Holding a member chooses the ratio

A gearset with two degrees of freedom and one shaft turning is not yet transmitting anything. Its output can still do whatever it likes; the mechanism has an opinion only about the combination. To get a ratio, something has to remove the second freedom, and the hardware has exactly two ways of doing it:

• a brake, which holds one member to the case;
• a clutch, which locks two members to each other.

Both are one linear equation, both are homogeneous, and each therefore selects a line out of the plane of permitted motions. That is the whole of what a shift is: not a re-arrangement of the gears, which do not move, but a change of which line is being used.

So the question what ratios does this gearbox have becomes a finite enumeration. List the members; try every brake; try every clutch; try each possible input; see what comes out. It takes no cleverness and it is not how gearboxes are usually presented, and doing it turns up three things worth an essay.

Three outcomes, and only three #

Add one row to a gearset whose input is already fixed and the null space’s dimension is 1 or 0. In practice three things happen.

A ratio. The usual case. One line is left and the output turns at a fixed proportion of the input.

The output is held. A brake on the output member gives a perfectly good line with the output’s entry zero. The input turns, the mechanism is happy, and nothing comes out. On the Ravigneaux gearset three of the twenty-seven combinations are of this kind — one for each possible input, all of them being hold the carrier, the carrier being the output. This is not a fault and it is not neutral. It is the gearbox with the output shaft locked to the case, which is what a parking pawl does from outside and what some transmissions do from inside during a hill hold.

Nothing can turn at all. Two brakes on a two-freedom gearset take both freedoms, so the whole train is a structure and the input is stationary. The enumeration here never produces it, because it applies one element at a time; a real transmission produces it constantly, whenever a shift overlaps, and the resulting state is called a tie-up. It is the reason shift elements are timed rather than switched.

All six clutches are the same gear #

The first surprise, and it is a consequence of the lever’s own foundation.

A gearset whose frame carries no teeth admits the motion in which every member turns at once. Now impose a clutch — any clutch, between any two members. That is one equation, $\omega_a = \omega_b$, and the block rotation satisfies it. So does no other motion, generically. Therefore locking any two members of a gearset locks all of them, and every clutch in the enumeration returns exactly 1.

For the Ravigneaux that is eighteen of the twenty-seven rows: six possible clutches times three possible inputs, all returning the same fraction $1/1$.

The engineering consequence is direct. A transmission fits one direct-drive clutch, not six, and which two members it joins is a packaging question — which drums are next to each other, where the oil can get to — rather than a kinematic one. Every choice gives the identical gear. That is a rare thing in mechanism design: a decision that genuinely does not matter, and it is worth knowing which decisions those are.

Neutral is not a gear #

Which leaves the state a gearbox spends a good deal of its life in, and which does not appear in the table at all.

Neutral is the gearset with nothing engaged. No extra row, two degrees of freedom, the input turning and the output free. It is not a ratio, not a line, not a choice — it is the mechanism in the condition it is naturally in, and every gear is a restriction of it.

That is worth stating plainly because the usual mental model runs the other way round: a gearbox is imagined as a device that is normally connected and is sometimes disconnected. Kinematically it is the opposite. A planetary gearset connects nothing. Its members are free to windmill against each other in a one-parameter family for every speed of the input, and what a transmission does is spend hydraulic pressure to forbid all but one of those motions at a time. Take the pressure away and the mechanism goes back to permitting everything, which is why an automatic transmission with no oil pressure freewheels rather than seizes.

The site’s own mobility vocabulary makes the point exactly: a mechanism with two freedoms and one input is not a mechanism that has failed to be determinate. It is a mechanism whose determinacy is somebody else’s job.

One gearset, six numbers #

The simplest case makes the counting visible. A single planetary has three shafts, so there are three choices of which member is the output, and for each of those the enumeration returns one ratio per brake plus direct drive. Six non-unity ratios in all, and every one exact:

Three of them are reductions, three are overdrives, and they pair off as reciprocals because swapping the input and the output of a one-freedom train inverts its ratio and does nothing else. The two carrier-held cases run backwards, for the betweenness reason rather than because of a minus sign anybody has to remember.

The Simpson gearset, against a real one #

A single planetary gives one useful reduction per output choice, and a car needs several without changing which shaft comes out of the box. The Simpson arrangement is the classical answer and it is the arrangement in more automatic transmissions than any other: two simple planetaries sharing one sun, with the front carrier joined to the rear ring, both being the output.

Six members, and eighteen combinations across the two inputs a real unit provides. They give five distinct ratios. With a sun of 34 and both rings at 74:

A three-speed automatic built on this gearset was sold with 2.45, 1.45, 1.00 and a reverse of 2.20. Three of those four are reproduced to three figures from two tooth counts, and the fourth to within two per cent — which is the tolerance on my guess at the tooth counts rather than on the arithmetic. The point is not the coincidence; it is that the shift pattern of a real transmission is an enumeration of one matrix’s extra rows, and the numbers on the side of the box are the entries of a null space.

The fifth ratio is the interesting one. 3.1765 is a perfectly good reduction, deeper than first gear, available from the same casting with a brake the gearbox already has on a member it already has — except that reaching it requires driving the sun and holding the front ring, and the unit is not built to drive the sun in forward. It is a gear the mechanism has and the hardware does not buy.

Where the ratios sit in the plane #

The enumeration is a list, and a list hides the geometry it came from. The same content drawn as a picture makes the counting obvious.

Put two members’ speeds on a pair of axes. Every motion the gearset permits is a point; the whole plane is available because the gearset has two freedoms. Now each shift element is a line through the origin, and a gear is the direction of that line — the slope, which is the ratio.

Three things become visible at once. The clutches are all the same line, at forty-five degrees, because “these two members turn together” is one condition however it is applied and the diagonal is where every member’s speed is equal. The brakes are different lines, one per member, fanning out around the origin. And the sign of a gear is which quadrant its line runs through: a line in the first and third quadrants is a forward gear and one in the second and fourth runs backwards, so reverse is not a special case of anything, it is a line that happens to have a negative slope.

That picture is also the cleanest statement of why neutral is not on the list. Neutral is not a line in the plane. It is the plane.

What each gear costs #

This is where the enumeration stops being a curiosity and starts being the design problem.

A gear costs one element, and an element is a clutch pack or a band with hydraulic pressure behind it, a piston, a feed circuit and a control. A gearset offers more lines than a gearbox can afford, and a transmission is the subset somebody chose. Reading the Ravigneaux’s table with that in mind:

• 2.4800, 1.4600, 1.0000 and 0.6892 make a usable four-speed ladder, and reverse at −2.2174 comes free from an element already fitted for one of the forward gears.
• 1.6757 sits between second and third and would make the ladder better, and needs an input clutch to the ring that the unit does not have.
• 3.1739 is a deeper first gear, and needs the small sun driven, which is a different input clutch again.

Both leftovers need an input element rather than an output one, which is the general shape: adding a brake gives one more gear on the existing input, and adding an input clutch gives a whole family. That is why a four-speed automatic has two input clutches and a three-speed has one, and why the step from three gears to four is a bigger change to the hardware than the step from four to five.

The same enumeration on a differential #

One more case, because it shows the enumeration answering a question nobody would think to pose it.

A differential is a two-freedom gearset like any other, and the enumeration applies unchanged. Its members are the two wheels and the cage; there are three brakes and three clutches, and with the cage as input and one wheel as output:

Five outcomes, and every one of them is a thing that happens to a car. The diff lock is a clutch in exactly the sense of this essay — it is the same object as a gearbox’s direct-drive clutch, applied to a mechanism nobody calls a gearbox — and it takes the axle from two freedoms to one. Locking either wheel to the cage does the same job, which is the all-clutches-are-one-gear fact again and is why a lock can be fitted wherever there is room.

What the enumeration does not know #

Two limits, since a complete list of a gearset’s ratios reads like a complete list of its possibilities.

It does not know which pairs of elements can be applied at once. Every row here applies exactly one. A real transmission applies two or three simultaneously in the higher gears of a six-speed, and a pair of elements on a two-freedom gearset generally takes both freedoms and locks the input — which is why the ones that do not are worth finding and why the enumeration would have to be extended to pairs to find them.

It does not know which gears are usefully ordered. A gearbox needs its ratios in a sensible sequence with a shift pattern that changes one element at a time, and the enumeration returns an unordered set. Two of the Ravigneaux’s seven sit between gears the transmission uses, and fitting them would mean a shift that releases two elements and applies two others — mechanically possible and, in a hydraulically controlled box, a much harder thing to time than a single swap.

Both limits are about the controller rather than the mechanism, which is the right place for them: this essay’s claim is about what the gearset offers, and what a gearbox takes up is a separate decision made by somebody with a different set of constraints.

The gears are vertices and the shifts are edges #

The enumeration produces a set of ratios, and a gearbox is not a set of ratios — it is a sequence of them with shifts between. The list above supplies the vertices of that problem, and one more piece of bookkeeping supplies the edges.

Each gear in the table is one element applied. A shift from one gear to the next means releasing what is held and applying something else, and the number of elements that change decides how hard the shift is to control. Changing one element — release a band, apply a clutch, with the two overlapping briefly — is a single-transition shift, and it is the only kind a hydraulic control can execute smoothly without a third element to carry the load through the middle. Changing two means the gearbox passes through neutral or through a locked state on the way, and both are audible.

So put an edge between two gears when they differ by one element, and a usable ladder is a path through that graph whose vertices appear in ratio order. That is a genuine combinatorial condition on the design, it is decidable from the same table the enumeration produces, and nothing in the ratios themselves reveals it: two gearsets with identical ladders can differ entirely in whether their gears can be shifted between in order.

That reframes what “adding a gear” costs. An extra ratio is an extra element, which the essay above already prices in hardware. What the graph adds is that the extra ratio must also sit adjacent in the shift graph to its neighbours in the ladder, or it is a gear that exists and cannot be reached from the gears on either side of it. A ratio like that is not useless — it can be entered from somewhere else, or used as a fixed range — but it is not a rung of the ladder, and counting it as one overstates what the gearbox has.

It also explains a fact about real automatics that the ratio table makes look arbitrary. Their gear ladders are not the seven or eight best-spaced ratios the gearset offers; they are a subset chosen so that consecutive gears are single-transition shifts, which frequently means passing over a ratio that would have spaced the ladder better. The compromise is between a quantity this essay computes and a constraint this essay does not, and knowing that the second exists is what keeps the first from being read as a design.

The check that has to be able to lose #

The gate asserts that a gearset offers more distinct ratios than the transmission is sold with, and that at least one of them runs backwards. Both halves can fail: a gearset whose enumeration collapsed to the sold list would report it, and so would one whose reverse had gone missing because a sign had flipped somewhere in the elimination.

The comparison of ratios is by exact fraction, which is the part that would be silently wrong in floating point. Two combinations giving 1.4599999999999999 and 1.46 are the same gear or they are not, and the enumeration has to know; here they are $50/73$ and $50/73$, and the question does not arise.

The next essay takes the same table and asks the question a designer asks: given that the ladder is a consequence of two tooth counts, how much of it is actually chosen?

There is a third structure on the same vertices worth naming, since it costs nothing once the graph is drawn. Two gears sharing an element can be brought about by applying a second element while the first stays on, which is the arrangement a controller uses to hold torque through a shift rather than releasing everything and catching it again. So the edges are not merely a count of differences; they carry a direction, and a shift that adds an element behaves quite differently from one that removes it. None of that is visible in a ladder of four numbers, and all of it is decided by the same table of which member each gear holds.