. 

As the splice begins to slide apart, the friction on the screw link forces the two legs of the eye to bear unequal forces, as argued above. The ratio of the two forces is .48, and the sum of the two forces is the full 1000# applied by the car. Thus the cover leg, the one generating squeeze force, will be under 676# tension and the core will experience the rest of the 1000#, or 324# tension. 

In the previous experiment (Post #1) the coefficient of friction of Tenex on Tenex was measured at .236 (which is coincidentally the same as Tenex on steel). For every pound of squeeze force, .236 pounds of holding friction is generated. We need 324 lbs of holding friction to keep the splice from failing (which we had for several seconds!). 1373 lbs multiplied by .236 gives 324, so 1373 is the squeeze force. This much force was generated over a spice length of 1.5 inches. Dividing by 1.5, we get 915 lbs of squeeze force for each inch of splice length.

The 1000 lbs figure was just an arbitrary guess. We can eliminate it and state: when the rope is under a tension of T pounds, the squeeze force on each inch of splice will be .915T pounds.

In slightly simpler form, the squeeze force in every inch of splice equals 9/10 of the tension in the rope.


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## ClimbinArbor (Jun 11, 2008)

*grizzly splice*

i think i might have been wrong before moray. i found an old article while i was looking for the one i spoke of above, when i found this one. it is in sherrill 07. it says 90% stronger than knots. 

this may be what i was thinking of. which may make more sense huh


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## moray (Jun 12, 2008)

Now for the computational method. I like and trust this method more for two reasons. One, one of the measurements we made before is no longer necessary: we don't need the coefficient of friction of Tenex on Tenex. Removing a messy experimental variable should give us more accuracy. Two, we avoid the messy irregularities of the splice itself. Trying to find the squeeze force from an extremely short failing splice sounds dicey from the get-go.

This method assumes a long uniform rope with a long uniform core that is being squeezed due to tension in the rope. In other words, we are dealing with a pure Chinese finger cuff.

The photo below shows a section of splice containing full-diameter core. There are 12 strands (yarn pairs) forming 6 right-hand spirals and 6 left-hand spirals. Each spiral should act just like a bull rope wrapped spirally around a tree. If we can compute the squeeze force of a spiral around a tree, we can compute the squeeze force of a spiral around a rope core. Multiplying the squeeze force generated by one spiral by 12 should give us the squeeze force for the entire rope.






The major assumptions are:
1. The rope construction does not interfere with the squeezing action.
2. The 12 strands form perfect spirals.

How good are these assumptions? Anyone who has handled Tenex knows that the _relaxed_ form of the rope is the constricted form. It will constrict around anything you put inside it without outside help or effort. If it interfered in any way with the squeezing action, it would probably be totally negligible.

Assumption #2 is a little trickier. The 12 strands clearly do not form perfect spirals. Each strand is braided with the others such that half the time it is on the outside, covering two strands below it, and half the time it is on the inside against the core, covered by two strands above it. As it oscillates up and down on its spiral path around the rope, it may also, to a much lesser degree, oscillate "left" and "right", changing the pitch of the spiral. The up-and-down deviation is by far the main departure from the ideal spiral and certainly needs to be accounted for, but my instinct tells me we can find a suitable correction that will largely eliminate any error from assumption #2.

The angle P in the photo I am calling the pitch of the spiral, and this angle is vital for the calculations. Each strand makes one complete revolution around the core in 1.75 inches of rope length, as shown for the strand labelled Y-Y. There are difficulties in trying to measure the pitch angle directly, as I found out, and using the 1.75-inch spiral length will help us calculate it rather than measure it directly.

We need 2 more numbers: the rope diameter (about .53 inches, and not so easy to measure!) and the diameter of a single yarn (about 1/16 inch, the size of the up/down oscillation). We should be ready...


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## moray (Jun 13, 2008)

A rope under tension wants to be straight. It takes force to put a bend in it; this force is the squeeze force. This applies to much more than just holding a splice together--it is the force that makes a figure eight or other rappel device work; it is the basis for natural crotch or any other kind of rigging; it applies to a capstan winch; for that matter it is what holds knots together. It is everywhere.

The relationship is simple: squeeze force is proportional to the angle of bend. Take the simple case of a rope over a limb. A 200-lb climber is tied in at the bottom such that each leg of the rope supports 100 lbs. We have a rope with 100 lbs tension throughout going through a bend of 180 degrees. The _net_ force on the limb, as everyone knows, is 200 lbs pointing straight down. The _squeeze_ force on the limb is 314 lbs. This force is spread out uniformly along the contact zone between the limb and the rope, and the direction of the force at each point is always perpendicular to the contact surface. (Those who remember some high-school math will recognize the number pi in that 314 lbs, and indeed pi is involved.)

The simple case just described is what we might call a flat bend. A spiral bend, as it turns out, always produces less squeeze force than a flat bend, and the reason is simply that the rope itself doesn't bend as much. As an example, if a rope under 100 lbs tension is wrapped 180 degrees around a vertical pole, the pole will experience 314 lbs of squeeze. If we now keep the tension the same and still wrap 180 degrees of the pole, but incline the rope 45 degrees to make a uniform spiral, the squeeze will drop by about 30%. This implies-- and it is true--that even though the spiral rope has wrapped around 180 degrees of pole, the rope itself has bent 30% less than that, or about 127 degrees. The angle P in the earlier diagram is what we need for the correction: multiply the sine of P (which is always less than or equal to 1) by the number of pole degrees to get the number of rope degrees. A nice tidy little relation.

There is a similar adjustment that must be made for tension. If we were to load up our 12-strand rope with a 120-lb weight, we might expect each of the 12 strands to take its proper share of the load, or 10 lbs. This won't be right. The strands can only carry a load along their length. This is the same trigonometry that applies to ziplines, traverses, clotheslines, and so on, where the direction of the load force is at some angle to the direction of the supporting rope. The tension in the rope is always greater in such cases, and the angle between the load force and the rope direction directly directly specifies the correction factor.

In our case, the angle P is the angle between the load force and the actual trend of the strands. We divide the nominal load per strand by the cosine of P to get the true tension in each strand. Another gratifyingly tidy little expression.

Before we can start crunching numbers, we need know the size of P, which is the key to everything. Why not just use a protractor and measure it from the picture? I tried that and got readings that bounced around a lot more than I liked. A builder doesn't use a protractor or even a framing square to make sure the corners of the floor are square. Instead the two diagonals of the floor are measured to make sure they are equal. We can do something similar to indirectly but accurately measure the angle P.


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## moray (Jun 15, 2008)

Imagine that the cylindrical rope is a paper tube that we can unroll. On the surface of the paper tube draw the strand Y-Y, shown in the photo, as it winds around one full revolution of the rope. Also draw a circumference line at one of the Y's. From the other Y draw a line parallel to the tube axis. Now unroll the tube, and you have the figure shown below.






This is a right triangle, and we know the lengths of both legs. The rope length of one revolution of the spiral, 1.75 inches, was measured earlier. The diameter of the rope, averaging 7 measurements, is .469 inch, which gives a circumference of 1,473 inch, which is the vertical line C on the diagram.

Knowing both legs of the right triangle, we can directly calculate the angle P as 40 degrees.

So far so good, but not so fast. This angle corresponds to the spiral inclination of a fiber on the surface of the rope. Any particular fiber actually moves in and out with respect to the axis of the rope as its strand weaves in and out of the other strands. The _radius_ of the fiber, i.e., the distance of the fiber from the rope axis, increases and decreases as the fiber sprials around the core. Since there are two layers of strands, we could take as a sort of average radius the midway point between these two layers. We could then assume all fibers in all strands lie on the surface of this new cylinder and make perfect spirals around the cylinder. If we do this, knowing that the diameter of one strand is just over 1/16 inch, the adjusted diameter will be 1/8 inch less than the measured diameter, or .344 inch, and the new circumference is 1.081 inch. The value for P becomes 31.7 degrees. This first-order adjustment will reduce our estimate of squeeze force by about 27%, a big improvement over the unadjusted estimate. There are other finer adjustments that could be made if one could really understand the shape of the rope strands, but the improvement would probably be much less than the 27% we already have.

So what is the squeeze force? It is 2.2 times the rope tension per inch. In the case of the splice that pulled apart, where the calculated tension in the cover was 676 lbs, the squeeze force per inch would be 1500 lbs. 

This is about 65% higher than the number derived from the splice failure experiment. What is going on?


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## moray (Jun 18, 2008)

*Getting closer...*

For the calculated squeeze to be 65% higher than the number derived from the splice failure experiment may be entirely plausible, but it also suggests something is wrong with the argument above. The calculations apply to a long uniform stretch of rope with another rope inside.

The schematic diagram below shows the situation.







At the points marked "A" we have the undisturbed ordinary Tenex. When the rope is put under tension, the full tension would appear at both points "A". Likewise, the full tension would appear at "B", where both core and cover are present. But when tension is applied, the rope stretches. It has to. We know that the cover is going to clamp down on the core with an iron grip, and the resulting friction is going to lock the two together. Therefore, as tension is applied and the cover stretches, it forces the core to stretch as well. If the core is stretched, it is under tension.

This is the essential element missing from the analysis above. In the section of rope where the core is present, the core and cover share the load. The cover tension that produces the squeeze force on the core is not the full tension on the rope, but something much less.

How do these two share the tension between them? Noting that the cover and core are both the same thing, 3/8 in Tenex, we could assume, as a first approximation, that they share the tension equally. Since we know the two are clamped together by friction, and therefore stretch the same amount, we can use that fact to get a better estimate for how the load is partitioned between core and cover.


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## moray (Jun 18, 2008)

*The Answer*

If you have two different kinds of rope supporting a load together, the way in which they share the tension is fairly straightforward. The stiffer, less stretchy rope will carry the bulk of the load. Mathematically, there is a direct relation between stiffness and amount of load, and this comes from the fact that the rope behaves roughly like a linear spring.

In our situation, even though the ropes are the same (Tenex), they are _structurally_ different because one is fat and distorted from swallowing the other. What we need to know, for both ropes, is the angle P, and from that we can get the stiffness.

This angle, for the cover, has already been determined: 31.7 degrees.

I went through the same drill to find this angle for undisturbed Tenex under light tension: measure the diameter (.373 in), adjust downward because the rope is 2-layered (.213 in), measure the length of a complete spiral (2.11 in), and calculate the angle (5.76 deg).

When the rope stretches, assuming all the constructional slack is gone, the indivdual strands have to stretch. But they are not aligned with the rope, and so they have to stretch more then the rope does. This is where the angle P comes in. For the core strands, with a very small angle P, the difference is very slight: the ratio of strand to rope stretch is 1.005. It is another matter for the cover, with much larger P. There the ratio is 1.175. These two numbers represent the relative stiffness of the two "types" of rope. Their relative shares of the tension are in the same ratio. The cover, because it is stiffer, takes the bulk of the tension (53.9%) and the core takes the rest (46.1%).

Now we are getting somewhere! In the earlier calculations, we came up with a squeeze force of 2.2 times the tension for each inch of rope. Multiplying 2.2times 53.9% we get 1.186 times the cover tension per inch of length.

The same figure derived from the splice-pull experiment was .915, so the experimental value is 77% of the calculated value. 

Considering that these were two independent ways of getting at the same number, and the experimental method was quite indirect, I consider this to be a remarkable degree of agreement. It doesn't prove anything--there could still be errors in reasoning or in the math--but it gives one some confidence in the result.

We can take a rough average of the two results and summarize: every inch of a Tenex splice experiences a squeeze force equal to the tension in the cover. Also: every inch of a Tenex splice has a holding strength equal to 1/4 the tension in the cover. Nice!


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## moray (Jun 27, 2008)

*Field Repair*

One day you discover your 3/8" Tenex rope has 3 of its 12 strands completely severed from some unknown or unremembered mishap. This particular line is used to lift heavy loads, and does not need to wind around any sort of pulley or drum. To repair it, you decide to cut it in half where the strands are broken, then do an end-to-end splice. This will restore the rope to full strength while shortening it about 3 feet in the bargain.

But this is Tenex, and there is a much better way. Just find a foot or so of spare rope, 3/8" or smaller, and pull it into the damaged rope at the site of the broken strands. Repair is done! No spare rope handy? Then use something else with good tensile strength, like a steel rod. 

When the repaired rope is put under tension, the cover squeezes down on the core material, and the cover and core lock together from the friction. The two therefore have to stretch together from the load, and together they bear the load. As long as the combination of core and cover has as much tensile strength as the original rope, the repaired rope is as strong as the original.

Even though this should work as described, the description is an oversimplified first approximation of the problem; there are at least 3 issues that significantly affect the outcome. 

1. The stiffness of the core and the cover may be different.
2. The minimum length of repair material will depend on #1.
3. The broken strands must not be allowed to unravel. 

More to come...


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## moray (Jun 29, 2008)

*The Steel Rod*

The case of the steel rod holds a suprise or two, and has relevance not only for the behavior of splices, but of knots as well.

The diagram shows the situation, but is merely schematic--it has no particular scale. Let's start with a few reasonable assumptions so we have the means to make an argument.






1. We assume that the rod has been placed so that the broken rope strands are over the center of the rod. 
2. Assume further that the rod is very long so that there is no question that the rope under tension will produce plenty of squeeze friction to lock onto the rod.
3. Assume that for any given load the rope is 10 times as stretchy as the steel. 
4. Assume that the coefficient of friction of rope on steel is 0.2.

As the rope is put under tension, it stretches. Because it locks onto the steel rod, the rod stretches as well, and by exactly the same amount. A normal splice behaves this way (but a knot does not!). Since it takes 10 units of force to produce a unit of stretch in the steel, but only 1 unit to produce the same stretch in the rope, The steel will at all times bear 10 times more tension than the enveloping rope. The entire length of the rod, including the eveloping rope, becomes in effect a new stretch of rope 11 times as stiff as the orginal (but not necessarily 11 times as strong!). 

At this point we should remember that the _source_ of the tension is somewhere off to the sides, not on the diagram. The undisturbed rope outside of the two points marked "A" is under full tension, and it is this tension that is somehow transferred to the rod, forcing it to stretch. How? By friction generated by the squeeze force.

Just imagine trying to pull the rod apart with your two hands; the situation is completely analagous. Your hands squeeze the rod, the pulling force developed in your arms ends up as tension in the rod. What seems peculiar and non-intuitive is the distribution of tensions: you have the enveloping rope under very light tension, yet it produces very high tension in the steel rod.

The way rod stiffness figures into everything produces another odd result. If our current steel rod is just barely long enough so that the rope can lock onto it with squeeze friction, what happens if we replace the rod with another of the same length but twice the stiffness? We know it will now carry 20 times as much load as the enveloping rope. Alas, because the enveloping rope tension has just dropped by half, so has the squeeze force and resultant friction. The rope can no longer lock onto the rod!

This is beginning to get interesting...


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## moray (Jul 8, 2008)

*Describing the standard splice*

Having developed the idea of load sharing between two elements of the rope, namely the cover and whatever load-bearing element has been inserted to form a core, we can now give a fairly complete description of the behavior of a normal hollow-braid splice.

We know that for the full length of the bury, the cover is everywhere stiffer than the core, with the greatest discrepancy where the splice is fattest, and the least discrepancy at the end of the taper. The reason for this, as described earlier, is simply that the core strands are oriented more nearly parallel to the load. The other ingredient leading to overall stiffness is the number of strands. It takes twice as much force to stretch two strands a given amount as to stretch one; therefore two strands are twice as stiff as one.

Let's ignore the effect of stand orientation; it is relatively insignificant compared to the effect of the number of strands. We are talking about 12-strand rope, and we are describing a splice whose bury consists of two parts, a length of full diameter rope, and a second length that tapers down to a point. It tapers because strands are cut away at regular intervals towards the tip until only one strand remains. 

Since we are describing a standard splice that is much longer than necessary to generate plenty of compression friction to lock cover and core together, the cover and core stretch together. At any given point, the core and the cover together bear the full tension of the rope. They _share_ the load in proportion to the number of strands in each. Though the load is the same everywhere, the stretch is not. The native rope will stretch the most, as it has only 12 strands; the untapered part of the splice will stretch half that much, as it has 24 strands; the amount of stretch in the tapered section will smoothly vary between these two extremes.

How does the splice carry the load? In the section of splice where the bury is full diameter, the core and cover share the load equally because each element has 12 strands. The tapered section is a different story. If we imagine traversing along the tapered section from the thick end to the tip, taking measurements as we go, we would discover several things. There would be progressively fewer strands in the core. There would be increasing stretch in the strands. Finally, the core's share of the total load would progressively diminish. In effect, the core is smoothly handing off its share of the load to the cover.

There is an interesting twist in this (which is why I love this stuff!). Even though the core's share of the load diminishes along the taper, the load on each individual core strand actually increases. This is obvious when you think about it. In the full-diameter section of the core, there are 24 strands in all, so each core strand carries 1/24 th of the load. Out at the tip of the taper, where only one core strand remains, that strand must carry 1/13 th of the load. The hardest working part of the core is at the very tip of the taper! Cool!

There is at least one more experiment in the works...


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## moray (Jul 29, 2008)

What about the situation where the entire load is hanging on the buried core? This is the worrisome worst of all possible cases and the one I really wanted to test. How does the splice compression force develop in such a case, and how safe is the standard splice in this situation? It is worth repeating (again) that it is next to impossible to dream up a scenario in which 100% of the load is on the bury, but let's press ahead and see what we come up with.

The first photo shows the test rig, in 3/8" Tenex, that I came up with. 






There are two entirely separate ropes. The one on the left is the core rope, with an eye at one end and an untapered, reinforced tip at the other. The cover rope, with the hook, has an open end that has been stabilized with tape. The opening is just slightly bigger than the core rope so that the opening itself doesn't actually squeeze the core. The next photo is a closeup of the opening with the core in place. 






For the first test I buried the core a distance of 13 inches (16 is standard) with no stitching of the throat and hooked it up to the car for pulling. I was quite sure it would survive the test, as all other tests of conventional eye splices had survived even when the buries were much shorter than regulation. But as soon as the rope became taut the splice simply slid apart! I later confirmed this result by burying the core over 17 inches and then pulling the thing apart by standing on one end while pulling on the other. At a rough guess, it took about 50 lbs to pull it apart. Surprised as I was, still this result was a terrific clue to what was really going on. The next experiment was a logical followup.

The next splice was a little shorter--16 inches--to make it even easier to pull apart, but this one was stitched. But I did not use the heavy yarn specified by rope manufacturers, and I used only half as many stitches. The "yarn" I used was very light sewing thread. I purposely picked the lightest stuff I could find, and I tested the tensile strength at about 2 lbs. The needle was passed through the throat area exactly 7 times and the two ends were tied together. Though I did make some effort to make the stitching and the final knot snug, with such weak material it was difficult to know if the thread was under any tension or not. The photo below shows the throat area and the fine blue stitching thread.






When I pulled this splice with the car, it survived! The force involved was probably 500 lbs or a bit more. A very minor strength reinforcement to the throat area has increased the strength of the splice many times over.

I intend to present an explanation for this behavior later. For now there is a clear warning and lesson in these results for anyone who uses spliced eyes in hollow braid rope: THE STITCHING IS VITAL! It is not just there to keep the splice from bouncing apart in your rope box, it also protects the splice in the unlikely event that 100% of the load ends up on the core leg of the eye.


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## moray (Jul 30, 2008)

How can a few stitches of very weak thread add hundreds of pounds to the strength of a splice, as demonstrated in the previous experiment? Before trying to explain this cool result, it is worth doing a test or two on the stitching. Since I wanted to minimize or eliminate the effect of splice compression while testing the thread, I decided to eliminate the splice and use two pieces of 1" nylon webbing for the test. The photo shows the setup.






Eight stitches were taken in a crosswise direction, more or less in a straight line. In this configuration there should be no tendency for the stitching to force the webbing pieces together when put under tension, and therefore no tendency to produce friction between the layers. I ran the test twice, measuring 15 lbs. break strength the first time and 16 lbs. the second.

Then I changed the direction of the stitching, running eight stitches lengthwise along the webbing. I really expected to get a different result this time, but again the stitching broke at 16 lbs.

From this I think we can conclude that the stitching in the splice experiment, by itself, was adding about 14 lbs. of strength to the splice. This is chump change compared to the observed change in splice strength, so we are left with a very interesting puzzle to explain.


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## moray (Jul 31, 2008)

The fact that a few tiny stitches supplying maybe 14 lbs. of strength to the splice actually increases the splice strength by many hundreds of lbs. suggests some sort of exponential relationship is involved. But there is a wrinkle in the analysis that requires a brief detour to describe some problems that are mathematcally solvable and some that are not. The relevance of this to our splice problem will become apparent. Anyone can read this--there won't be any equations.

First take the famous case of the 4-legged stool. The stool is standing on a level floor, and all four legs are contacting the surface. You place a weight at some arbitrary position on the stool. Knowing the exact position and size of the weight, can you calculate the load borne by each of the four legs? It would seem that you know everything you could possibly need to calculate this simple problem, but you don't. The problem cannot be solved. There is a whole range of possible solutions and there is no way to pick the true one out of this range. Now remove one leg. Same problem, 3-legged stool. This _is_ a simple problem with a single solution.

Here's another. You are a building contractor just finishing up a bedroom addition. As one of your last tasks you hang the door to the new room. After carefully mortising the hinge plates into the jamb, you fasten everything down, slide the door into place, and drop in the hinge pins. Perfect! You walk away believing that the two hinges in your perfectly hung door are equally supporting the weight of the door. Not so fast. It is a mathematical certainty that the two hinges are unequally loaded. There is a whole range of possible distributions of weight, from 100% on the upper hinge to 100% on the lower, and you have no way of calculating where the actual value lies.

Here's one from the arborist world, quite relevant to our case. You have tossed a rope over a limb in your back yard and given it an extra wrap. From previous experience you know that 1.5 wraps of this rope on this limb gives you precisely a 10:1 force ratio, meaning you can control the descent of a 100-lb. weight by applying 10 lbs. of force on the other leg. You discover one day that someone has installed your rope over the limb and hung a weight on each leg. The rope is not moving, and one of the weights is clearly marked as "10 Pounds". How big is the other weight? You cannot know. There are an infinite number of possible values for the mystery weight, ranging from 1 lb. to 100 lb., all of which are stable and will cause no motion of the rope. It is at the two ends of this range that we essentially kick away the 4th leg of the stool and the problem transforms into one with a unique solution.

The splice problem will prove to be just like this. It is just when the splice is on the verge of slipping that the range of possible solutions collapses down to just one.


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## moray (Aug 1, 2008)

Understanding the behavior of the hollow-braid splice when the entire load is on the core--the worst case scenario--is now within our grasp. The analysis is a mathematical argument, but we can do it with pictures!

The first picture shows the standard rope-around-a-post friction setup. The red arrow in all the photos indicates the direction of major pull and the direction of rope movement. A much smaller resisting force is pulling in the opposite direction at the other end of the rope. Friction around the post makes up the difference and opposes the major force. In all photos it is assumed the post is immobile.






In the next picture the post has been rotated 90 degrees. The situation is the same as before, only now the wraps around the post are in the form of a spiral. This was analyzed earlier in this thread, the only real difference being that the wraps are less effective the more severe the spiral.






Who says the post has to be made of wood? In the next photo the "post" is a piece of rope. The physics will be the same as before, but the coefficient of friction may be slightly different.






Let's wrap two ropes. Each rope has a major force pointing right and a minor force pointing left. The physics is still the same as the previous case, and we are still dealing with a case of rope around a post.


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## moray (Aug 1, 2008)

The next step is to change the way we supply the minor force. We have only assumed, so far, that the wrapping rope is moving slowly to the right and that the major force, minor force, and friction force are all in balance. As far as the physics goes, it makes no difference where the minor force comes from. Anything attached to the rope that applies a force opposed to the major force is a component of the minor force. How about some stitching? The next photo shows the setup.






The stitching, of course, will stop the motion. Does this screw up the analysis? Not really. Refer back to the discussion of solvable and unsolvable problems. The strength of the stitching will define one of the end points where a problem with an infinite number of solutions collapses down to one.

Imagine we have the wrapping rope slowly moving to the right, all the forces in balance, and no stitching. Let's say we measure the minor force at exactly 10 lbs. Now we magically and instantly apply stitching with a breaking strength of 10.001 lbs. The motion will stop. If the pulling force (major force) increases a small amount, the stitching will fail and motion will commence. We are at the end point. Whether we apply 10.001 lbs. of minor force by pulling with our hands, or we apply it with stitching, it will take the same amount of major force to start the wrapping rope moving. The formula that describes the rope and post friction problem tells us that the _ratio_ of the major force to the minor force is determined by the number of wraps. Just a few wraps could easily give us a ratio of 100 to 1 or more. If we know the ratio is 100 to 1, then we also know that stitching worth 20 lbs. will prevent any motion until the major force reaches 2000 lbs. This is what we are after.

In the next photo we have increased the number of wrapping ropes to 12. 






We started with a rope wrapped around a post, but we are now looking at a hollow-braid splice in which the entire load is on the core! They are two flavors of the same problem!

Again we supply the minor force with some stitching. As just discussed, when motion begins the ratio of the major and minor forces is some fixed number determined by the number of wraps. The force required to pull the cover off the core (i.e., pull the splice apart) is directly proportional to the strength of the stitching! 

This is a gratifyingly tidy result, for sure. It shows theoretically, where we earlier showed it experimentally, that a small amount of stitch strength can produce a very much larger splice strength.
So much for doing math with pictures. I'll wind this up later by calculating some actual numbers.


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## OLD CHIPMONK (Aug 2, 2008)

Don't stop ! This is getting interesting .


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## moray (Aug 8, 2008)

The buried part of the hollow-braid splice consists of a full-diameter section and a tapered section. We need to know a little more about the tapered section before we can attempt some final calculations.

When 3/8" Tenex is under tension, does the internal cavity completely close up, or is there always a little space in there? What we want to know is actually this: what is the the minimum number of strands where the normal squeeze of the rope comes into full play? That will be the minimum number we can use in our calculations.

I used my standard rig to run 3 tests. I used one, two, or three strands cut from another rope. In each case the strand(s) were buried 18 inches, and the throat was stitched with about 8 stitches of the 2-lb.-test thread described previously. The picture below shows the setup with 3 strands in place.






The rope is Tenex TEC (two-end carrier), meaning each of the 12 strands of the rope is actually a pair of strands. This is the so-called "sling grade" rope. In my tests I am using single strands, not strand pairs.

In the single-strand test, the strand easily broke the stitching and pulled out. I have done a lot of these by now, and I had no sense that the main rope was grabbing the single strand at all.

In the two-strand test, it was much harder to break the stitching and pull the strands out, and it seemed very clear that the main rope was applying a significant amount of compression to the core.

In the three-strand test, I applied 70 or 80 lbs. by hand, but I could not pull it apart. From this I conclude that Tenex TEC goes into full compression mode when the core consists of at least 3 strands.


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## moray (Aug 9, 2008)

If we have a uniform taper 8 inches long, only the last inch of the taper would drop below 3 strands thickness. We'll ignore that for the calculation of splice holding ability, and just to be conservative, we'll ignore the second-to-last inch as well. We now have 6 inches of uniform taper, the small end of which is 6 strands thick. We know from the previous experiments that this is plenty thick for the cover to grab it securely when the cover is under tension.

Is it necessary to make some sort of special calculation for the tapered section of the core, or does it behave just like the untapered part? Surprisingly, perhaps, it behaves exactly like the full-diameter core. The reason is the standard treatment of friction is only concerned with force and coefficient of friction, not area. For a given compression force, the size of the core, within broad limits, does not matter. Thus an inch of tapered section will behave just like an inch of untapered core. The analogous problem of the rope around a post is similar: neither the size of the rope nor the size of the post are relevant.

This is good news for the math weary: our problem is now as simple as it's going to get--figure out the numbers for a simple splice with a 14-inch bury of full-diameter rope.

For anyone who makes their own splices this is a valuable piece of info. The taper in the buried core does not weaken the splice, and it is not there just to give a smooth transition from fat splice down to normal rope. All of it, except for the last inch, is a fully functioning load bearing part of the splice.


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## moray (Aug 10, 2008)

A splice that is about to slip apart is analogous to a rope around a post that is about to slip. In the case of the post, the large force towards which the rope is about to move is resisted by a much smaller force at the other end of the rope. The tension grows at a constant rate, some percentage per degree of wrap, from the low- to the high-tension ends.

In the case of the splice, the buried core is analogous to the post, and the cover surrounding the core is analogous to the rope. The tension in the cover varies from a minimum at the throat of the splice to a maximum at the end of the bury. Between these two points it grows at a constant percentage rate per inch. (There is nothing special about using English units for this--any units will do.)

From the numbers derived earlier for squeeze force per inch and coefficient of friction of polyester on itself we can derive the number we want: for each inch of cover between the throat and the bury end the tension in the cover grows by 30%. Over the full length of our 14-inch bury, the tension will increase by a factor of 39. To say it another way, the cover tension, when the splice is about to pull apart, is 39 times as great at the bury end as at the throat.

At the rated strength of the rope, 6000 lbs., the cover tension at the throat will be 154 lbs. For a more ordinary load of 200 lbs., the required throat tension would be only 5 lbs.

To make sure the rope would break before the splice would fail, we need to somehow supply 154 lbs. of cover tension at throat.

Where do we get the 154 lbs. of tension? Normally, of course, the eye is symmetrically loaded, and half the total load is on the cover leg of the eye. This would be 3000 lbs., vastly more than the 154 we need. The only plausible scenario in which the eye itself can't supply the necessary tension is one in which the core leg of the eye has somehow been snagged or impaled on a spike of some kind. In that case, the stitching at the throat supplies the force. If the stitching is strong enough to resist a 154-lb. pull then the splice will hold.

Samson, the manufacturer of Tenex, suggests using a yarn the size of one in the rope for the stitching, and they call for installing at least 12 stitches. One yarn from 3/8" Tenex has a tensile strength over 500 lbs since the rope is comprised of 12 yarns and has a tensile strength of 6000 lbs. Twelve stitches of 500-lb.-test yarn is going to supply vastly more than 154 lbs. of strength at the throat! Personally, I use a yarn of about 70 lbs. test for stitching, but I do take 12 stitches or more. This should give me a very comfortable margin beyond the 154 lbs. needed. Remember, we are talking about the 6000-lb. load needed to break the rope. AND we are talking about the somebody's-bad-dream case of the splice core impaled on a spike. You are more likely to get struck by lightening and bit by a shark in the same day than to have that happen to your rope. Nevertheless, decent stitching has you covered.


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## moray (Aug 11, 2008)

*Conclusion*

Like many others, I was quite skeptical about the reliability of splices when I first started using them. None of my intuitions about mechanical things seemed to apply, so to hang from a splice far off the ground was pretty much a leap of faith.

No longer. A lot of hands-on experience and numerous (admittedly primitive) experiments have shown me how over-engineered the standard hollow-braid splice really is. If built according to manufacturer's specs, which is easy to do, the splice has a large margin of safety even under the most extreme circumstances. (This statement needs an asterisk. I don't know if a shock-loaded splice would perform the same as one under a steady pull. Maybe I'll try to test this...) 

If the information I developed in the course of this project could have been found on a Web page somewhere, I undoubtedly would have done something else instead. I don't mind reinventing the wheel now and then, but this was a lot of work! Anyway, I found what I was looking for and a lot of things I wasn't looking for and had a lot of fun in the bargain. If some splice-averse person out there reads all this and decides to give the hollow-braid splice a try, that will be a bonus.


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## Adkpk (Aug 12, 2008)

I need an eye to eye prusik loop for a foot locking loop I spliced. My lanyard splicing is still hard as a rock. What cord do you think I should use? I am using tenex for my short lanyard and beeline for my longer one. I would like to try one of those splices you used on that amsteel blue.


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## moray (Aug 13, 2008)

Adrpk, I have been using 5/16 Tenex for that purpose--it is fairly small, easy to splice, and seems to work just fine. I actualy use it in the form of a Whoopie sling so I can adjust the total length to suit. There is no prusik in my setup--I use an ascender instead. I tried it with a prusik for awhile but found it too hard to advance the knot.


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## Adkpk (Aug 13, 2008)

I will also need a whoopi sling at some point. I will combine the length of the two when I order it. Thanks and I be back with questions probably when I get the rope.


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