The convoy that keeps the deck's time

The deck that forgets its own rhythm took a 120 m stay at 3,500 kN, whose first mode is at 1.01 Hz, over a deck whose mode at its anchorage is at twice that, 2.01 Hz. A deck moving at twice a stay’s frequency swings the stay’s tension in step with its swinging, and a stay shaken along its own length grows without limit once a quarter of the swing beats its damping. That essay replaced the deck’s sine with a random motion — one of the deck’s own lightly damped modes pushed by white noise — and found that a random deck asks for far less: not a quarter of the swing but its square over sixteen times the deck’s damping, 0.06 per cent of damping for a deck moving 7.1 mm rms, against 0.35 per cent for a sine of the same rms.

It ended on a gap in its own model. White noise has no rhythm at all, which is the right model for wind. Traffic is made of vehicles, and vehicles arrive one after another at intervals that have a mean and a scatter. Whether a stream of them keeps the deck’s timing from one vehicle to the next — and so behaves, for the stay, like the sine rather than the noise — is a question about the spacing of the vehicles in time against the deck’s period.

Each vehicle is a kick #

The model is the simplest that has the right character. Each vehicle, as it passes some fixed point of the deck — an expansion joint, a step in the surfacing, the anchorage zone itself — gives the deck’s mode the same impulse, a sudden change of velocity. Between vehicles the mode rings down at its own 1 per cent damping. The stay is the earlier essays’ stay, its tension swung by whatever the deck is doing.

The impulse is calibrated so that the comparison is clean. A stream of these vehicles arriving at random — a Poisson stream, every 2 s on average — gives the deck an rms movement of 7.1 mm, and a Poisson stream of identical kicks is shot noise, whose spectrum near the deck’s frequency is flat. That stream is the earlier essay’s random deck exactly: the same rms, the same requirement of 0.06 per cent. Every other stream in this essay is the same vehicles, arriving with some other rhythm.

The deck’s period is 497 ms, and the deck forgets its motion over about $1/(\zeta_d\Omega_d)$, 7.9 s. Between two vehicles 2 s apart it keeps $e^{-\zeta_d\Omega_d H} = 0.78$ of its motion — enough for each kick to land on a deck still ringing from the last three or four. The stay’s own clock is the one the force read off a frequency used to weigh it: 1.01 Hz, set by its tension and its mass, and the deck’s mode sits at twice that.

The same vehicles, on time #

Let the vehicles arrive every 1.987 s exactly, four periods of the deck. Each kick lands when the deck is at the same point of its swing as it was at the last kick, so it adds to the motion instead of sometimes cancelling it. The deck grows until the decay over one headway equals one kick, and then holds a steady swing of 28 mm amplitude, 20 mm rms — nearly three times the random stream’s, from the same vehicles at the same flow.

That is the train that arrives in time with itself in a different form. A train’s axles are coupled to each other by steel, so their spacing is exact, and where a multiple of the axle passing frequency lands on a bridge’s frequency each coach arrives in step with the motion the last one left. Road vehicles are not coupled. Whether they ever arrive in step closely enough to matter is the question, and before it can be asked, how close is close enough has to be known.

A comb of headways #

A perfectly regular stream makes the deck’s motion periodic, with harmonics at every multiple of the passing frequency. The deck only responds strongly to the harmonic that lands on its own frequency, and the stay only answers a deck motion at twice its own frequency, so the stream matters only at headways that are a whole number of deck periods: three periods, 1.49 s; four, 1.99; five, 2.48; six, 2.98. At those headways the deck’s response is a sine at exactly the frequency the stay is tuned to, and the stay needs a quarter of that sine’s tension swing — 1.3 per cent at three periods, 0.99 at four, 0.66 at six.

Between the teeth, nothing. This is the opposite of the resonance that ran out of time, where a machine sweeping through a frequency spent too little time there to build up: a regular stream sits at its frequency indefinitely, and the question is only whether it sits at the right one. The deck still moves, but its motion is at the passing frequency’s harmonics and none of them is at twice the stay’s frequency, so the stay is not pumped. The teeth are narrow. At four periods the stay’s own 0.1 per cent of damping is enough everywhere except between 1.972 and 2.002 s: a regular stream 1.5 per cent too fast or too slow has already missed.

Two things set the width. The deck’s own resonance has a width of about twice its damping, so its harmonic falls away on either side; and the stay is a parametric oscillator, which only grows when the pumping frequency is within $\varepsilon/2$ of twice its own. A regular stream at the right headway, in other words, has to be right to within both bandwidths at once.

Two per cent of scatter #

A perfectly regular stream of road vehicles does not exist. The next question is what a little irregularity does — and the answer is not obvious, because a stream whose headways scatter by 2 per cent is still, to the eye, a very regular stream.

The integration knows nothing about combs or bandwidths. It runs the deck and the stay together, kicks the deck as each vehicle arrives, draws the headways from a distribution with the stated mean and scatter, and measures how fast the stay would grow with no damping at all; that rate over the stay’s frequency is exactly the damping that would hold it still, because damping in this equation only ever subtracts. At no scatter it reproduces the comb’s 0.99 per cent to within 1 per cent, which is a check on both.

Then it falls, and steeply. At 1 per cent scatter the requirement is 0.86 per cent; at 2 per cent, 0.58; at 3 per cent, 0.30; at 5 per cent, 0.13; and from about 7 per cent it is the random stream’s 0.05 per cent. Half of the regular stream’s danger is gone at 2.2 per cent of scatter, and all of it by about seven. A stream of lorries keeping a two-second gap to within a seventh of a second — 7 per cent — is, as far as the stay can tell, random.

The histories make the same point in time. The exactly regular stream grows the stay along a straight line on these axes, tenfold every 42 s; the 2 per cent stream grows it more slowly and raggedly, with the steps and pauses that mark a pump that keeps losing its timing; the random stream lets it wander below its starting millimetre. A metre is far past the point at which the stay’s own stretching, left out of this linear count, would have stopped the growth, as the stay shaken along its own length found — but a stay held at a few hundred millimetres by its own nonlinearity is already a stay in trouble.

Time kept to a tenth of a period #

The obvious guess about the scatter a stream may have is that it depends on the deck’s damping: a deck that forgets quickly remembers fewer vehicles, so it should be harder for scatter to accumulate across them. The guess is wrong, and the way it is wrong is the result of the essay.

Measured in time rather than as a share of the headway, the scatter that halves the danger is nearly a constant: 28 to 50 milliseconds, against a deck period of 497. At one deck period between vehicles and at eight, on a deck damped at a half per cent and at two per cent, the vehicles have to keep time to about a tenth of the deck’s period for their rhythm to survive. The deck’s damping decides how dangerous a regular stream is. It does not decide how regular a stream has to be.

The reason is that what matters about each kick is the point in the deck’s swing at which it lands, and that point is set by the timing error over the deck’s period, whatever the headway. A heavily damped deck remembers only the last vehicle or two, so it is pumped by a short run of kicks, and one kick a tenth of a period out of place spoils a large share of the run. A lightly damped deck remembers a dozen, and the errors of a dozen kicks partly average; but it also builds a much larger coherent swing that a mistimed kick spends. The two effects nearly cancel, and what is left is an angle of the swing: about 0.4 to 0.6 of a radian of scatter per vehicle, or a tenth of a period, is where coherence goes.

A share of the headway is the wrong measure for the same reason. A 2 per cent scatter is 10 ms on a headway of one deck period and 80 ms on a headway of eight. A stream of closely spaced vehicles can afford proportionally more scatter than a stream of widely spaced ones, because it is the absolute timing that has to be kept.

Ordinary traffic is counted, not timed #

Human drivers do not keep time to 40 milliseconds; a driver’s reaction time alone is many times that. Even a queue at a steady speed lets its gaps wander by far more than a tenth of a deck period, and free-flowing traffic scatters its headways by something like half their mean. All of it is far past the scatter at which a stay can tell a stream from noise.

For that traffic the random deck’s rule applies unchanged, with one input taken from the traffic: the deck’s variance, which for a stream of kicks is in proportion to how many arrive per second. Twice the flow is twice the variance, and since the stay’s requirement goes as the square of the deck’s rms, twice the requirement. A vehicle every second asks 0.10 per cent of this stay — its own damping — and a vehicle every ten seconds asks 0.02. Ordinary traffic matters to a stay by its flow, which sets how much the deck moves, and not by its rhythm, which it does not have.

That is reassuring in a specific way. The earlier essay’s random deck was a model chosen because wind has no rhythm; it now turns out to be the right model for traffic too, for the reason that traffic’s rhythm, at the precision a deck can hear, is noise.

The step soldiers are told to break #

The case in which vehicles do keep time is the one in which they are made to. Soldiers marching in step put a regular rhythm into a bridge, and the order to break step on a bridge, given since the collapse of the Broughton suspension bridge in 1831 under a column marching in step, is an order to add scatter: a column out of step is a crowd, and a crowd is noise. The crowd that is also the structure found the same filter in a grandstand, where the scatter of a crowd’s timing strips the higher harmonics of its jumping. And the bridge that was pushed by its own sway found the case where a crowd’s timing is not its own: walkers who adjust their footing to a deck already swaying push it the way it is going, a rhythm taken from the deck rather than brought to it.

Lorries driven in a platoon by a controller that holds a fixed time gap are a convoy made to march. The purpose of the controller is that the gap does not wander; a controller that held it to a few hundredths of a second would be keeping exactly the time this essay asks for, and a platoon whose gap happened to be a whole number of a stay deck’s periods would be a perfectly regular stream on that deck. The comb says the danger is confined to narrow bands of headway, so it would also be easy to avoid — a gap specified to stay out of the bands, or a deliberate scatter in it of a tenth of a period, as a column is told to break step. The defence against a convoy that keeps time is to make it keep the wrong time, or not quite keep it.

Whether real platoons are that precise, and whether their gaps are chosen with any bridge in mind, is outside this calculation. What it says is that the number to know is not the gap, nor its scatter as a share of the gap, but its scatter in milliseconds against the period of the deck it is crossing.

The comb by hand #

A Poisson stream of kicks $\Delta V$ arriving every $H$ seconds on average gives the deck’s mode, by Campbell’s theorem, a variance $\sigma^2 = \Delta V^2/(4\zeta_d\Omega_d^3 H)$. With $\sigma = 7.1$ mm, $\Omega_d = 12.65$ rad/s, $\zeta_d = 0.01$ and $H = 2$ s, $\Delta V = \sigma\Omega_d\sqrt{4\zeta_d\Omega_d H} = 0.0071 \times 12.65 \times 1.006 = 0.090$ m/s per vehicle.

A perfectly regular stream of the same kicks every $H$ seconds has a harmonic at each multiple of $2\pi/H$ whose amplitude in force is twice the mean force $\Delta V/H$. At resonance the deck answers it with $1/(2\zeta_d\Omega_d^2)$ of the force per unit mass, so the harmonic’s amplitude is

$$a = \frac{2\Delta V/H}{2\zeta_d\Omega_d^2} = \frac{2 \times 0.0451}{2 \times 0.01 \times 160} = 0.028\ \text{m},$$

an rms of 20 mm. The stay’s tension swings by 1.39 per cent for every 10 mm at the anchorage, so the harmonic swings it by $\varepsilon = 0.0396$ and the stay needs $\varepsilon/4 = 0.99$ per cent. The random stream’s 7.1 mm swings it by an rms of 0.99 per cent, and needs $\sigma_\varepsilon^2/16\zeta_d = 0.0099^2/0.16 = 0.061$ per cent. The ratio of the two, sixteen, is what landing every kick in step is worth on this deck: the regular stream’s requirement is proportional to the harmonic’s amplitude, the random one’s to the square of an rms, and at the comb’s tooth the first is the larger by that factor.

What the stream of kicks leaves out #

Every vehicle kicks once, at one point, with the same impulse. A real vehicle crosses the whole span and loads the deck’s mode in proportion to the mode’s shape under it, so its force on the mode is a slow pulse with a wave in it, and the stream’s harmonics are filtered by that shape before the deck sees them. That changes the size of the harmonic at the deck’s frequency and not the timing argument: a pulse that arrives a tenth of a period late is a tenth of a period late whatever its shape.

The vehicles weigh the same. A stream of lorries of different weights is a stream of kicks of different sizes, which scatters the amplitude of each kick and not its timing. Amplitude scatter does not spoil a rhythm the way timing scatter does: a kick of the wrong size still lands in step.

The vehicles do not bounce. Every road vehicle is a mass on springs with a bounce frequency of its own, typically between one and a half and three hertz for a lorry — which is exactly the range of the deck’s mode here. A vehicle bouncing at the deck’s frequency is not a kick but a small oscillator coupled to it, and it gives the deck energy or takes it away depending on where it is in its own bounce. That interaction is the large thing the model leaves out, and it is the subject of the question below.

One lane, one deck mode, one stay. Two lanes of traffic in opposite directions are two streams, independent of each other; their sum has the rhythm of neither. A deck has many modes and a cable-stayed deck has many stays, and a stream that hit a tooth of the comb for one pair would miss it for the next.

Still open: the lorry that bounces at the deck’s frequency #

Every vehicle here is a kick and nothing more: it adds velocity to the deck as it passes and then plays no further part. A real lorry rides on springs, and its body bounces at a frequency of its own — often within a few tenths of a hertz of a deck mode like this one — while it crosses. A vehicle bouncing in step with the deck pushes on it at the deck’s frequency for the whole of its crossing, which is a far larger rhythm than a single kick, and a bouncing vehicle is also a tuned mass on the deck, which the line of ties that stops short and the damper that is too near the end would both recognise as a damper of a kind. Whether a stream of lorries tuned near the deck’s mode feeds it or damps it — and whether, for the stay above it, a deck carrying its own traffic as moving tuned masses is quieter or louder than one carrying kicks — is the question the vehicles’ suspension asks of the deck’s rhythm.