A blade leaves loose fibre in every tuft

Hair, nap and pile are one construction put the three surfaces in one sequence and found one property running the wrong way along it: uniformity. A yarn’s hairs are exponential in length, a nap is the same population multiplied, and a cut pile is one length to the accuracy of the blade. So a cut pile cannot pill and cannot feather a print, it concluded, because none of its fibres reaches past the others.

That is true of the pile’s tips, and the conclusion needs one more thing to be true of its fibres. A tuft is not a bundle of fibres each exactly as long as the pile; it is a short length of spun yarn, cut from a longer one, and the fibres in that yarn were a few centimetres long before the blade arrived. The blade cuts the yarn to one length; it cuts the fibres into pieces of many lengths, and some of those pieces are held by nothing the draft decides.

What the blade actually cuts #

A cut-pile tuft is a length of pile yarn bent into a U: up one leg to the blade, down under the binding pick, up the other leg to the blade. Call the whole length $L = 2H + b$, with $H$ the pile height and $b$ the stretch under the binding. How a tuft is held is about that U: the capstan wrap round the pick that stops the yarn pulling out.

The capstan holds the yarn, and a yarn is held only where its fibres are. A staple yarn is $n$ fibres in its section, each of length $\ell$, overlapping end to end along it; wherever a fibre ends another is already running. Before cutting, every fibre is gripped along its whole length by the twist of the yarn round it. After cutting, the yarn in one tuft is a piece $L$ long, and the fibres in it are pieces of fibres, cut wherever the blade happened to cross them.

A piece is anchored if it passes under the binding. Then the capstan grips it through the yarn, as it grips the tuft. A piece that lies wholly in one leg is not anchored: it runs from a fibre end somewhere in the leg up to the blade, and what holds it is the yarn’s own twist and friction and nothing else — the grip what grips the end of a fibre computed for a fibre end inside a yarn under load.

The count is exact and short #

Suppose the fibre ends are scattered uniformly along the yarn, which is what drafting and spinning are designed to achieve. Then two facts about a yarn of staple fibre are exact:

• $n$ fibres cross any point, since that is what $n$ fibres in the section means;
• $n(x + \ell)/\ell$ fibres touch any stretch of length $x$, the ones crossing its start plus the ones whose left ends fall inside it.

Apply them to half a tuft, from a tip to the middle of the base. The fibres touching that half number $n(L/2 + \ell)/\ell$; the fibres crossing the middle number $n$; the difference, $nL/2\ell$, is the pieces that touch the half and do not reach the middle. Both halves together:

$$\text{anchored} = n, \qquad \text{loose} = \frac{nL}{\ell}, \qquad \text{loose share by count} = \frac{L}{L + \ell}.$$

A loose piece runs from a fibre end to the tip, and the end is anywhere in the half with equal likelihood, so while the staple is longer than half a tuft a loose piece’s length is uniform from nothing up to $L/2$, and the loose share of the tuft’s fibre by mass is $L/4\ell$. When the staple is shorter than half a tuft, some fibres lie whole inside a leg and the mass share is $1 - \ell/L$.

A simulation — fibres laid at random along a long pile yarn, the yarn cut into tufts, every piece classified by whether it covers a tuft’s middle — agrees with the three formulas at every construction tried, to within its own sampling, and never finds a loose piece longer than half a tuft.

How much is loose, construction by construction #

The loose share is a race between the tuft’s yarn length and the staple length. A short pile on a long staple is nearly all anchored: a velveteen a little over a millimetre tall on a 25-millimetre cotton loses a ninth of its pieces and three per cent of its fibre. A tall pile on a short staple is mostly loose: a ten-millimetre cotton pile has nearly half its pieces and a fifth of its fibre held by nothing but friction.

The constructions line up as the formula says. Cotton velvet at two and a half millimetres on a 28-millimetre staple is 17.6 per cent loose by count and 5.4 by mass. A wool carpet at twelve millimetres on a 90-millimetre worsted is 22.4 and 7.2. A nylon staple carpet on a much longer staple, 190 millimetres, is 13.6 and 3.9. The two filament piles — a silk velvet, a nylon filament carpet — are at nought, because a filament has no ends: every piece in the tuft runs from one tip to the other through the base.

A pile cut from floats is a tuft too #

The same count applies to the piles that are not a third thread system at all. Corduroy is a cut float: a weft float lying across several ends is cut along its middle, and each half stands up as a tuft whose yarn length is half the float, anchored where the float was bound. Velveteen is the same construction with the floats scattered rather than lined up in wales.

For those the tuft’s yarn length is set by the float, not by a pile height chosen at the loom, and a float is short: a few ends at a close sett is a millimetre or two of yarn. On a 25-millimetre cotton staple that is a loose share of four to seven per cent by count and one or two by mass — the reason a cotton corduroy barely sheds while a cotton cut-pile rug of the same fibre sheds for a season. The fibre, the twist and the blade are the same; the length of yarn the blade leaves between two binding points is not.

Where the loose fibre goes #

Every one of these numbers is the reservoir a new cut pile draws on, and the trade knows the reservoir well without having counted it: a new cut-pile carpet sheds, sometimes for months, and a filament carpet does not. The shed fibre is exactly what the count predicts — short pieces, none longer than the pile, mostly from the tips’ end of the tuft.

A tuft is set so it cannot untwist supplies the mechanism that turns a loose piece into a shed one. At a free end the pressure the twist puts on its fibres falls to nothing, over a length that essay computed; setting the tuft slows that. A loose piece is gripped only by that pressure, so the loose pieces nearest the tip — in the zone where the twist has run out — are held by almost nothing at all and come out at the first vacuuming, while those lower down stay until wear opens the tuft.

And shed fibre does not simply leave. A pill is anchored, not made: a pill is a tangle of loose fibre held to the surface by a few fibres that are still attached, and it persists as long as those fibres hold. A cut pile of staple has both ingredients — a reservoir of loose pieces that work up to the tips, and a forest of anchored tips to tangle them round. So a cut pile of staple can pill; what it cannot do is pill from its tips’ own length, which is the half of the earlier claim that stands.

The tail the blade does not remove #

The earlier account’s point was that a blade collapses a length distribution’s variance, and that every property living in the tail goes with it. The pieces the blade leaves have a distribution of their own, and it is worth setting beside the one it replaced.

The loose pieces are flat from nothing to half a tuft’s yarn — 3.5 millimetres here — and stop there, because no piece can be longer than the leg it lies in plus half the base. A hair population with the same mean length, exponential as a yarn’s surface is, puts an eighth of its members beyond that length and a few per cent beyond twice it.

So the earlier claim survives in its sharpest form and fails in its broadest. A cut pile has no member reaching past the others from the surface: every tip is at the blade, and every loose piece is shorter than the tuft it sits in. What it has that the claim missed is a population of short pieces inside the surface, free to migrate out. The difference matters for exactly the properties the earlier essay listed. A print is as sharp as the hairs are long, and a cut pile’s print edge is set by its tips, so it is sharp; a pill is anchored by a tail, and a cut pile’s pills are anchored by its tips and fed from its reservoir, so they form.

What decides the reservoir, and what does not #

The two formulas have only two lengths in them, and the design choices that move them are few and plain.

• A longer staple empties the reservoir in proportion: the loose share by mass is $L/4\ell$, so doubling the staple halves it. Worsted-spun carpet yarn, from long combed wool, is loose by a smaller fraction than woollen-spun yarn of the same pile height.
• A shorter pile empties it in the same proportion, which is one reason velvet, at two or three millimetres, is far less of a shedding problem than a shag carpet at twenty-five.
• Filament empties it completely, and nothing else does.
• The twist does not appear, and the binding appears only through the length of yarn it takes. A harder twist holds the loose pieces more firmly and a W fastening holds the anchored ones more firmly, but neither changes how many pieces are loose — except that a W’s yarn runs under three picks rather than one, so its base is longer and its tuft’s yarn with it. On a three-millimetre pile of 25-millimetre cotton, going from a V to a W with a base twice as long moves the loose share from 22 to 24 per cent of the pieces: a W holds its anchored pieces three capstan wraps harder and leaves a few more pieces unanchored in exchange.

That trade is the whole of what the binding can do to the reservoir. The count is a property of where the fibre ends fall, which is set when the yarn is spun, and of how much yarn the blade leaves between two tips, which is set when the pile height is chosen. Everything after that — the fastening, the twist, the set, the finish — decides how firmly the two populations are held, not how large they are.

That last point is the useful one. Shedding is decided at the spinning frame, by the staple, and at the loom, by the pile height, and nowhere later. Finishing — shearing the tips, brushing, a latex back that glues the base — removes or locks some of the reservoir, and it does so by treating the symptom: none of it changes how many fibre ends a blade left in a leg.

How the count was taken #

The model is a staple yarn with fibre ends uniformly scattered: $n$ fibres in the section, each $\ell$ long, cut into tufts of yarn length $L = 2H + b$. A piece is anchored if it covers the middle of the base, which is where the binding pick grips it; every other piece is loose. The counts follow from the two exact facts about uniformly scattered ends, and the lengths from where a fibre end can fall in a half-tuft.

The simulation is independent of the formulas. Fibres are placed at uniformly random positions along a yarn four hundred to six hundred tufts long, as many as keep $n$ in the section on average; the yarn is cut into tufts and every piece classified. Five constructions — including one whose staple is shorter than half a tuft — are required to agree with the formulas by count and by mass within two per cent, to give one anchored piece per fibre in the section, never to produce a loose piece longer than half a tuft, and to give loose pieces a mean length of a quarter of the tuft wherever the staple exceeds half of it. A first version laid fibres end to end on fixed tracks, and when the staple was a whole number of tufts long every end fell in the same place in every tuft; the mean-length check caught it.

The constructions’ dimensions are illustrative. Pile heights and staples vary widely between makers; the chart is there to show where ordinary constructions fall on the formula, not to rate any product.

What the count leaves out #

Fibre ends are not perfectly uniform. Drafting leaves a yarn’s ends slightly clustered, and a yarn’s irregularity — a cloth is a population, not a thread — means some tufts are fuller of ends than others; the formulas are means over a pile, not a statement about any one tuft.

A staple is not one length. Real staple has a distribution, and the short fibres in it contribute more loose pieces than their share of the mass, since the loose count goes as one over the length. The mean over a real staple distribution is the average of $L/\ell$, which is larger than $L$ over the average $\ell$ — so every loose share above is a slight underestimate.

And friction is not in the count at all. Whether a loose piece is shed, stays or migrates to the surface depends on the grip the twist gives it along its length, which the count does not model; the count says how much is at risk, not how fast it goes.

Who found which part #

Carpet shedding is known to every maker and every buyer, and is conventionally put down to “short fibres” and “loose fibres from the cutting”. What this essay adds is the count: exact formulas for how many pieces and how much fibre a blade leaves loose, from two lengths and nothing else.

The correction is to this account’s own earlier claim, which read the pile’s one tip length as one fibre length. The blade makes the first; spinning makes the second impossible for any staple yarn, and the two are different by a share of a tuft that a velvet keeps small and a shag carpet cannot.

Still open: how fast the reservoir drains #

The count says how much fibre in a new cut pile is loose. It does not say how quickly it leaves, and the two are different questions with different inputs.

A loose piece leaves when the grip on it fails, and what grips the end of a fibre gives that grip as a length: a fibre end gripped over less than a critical length slides out, over more it breaks. A loose piece is gripped only over its own length, and only where the twist is still holding — below the untwisted zone at the tip. So the pieces shorter than the critical length plus the untwisted zone are the ones free to go, and the reservoir should drain in two stages: those at once, the rest as wear opens the tuft and lengthens the untwisted zone. The calculation that would put numbers on it is the critical length for a carpet yarn’s twist and fibre, and the untwisting length from the setting essay — both already computed for other constructions, and neither yet applied to a cut tuft.