A clock is a factorisation

An escapement lets a train advance by a fixed amount and stops it again. What that is for is a count: a wheel further down the train has turned by an amount proportional to the number of beats, and if the beats are regular the wheel is a clock.

Which makes the design of the train an arithmetic problem rather than a mechanical one, and an unusually pure sort of arithmetic problem, because gear ratios are ratios of whole numbers and nothing else.

The going train’s number #

A seconds pendulum beats once a second. A thirty-tooth escape wheel advances half a tooth per beat, so it turns once a minute — sixty beats to the revolution — and the centre wheel that carries the minute hand turns once an hour.

So the train between them must multiply by exactly sixty, and the question is whether sixty can be made out of wheels and pinions a wheel-cutting engine will cut.

In one pair it cannot. A single wheel and pinion give $W/P$, and with a pinion of at least six leaves a ratio of sixty needs a wheel of at least 360 teeth, which is a wheel nobody makes for a clock. The count of exact one-pair trains inside the plausible range is zero.

In two pairs it is easy: 404 exact solutions, with wheels between 40 and 120 and pinions between 6 and 16. In three pairs there are 2,067. The mechanism’s designer is not solving a problem, they are choosing from a list.

Exactness has to be decided in integers #

There is a way to write this search that returns answers which are merely close, and it is the obvious way.

Compute $W_1 W_2 / (P_1 P_2)$ in floating point and compare it against 60 with a tolerance. That admits $59.999999999999996$, and there are ratios in the search space that land there — not because they are wrong but because binary floating point cannot represent every rational. Widen the tolerance to be safe and inexact trains start arriving; narrow it and exact ones start being rejected.

The search here does not compare ratios at all. It asks whether

$$W_1 W_2 \cdot 1 = P_1 P_2 \cdot 60$$

in integers, which is a question with no tolerance in it and no rounding. Every train the search returns satisfies that identity, and the assertion behind it re-derives the products and checks them one by one rather than trusting the search.

That is a small discipline and it is the same one this site applies to a gear train’s ratio: a ratio built from tooth counts is a rational number known exactly, and the moment it is evaluated as a decimal the exactness has been thrown away for no gain.

The other kind of problem #

Now a ratio that does not factorise, and the difference is total.

A star returns to the same place in the sky in 23 hours 56 minutes 4 seconds — a sidereal day — because the Earth has gone once round its axis and a three hundred and sixty-fifth of the way round the Sun. A clock that shows sidereal time has to run fast by the ratio

$$\frac{366.2422}{365.2422} = 1.0027379$$

and the whole of the mechanism is a pair of gears somewhere in the train.

Search every pair of wheels between 40 and 150 teeth — 12,321 candidates — and the closest is one to one. Not a near miss: no gearing whatever is a better approximation to 1.0027379 than any single pair of cuttable wheels, because the nearest useful convergent is 366 to 365 and 366 teeth is not a wheel. The best single pair is out by 236 seconds a day, which is four minutes.

Go to two pairs and 1,413,721 candidates, and the exact count is still zero — the ratio in lowest terms needs 366 and 365 somewhere, and neither fits in one wheel. What two pairs buy is the ability to factorise those numbers instead: $366 = 6 \times 61$ and $365 = 5 \times 73$, so

$$\frac{48}{40} \times \frac{61}{73} = \frac{2928}{2920} = \frac{366}{365}$$

and the error against the true sidereal ratio is 0.157 seconds a day — one second in six days, one day in fifteen hundred years.

Why the two problems look the same and are not #

Both are searches over tooth counts. Both return a list ranked by error. The difference is in the first column of the result.

For sixty, the exact count is 404 and the error of every one of them is zero. Choosing between them is a mechanical question — shaft positions, wheel diameters, how the depths come out — and arithmetic has stopped having an opinion.

For the sidereal ratio, the exact count is zero however many wheels are thrown at it, and every candidate has an error. Choosing between them is arithmetic and nothing else, and the answer is a best approximation rather than a solution.

The distinction is the same one this site has been drawing since Peaucellier and Watt: some problems have exact mechanical answers and some have only good ones, and knowing which kind a problem is before starting is most of the work. A straight-line linkage and a sidereal train are the same situation in different centuries.

The moon wheel, and the number everybody quotes #

The most famous approximate train in horology is a single wheel and it is worth putting through the same search, because the error is quotable.

A lunar month is 29.530589 days. A clock’s moon dial is driven once a day, so the wheel needs 29.530589 teeth, and it is given 59 driven two teeth at a time — a ratio of 29.5.

That error is 0.030589 days a month, which accumulates to one whole day in 2.64 years. The traditional statement is “a moon dial gains a day in about two and a half years”, and it is the same number arrived at by dividing rather than by observing.

What is interesting is that 59 is very nearly the best a single wheel can do, and the next candidates are much worse: 89 teeth in 3 gives 398 seconds a day and 88 in 3 gives −577. The lattice is coarse at that end, and 59/2 is the one good point on it — which is presumably why every moon dial in the world has 59 teeth.

Which numbers are easy #

The count of exact trains is not a smooth function of the target, and its variation says something about which clocks are easy to build.

Sixty is the easiest number in the range by a factor of two, and the reason is that it has twelve divisors. Every way of splitting 60 into two factors that both land inside the box gives a family of trains, and 60 splits as $2 \times 30$, $3 \times 20$, $4 \times 15$, $5 \times 12$, $6 \times 10$ and their reverses. A prime like 61 splits only as $61 \times 1$, so every train has to reach 61 in one stage and merely pass the ratio through in the other, and there are far fewer ways to do it.

This is the arithmetic showing up in the machine. A clock divided into sixty minutes and sixty seconds is easy to gear because sixty is a highly composite number, and a clock divided into sixty-one of anything would be markedly harder — not slightly harder, six times harder by this count. The Babylonian choice of sixty was made for reasons that had nothing to do with gear trains and it has been paying dividends to clockmakers for four hundred years.

Which box #

The box matters as much as the number, and it is worth measuring rather than asserting.

Allowing four-leaved pinions adds a quarter as many solutions again. Restricting to the eight-to-twelve range a good clock actually uses cuts the list by three quarters. And narrowing the wheels from 40–120 to 40–90 leaves 165 of the 404.

Each of those restrictions is a mechanical judgement — a small pinion runs badly, a large wheel is expensive — and each of them changes an arithmetic answer. Which is the clearest statement available of what a wheel train is: an arithmetic problem whose feasible set is drawn by a machine shop.

Three pairs, and why they are not the answer to everything #

Sixty in three pairs has 2,067 exact trains against 404 in two, so more stages always help — and for the sidereal ratio more stages do not help at all, which is the distinction this essay is about, seen once more.

A third pair multiplies the search space by about a thousand and multiplies the exact count by about five, because most of the new candidates are the old ones with a redundant 1 : 1 stage in them. For a ratio that factorises inside the box, the extra stage buys choice rather than possibility.

For a ratio that does not, the extra stage buys a better approximation and nothing else, and the improvement is real: one pair gets the sidereal ratio to 236 seconds a day, two pairs to 0.157. A third would do better again. But the exact count stays at zero however many stages are added, because 366/365 in lowest terms needs the factors of 366 and 365 to be distributed among the wheels, and once they have been — as they are in $48 \times 61 / (40 \times 73)$ — the answer is not exact anyway, since the true ratio is 1.0027379 and 366/365 is 1.00273973.

Which is the last thing to notice about the sidereal problem: the best train is exact for the wrong number. It gears 366/365 perfectly, and 366/365 is itself an approximation to the ratio of the two days. The mechanism’s error is inherited from an arithmetic simplification made before any wheel was cut, and the 0.157 seconds a day is the residue of that rather than of anything mechanical.

What a search for an inexact ratio is really doing #

The two problems are described above as different in kind, and the second one has a name and a shape worth giving it, because the shape says what an extra stage is worth before anybody runs the search.

A target that is a rational number with a factorisation inside the box has exact answers, and the count of them is a fact about the integer. A target that is not — the sidereal ratio, a lunar month, anything measured rather than defined — has no exact answer at any number of stages, and the search is therefore not looking for a solution at all. It is looking for the best rational approximation to a number, with the numerator and denominator each constrained to be a product of counts a wheel-cutting engine will cut.

That is a Diophantine problem rather than a factorisation, and it behaves in the way those do. The accuracy available is governed by how large the denominators may get, so each additional stage multiplies the reachable denominators by roughly the size of the box and buys about that factor in accuracy. Going from one pair to two took the candidate count from twelve thousand to one and a half million, and the best error fell correspondingly — not because the search looked harder but because the set of representable ratios got denser.

Which gives the figure of merit that a clockmaker actually needs, and it is not the error. It is the error against what another stage would buy. A two-pair train at 0.157 seconds a day is worth accepting if a third pair would take it to 0.14; it is not if a third pair would take it to 0.002. The first is a mechanism at the limit of the arithmetic and the second is a mechanism at the limit of somebody’s patience, and only the comparison distinguishes them.

It also explains why approximate trains in horology cluster on a small number of famous values. The best approximations to a given target with bounded parts are a short and rapidly improving sequence, so there is usually one obvious answer, one that is much worse, and nothing in between. The moon wheel’s 59 is that: not a compromise somebody settled on but very nearly the only candidate, with the next one out by more than twice as much. A tradition that quotes one number for such a train is not being lazy — there is one number.

And it sharpens what the box is doing. For an exact target the box decides whether there is an answer; for an inexact one it decides how good the answer is, continuously. So widening the range of cuttable pinions has completely different consequences in the two cases — in the first it turns a zero into a count, and in the second it improves an error by a factor. A workshop’s capability shows up as a discrete gain on one kind of problem and a continuous one on the other.

That division runs right back through the field. An exact ratio is a count and an inexact one is a measurement, and every consequence in this essay follows from which of those the clock is being asked for. A clock that divides a day is doing arithmetic; a clock that tracks a star is doing approximation, and no amount of gearing will make it do the first.

The count is exact and the beat is not #

One boundary, because this essay’s arithmetic is exact and the clock it describes is not.

Everything above is a statement about counts: how many turns of one wheel correspond to how many of another. The counts are integers, the ratios are rationals, and the errors are exactly computable to as many digits as anybody wants. Nothing in it depends on how well the wheels are cut, and nothing on how heavy the weight is.

What decides whether the clock keeps time is the beat, and the beat is a period of oscillation, which is dynamics. A perfect train on a bad pendulum is a bad clock; a mediocre train on a good pendulum is a clock that is wrong by a known and constant amount, which is a very different sort of wrong and can be corrected by moving a bob.

That division is why the two halves of clockmaking were historically separate trades, and it is exactly this site’s boundary again. The train is geometry and arithmetic. The oscillator is not.

What a wheel-cutting engine decides #

One last observation, because it explains a pattern in old clocks that looks like superstition.

The search ranges above — wheels of 40 to 120 or 40 to 150, pinions of 6 to 16 — are not arbitrary. A pinion of fewer than six leaves undercuts badly and runs poorly; a wheel of more than about 120 is large and expensive and hard to divide accurately. Those two bounds are what turns the arithmetic from find any factorisation into find a factorisation inside a box, and the box is where the difficulty lives.

Widen the box and problems get easier: with pinions down to 4 leaves, ratios that had no exact train acquire one. Narrow it and they get harder. So a clockmaker’s stock of dividing plates — the physical plates with holes drilled at particular counts — decided which ratios were available to them, and two workshops with different plates had genuinely different sets of solvable problems.

That is a rare case of a manufacturing constraint reaching all the way into the arithmetic rather than merely into the tolerance, and it is why the tooth counts in old clocks repeat: not tradition, but the plates.

It also explains why the escapement sits where it does in the train. The escape wheel is at the fast end, turning once a minute against the centre wheel’s once an hour, and everything between is a multiplication by sixty. Putting the escapement anywhere else would change the number to be factorised — a clock whose escape wheel turned once every two minutes needs a ratio of thirty, which has fewer divisors than sixty and therefore fewer trains — so the choice of where in the train to escape is itself an arithmetic decision dressed as a mechanical one.

And it is the point at which this essay meets the field it came from. An escapement’s half-tooth-per-beat is a counting fact, established without reference to any dimension; a train’s ratio is a counting fact for the same reason. The two counts multiply, and what comes out is a clock. Everything in between — the arcs, the draw, the drop, the recoil — decides whether the count is reliable, and none of it decides what the count is.