A strand in a tube

Every strand in this field so far has been routed by the bodies it wraps. A member with no length of its own set out the premise: a belt, a rope or a tendon has no shape of its own, and where it runs is decided by the pulleys, pegs and drums it touches. Between them it is straight; on them it is an arc; and the length that constrains it is the sum.

One strand over many joints routed a tendon over idlers on an arm and noted, in passing, the strand it had deliberately left out. A tendon often runs in a sheath, and a Bowden cable — the brake and gear cable of a bicycle, the throttle cable of a motorcycle, the release on a camera — is nothing else: an inner strand inside a flexible tube whose ends are fixed to the frame at the lever and at the brake. Bending the arm changes the cable’s effective length even when nothing at either end has moved, and that essay flagged it as a different drive.

It is a different object for one reason. The strand’s route is decided not by bodies it wraps but by a tube it runs inside, and the tube has a length of its own. This essay asks what that length does to the strand, and the answer is shorter than the setting suggests.

The tube, and the strand inside it #

A Bowden sheath is built to be incompressible along its length and flexible across it — a close-wound coil of wire, usually lined — so its centreline keeps its length however it is routed. Its bore is a little larger than the inner cable. Call the radial clearance, half the difference between bore and cable, $c$.

Pulled, the inner is in tension, so it is taut, and a taut strand takes the shortest path its route allows. Inside a tube, the shortest path presses against the inside of every bend: the inner hugs the wall nearer the bend’s centre, on an arc of radius $R - c$ about it, and runs straight between bends. That is precisely a strand over a row of pulleys — one pulley at each bend’s centre, of radius $R - c$, wrapped in the bend’s own sense — and the strand field’s closed form for a taut run computes its length with no new mathematics.

Pushed, the inner is in compression and has nowhere to go but the other wall: at every bend it is pressed against the outside, on an arc of radius $R + c$, and its route is the same list of centres with that radius instead.

The first figure draws both, with the clearance exaggerated to 7 on a route a few hundred units across so that the two paths separate on the page. The pulled inner cuts the corners of both bends; the pushed inner bulges round them. The ends of the inner are fixed where the sheath’s ferrules are, on the centreline.

Clearance times angle, with no radius in it #

The length of an arc is its radius times its angle. The centreline’s arc at a bend of angle $\varphi$ is $R\varphi$; the pulled inner’s is $(R - c)\varphi$; the difference is

$$c\,\varphi,$$

and $R$ has cancelled. The straights between bends are the same length for the inner and the centreline to first order in $c$, because a line tangent to the inside of two arcs is parallel to the line tangent to their centres’ arcs. So a sheath that turns through bends of angles $\varphi_1, \varphi_2, \ldots$ leaves its pulled inner short by

$$\Delta_{\text{pull}} = c \sum |\varphi_i|,$$

pushed long by the same, and with backlash — the travel at the lever between taking up the inner one way and the other — of $2c\sum|\varphi_i|$.

That is the whole of the argument, and it makes three predictions that are easy to test and not obvious to a person routing a cable: the bend radii do not matter, the order and shape of the bends do not matter, and what counts is the angle turned without its sign.

Why the straights do not count, and when they start to #

The arc argument accounts for the bends. The straights need a sentence of their own, because the inner does not run along the centreline between bends: it runs from the inside wall of one bend to the inside wall of the next, and on an S-bend those walls are on opposite sides of the tube.

Where two consecutive bends turn the same way, the inner’s straight is tangent to two arcs of radius $R - c$ about the same centres the centreline’s straight is tangent to at radius $R$. Two lines tangent to the same pair of circles, one at radii shifted inward by the same amount, are parallel and the same length. So the straight contributes nothing at any clearance.

Where the bends turn opposite ways, the inner’s straight is the crossed tangent between the two inside walls, and there the accounting is less simple. The crossed tangent between two circles of radius $R - c$ is longer than the one between circles of radius $R$, by an amount proportional to $c$; but it also leaves each circle at a shallower angle, so the inner wraps each bend through slightly less than the bend’s own angle, and that shortens the arcs by an amount proportional to $c$ as well. The two first-order changes cancel exactly — the line and its tangency points move together — and what is left is of order $c^2/L$, largest where the straight between the bends is shortest. The ferrules at the ends behave the same way: the inner leaves the centre of the bore and reaches the inside wall of the first bend along the first straight, at a cost of the same order.

That is the pattern the six routes show. The routes whose departure from the law is largest at a clearance of 1 are the S-bend and, most of all, the four alternating bends, which have the most crossings over the shortest straights; and every departure, divided by $c\sum|\varphi|$, falls in proportion to the clearance, which is what a remainder of order $c^2$ does.

What it amounts to on a bicycle #

The law is small in each bend and adds up over a route. A bicycle brake inner is typically 1.5 mm in diameter in a liner of about 1.9 mm bore, a radial clearance of 0.2 mm. A rear brake cable on a road frame runs from the lever round the handlebar, under the tape, out along the top tube and down to the brake, and between them its sheaths turn through something like a full turn in total.

At $c = 0.2$ mm and $\sum|\varphi| = 2\pi$, the pulled inner is short of its sheaths by 1.26 mm and the backlash between pushing and pulling is 2.5 mm. That is travel at the lever that moves nothing at the brake — part of the dead band a rider feels before the pads touch — and it is set by the liner’s bore and by how much the cable turns, not by how carefully its radii were chosen. A cable routed internally through a frame, which typically adds turning, adds to it; a shorter route with the same number of bends does not reduce it.

Six routes, one law #

Six routes were built, each as a list of straights and bends, and the inner’s path computed as a taut run for each. Five of them turn through a total of 180° — one U-bend, two quarter-turns the same way, an S of a quarter-turn each way, a U on a radius a third as large, and four eighth-turns alternating left and right — and the sixth through 270°.

At a clearance of 1, on routes some four hundred units long, the shortfall divided by $c\sum|\varphi|$ ranges from 0.948 to 0.998 and the pushed excess from 1.002 to 1.055. As the clearance falls by factors of three and ten the ratios close on one, and at a clearance of 0.01 every route is within 0.05% of the law, pulled and pushed. The departure falls in proportion to the clearance — the second-order term — and is largest on the route with the most, shortest straights, where the inner cutting from the inside of one bend to the inside of the next is furthest from parallel to the centreline.

The radius does not enter #

The claim that is most useful to someone routing a cable is also the most counterintuitive: a tight bend costs no more than a generous one. It is the same separation a drum’s size from a wrap’s shape makes for a strand on the outside of a body: angles decide, lengths scale. One 90° bend was built on six radii from 15 to 140, a ninefold range in which the centreline’s arc goes from 23.6 to 219.9, with a clearance of 0.25.

The pulled inner’s shortfall is 0.392282 on the tightest bend and 0.392283 on the widest — a spread of $2 \times 10^{-7}$. The law $c\varphi$ gives 0.392699. The remaining 0.0004 is not in the bend at all: it is where the inner leaves the ferrules at the two ends, which are on the centreline, and has to cross from the middle of the bore to the inside wall. That crossing depends on the length of the straight run, and not on the bend.

So what makes a Bowden cable spongy is not a tight radius. A tight radius has other costs — friction on the wall rises sharply with the wrap angle and the pressure of a small radius, and a coil sheath compresses more under a tight bend — but its geometric lost motion is the same as a gentle one’s through the same angle.

An S-bend is not free #

The sum in the law is of absolute angles, and the reason is visible in the route figure. At a left-hand bend the inner hugs the left wall; at a right-hand bend it hugs the right wall; each bend shortens the path by $c$ times its own angle, and the two shortenings add. They do not cancel, because the inner is on the inside of whichever bend it is in.

The table makes it concrete. The S-bend, a quarter-turn left then a quarter-turn right, has a net turning of nought — the sheath leaves in the direction it arrived — and its pulled inner is short by 0.7832, against 0.7849 for the U-bend that turns back on itself. The four alternating eighth-turns, net nought again, lose 0.7751. A law written with the sheath’s net turning would predict nothing lost for either.

That separates the two quantities the strand field already has names for. The turning number of a closed strand’s wraps is a signed sum and a topological invariant: it cannot change without the route changing its type. The lost motion of a strand in a tube is an unsigned sum and changes continuously with every bend. One counts how many times a route goes round; the other counts how much it bends.

Steering takes up or releases the brake #

The sheath’s length does not change when a handlebar turns, but its route does, and so does the angle it bends through.

The route in the figure has a 60° bend near the bar, which opens and closes as the bar steers, and a fixed quarter-turn the other way further down. As the bar turns through ±40° the first bend’s angle changes by the steering angle, the total turning changes with it, and the pulled inner’s shortfall changes by $c\,\Delta\varphi$: 0.1745 at 40° with a clearance of 0.25, matched by the law to $10^{-13}$, because nothing but one arc’s angle has changed.

The consequence is familiar to anyone who has set up a front brake with a tight cable. Turning the bar one way adds bend angle, the inner needs less length to reach the brake, and the brake slackens by $c\,\Delta\varphi$; turning the other way removes bend angle and takes it up. With a clearance of a quarter of a millimetre and a steering angle of 40°, that is 0.17 mm at the brake — small on a rim brake, and enough on a mechanical disc brake set up with a small pad gap to make the lever’s biting point move perceptibly as the bar turns.

What this model leaves out #

Friction. A pulled inner presses on the inside of every bend with a force proportional to its tension and wraps through the bend’s angle, so the capstan equation applies and the tension lost to friction grows exponentially with the total angle turned. That is the Bowden cable’s other defect and usually the larger one, and it is a force, which is outside this field; what is notable is that it too depends on the angle and not on the radius.

Sheath compression. A coil sheath shortens under the compressive load the tension puts on it, and more so where it is bent. Real lost motion is the geometric term here plus that elastic term, and a lined, compressionless sheath exists precisely to remove the second.

The third dimension. A real cable route is not planar. A helical sheath route — a cable wrapped round a frame tube — turns through an angle that is the sum of its curvature along its length, and the inner’s shortfall is still $c$ times that total by the same argument applied to the osculating plane at each point, which the wrap that walks along the axis makes plausible and this essay does not measure.

A clearance that varies. The law uses one $c$; a cable whose inner is worn thinner in places has a clearance that varies along its length, and the shortfall becomes the integral of clearance against turning.

What a designer takes from it #

The law puts one number on a routing drawing: the total angle the sheath turns through, added up without sign. Two cables with the same clearance and the same total turning have the same geometric lost motion whatever their radii and shapes. To reduce it, turn less — not more gently. And to stop steering moving the brake, route the sheath so the bend that moves with the steering is balanced by one that moves the opposite way, which leaves the total turning unchanged as the bar turns — something no amount of generous radius achieves.

It also completes a picture the field has been drawing. A ratio that is a derivative of a length found that a tackle’s advantage is how fast a strand’s length changes with a body’s position; here the body is the sheath’s own route, and the strand’s length changes with it at a rate of $c$ per radian of bend.

Still open: the active set when the strand does not touch every wall #

The model here assumes the pulled inner touches the inside of every bend. On a route with two bends the same way close together, the taut path from the inside of the first to the inside of the second may lift off the wall between them and pass straight through the middle of the bore — the tube holds it only where it would otherwise leave. Where a strand stops touching found that a body joining a run costs length as the square of how far it intrudes, and here the question is which walls are in contact at all.

Its distinct argument would be that active set: for a given route and clearance, which bends’ inside walls the taut inner touches, which it bridges, and how the lost motion departs from $c\sum|\varphi|$ once a bend is bridged. Two things would come out of it. A condition, in terms of the straight length between two bends and the clearance, below which two bends behave as one — so that closely spaced small bends cost less than the law says, which is the departure the four-bend route already shows at large clearance; and whether the set of touched walls changes monotonically as the clearance grows, or whether, as for the taut path with more than one answer, a strand in a tube can have two shortest routes and jump between them.