How many wiggles

A wheel’s two permitted directions span a plane, and their bracket points out of it into the third dimension. Three dimensions, filled. The process stops because there is nowhere left to go.

Most mechanisms are not that lucky. A car has four coordinates and two controls, so even after one bracket there are only three directions accounted for and a fourth to find. It is found by taking the bracket again — of the bracket with one of the original fields — and the question of how many times that has to be done is the subject here.

The construction #

Write $\Delta_1$ for the span of the permitted directions at a configuration — the plane, for a wheel; also a plane, for a car, since a car has two controls too.

Write $\Delta_2$ for the span of $\Delta_1$ together with every bracket of a $\Delta_1$ direction with a $\Delta_1$ direction. Then $\Delta_3$ for the span of $\Delta_2$ together with every bracket of a $\Delta_2$ direction with an original field. And so on.

Each $\Delta_k$ contains the one before it, so the dimensions can only rise or stay put, and once they stay put they stay put forever. The sequence of dimensions

$$\dim\Delta_1,\ \dim\Delta_2,\ \dim\Delta_3,\ \dots$$

is the growth vector, and there are exactly three things worth knowing about it.

It starts at the number of independent controls — what the constraint rows leave.

It ends at the dimension of the reachable set. That is Chow’s theorem, and it is the piece of mathematics this whole field leans on: if the brackets eventually fill the space then the mechanism can be driven anywhere in it. The manoeuvre of the previous essay is the constructive half — a bracket direction is not merely spanned, it is travelled.

Its length is how many nested manoeuvres are needed, and that is the number this essay is really about, because it is the one with a physical consequence.

What the mechanisms return #

A wheel: 2, 3. Two controls, one bracket, done. The bracket is the forbidden direction itself, so one manoeuvre reaches everything.

A car: 2, 3, 4. The controls are drive and steer, where steer changes the steering angle rather than setting it. Their bracket is not the sideways direction at all — it is a pure change of heading, a turn on the spot, which is not something a car can do directly. Taking the bracket again, of that with the drive field, gives the sideways direction. So a car reaches sideways only at the second level.

That is the formal statement of something every driver knows and no driver would put this way: a car cannot turn on the spot, and it cannot go sideways, and those are different degrees of difficulty rather than two instances of the same difficulty. Turning on the spot is one manoeuvre away; going sideways is two.

A car and trailer: 2, 3, 4, 5. Five coordinates, still two controls, three brackets. Each additional towed unit adds a coordinate and a level, so a lorry with two trailers needs four and a road train needs five. Nothing about the mechanism gets harder except the depth, and the depth is what costs.

A ball on a plane: 3, 5. Five coordinates and three controls, because a ball resting on a plane may be spun about the vertical through its contact point without sliding. Three directions, one round of brackets, and the remaining two appear at once.

A ball that may not be twisted: 2, 3, 5. Forbid the spin — which is what happens to a ball held between two plates, or under a fingertip — and the same object has two controls instead of three, needs two rounds of brackets instead of one, and the second round produces two new directions rather than one.

That last pair is the sharpest thing in the table. The two mechanisms are the same ball on the same plane. Nothing has changed about the geometry, the radius, the contact or the coordinates. One constraint has been added, the count of freedoms has gone from three to two, and the growth vector has gained a level. The difficulty of manoeuvring a ball is not a property of the ball.

The second route, which knows nothing about ranks #

Everything above is linear algebra on derivatives. The site’s standing habit is that a number computed one way is worth what a second computation says about it, and there is a second computation here that shares nothing with the first.

Fly the four-leg manoeuvre and fit the exponent of each coordinate separately against the amplitude.

For a wheel: the sideways coordinate comes out at $1.99$.

For a car: the heading comes out at $2.02$ and the sideways coordinate at $3.02$.

Those are the depths, and the fit was given no integers and told nothing about brackets. A direction that first becomes available at level $k$ of the filtration moves by $\varepsilon^k$ under a manoeuvre of amplitude $\varepsilon$. A wheel’s sideways coordinate is new at level 2 and moves as $\varepsilon^2$; a car’s heading is new at level 2 and does the same; a car’s sideways coordinate is new at level 3 and moves as $\varepsilon^3$.

The rule holds for the directions that are new and not for the others, and the exception is worth stating because leaving it out would make the correspondence look tidier than it is. A wheel’s along-track coordinate is available at level 1 — driving forward is exactly that — and it does not move as $\varepsilon^1$ under the manoeuvre. It moves as $\varepsilon^{2.99}$, and a car’s moves as $\varepsilon^{5.05}$.

The reason is that the manoeuvre is built to cancel the directions it starts with. Its four legs undo each other in that direction by construction, so what survives there is not a bracket at all but the leftover of a deliberate cancellation, and its order is a fact about the manoeuvre’s symmetry rather than about the mechanism. The exponents that mean something are the ones belonging to directions the manoeuvre had no way of producing directly, and those are exactly the new ones at each level.

The two routes agree, and they are as independent as any pair on this site. One differentiates fields at a point and takes ranks; the other integrates trajectories over finite intervals and fits a slope in logarithms. Neither can be derived from the other’s output, and a mistake in either shows up as an integer that does not match.

What the depth costs #

Reading the growth vector as a cost is the reason to have it, and the arithmetic is one line.

A manoeuvre of amplitude $\varepsilon$ costs about $4\varepsilon$ of driving and wins about $\varepsilon^k$ in a direction of depth $k$. To gain a fixed distance $\delta$ in that direction therefore takes about $\delta/\varepsilon^k$ manoeuvres and about $4\delta/\varepsilon^{k-1}$ of driving.

For $k = 2$ the driving grows as $1/\varepsilon$: halve the room and drive twice as far. For $k = 3$ it grows as $1/\varepsilon^2$: halve the room and drive four times as far. That is the difference between a trolley in a supermarket aisle and a car in a tight bay, and it is a difference of one in an integer nobody measures directly.

The next essay turns this into metres, because the manoeuvre used here — drive, turn on the spot, drive back, turn back — is not one a car can perform, and the version built from arcs has a closed form worth having.

What happens in between #

The construction says nothing about mechanisms whose brackets stop growing before they fill the space, and it is worth saying what that case looks like rather than leaving it as an unmentioned possibility.

If the sequence stops at $r < n$, the mechanism is confined to a family of $r$-dimensional surfaces — not to a single one, as the trolley on a rail is, but to whichever member of the family it started on. Frobenius’ theorem is the case $r = \dim\Delta_1$: the sequence never grows at all. Everything between is possible in principle, and none of it occurs in this field, where every mechanism either fills its space or is the trolley.

That is not a coincidence, and it is not a claim that partial cases are rare in general. It is a consequence of what the field is made of: rolling constraints on a plane, whose fields turn with the mechanism, and turning is what makes brackets point somewhere new.

The step, which had to be chosen against itself #

A bracket is a derivative, and a bracket of a bracket is a derivative of something that already carries a derivative’s error. That matters here more than anywhere else on this site, because the deepest mechanism in the table needs three nested rounds and the naive approach collapses well before that.

Central differences at $10^{-5}$ give a first bracket accurate to about $10^{-10}$. Differencing that at $10^{-5}$ divides the inherited error by $10^{-5}$ and returns $10^{-5}$ of noise, which is survivable; doing it once more returns numbers with no digits in them at all. Three levels at a single step size is not a computation.

So the step grows with the level — $10^{-5}$, then $2\times10^{-3}$, then $2\times10^{-2}$ — each chosen so that the truncation error the level adds and the amplified error it inherits are about the same size. That is a compromise and it is worth saying so plainly rather than presenting the ranks as exact.

What makes the ranks trustworthy despite it is that a rank is a decision about directions rather than about sizes, and the decisions here are not close. Every vector is normalised before the rank is taken, and at every level of every mechanism in the table the rows that were rejected were rejected because they were exactly zero — brackets that vanish identically, of which there are many — rather than because they were small. There was no marginal call anywhere in the table, which is the property that makes a coarse step acceptable and would make it unacceptable if it were not true.

The independent confirmation is the exponents, which are computed with no differencing at all.

Where the growth vector does not change, and why that is the finding #

The obvious place to look for a growth vector that collapses is a configuration where a mechanism is in trouble, and the obvious candidate is a car and trailer folded up.

It does not collapse. The growth vector of a car and trailer is $2, 3, 4, 5$ at a hitch angle of zero, at forty-five degrees, at ninety degrees, at a hundred and thirty-five, and at a hundred and eighty — the rig folded back on itself. Every configuration is still reachable from every other, and the mechanism is exactly as manoeuvrable, in the sense this essay measures manoeuvrability, as it is when straight.

Whatever a jackknife is, it is not a loss of reachability, and the essay that measures it has to find a different explanation — which turns out to be about the room the rig has rather than about its constraints.

What the first number means, and what it does not #

The growth vector’s first entry is the count of velocity freedoms, which is the number the mobility field spent its whole argument on. It is worth being careful about what the rest of the row does to that argument, because it would be easy to read this field as saying the mobility count was wrong.

It was not wrong. A car really does have two velocity freedoms; at any instant there are two things it can do and no more. What the row adds is that the count answers a question about instants, and there is a second question about intervals whose answer is a different number.

The two questions coincide on every mechanism outside this field, which is why the site never had to separate them. A four-bar’s velocity freedoms are 1 and its growth vector is just $1$: nothing to bracket, since one field has no bracket with anything, and the reachable set is the one-dimensional curve the loop closure defines. Every closed-loop mechanism on this site has a growth vector of length one, and that is the same statement as its constraints are position constraints, which is the same statement as it has a configuration space that can be solved for.

The bound nobody needs #

There is a natural worry about the construction: might it go on forever? Could a mechanism need ten brackets, or a hundred?

No, and for a boring reason: each level either adds at least one dimension or ends the sequence, and there are only so many dimensions. A mechanism with $n$ coordinates and $m$ controls needs at most $n - m + 1$ levels. A car and trailer has $5 - 2 + 1 = 4$, and takes exactly 4. A ball that may not be twisted has $5 - 2 + 1 = 4$ and takes only 3, because its last level supplies two dimensions at once.

The bound being tight for the trailer and slack for the ball is not a defect in either. It says something about how the brackets are arranged, and the ball’s arrangement — two new directions arriving together — is what makes it the object it is. That arrangement has a name and a history in the literature on distributions, and what matters here is that it is measured on the ball this site draws rather than quoted about a ball in general.

The increments say more than the length #

The vector’s length is the number the essay is named for, and reading the increments rather than the length turns the same measurement into a statement about symmetry.

Each bracket level adds some number of dimensions. For a wheel the sequence is 2, 3 — one added. For a car, 2, 3, 4 — one, then one. For a car and trailer, 2, 3, 4, 5 — one at every level. But for a ball that may not be twisted it is 2, 3, 5: the first bracket adds one and the second adds two.

An increment of one means the directions arrive in a strict order: there is a first forbidden direction, reached at depth one, then a second at depth two, and the mechanism’s difficulty is graded. An increment of two means two new directions arrive together, at the same depth, with nothing in the geometry to separate them.

That is a symmetry, and on the ball it is visible. The two directions that arrive at the second bracket are the two horizontal components of the ball’s orientation — tip forward and tip sideways — and the ball is round, so nothing distinguishes them. A mechanism with a symmetry acting on its forbidden directions delivers them in an orbit rather than one at a time, and the increment counts the orbit.

So the vector carries two different pieces of information and the essay has been reading one. The length says how deep the deepest manoeuvre is, which is the cost exponent. The increments say how the new directions are grouped, which is where the mechanism’s symmetries are — and the two are independent: a mechanism could have length four with increments 1, 1, 1 or length three with 1, 2.

The practical reading is about what a manoeuvre buys. Where an increment is one, a manoeuvre at that depth wins one new direction and the others have to be reached separately. Where it is two, the same manoeuvre reaches a two-dimensional set of new directions, and which of them it wins is chosen by the manoeuvre’s own orientation rather than by its depth. A ball can be tipped either way at the same price; a trailer’s four coordinates cost four different prices.

Which is why the trailer’s sequence is the one the field returns to. Every level adding exactly one is the case with no symmetry anywhere and a strict ordering of difficulty, and it is the shape in which the growth vector’s arithmetic is most nearly a schedule of what each coordinate costs.

Reading the vector #

Four short readings, which between them cover every use the rest of this field makes of it.

Length 1 means confined. The mechanism has a configuration space of the dimension its count says, and everything the rest of the site does applies.

First entry equals last means the same thing, said in the vector’s own terms.

Length $\ell$ means the deepest direction costs $\varepsilon^{\ell}$, which is the practical reading and the one that turns into metres.

Where the vector is measured matters. It is a fact about a configuration, not about a mechanism, and it can in principle differ from place to place. On every mechanism in this field it does not — the trailer’s constancy through the jackknife is the strongest instance — but that is a measurement rather than a theorem, and the essays that need it say where they measured.