Four speeds from two numbers

The Ravigneaux gearset has five gears in it: two suns of different sizes, a set of short planets, a set of long planets, and a ring. The short planets mesh the large sun and the long planets; the long planets mesh the small sun and the ring; everything is carried on one carrier. It is the arrangement in a great many four-speed automatics, and drawn in section it looks like a great deal of gearing.

Five tooth counts. Ask the enumeration what ratios it offers and seven come back, all exact fractions. Then ask which of the five counts each ratio depends on, and the answer is unexpected enough to be the essay.

The short planet does not appear #

Sweep the short planet’s tooth count across every size the geometry admits — from 12 up to 48, wherever the two orbit radii and the planet-to-planet centre distance still close as a triangle — and recompute the whole table each time.

Nothing moves. Not approximately: the seven fractions come back identical, $74/51$, $1$, $50/73$, $37/62$, $25/62$, $23/73$, $-23/51$, at every admissible size.

The reason is that the short planet is an idler. It sits between the large sun and the long planet and transmits between them; an idler contributes a reversal of sense and no proportion at all, because its tooth count enters the two meshes it takes part in once on each side and cancels. The site has met that before, in an ordinary train where an idler changes the direction and not the ratio, and the enumeration reproduces it here without being told.

So a designer choosing the short planet is choosing a packaging constraint — how far apart the two planet axes sit, how much room the pinion pins have, whether the assembly will go together — and is not choosing anything about the gears the car will have. That is a genuinely free parameter, and knowing which parameters are free is worth as much as knowing what the constrained ones do.

Nor does the size of anything #

The second reduction is a scale argument and is easy to check: double every tooth count in the gearset — suns 46 and 100, ring 148, planets 24 and 60 — and ask again.

The identical seven fractions. Which says the ratios are functions of the proportions only, and since the coaxial condition ties the long planet to the small sun and the ring, there are exactly two independent proportions in the whole gearset:

$$K_1 = \frac{z_{\text{ring}}}{z_{\text{sun}_1}}, \qquad K_2 = \frac{z_{\text{ring}}}{z_{\text{sun}_2}}.$$

Here $K_1 = 74/23 = 3.2174$ and $K_2 = 74/50 = 1.48$.

Why an idler cancels #

The short planet’s disappearance is worth deriving rather than asserting, because the cancellation is the reason the whole reduction works and it is visible in one line.

The short planet takes part in two meshes, one with the large sun and one with the long planet, and both are carried by the carrier. Write the two rows:

$$z_{S_1}(\omega_{S_1} - \omega_C) + z_{sp}(\omega_{sp} - \omega_C) = 0, \qquad z_{sp}(\omega_{sp} - \omega_C) + z_{lp}(\omega_{lp} - \omega_C) = 0 .$$

Subtract, and $z_{sp}(\omega_{sp} - \omega_C)$ goes. What is left is

$$z_{S_1}(\omega_{S_1} - \omega_C) \;=\; z_{lp}(\omega_{lp} - \omega_C),$$

with no $z_{sp}$ in it and no $\omega_{sp}$ either. The short planet has been eliminated entirely, and what remains is the relation an internal mesh between the large sun and the long planet would have given — which is what an idler does: it changes the sense and nothing else.

That is the general statement, and it is why an idler’s tooth count is famously irrelevant to a train’s ratio. What is new here is that the same fact survives into a gearset with two degrees of freedom, where “the ratio” is not one number: the idler drops out of the relation, so it drops out of every ratio the relation can be cut into.

The ladder, in closed form #

With those two numbers the whole table can be written out, and every entry is a short expression:

Seven expressions in two variables. That is the whole kinematic content of a four-speed automatic transmission’s gear set, and it is why two gearboxes with completely different tooth counts can have the same gears and why a manufacturer’s ratio table is so much shorter than its parts list.

The forms are checked against the null space at three different gearsets, including two that appear in no figure, and they agree to the last digit available. Two routes: the closed forms know which member is a sun and nothing about matrices; the elimination knows about matrices and nothing about suns.

A three-speed is one number #

The Simpson gearset makes the point harder. Two simple planetaries share one sun; in the classical unit both rings are the same size, so there is only one proportion in the entire gearbox:

$$K = \frac{z_{\text{ring}}}{z_{\text{sun}}}.$$

And the ladder is

Read the first two rows again: first gear and second gear differ by exactly one. Not approximately, not by design intent — the two expressions are $2 + 1/K$ and $1 + 1/K$, so their difference is 1 for every gearset of this arrangement ever made. A Simpson three-speed’s first and second gears are always a whole number apart, and 2.45 with 1.45 is not a coincidence of that particular gearbox.

That is the kind of statement the closed forms are for. From a table of ratios it is an observation about one transmission; from the formula it is a property of the arrangement.

Three identities the ladder cannot escape #

Seven ratios written in two numbers means five relations among them, and the relations can be written down without knowing either number. They are the sharpest form of this essay’s claim, because they are testable against any Ravigneaux ever built without measuring a single tooth.

Take the ratios in the order given. The first is $1 + K_2$ and the last is $1 + 1/K_2$, so subtracting one from each leaves $K_2$ and $1/K_2$, whose product is one. The second is $1 + K_2/K_1$ and the sixth is $1 + K_1/K_2$, and the same subtraction leaves a reciprocal pair again. The fourth is $1 - 1/K_1$ and the reverse is $1 - K_1$, so one minus each leaves $1/K_1$ and $K_1$, and their product is one as well.

Three exact statements, each involving two of the seven ratios and neither of the two design numbers:

First and seventh. $(r_1 - 1)(r_7 - 1) = 1$.

Second and sixth. $(r_2 - 1)(r_6 - 1) = 1$.

Overdrive and reverse. $(1 - r_4)(1 - r_5) = 1$.

The third of those is the one with teeth, because both of its members are gears a real gearbox actually uses. It says the reverse ratio of a Ravigneaux is determined by its overdrive, exactly, with no freedom left. An overdrive of 0.6892 forces a reverse of 2.2174 and nothing else, and a designer who wants a deeper reverse must accept a taller overdrive to get it — not as a trade-off to be balanced but as an equality.

Reading it against the published transmission is instructive and slightly awkward, which is why it is worth doing. That gearbox is quoted at 0.67 overdrive and 2.20 reverse. Put 0.67 into the identity and it demands a reverse of 2.0303; put 2.20 in and it demands an overdrive of 0.6875. Neither published figure sits on the curve the other implies, so the two quoted numbers cannot both be exact ratios of one Ravigneaux gearset. The likeliest explanation is the ordinary one — published ratios are rounded for a brochure — and the identity is what makes the rounding visible, since 0.67 and 2.20 are exactly the shapes a rounded number has.

One caution about how far the identities reach. They are statements about the gearset’s ratios and not about the gearbox’s gears, and a real automatic does not necessarily use all seven. A design that takes only four of the seven to the road is still bound by every identity connecting the four it uses, and is bound by nothing at all connecting a used ratio to an unused one — so a claim that two published gears must satisfy a relation needs the relation to hold between those two gears specifically, which is why the overdrive-and-reverse pairing is the one worth quoting and the other two mostly are not.

That is the general use of a relation like this. A closed form in two parameters can be fitted to any four numbers and will report a residual; an identity involving no parameters at all can be checked against two numbers and reports a contradiction. The second is the stronger instrument and it is available only because the reduction was carried all the way down: seven ratios from five tooth counts would have no such identities, seven from two do, and the identities are what two-ness looks like when it is written on the ratios themselves rather than on the gears.

The Simpson set carries the same structure one size down. Its ladder is a function of one number, so its three ratios satisfy two relations, of which the essay has already named one — first and second differ by exactly one. That is an identity of precisely this kind, arrived at from the other end, and it is why it can be stated as a fact about every Simpson three-speed rather than about the one in the figure.

Choosing the two numbers #

Now the design problem, which is what makes the reduction interesting rather than merely tidy.

A four-speed wants a bottom gear deep enough to move the car, a top gear tall enough to be an overdrive, a reverse of about the same depth as first, and steps between them that are not wild. That is four wishes and two numbers. Something has to give.

Work it through with the Ravigneaux’s forms. Ask for first gear at 2.48: that fixes $K_2 = 1.48$ and there is nothing further to say about the small sun. Ask for reverse at −2.22: that fixes $K_1 = 3.22$ and there is nothing further to say about the large sun. Both numbers are now spent, and second gear is $1 + K_2/K_1 = 1.46$ and the overdrive is $1 - 1/K_1 = 0.689$ whether anybody wanted those or not.

That is what the title means. The four-speed ladder 2.48 : 1.46 : 1.00 : 0.689 is not four decisions. It is two decisions and two consequences, and the consequences are exactly where the ladder is at its least satisfactory — the step from first to second is 1.70 while the other two steps are 1.46 and 1.45, which is the next essay’s subject.

The alternative reading is available too and is worth stating, because it is how a real gearbox is designed. Fix the two steps that matter most and accept whatever first gear and reverse come out. The forms invert perfectly well; there is simply no assignment of two numbers that makes all four wishes come true, and the transmission engineer’s job is deciding which wish to break.

What the search actually was #

The tooth counts here were not looked up. They were found, and how they were found is worth recording because it is the same shape of problem the clock trains essay met in the timing field.

A published four-speed on this arrangement has 2.46, 1.46, 1.00, 0.67 and a reverse of 2.20. Turning those into tooth counts is: find integers whose $K_1$ and $K_2$ reproduce the ladder, subject to the ring and the small sun differing by an even number so the long planet is whole, and to the triangle inequality on the two planet orbits.

Inverting the forms gives $K_2 = 1.46$ and $K_1 = 3.20$ directly, and the rest is search. A ring of 74 with a small sun of 50 gives $K_2 = 1.48$; a large sun of 23 gives $K_1 = 3.2174$. Second gear then comes out at exactly $1.4600$ — the published figure to four figures, from a fraction $50/73$ that was never asked to be anything in particular.

The residual disagreements are honest and are worth quoting rather than tidying: first gear 2.4800 against 2.46, the overdrive 0.6892 against 0.67, reverse −2.2174 against 2.20. Those are the gaps between the real gearbox’s tooth counts and my guess at them, not gaps between the arithmetic and the world. The arithmetic has no error in it at all: every one of those numbers is an exact fraction.

Reading the same result off the lever #

The closed forms are one route and the null space is another, and there is a third that needs no arithmetic at all.

The Ravigneaux’s lever has its four shafts at 0 (small sun), 0.3151 (carrier), 0.5280 (ring) and 1 (large sun). Every one of the seven ratios is a reading of that line, and the two dimensionless numbers are exactly what fixes the two interior positions: the carrier at $K_2/(1+K_2)$ of the way from the small sun to the ring, and the ring wherever $K_1$ puts it.

So “four speeds from two numbers” has a geometric statement: a four-station lever has two interior positions, and the ladder is a function of where they are. The ends are the two extreme shafts and can be scaled away; the two interior coordinates are the whole of the design. A gearset with five shafts would have three interior positions and three numbers, which is why a six-speed automatic is built by adding a planetary rather than by adding a clutch.

What the reduction is an instance of #

It is worth naming the general shape, because the same argument answers a question that keeps coming up in this field: how much of a mechanism’s behaviour is decided by how much of its description.

A gearset is described by a handful of integers. Its behaviour — the set of ratios it offers — is a function of those integers. The reduction says that the function factors through a much smaller set of numbers: two, here, and one for the Simpson. Everything else in the description is either an idler, whose count cancels, or a scale, which the ratios are blind to.

That is a dimensional analysis of a sort, and it has the same payoff as one: it says what a designer is actually choosing, and it says which experiments are the same experiment. Two Ravigneaux gearsets with the same $K_1$ and $K_2$ and completely different tooth counts are the same gearbox as far as any ratio measurement can tell, and a test programme that measured both learned one thing.

The site has done this before in a different currency. The pawl’s holding criterion turned out to be a comparison of directions, so it gave the same critical angles to twelve digits on wheels of 0.13, 1, 7 and 55 units of radius: the verdict does not know how large the mechanism is. Here the ratios do not know how large the gearset is, for the same structural reason — the quantity being computed is a ratio of like quantities and there is nowhere for a scale to enter.

What differs is which parameters cancel. There, it was every length at once. Here, it is one tooth count entirely and the overall scale of the rest, and the two that survive do so because the coaxial condition ties the remaining counts together and leaves exactly two independent proportions.

Where the two numbers stop being enough #

Two remarks on the boundary of the reduction, since it is easy to over-read.

Adding a gear does not add a number. A five-speed built on the same gearset would still be a function of $K_1$ and $K_2$; the extra gear comes from an extra element, and its ratio is one of the two unfitted forms above. That is why five- and six-speed automatics are built by putting a whole extra planetary in front of the Ravigneaux rather than by re-cutting it: an extra gearset brings an extra number, and an extra clutch on the existing gearset does not.

The tooth counts are not free even once $K_1$ and $K_2$ are chosen. The ratios depend on the proportions, but whether the gearset can be assembled depends on the integers themselves — how many planets will fit, whether the counts divide, whether the planets clear each other. A ratio target picks a line in the $(z_{\text{sun}}, z_{\text{ring}})$ plane and the assembly conditions pick a lattice; a real gearset is a point on both, and there are far fewer of those than the ratio arithmetic suggests. That is the reductions a planetary cannot give, and it is the phase’s most surprising number.