Machine Kinematics
solved, then drawn
Concept

Gear train — where it appears

A set of meshing wheels transmitting rotation, whose overall proportion is decided by the tooth counts and nothing else. Its proportion is exact in integers, which is the property clock trains are chosen for and the reason a train can be searched for exhaustively.

Named by 12 essays across 3 fields — each of them below, with the objects they name alongside it.

404 ways to gear 60 to 1. Each dot is a pair of wheels and a pair of pinions whose ratios multiply to exactly 60, with wheels of 40 to 120 teeth and pinions of 6 to 16 leaves. The line is the hyperbola every exact answer must lie on; the dots are the ones that lie on it at whole numbers of teeth, and there are 404 of them. In a single pair there are 0 — a ratio of sixty needs a wheel of 360 teeth against a pinion of six, and nobody cuts that. Exactness is decided in integers here and never by comparing floating-point ratios, which is the one way a search like this quietly returns answers that are merely close.

A clock is a factorisation

A going train's job is a ratio and its parts are tooth counts, so whether a clock can be built is whether a number factorises inside the counts a wheel-cutting engine will cut. Sixty has 404 answers in two pairs and none at all in one. The ratio between a sidereal day and a mean one has none in any number of pairs, and the best two-pair train is out by a sixth of a second a day.

gears · Tooth Which teeth one tooth ever meets: 20 on 40. Follow one tooth of the pinion round and mark every wheel tooth it touches. It does not touch them all. It touches 2 of 40, which is z₂ divided by the greatest common divisor of the two counts — 20 here — and it goes on touching the same ones for as long as the gears are in mesh. The pattern repeats after 2 turns of the pinion. Adding one tooth to the pinion makes the counts coprime and takes the count from 2 to 40. The marks are produced by walking the mesh, and the count they give is compared with the gcd rather than derived from it.

Which tooth meets which

A tooth on a pinion does not meet every tooth on its wheel. It meets z₂ divided by the greatest common divisor of the two counts, and it meets the same ones for the whole life of the drive. On the default planetary used throughout — sun 24, planets 24 — a planet tooth touches exactly one sun tooth and never touches another, and nothing in the drawing says so.

gears · Tooth Reverted trains of 12 : 1. A reverted train has its input and output shafts in line, which means the two stages share one centre distance — so the tooth counts must satisfy z₁ + z₂ = z₃ + z₄ and give the ratio asked for. Two equations in four integers, and there is no reason for a solution to exist. For 12 : 1 there is none at all until the wheels are allowed to reach 63 teeth; below that the coaxial condition and the ratio simply cannot both be met. This is the same kind of arithmetic as the clock trains of the timing field, with one extra equation, and the extra equation is what a shaft position costs.

Two shafts that must be in line

Asking a two-stage gear train for a ratio is easy. Asking it for a ratio and for its input and output shafts to be coaxial is asking for a solution of two equations in four integers, and there is no reason for one to exist. A twelve-to-one reverted train needs a sixty-three-tooth wheel before it has any solution at all — while sixteen to one, a larger ratio, manages with fifty-six.

transmission · Transmission A variator that spans under four to one, driving a machine that spans everything. The whole machine's ratio against the variator's own, for an engine split between a variator and a straight path and summed by a planetary with 50 sun teeth and 70 ring teeth. The variator's travel runs from 0.544 to 2.045, a span of 3.76 to one, and the output's ratio is (1 + K) / (K − v) with K = 1.4: a hyperbola with a pole at v = 1.4. Left of the pole the output turns one way, right of it the other, and the vertical scale is cut at ±12 because nothing else would fit. The measured extremes over the travel are 2.80 and -3.72, and between them the ratio passes through every value there is.

A bounded ratio made unbounded

Split an engine between a variator and a straight path and add the two with a planetary, and a variator that spans 3.76 to one becomes a machine whose ratio passes through infinity. The setting at which the output stands still is the planetary's own tooth ratio and nothing the belt does moves it. The price is a sensitivity that grows as the reciprocal square of the distance to that setting, and the variator's one and a half per cent becomes a hundred and ten before the setting is reached.

transmission · Transmission Sliding the variator's travel across the pole. The variator of the power split — travel 0.544 to 2.045 — geared by a fixed ratio k ahead of a planetary with K = 1.4, so the planetary sees k times the variator's ratio. At each k the travel is trimmed wherever the output's ratio is uncertain by more than 10%, and what is left gives a forward span and a reverse span. Solid lines use the variator's own tolerance, which grows from 0.98% at one end of its travel to 2.24% at the other; dashed lines use one tolerance of 1.6% everywhere. With one tolerance the forward span peaks sharply, at 5.59 where the travel's top meets the trim, and falls as the pole moves into the travel. With the variator's own it peaks at 4.81 at k = 1.00 and stays within a tenth of that from k = 0.6 to 1.2, while the reverse span climbs from one, meeting the forward span at k = 1.26.

Sliding the travel across the pole

A power split's ratio has a pole the variator's own tolerance makes unusable, so the question is where to put the variator's travel relative to it. With a tolerance that is one number, the best forward span comes where the travel's top just meets the trim — 1 + τ(1 − r)/p, with no gearset in it. With the variator's real tolerance, which grows along its travel, that peak flattens into a plateau: the pole can be moved well inside the travel, buying reverse, for under a tenth of the forward span.

transmission · Transmission A ball bearing turned by its inner race. A bearing of pitch diameter 40 with 8 balls of diameter 8, its outer race held and its inner race turned 60°. Rolling without slipping at both contacts leaves the cage — the ring that keeps the balls apart, marked by the dark tick — turned 24.00°, a share of 0.4000 of the race, and each ball spun 120.0° backwards about its own centre, marked by its own tick. The share is (1 − d/D)/2, less than a half by d/2D; the gear-train solver, handed a planetary with a sun of 32 teeth and a ring of 48, returns 2/5 for its carrier. Dragging turns the inner race.

A bearing is a planetary with no teeth

Roll a ball between two races and nothing but the two rolling constraints decides how fast its centre goes round. Solved, they put the cage at (1 − d/D)/2 of the inner race's speed — always less than half — and the gear field's train solver, handed a planetary with a sun of D − d teeth and a ring of D + d, returns the same fraction exactly. And a tapered roller rolls without slipping along its whole line only if its axis meets the bearing's at the apex, the wheel field's concurrency one dimension up.

rolling · Rolling What each branch carries, against what the engine delivers. The power in the variator's branch and in the straight path, as multiples of the engine's own, for a planetary of K = 1.40. Both are ratios of powers, so the load cancels and nothing here is a force: the split comes from requiring the gearset to be lossless at every admissible set of speeds, which fixes the torques at 1 : K : −(1+K). The variator carries v/(K − v) and the straight path K/(K − v), and both run away at the pole. The variator first carries the engine's whole power at v = 0.700, which is exactly half the way to the pole — and the straight path is already carrying more than the engine everywhere past nought, flowing the other way through the planetary. That excess is the circulation.

The power that goes round twice

Sliding a power split's travel towards its pole buys ratio span for nothing, on the kinematics. It is not for nothing. The variator's own branch carries v/(K−v) of the engine's power and overtakes it at exactly half the way to the pole, and the tolerance trim the span was computed from is not reached until the variator is rated for six times the engine — which no machine is.

transmission · Transmission Every power split, on one curve. 1,994 power splits: all six ways of putting an engine, a variator and an output on a simple planetary's three shafts, with the variator coupled at the input or at the output, turning either way, on nine gearsets, with its travel placed at eleven positions. Every variator spans 3.333, as the variator of the previous essay does. Each dot is one split: the largest share of the engine's power its variator carries anywhere on the travel, and the span its output covers. They fall on three curves and nowhere else, to 7e-16: with no circulation, span R/(R(1 − S) + S), never more than the variator's own; with the variator's branch running backwards, 1 + S(1 − 1/R); with the variator carrying more than the engine, 1 + S(R − 1). 1188, 394 and 412 splits lie on the three. Which curve a split is on is decided by where its travel sits, not by which shafts it uses: every one of the twelve attachments appears on all three.

Every power split is on one curve

There are twelve ways to attach an engine, a variator and an output to a planetary's three shafts, and asking which is best turns out to be the wrong question. On 1,994 splits across all twelve and nine gearsets, the output's span is (1 + R·a)/(1 + a) to 5 × 10⁻¹⁵, where R is the variator's own span and a is the power the variator carries over the power the gears carry. Every split lies on one of three curves of span against the variator's largest share. The previous essay's 1.70 is 2 − 1/R on the worst of them, and a split beats its variator's own span only by making the variator carry more than the engine.

transmission · Transmission A second planetary buys span at every variator share, if its lever can be built. For a variator spanning 3.333, the most output span a power split reaches when its variator carries at most a given share of the engine's power. On one planetary it is R/(R(1 − S) + S) (dashed). On a four-shaft gearset with the variator between two of its shafts — the compound split — it is the square of (√R + k)/(1 + k√R), k = (1 − S)/(1 + S) (solid): 1.124 against 1.075 at a share of 0.1, 1.802 against 1.538 at 0.5, and both reach R at 1. The dots are the best compound split on four real Simpson and Ravigneaux gearsets at each share: ahead up to 0.3 (1.320 against 1.266), behind from 0.4 (1.423 at 0.5).

A second planetary lets the variator carry a difference

On a single planetary every power split lies on one of three curves of span against the variator's share of the engine's power. Build the split on a four-shaft Simpson or Ravigneaux instead and nothing changes while the fourth shaft is free: 3,588 splits, all on the curves. The one new arrangement is the variator between two of the gearset's shafts. That compound split is an input-coupled split and an output-coupled split in series, and its variator carries only the difference of their shares — so at every share below one it can span more than any single planetary, the square of (√R + k)/(1 + k√R). It needs a lever with the variator's shafts exactly its share apart, which real gearsets do not have beyond a share of about a third.

transmission · Transmission A compound split sees its gearset through one cross-ratio. The four shafts of the Ravigneaux 69/74/100, planets 13 and 13 on its lever, each at the position that makes every shaft's speed a straight-line function of it: sun2 at 0.000 (engine), carrier at 0.483 (variator), ring at 0.840 (variator), sun1 at 1.000 (output). The best compound split puts the variator between the two shafts shaded together and the engine and output on the other two. What the split can do depends only on the cross-ratio of the four positions, here 5.613, and a split with that cross-ratio beats every split on a single planetary up to a variator share of 0.542. A share of a third needs a cross-ratio of 2.52 and a half needs 4.67.

One cross-ratio decides the compound split

A compound power split, its variator between two shafts of a four-shaft gearset, can beat every split on a single planetary, but four real gearsets let it do so only up to a variator share of about a third. Whether teeth can do better turns on one number: the cross-ratio of the gearset's four shaft positions on its lever. A third needs 2.52 and a half 4.67. Every Simpson inside a ring of 100 teeth stops at 3.87, a share of 0.454. A Ravigneaux with large suns and 13-tooth planets reaches 5.61, a share of 0.542, and solved on its lever it passes a half. With planets of 17 teeth, neither family does.

transmission · Transmission Two planetaries joined at two places make one four-shaft lever. The sun–carrier, carrier–ring join: planetary 1, a sun of 26 in a ring of 52 (ratio 2.00), and planetary 2, a sun of 41 in a ring of 71 (ratio 1.73), each on its own lever with sun, carrier and ring at 0, K and K + 1, laid so that the joined members coincide. The four shafts below are the two joints and the free member of each. Their cross-ratio is 4.732, enough for a compound split to beat every split on one planetary up to a variator share of a half; its best split there spans 1.544, with the engine, output and variator where marked, and no member of either planetary carries more than 0.80 of the engine's power.

The compound split has gearsets of its own

A compound power split depends on its gearset through one number, the cross-ratio of the four shafts on its lever, and a variator share of a half needs 4.67. Simpsons stop at 3.87 inside a ring of 100 and Ravigneaux reach it only with 13-tooth planets. Two simple planetaries can be joined into four shafts in eighteen ways, twelve of them distinct gearsets, and ten of the twelve reach a half with ordinary ring-to-sun ratios. The cross-ratio is then no longer the question. Three joins reach any cross-ratio by setting two shafts nearly together and pay with 2.7 to 4.2 times the engine's power inside, and one, carrier to ring both ways, never reaches a half. A sun-to-carrier, carrier-to-ring join builds the lever with planetaries of 26/52 and 41/71, and nothing inside it carries more than 0.80 of the engine's power.

transmission · Transmission Each fixed gear is a straight speed line through a braked shaft. The sun–carrier, carrier–ring join of a 26-tooth sun in a 52-tooth ring and a 41-tooth sun in a 71-tooth ring, its four shafts at their lever positions: sun₂ at 0.000, sun₁ + carrier₂ at 0.536, carrier₁ + ring₂ at 0.845, ring₁ at 1.000. The output stays where the compound split has it, on sun₁ + carrier₂. In the gear shown the engine drives carrier₁ + ring₂ and sun₂ is braked, so the shafts' speeds lie on a straight line through nought at the brake and one at the engine; the output turns at 0.634 of the engine's speed, a ratio of 1.58.

The winning join is a gearbox too

The sun–carrier, carrier–ring join of two simple planetaries builds a compound power split's half-share lever with ordinary teeth. Locked out of the split and given clutches and brakes, the same four shafts give five forward ratios below the split's range, from 2.73 to direct drive, and an overdrive of 3 above it. The two gears that brake a variator shaft sit 1.76 and 1.74 outside the split's ends, and three gearsets with the same cross-ratio put them at the same steps. Its planets space evenly, three and four to a set, and its meshes lose 0.41% of the engine's power along the split where the Ravigneaux that builds the same lever loses 1.01%. What none of these gearsets gives, with the output where the split needs it, is a usable reverse.

transmission · Transmission

Named alongside it

The objects these essays reach for when they reach for this one.

Design ruleEpicyclicRatioVariatorContinuously variableEnumerationVelocity ratioEfficiencySensitivityToleranceWheel trainDifferential

All concepts