A slow fibre starts its ring deep

Three essays have now followed indigo into a yarn, and all three made the same simplification. A dip dyes a depth, not a share let the liquor diffuse through the pores between fibres, slowed by the fibres taking dye out of it. A fibre that fills dyes deeper with every dip let those fibres run out of room. Indigo is too little to close the pores it dyes through let the pigment narrow the pores. In every one, a fibre reached by liquor took its share at that instant.

A cotton fibre is not a surface. It is a porous solid about a dozen micrometres across, and dye has to diffuse into it. If that takes as long as the dip, the assumption fails, and the plausible consequence was stated at the end of the last essay: a fibre that takes up slowly lets the liquor run ahead of the fixed dye, which should deepen the ring.

The calculation says that is half of it, and the wrong half to expect first. A little slowness makes the ring shallower. Past a clock of about the dip’s own length it runs deeper. And no single measurement of depth can say which side of that a yarn is on, while two sections of it can.

Two scales in one yarn #

The yarn now holds two things that can differ at one place: the concentration of the liquor in the pores, $c$, and the dye held by the fibres there, $s$. The liquor diffuses as before, through a liquor fraction $\varepsilon$ with Archie’s tortuosity, and loses what the fibres take:

$$\varepsilon\,\frac{\partial c}{\partial t} = \nabla\cdot\left(D_0\varepsilon^{m}\nabla c\right) - (1-\varepsilon)\,\frac{\partial s}{\partial t}.$$

The fibres move towards what they would hold in equilibrium with the liquor around them, at a rate set by a single time:

$$\frac{\partial s}{\partial t} = \frac{K(1-q)\,c - s}{\tau}.$$

This is Glueckauf’s linear driving force, the standard stand-in for diffusion into a small particle, and for a cylinder of radius $a$ with an internal diffusivity $D_f$ it gives $\tau$ of about $a^2/15D_f$. Everything else is the earlier model: the same 50 tex yarn, an affinity $K$ of ten, a twenty-second dip, and airing that fixes what the fibres hold when the yarn leaves the bath. The surface sits in the bath itself, so it has no liquor to wait for and fills on its own clock alone.

With $\tau$ at nought the model has to be the old one, and that is a real test, since the calculation solves the exchange in each shell exactly rather than taking the instant case as a separate branch. It returns 13.99 per cent of the radius against Crank’s cylinder’s 13.99, and it follows the saturating model’s eight dips to within a hundredth of the radius at every dip.

How slow a fibre might be #

Nothing in this essay measures $\tau$ for indigo in cotton, and the honest position is that it is unknown here. An estimate of its size is still worth making, since it decides which part of the following matters. A fibre of six micrometres’ radius and an internal diffusivity of the order quoted for small dyes in swollen cotton, $10^{-14}$ to $10^{-13}$ square metres a second, gives a $\tau$ between about 24 and 240 seconds. Those are not measurements of leuco-indigo, a small molecule of low affinity that may well move faster than that.

So a dip of twenty seconds is not safely long against the fibre’s clock, and it is not safely short. The calculation is run with $\tau$ swept across the whole of that range and well beyond it, and the question is what each value would look like.

The ring gets shallower before it gets deeper #

The first thing the figure refutes is the sign. At a clock of two seconds the ring is 13.0 per cent of the radius, at five it is 11.9, at eight it is 11.5. Up to about twenty-six seconds a slow fibre gives a shallower ring than an instant one, and the most it loses is nearly a fifth.

The reason lies in what a ring’s depth means. The depth is where the fixed dye falls to half its value at the surface, as it has been in every essay here, so it is a statement about the shape of the profile and not about how far anything has travelled. A clock of a few seconds does very little to the surface, since a fibre in the bath itself has twenty seconds to fill and manages 98 per cent of its share at a clock of five. Further in, the liquor has only recently arrived, and a fibre there has had only a few seconds of it. So the deep fibres lag behind what the liquor would allow while the surface fibres do not, and the half-strength point moves outward.

Then the other effect takes over. With a longer clock, the surface itself is no longer full at the end of the dip. It drops from 98 per cent at five seconds to 63 at twenty and 18 at a hundred. Meanwhile the liquor is taken up more slowly, so it travels further before it is used. The profile has a lower top and a longer reach, and past twenty-six seconds the half-strength point lies deeper than an instant fibre’s would.

That is the sense in which the original guess was right. The liquor does run ahead. It simply takes a clock longer than the dip before that outweighs the fibres lagging.

What the shape does #

The profile changes kind as well as depth. An instant fibre’s curve reaches nearly nothing by 45 per cent of the radius: it has a front, with undyed cotton behind it. A fibre at twenty seconds crosses a half in almost the same place, but its steeper start gives way to a longer tail. At a hundred seconds a quarter of the surface value is still present at half the radius, and by four hundred the curve is the liquor’s own diffusion profile drawn at low strength, since the fibres have taken so little that the liquor barely noticed.

So the same half-strength depth can belong to very different cloths. The instant fibre at 14.0 per cent and a fibre at twenty-six seconds at 14.0 agree on depth and disagree on everything else. The second holds 61 per cent as much dye, with 54 per cent of the surface shade, and a tail where the first has a clean edge.

A plateau needs its two times apart #

There is a tidy limit worth finding, and finding where it fails. Once the liquor has settled and before the fibres have filled, the liquor is steady and its profile contains a length and no time:

$$\ell^2 = \frac{D_0\varepsilon^{m}\tau}{(1-\varepsilon)K}, \qquad c = \frac{I_0(r/\ell)}{I_0(a/\ell)}.$$

The fibres fill in that fixed shape, so in that regime the ring would have a depth that does not grow with the dip at all, and a surface that darkens in proportion to it. It looks like the perfect signature: one account predicts depth growing as the root of time and the other predicts it standing still.

The trouble is the phrase “once the liquor has settled and before the fibres have filled”. Those two times are about $\tau\varepsilon/((1-\varepsilon)K)$ and $\tau$, and they differ by $(1-\varepsilon)K/\varepsilon$, which at an affinity of ten is only fifteen. There is hardly any room between them. At an affinity of two hundred, the plateau is there: the ring sits within a tenth of the Bessel form’s value, and its depth barely responds to the dip. At the affinity these essays have used it is a shoulder and not a plateau. The depth’s local exponent in time falls from 0.55 for an instant fibre to 0.22 at a clock of fifty seconds, and it comes back up to 0.58 at four hundred.

So the appealing test is weak. Across a factor of four in the dip, an instant fibre grows its ring by 2.15 times and a four-hundred-second fibre by 2.23, for different reasons: the first is a front being pushed along and the second is free liquor spreading before the fibres have noticed it. A plot of depth against dip time would have to be read at the middle of the range to say anything, and it would need the affinity known independently, since that moves the depth by more than the clock does. An affinity of five gives 20 per cent where ten gives 14 and twenty gives 9.8.

The surface has no such ambiguity #

What the depth cannot separate cleanly, the shade at the surface does, and the reason is that no diffusion stands in the way there.

A surface fibre sits in the bath, sees the bath’s full strength from the first moment, and fills as $1 - e^{-t/\tau}$. So the surface’s dependence on the dip’s length reads the clock directly, with no affinity and no pore structure in it. An instant fibre’s surface does not change with the dip at all, an exponent of nought in time. A slow one’s rises in proportion to it, an exponent of one. At a clock of twenty seconds the exponent is 0.58, and at fifty it is 0.81.

This has a consequence for the earlier reading of the share. The essay on pores found that a yarn’s surface reads the share a dip fills exactly, and it does, on the assumption that the surface fibres finish each dip full. With a clock in the fibre, what the surface reads is the product of that share and $1 - e^{-t/\tau}$. Those two cannot be separated from one dip’s surface, and the earlier inference of a share from the surface is a lower bound on the room the dip could have used. Two dip lengths separate them at once, and a rope range’s speed is a dip length that a dyer can change.

An instant fibre earns its ring, and a slow one starts there #

The clearest separation comes from the thing denim is actually made by, which is dipping repeatedly.

The instant fibre’s ring deepens by nine and a half points across eight dips, for the reason the saturating essay found: its skin fills, so each later dip passes the skin at no cost and spends its dye further in. A slow fibre never fills its skin. At a clock of a hundred seconds the surface takes 18 per cent of its share per dip, so eight dips leave it about a quarter full against 83 per cent, and most of the room the deepening depended on is still empty. So its ring deepens by three points, and at four hundred seconds by one.

This is the reading that does the work. A section after the first dip and one after the last tell the accounts apart with no other measurement, since the same final depth can arise two ways. A 23.5 per cent ring reached from 14 is a filling fibre. A 26 per cent ring reached from 23 is a slow one. And the shade separates them again, since one of those skins is nearly saturated and the other is nowhere close.

A fibre at twenty seconds is an awkward middle case, since both effects are there at once: it deepens by less than the instant fibre and from a shallower start, and it ends the shallowest of the four.

What slowness costs in dye #

The total held after one dip falls continuously with the clock: to 92 per cent of an instant fibre’s at five seconds, 68 at twenty, 29 at a hundred and 10 at four hundred. A range that wanted a given shade from a slow fibre would need more dips or a stronger bath, or a longer one, and the first two are what real indigo ranges do at some cost. Twenty-second dips are a compromise between that and the speed of a rope.

Whether a deep pale ring is what a dyer wants is a separate matter, and a ring-dyed yarn whitens all at once says it is not. That essay’s sudden whitening came from a ring with a front: nothing shows until the abrasion cuts through it. A profile with a long tail has no sharp point to cut through, so a slow fibre’s yarn should fade gradually and show less of the hard contrast that makes a float fade late and hard and a crease fade in dashes. That hard contrast is itself evidence about the clock: denim that fades as sharply as it is known to, with a pale core appearing quickly, is hard to reconcile with a fibre whose clock is much longer than the dip. It is a qualitative inference and has not been run through the wear calculation, but it points at the short end of the range.

A thin yarn and a thick one #

The clock’s time does not scale with the yarn, and the yarn’s own diffusion time does, so a count changes where a yarn sits in all this. At a clock of a hundred seconds a 20 tex yarn’s ring reaches 46.5 per cent of its radius, against 23.6 for an instant fibre; a 100 tex yarn’s reaches 15.2 against 9.6. The instant ring shallows with count because the radius grows, as the first of these essays found, and the slow one shallows with it, more steeply.

So a fine yarn with a slow fibre is the case nearest to being dyed through, and the depth ratio to the instant case is 1.97 at 20 tex and 1.58 at 100. A comparison of ring depth across counts in one bath would be one more reading of the clock, needing only sections and needing no change to a range. It would also run into the affinity, which moves both.

The surprising part is where the ambiguity sits #

The earlier essays in this series each added one physical effect and found it moved the ring by a modest and calculable amount, and each read the result off the ring’s depth. This one breaks that habit: the same physical effect moves the depth in both directions according to how much of it there is, so depth is not a monotone reading of the clock at all. A measured ring of 14 per cent of the radius is consistent with an instant fibre and with a clock of twenty-six seconds, with nothing in the depth to say which.

That has the same shape as the stiffness of a yarn being a bracket and not a number, in which a single measure of bending stands for two mechanisms and a test at one radius could not separate them. Here a test at one dip cannot. A quantity with a turning point in it cannot be inverted from one reading, and a calculation that takes the expected sign for granted would have read it as a bound.

The model named #

Radial diffusion of liquor through a cylinder’s pores at $D_0\varepsilon^m$, Archie’s law with exponent 1.5, coupled in each of 120 shells to the fibres by a linear driving force with time $\tau$. The exchange in each shell over a step is solved in closed form, which keeps it stable at small $\tau$. A fibre’s room is $K(1-q)$, with $q$ the saturation from earlier dips, fixed by airing at the end of each. $D_0$ is $4\times10^{-10}$ square metres a second, the yarn is 50 tex at a packing of 0.6 and the dip twenty seconds.

What was counted #

Nine clocks from nought to four hundred seconds at one dip, and a further fourteen for the depth curve; twelve dip lengths on each of four clocks; eight dips on each of four; three counts and four affinities. The checks: agreement with Crank’s cylinder at $\tau = 0$ to 0.01 of the radius; agreement with the saturating model at each of eight dips; agreement with the Bessel form within a tenth at an affinity of two hundred, where its conditions hold; a surface exponent of nought at $\tau = 0$ and above 0.95 at 400 s; a ring that is shallower at five seconds and deeper at a hundred; uptake falling with every increase of the clock; and a slow fibre starting deeper and deepening less than half as much. The figures’ four-way comparisons use the same runs. A negative clock, a dip of no length, a fibre of no affinity and a dip filling nothing are refused.

What the model cannot show #

$\tau$ is one number for all fibres, at all depths and at all saturations. Real cotton has a distribution of fibre diameters and maturities, and a cloth is a population and not a thread in this as in everything, so a real yarn has a spread of clocks that would blur the turning point. Mercerising swells the fibre, which should shorten its clock as well as changing the packing, and that is not here either. A fibre that already holds pigment may also take up more slowly than an empty one, which would make the clock lengthen dip by dip, and nothing here moves it.

The linear driving force is itself an approximation to diffusion into a cylinder, which has a square-root start: a fibre takes up fast at first and slowly later. Over dips short against $\tau$ the true uptake would be larger than the linear form gives, in proportion to the root of the dip rather than to the dip. The surface exponent of one is therefore the model’s, and a real fibre’s could lie nearer a half. That would still differ from an instant fibre’s nought, but it changes what number a test should expect.

Airing is taken to fix what the fibres hold and nothing more. Liquor left in the pores after squeezing is lost to the model. In a real range some of it continues into the fibres before it oxidises, and that gives a slow fibre extra time beyond the bath’s twenty seconds. It may be a large correction, since a slow fibre’s whole problem is shortage of time, and the airing is usually longer than the dip.

The clog and the fill are not combined with the clock, except that room runs out. The affinity is taken as ten, as it has been in the earlier essays, and it is not itself a measurement.

Who worked out which part #

The linear driving force is Glueckauf’s, from 1955 and from the theory of chromatography, where its factor of fifteen for a sphere was derived. Two-scale diffusion in porous particles and the reaction length with its Bessel profile are the Thiele modulus of catalysis, from 1939 and still the standard treatment. Indigo’s ring dyeing and the short dip are a dyer’s working knowledge, and the kinetics of its uptake on cotton have been studied by Etters and others, whose work is what any real value of the clock should be drawn from.

Running a fibre’s clock through this dip model, finding that the ring moves the wrong way first, and reading the clock from the surface’s response to a dip’s length and from the ring’s first and last sections are calculated here and have not been tested.

Still open: what the airing gives a slow fibre #

The largest thing left out is the time after the bath. A rope dyeing range squeezes its yarn and airs it for a minute or more, and oxidation does not happen at once throughout the section. Liquor held between the fibres in that time goes on diffusing inward and being taken up while the outside oxidises. For an instant fibre the airing changes little, since the fibres already hold what they can. For a slow one it may matter more than the dip does.

That is a further stage with its own clock, the oxygen’s, running inward from the surface against the dye running in the same direction. The model needs a third field, oxidised against unoxidised dye, and an oxygen front to say whether a slow fibre’s ring is set in the bath or in the air, and whether a longer airing can do what the longer dip does at none of its cost in speed. Airing time is a thing a range can vary directly, so the calculation would have a test ready for it.