Every power split is on one curve

The power that goes round twice took one power split and followed the power through it. The engine turned the ring, a variator turned the sun, and the output came off the carrier. As the variator’s setting moved towards the pole, where the output stands still, the variator’s branch carried more and more of the engine’s power and overtook it halfway there. A variator built to carry the engine’s power could therefore use only part of its travel, and the span left was 1.70. The kinematic arithmetic in sliding the travel across the pole had promised more than five.

That essay ended by pointing out that it had measured one arrangement. The engine, the variator and the output can go on the planetary’s three shafts in six ways. The variator can also sit on the output side, driven by the output rather than by the engine, which gives six more. Perhaps one of the twelve keeps its variator below the engine’s power over a span worth having.

The answer is that the arrangement is not the variable, and the reason is short.

Twelve attachments, one number #

A simple planetary satisfies $\omega_s + K\omega_r - (1+K)\omega_c = 0$. Asking that it be lossless for every admissible set of speeds fixes its three torques in the same proportion as those coefficients, $1 : K : -(1+K)$. The lever that is the gearset draws that fact. The power on each shaft is its coefficient times its speed, times one common torque scale that cancels from every ratio below.

In an input-coupled split the engine turns one shaft directly, the direct path, and turns a second through the variator at ratio $v$. Call $a$ the power the variator’s branch carries divided by the power the direct path carries. Both branches carry torques in fixed proportion, so their powers are in the proportion of their speeds. The direct shaft turns at the engine’s speed, and the variator’s shaft at $v$ times it, so $a$ is proportional to $v$. The variator’s travel runs over a span $R$, from its lowest ratio to its highest, and $a$ runs over the same span, from $a$ at one end to $R a$ at the other.

The output’s power is the engine’s, and the engine’s power is the direct path’s plus the variator’s, which is the direct path’s times $1 + a$. The direct path’s power is fixed, because its speed is the engine’s and its torque is a fixed fraction of the output’s. For a given output torque, then, the output’s speed is proportional to $1 + a$, and over the travel it runs from $1 + a$ to $1 + Ra$. The output’s span is

$$\text{span} = \frac{1 + R\,a}{1 + a},$$

or its reciprocal, whichever exceeds one. There is no $K$ in it, and nothing says which shaft is the sun. With the variator on the output side the same steps give the same expression, with $a$ measured at the other end.

The argument is the planetary used as an adder. Two inputs and one output showed a differential’s cage turning at the mean of its two wheels, so that neither input alone determines the output. A power split is the same device with its inputs supplied from one engine by two routes. What the identity adds is that the only property of the two routes that matters is how their powers compare. Their speeds, their torques and which shafts they reach all cancel.

The variator enters through one number, $R$. A ratio with no steps in it solved a belt variator’s pulley radii from its fixed belt length and found its ratio law departing from the rule of thumb by 7.4% at full shift. None of that departure reaches the output’s span, which sees only where the travel starts and where it ends. The law in between decides how the output’s ratio is spread along the travel, not how far it reaches.

Read as a function of $a$, the identity has two places where it cannot be used and three stretches where it can. A travel that starts between −1 and −1/R ends beyond −1, so $1 + a$ passes through nought on the way and the output stops. That band is shaded in the figure, and approached from either side the span grows without bound. Towards $a = 0$ from either side the span falls to one: a variator that carries almost none of the power moves the output almost not at all, however far its own ratio travels. The span reaches the variator’s own R only far out on either side, where $a$ is so large that the direct path hardly matters.

The identity is checked on a census rather than taken from the argument. Every attachment was built with the variator coupled at the input and at the output, turning either way, on nine gearsets from K = 1 to K = 2.25, with the variator’s travel placed at eleven positions. Each was solved from the planetary’s speed relation. Travels that passed the output through standstill were dropped, since a span through a pole is not a span, and 1,994 splits remained. Every one satisfies the identity to 5 × 10⁻¹⁵. The planetary’s three powers sum to nothing at every operating point, to rounding, which is the lossless condition doing its job.

Three branches, and where a split falls #

The identity has three regimes, and the sign of $a$ divides them.

If $a$ is positive, both branches carry power forward from the engine. The variator’s share of the engine’s power is $a/(1+a)$, between nought and one at every setting, so the variator never carries as much as the engine. This is a split with no circulation, and its span lies between one and $R$. It never beats the variator it is built round.

If $a$ is between −1 and nought, the variator’s branch runs backwards. It takes power from the planetary and returns it to the engine’s shaft, and the direct path carries more than the engine delivers. This is the circulation of the previous essay, with the variator’s branch backwards.

If $a$ is below −1, the roles reverse. The variator carries more than the engine and the direct path runs backwards, returning the excess through the gears. Here the variator overtakes the engine everywhere on its travel.

A travel crossing $a = -1$ passes the output through standstill, and one crossing $a = 0$ passes the variator’s branch through zero power. Neither is a split over a whole travel, so every split sits wholly on one branch.

The three placements above use one attachment, one gearset and one variator: the previous essay’s machine. Turned backwards over a low stretch of its travel, from −0.21 to −0.70, it is the previous essay’s split, with its share running from −0.18 to exactly −1 and an output span of 1.70. Turned forwards over a high stretch, from 1.68 to 5.60, it is a split with no circulation, its share between 0.55 and 0.80, spanning 2.27. Turned backwards over a much higher stretch, from −2.2 to −7.33, its share runs from 2.75 down to 1.24 and it spans 7.42.

Nothing changed between the three but the direction of the variator’s turning and where its travel sits. The table shows the same for every attachment: each of the twelve reaches all three branches. So the previous essay’s question — whether some other attachment keeps its variator below the engine — has the answer that every attachment does, if its travel is put where $a$ is positive.

The curve #

On each branch the output’s span and the variator’s largest share are both functions of $a$, so the span is a function of the share alone. Write $S$ for the largest share of the engine’s power the variator carries anywhere on its travel. Then:

$$\text{no circulation: } \frac{R}{R(1-S) + S}, \qquad \text{backwards: } 1 + S\left(1 - \tfrac{1}{R}\right), \qquad \text{overtaking: } 1 + S(R - 1).$$

The figure plots all 1,994 splits on those axes. They lie on the three curves to 7 × 10⁻¹⁶, with nothing between them. The attachment moves a split along its curve and the gearset does not appear at all. The whole design space of single-planetary power splits, for a variator of a given span, is three curves.

The previous essay’s number is on the second. Its variator was rated for the engine’s power, so $S = 1$, and the backward branch gives $1 + (1 - 1/R) = 2 - 1/R$. For its variator, $R = 1/0.3$, that is 1.70 exactly. The previous essay reached it by following one gearset through a crossing at half the pole. The figure shows it is a property of the variator’s span and the branch, with no planetary in it.

The split whose variator never overtakes #

Read at a fixed rating, the three curves say which branch to use.

Up to a share of one, the best branch is the one without circulation. At every share, a split with no circulation spans more than one with its variator’s branch running backwards. With the previous essay’s variator and its share held below 0.9, a split with no circulation spans 2.70, against 1.70 for the backward split at a full unit of share. As the share approaches one, the span with no circulation approaches the variator’s own 3.33. At that limit the direct path carries almost nothing and the “split” is nearly a variator alone, which is the honest price of the last part of the span.

So the split the previous essay asked for, “the split whose variator never overtakes”, exists for every attachment and every gearset: it is every split on the first branch. Its limitation is not the variator’s power. It cannot beat the variator’s own span, and it approaches it only by making the variator do nearly all the work.

Beating the variator’s own span #

A split exceeds $R$ only by circulating, and on either circulating branch the variator must then carry more than the engine. On the overtaking branch the span is $1 + S(R-1)$, so each unit of share above one buys $R - 1$ of span. On the backward branch the gain per unit of share is smaller, $1 - 1/R$, and passing $R$ there needs a share above $R$.

The census confirms that no split with no circulation spans more than its variator, and that every split spanning more has a variator carrying more than the engine. Of the 1,994 splits, 470 beat their variator’s span. Every one of them circulates, and those on the backward branch have shares above R.

That turns a design target into a rating directly. To make a variator spanning 3.33 give a span of six, a single split must put it on the overtaking branch at a share of $(6 - 1)/(R - 1) = 2.14$. The variator must be built for twice the engine’s power, and the gears must carry the returning excess. The previous essay found a related figure, that the variator’s tolerance stops binding only at ratings of six times the engine. Both say that the span a split’s pole seems to offer is paid for in variator rating, and the curve says at what rate.

Span from modes instead #

A variator rated below the engine’s power therefore bounds a single split’s span below its own. Real continuously variable transmissions escape this without circulating, by changing attachment part way along the travel. In one mode the variator sweeps across its whole travel. At its end the machine engages a clutch that puts the same variator in a different split, at a matching output ratio, and the variator sweeps back. Each mode is a split on the curve, and their spans multiply.

With the share held to 0.75 in every mode and the same variator, one mode spans 2.11, two span 4.43 and three span 9.33. Matching two modes with a single split would need the variator on the overtaking branch at 1.47 times the engine’s power. The modes reach it with a variator that never carries three-quarters of the engine’s. The price is a clutch per mode and a shift that has to be synchronised, which the steps are not free describes for fixed gears. Here the ratio is continuous across the shift, and the cost is in the hardware that carries it out.

The clutches that change mode are the shift elements of an automatic gearbox. Holding a member chooses the ratio counted what brakes and clutches do to a gearset: each one adds a linear condition, and a gear is a line in the gearset’s plane of motions. A multi-mode split uses the same elements to move the variator between attachments. A compound gearset of the kind four speeds from two numbers describes can offer several attachments on one set of shafts. That is why real multi-mode machines are built round compound gearsets and not stacks of simple ones.

This also changes what the pole is for. A bounded ratio made unbounded showed that a split’s ratio passes through infinity, and a split whose travel sits near the pole has a large span on paper. The curve shows that span comes only from circulating power. A machine built for span without circulation stays on the first branch in every mode and never goes near a pole.

A hybrid is the same machine #

A power-split hybrid drivetrain replaces the belt variator with an electrical path: a generator on one shaft of a planetary feeds a motor that drives the output. The electrical path is a variator whose ratio is set by the generator’s speed, and nothing in the identity cares how the variator works. Its power is proportional to its speed because the planetary’s torque split fixes its torque, and the three branches follow.

The regime such a machine’s operators call power recirculation, in which the generator turns backwards and runs as a motor while the output’s motor runs as a generator, is the backward branch. The variator’s branch carries power from the planetary back towards the engine’s side of the machine, and the gears carry more than the engine delivers. The curve says what that regime buys: a span on the backward branch of $1 + S(1 - 1/R)$, the least span per unit of electrical power of the three branches. It also says why such machines are arranged to spend as little of their driving on that branch as they can. The electrical path’s losses are larger than a belt’s, which makes the ordering of the branches more severe, not less.

What the curve does not say #

Losses. The identity is lossless, and circulation is where losses concentrate. A split on the overtaking branch at a share of two passes twice the engine’s power through the variator, the least efficient element in the machine, and its losses scale with the power it carries rather than the power delivered. With losses included the branches would separate further than this lossless account shows, and in the same order.

Which gearset. $K$ has dropped out of the span, but it still decides what speeds the planets, the sun and the variator run at to give a particular output ratio. It also decides where along the travel a given value of $a$ falls, which is the practical question when the variator’s own ratio range is fixed. The attachment and gearset are packaging and speed choices once the branch is chosen.

Two planetaries. Everything here is one simple planetary. A compound gearset has more shafts and more ways to attach three roles, and whether the identity survives it is not established.

Still open: a split with a second planetary #

With two planetaries, as in a Simpson or a Ravigneaux gearset, a split has a two-freedom gearset. The engine, the variator and the output leave one shaft over, which can be held, coupled or left free. The torque split is then no longer one fixed proportion, and the argument that made the variator’s power proportional to its speed needs the extra shaft’s torque to vanish or to be fixed.

The distinct argument there would be whether a two-planetary split still lies on the three curves. The census would repeat with a two-freedom gearset and every assignment of the four shafts. If every split with the fourth shaft free lies on the curves, the identity is a fact about power paths and not about gearsets. If some split lies off them, the fourth shaft is doing something a single planetary cannot, and the curve would show what that costs in variator share.