Which walls a strand is held by

A strand in a tube found a law with almost nothing in it. A Bowden inner pulled through a sheath is shorter than the sheath’s centreline by $c\sum|\varphi|$ — the radial clearance times the total angle the route turns through — with no bend radius, no route shape and no spacing anywhere in it. An arc of radius $R - c$ through $\varphi$ is shorter than the centreline’s arc by $c\varphi$ whatever $R$ is, the straights between bends are unchanged to first order, and the sum is the whole answer. A length error undone by its own size is the constraint field’s version of the same surprise: a clearance that turns out to be the whole of a quantity rather than a correction to it.

The derivation has one assumption in it and the essay said so: the inner is taken to touch the inside wall of every bend. That is where the $c\varphi$ per bend comes from. It is a statement about a contact pattern rather than about an inner, and nothing in the calculation checks it. Where a strand stops touching is the same question asked of a body joining a run, and it found that the answer is a threshold rather than a proportion.

It is not always true, and the sentence that shows why needs no arithmetic. The inner enters the sheath at a ferrule, on the centreline. To reach the inside wall of the first bend it has to cross the bore, and crossing costs turn — over a straight approach of length $L$ it costs about $\arcsin(c/L)$ of direction. A bend that turns through less than the inner needs to spend getting there is a bend the inner never reaches: it passes down the middle of the bore and the tube holds it nowhere.

Solving for the inner instead of for the contact #

The repair is to stop assuming the pattern and compute it, which means solving a different problem from the one the wrap model solves.

The inner is the shortest path from one ferrule to the other that stays inside the bore, and the bore is the set of points within $c$ of the centreline. Which walls it touches is an output of that problem. Written that way there is no list of pulleys to supply, no sense to get right, and nothing to assume.

The bore is sliced into cross-sections, each a segment from one wall to the other, and a path inside the bore is a path crossing all of them in order. The shortest such path is what Lee and Preparata’s funnel finds in a single pass: it carries an apex and two taut sides, tightens them against each cross-section in turn, and emits a corner whenever the funnel closes on itself. Every corner it emits is a cross-section endpoint, which is a point on the wall — so the contacts come out of the answer rather than going into the question. That is the opposite arrangement from a strand over a row of pulleys, where the bodies are given and the tangents follow.

It is worth recording what that replaced, because the failure was the kind that looks like success. Writing the path as one offset per cross-section and sweeping each offset against its two neighbours is a correct method and it converges at one cross-section per sweep, because that is how far news of a change travels in a Gauss–Seidel pass along a chain. At three hundred sections and four hundred sweeps it looked converged and was. At two thousand four hundred sections the same four hundred sweeps had relaxed a sixth of the path and reported a lost motion of 0.67 against a true 6.23 — a number that moved the right way as the sampling was refined and was wrong by a factor of nine. The funnel has no iteration count to get wrong.

Where the two models are both valid they agree, and the agreement is what licenses everything after it. On a route of two right-angle bends, where the inner is held over the whole of both, the shortest path in the bore, the wrap over two pulleys of radius $R-c$, and the closed form $c\sum|\varphi|$ come out within 0.31% of each other at the worst of three clearances — and that residue is the cross-sections’ own spacing, falling as the square of it. The three know different things: the closed form knows only the total turning, the wrap knows the radii and is told which walls to touch, and the shortest path knows neither and is told nothing.

Why the crossing costs a turn and not a length #

The crossing angle is the whole of the mechanism and it is worth deriving rather than quoting, because the form it takes is what makes the rest of the argument scale-free.

An inner leaving a ferrule on the centreline and arriving at the inside wall of a bend a distance $L$ away has moved $c$ sideways over $L$ along. If it does that as a straight line — which is what a taut path does over a straight stretch of tube — its direction differs from the centreline’s by $\arcsin(c/L)$, and that difference has to come from somewhere. It comes out of the bend. The bend turns the tube through $\varphi$; the inner arrives already turned by the crossing and leaves having to turn back by it again, so what is left to be held against the wall is $\varphi - 2\arcsin(c/L)$.

Two things follow that a length-based intuition does not give. The first is that the bend’s radius does not appear. A tight bend and a gentle one of the same angle lose the same amount to the crossing, which is why the original law had no radius in it and why its failure mode has none either. The second is that the quantity being compared to the bore is a length along the route, not a length across it: doubling every dimension of a route while holding the clearance fixed halves the crossing angle and makes the law more accurate, and a small mechanism with the same housing is the one where the law breaks first.

There is a third reading, and it is the one a designer can act on. $\arcsin(c/L)$ is small whenever the straights are long compared with the bore, so the regime where the law fails is the regime of short straights — a cable turned twice within a few bore-widths, which is what a tight routing looks like. The failure is not caused by the bends being shallow in themselves. It is caused by their being shallow relative to how close together they are, which is a property of the route as a whole and not of any bend in it.

The share of a bend that still holds #

Run the shortest path against a widening bore and the contact does not switch off; it erodes.

On two 8° bends of radius 40, at a clearance of 0.5 the inner is in contact with 95% of each bend’s turn and loses 97% of what the law predicts. At 2 it holds 76%, at 4 it holds 54%, at 6 it holds 31%, at 8 it holds 8%, and at 10 it holds none of either bend and the tube is not touching the inner anywhere along its whole length.

The shape of that decline is the crossing angle again, and it can be written down. The inner arrives at the bend having spent $\arcsin(c/L)$ of turn crossing the bore, and it must spend the same again leaving; what is left of the bend’s own $\varphi$ to be held over is $\varphi - 2\arcsin(c/L)$. Divided by $\varphi$ that is the held share, and it is drawn against the measurement without a fitted constant in it. The two track each other down to the point where the bend is lost entirely.

That point is a prediction with a number in it, and it is the one the earlier reading reached from the other direction. Asking the model for each bend angle in turn and taking the shallowest it can still answer gives a threshold, and against $\arcsin(c/L)$ over an approach of 60 the two agree to a quarter of a degree from a clearance of 0.5 to one of 8. A bend’s own turn is what it has to pay the crossing with. A bend that turns less than the crossing costs is passed straight through, and two such bends in a row behave as one long straight — which is the departure the four-bend route showed at large clearance and could not explain. A wrap’s angle is decided by the geometry either side of it rather than by the body it is on, and this is that rule at its limit, where the geometry either side leaves no angle at all.

The law is an upper bound #

Once the held share is falling, the closed form is not merely inaccurate. It is wrong in a stated direction, and it stays wrong that way.

Each bend contributes $c\varphi$ only over the part of its turn the inner is actually held through, so a bend held over a fraction $h$ of itself contributes something like $c h \varphi$ and the law over-counts by the rest. The measurement bears that out at every clearance tried: the ratio of what is lost to what the law says falls monotonically — 0.97, 0.88, 0.77, 0.65, 0.54, 0.43 — and never once rises. The closed form never under-estimates. For a designer that is the useful half: a lost-motion budget computed from $c\sum|\varphi|$ is conservative, and it is conservative by a factor that grows with the bore and shrinks with the bend angles. Backlash is an allowance rather than a defect, and an allowance known to be an over-estimate is the right kind to carry.

The direction is worth stating plainly because the intuition runs the other way. A wider bore sounds like more lost motion, and it is, up to a point; what it also does is let the inner stop following the route, and an inner that is not following the route is not paying for it.

The ceiling, which has no clearance in it #

The other half of the same fact is a hard bound, and it comes from a sentence rather than a sweep.

A taut inner is a path from one ferrule to the other. No path between two points is shorter than the straight line between them. So however wide the bore, the inner is at least as long as that line, and the most the sheath can ever swallow is

$$\text{ceiling} \;=\; \text{centreline length} \;-\; \text{the chord between the ferrules}.$$

There is no $c$ in it and no $\varphi$. On the two-bend route it is 1.204, and the measured loss flattens onto exactly that: 1.1957 at a clearance of 8 and 1.2041 at 10, where the inner has stopped touching anything and runs straight. The closed form passes the ceiling at a clearance of about 4.3 and carries on rising, which is the sharpest way to say that $c\sum|\varphi|$ is not a law about a cable but a first-order expansion of one. It is the same shape as an error that accumulates rather than bounding, read backwards: there a quantity that looked bounded grew without limit, and here one that looks unbounded does not.

The ceiling is a bound on every route and an attainment on only some, and separating the two is a question about whether a chord fits in a tube. A shallow route’s ferrule-to-ferrule line lies inside its own bore once the bore is a few units wide, and the ceiling is reached. An S of two right angles at radius 60 has a centreline that wanders 88 units further than its chord, and that chord leaves the tube long before the bore is wide enough to contain it — so the inner is still held somewhere at the widest bore its bends allow, and stops 4.42 short of its own ceiling.

What the two numbers are for #

Put together, the route’s lost motion has a small regime and a large one and the boundary between them is computable in advance.

Below the crossing angle — every bend turning through more than $2\arcsin(c/L)$ of its adjacent approaches — the inner is held all the way round every bend and $c\sum|\varphi|$ is right to a per cent or two. That is the regime a real cable is designed into: a brake inner of 1.5 mm in a housing of 2.0 mm has $c = 0.25$ — the clearance the original measurement used — and over a 60 mm approach the crossing angle is 0.24°, so every bend worth calling a bend is fully held. The law is not fragile in practice.

Above it, the two numbers that matter are the ceiling and the held shares, and neither is the law. A route of many shallow bends — a cable snaked round obstructions rather than turned deliberately — is exactly the case where the bends fall under the crossing angle one at a time, and the budget computed from the total turning is an over-estimate that grows as the housing wears. A worn housing does not lose motion in proportion to its wear. It loses less than that, and eventually it stops losing more at all.

What is not modelled #

The inner has no stiffness. Every path here is the taut limit, which a pulled cable approaches and a pushed one does not. A real inner is a rod that resists bending, so at low tension it lies somewhere between the centreline and the taut path, and at any tension it cannot turn the corner the funnel’s corners turn. Whether the path is even unique is a separate worry: a taut path can have more than one answer, and the funnel returns one of them without saying so.

The route is planar. A real cable is routed in three dimensions and its turning is a total curvature rather than a sum of signed plane angles. The crossing-angle argument carries over unchanged — it is about a distance across a bore and a turn — but the ceiling’s chord is a three-dimensional chord and nothing here computes one.

Friction is absent, as everywhere in this collection. The taut path is the shortest path, not the path a cable under friction actually takes; a real inner loses more than its geometry says because the contacts it is held at are the contacts that resist it.

The corridor must be a sleeve. The model needs the bore narrower than the tightest bend it turns through, so that the inside wall does not collapse through the centre and the cross-sections do not cross each other. Asked for a clearance past that, it refuses: the funnel returns a number there — on a right-angle pair at a clearance of ten thousand, a lost motion of −19,774 — and the number means nothing.

The contact set is read at a tolerance. A cross-section counts as held when the path reaches within a millionth of the wall. The held shares quoted are therefore sampled quantities; the threshold and the ceiling are not, and they are the two that carry the argument.

One bend angle and one radius carry the sweep. Two 8° bends of radius 40 on approaches of 60. The threshold is checked across six clearances and the ceiling across two routes; how the held share behaves for bends of very different radii in one route is not charted.

Still open: the bore that is not round #

Everything above gives the bore one number. A real housing is a coil of wire with a liner, and what it presents to the inner is not a circle of radius $c$ but a cross-section that is stiffer across one diameter than another and that changes as the housing is bent.

Its distinct argument would be the same shortest path in a bore whose half-width varies along the route — which the funnel handles without modification, since each cross-section carries its own endpoints and nothing in the algorithm assumes they are the same width. Two things would come out of it. Whether a housing that pinches at its bends and opens on its straights loses more or less than a uniform one of the same mean clearance, which is a question about where in the route the contact is worth having; and whether the ceiling survives, since the chord’s fitting inside a varying bore is a different test from its fitting inside a uniform one and a route could have a chord that fits at every straight and leaves at one bend.