The steps are not free

There is a design rule about gearbox ratios that everybody states and nobody quite justifies in the same words. The steps between the gears should be equal in ratio, so that a shift always drops the engine by the same proportion and the driver — or the controller — is always handed back the same operating point. Equal ratios between consecutive gears means the ladder is a geometric progression, and the number quoted for a gearbox, its spread, is the top gear divided by the bottom one.

It is a good rule. It is also, for a planetary automatic, mostly unavailable, and the reason is the reduction from the previous essay: the ladder is a function of one or two numbers. A three-speed Simpson gearbox has one proportion to spend on three gears. A four-speed Ravigneaux has two to spend on four. Choose the ends and the middle is decided.

How much is actually chosen #

Work the accounting through explicitly, because it is the whole argument.

A Ravigneaux’s ratios, with $K_1$ and $K_2$ the two ring-over-sun proportions, are $1+K_2$, $1+K_2/K_1$, $1$, and $1-1/K_1$. Four gears; two parameters; and the third gear is 1 whatever anyone does, because direct drive is a clutch and a clutch does not know about tooth counts.

So really there are two free numbers and two free gears, and the other two are determined. Ask for first at 2.48 and $K_2 = 1.48$ is fixed. Ask for reverse at −2.22 and $K_1 = 3.22$ is fixed. Second is now $1.46$ and the overdrive is $0.689$, and it does not matter in the slightest what anybody thinks of them.

The three steps that come out are

$$\frac{2.48}{1.46} = 1.699, \qquad \frac{1.46}{1.00} = 1.460, \qquad \frac{1.00}{0.689} = 1.451,$$

against a spread of 3.598, which over three steps asks for $3.598^{1/3} = 1.5324$ each. The bottom step is 10.8% too large and the top two are about 5% too small, and no amount of re-cutting the gears inside this arrangement moves the shape — only where the whole ladder sits.

A Simpson three-speed is worse off still, having one number for two steps. Its ladder is $2+1/K$, $1+1/K$, $1$, and the steps are

$$\frac{2K+1}{K+1} \quad\text{and}\quad \frac{K+1}{K} .$$

At $K = 74/34$ that is 1.6852 and 1.4595, against an ideal of 1.5683 — 7.5% and 6.9% out, in opposite directions.

Asking for equal steps is a quadratic #

Since the steps are functions of $K$, “make them equal” is not a design goal but an equation. For the Simpson:

$$\frac{2K+1}{K+1} = \frac{K+1}{K} \;\Longrightarrow\; K(2K+1) = (K+1)^2 \;\Longrightarrow\; K^2 - K - 1 = 0,$$

whose positive root is

$$K = \frac{1+\sqrt5}{2} = \varphi = 1.6180\ldots$$

The golden ratio, arriving unbidden out of a gearbox. And it is not a coincidence of that arrangement: doing the same for the Ravigneaux’s three steps, with $u = 1 + K_2/K_1$, gives $u^2 - u - 1 = 0$, the same quadratic, so $u = \varphi$, $K_2 = \varphi$ and $K_1 = \varphi^2$. The four ratios then come out as

$$\varphi^2 : \varphi : 1 : \varphi^{-1} \;=\; 2.618 : 1.618 : 1.000 : 0.618,$$

with a spread of $\varphi^3 = 4.236$ and every step exactly $\varphi$. Reverse, which was not asked to join in, is $-(K_1-1) = -\varphi$.

The reason $\varphi$ shows up is not mysterious once it is written down. The ladders are built from expressions of the form one plus a reciprocal, and the fixed point of “one plus a reciprocal” is the equation that defines the golden ratio. Any gearset whose members compose that way will produce it.

The tooth counts are Fibonacci numbers #

$\varphi$ is irrational and a ratio of tooth counts is not, so the question is how close the integers get. The best rational approximations to $\varphi$ are ratios of consecutive Fibonacci numbers, and that is what the gearsets are:

• a Simpson with a sun of 55 in a ring of 89. $K = 89/55 = 1.61818$, planets of 17, ladder $2.61798 : 1.61798 : 1.000$, and the two steps are 1.61806 and 1.61798 — different from each other by 0.0024%, or one part in forty thousand.
• a Ravigneaux with suns of 34 and 55 in a ring of 89. $K_1 = 89/34$, $K_2 = 89/55$, long planets of 17, and the four ratios $2.61818 : 1.61818 : 1.00000 : 0.61798$, whose three steps differ by 0.0084%.

Two things are worth checking before this becomes a party trick, and both check out.

It assembles. A sun of 55 and a ring of 89 add to 144, which is divisible by 3 and by 4, so the gearset takes three or four equally spaced planets; the planets are 17 teeth, comfortably above the undercut limit, and they clear each other with room to spare. The assembly conditions are the ones that kill most nice-looking tooth counts, and these survive them.

The ratios are usable. 2.62 : 1.62 : 1.00 : 0.62 with a spread of 4.24 is an entirely ordinary four-speed ladder — deeper than some, taller than others, nothing out of range.

Two of the seven ratios coincide there #

The golden gearset is a single point in the plane of the two proportions, and something happens to the Ravigneaux’s full ledger at that point which does not happen anywhere else in it.

The seven exact ratios are $1+K_2$, $1+K_2/K_1$, $1$, $1-1/K_1$, $-(K_1-1)$, $1+K_1/K_2$ and $1+1/K_2$. Equal steps forced $K_2 = \varphi$ and $K_1 = \varphi^2$, and $\varphi^2 = \varphi + 1$, so every one of those expressions collapses onto a power of the golden ratio. The first becomes $1+\varphi = \varphi^2$. The sixth becomes $1 + \varphi^2/\varphi = 1 + \varphi = \varphi^2$ as well. The second becomes $1 + \varphi/\varphi^2 = 1 + 1/\varphi = \varphi$, and so does the seventh, $1 + 1/\varphi$.

So two pairs of the seven coincide. The gearset has seven ways of being driven and, at this one setting of its proportions, only five distinct ratios among them: $\varphi^2$, $\varphi$, $1$, $1/\varphi$ and a reverse of $-\varphi$. Two of the seven arrangements deliver a ratio another arrangement already delivers, which is a degeneracy that exists at no other point in the parameter plane.

The reverse is the neatest of them. $-(K_1 - 1) = -(\varphi^2 - 1) = -\varphi$, so the geometric Ravigneaux reverses at exactly the magnitude of its own second gear. That is a real and slightly startling property of a real gearbox layout — reverse and second identical in ratio, differing only in sign — and it is not a coincidence so much as the same identity $\varphi^2 = \varphi + 1$ turning up in a third place.

Nothing about that is a design argument, and it is worth saying so plainly rather than letting the tidiness carry it. Coincident ratios are not useful; a transmission with two ways to produce $\varphi$ has a redundancy rather than a feature, and a designer would not pay anything for it. What the coincidence is worth is as a check, and it is a strong one. The seven closed forms were derived independently of the quadratic that produced $\varphi$, so a computation that puts $K_2 = \varphi$ and $K_1 = \varphi^2$ into all seven and does not get two coincident pairs has an error in one of the two derivations. That is a test with a sharp verdict and no tolerance in it, arriving free from a fact about the golden ratio.

It also says something about why $\varphi$ turns up at all, which the essay above declined to call mysterious and is worth finishing off. The equal-step condition asks for a ladder in which each rung is the same multiple of the last, and the gearset’s forms are built from $1 + x$ and $1/x$ shapes. A number that is its own reciprocal plus one is the fixed point of exactly that pair of operations, and there is only one positive such number. The gearbox is not producing the golden ratio because of anything about gears; it is producing it because the ladder’s own recurrence and the gearset’s algebra are the same recurrence.

What a step is for #

Before deciding whether the geometric rule is worth obeying, it is worth being precise about what a step does, since the usual justification is stated in a hurry.

An engine has a useful band of speeds. Driving along at a given road speed, the transmission’s ratio decides where in that band the engine sits. Shifting up multiplies the road-speed-to-engine-speed relation by the step, so the engine drops by exactly the step and the driver — or the controller — is handed the engine at a new operating point.

If every step is the same size, then every shift drops the engine by the same proportion, so a single strategy works at every gear: run up to the top of the band, shift, land at the same place, repeat. That is the whole of the argument, and it is exactly the argument for a geometric progression: equal ratios between neighbours means the ladder is $r_0, r_0/s, r_0/s^2, \ldots$

Its assumption is that the engine’s band is the same in every gear and that the object is to stay inside it. Both are true for a truck climbing a hill and for a machine tool cutting at a fixed surface speed; neither is quite true for a car, where the bottom gear is used from rest and the top ones at part throttle.

So why does nobody build it #

This is where the finding has to be handled honestly, because “the mathematically perfect gearbox exists and industry ignores it” is the kind of sentence that is nearly always wrong.

The equal-step rule is not what a transmission engineer is actually trying to achieve, and the departures in real gearboxes are in a consistent direction that says so. Look again at the four-speed measured above: bottom step 1.699, then 1.460 and 1.451. The bottom step is deliberately the biggest and the top steps are deliberately close together. That shape has a name — a progressive ladder — and the reasons for it are about what the gears are for:

• The bottom gear exists to get a heavy vehicle moving from rest, and how far the engine falls on the shift out of it matters much less than how much torque multiplication it provides. Making first deeper costs almost nothing in driveability.
• The top gears are used at cruise, where the engine is near its efficient band and where a large step means either revving needlessly or labouring. Close spacing at the top is worth paying for.
• A torque converter blurs the bottom of the ladder anyway, multiplying torque continuously below its coupling point, which further reduces the value of a precise first-to-second step.

So the geometric ladder is a rule of thumb aimed at a use — hold the engine in a band, all the way up — that fits a truck, a racing car or a machine tool much better than a passenger car. The real result of this essay is not that manufacturers are wrong. It is the structural one: whether a gearbox’s ladder is geometric or progressive, it has at most two numbers in it, so its shape is chosen once for all the gears at once and cannot be tuned gear by gear. What a designer picks is a family, not a list.

And the family that is geometric is a specific point in the parameter plane with a closed form, and the tooth counts nearest to it are Fibonacci numbers. That is worth knowing even if nobody wants to be there, because it says exactly how far from geometric an arrangement is and in which direction.

What it would take to fix #

Suppose the geometric ladder were wanted after all. What is the cheapest change to a Ravigneaux four-speed that gets it?

Not re-cutting the gears within the arrangement, because the arrangement’s two numbers are already spent — and this is where the closed forms earn their keep. The equal-step condition fixes $K_1 = \varphi^2$ and $K_2 = \varphi$, and those two numbers then fix first gear at 2.618 and reverse at −1.618, whether anybody wanted them or not. If a designer needs a first gear of 2.9 and a reverse of −2.6, the equal-step ladder is unavailable at any tooth count; there is nothing left to spend.

So the fix is a third number, which means a third proportion, which means another gearset. A simple planetary in front of a Ravigneaux — the arrangement used in most six-speed automatics — brings its own ring-over-sun and takes the design space from two dimensions to three. With three numbers and six gears the ladder is still not free, but there is now enough room to hit a spread, a first gear and a step profile at the same time.

That is a useful way to read the history of automatic transmissions. Three speeds from one number, four from two, six from three: each generation adds a gearset, and what the gearset buys is not gears — those come from clutches — but a degree of freedom in the ladder’s shape.

What “spread” hides #

One more consequence, since spread is the number gearboxes are advertised with.

Spread is top over bottom and says nothing about what is between. Two four-speeds with a spread of 4.24 can have completely different ladders: the geometric one with three steps of 1.618, or one with a first step of 2.0 and two of 1.46. The advertised number is the same and the cars drive differently.

For a given arrangement, though, spread is more informative than it looks, because it is a function of the same one or two parameters as everything else. A Simpson’s spread is exactly its first gear, $2 + 1/K$, since third is 1 and there is no overdrive; so a Simpson advertised with a spread of 2.46 has declared $K$, and therefore its second gear, and therefore both its steps. One number quoted, and the whole gearbox is determined. That is not true of a countershaft manual gearbox, where every pair is independent, and it is one of the ways in which planetary automatics are a different sort of object.

The other ladders in this field #

Two of the field’s mechanisms have no ladder at all, and putting them beside the argument sharpens what a ladder is.

A continuously variable transmission has no steps, so the whole geometric-progression question evaporates: at every road speed the controller picks the ratio it wants rather than the nearest of four. That is the mechanism’s whole selling point, and what it costs is that the ratio stops being a count and starts being a quotient of two solved lengths.

A reverted train has one ratio and no ladder, and its difficulty is of a completely different kind: getting a single exact ratio out of it while keeping its shafts coaxial turns out to be a Diophantine problem with no solutions at all below a certain size.

Between the three, the field has a nice spread of what “a ratio is hard to get” can mean. For a gearset the ratios are easy individually and constrained collectively — any one of them can be had, and not all four at once. For a reverted train one ratio is hard because two integer conditions have to hold together. For a variator every ratio is available and none of them is exact.

The check #

The gate asserts four things separately, and the third is the one that keeps the others honest.

That $\varphi$ solves the quadratic the ladders produce. That the Fibonacci gearsets’ steps are equal to within a hundredth of a per cent. That a catalogue four-speed’s steps are not — the check fails if the ordinary gearset ever comes out closer than five per cent — so the first two are a discovery rather than a property of the measurement. And that the Fibonacci gearsets pass the assembly conditions, without which the whole result would be a piece of arithmetic about numbers rather than a statement about a gearbox that could be cut.

One last consequence of reading the ladder as a recurrence rather than as four numbers. A geometric ladder is the only one whose shape is unchanged by adding a gear at either end: append a rung above the top and the ratios are still in geometric progression, with the same step, and every strategy built on the ladder still applies. Any other ladder has to be redesigned when a gear is added, because the steps that were tuned around a particular top gear are now interior steps with a different job. That is a property nobody asks a gearbox for and it is quietly the reason the rule survives — a designer who follows it has a family of transmissions rather than a design, and the eight- and nine-speed automatics that came later were extending a ladder rather than replacing one.