The air moves a ring outward and adds no dye

A slow fibre starts its ring deep ended at the bath’s edge. A rope-dyeing range squeezes its yarn and airs it for a minute or more, and the oxidation that fixes indigo does not happen everywhere at once. Liquor stays in the pores while the outside takes up oxygen, and a fibre with a clock of its own could go on taking up that liquor long after the dip is over. That essay left the question open. Whether a slow fibre’s ring is set in the bath or in the air, and whether a longer airing could do what a longer dip does at none of its cost in speed, is what a range could test directly by turning one dial.

It seemed a question with an obvious direction: more time, more diffusion, a deeper ring. It has the opposite answer, and the reason is one thing that was left out of that picture. The air is not only a clock. It is a sink. Where oxygen has reached, the dye in solution stops being in solution, so the liquor at that place is empty, and liquor behind it runs towards the empty place.

What the airing adds to the model #

Everything the dip had stays in place. Liquor in the pores diffuses, and each shell of fibres takes up from it with a linear driving force and a clock of $\tau$ seconds, solved exactly over a step, so an instant fibre is the same code with $\tau$ at nought. Those equations belong to the earlier essay, and two things change at the moment the yarn leaves the bath.

The first is that the surface supply stops. There is no more bath, and a squeezed yarn holds what liquor it holds, so nothing crosses its surface.

The second is the oxygen. It comes in from outside as a front at a share of the radius $\sqrt{t/T}$, so that $T$ is how long it would take to reach the axis. Behind the front, dye held in a fibre is fixed there. So is dye in solution in a pore, oxidised where it stands, and it now counts as pigment in that pore. Ahead of the front the dip’s own equations go on running, with a boundary condition that was not there before: the liquor concentration at the front is nought.

$T$ is not measured here. It is swept across three and a half decades, and the two ends of the sweep are checks. A front that arrives at once fixes the ring that left the bath. No airing at all also leaves it as it was. In between there is an estimate to orient it. With a rule-of-thumb oxygen in water, a typical vat strength, and a yarn’s fibres holding the working affinity’s share of leuco dye to be oxidised, a stoichiometric front on a 50 tex yarn should take of the order of a thousand seconds to reach the axis, which is a guess about one factor of three each way. It puts the front through the ring’s own depth, about a seventh of the radius, in a few tens of seconds. So the ring’s outer edge is being overtaken by oxygen during a real airing, and neither is far ahead of the other.

Every aired ring is shallower #

At every front computed, an aired ring is shallower than the ring that left the bath. An instant fibre leaves the bath at 14.0 per cent of the radius. With a front of 300 seconds it ends at 9.0, and with a front of 1,000, at 6.5. A fibre with a 20-second clock goes from 13.2 to 10.8 and 10.4. A fibre with a 60-second clock, which leaves the bath deepest of all at 19.6 per cent, ends at 10.7 and 7.2.

So the slowest fibre has the largest change, as the earlier essay guessed, and it goes the other way. A slow fibre’s depth on leaving the bath is the shallowest kind of fact: most of it is liquor that has not yet been taken up, and a liquor stands free to move. An instant fibre has already fixed its dye to fibres, by the time it leaves the bath, in a form that holds the ring where it is.

The total, meanwhile, does not change. The dye carried out of the bath in fibres and pores together is 0.304 for an instant fibre, in units of a full fibre over the section, and 0.302 after airing at the 300-second front; 0.216 and 0.215 for the 20-second fibre; 0.146 and 0.145 for the 60-second one. Airing does not make a yarn take up any more dye than it carried out, and it can’t, as there is no more dye to be had. A longer airing cannot do what a longer dip does, because a dip brings dye in from a bath and an airing has only a yarn’s own contents to rearrange.

A slow fibre’s shade therefore stays short even after airing, and all of the slow fibre’s lighter surface survives except for what the skin picks up from behind it.

Why dye runs the other way #

The obvious picture has liquor going inward after the bath, since the middle of the yarn is empty and the outside is full, and for the first instant that holds. Once oxidation starts, there are two empty places, and the one at the skin is far nearer.

A liquor profile leaving the bath is steep. It has its full bath strength at the surface and has fallen by half within a seventh of the radius. Oxidise the outer layer and the profile suddenly has a nought on its outside as well as a low value on its inside. The distance to the outer sink is a small part of the distance to the inner emptiness, and diffusion’s time goes as the square of distance, so nearly all the liquor in the ring goes out rather than in. It deposits at the front, in a sharp band, and the band is the new skin.

A skin that holds more than a full fibre’s load is a real prediction. The dye that arrives at the front is liquor, and liquor oxidising in a pore does not need room in a fibre. That is the same pigment in the pores a clogged skin reasoned about, and it arrives here by a route that essay did not have. In that account the pores filled over many dips and held the ring back; here a single airing fills them on a slow front. Its bound on how much could lie in the pores would apply here as well, and a skin whose pores fill by this route is one more thing that account’s bound would have to include.

It also means how an indigo stays on a surface depends on the air. Pigment in the pores of the skin is less firmly held than dye inside a fibre, so a slow-oxidising range would crock more. That is qualitative and it is a direction. A slower airing front means more dye in the pores and a sharper ring, at one shade, and a dyer’s reading of crocking against airing conditions is a test of it.

A longer airing does nothing #

The sum has one more surprise, and it goes straight at what the earlier essay asked a range to try.

Almost all the change comes in the first five seconds. With a front of 300 seconds, an instant fibre’s ring has gone from 14.0 to 9.0 per cent of the radius by five seconds and is 9.0 at three minutes. A 20-second fibre is at 10.8 by five seconds and stays there. A 60-second fibre is at 10.7 and stays.

It is not hard to see why. The time for liquor to cross the ring, its thickness squared over its diffusivity in the pores, is of the order of a couple of seconds. So the ring’s liquor has run to its nearest sink almost as soon as a sink exists, and what is left is fibres that hold what they hold. The fibres do then release dye slowly, and a little more makes it across; that is too little to move a half-strength depth. The dial a range can turn most easily therefore does nothing to the ring’s depth after the first seconds, though it certainly goes on doing something: an airing has to run until the axis-side dye is oxidised, for a dyed yarn must be fully oxidised before its next dip. It is a question about completing a reaction, and on this arithmetic it is not a question about shade.

That would be a very direct test. Take a yarn from one dip through airing times from five seconds to three minutes, and section it. On this account the ring depth and the surface shade should agree to within a measurement’s scatter across that whole range. If they do not, oxidation runs slower at the outset than a front that starts at once, which is the first thing the model assumes.

What a slow fibre gets from this #

The earlier essay found that a slow fibre’s ring is a deceptive object. It starts shallow and then deepens, and it crosses the instant fibre’s depth around 26 seconds. A 60-second clock leaves the bath 40 per cent deeper than an instant fibre does.

That is now almost entirely undone in the air. The depth a 60-second fibre takes out of a 20-second dip is 19.6 per cent of the radius, and the depth it keeps is 7 to 11, about the same as or less than the fibres that took it up at once. A depth measured on wet yarn, or one computed from a bath’s contact alone, reads a stage that does not survive the next few seconds. It also changes what is worth reading. A slow fibre’s depth has a turning point and cannot be inverted, as that essay said; once aired, across this range, its depth and its τ hardly separate at all at a slow front, and only its total, the shade, still reads the clock.

So the surface reading, which that essay recommended, is the one that survives the airing, in the sense that total dye goes through unchanged. A surface shade is a mixture of what was there and what has come out to the skin, and both of those scale with the total. It would be a mistake to read the clock from the skin’s shade alone, since the skin is now also reporting the front, but the total is safe.

What a fading denim would inherit #

A sharper ring is not a detail for the cloth this all ends up in. A dip dyes a depth and not a share, so a shade is a depth times a surface, and the airing moves those two against each other at a fixed total. That matters most at the wear stage, where a ring-dyed yarn whitens all at once: the wear at which white first appears is about the ring’s thickness. A ring brought from 14 to 9 per cent of the radius by its first airing should reach that point sooner, at the same shade when new.

So two cloths dyed to one shade on two ranges that differ only in how freely air reaches a rope should fade at noticeably different wear. Floats and crossings wear differently, and a crease along a twill wears its line at one rate, but those are all ratios on a ring depth. Every absolute wear figure in those accounts was taken from a depth the bath set, and on this result the air sets it. The figures would all shift together, and their ratios would not.

How it was computed #

The dip is the earlier essay’s: a cylinder of 50 tex cotton at a packing of 0.6, a Crank-form pore diffusion with Archie’s correction, a working affinity of ten, 120 radial shells, a 20-second dip, and a fibre’s linear driving force with τ of nought, 20 or 60 seconds. The airing continues from the bath’s last state with no flux across the surface. A front moves inward at $\sqrt{t/T}$, and each shell it reaches is fixed: its fibres and its pores’ liquor together, in units of a full fibre, per cell. Whatever crosses from the unfixed shell into the fixed one is added to the fixed one, and that term is what gives the sink its dye.

The depth is read at half the peak, the ring’s own usual measure, and it can jump when a deposit changes what the peak is. What is required of it, not shown: that no airing and an instantly arriving front each return the bath’s own ring; and that the dye fixed at the end be within three per cent of the dye that left the bath, at two fronts and two clocks.

What the front leaves out #

The front’s speed is imposed. A real front is set by how much oxygen arrives and how much dye it meets, and so it depends on the same liquor concentrations being calculated here; a real oxygen-limited front stalls at the surface while dye is arriving faster than oxygen can dispose of it. A consistent Stefan problem would slow the front where the liquor is strongest. That would change these numbers in the direction of a slower front, and it has been swept across that direction, but it has not been solved.

The sink has no reaction time. Oxidation in air is quick, and here it is taken as instant at the front. Liquor is treated as staying. A squeezed yarn loses some of its liquor to the squeeze roll, and a wet yarn drains in air, so a real start is a little lower than the bath’s own last profile. The yarn’s own diffusion is unchanged by its contents. A pore filling with pigment would slow the liquor behind it, and the skin here is getting past a full fibre’s load without any such feedback. That effect would hold the ring deeper than these figures show, and it is the most likely place for the argument to need correcting.

The cylinder holds no twist, and a real yarn is twisted, porous and not round. And there is only one dip. The next dip goes into a section with this sharper ring. What a sequence of dips does with it, once each carries its own airing, is not computed here.

Who worked out which part #

That indigo oxidises in air, and that rope dyeing owes its ring to short dips and repeated airings, are the dye-house’s own knowledge. Stefan problems, of which an oxidation front is an instance, are standard diffusion.

What is added here is that the front is a sink which reverses the liquor’s direction; that an aired ring is shallower than the ring that left the bath; that a slow fibre’s bath depth is the largest and the least durable; that dye is conserved through the airing, so a longer airing cannot substitute for a longer dip; and that the change is complete within seconds.

Still open: whether a shortage of oxygen holds the front at the skin #

The largest assumption above is that a front runs at a rate given in advance. Oxygen does not behave so: it arrives by diffusion from the air through a wet layer, and it is spent on the first leuco dye it meets. When liquor is arriving at the skin faster than oxygen can dispose of it, the front has nowhere to go, and it stands at the skin until the arriving flow has fallen below what oxygen can meet.

That is a free boundary, and it would give a real $T$, with no sweep. It predicts a stand-still of its own, a first interval in which the skin takes up nearly everything and the front does not move at all, and its length sets the skin’s darkness. If that is what happens, the airing’s first second is all of it, and a yarn hanging in still air against a fan would differ only in how long the stand-still lasted. It is the same two diffusions with one extra equation at the boundary, so the calculation is near at hand and the two conditions on either side of a stand-still can both be measured from a section.