Parking is an exponent

The manoeuvre of the last two essays uses a turn on the spot, and a car cannot turn on the spot. Everything said about brackets and exponents is true of it anyway — the exponents were measured on a car with a steering rate, and they came out at 2 and 3 — but the manoeuvre itself was a shopping trolley’s rather than a vehicle’s, and its amplitude was a mixture of a distance and an angle rather than something a driver can point at.

The version made of arcs is what a car actually does, and it turns out to have a closed form.

The four legs #

Take a car whose tightest turn has radius $R$ — five metres for an ordinary saloon — and let $s$ be how far it can go before the space runs out. The manoeuvre is:

1. Forward $s$, steering on left lock.
2. Forward $s$, steering on right lock.
3. Backward $s$, steering on left lock.
4. Backward $s$, steering on right lock.

Each leg is a circular arc of radius $R$ turning through $\varphi = s/R$, and the four are chosen so that the heading changes by $+\varphi$, $-\varphi$, $-\varphi$, $+\varphi$ — back where it started. That much is arithmetic on angles.

The positions are the interesting part. Because the four legs are a symmetric set, the displacement along the road cancels exactly and the displacement across it does not. Not approximately, and not to leading order: the along-road displacement and the heading change are both exactly zero, at every leg length, at every turning radius, in the drawn integration to the last bit of double precision.

What is left is a pure sideways translation of

$$4R \sin\varphi \, \tan\frac{\varphi}{2}, \qquad \varphi = \frac{s}{R}.$$

Checking the closed form #

The formula is a formula and the site’s habit is to distrust one until something has had a chance to disagree with it. So the manoeuvre is flown — four arcs, stepped exactly, with the arc step written from the closed form of a circular arc rather than integrated — and the result compared with the expression above at seven leg lengths from 1.5 m down to 0.125 m.

The largest relative departure over the seven is $4\times10^{-16}$. The two agree to the last digit that double precision has, which is what should happen when a closed form is right and its two derivations share nothing but geometry.

The second check is the exponent. Fit the measured gains against the leg lengths in logarithms and the slope comes out at

$$1.999,$$

against the 2 that the small-angle expansion of the closed form predicts:

$$4R\sin\varphi\tan\frac{\varphi}{2} \approx 4R \cdot \varphi \cdot \frac{\varphi}{2} = 2R\varphi^2 = \frac{2s^2}{R}.$$

That is the growth vector’s promise, in metres. A car’s sideways direction is three brackets deep when the controls are drive and steering rate; with the steering allowed to be set between legs — which is what a real parking manoeuvre does, because a driver spins the wheel while stopped — it is two, and the gain is quadratic rather than cubic. Both are measurable and the difference between them is exactly the difference between shuffling with the wheel already turned and shuffling while turning it.

The numbers, since they are checkable against experience #

For $R = 5$ m, one cycle of four legs wins:

A car with 1.5 m of shuffling room moves nearly a metre sideways in one cycle, which is why an ordinary parallel park takes two or three movements and not twenty. A car with 30 cm of room needs fourteen cycles and sixteen metres of driving to gain the same half-metre that four cycles and ten metres of driving win at 60 cm.

Both of those are recognisable. The first is a normal parking manoeuvre; the second is the situation everybody has been in, where the car in front and the car behind have left a space that is nearly the right size and the job takes an absurd amount of shuffling for a very small gain. The absurdity has an exponent of two in it and nothing else.

What the closed form says that the exponent does not #

The small-angle law $2s^2/R$ is the useful one and it hides a fact worth having.

$\tan(\varphi/2)$ blows up as $\varphi$ approaches $\pi$. A car that could drive a half-circle on each leg would win an unbounded amount sideways — which is nonsense as a statement about parking and is correct as a statement about the formula, and the resolution is that at that point the manoeuvre is no longer a shuffle. Turning through 180° on each of four legs is a three-point turn and then some, and the sideways displacement it produces is the ordinary displacement of driving round in circles rather than anything to do with brackets.

The honest reading is the reverse one: the second-order law is exact only in the limit, and it is already 0.3% optimistic at $s = 1$ m and 3% at $s = 3$ m. The closed form is what the figures are drawn from and the square law is what the sentences are about, and the essays say which is which.

Why a real park is not four legs #

A parallel park is not the manoeuvre above and it is worth saying what the difference is, because the difference is not in the physics.

A real park exploits the fact that the space is behind the car rather than to one side. The standard manoeuvre — reverse on full lock until the car is at forty-five degrees, then reverse on the opposite lock until it is straight — is two legs rather than four, and it produces a large sideways displacement together with a large backward one. It works because the driver does not want a pure sideways translation: getting into the bay is going backwards and sideways at once, and a manoeuvre that moved the car sideways without moving it back would be the wrong manoeuvre.

The four-leg cycle is what is needed when the sideways displacement has to be pure — the car is already alongside the space and needs to be closer to the kerb, or is between two others with nowhere to go. That is the expensive case and it is the one the exponent is about.

The turning radius, which is the other lever #

$4R\sin\varphi\tan(\varphi/2)$ with $\varphi = s/R$ depends on $R$ twice, and the two dependencies partly cancel. In the small-angle limit the gain is $2s^2/R$, so a tighter turning circle wins more per shuffle — inversely, so a car with half the turning radius gains twice as much from the same amount of room.

That is why a small car is easy to park and a long one is not, and it is a more specific statement than small cars are easier. The relevant quantity is not the car’s length; it is its turning radius, which is the wheelbase divided by the tangent of the steering lock. A long car with a lot of lock can have a tighter circle than a short car with little, and does: a London taxi’s turning circle is famously tighter than most family cars’, at a considerably greater length, and the reason is entirely in the lock angle.

The ledger of steering geometry is where the turning radius comes from, and it is one line: $R = L/\tan\delta$ for a wheelbase $L$ and a steer angle $\delta$, measured to the rear axle’s centre.

The trailer, which is a level worse #

A car with a trailer has a growth vector one entry longer than a car’s, and the practical reading of that entry is what happens when the trailer has to end up somewhere particular.

Moving the trailer sideways by $\delta$, with the car ending where it started, is a depth-three problem: the gain per shuffle is cubic in the room rather than quadratic. At $R = 5$ m and a 6 m trailer the constant is small and the exponent is what does the damage — halving the room now costs eight times the number of shuffles rather than four, and twice the distance driven where a car pays the same.

That is the arithmetic behind the thing everyone knows about reversing a trailer into a gap: it is not a little harder than reversing a car, it is a different order of difficulty, and the extra order is exactly the extra coordinate the trailer’s heading contributes. Nothing about the trailer being heavy, or long, or hard to see is in that statement.

The other half of a trailer’s difficulty is not in this field’s arithmetic at all — it is that the trailer’s angle runs away while reversing, doubling every 4.16 m for a six-metre trailer, so a manoeuvre that takes many shuffles is also a manoeuvre during which the thing being manoeuvred is trying to fold up. The two difficulties are independent and they compound.

What a sideways metre costs a car #

The closed form gives the gain per cycle and the shuffle count follows, but the quantity a driver actually spends is distance, and it comes out of the same two lines with a shape worth having.

Working in the small-angle law, one cycle wins $2s^2/R$ and costs $4s$ of driving. Moving sideways by $\delta$ therefore takes $\delta R / 2s^2$ cycles and

$$4s \times \frac{\delta R}{2s^2} = \frac{2\delta R}{s}$$

metres of driving. Half a metre sideways, in a saloon with $R = 5$ m: with 1.5 m of room, three and a third metres of driving and a single cycle; with half a metre of room, ten metres and five cycles; with a quarter of a metre, twenty metres and twenty cycles. The distance doubles every time the room halves, and the shuffle count quadruples, which is the same pair of statements the exponent makes and is easier to argue with from a driving seat.

The turning radius enters once rather than twice here, and that is the difference between this expression and the gain it came from. In the gain, $R$ appears in $\varphi = s/R$ as well as in the prefactor and the two partly cancel; in the total distance the cancellation is gone, and the cost is simply proportional to $R$. A van with a seven-metre turning circle needs fourteen metres of driving where the saloon needs ten, in the same bay, for the same displacement — a 40% penalty that follows from the turning radius alone and has nothing to do with the van’s length or its mirrors.

The trailer’s cubic law can be read the same way and the reading is bleak. If the gain per cycle goes as $s^3$ rather than $s^2$, the number of cycles for a fixed displacement goes as $1/s^3$ and the distance driven as $1/s^2$. Halving the room then multiplies the driving by four rather than by two, and the shuffle count by eight rather than four. That is the depth of the growth vector showing up as an exponent on a bill, and it is why the difference between reversing a car and reversing a trailer is not a matter of practice.

There is a third reading of the same expression that is worth a sentence, because it is the one a designer of car parks would want rather than a driver. The cost is proportional to $\delta R / s$, so the three quantities trade against each other freely: a bay one metre longer is worth exactly as much as a turning circle one fifth tighter, and both are worth exactly as much as needing to move twenty centimetres less sideways. None of the three is privileged, and the only reason the turning circle feels like the fixed one is that it belongs to the car rather than to the space. A bay length is a decision somebody makes with a paint roller, and this expression says what it is worth in metres of manoeuvring.

Two qualifications keep the arithmetic honest, and both run in the same direction.

The small-angle law is optimistic and the exact form is worse, not better. $\tan(\varphi/2)$ grows faster than $\varphi/2$, so the true gain per cycle exceeds $2s^2/R$ — which makes the distance above an over-estimate in a large bay, by the 0.3% and 3% already quoted at one and three metres. In a small bay, where the arithmetic matters, $\varphi$ is small and the law is very nearly exact.

And the cycles are assumed to compose. Each one returns the heading and the along-road position exactly, so they do compose, and that exactness is what licenses multiplying a per-cycle gain by a count. A manoeuvre that returned the heading only approximately would accumulate a drift that had itself to be corrected, and the total would not be a product at all. The symmetry argument that makes the return exact is therefore not a decoration on the result: it is what makes there be a result to state in metres.

Two ways to spend the same room #

Something the arithmetic makes obvious and experience does not is that the shuffle is not the only way to use the space.

Shuffling wins $2s^2/R$ per cycle and costs $4s$ of driving. Doing it $n$ times wins $2ns^2/R$ for $4ns$ of driving, so the total distance driven to gain $\delta$ is $2\delta R/s$ — which depends on how much room there is but not on how many cycles are used, since more room means fewer, longer cycles for the same total.

Driving out and coming back is available whenever there is somewhere to go. If the car can leave the space, drive round, and come back in from a better angle, the whole problem disappears — and this is what a driver does whenever the shuffle starts to look unreasonable. The next essay is about what that costs when reversing is not allowed at all, and the answer is a complete circle, which is worth exactly $2\pi R$ regardless of how far sideways the car needed to go.

Comparing the two gives the crossover. Shuffling to gain $\delta$ costs $2\delta R/s$; going round costs $2\pi R + \delta$. At $R = 5$ m and $s = 0.5$ m, shuffling costs $20\delta$ and going round costs $31.4 + \delta$, so shuffling is cheaper only for offsets below about 1.65 m. Any larger gap and it is shorter to drive round the block — which is, roughly, the rule everybody uses, arrived at without anybody computing $4R\sin\varphi\tan(\varphi/2)$.

What the exactness is worth #

Two of the claims in this essay are exact and the rest are measured, and the distinction is worth drawing because exactness here is doing real work.

The heading returning and the along-road position returning are exact by symmetry: the four legs form a set that is invariant under the transformation that reverses time and reflects the road, so those two components have to cancel. Nothing about $R$ or $s$ enters, and the integration confirms it to the last bit at every leg length tried.

That matters because it makes the manoeuvre repeatable. A shuffle that returned the car to a slightly different heading each time would accumulate a heading error, and after fourteen cycles of a tight park the car would be pointing somewhere unhelpful. It does not, and the reason it does not is a symmetry rather than a control loop.

The sideways gain is exact too, and its exactness is of a different kind: it is a closed form, checked against an integration that arrived at it by adding up arcs. What is not exact is $2s^2/R$, which is a limit and is quoted as one.

The general shape of the result #

Strip out the car and what is left is a statement about any mechanism with a forbidden direction.

Let $\varepsilon$ be the amplitude of whatever manoeuvre the mechanism can perform, and $k$ the depth of the direction wanted. One manoeuvre wins $c\varepsilon^k$; a distance $\delta$ takes $\delta/(c\varepsilon^k)$ of them and $4\delta/(c\varepsilon^{k-1})$ of travel. Every practical question about manoeuvring a constrained mechanism is an instance of that, with the constant and the exponent supplied by the mechanism.

A car’s constant is $2/R$ and its exponent is 2. A car with a trailer, asked to move the trailer sideways, has an exponent of 3, and the arithmetic says what everybody who has manoeuvred a trailer already knows.