The reductions a planetary cannot give

Ask what reductions a planetary gear set can give and the formulas look accommodating. With the ring held and the sun driven it is $1 + z_R/z_S$; with the sun held and the ring driven it is $1 + z_S/z_R$; with the carrier held it is $-z_R/z_S$; and the other three are the reciprocals. Six expressions in one variable $k = z_R/z_S$, which is a positive number a designer chooses.

Nothing in that suggests a restriction. But $k$ is not a free positive number. It is a ratio of two integers that must also satisfy three conditions having nothing to do with ratios, and once they are imposed, the set of reachable reductions is a finite set with a floor, a ceiling and holes in it — one of which is large, is exactly where the closed form says it is, and contains the single most useful reduction there is.

The three conditions, and none of them is about ratios #

The coaxial condition. A planet sits between the sun and the ring, so $z_R = z_S + 2z_P$. The immediate consequence is that $z_R > z_S$ always, so $k > 1$ always, so the ring-held reduction $1 + k$ is greater than 2 on every planetary ever made. That bound needs no catalogue and no computation. It is the first thing the assembly geometry says, and it already contradicts the reading that any reduction is available.

The undercut limit. A planet cannot be arbitrarily small. Below about seventeen teeth at a twenty-degree pressure angle the cutter removes part of the involute near the root and the flank stops being conjugate, which is the seventeen-tooth rule the teeth field established. So $z_P \ge 17$, and therefore $z_R \ge z_S + 34$, and therefore

$$k \;\ge\; 1 + \frac{34}{z_S}.$$

A limit on the sun. The sun cannot be arbitrarily large either, because at a fixed module a large sun is a large gearbox. Take 60 teeth as the biggest that belongs in this catalogue and $k \ge 1 + 34/60 = 1.567$; the largest admissible value that also divides properly is $k = 92/58 = 1.5862$.

Put those together and the ring-held reduction cannot be below $1 + 1.5862 = 2.5862$, which is more than a quarter above the naive floor of 2. The three conditions that produce it are about cutting, about assembling, and about size — and none of them mentions a ratio.

Sweeping the catalogue #

Enumerating is easy once the conditions are written down. Run the sun from 17 to 60 and the ring up to 140; keep the pairs whose difference is even and whose planet clears the undercut limit; keep those whose counts sum to a multiple of three, so that three equally spaced planets will actually go in; and keep those whose planets do not touch each other.

506 designs survive. Their ring-held reductions run from 2.5862 to 9.1765 — the floor exactly as predicted, and a ceiling set by how big a ring the catalogue allows.

Now do the same for the other five configurations. The bands come out as:

The two forward-reducing bands are the first two, and they do not meet. One stops at 1.6304 and the other starts at 2.5862, and there is nothing at all between them.

The gap, and its closed form #

The gap is not an artefact of the catalogue’s edges; it has an exact description. The lower band’s top is $1 + 1/k$ at the smallest $k$, and the upper band’s floor is $1 + k$ at the same smallest $k$. So the gap runs from $1+1/k_{\min}$ to $1+k_{\min}$, and its width as a factor is

$$\frac{1 + k_{\min}}{1 + 1/k_{\min}} \;=\; k_{\min}$$

exactly, by one line of algebra. Measured over the sweep, the widest hole has a factor of 1.5862069 and $k_{\min}$ is 1.5862069. Two routes, one of them an enumeration of thirty thousand tooth-count pairs and the other a cancellation.

Which means the gap is not a curiosity of a particular catalogue. It is present for every choice of undercut limit and sun limit, and its width is exactly the smallest ring-over-sun ratio those limits allow. Relax the undercut limit to twelve teeth and $k_{\min}$ falls to about 1.41 and the hole narrows accordingly, and the check asserts that it does — because a hole that did not respond to the condition that supposedly causes it would be evidence that something else was causing it.

And 2 is inside it #

The reduction that falls in the hole is 2. Halving a speed: the most ordinary thing a gearbox is ever asked for, the ratio of the second hand to nothing in particular, the ratio a designer reaches for when a motor turns twice as fast as the thing it drives.

A single planetary gear set cannot do it. Not with any tooth counts, in any of its six configurations, at any size.

That is worth sitting with, because it is the opposite of what the formula suggests and the opposite of what intuition suggests. A 2:1 reduction is available from a plain pair of spur gears with 20 and 40 teeth and no thought at all; it is available from a chain, a belt or a worm; and it is unavailable from the mechanism that is specifically used when a large reduction is wanted in a coaxial package. Anyone who needs 2:1 in a coaxial epicyclic must use two stages, or a compound gearset, or accept 1.63 or 2.59.

Reading it as a design procedure #

Turned round, the sweep is a usable procedure rather than a curiosity, and it is worth writing down because the usual one is wrong.

The usual procedure is: pick the reduction, solve $1 + z_R/z_S = R$ for the ratio of the counts, choose a convenient pair. That produces a design that satisfies the ratio and has a good chance of failing one of the other two conditions, and the failure is discovered when the second planet will not go in.

The procedure that works runs the other way. Enumerate first, then choose. The 506 designs are a small enough list to hold, they all assemble, and picking from them means picking a reduction that exists rather than one that ought to. In practice the list is generated for the planet count and size limits of the job in hand — five minutes of arithmetic — and then read.

Two things fall out of having the list that do not fall out of the formula. The nearest achievable reduction to a target is immediate, including how far away it is; and the cost of moving the target is visible, because the local spacing of the reachable set is what a per cent of ratio actually costs in tooth counts. Near the floor the set is dense, at 0.5% spacing, so a specification of 2.6 can be met almost exactly. Near the ceiling it is sparse at 4%, so a specification of 9.0 cannot be met at all and 8.82 or 9.18 are the options.

That asymmetry is the practically useful half of the essay. A deep reduction is not merely harder to package; it is harder to hit, and the tolerance a designer must accept on the ratio itself grows with the ratio.

Where the holes at the top come from #

The gap in the middle is structural. The other four holes the sweep reports — at 5.79 to 6.00, 6.00 to 6.23, 8.47 to 8.67 and 8.82 to 9.18 — are of a different kind, and the difference is worth naming.

They are discreteness, not exclusion. High reductions need $k$ large, which needs a small sun in a large ring, and there are simply fewer integer pairs up there: the lattice of admissible $(z_S, z_R)$ thins out as the sun shrinks, because the divisibility condition removes two thirds of the pairs and the ring limit caps the other axis. The holes at the top are between 2% and 4% wide, which is the local spacing of the lattice rather than a forbidden region.

The distinction matters for design. A hole of the first kind cannot be got round by relaxing anything except the physics — it is there for every catalogue. A hole of the second kind disappears the moment the catalogue is widened, by allowing a bigger ring or a smaller module or a fourth planet.

What widening the catalogue actually buys #

Since two of the three conditions are engineering limits rather than laws, it is fair to ask how much is bought by moving them.

Going from seventeen teeth to twelve — which means accepting undercut planets, or shifting the profile to cure the undercut — moves the floor from 2.586 to 2.400 and adds seventy-four designs. It does not reach 2. Nothing reaches 2, because the coaxial condition alone forbids it: $1 + k$ with $k > 1$ is greater than 2 whatever any cutter does.

That is the cleanest statement of the result. Two of the three conditions set where the floor is. The third — the coaxial condition, which is the definition of a planetary and cannot be relaxed at all — sets that there is a floor, and puts it above 2.

The bands as a picture of the six configurations #

It is worth looking once more at the six bands as a whole, because their arrangement says something the individual numbers do not.

They come in reciprocal pairs. The ring-held sun-in band, 2.586 to 9.177, is the reciprocal of the ring-held carrier-in band, 0.109 to 0.387; the sun-held pair are 1.122–1.630 and 0.613–0.891; the carrier-held pair are the two negative ones. That is not a discovery — swapping input and output inverts a one-freedom train’s ratio — but it means the reachable set is symmetric under inversion, so a hole above 1 has a mirror image below it. The gap from 1.630 to 2.586 has a twin from 0.387 to 0.613, and a designer wanting an overdrive of 0.5 is as stuck as one wanting a reduction of 2.

It also means the whole set can be described by three bands rather than six, which is the honest way to count what a planetary offers: three magnitudes, each available in both directions, two of them forward and one reversed.

The hole in overdrive, which is the same hole #

The six bands were noted above to come in reciprocal pairs, and that observation has a consequence about the gap that is worth following, because it doubles the result at no cost.

If the reachable reductions have a hole from 1.630 to 2.586, then the reachable overdrives have a hole from $1/2.586$ to $1/1.630$ — from 0.387 to 0.614 — because an overdrive is a reduction read from the other shaft and the pairing maps one band onto another exactly. The gap is not a feature of the reducing configurations; it is a feature of the design set, and it appears once in every reciprocal reading of it.

Which puts a second entirely ordinary number inside a hole. A reduction of 2 is unavailable, and so is an overdrive of 0.5 — the shaft turning at twice the input, which is the most obvious thing anybody would ask an overdrive stage for and the direct counterpart of the reduction that started this essay. Both are excluded, both for the same reason, and neither exclusion is visible in any formula for a planetary’s ratio.

The pattern is worth stating in the general form, since it applies to whatever the endpoint values turn out to be after a designer has argued about the undercut limit. Any ratio excluded by the gap has its reciprocal excluded too, so the hole is symmetric under inversion when the ratios are plotted logarithmically, and it straddles the point where a reduction and its own reciprocal are equidistant. That point is 1, and the two edges of the hole are the two ratios nearest 1 that a single planetary can reach in either direction. Read that way the result is a statement about how close to direct drive a planetary can get without being direct drive — and the answer is not very close, in either direction, from any tooth counts.

It also sharpens the design advice. A stage wanting a ratio near unity is the wrong job for a single planetary, whichever side of unity it is on, and the reasons a designer would give for choosing one — compactness, coaxial shafts, load sharing — are exactly the reasons that make the alternative awkward. That is the trade the reverted train wins on, and it wins on it at both 2 and 0.5 for the same arithmetic.

Two ways out of the hole, and what each costs #

A designer who needs 2 : 1 coaxially has three options and it is worth pricing them, since the essay is otherwise only a refusal.

Two stages. Two planetaries in series, each at $\sqrt2 = 1.414$ — which is below the floor, so that does not work either. Two at 2.586 and 0.773 does, since the second is an overdrive from the sun-held-carrier-in band; total 2.000. Two gearsets, two sets of planets, two carriers.

A compound gearset. A Simpson or Ravigneaux offers ratios its component planetaries do not, because its members are tied together in ways a single set’s are not — the Ravigneaux’s 1.4600 and 1.6757 both sit inside the forbidden band’s lower half, and its 3.1739 above it. Compounding buys ratios, and the reason it does is precisely that the gearset is no longer a single planetary and the argument above no longer applies.

A reverted train. Two shafts in line gives 2 : 1 with wheels of 14, 21, 15 and 20 teeth, exactly, coaxially, and it is one of a hundred and eighteen solutions. The layshaft is offset, which is the whole difference: a planetary keeps everything on one axis and pays for it with the coaxial condition, and a reverted train puts one shaft to the side and does not.

The third is much the cheapest and it is what anybody would build. What the essay establishes is why the first thing a designer reaches for — a single planetary, because it is the compact coaxial reducer — is the one arrangement that cannot do it.

What this is an instance of #

The site has been building towards this shape of result for several fields and it is worth naming.

A mechanism’s specification lives in a continuum: reductions, lengths, angles, positions. The mechanism itself lives in a much smaller set: integer tooth counts, lengths that assemble, links that reach. The design problem is the intersection, and the interesting failures are the specifications that look reasonable in the continuum and have no representative in the set.

The synthesis field met it as exact syntheses that cannot be built — 1,176 exactly correct three-position four-bars of which 111 were usable. The timing field met it as exact gear trains for a target ratio, of which sixty has four hundred and four and the sidereal ratio has none. Here it is a whole interval of reductions with no representative at all, and the interval is where the most useful number lives.

The next essay puts one more equation on the same set — that the input and output shafts be in line — and the intersection nearly disappears.

There is one reading of the hole that is worth resisting, because it is the natural one and it is wrong. The gap looks like a property of the ratio formula — some algebraic obstruction that keeps a quotient away from a region — and it is nothing of the kind. Every value in the hole is produced by the formula perfectly happily for tooth counts that satisfy the coaxial condition and nothing else; a sun of 24 with a ring of 24 gives a ring-held reduction of exactly 2, and the formula raises no objection at all. What excludes it is that such a gearset has no room for a planet, since the coaxial condition then demands planets of zero teeth.

So the hole is a manufacturing fact wearing an algebraic disguise, and that is why its edges move when the undercut limit moves and why the closed form for its width contains the minimum tooth count. It also explains why the exclusion is invisible to everybody who reaches for the formula first: the formula is a relation among three counts, the constraint is that all three must be gears, and nothing in a quotient of integers remembers that its numerator and denominator have to be cuttable.