A second planetary lets the variator carry a difference

A continuously variable transmission built as a power split sends part of the engine’s power through a variator, a device whose ratio can be set anywhere across a span R — a ratio with no steps in it — and the rest through gears. A planetary recombines the two, the adder that two inputs and one output found in a differential. Every power split is on one curve took every way of attaching an engine, a variator and an output to a simple planetary’s three shafts and found one identity under all of them. The output’s span is $(1 + Ra)/(1 + a)$, where a is the variator’s power over the gears’ power at the start of its travel. So every split lies on one of three curves of span against the largest share of the engine’s power the variator ever carries. The gearset, and which shaft is the sun, never appear.

That essay ended with a gearset it could not answer for. A Simpson or a Ravigneaux has two planetaries and four shafts that a designer can reach, and a split built on one has a shaft left over after the engine, the variator and the output have taken theirs. The fourth shaft can be left free, held, or given a job. The question was whether every such split still lies on the three curves. If it does, the identity is a fact about power paths. If not, the fourth shaft is doing something one planetary cannot, and the curves would show what that costs in variator share.

Both answers turn out to be true, for different jobs. Leave the fourth shaft free and nothing changes. Give it the variator’s other end, so that both ends of the variator are on the gearset, and the split leaves the curves, mostly below them but, on the right gearset, above them at every share.

A four-shaft gearset is a lever with four points #

The lever that is the gearset showed that any gearset whose frame carries no teeth, and which has two freedoms, is exactly a lever. There is a position x for each shaft such that at every operating point the shafts’ speeds lie on a straight line, $\omega = p + q\,x$. A simple planetary has three such points; a Simpson or a Ravigneaux has four. The positions come out of the gearset’s tooth counts, exactly: a Simpson of 34, 74 and 74 teeth has its four shafts at 0, 0.469, 0.685 and 1; a Ravigneaux of 36, 30 and 78, at 0, 0.545, 0.755 and 1.

Losslessness does for torques what the straight line does for speeds. If no power is lost at any operating point, the torques the outside world applies must satisfy $\sum T = 0$ and $\sum T x = 0$, so that $\sum T\omega = p\sum T + q\sum Tx$ vanishes whatever p and q are. That is two conditions on four torques. With three shafts, as on a simple planetary, it fixes the torques in one proportion, which is where the first essay’s identity came from. With four there is one more freedom to spend.

Everything below was measured on four real gearsets, two Simpsons and two Ravigneaux, as their exact levers. The engine, the variator and the output were put on every choice of shafts, with the variator turning either way over a travel of span 1/0.3 started at eleven places. Each operating point was solved from the lever: speeds from the two kinematic conditions, torques from equilibrium and the variator’s own power balance. At every point of every split the engine’s power equals the output’s to $8 \times 10^{-12}$ of it.

A free fourth shaft changes nothing #

Put the engine, the variator and the output on three of the four shafts, the variator coupled at the input or at the output as on a simple planetary, and leave the fourth shaft free. It carries no torque, so its torque condition is $T = 0$. The two lossless conditions then fix the other three torques in one proportion, as on a simple planetary, and the speed of the free shaft is simply whatever the lever says it is. The machine is a simple planetary with a spare shaft spinning idly.

The census confirms it. On the four gearsets, 1,794 input-coupled and 1,794 output-coupled splits have a free fourth shaft, and every one of the 3,588 lies on one of the three curves to $3 \times 10^{-14}$ of its span. The identity survives a second planetary as long as the second planetary has nothing to do.

Hold the fourth shaft instead and there is no split at all. The lever has two freedoms; holding one shaft at nought takes one, and the engine’s speed takes the other, so every shaft’s speed is fixed and the variator’s ratio is forced. That is a fixed gear, not a continuously variable one.

The one new arrangement: both ends of the variator on the gearset #

A simple planetary has no room for one arrangement a four-shaft gearset allows. Put the engine on one shaft and the output on a second, and connect the variator between the third and the fourth. Neither of the variator’s ends touches the engine or the output directly. Transmission designers call this a compound split. There are twenty-four ways to place it on four shafts, and on the four gearsets 1,562 of the travels tried are splits in the full sense, with the output never stopping and the variator’s power never changing sign. A travel through the output’s standstill is the geared neutral of a bounded ratio made unbounded, and a span measured through it is not a span.

Not one of them lies on the three curves.

Most of the compound splits fall below the curves, some well below. But 704 of them lie above the best single-planetary curve, spanning more for the same variator share than any split a single planetary can build. On these four gearsets every one of those 704 has a variator share under 0.25.

Two splits in series #

The compound split’s output can be written down. With the engine turning at 1 and the variator’s ratio v, the two kinematic conditions give the lever’s slope, and the output’s speed comes out as

$$\frac{\omega_o}{\omega_e} = C\,\frac{1 + A\,v}{1 + B\,v},$$

with A and B set by the four shafts’ lever positions. A simple split gives only half of this. An input-coupled split’s output is linear in v, the numerator alone, and an output-coupled split’s is one over a linear function, the denominator alone. The compound split has both, so it is an input-coupled split and an output-coupled split in series, and its span is the product of theirs:

$$\text{span} = \frac{1 + R A'}{1 + A'} \cdot \frac{1 + B'}{1 + R B'}, \qquad A' = A v_0,\ B' = B v_0,$$

with $v_0$ the start of the variator’s travel. The variator’s share of the engine’s power, solved from the torques at every point, is the difference of the two halves’ shares:

$$\frac{P_{\text{variator}}}{P_{\text{engine}}} = \frac{A v}{1 + A v} - \frac{B v}{1 + B v}.$$

Both forms were fitted from three points of each compound split’s travel and tested at three others. Across all 1,562 they hold to $2 \times 10^{-14}$ in the output’s speed and $5 \times 10^{-14}$ in the variator’s share, and the product gives the span to $6 \times 10^{-14}$.

That difference is the whole of the finding. Each half of a compound split moves the output as a simple split would, with its own share of the power going round its own variator branch. But the two halves have one variator between them, so the variator carries only the part on which they disagree. Two halves that each carry a large share and nearly agree move the output a long way through a variator that carries little.

Why most compound splits are worse #

The same formula says why most of the census falls below the curves. A difference of two shares is small only if the shares have the same sign. When A and B have opposite signs, one half’s variator branch runs forward and the other’s backwards, and the difference of their shares is the sum of their sizes. The variator then carries more than either half would on its own.

The census divides exactly along that line. Of the 1,562 compound splits, 498 have halves of opposite sign, and every one of them lies below the best single-planetary curve. The other 1,064 have halves of the same sign; 704 of those lie above the curve and 360 below it. Every compound split that beats a single planetary is one whose halves both push power the same way, so that their shares can cancel in the variator. Opposite halves add to the variator’s load instead of relieving it: one of them is sending power back round, which is the circulation the power that goes round twice priced on one planetary, arriving here through the other half’s variator.

The frontier, in closed form #

How far that can go is a question with two numbers in it, A′ and B′. The most span for a largest variator share S was searched over both, and the answer came out with a structure simple enough to state exactly.

The best compound split at every share is built on a symmetric lever. The engine is at one end and the output at the other, and the variator’s two shafts sit between them at $(1 - S)/2$ and $(1 + S)/2$ — exactly its largest share apart, centred on the middle. On that lever $A = -k$ and $B = -1/k$, with $k = (1 - S)/(1 + S)$. The best travel runs from $-1/\sqrt{R}$ to $-\sqrt{R}$, and the span is

$$\text{span} = \left(\frac{\sqrt{R} + k}{1 + k\sqrt{R}}\right)^{2}.$$

The searched frontier agrees with this to $10^{-6}$ at every share tried from 0.1 to 0.9. Each frontier design was then built as its lever and solved point by point through the travel like any split in the census, and it spans what the closed form says at the share it says.

Against a single planetary’s best, $R/(R(1 - S) + S)$, the compound frontier is ahead at every share below one, for the variator of span 3.33 used throughout: 1.124 against 1.075 at a share of 0.1, 1.421 against 1.266 at 0.3, 1.802 against 1.538 at 0.5, 2.937 against 2.703 at 0.9. At a share of one both reach R, the variator’s own span, because a variator that carries all the power is simply the variator. At nought both give one. Beyond a share of one, on the overtaking branch, the search found the compound frontier ahead as well, 5.48 against 4.50 at 1.5, though a variator carrying more power than the engine is a machine built around its variator rather than one sparing it.

The lever at a share of 0.5 shows how the travel works. The variator’s shafts sit at 0.25 and 0.75. Its ratio runs from −0.548 through −1 to −1.826: at the middle of the travel its two shafts turn equal and opposite, the lever pivots about its centre, and the output turns exactly against the engine. Moving the variator either way tilts the lever about a point off-centre, and the output’s speed runs from 1.342 of the engine’s to 0.745, a span of 1.802.

A negative variator ratio means only that the two shafts turn in opposite senses. A belt variator’s pulleys turn the same way, so a real one would reach this lever through one more pair of gears, which changes nothing in the kinematics.

What each half carries #

Read as two splits in series, the frontier design at 0.5 has an input half whose variator share would run from 0.154 to 0.378 and an output half whose share would run from 0.622 to 0.846. Their difference is 0.467 at both ends of the travel and 0.500 at its middle, and never more. The variator’s share is not flat, but it is close to flat, peaking at the middle rather than at either end. On a single planetary the share always peaks at one end of the travel, and the variator has to be rated for that end while it carries less everywhere else.

That near-flatness is one way to see the gain. A single-planetary split’s variator is underused over most of its travel, since its share rises or falls monotonically between the two ends. The compound’s variator carries close to its rating everywhere, and so buys more span for the same rating.

What real gearsets reach #

The frontier is exact, but it asks for a particular lever, and a lever is not free. Its positions are set by the gearset’s tooth counts, and the four shafts of a real Simpson or Ravigneaux fall where the geometry of its planets puts them.

The four gearsets measured have their inner shafts about 0.22 apart, centred between 0.58 and 0.65 rather than at 0.5: 0.469 and 0.685 on the first Simpson, 0.545 and 0.755 on the first Ravigneaux. A share of 0.22, centred, would suit that spacing; nothing larger fits. On them, with the variator’s travel placed anywhere over six decades either way, the best compound split is ahead of the single planetary up to a share of 0.3: 1.093 against 1.075 at 0.1, 1.203 against 1.163 at 0.2, 1.320 against 1.266 at 0.3. From 0.4 it is behind: 1.327 against 1.389, then 1.423 against 1.538 at 0.5, and at 0.9, 1.844 against 2.703.

So the second planetary’s advantage is real and narrow on the gearsets a catalogue offers. It buys span for a variator carrying a small share of the power, which is the regime a designer protecting a low-rated variator would be in. For a variator carrying half the power it needs a lever with its inner shafts at a quarter and three quarters, and for 0.9, at 0.05 and 0.95. Those are shafts crowded towards the engine’s and the output’s ends, which no two-planetary gearset of ordinary proportions has. The reductions a planetary cannot give found the same kind of limit on a single planetary: the lever positions a design wants and the positions teeth allow are different sets.

What the four shafts leave unmeasured #

Losses. Every split here is lossless, which is what makes the lever’s torque conditions exact. A real compound split’s gears carry the two halves’ shares in their meshes even though the variator carries only the difference, and the meshes lose a little of what they carry. Whether that erodes the gain at small shares is not measured.

Which gearsets could reach the frontier. The frontier’s lever is stated as positions, not tooth counts. Whether some Ravigneaux, or a gearset with three planetaries, can put two shafts at a quarter and three quarters with the engine and output at the ends was not searched.

The variator’s torque. A variator is rated in torque as well as power, and the share measured here is of power. Along a frontier travel the variator’s shafts turn at 0.3 to 0.76 of the engine’s speed, so its torque is correspondingly larger than its power share suggests. That trade is not priced.

Still open: tooth counts for the frontier’s lever #

The frontier needs a four-shaft lever with the variator’s shafts at $(1 \mp S)/2$, and the question it leaves is whether teeth can build one. Four speeds from two numbers found that a Ravigneaux’s every ratio is a function of two tooth ratios alone, which means its two inner lever positions are too, over whatever range its planets can close as a triangle.

The distinct argument would be the map from a Ravigneaux’s and a Simpson’s tooth ratios to their inner lever positions, and the region of that map the frontier’s positions fall in. The measurement would sweep both families over every tooth count that assembles, place their inner shafts on the lever, and read off the largest share at which the best compound split each one allows still beats a single planetary. If some real gearset reaches a share of one half, the frontier is a design target. If none passes about a third, the compound split is a low-share device, and the three curves remain the design space for every variator that carries much of the power.