Three wheels read the road they measure

A real road sets its own power rolled rigid wheels over roads built to the international standard’s spectrum and found two things. Over the band the standard describes, every practical wheel counts the road’s whole length, a floor no size of wheel gets under. Below that band, in the texture, the overcount falls with the wheel’s radius as a power the road’s spectrum sets, two thirds on an average road, where the one-sine road had given two. So the recipe of a second wheel cannot correct the first, which extrapolated two wheels’ counts on an assumed inverse square, removed only a sixth of the error.

That essay ended with the obvious step. If the power is the road’s and unknown, fit it. A third wheel gives a third count, and three counts are enough to read the power, the level and the distance at once. The question was how well, and whether the floor would stay out of reach.

It reads them well, and the floor turns out not to be a floor. What limits the method is not the long waves but the short end of the texture.

Three counts, three unknowns #

A wheel that rolls without slipping turns through its centre’s path length over its radius, so a counted distance is the true distance $D$ times $1 + e(R)$, with $e$ the excess of the centre’s path over the road’s horizontal length. The earlier essay found $e$ falling as a power of $R$ over the texture the wheel bridges. Write it with room for whatever the power law does not reach:

$$C_i = D\,\big(1 + e_0 + c\,R_i^{-p}\big).$$

The distance is the thing being measured, so it is not known. Only ratios of counts are free of it, and three wheels give two independent ratios. Those fix the power $p$ and the level $c$, after which any one count gives $D(1 + e_0)$, the distance the law extrapolates to as the wheel grows without limit. The fit is exact rather than a least-squares compromise: the ratio of the two ratios depends on $p$ alone and is found by bisection, and $c$ and the distance then follow in closed form.

Two wheels cannot do this because they have one ratio for two unknowns, and so must be given the power. The earlier recipe gave it the inverse square, and on a road whose power is two thirds it extrapolated to the wrong place.

The wheels were rolled over roads built as before, sums of sines of random phase with a displacement spectrum falling as $n^{-2}$, but with the texture carried down to 1 cm instead of 5. That needs the road sampled every 0.4 mm. A wheel’s centre path is made of arcs of its own radius between contacts, so the centre was sampled more coarsely on larger wheels, at a five-hundredth of the radius, where chords lose under a millionth of the path’s length. Each road is 150 m long, and every class from A, the smoothest the standard names, to E was rolled in two realizations.

The floor is on the law #

The first surprise is in the shape of the curve the three wheels sample.

On the class C road the three wheels count excesses of $9.97 \times 10^{-4}$, $6.32 \times 10^{-4}$ and $3.98 \times 10^{-4}$. Fitted, the power is 0.642 and the extrapolated distance is short of the truth by $1.9 \times 10^{-5}$ of it, 2.9 mm in 150 m. Carried out to a wheel of 4.8 m, which was rolled but not used in the fit, the law predicts $1.52 \times 10^{-4}$ against $1.54 \times 10^{-4}$ measured. On the class E road the same check gives $9.9 \times 10^{-4}$ against $1.02 \times 10^{-3}$, and on class A $1.7 \times 10^{-5}$ against $2.3 \times 10^{-5}$, the smooth road’s law already visibly bent.

The road’s standard-band excess, the earlier essay’s floor, is $1.51 \times 10^{-4}$, and the fitted law runs straight through it. That seems to contradict the earlier finding that no wheel under a metre and a half bridges the band at all. It does not, and the reason is worth having. A wheel follows every wave whose troughs are wider than it and bridges the rest, so its excess is the sum of $\pi^2A^2/\lambda^2$ over the waves longer than some $\lambda^*(R)$. For a spectrum falling as $n^{-2}$ that sum is $1/\lambda^* - 1/\lambda_{\max}$, one expression whether $\lambda^*$ falls in the texture or in the band. The band’s excess is the part of the sum from 0.35 m up, and every practical wheel’s $\lambda^*$ lies below 0.35 m, so every practical wheel counts it in full. But the fitted law does not need a wheel to reach it. It extrapolates the sum, and the sum’s band part is simply where the law continues.

So the earlier essay’s floor is real for a road whose spectrum stops at the band’s edge, and not for one whose spectrum goes on. A road with texture has no floor in this sense. Its wheel excess falls towards nought as the wheel grows, and three wheels can see the whole of the fall in the part of it they reach.

How much the third wheel buys #

The distance errors set the method beside the alternatives.

On every road the three wheels beat every simpler correction, and by a factor that grows with the road’s roughness. On the two class E roads the 0.3 m wheel is out by $6.4 \times 10^{-3}$ and the three wheels by $1.8 \times 10^{-5}$ and $1.2 \times 10^{-5}$, about 350 and 550 times better. On class C they are 18 and 51 times better, and on class A 5.5 times. The inverse square, for comparison, never does better than the larger of its two wheels by more than a quarter.

The pattern has a geometric reason. A rough road’s troughs are tighter, since its amplitudes are larger at every wavelength, so a wheel of a given size bridges only longer waves on it. Its $\lambda^*$ sits further up the spectrum, away from the texture’s short end at 1 cm, where the law holds cleanly. On a smooth road the same wheels follow much shorter waves and sit near that end, and the law the three counts see is bent by it. A class A road’s three wheels read a power of 0.50, a class E road’s 0.67 and 0.68.

The level is read from the fitted $c$. For the $n^{-2}$ spectrum it should grow with the road’s spectral level $G_0$ as $G_0^{2/3}$, since a rougher road both has more texture and bridges less of it, and one road, class C in its first realization, calibrates the constant. Read through that calibration every one of the ten roads comes out in the class it was built at. The level estimates lie within a factor of 1.5 of the truth, worst on class A, and the classes are a factor of four apart, so the reading has room to spare.

Why the error is nearly always short #

Of the ten roads, nine read a distance shorter than the truth, and the one long reading is the roughest. The sign comes from the shape of the bend. Near the texture’s short end the smallest wheel’s count stops rising as fast as the law, because no wheel can count more than the whole road and the smallest is getting close to doing so: its count is therefore lower than the law would make it, and the three counts look as though they fall more slowly with radius than they do. A shallower fitted power explains the same spread of counts with a larger share of each count left over at an infinite wheel. That leftover is read as distance, so the reading comes out short.

That says something useful about the error before any road is measured. It is not noise but a bias with a known sign and a known cause, and it shrinks as the wheels grow away from the texture’s end. A trolley built for a known class of road can be given wheels large enough that the bias is below whatever the distance is needed to. Its size can be estimated from the reading itself, since a power read well below two thirds on a road that is supposed to be ISO-shaped is the signature of wheels too near the end.

Where the wheels must sit #

That bending is the method’s real limit, and it was measured directly by moving the three wheels along the radius axis.

On the road textured to 1 cm every set reads close to the law: 0.616, 0.642, 0.638 and 0.694, with distance errors from $5.4 \times 10^{-5}$ down to $3.8 \times 10^{-6}$ as the wheels grow. On the earlier essay’s road, whose texture stops at 5 cm, the smallest set finds no power at all, and the next reads 0.329 and puts the distance out by $3.8 \times 10^{-4}$. That is no better than the 1.2 m wheel on its own: wheels of a few tenths of a metre bridge almost all of a texture that stops at 5 cm, their counts crowd together near the road’s full length, and the fit reads that crowding as a shallow power. Only the set reaching 4.8 m, whose bridged waves lie well inside the spectrum, reads 0.622 there.

The rule for a designer is the rule the derivation states. The three wheels must be large enough that the waves they just bridge lie inside the road’s texture and not at its end, which on a smooth road means larger wheels. The end of a real road’s texture is not a fixed length either. Below a centimetre or so a rubber tyre envelops the texture rather than bridging it, and only a rigid wheel sees what is modelled here.

Spectra other than the standard’s #

The law the method rests on, $p = 2(3 - w)/(5 - w)$ for a spectrum falling as $n^{-w}$, was the earlier essay’s. It was checked again here with the distance known, and the three wheels were asked to read it without.

With the distance known, the slope of the excess between 0.3 and 1.2 m is the law’s to 0.007 on every spectrum: 0.663, 0.543, 0.393 and 0.225 against 0.667, 0.545, 0.400 and 0.222. Without it, three wheels read the steeper spectra low, 0.462 for $n^{-2.25}$ and 0.257 for $n^{-2.5}$, and find no power at all for $n^{-2.75}$. A steeper spectrum has less of its length in the texture and more in the long waves, so the counts differ by less, and the same short end of the texture bends a smaller signal further. The distance still improves, by 8 and 2.5 times, but the slope it implies, 2.40 and 2.71, is not to be trusted.

A road whose texture falls at a different slope from its band carries a bend in its law at 35 cm. The earlier essay could not say what that does, and the measurement says it matters in one direction. A texture flatter than the band, $n^{-1.5}$, puts more of the road’s length into the bridged waves, and three wheels then extrapolate the distance to $1.5 \times 10^{-6}$. A steeper one, $n^{-2.5}$, starves them, and the error rises to $1.9 \times 10^{-4}$, four times better than one wheel and no more. Across the five texture slopes the error grows steadily with the steepness.

Counting to a part in a million #

Everything read here comes from differences between counts that agree to a few parts in ten thousand, which asks a great deal of the counting.

A count error of a part in a million, 0.15 mm in 150 m, moves the power by 0.02 and the distance by $2.2 \times 10^{-5}$, as much as the method’s own residual on this road. Both move in proportion to the count error, so a trolley that counts to a part in ten million reads the power to 0.002 and leaves the distance to the method. The power is twenty thousand times as sensitive as the counts are wrong, because it is read from their differences. A wheel encoder that resolves a fraction of a millimetre over a run of a hundred metres is ordinary, so the demand is met by instruments that exist. It is not met by a turn counter.

This is the familiar shape of an identification. A calibration is a synthesis with more equations set out the general case, and the instrument’s error, multiplied measured how a calibration amplifies the error in what it reads. Here three equations meet three unknowns exactly. There is no redundancy, so the counts’ errors pass straight through to the reading, amplified by how nearly the three ratios coincide.

What the three wheels say about a wheel #

A wheel that cannot report its radius found that rolling ties an angle to a size: a wheel’s turning says how far it has gone only if its radius is known. What a wheel counts on a rough road found the size telling the angle something else, how much of the road the wheel bridged. Three wheels turn that second fact into an instrument. The differences between their counts are a measurement of the road’s spectrum, its power and its level, made by nothing but turning, and the distance follows from the reading.

That is close to what a machine that measures itself described for a mechanism: an instrument whose errors carry enough structure to be estimated from its own readings. The structure here is the power law, and the method is exactly as good as the law. Where the wheels sit inside the road’s spectrum it is very good. Where they reach an end of it, or where the spectrum bends, the law is the wrong model, and the extrapolation carries the error as a parameter the model has not got carries one. The residual cannot show it, since three points determine three parameters with nothing left over.

What the trolley does not settle #

A fourth wheel. Three counts fit three unknowns exactly, so nothing in them says whether the law fits. A fourth wheel would leave a residual. Reading a residual is the method for what such a residual means, and it would flag the bent spectra and the texture’s end before they spoiled the distance. It was not added.

Rigid wheels. Every wheel here is a rigid disc. A pneumatic tyre absorbs texture below a centimetre or so, which moves the effective end of the spectrum to wherever the tyre’s envelopment sets it and changes the wheel sizes at which the law holds.

Gaussian roads. The roads are sums of sines of random phase, close to Gaussian. Potholes, joints and repairs are not in a spectrum. A wheel bridges or drops into each one separately, and the counts’ differences then carry events rather than a law.

Separate tracks. The three wheels roll in one track here. On a real trolley they would be side by side, on three realizations of the same spectrum, and the realization-to-realization spread, about 2% of the excess between the roads here, would enter the differences the method reads.

Still open: wheels of one size at three loads #

The method needs three different radii because the radius decides which waves a wheel bridges. A wheel’s effective radius on texture can also be changed without changing the wheel, by changing how hard it is pressed into the road, since a deformable tyre’s contact patch grows with its load and bridges more as it grows.

The distinct argument would be a single wheel run three times, or three identical wheels on one axle, at three loads. Each load would set a contact length, and the contact length would play the part the radius plays here. The measurement would model a tyre’s contact patch against its load, find the effective bridging length for each load, and ask whether three loads read the power and the distance as well as three radii do. It would also ask whether the patch’s softening near its edges smears the bridging enough to bend the law the way the texture’s end did. If it works, a profiler needs one wheel and a spring rather than three wheels of very different sizes.