One cross-ratio decides the compound split

A second planetary lets the variator carry a difference found the one new thing a four-shaft gearset adds to a power split. Put the variator between two of the gearset’s shafts, with the engine and the output on the other two, and the split becomes an input-coupled split and an output-coupled one in series. Its variator then carries only the difference of their two shares. At every variator share below one, such a compound split can reach more output span than any split on a single planetary. Every power split is on one curve had shown that the single-planetary splits all lie on three curves, so this was a new frontier.

The catch was the lever. The frontier needs the variator’s two shafts exactly its share apart on the gearset’s lever, centred between engine and output. The four real gearsets that essay tried, two Simpsons and two Ravigneaux, put their inner shafts about 0.22 apart and off centre, and their compound splits beat a single planetary only up to a variator share of about a third. It ended by asking whether any tooth counts can do better, and whether any reach a share of a half.

The question looks like a search over four-dimensional tooth-count spaces. It is really a question about one number.

The travel’s start makes one number of two #

A compound split’s output ratio, as the earlier essay derived, is $C(1 + Av)/(1 + Bv)$, with $v$ the variator’s ratio running over its span $R$ and $A$, $B$ fixed by the gearset and the attachment. The travel’s starting point $v_0$ is free. The split’s behaviour over the travel depends on $a = Av_0$ and $b = Bv_0$, and as $v_0$ is varied the pair $(a, b)$ runs along a ray through the origin. So of the gearset’s two numbers only their ratio $\rho = B/A$ matters, and $\rho$ and $1/\rho$ are the same ray run from the other end.

Written in the lever positions of the four shafts, $\rho$ turns out to be a cross-ratio: with the engine at $e$, the output at $o$ and the variator between $x$ and $y$,

$$\rho = \frac{(x - e)(y - o)}{(y - e)(x - o)}.$$

That is reassuring before it is useful. The lever that is the gearset places shafts on a line whose datum and unit are arbitrary, and a cross-ratio is exactly the quantity that does not change when a line is shifted and stretched. The split could not have depended on anything else.

Four points can be split into two pairs in three ways. The interleaved pairing gives a negative cross-ratio, and a split built on it has a variator whose share changes sign during the travel, which is not a split over its whole travel. The other two pairings give $\Omega_1$ and $\Omega_1/(\Omega_1 - 1)$. With the four shafts at $0 < p < q < 1$ on the lever, $\Omega_1$ is the nested pairing, engine and output on the ends and the variator between the inner two:

$$\Omega_1 = \frac{q/(1-q)}{p/(1-p)},$$

the ratio of the inner shafts’ odds. Its partner puts engine and output on one adjacent pair and the variator on the other. The gearset’s number is the larger, $\Omega = \max(\Omega_1, \Omega_1/(\Omega_1 - 1))$, never less than two. It is large when the inner shafts are far apart, near the ends, or when they are nearly together.

What the number buys #

On a ray the best span grows with the variator’s share until the ray meets the frontier, at a share of $(\sqrt\Omega - 1)/(\sqrt\Omega + 1)$, and does not grow after that. The frontier’s own closed form then gives the span there as $F = ((\sqrt{R\Omega} + 1)/(\sqrt\Omega + \sqrt R))^2$. The single planetary’s best at share $S$ is $R/(R(1-S) + S)$, so the compound split beats it up to the share where the two are equal:

$$S_\times = \frac{R\,(1 - 1/F)}{R - 1}.$$

The law was checked by doing what it summarises. For six cross-ratios, the ray was swept over eight hundred starting points in each direction, every travel that kept its share of one sign was kept, and the best span at each share was compared with the single planetary’s. The searched crossovers agree with the law to the search’s grid of 0.005. With the variator span of the earlier essays, $R = 1/0.3$, a share of a third needs a cross-ratio of 2.52 and a share of a half needs 4.67.

The four gearsets of the earlier essay have cross-ratios of 2.46, 2.64, 2.57 and 2.48. That is why they stopped at a third: they sit either side of the number a third needs. The question of tooth counts is now the question of which teeth give a cross-ratio above 4.67.

The teeth, in two formulas #

Four speeds from two numbers found that a Ravigneaux’s every ratio depends on two tooth ratios alone, and the same is true of a Simpson, so the lever positions and the cross-ratio are functions of two numbers for each. Writing $K$ for a ring’s teeth over a sun’s, the lever gives them directly.

A Simpson has one sun shared by two planetaries, the first carrier joined to the second ring. Its first carrier sits at $K_1/(1 + K_1)$ of the way from sun to first ring, and its second carrier at that times $K_2/(1 + K_2)$. Its nested cross-ratio comes out as

$$\Omega_1 = 1 + \frac{1 + K_1}{K_2},$$

so its number is $1 + \max(u, 1/u)$ with $u = (1 + K_1)/K_2$. A large first ring on a small sun and a small second ring push $u$ up. Nothing else does, and the ring a sun can carry is limited by the size of the gearset.

A Ravigneaux has two suns, a ring, short planets meshing the first sun and long planets meshing the second sun and the ring. With $K_1$ the ring over the short planets’ sun and $K_2$ the ring over the long planets’ sun, its carrier sits at $K_2/(K_1 + K_2)$ and its ring at $(1 + K_2)/(K_1 + K_2)$, and

$$\Omega_1 = \frac{K_1\,(1 + K_2)}{K_2\,(K_1 - 1)}.$$

This one has no ceiling. As $K_1$ approaches one, the short planets’ sun approaching the ring in size, the denominator vanishes and the cross-ratio grows without bound. What stops it is the planets. A sun nearly as large as its ring leaves room only for small planets.

Both formulas were checked against the exact lever, computed in rational arithmetic from the gearsets’ own meshes, on 246 tooth counts. They agree to $10^{-15}$.

Why the four gearsets sat at a third #

The formulas also explain the earlier essay’s result without a census. The commonest Simpson has its two rings the same size, since one ring blank then serves both planetaries. With $K_1 = K_2 = K$ its cross-ratio is $1 + (1 + K)/K = 2 + 1/K$. A ring has to be larger than its sun, so $K > 1$ and the cross-ratio is below three, whatever the teeth. By the law, three buys a share of 0.385. An equal-ring Simpson can never carry a compound split past about 0.38 of the engine’s power, and at the usual ring-to-sun ratio of a little over two, 74 teeth on 34, it sits at 2.46.

The Ravigneaux of an automatic gearbox is similar for a different reason. Its ratios are chosen for the stepped speeds it was designed to give, and they put both suns between a third and a half of the ring. With $K_1 = 78/36$ and $K_2 = 78/30$, as in the earlier essay’s first Ravigneaux, the formula gives 2.57. Nothing about a gearbox’s set of fixed speeds asks for a large short-planet sun, so nothing puts a gearbox Ravigneaux near the half.

So the third was not a property of four-shaft gearsets. It was a property of gearsets proportioned for four fixed speeds, and a gearset proportioned for a compound split looks different: two suns nearly as large as the ring they sit in.

Where each gearset is at its best #

A ray meets the frontier at one share, $(\sqrt\Omega - 1)/(\sqrt\Omega + 1)$, and that is where the gearset is being used as well as any four-shaft gearset could be. For the best Ravigneaux it is 0.406. Below it, the split has share to spare and could be given a longer travel. Above it and up to 0.542, the split is no longer at the frontier but is still ahead of every single planetary. So a designer choosing this gearset for a variator that must carry at most a half would run it past its own optimum, and a designer who can live with 0.41 would get the most span per unit of variator share it offers.

The same reading applies to any gearset. Its cross-ratio fixes both its best share and its last useful one, and the two are always some way apart. That is the practical difference from a single planetary, whose three curves have no such plateau. A compound split has a share it is built for and a band beyond it where it still pays.

Every buildable gearset up to a ring of 100 #

With formulas this cheap, every buildable tooth count can be tried. Rings run up to 100 teeth and every planet has at least 13. A Ravigneaux must also be buildable in the plane, which the lever does not check. Its long planet must come out whole, its two planet orbits and their mesh must close as a triangle, and its short planet must clear the ring’s tips, and its long planet the large sun’s, by a quarter of a module.

There are 20,832 Simpsons. 7,846 reach a share of a third, far more than the four the earlier essay happened to try, and none reaches a half. The best has a sun of 13 teeth, a first ring of 99 and a second of 39, with a cross-ratio of 3.87 and a share of 0.454. The map shows why. The half contour on the Simpson panel runs where the first ring is over ten times its sun, off the edge of anything a ring of 100 teeth allows.

There are 221,065 buildable Ravigneaux. 65,768 reach a third, and 176 reach a half. The best has suns of 69 and 74 teeth in a ring of 100, with both sets of planets at 13 teeth, a cross-ratio of 5.61 and a share of 0.542. It is the shape the formula asked for: two large suns, the short planets’ sun about seven-tenths of the ring, and planets as small as the census allowed. Its inner shafts sit at 0.483 and 0.840 of its lever, one near the middle and one near the end, which is the nested pairing’s way of being far apart.

Solved on the lever #

A cross-ratio and a closed form are two steps removed from a machine. The check that trusts neither is the earlier essay’s own instrument: attach the engine, the output and the variator to the gearset’s shafts, sweep the travel’s start, and solve every operating point from the lever. Speeds come from its two kinematic conditions and torques from equilibrium and the variator’s balance of power, and the span and share are read off the solved travel.

The two gearsets behave as the law says. Each gearset’s span climbs to meet the frontier at the share its cross-ratio sets, then holds level while the single planetary’s best climbs past. The Simpson holds 1.466 from a share of 0.35 to 0.5, and the single planetary passes it just after 0.45, which is 1.460 there and 1.538 at a half. The Ravigneaux holds 1.612 from 0.45 to 0.6, above the single planetary’s 1.538 at a half, and is passed near 0.54. At a share of a half, the best Ravigneaux inside a ring of 100 does what the earlier essay’s gearsets could not.

The planets set the limit #

The census allowed planets of 13 teeth, and the best Ravigneaux uses exactly 13 in both sets. That is not a coincidence. The formula pays for a large short-planet sun, and the triangle and the clearances then leave room only for the smallest planets allowed. The census was run again with the planets at 17 teeth or more, the count below which a standard 20° pinion undercuts without correction.

With 17-tooth planets the best Simpson falls to 3.27, a share of 0.409. The best Ravigneaux falls to 4.26, a share of 0.478, with suns of 61 and 66 in a ring of 100. Neither family reaches a half. So the half is reachable, but only through planets small enough to need the cutter moved out: profile-shifted pinions of 13 teeth, running in a gearset whose short planets clear the ring by a quarter of a module. A compound split for half the engine’s power is a design target, though a demanding one.

What a designer can take from it #

The earlier essay’s frontier was stated as positions on a lever, and this essay turns those positions into one number and the number into teeth.

Compute one cross-ratio. For any four-shaft gearset, place its shafts on the lever, take the cross-ratio of the nested pairing and its partner, and keep the larger. That single number fixes the largest variator share at which a compound split on that gearset beats every single-planetary split. Tooth counts, attachments and the travel’s start matter only through it.

A third is easy, a half is not. Over a third of all buildable Simpsons and nearly a third of all buildable Ravigneaux clear 2.52. Clearing 4.67 takes a Ravigneaux with two large suns and the smallest planets its teeth can take, and no gearset of either family inside a ring of 100 does it with unshifted planets.

The Simpson is the wrong family for a large share. Its cross-ratio is bounded by the ratio of its first ring to its sun. The Ravigneaux’s grows without bound as its short planets’ sun approaches its ring, so for the compound split the Ravigneaux’s awkward geometry is an advantage.

What the teeth do not settle #

Losses. Every split here is lossless. A compound split circulates power through its gears, as the power that goes round twice measured for simpler splits, and a gearset with 13-tooth planets running near the frontier may circulate a great deal. The cross-ratio says nothing about mesh losses, and they may take back part of the span.

Equal spacing and more planets. The census builds one planet of each kind. A real gearset carries three or four equally spaced, which adds an assembly condition on the tooth sums and a condition that neighbouring planets clear each other. Both remove tooth counts, and neither is applied.

Strength. A 13-tooth planet between a 69-tooth sun and a 100-tooth ring carries the gearset’s torque on a small tooth. Whether it can is a question of module and face width, which the lever does not see.

Other gearsets. Only Simpsons and Ravigneaux are counted. Other four-shaft gearsets, two simple planetaries joined in different ways, have their own two-number formulas and may reach further.

Still open: the four-shaft gearset built for the lever #

The census asked what two named families can reach. Its formulas suggest the question can be turned round. A cross-ratio is a single number, and two simple planetaries can be joined into a four-shaft gearset in several ways besides Simpson’s and Ravigneaux’s.

The distinct argument would be to enumerate the ways two simple planetaries can be joined into four shafts, derive each join’s cross-ratio as a function of its two tooth ratios, and find which join reaches a cross-ratio of 4.67 with the least extreme teeth. The measurement would take each join over the same census of buildable tooth counts, with unshifted planets, and ask whether any family puts the frontier at a share of a half within ordinary proportions. It would also take the winner’s circulating power at the frontier. If some join does it, the compound split has a gearset of its own. If none does, the share of a half belongs to the Ravigneaux with small planets, and the circulation decides whether it is worth it.