The path a towed wheel takes

Attach an axle to a point that is being dragged along, with a rigid rod between them, and let the axle roll without sliding. That is a trailer, a caster, a bicycle’s back wheel, the rear axle of an articulated lorry, and the toy a child drags on a string, and all of them do the same thing: the towed wheel cuts every corner the towing point turns.

The whole mechanism is one equation, and the interesting part of this essay is which of the things about it are consequences rather than assumptions.

The equation #

Let the hitch be at $h(t)$ and the axle at $a(t)$, with the rod of length $L$ between them, and let $\theta$ be the axle’s heading. The rod is rigid and points from the axle to the hitch, so

$$a = h - L\,(\cos\theta,\ \sin\theta).$$

Differentiate and impose the rolling condition — the axle’s velocity must lie along its own heading, since it may not slide sideways. Resolving the result along the heading and across it gives two statements that are each one line:

$$\dot{\theta} = \frac{\dot h \cdot \hat\theta^{\perp}}{L}, \qquad v = \dot h \cdot \hat\theta,$$

where $\hat\theta$ is the axle’s heading and $\hat\theta^{\perp}$ is across it. The heading turns at the hitch’s velocity across the rod, divided by the rod’s length; the axle’s own speed is the hitch’s velocity along the rod. Everything anybody says about off-tracking, cut-in and swept width is a consequence of those two lines, and none of it needs a force.

What is integrated, and what is not #

The integrator carries three numbers: the axle’s two coordinates and its heading. It does not carry the rod length, and it is never told to keep it.

That is deliberate and it is the difference between a computation and a construction. The obvious way to draw a trailer is to put the axle exactly $L$ behind the hitch at every step, in which case the rod is the right length because it was placed there and the picture proves nothing. Here the rod length is whatever the equation leaves it at.

What it leaves it at, over a 6 m wiggling path at 120 samples with 24 sub-steps each, is $L$ to within $9\times10^{-14}$ m. The rod length is a conserved quantity of the differential equation — differentiating $|h - a|^2$ and substituting the two lines above gives zero identically — and the number quoted is the accumulated rounding of a few thousand fourth-order steps.

That matters for the same reason the site cares about it everywhere: a figure whose defining property was imposed cannot fail, and a check that cannot fail is not a check. Here the constant length is a result, and if the heading equation had a sign wrong or a factor of $L$ misplaced the rod would visibly grow or shrink along the path.

The tractrix #

Give the hitch a straight line to travel along, starting with the rod perpendicular to it, and the curve the axle traces is the tractrix — the curve whose tangent, extended to a fixed line, always has the same length. It is the same curve Huygens studied, the one whose surface of revolution is the pseudosphere, and it arrives here not because anybody went looking for it but because it is what the equation produces.

It has one property worth having and it is the defining one: the tangent length is constant. That is exactly the statement checked above, so the numerical result and the classical description are the same fact.

Give the hitch a circle instead and the axle settles onto a concentric circle. Which one is decided by a right triangle: the rod is tangent to the inner circle, so

$$R_{\text{rear}} = \sqrt{R^2 - L^2}.$$

No calculus at all. The rod, the inner radius and the outer radius are the three sides.

The residual, which is not what it looks like #

The integration and the closed form disagree, and the disagreement is worth an argument rather than a tolerance.

Take $R = 12.5$ m and $L = 6$ m. Pythagoras gives $R_{\text{rear}} = 10.965856$ m. The integration, run for three turns and averaged over the last fifth, gives $10.965809$ m — off by $4.7\times10^{-5}$, which is far above the $10^{-13}$ the rod length is conserved to and would ordinarily be read as a defect in the law.

It is not the law and it is not the integrator. It is that the towing point is given to the integrator as a polygon, sampled at a few thousand points around the circle, and an inscribed polygon of $n$ sides has a radius short of the circle’s by $1 - \cos(\pi/n)$. That deficit propagates to the inner radius through the square root, and it lands exactly where the residual is.

The test that settles it is a refinement rather than a threshold. Double the sampling and a polygon deficit falls by four; an integration error of a fourth-order scheme falls by sixteen; a modelling error does not fall at all. Measured:

$$\frac{\text{error at }1{,}500\text{ points}}{\text{error at }3{,}000\text{ points}} = 3.99995.$$

Four, to five figures. The residual is the polygon, and the law is exact.

That is a small result and it is the kind this site keeps having to make, because the alternative is a tolerance chosen to make a check pass — which converts a measurement into a decision about what counts as close enough, and hides whatever else might be wrong.

Chains: a lorry is the same line, three times #

Each unit’s hitch is the unit in front of it, so a tractor with a trailer is the same equation applied twice and nothing is added at each stage except another sine. The settled radii are nested square roots.

For a tractor unit whose kingpin sits 5.4 m behind its front axle, pulling a semi-trailer with 7.7 m from kingpin to axle, on the 12.5 m outer circle:

$$12.500 \;\to\; 10.489 \;\to\; 7.122 \text{ m},$$

a total cut-in of 5.38 m. The integrated radii and the closed-form ones agree to a hundredth of a millimetre once the sampling is fine enough to stop the polygon showing.

The comparison with the regulation is worth being careful about, because the numbers being compared are not quite the same numbers. The regulation constrains the outermost and innermost points of the vehicle; the computation above gives the paths of its axles. A trailer 2.55 m wide has its inner edge 1.275 m inboard of its axle centre, which puts the innermost swept point at about 5.85 m against a regulation floor of 5.3 m — inside the limit, with the margin the designer of the trailer chose by putting the axles where they are.

What the arithmetic does say without qualification is where the sensitivity is. The cut-in depends on the kingpin-to-axle distance through a square root, so lengthening a trailer costs more cut-in per metre the longer it already is, and moving the axles back on a trailer — which is done for load distribution — is paid for in swept path. That is a real design trade and it is one line of algebra.

The sum of squares, and why a lorry is jointed #

The nesting is worth carrying one step further than the two units above, because the general form is short and it settles a question about vehicle layout that looks as though it ought to need a simulation.

Each towed unit takes the circle it is given and returns a smaller one, $\sqrt{R^2 - L^2}$. Apply that twice and the inner square roots collapse: a unit of length $L_1$ followed by one of length $L_2$ settles on $\sqrt{R^2 - L_1^2 - L_2^2}$, and $n$ units settle on $R$ with the sum of the squares of every unit length taken out from under the root. Nothing survives of the individual lengths but their squares, and nothing survives of their arrangement at all.

Two consequences, and both are exact rather than approximate.

Order does not matter. A long trailer behind a short tractor tracks on precisely the same circle as a short trailer behind a long tractor, provided the two lengths are the same pair. That is not obvious and it is not what a swept-path drawing looks like — the intermediate circles differ, the transient behaviour differs, and the steady state does not.

And splitting a length is an enormous improvement. Sum of squares is not sum, so dividing one long unit into two shorter ones with the same total reduces what comes out from under the root. Take this essay’s own rig on its 12.5 m circle: axle to kingpin 5.4 m and kingpin to trailer axle 7.7 m give $\sqrt{156.25 - 29.16 - 59.29} = 8.23$ m. Weld the joint and the same 13.1 m becomes one rigid wheelbase, the quantity under the root is $156.25 - 171.61$, and there is no answer at all — the vehicle cannot hold that turn. The rigid version does not track worse; it does not exist.

That is the arithmetic reason a long road vehicle is articulated, and it is a good deal sharper than the usual account about manoeuvrability. A rigid vehicle’s steady turn requires its wheelbase to be shorter than the radius, which is the floor the hitch-angle essay reaches from the other direction; a jointed one of the same total length faces the sum of squares instead, and the sum of squares of two halves is half the square of the whole. Halving is the best case, when the joint is in the middle, and every other position is somewhere between.

It also says where a designer’s freedom actually is. The total length is set by what has to be carried and is not negotiable. The number of joints and where they fall is negotiable, and the expression says exactly what each choice is worth — a second joint dividing a 13.1 m rig into three roughly equal units puts about 57 m² under the root instead of 88, which on the same 12.5 m circle moves the inner axle from 8.23 m out to 9.97 m. Two joints buy nearly two more metres of clearance on a roundabout, from the same vehicle carrying the same load.

None of that is a claim about the swept band a real lorry needs, which is a question about corners and overhangs rather than about axles, and the comparison with any regulation has the qualifications already made above. It is a claim about where the axles run in a steady turn, it is exact, and it comes from applying one line twice.

What the equation says about the straight parts #

Two consequences fall straight out of $\dot\theta = (\dot h \cdot \hat\theta^\perp)/L$ and are worth stating because they are the ones that get misremembered.

Off-tracking is not a lag. A trailer does not simply arrive late at where the tractor was; it takes a different route. On a straight section the two paths coincide exactly, and the difference appears only where there is curvature. A model in which the trailer follows the tractor’s path with a delay would predict identical paths on a curve too, and it would be wrong by metres.

Off-tracking has no memory once the curve is steady. On a circle the axle settles onto its own circle and stays there, at $\sqrt{R^2 - L^2}$ regardless of how long the turn continues. The settling is exponential and its length scale is the rod’s own length, which is the subject of another essay because the same exponential runs the other way in reverse.

And nothing about speed appears anywhere. The heading equation has $\dot h$ on the right and $L$ underneath, so scaling the speed scales $\dot\theta$ by the same factor and the path is unchanged. A lorry taking a roundabout slowly sweeps exactly what it sweeps taking it quickly. Whatever changes with speed is a force question and is not in this field.

Why the nesting is square roots and not subtraction #

The nested form is worth one more paragraph because the wrong version of it is the one most people carry.

A plausible guess is that each towed unit cuts in by an amount proportional to its own length, so two units cut in by the sum. That guess is linear and the truth is not. Each stage takes a square root of the difference of two squares, so the cut-in of a stage depends on where the stage starts as well as on how long it is: the same 7.7 m trailer costs 3.37 m of cut-in when hitched behind a tractor already at 10.49 m, and would cost only 2.65 m if it were hitched to something running at 12.5 m.

Two consequences follow and they point in opposite directions. Adding units to the back of a road train costs more each time, because each new unit starts on a tighter circle than the last. And a rig taking a wider turn cuts in by less — not proportionally less, but much less, since $R - \sqrt{R^2 - L^2} \approx L^2/2R$ for a long enough turn. Doubling the turn radius roughly halves the cut-in, which is why a lorry driver’s answer to a tight corner is to take it wide, and why a mini-roundabout on a lorry route has a mountable centre.

Where the sine goes #

Written in terms of the angle rather than the vectors, the heading equation is

$$\dot\theta_1 = \frac{v}{L}\sin(\theta - \theta_1),$$

with $\theta$ the tractor’s heading and $\theta_1$ the trailer’s. Three readings of that sine are worth having.

Zero at zero. In line, the trailer’s heading does not change, which is why a straight tow is stable and requires nothing of the driver.

Maximum at ninety degrees. With the rig folded to a right angle the trailer’s heading changes as fast as it can — $v/L$ per unit distance — and no steering input can make it change faster. That is the geometric fact underneath a jackknife and it is not the same as being stuck.

Zero again at a hundred and eighty. A trailer folded exactly back on itself has no tendency to fold further or to unfold. It is an equilibrium, and reversing makes it a stable one, which is why a rig that has jackknifed while reversing stays jackknifed.

The caster, which is the same mechanism upside down #

A caster wheel — the swivelling wheel on a trolley, a chair or a hospital bed — is a towed axle with a very short rod. The rod is the trail: the offset between the swivel axis and the wheel’s contact point, and it is why a caster points itself the way it is being pushed.

Everything above applies with $L$ a few centimetres instead of a few metres. The settling length scale is the trail, so a caster with 30 mm of trail lines itself up within a few centimetres of travel; the off-tracking is $\sqrt{R^2 - L^2}$, which for a trail that small is indistinguishable from $R$; and the reversing instability is present and is exactly why a trolley pushed backwards has its casters flap round.

The design consequence is the one a hospital bed’s designer cares about: more trail means straighter tracking and a bigger swept path, and the two cannot be separated because they are the same length. A caster with no trail at all does not swivel to follow at all — it is a wheel bolted on, and the mechanism disappears.

What the towed axle leaves behind #

The tracks in the pictures above are not decoration; they are the object of the next essay. A towed axle’s track is a curve that stands in a very specific relation to the towing point’s track — the tangent, extended by the rod length, lands on it — and that relation is strong enough to answer a question about a mechanism that has driven away and left nothing behind but marks.

It is the closest this site comes to an inverse problem in the ordinary sense: not given the lengths, find the motion and not given the motion, find the lengths, but given the marks, find out what made them.