A cable weighs its singles at the twists the finished thread keeps

A two-fold yarn weighs what its notation says. Two thirds is where a fold weighs its singles found the reason: folding changes two lengths, the fold’s helix adds single to every metre and the singles’ untwisting takes fibre out of every metre of single, and the two cancel at a folding ratio of exactly two thirds in the small-angle limit. A fold folded where the trade folds it is within a per cent and a half of the sum of its singles, inside the spread of the count test itself.

A cabled yarn is a fold of folds. Two or three folds are twisted together again, the other way from the folds and so the same way as the singles, and the cabling rule sets both twists from two integers: folds at one over the root of the number of singles in each, the cable at one over the root of the number of singles in all. It is the same arithmetic one level up, so the obvious guess is that the same cancellation happens one level up and a cable weighs its notation too.

It does not, and the reason is clean enough to state in a sentence. Cabling untwists every fold and hands that twist straight back to the singles inside it, so the fold and the singles beneath the cable behave as a single fold folded at the difference of the two twists — and on top of that the cable has a helix of its own, which nothing below it can pay for. Twisted where the rule is quoted, a two-by-two cotton cable is 3.5 per cent heavier than its four singles.

Cabling gives the singles back what it takes from the fold #

Folding is untwisting because of how a folding frame works: it winds each single round its neighbours without letting it rotate freely, so a single held that way turns about its own axis once for every turn of the fold. A Z single spun at $T_s$ and folded S at $T_f$ keeps $T_s - T_f$ of its own.

A cabling frame does the same thing to folds. Each fold is wound round the others, so a fold set S at $T_f$ and cabled Z at $T_c$ keeps $T_f - T_c$. That is the first half of the arithmetic, and it is the obvious half.

The second half follows from the first. What a single keeps is its spun twist less whatever twist the fold round it still holds. Once the cable has taken $T_c$ out of the fold, the fold holds less, and the single gets that much back: $T_s - (T_f - T_c)$. The trade says this in words — cabling restores twist to the singles — and it is the reason cabled threads alternate their hands at every level.

Put the two halves together and the singles and the fold have no way of knowing whether they were folded at $T_f$ and cabled at $T_c$ or simply folded at $T_f - T_c$ in the first place. Their geometry is identical. Beneath its own helix, a cable is a fold folded at the difference of its two twists. The calculation that priced a two-fold applies without a single change, at a folding ratio that is no longer the one set on the folding frame.

For a 20-tex cotton spun at 850 turns a metre — a twist factor of 3,800, an ordinary weaving twist — and taken through the rule as quoted, the folding frame puts in 601 turns, the cabling frame takes out 425, and the folds are left with 176: a fifth of the singles’ spun twist, where a plain two-fold would keep seven tenths. The singles get back to 674 of their 850.

The cable’s helix has nothing to cancel it #

There are now three lengths, and the figure below draws them separately.

The fold’s helix is the length a plain fold adds by carrying each single round its axis. At the folding frame it is 4.86 per cent. Cabling unwinds the fold to a fifth of the singles’ twist, and the helix flattens to 0.43 per cent.

The singles’ saving is the length a single gains by losing twist: fewer turns, straighter fibres, more single for the fibre it holds. Folded at 0.707 a single saves 4.19 per cent. Cabled, it gets most of its twist back and saves only 1.67.

Those two fell together, by 4.43 and 2.52 points, and nearly cancelled each other’s fall — the fold went from 0.46 per cent heavy to 1.26 per cent light. That is the two-fold arithmetic doing exactly what it did before, at a smaller ratio.

The cable’s helix is new. Each fold’s centre line runs round the cable’s axis at a radius set by how the folds lie, half a fold’s diameter off for two side by side, and there is more fold than cable by the secant of that helix’s angle. At a cable twist of half the singles’, the helix stands at 17.5 degrees and adds 4.86 per cent — exactly what the fold’s own helix added before it was cabled. That is no accident of the numbers: a fold is $\sqrt2$ singles across and the rule cables it at $1/\sqrt2$ of the folding twist, so the radius times the twist — which is what sets a helix’s angle — is the same at both levels.

Nothing in the cable opposes that length. In a two-fold the helix that lengthens the singles comes with an untwisting that shortens their fibres; in a cable the untwisting the cabling does falls on the fold, whose own helix it flattens, and the twist it removes goes back into the singles and makes them heavier, not lighter. The product of the three is 3.54 per cent heavy: 82.83 tex against a notation of 80.

The most a fold can ever pay #

The heavy cable is not a property of the rule’s particular numbers. It can be bounded, and the bound is the useful thing to know.

Because the cable’s helix depends only on the cable twist and the rest depends only on the twist the folds keep, the two are separate factors. A cable weighs its sum only if the fold beneath it is light enough to cover the helix on top — and a fold’s lightness has a ceiling. Written to second order, with $u$ the fold’s kept ratio and $y$ the cable’s, a cable’s excess in units of the spun single’s $\tan^2\alpha$ is

$$\tfrac14\left(u^2 - 2u\right) + 2\left(\tfrac{o_n}{d}\right)^2 u^2 + 2\left(\tfrac{o_m}{D}\right)^2 \left(\tfrac{D}{d}\right)^2 y^2,$$

where $o_n$ is how far a single’s centre sits from the fold’s axis and $o_m$ how far a fold’s centre sits from the cable’s. The first two terms are the fold, and for two singles side by side they are smallest at $u = \tfrac13$, where the fold is lighter than its sum by exactly $\tfrac1{12}\tan^2\alpha$. The third term is the cable and is positive for every cable twist there is.

For a two-by-two the third term is $y^2$ exactly, so the largest cable twist any fold can cover is $\sqrt{1/12}$ of the singles’ twist, which is $1/(2\sqrt3)$, 0.289. The rule as quoted asks for a half. That is $\sqrt3$ times the most that any fold twist can pay for, and the ratio is exact in the small-angle limit.

At a weaving twist factor the bound tightens a little, because the singles’ saving levels off sooner at real angles: 0.273 for two-by-two, with the folds keeping 0.327 of the singles’ twist; 0.236 for three folds of two; 0.200 for two folds of three; 0.173 for three of three. The rule as quoted asks 0.500, 0.408, 0.408 and 0.333 — between 1.7 and 2.1 times the bound in every case.

The map below puts every pair of twists a two-by-two can hold on one page. Across is what the folds keep, up is the cable, and the cable weighs no more than its sum only inside the lobe.

The lobe’s top is low and far to the left. A cable that weighs its notation needs its folds to keep about a third of the singles’ twist and its cable twist under three tenths. Set on the frames, that is a folding twist of 0.600 and a cable twist of 0.273 — the folding frame at the bottom of the trade’s own two-fold bracket.

Three ways to read one rule #

So far the rule has been read the way a process sheet would write it: the twist the folding frame puts in and the twist the cabling frame puts in. That is not the only reading, and the rule’s own derivation suggests it is not the right one.

The folding rule is a surface angle. It says a fold should present the angle its singles present, so that a fold $\sqrt n$ singles across carries $1/\sqrt n$ of their twist. Applied twice, it says the same of a cable. But a surface angle is a property of a yarn as it lies, and inside a cable every level’s twist has been changed by the level above it. A rule about surfaces is a rule about the finished thread.

Read as quoted, it leaves the finished thread’s three surfaces at three different angles. The singles end at 19.5 degrees, the folds at 7.4, and the cable at 24.0 — exactly the singles’ spun angle, which is what the rule was written to match, matched only at the top level, the one nothing above it has untwisted.

Read as the twists the folds keep, the rule would hold them at $1/\sqrt2$ inside the cable. The cabling frame then takes out a half on top of that, so the folding frame would have to put in 1.207 times the singles’ twist. That is impossible: past the singles’ own twist a folding frame reverses them. It is equally impossible for three folds of two, at 1.115. Only the cables of three-folds can be made this way, and both are heavier still: at a twist factor of 3,800, 5.3 per cent for two folds of three and 4.8 for three of three.

Read in the finished thread — every level’s surface at one angle, after every level has been changed by the next — the conditions are three equal tangents. A single keeps $T_s - T_k$, a fold keeps $T_k$ and is $\sqrt n$ singles across, the cable keeps $T_c$ and is $\sqrt{nm}$ across, so

$$T_s - T_k = \sqrt{n}\,T_k = \sqrt{nm}\,T_c .$$

For two singles the fold keeps $\sqrt2 - 1$ of the spun twist and the singles $2 - \sqrt2$; the cable takes $1 - 1/\sqrt2$, 0.293; and the folding frame is set at their sum. The sum is exactly $1/\sqrt2$. It is a general result for a square cable, n folds of n singles: the frame is set at $1/\sqrt n$, precisely where the rule as quoted sets it and precisely where a plain fold would be. The two readings agree about the folding frame and disagree only about the cable — half the singles’ twist as quoted, 0.293 in the finished thread.

Every level then sits at 14.6 degrees, and the count is 0.32 per cent over the notation.

The cable comes back, and lands where the fold did #

That 0.293 sits just above the lobe’s top at 0.273 — the same relation the two-fold had, where the surface rule’s 0.707 sits just above the weight rule’s 0.658. The finished thread’s reading of the rule lands near the edge of the lobe in the same way the plain fold’s rule landed near its crossing.

Across twist factors 3,000 to 4,500 the rule as quoted leaves the two-by-two 2.2 to 5.0 per cent heavy, three folds of two 1.8 to 4.0, two folds of three 2.4 to 5.3, three of three 1.9 to 4.4. Read in the finished thread, the same four come to +0.16 to +0.54, +0.05 to +0.28, +0.56 to +1.42 and +0.43 to +1.12. A plain two-fold at its own surface rule is +0.28 to +0.66. The cables of two-folds weigh their notation as well as a two-fold does; the cables of three-folds, a little less well, because a three-fold’s own saving is smaller and its folds are fatter, which widens the cable’s helix.

That is the answer to the question the two-fold left. A cable’s three levels cancel as a fold’s two do — but only when its twists are the ones the finished thread keeps. Twisted to the rule as it is usually written down, the cable is heavy the way a same-way fold is heavy, and for a related reason: as far as the singles are concerned the cable twist runs their way, and folded the same way nothing cancels. It is less heavy than a true same-way fold, which is 8.0 per cent over at half the singles’ twist, because the fold is still in between and still pays part of the bill.

The count, the fibre and the twist still drop out #

Everything that made the two-fold’s result clean survives one level up. Each length is a function of an angle, each angle is fixed by a twist factor and a ratio, and the count enters only through the twist factor: four cables of one construction spun from 10, 20, 40 and 80 tex at one twist factor differ from their notations by the same fraction to the twelfth figure. The small-angle bound, $1/(2\sqrt3)$ for a two-by-two, has no count, no twist and no fibre in it. The finished-thread twists — $1/\sqrt2$, $\sqrt2 - 1$, $1 - 1/\sqrt2$ — have nothing in them but the arrangement.

What the twist factor does decide is the size of the error. At 3,000 a cable twisted as quoted is 2.2 per cent heavy; at 4,500, 5.0. A sewing thread’s singles are harder than a weaving yarn’s, so the error is largest in exactly the yarns that are cabled most.

What a heavy cable costs a reel #

A sewing thread is a different animal, specified by a ticket that is closer to a statement about diameter than about mass, and sold on a reel that is labelled by length. Neither label is the count. But the mill that makes it buys fibre by the kilogram and costs a reel from the notation: four singles of 20 tex make a thread of 80, so a 500-metre reel holds 40 grams.

At the rule as quoted it holds 41.4. Over a year’s production that is three and a half per cent more cotton than the costing allows, a difference that sits between a lea test’s own spread and the margin on a commodity thread. It is invisible on any one reel and systematic over all of them, which is the kind of error a mill finds in its stock-taking rather than in its laboratory — and attributes to waste, because waste is where unexplained cotton usually goes.

In the other direction, a reel wound by mass to a nominal length holds 3.4 per cent less thread than its label says. Whether any mill winds that way is not something this calculation can say; the arithmetic only says which way the error runs if one does.

What a twist tester should see #

The three readings make different predictions about something a mill can measure with the instrument it already has. A twist tester untwists a yarn and records how much it lengthens as it does — the contraction the twist had taken up. For a cable it can do that twice: untwist the cable to its folds and note the lengthening, then untwist one fold to its singles.

Twisted as quoted, a two-by-two should lengthen 4.86 per cent when the cable comes out and only 0.43 per cent when a fold does — eleven times as much at the first step, because the folds inside have kept so little of their twist. Twisted to the finished thread’s reading, the two steps should give 1.69 per cent each: equal, because every level’s surface is at one angle and in a square cable the centre lines are then at one angle too. The other constructions run from five to thirteen times as quoted, and within about an eighth of equal in the finished thread.

This is a better test than weighing. A count error of three per cent is near the edge of what one lea can resolve; a ratio of eleven to one against one to one is not near any edge. A thread maker’s own process sheet, if it records twists as set on the frames, could be checked against it in an afternoon.

How the three lengths were computed #

The fold and the singles are the two-fold’s model unchanged: each single the ideal helical yarn of twist is one angle, fibres spread evenly over a round section at the volume arithmetic’s diameter for a 0.6 packing, its retraction the closed-form mean secant over the section; the fold’s centre lines touching circles round its axis, half a diameter off for two singles and $d/\sqrt3$ for three. The fold’s kept twist is the set twist less the cable’s, the singles’ is the spun twist less the fold’s kept twist, and the fold’s part of the count is computed at that kept ratio.

The cable adds one factor, the secant of the folds’ own helix. A fold is taken as a round object $\sqrt n$ singles across — the diameter the surface rule itself uses — and the folds as touching circles round the cable’s axis by the same arrangement.

The calculation is required to give each single back exactly what it takes from the fold; to give the same yarn whether it is read by its set or its kept fold twist; with the cable’s helix divided out, to equal the two-fold calculation at the kept ratio to the fourteenth figure; to return the plain fold as the cable twist vanishes; to drop the count to the twelfth figure; and, as the twist factor falls to 150, to reach the small-angle expression above within two parts in a thousand for every construction. It must find the finished thread’s reading leaves all three surface tangents equal to twelve figures and sets a square cable’s folding frame at exactly $1/\sqrt n$, must find the bound at a weaving twist with the lightest cable on the sum to nine figures, and must refuse a cable of one fold or five, folds of five singles, a cable twisted at or past its folds, and any reading that sets a folding frame at or past its singles.

Where three ideal helices stop #

The fold is not round. Two singles pressed together are closer to a figure of eight than a circle, and a cable presses its folds against each other harder than a fold presses its singles. The cable’s helix is computed for round folds $\sqrt2$ singles across; taken instead as the smallest circle round two singles, the length the helix adds about doubles, and the rule as quoted moves from 3.5 to 8.1 per cent heavy. A flattened fold sits nearer the axis and moves the other way. The true helix is somewhere between, and this calculation does not place it.

The three helices are taken in series. A fibre in a cable follows a helix round its single, which follows a helix round its fold, which follows a helix round the cable. The product of three secants is the first approximation to that path and the correction is of higher order; at the angles here it is small, and neither its size nor its sign is computed.

Migration, and the tension the twist went in under, are omitted as they were for the two-fold, and for the same reasons: a fibre that wanders between radii adds radial length the ideal helix lacks, and a thread cabled under tension relaxes on the reel.

And the strength is not computed at all. The lobe where a cable weighs its notation is a lobe of lightly twisted folds, and a fold’s strength depends on the twist its singles keep, which is a separate account with its own price for every turn taken out. A cable that weighs right is not thereby a cable worth making; it is only one whose ticket arithmetic is honest.

So of the four omissions, one moves the cable heavier, one lighter, and two are of higher order. The finished thread’s reading puts a two-by-two within half a per cent on round folds; on real folds it could be a little either side, and the rule as quoted cannot be brought inside the lobe by any of them.

Who measured what #

Cabled sewing threads, and the habit of reversing the hand at every level, are much older than any arithmetic for them; the reason is given in the trade as torque and liveliness, and what a balanced yarn is balanced about separates the several things that word means. The untwisting identity — a component wound round others loses one turn of its own per turn of the winding — is classical, and Treloar gave the geometry of multi-ply and cabled yarns in the 1950s, including the compound helix this calculation treats as a product. The retraction of an ideal helical yarn and its closed form are in Hearle, Grosberg and Backer’s collected structural mechanics of 1969.

What is added here is the observation that cabling makes the fold and singles beneath it identical to a fold at the difference of two twists; the cable’s helix as an uncancelled length, and the bound on what any fold can pay for it, $1/(2\sqrt3)$ for a two-by-two in the limit; the observation that the cabling rule read as kept twists cannot be made for cables of two-folds; the finished-thread reading, with its folding frame at exactly $1/\sqrt n$ for a square cable and its cable twist at $1 - 1/\sqrt2$ for a two-by-two; and the twist-tester ratio that tells the readings apart.

Still open: what a real cable’s process sheet says #

The whole question turns on one fact this calculation cannot supply: where the trade actually sets its cabling frames. If thread makers twist to the rule as quoted, their cables are three or four per cent heavy and their folds keep a fifth of the singles’ twist; if they twist to the finished thread, the cable twist is about 0.29 of the singles’ and the yarn weighs its notation. A process sheet for a cabled cotton sewing thread would settle it, and so would the twist tester’s two contractions on a commercial reel.

The second open question is the fold’s shape. The cable’s helix is the largest single length in the calculation and it is computed for round folds, which a cabled thread under tension does not have. A cross-section of a two-by-two cut, set in resin and measured would give the folds’ real centre distance, and with it where the fold sits between its two diameters — the one its count gives and the one that encloses its singles. That distance is still a bracket, as it has been for every folded yarn so far.

And the cancellation found here suggests a third. A cable at its finished-thread twists has every surface at one angle and weighs its notation; the two conditions seemed to arrive together for a two-fold and again for a cable, and neither was put in by hand. Whether a cable of cables — the three-level cords of rope and tyre yarn — repeats it a third time is the same arithmetic with one more factor. It does not: the singles pay once and every level adds a helix finds that equal angles fix the singles’ saving at the first level and add one equal helix per level above it, so the cable’s agreement is particular to two levels, and the two-fold’s was reached under the other reading of the rule.