A bounded ratio made unbounded

A ratio with no steps in it measured what a variator is and what it costs. Its ratio is a quotient of two solved lengths rather than a quotient of two integers, so it has a tolerance table where a gear train has none — a belt long by three parts in a thousand, a centre distance out by two tenths and a sheave a seventh out of position together leave the ratio uncertain by about one and a half per cent. And its range is bounded: the sheaves can only move so far, and this one spans 3.76 to one.

Two inputs and one output is the other half of what follows. A mechanism with two inputs has no ratio at all until something says how the two are related; its kinematics is one linear condition on three speeds, and the output is a sum rather than a function.

Put the two together — feed one input of a planetary from a variator and the other straight from the engine — and the bounded thing becomes unbounded.

One subtraction #

The planetary’s relation is exact and it is short. With $K$ the ratio of ring teeth to sun teeth, the three speeds satisfy

$$\omega_{\text{sun}} + K\,\omega_{\text{ring}} = (1 + K)\,\omega_{\text{carrier}} ,$$

which is the same relation three mechanisms share and the reason a differential’s carrier turns at a weighted mean of its two side gears.

Drive the ring straight from the engine and the sun from the variator, taken in the opposite sense, and the carrier — the output — turns at

$$\frac{\omega_{\text{out}}}{\omega_{\text{in}}} = \frac{K - v}{1 + K},$$

with $v$ the variator’s own ratio. That is linear in $v$ and it passes through nought at $v = K$. So the machine’s overall ratio, input turns per output turn, is $(1 + K)/(K - v)$: a hyperbola with a pole inside the variator’s travel.

Left of the pole the output turns one way and right of it the other. There is no step, no clutch and no idler gear anywhere in the description, and the machine delivers forward, a stop, and reverse.

The relation itself is not new here and its exactness is worth leaning on. A ratio is a null space: the gearset’s kinematics is a matrix of tooth counts, its constraint rows are one per mesh, and the solution space of a two-freedom set is a plane. Every gear a conventional gearbox sells is that plane cut by one more row — a brake or a clutch — and holding a member chooses the ratio is the whole of how that works. A power split cuts the plane with a row of a different kind: not “this member is still” but “these two members stand in the ratio $v$”, with $v$ a real number the driver can move. The plane is the same plane; what has changed is that the cutting line sweeps.

That is why the output’s ratio is a hyperbola and not a list. The steps are not free argued that a gearbox’s ratios are constrained collectively — any one is available and not all four at once — because each is a different cut of one plane. Here the cuts are a continuum, and the line the cuts trace out in the plane passes through the direction in which the output is stationary.

The neutral belongs to the gearset #

The pole is at $v = K$, which is a statement about tooth counts. The obvious check is whether the variator can move it, since the variator is the only thing in the machine anybody adjusts.

In every row the setting is the planetary’s own $K$, to within 0.0033, and lengthening or shortening the belt by two per cent — which moves the variator’s whole travel, both ends of it — does not move the neutral. That is what the relation says and it is worth measuring anyway, because the belt is the part of this machine whose length is not exactly known and it would be easy to assume a quantity that depends on the belt somewhere.

It also sets a condition on the gearset that is not obvious. A simple planetary’s $K$ is always above one, because its ring is always larger than its sun. This variator’s travel runs from 0.544 to 2.045. So the neutral is reachable only when the ring is less than about twice the sun — and a planetary chosen for any other reason, with a ring three or four times its sun, gives a machine with no neutral and a perfectly ordinary bounded ratio.

One more thing about the sign, because it is the only place the arrangement is delicate. The variator’s path is taken in the opposite sense to the straight one — through an idler, or by taking the variator’s output from the other side of its second sheave — and without that reversal the output is $(K + v)/(1 + K)$, which is positive over the whole travel and has no pole in it at all. A split built with both paths the same way round is an ordinary continuously variable transmission with a slightly compressed range, and it is the same four shafts and the same gearset. The whole of what is described here turns on one reversal, and a machine built without noticing it would work, would be quiet, and would have none of the properties this essay is about.

What the span becomes #

The span is the point of the arrangement and it is worth quoting as a number.

The variator spans 3.76 to one. Over the same travel the machine spans 56 to one going forward, from 2.80 at one end to 156 near the pole, and 756 to one going backwards. Those two numbers are not the machine’s: they are where the sweep stopped. The hyperbola has no largest value, and the only thing bounding the span is how close to the pole the sweep was asked to look.

Reverse arrives with it, at no cost in parts. A conventional gearbox buys reverse with an idler gear and a selector; here it is the far side of a pole that was already there.

It is worth setting that against the field’s other unreachable ratios. The reductions a planetary cannot give found whole stretches of the positive line unavailable to any epicyclic that can actually be cut and assembled, because the tooth counts are integers and the assembly conditions are arithmetic. Two shafts that must be in line found a Diophantine problem with no solutions at all below a certain size. Both are gaps in the reachable set. What a split produces is the opposite: a reachable set with no gaps and no ends, obtained from a variator whose own set has both.

The price, and it is a square #

An unbounded span from a bounded variator is arithmetic rather than magic, and the arithmetic says where the cost went.

Differentiating the hyperbola,

$$\frac{\mathrm{d}}{\mathrm{d}v}\!\left(\frac{1+K}{K-v}\right) = \frac{1+K}{(K-v)^2},$$

so how sharply the machine’s ratio answers the variator is the reciprocal square of the distance to neutral.

The two routes agree to $10^{-8}$ at every gap, and the slope is two: 15 at a gap of 0.4, 60 at 0.2, 240 at 0.1, and 24,000 at 0.01. Halving the distance to neutral quadruples what a given error in the variator is worth at the output.

The tolerance table, carried through #

The variator already has a tolerance table, so the sensitivity can be spent rather than admired.

The variator is uncertain by between 1.06% and 1.60% across the travel — its own stack, unchanged by anything the planetary does. The output’s uncertainty is 1.1% at a gap of 0.7, 3.2% at 0.4, 8.6% at 0.2, 19.8% at 0.1, 42.4% at 0.05, and 110.3% at a gap of 0.02.

That last row is the result. Before the machine gets within a fiftieth of its neutral setting, the output ratio is uncertain by more than its own value: the control knows the ratio is large and does not know whether it is 120 or 250 or negative.

So geared neutral is not a ratio a machine can be asked to hold. It is a boundary, and the usable travel stops short of it by an amount the gearset’s $K$ and the variator’s own tolerance decide together. A machine that has to stand still with its engine turning still needs a clutch or a brake; what the split buys is that the clutch is released at a ratio the driver does not notice rather than at a standstill.

The shape of that table is the same shape two routes to a sensitivity is about, one field over: a derivative computed two ways and then spent on a tolerance. What is unusual is the size. A sensitivity of 15 is a design number; a sensitivity of 24,000 is a statement that the quantity has stopped being measurable, and the tolerance column is where that crossing is located rather than guessed at.

An exact pole a machine cannot find #

The two halves of this machine are exact and inexact in different senses, and the pole sits exactly on the join.

The planetary’s $K$ is $70/50$, which is $7/5$ and is that rational number with nothing rounded: tooth counts are integers, the constraint matrix is integral, and the field computes its null spaces in exact rationals for that reason. So the statement “the output stands still at $v = 7/5$” is exact in the strongest sense the subject has — it is a fact about two whole numbers, and the sweep above agreeing with it to 0.0009 is the sweep’s resolution rather than the fact’s.

The variator’s $v$ is nothing of the kind. It is a ratio of two radii, one of which is solved from a transcendental belt-length equation, and it is uncertain by one and a half per cent for reasons that have nothing to do with arithmetic.

So the machine has an exactly located pole and an instrument that cannot be set to it. That is a shape worth naming because it is the opposite of the usual one: elsewhere in this field the trouble is that a required ratio is not available — a Diophantine condition with no solutions, a reduction no assemblable planetary reaches. Here the ratio is available, is available exactly, and is the one ratio the machine cannot be commanded to. The gap the control has to leave around it is therefore not a manufacturing allowance on the gearset, which has none to give, but a working allowance on the belt, and it is the only number in the machine that has to be chosen rather than computed.

Where the cost is not #

Three things are worth saying about what this measurement does not show, because they are what the arrangement is usually criticised for and they are not kinematics.

Power circulates. Near neutral the two paths carry much more power than the engine delivers, one forwards and one backwards round the loop, and the variator has to carry its share of a quantity larger than the whole. That is a statement about torques, and no torque is computed here.

Efficiency falls with it, because the circulating power passes through the variator’s belt twice and pays its losses each time.

And the variator’s own speed does not diverge. Its ratio is $v$ and $v$ stays inside its travel; what diverges is the output’s ratio, which is a quotient with a small denominator. Nothing in the machine turns arbitrarily fast — the arithmetic’s pole is in a ratio, not in a speed, and the two are easy to confuse.

What the machine is, as a list of parts #

It is worth writing the machine out, because the kinematic argument above is short enough to sound like a trick and the parts list is what says it is not.

An engine shaft. A belt variator on it, of the kind the field already has — two pairs of conical sheaves, a belt, a controller that moves one pair. A fixed reduction from the variator’s output to the sun of a planetary. A second fixed path from the engine shaft to the ring of the same planetary. The carrier is the output. Four shafts, one belt, one gearset, and no clutch anywhere in the power path.

Against a conventional automatic that is fewer parts, not more. A Simpson gearset delivers four speeds and reverse from two planetaries and five friction elements, every one of which has to be applied and released in a sequence. Here nothing is applied and nothing is released; the ratio is wherever the sheaves are.

What the arrangement buys, then, is not a ratio range that could not otherwise be had — the Simpson’s range is perfectly adequate — but a range with no steps in it and no shift, at the cost of a belt that has to carry a power larger than the engine’s near one end of its travel. That is the trade, and everything above is the kinematic half of it.

What this does not settle #

One variator and four gearsets. The belt variator measured here is the one the field already has, with its own centre distance, sheave travel and tolerance stack. A toroidal or a chain variator has a different span and a different stack, and both enter the answer.

Both senses are not swept. The reversal is stated above and its consequence written out; the same-sense machine is not measured, so the claim that it has no pole rests on the sign of a sum rather than on a sweep.

The split is one of several. Sun from the variator and ring from the engine is one arrangement; the other five permutations of which member takes which path, and which is the output, give different values of $K$ and different places for the pole, and the input-coupled and output-coupled families are genuinely different machines. Only one is measured.

The tolerance stack is worst-case-and-root-sum-square. It is the convention the field uses, and it treats three independent contributions as independent. A real machine’s belt length and centre distance are not independent once the belt is under tension.

A hundred to one is available two ways. A difference of one tooth gives an epicyclic an enormous reduction from a small difference in the denominator, which is arithmetically the same manoeuvre as this one — a quotient with a small denominator — and it has the same amplification of tooth errors as its price. Nothing here compares the two, and the comparison would be a useful one, because the first achieves it with integers and the second with a length.

Nothing here is a control. The statement that the machine must stop short of neutral is a statement about what the ratio is known to be, not about what a controller can do with feedback from the output shaft — which is exactly what a real transmission has, and which changes the problem from measuring the ratio to regulating it.

Still open: where the travel should be spent #

The pole’s position is fixed by $K$, and the variator’s travel is fixed by its sheaves — but where the travel sits relative to the pole is a free choice, made by whatever fixed gearing stands between the variator and the planetary. Slide the travel and the machine’s reachable band of ratios slides with it, and so does how much of the travel is spent in the region where the output’s uncertainty is unacceptable.

Its distinct argument would be that choice, priced: for a fixed variator and a fixed gearset, the reachable band of output ratios as a function of where the travel is placed, with the unusable part near the pole trimmed off at a stated uncertainty rather than at the pole itself. Two things come out of it. There is a placement that maximises the usable forward span, and it is not the one that puts the pole in the middle; and the trimming means the usable span is finite and computable, which turns the unbounded number quoted above into the one a designer could put on a drawing.