Sliding the travel across the pole

A bounded ratio made unbounded put a variator and a straight path into a planetary and got a machine whose ratio passes through infinity. With $K$ the planetary’s ring-to-sun ratio and $v$ the variator’s, the machine turns its input $(1 + K)/(K - v)$ times for each turn of its output: a hyperbola with a pole at $v = K$, forward on one side, reverse on the other, and a standstill with the engine running at the pole itself.

It then priced the pole. The machine’s ratio answers the variator at $(1 + K)/(K - v)^2$, the square of the reciprocal distance to neutral, and a variator uncertain by a per cent or two delivers an output uncertain by more than its own value well before the pole — 110% at a gap of 0.02. The unbounded span is unusable near its middle, and the essay ended on the design question that leaves: the variator’s travel has a fixed length, the pole is fixed by the gearset, and where the travel sits relative to the pole is a free choice. Where should it be spent?

The choice, and what is trimmed #

A fixed gear pair of ratio $k$ between the variator and the planetary — its ratio exact, as a gear train’s always is — multiplies everything the variator delivers: the planetary sees $kv$ instead of $v$. The variator’s travel, 0.544 to 2.045 on the machine the earlier essay built, lands at the planetary between $0.544k$ and $2.045k$. The pole stays at $K = 1.4$. So $k$ slides the whole travel along the axis past a pole that does not move, and nothing about the variator or the gearset changes.

What is kept of the travel is decided by a trim. With the variator uncertain by a fraction $p$ of its own ratio, the output ratio is uncertain by

$$\frac{p\,kv}{|K - kv|},$$

the earlier essay’s sensitivity carried through as a share. A machine that must know its ratio to within τ can use only the settings where that is at most τ. Everything else — a band around the pole, wider where the variator is less certain — is travel the machine owns and cannot use.

At $k = 0.95$ the travel reaches the planetary from 0.516 to 1.943, straddling the pole. The forward stretch that survives a 10% trim runs from an overall ratio of 2.72 to 12.95, a span of 4.77; the reverse stretch from 4.42 to 6.49, a span of 1.47. A quarter of the travel sits in the unusable band and is thrown away.

Dragging the travel across the pole shows the trade the rest of this essay prices. Moving it left, away from the pole, keeps the forward stretch whole and loses reverse; moving it right, into the pole, gains reverse and moves the forward stretch’s low end up.

With one tolerance, the answer is a formula #

Suppose first that the variator’s uncertainty is one number, $p$, everywhere along its travel. Then the forward stretch’s top is wherever the trim bites, $kv = K\tau/(\tau + p)$, and there are two cases.

If the travel’s top is short of that point, the whole forward travel is usable and the span is $(K - kv_{\text{lo}})/(K - kv_{\text{hi}})$, which grows as $k$ grows. If the travel’s top is past it, the top is cut off at the trim and the span is $(K - kv_{\text{lo}})/(K - K\tau/(\tau+p))$, which shrinks as $k$ grows, because only the low end moves. So the best placement puts the travel’s top exactly on the trim, and the span there is

$$1 + \frac{\tau\,(1 - r)}{p}, \qquad r = \frac{v_{\text{lo}}}{v_{\text{hi}}}.$$

$K$ has cancelled. The gearset decides where the best placement is and not how good it is; the variator’s own span, through $r$, and the ratio of the trim to the tolerance decide the rest.

Swept at 1,201 values of $k$, the best placement is within 1% of the travel’s top meeting the trim at all nine combinations, and the best span agrees with the closed form to the sweep’s resolution: 3.28 against 3.29 at a 5% trim, 5.55 against 5.59 at 10%, 10.14 against 10.18 at 20%, the same for all three planetaries at each trim.

The formula is also a statement about what the pole’s unbounded span is worth once a machine must know its ratio. A variator spanning 3.76 to one, uncertain by 1.6%, trimmed at 10%, gives a machine whose usable forward span is 5.6 — better than the variator alone, and nothing like unbounded. Doubling the trim to 20% nearly doubles it; halving the variator’s tolerance does the same.

Why the gearset drops out #

That $K$ cancels from the optimum is the most useful property of the formula and deserves a sentence of explanation, because it is not obvious that the planetary should have no say in how much usable span the machine gets.

Everything in the problem scales with $K$ once the fixed ratio is chosen to match it, which is the same reason a planetary’s lever diagram can be drawn at any size. The trim’s position, $K\tau/(\tau + p)$, is proportional to $K$; the best fixed ratio, which puts the travel’s top there, is proportional to $K$; and so the whole travel, at the planetary, is $K$ times a picture that does not involve $K$. The forward span is a ratio of two distances from the pole, and when every distance is scaled by the same factor their ratio is unchanged. The planetary’s choice of tooth counts therefore moves the best placement and nothing else — which is why, in the table, the three planetaries give identical spans at each trim and differ only in the fixed ratio they need.

The practical consequence is a division of labour. The gearset is chosen for reasons of packaging, torque and the teeth that can be cut; the fixed pair is chosen to put the travel where the formula says; and the span the machine delivers is settled by the variator and the trim alone.

What the trim costs in the first place #

The earlier essay’s table is where the trim’s cost comes from. The variator’s own uncertainty is modest everywhere; what makes the output’s grow without bound is the distance to the pole in its denominator, so a trim is always a band around the pole, never a limit at an end of the travel. How wide the band is at a 10% trim can be read straight off the table: the output crosses 10% of its own value somewhere between a gap of 0.2 and a gap of 0.1, which on this planetary is a band roughly 0.4 wide in the ratio the planetary sees, centred on the pole — about a quarter of the variator’s whole travel, spent whether the pole is inside the travel or just beyond its end.

That quarter is the price of the unbounded span, paid once, and the placement question is only about which quarter. Put the pole outside the travel and the band is spent beyond the travel’s end, where it costs forward span directly; put it inside and the band is spent in the middle, where it splits the travel into a forward piece and a reverse piece. With a tolerance that is the same everywhere the first is better. With this variator’s, the band in the middle is narrower because the middle of the travel is better toleranced, and the difference almost pays for the split.

The variator is not equally certain everywhere #

That analysis assumed the tolerance was one number, and the variator the earlier essay built does not have one.

Its relative uncertainty rises steadily from 0.98% at the low end of its travel to 2.24% at the high end. The reason is geometric: at a high ratio the driven sheave’s pitch radius is small, and the same absolute errors in belt length, centre distance and sheave position are a larger share of a small radius. The high end of the travel — the end that reaches furthest towards the pole when the travel is placed below it — is the least certain part of the variator.

With the real tolerance, the peak becomes a plateau #

Put that tolerance into the same sweep and the answer changes character.

With a constant 1.6%, the forward span peaks at 5.55 at $k = 0.59$ and falls away on both sides: by $k = 1.2$ it is 0.69 of its best. With the variator’s own tolerance, the forward span peaks lower, at 4.81 near $k = 0.98$ — the less certain high end of the travel costs about a seventh of the span — but across the whole range from $k = 0.6$ to $k = 1.2$ it never falls below 0.91 of that best.

The flatness has a clear cause. Sliding the travel towards and past the pole does two things. It lifts the forward stretch’s low end, which costs span. And it changes which part of the variator sits next to the pole: at small $k$ the pole is approached by the variator’s high end, uncertain by over 2%, and the trim bites early; at large $k$ it is approached from below by the middle of the travel, uncertain by about 1.4%, and the trim bites late. The second effect nearly cancels the first. Moving the pole into the travel brings the variator’s better-toleranced part to the one place where tolerance matters most.

And the move buys something the constant-tolerance machine also gets but pays for heavily: reverse. At $k = 0.6$ the travel is entirely below the pole and there is no reverse; at $k = 1.2$ the reverse span is 3.86, with the forward span still at 0.91 of its best. A designer who wants a reverse range in the same machine can have most of it for under a tenth of the forward span, where a constant-tolerance analysis would say it costs a third.

What the spans mean on a machine #

Spans are ratios of ratios, and they read more concretely as speeds. Take an engine held at 2,000 revolutions a minute and the machine at the plateau’s reversing end, a fixed ratio of 1.25. Its usable forward stretch runs from an overall ratio of 3.33 to 15.2: the output from 600 revolutions a minute down to 130, a span of 4.6, with the engine speed unchanged throughout — the thing a stepped gearbox can only approach by spacing its gears. Its usable reverse stretch runs from 2.08 to 9.21, the output reversing at anywhere from 960 revolutions a minute down to 220. Between them lies the trimmed band, where the output turns slower than those lower limits and the ratio is not known to 10%.

That band is where a real machine uses a clutch or a closed loop on output speed, and its width is set mostly by the trim rather than by the placement, because it is defined by distance to the pole in the planetary’s own ratio. What the placement chooses is how the fast end of the travel is divided. At a fixed ratio of 0.6 all of it is forward and the output reaches 895 revolutions a minute forward with no reverse at all; at 1.25 forward tops out at 600 and reverse, lying on the side of the pole where the variator’s travel extends further, reaches 960.

A tractor, which spends its working life at low forward speeds, needs to reverse as often as it goes forward, and has a differential behind the transmission anyway, is the machine the plateau suits. A road vehicle, which needs a wide forward span and reverses slowly and rarely, is the machine the constant-tolerance optimum suits — and it is the case where the variator’s tolerance slope matters least, because the pole is kept outside the travel anyway.

The asymmetry is worth noticing because it is not what a designer drawing the hyperbola would expect. The pole sits at a single ratio, but the travel on either side of it is not the same length once it is multiplied by the fixed ratio, and the reverse side is where the variator’s high, less certain end lands. A machine that wanted a slow, precise reverse and a fast forward would place the travel the other way round — with a fixed pair that reverses the variator’s sense of travel as well as scaling it — and the chart would be read from the other side.

What the placement decides, and what it does not #

It decides the usable spans exactly, once a trim is chosen. Every number here is a count of variator settings whose output uncertainty is under the trim, converted into ratios. The trim is the designer’s; the machine does the rest.

It does not decide the trim. Ten per cent is an arbitrary line. A controller that closes a loop on output speed can live with a far larger kinematic uncertainty than one that sets a ratio open-loop, and for it the unusable band shrinks and the placement question softens. The closed form says how: the span grows in proportion to $\tau/p$.

It treats the variator’s tolerance as its only uncertainty. The planetary’s ratio is exact — tooth counts have no tolerance — and so is the fixed pair’s. That is true of the ratios and not of the backlash, which a machine running through its pole, reversing the torque on every gear, will certainly notice.

It says nothing about efficiency or power circulation. A power split that places its pole inside the variator’s travel circulates power through the variator near the pole, and how much depends on the branch torques rather than on any ratio here. The kinematics says which settings are usable; whether they are efficient is a separate question with a separate answer.

It rests on one variator. The plateau depends on the tolerance rising along the travel. A variator whose tolerance fell along its travel would sharpen the peak instead, and the design advice would reverse: keep the pole outside the travel. The measurement that decides it is the variator’s tolerance stack taken across its travel, which is cheap to make and should be made before the fixed ratio is chosen.

Reading the chart as a specification #

Turned round, the chart answers the question a transmission engineer brings to it. A machine that must reach a forward span of four with its ratio known to 10% can use this variator with any fixed ratio from about 0.6 to 1.3. If it must also reverse over a span of four, the choice narrows to about 1.2 to 1.3, where the forward span is still 4.5 or more. A machine that needs a forward span of five cannot use this variator at a 10% trim at all: the best available is 4.81, and the options are a looser trim, a better-toleranced belt or a variator with a wider span. And a machine that needs forward only should take a fixed ratio near 1.0, where the forward span is at its best and the travel reaches just past the pole.

That is the sense in which two routes to a sensitivity turns a derivative into a design: the square-law sensitivity of the earlier essay, spent on a trim, becomes a chart whose axes are the two numbers a drawing carries — a fixed ratio and a span.

Still open: power circulating through the variator #

Near its pole a power split’s output speed is small while its input speed is not, so the planetary’s two branches carry torques far larger than the output’s, and the variator branch can carry more power than the engine delivers, with the excess circulating back through the straight path. That is the reason real power splits keep their pole out of the working range despite the kinematic case for moving it in.

Its distinct argument would be that circulation computed from the same planetary relation — the branch speeds from the kinematics and the branch torques from the balance of the planetary’s three shafts — expressed as the ratio of the variator’s power to the engine’s across the travel, for the placements charted here. Two things would come out of it. The fixed ratio at which the variator’s load first exceeds the engine’s, which is a hard limit on the plateau found here; and whether the plateau’s cheap reverse survives once the variator’s power rating is imposed, which would decide whether the kinematic advice to place the pole inside the travel is advice a machine can take.