The motion left over by going nowhere

The last essay used the Lie bracket as a test: compute it, see whether it lies inside the plane of permitted directions, and the answer decides whether the mechanism is confined. That is a correct use of it and it makes the bracket sound like an instrument for classifying things.

It is not primarily an instrument. It is a motion, and the mechanism performs it.

The manoeuvre #

Take a wheel pointing along the $x$ axis at the origin. Do four things:

1. Drive forward a distance $\varepsilon$.
2. Turn on the spot through an angle $\varepsilon$.
3. Drive backward the same distance $\varepsilon$.
4. Turn back through the same angle $\varepsilon$.

Every leg is exactly undone by another leg. The distances cancel, the angles cancel, and the wheel is pointing exactly the way it started. It is not where it started.

The arithmetic is short enough to do in full. After the first leg the wheel is at $(\varepsilon, 0)$. The turn moves nothing. The third leg drives backward along the new heading, which is $\varepsilon$ rather than $0$, so it subtracts $\varepsilon(\cos\varepsilon, \sin\varepsilon)$ rather than $(\varepsilon, 0)$. The last turn moves nothing again. So the wheel ends at

$$\big(\varepsilon - \varepsilon\cos\varepsilon,\ -\varepsilon\sin\varepsilon\big) \;=\; \left(\tfrac{1}{2}\varepsilon^3,\ -\varepsilon^2\right) + O(\varepsilon^4).$$

The dominant term is $-\varepsilon^2$ in the $y$ direction — across the wheel, which is the direction the constraint forbids — and it is second order in the size of the manoeuvre.

That is the whole phenomenon, and everything else in this field is a more complicated instance of it.

What the leftover is #

Write the two permitted directions as fields on the configuration space:

$$F = (\cos\theta,\ \sin\theta,\ 0), \qquad G = (0,\ 0,\ 1).$$

The bracket is

$$[F, G] \;=\; \frac{\partial G}{\partial q}F - \frac{\partial F}{\partial q}G,$$

and since $G$ is constant the first term vanishes. The second is the derivative of $F$ with respect to $\theta$, which is $(-\sin\theta, \cos\theta, 0)$, so

$$[F, G] = (\sin\theta,\ -\cos\theta,\ 0).$$

At $\theta = 0$ that is $(0, -1, 0)$: pointing in $-y$, exactly where the manoeuvre went, and with the coefficient the expansion above produced. The theorem behind it is that the four-leg manoeuvre leaves $\varepsilon^2[F, G]$ plus a third-order remainder, and the two calculations here are the same statement arrived at from opposite ends — one by differentiating fields, one by composing flows.

Measured rather than differentiated #

The derivation above is a derivation, which on this site is not the same as a measurement. So the manoeuvre is flown: four legs of a fourth-order integrator, at seven amplitudes from 0.40 down to 0.05, with the net displacement recorded each time.

Two things are then checked, and they fail differently, which is why both are made.

The direction. The measured displacement is compared with $\varepsilon^2[F, G]$ as a vector. The angle between them is $1.43°$ at $\varepsilon = 0.05$ and $0.29°$ at $\varepsilon = 0.01$ — proportional to the amplitude, which is exactly what a third-order remainder against a second-order leading term does. A sign error anywhere in the bracket would show as an angle of 180° rather than as a slightly wrong size, which is a failure mode this site has met before: a velocity solve that had double-negated its right-hand side produced velocities of exactly the right magnitude pointing exactly backwards, and every figure still drew perfectly.

The size. The magnitudes are fitted against the amplitudes in logarithms. The slope comes out at

$$1.997,$$

against a predicted 2, on a fit that was never told what to expect. A first-order effect would give 1 and an integration error would give 4.

Why second order matters more than it sounds #

An exponent of two is the difference between a mechanism that can be nudged sideways and one that effectively cannot.

Suppose a wheel needs to move one metre sideways and has $\varepsilon$ metres of room to shuffle in. Each manoeuvre wins about $\varepsilon^2$ metres, so the count is about $1/\varepsilon^2$ and the distance driven is about $4\varepsilon \times 1/\varepsilon^2 = 4/\varepsilon$. With a metre of room that is four metres of driving; with ten centimetres it is forty; with one centimetre it is four hundred. The distance driven grows as $1/\varepsilon$ and the number of manoeuvres as $1/\varepsilon^2$, and both go to infinity as the room goes to zero.

So the constraint is not a wall and it is not nothing. It is a price that goes up steeply, and every practical fact about manoeuvring wheeled things is a statement about that price. Parking a car is the everyday one, and the car’s exponent is worse than the wheel’s for a reason worth an essay.

The step, and why the two routes need different ones #

Both routes are numerical and both have a step in them, and the steps have to be chosen against opposite failure modes. That is worth being explicit about, because the site’s habit of demanding two routes is worth nothing if the two routes are the same computation with different labels.

The differentiated bracket uses a central difference on the fields, at $10^{-5}$. Too large and the second derivative shows through; too small and the subtraction loses its digits to rounding. At $10^{-5}$ on fields whose scale is one, both errors sit near $10^{-10}$.

The measured bracket uses a step of $\varepsilon$, and $\varepsilon$ is not an error term at all — it is the manoeuvre’s actual amplitude, a physical quantity, and making it small is the opposite of what a mechanism wants. The measurement’s error is the third-order remainder, which is a property of the manoeuvre rather than of the arithmetic.

The two therefore disagree at exactly the rate they should, and the rate is worth quoting because it is so clean: the relative departure is $\varepsilon/2$, to three figures, over the whole range. 2.50% at $\varepsilon = 0.05$, 9.99% at $0.2$, 19.91% at $0.4$. An agreement that did not degrade with $\varepsilon$ would mean the two routes were secretly one route.

What one manoeuvre is worth, in metres #

The fitted law is $0.993\,\varepsilon^{1.997}$ for a wheel whose two controls are scaled so that driving is metres and turning is radians, and it is worth reading off a few of its values rather than leaving it as an exponent.

The right-hand column is the exchange rate — the gain divided by the $4\varepsilon$ of driving that bought it — and it is simply $\varepsilon/4$. Every halving of the room halves what each metre driven is worth and halves how many metres there are in a manoeuvre, which is where the square comes from.

The middle column is the one to keep. A shopping trolley with forty centimetres of aisle wins sixteen centimetres a shuffle; with five centimetres it wins two and a half millimetres. Anybody who has manoeuvred a trolley in a full supermarket has measured this quantity, unwillingly.

The other three coordinates #

The manoeuvre moves the wheel by $-\varepsilon^2$ across itself and by $\tfrac{1}{2}\varepsilon^3$ along itself, and the second of those is not a rounding error either — it is the next term in the same expansion, and it fits an exponent of 2.994.

That matters because it is the first sign of something the next essay is about. Different directions cost different powers of the amplitude. The sideways direction, reachable with one bracket, costs $\varepsilon^2$; the along direction, which needs the bracket taken again, costs $\varepsilon^3$. The exponent is not a property of the mechanism; it is a property of the direction, and it counts brackets.

What a metre sideways actually costs #

The exchange rate $\varepsilon/4$ is a rate per manoeuvre, and a rate per manoeuvre is hard to feel. The useful form is the total, and it follows from the fitted law with no further measurement.

One manoeuvre wins $0.993\,\varepsilon^{1.997}$ across the wheel, call it $\varepsilon^2$, and returns the heading exactly. Displacements therefore accumulate: $n$ repetitions win $n\varepsilon^2$, and a sideways metre needs $n = 1/\varepsilon^2$ of them. Each costs $4\varepsilon$ of driving, so the whole journey is

$$4\varepsilon \times \frac{1}{\varepsilon^2} = \frac{4}{\varepsilon}$$

metres driven for one metre gained. With half a metre of room to shuffle in, eight metres of driving. With a tenth of a metre, forty. With five centimetres, eighty. The cost is inversely proportional to the room available, and it is the room and not the skill that decides it — there is no cleverer sequence, because the second-order law applies to whatever manoeuvre is used and the largest legs the space allows are always the best buy.

That is the sharp form of what a nonholonomic constraint costs, and it is neither of the two things it is usually taken to be. It is not a prohibition: the metre is reachable from any starting room whatever. And it is not a mild inconvenience: the driving required diverges as the room closes, so a mechanism in a tight enough space is prevented by arithmetic rather than by geometry.

The along-track drift comes out of the same sum and runs the other way, which is worth noticing because it is a rare case of the small manoeuvre being better. Each repetition also moves the wheel $\tfrac{1}{2}\varepsilon^3$ along itself, so over the $1/\varepsilon^2$ repetitions of a sideways metre the accumulated drift along the wheel is $\varepsilon/2$ metres — five centimetres for a tenth-metre shuffle, and two and a half for a twentieth.

So the two costs of parking a wheel pull in opposite directions. Shorter legs mean more total driving and a cleaner result; longer legs mean less driving and a larger unwanted displacement along the wheel that has to be taken out afterwards. Neither is a defect in the manoeuvre. Both are terms in one expansion, read at the order each of them lives at, and the fact that they scale as $1/\varepsilon$ and $\varepsilon$ is the reason there is a trade at all rather than a best answer.

One assumption in that sum is worth making explicit, because it is what makes the repetitions add rather than average. The manoeuvre returns the heading exactly, so the wheel begins each repetition in the same orientation it began the last one, and the displacement each repetition wins points the same way in the world. A sequence of them is therefore coherent by construction: the gains sum instead of cancelling, and nothing has to be steered to keep them aligned.

That is not an accident of this particular manoeuvre, and it is what the antisymmetry of the bracket is worth in practice. Reversing the order of the legs reverses the displacement, so the same four legs run the other way round move the wheel the other way — which means a driver has both signs available, at the same price, from the same room. Sideways in either direction costs $4/\varepsilon$, and the choice between them is a choice of ordering rather than of manoeuvre.

The arithmetic also explains a familiar asymmetry. Getting a car out of a tight parking space feels disproportionately harder than getting it into a merely snug one, and the ratio $4/\varepsilon$ is why: halving the available room does not add half as much work again, it doubles the work, and the doubling continues without limit. Every driver has met the second-order law, and it is a good deal less abstract from behind a steering wheel than it looks as a bracket.

The bracket in mechanisms that are not wheels #

The bracket is not a wheeled-mechanism idea, and this site has already produced one instance of it without calling it that.

A redundant arm asked to take its tool round a closed loop brings the tool back and does not bring the joints back. The tool traverses a circle in the task space and returns exactly; the joints have moved a third of a degree along the arm’s self-motion curve and stay moved. That is the same structure: a closed loop downstairs lifts to an open path upstairs, and the discrepancy is a bracket of the fields that generate the loop.

The difference is what the discrepancy is for. In the arm it is a nuisance — the resolution is not cyclic, and a machine repeating a cycle drifts. Here it is the mechanism’s only way of reaching anything, and a machine repeating a cycle is parking.

The other instance is the one everybody has performed. A cat dropped upside down lands on its feet without any external torque, by changing its shape in a closed loop; the shape comes back and the orientation does not. A ball rolled round a closed loop on a table comes back turned by the loop’s area, which is the same thing again with an exact closed form attached to it.

What the bracket is not #

Three misreadings are worth heading off, because each of them makes the field sound either easier or more mysterious than it is.

It is not a violation of the constraint. Every instant of the manoeuvre satisfies the constraint exactly; the residual over the drawn history is at the floor of double precision. Nothing slides. The sideways displacement is produced entirely by motions that are not sideways, and the mechanism at no point does anything it is not allowed to do.

It is not a trick that gets something for nothing. The manoeuvre costs $4\varepsilon$ of driving to win $\varepsilon^2$ of displacement, so the exchange rate is $\varepsilon/4$ — bad, and getting worse as the manoeuvre gets smaller. Nothing here says a wheeled mechanism moves sideways cheaply. It says it moves sideways at a price that can be computed.

It is not an approximation. The $\varepsilon^2$ is the leading term of an expansion, but the reachability it demonstrates is exact: the manoeuvre really does end where it ends, and repeating it really does accumulate. The approximation is in the formula for how much, not in whether.

The order of the legs #

One detail of the manoeuvre repays attention because it is the part that is easiest to get wrong and hardest to notice.

The four legs are drive, turn, drive-back, turn-back. Reversing the order of the middle pair — drive, drive-back, turn, turn-back — gives a manoeuvre in which every leg is immediately undone by the next, and the mechanism comes home exactly. Nothing is gained. The gain comes from the interleaving, and specifically from the fact that the second drive happens at a heading the first drive did not have.

That is why the bracket is antisymmetric: $[G, F] = -[F, G]$, and doing the manoeuvre the other way round moves the wheel the other way. It is also why a mechanism with only one control gains nothing, however long it manoeuvres — there is no second thing to interleave with, and a single field has no bracket with itself.

Both of those are checkable statements about the drawn mechanism rather than facts about notation, and the second is the one that decides cases. A car with its steering clamped is a mechanism with one control, and it runs along an arc and stays there.

Where this goes #

The bracket answers whether the forbidden direction can be reached and how much one manoeuvre wins. It leaves two questions, and they are the next two essays.

The first is how deep the process goes. A bracket of two fields may point outside the plane they span; the bracket of that with a field may point outside the larger space; and the sequence of dimensions this generates is an integer signature of the mechanism. A wheel’s is short. A car with a trailer’s is longer, and every step of it costs another power of the amplitude.

The second is what any of it is worth in metres. The manoeuvre in this essay uses turning on the spot, which a car cannot do, so the arithmetic above is the arithmetic of a shopping trolley rather than of a vehicle. The version with arcs instead of pivots has an exact closed form, a measured exponent, and a count of how many shuffles a tight bay actually takes.