The power that goes round twice

Sliding the travel across the pole answered a design question in closed form. A bounded ratio is made unbounded by summing the variator with a straight path: the overall ratio is a hyperbola in the variator’s setting with a pole at $v = K$; a fixed pair between the variator and the planetary slides the whole travel relative to that pole without touching either; and the best placement puts the top of the travel exactly at the trim the variator’s own tolerance imposes, giving a forward span

$$1 + \frac{\tau(1-r)}{p},$$

with no $K$ in it — the pole’s position drops out. For a band $\tau$ of a tenth, a variator precise to 1.6% and a travel ratio $r$ of 0.3, that is 5.375.

Every step of that is kinematic. The essay ended by saying so and by naming what it had not asked: near the pole the output is barely turning while the engine is, so the planetary’s branches carry torques far larger than the output’s, and the variator may be asked to pass more power than the engine is delivering.

It is, and the setting at which it starts is a number with nothing in it but the gearset.

A power ratio needs no force #

The step from speeds to powers looks like it needs a force and it does not, and the reason is worth stating because it is what keeps this inside the collection’s own boundary.

A simple planetary satisfies $\omega_s + K\omega_r = (1+K)\omega_c$ with $K$ the ring’s tooth count over the sun’s. Require that the gearset be lossless — $\sum T_i\omega_i = 0$ — and require it for every admissible set of speeds rather than for one. Substituting the speed relation and collecting the two free speeds gives two equations, and they fix the torques up to a single scale:

$$T_s : T_r : T_c \;=\; 1 : K : -(1+K).$$

That is a statement about tooth counts. Nothing about a load, a material or a contact enters it, and every quantity below is one power divided by another, so the scale cancels. The lever that is the gearset is the same fact drawn: the three shafts sit at fixed positions on a line and the torques are the lever arms.

With the variator on the sun at $-v$, the straight path on the ring at 1 and the output on the carrier, the branch powers as multiples of the engine’s are

$$\frac{P_{\text{var}}}{P_{\text{eng}}} = \frac{v}{K-v}, \qquad \frac{P_{\text{straight}}}{P_{\text{eng}}} = \frac{K}{K-v},$$

the second flowing the other way through the planetary. Both are computed a second time by writing the speed relation and the power balance as a linear system and solving it with no formula in sight; the two routes agree to $10^{-16}$ and the three branch powers sum to nothing at every setting, which is the arithmetic’s floor rather than a small number. That balance is the same statement an assembled planetary has to satisfy in integers before it is a machine at all.

Why the branch grows and the output does not #

The two expressions differ by one term and it is worth seeing where the difference comes from, because it is the whole mechanism of the circulation.

The output’s speed falls linearly to nought at the pole. The variator’s speed does not — it is $v$, which is rising. And the torque in the variator’s branch is a fixed fraction of the output’s torque, which for a given load is a constant. So the variator’s power is a rising speed times a constant torque while the output’s is a falling speed times the same constant, and their ratio $v/(K-v)$ is a rising quantity divided by a falling one. It has to run away.

Said as a machine rather than as algebra: a power split near its pole is a device for turning a variator quickly while the output stands still. The engine is still delivering, the load is still there, and the only place the difference can be is inside the loop — so the loop carries it. The pole is not a place where the machine is doing less work. It is a place where it is doing the same work twice.

That is why the circulation is a property of the topology and not of the components. A variator that is bigger, more precise or more efficient still sees $v/(K-v)$, because that expression has only the tooth counts in it. The only way to reduce it is to change where the travel sits or to change which shaft the variator is on, and the first of those is exactly what the placement argument was doing — in the direction that makes it worse.

The straight path always carries more than the engine #

Read the second expression at $v = 0$ and it is exactly one: with the variator stopped, all the engine’s power goes through the ring and out of the carrier, and the sun carries torque and no power. Read it anywhere else in the forward travel and it is greater than one.

So the straight path is always carrying more than the engine delivers, from the first moment the variator moves. The excess comes back through the sun — it has to, because the three powers sum to nothing — which means a fraction of the engine’s output is going round the loop rather than to the wheels. That is the circulation, and it is present at every setting of a power split rather than only near the pole. What the pole does is make it unbounded.

There is a tidier way to say the whole of it, and the two expressions carry it between them: $K/(K-v)$ minus $v/(K-v)$ is $(K-v)/(K-v)$, which is one, identically. The straight path and the variator’s branch are the same curve shifted by exactly one unit. So the vertical gap between the two is the engine’s own power at every setting of the travel, and everything below that gap is loop. What grows towards the pole is therefore not the difference between the branches — that never moves — but the height at which a fixed difference sits, which is the clearest statement of the circulation available without a force in it. Past the pole both go negative together, because the output has reversed and both branches are carrying power the other way, and the gap between them is still exactly one.

The reading that matters for a machine is the variator’s branch rather than the straight path’s, because the variator is the expensive, wearing, power-limited component and the straight path is a gear train — a thing that costs a step and then carries whatever is asked of it. That branch carries $v/(K-v)$, which is nought at the bottom of the travel and grows without bound at the pole, and the question is where it passes one.

Half the way to the pole, and not a bit further #

Setting $v/(K-v) = 1$ gives $v = K/2$. The variator carries exactly the engine’s whole power at half the distance to the pole, and the number has nothing in it but $K$.

Measured across nine sun-and-ring combinations from $K = 1$ to $K = 2.25$, every crossing lies on that line exactly. The variator’s travel, its belt, its tolerance and the fixed pair in front of it do not appear, because the quantity is a ratio of two branch powers and those are decided by tooth counts. The overall ratio at the crossing is $K/2(1+K)$, which across the same nine gearsets runs only from 0.250 to 0.346 — so every power split in this family is overtaken at roughly the same gear, whatever its planetary.

There is a second way to see why it lands there, and it is the lever again. On the lever diagram the three shafts sit at fixed positions and the powers are the products of speed and lever arm; the variator’s arm is one and the carrier’s is $1+K$, so the variator matches the engine exactly when its own speed reaches $1/(1+K)$ of the sum — which is the same statement rearranged. The crossing is a point on a straight line, which is why it does not move when anything but the tooth counts changes.

For a rated component that is the line that cannot be crossed. A variator built to carry the engine’s power is a variator that must not be run past $v = K/2$, and for the reference machine, $K = 1.4$, that is $v = 0.700$.

The trim that is never reached #

Now the two conditions can be put against each other, which is the whole point of computing the second.

The tolerance trim — the top of the travel in the kinematic answer — sits at $K\tau/(\tau+p)$, which for the reference numbers is $1.207$. The power cap for a variator rated at the engine’s own power sits at $0.700$. The cap is lower, so it binds, and the travel stops well short of where the kinematic arithmetic wanted to put it.

The span that follows is 1.70 against the 5.375 the closed form promises. A power split built with a variator rated for its engine reaches less than a third of the span the placement argument computes for it. Put the other way round: of the 4.375 of extra span the arithmetic offered, 3.675 of it lies past a setting the variator cannot be run at.

And the two conditions swap over at a rating that has a closed form of its own. Setting $K\cdot\text{rating}/(1+\text{rating}) = K\tau/(\tau+p)$ and solving gives

$$\text{rating} = \frac{\tau}{p},$$

with the planetary cancelling out entirely. For $\tau = 0.1$ and $p = 0.016$ that is 6.25: the tolerance trim becomes the binding condition only when the variator is built to carry six and a quarter times the engine’s power. No machine is built that way, and the reason is that a variator is the component a power split exists to make small.

What this does to the advice #

The placement result is not wrong and it is worth saying exactly what survives.

The shape of the argument survives: the best place for the top of the travel is at whichever trim binds, because the span rises with the top until the trim and falls after it, and that reasoning did not depend on which trim it was. The closed form for the optimum span does not survive, because it was derived assuming the tolerance trim is the one reached.

The replacement is smaller and has a different variable in it. With a rating $\rho$, the top is at $K\rho/(1+\rho)$, the bottom at $rK\rho/(1+\rho)$, and the span is

$$\frac{K - rK\rho/(1+\rho)}{K - K\rho/(1+\rho)} \;=\; \frac{1 + \rho(1 - r)}{1},$$

which is $1 + \rho(1-r)$ — and $K$ drops out of this one too. Two different regimes, two closed forms, and the pole’s position is absent from both. At $\rho = 1$ and $r = 0.3$ it gives 1.70, which is the measurement.

That is a more useful sentence than the one it replaces, because $\rho$ is a number a designer chooses by buying a variator and $\tau/p$ is a number they are stuck with. The span of a power split is set by what its variator is rated to carry, not by what it is precise to. That is a different kind of sentence from the one it replaces, in the way a band with a direction in it is different from a band: it names which knob to turn.

The two answers, and which one a drawing needs #

It is worth putting the two closed forms side by side, because they are the same argument run against two different trims and the difference in what they depend on is the useful part.

The kinematic answer, $1 + \tau(1-r)/p$, depends on the band a designer is willing to allow and on how precise the variator is. Both are quantities that are discovered rather than chosen: the band comes from what the machine is for, and $p$ comes from what a belt-and-sheave variator can hold, which the ratio that has a tolerance measured and which does not improve much with effort.

The rated answer, $1 + \rho(1-r)$, depends on what the variator is bought to carry. That is a purchase. A designer who wants more span from a power split can buy a variator rated for twice the engine and get $1 + 2(1-r) = 2.4$ instead of 1.70, and the tolerance never enters until the rating passes 6.25.

Both have $K$ absent, which is the more surprising half and is the same surprise the placement argument produced. The planetary decides where the pole is and how fast the output falls towards it, and it decides neither the span the tolerance allows nor the span the rating allows — because both trims are proportional to $K$ and the span is a ratio of two distances from the pole. The gearset chooses the gear; the variator chooses the spread.

There is one thing the gearset does still decide, and it is the crossing itself: $K/2$ in variator ratio and $K/2(1+K)$ in output ratio. A designer who wants the overtaking to happen at a particular road speed chooses $K$ for that and takes whatever span the rating then gives.

What is not modelled #

No losses anywhere. The torque split comes from requiring the gearset to be lossless, and a real planetary is not. Efficiency matters twice over here: it makes the branch powers slightly different from the ones computed, and — far more importantly — a circulating power of several times the engine’s is being passed through gear meshes whose losses are a fraction of it rather than of the output. A split running near its pole can have an efficiency that collapses while every quantity computed here is still finite.

A rating is one number. A variator is rated for a power, a torque and a speed, and which of the three binds depends on where in the travel it is asked to work. Taking the rating as a single multiple of the engine’s power is the crudest version of that and is what makes the comparison legible; a real check would be against a torque at each setting.

The tolerance trim is taken from the previous result unchanged. It uses the variator’s relative uncertainty $p$ as a constant across its travel, which is an approximation the earlier measurement made and stated.

The topology is one topology. Variator on the sun, straight path on the ring, output on the carrier, with the variator’s path reversed. Other arrangements of the same three shafts give other relations, and the input-coupled and output-coupled splits differ in exactly the way this file would need to be rewritten for.

The circulation is not measured against a real gearbox. Every number here is for the reference machine — a 50-tooth sun, a 70-tooth ring, a variator of 1.6% precision and a band of a tenth. A production power split has a two-mode arrangement precisely to keep its variator out of the region this essay is about, and comparing against one would need that second mode modelled.

Nothing here says the circulation is bad. It is a fact about where the power goes, not a verdict. A split carrying twice its engine’s power through a gear train and out again is a perfectly ordinary machine if the train is sized for it; what the measurement establishes is that the variator cannot be sized out of the problem, because its share is what grows.

Still open: the split whose variator never overtakes #

Everything above measures one topology and finds a crossing at half the pole. That number came from the branch powers, and the branch powers came from which shaft each path is attached to — so a different attachment is a different crossing, and possibly a better one. Two shafts that must be in line is the constraint that decides which attachments are even available in one casing.

Its distinct argument would be the crossing over the arrangements rather than over the gearsets. There are six ways to attach an engine, a variator and an output to a planetary’s three shafts, and each gives its own pair of branch-power expressions from the same lossless condition; the question is whether any of them keeps the variator’s share below one over a travel that still reaches a useful span. Two things would come out of it. Whether the input-coupled and output-coupled splits differ in which branch runs away — which decides whether a split is limited by its variator or by its gears; and whether a two-mode machine, which changes attachment part way along its travel, can keep both branches under the engine throughout, since that is what a real continuously-variable transmission does and this subject has not asked why.