Machine Kinematics
solved, then drawn
Field

More than one input

Every mechanism here so far has had one input, and its output has been a function of it. A differential does not: its cage turns at the mean of two wheels, and neither of them decides anything alone. What a gear train has is not a ratio but a relation — a plane of permitted motions — and every number a gearbox is sold with is that plane cut by a brake, a clutch or a choice of which shaft is driven.
A bevel differential, turning. cage in, hold left, with every member's speed taken from the train's null space and every angular position that speed integrated. The teeth are marked at the pitch points rather than cut as involutes — the flank is the teeth field's subject — but the count is the tooth count and the positions are the solved ones, so what turns and how fast is real. Drag it and watch which way each member goes: left 0.000 · right 2.000 · cage 1.000.

Two inputs and one output

The mechanism whose output is not a function of its input: two degrees of freedom, one relation between three shafts, and why a gear train has a plane of permitted motions rather than a ratio.

 A simple planetary, as a graph. Members are vertices and meshes are edges. Each edge carries the two tooth counts and the body the two axes are stationary in — its carrier — and that third label is the whole of what makes an epicyclic different from an ordinary train. Write the mesh relation relative to the carrier and one formula covers both: an ordinary train is the case where every carrier is the frame. This train has 2 meshes across 5 members and 2 freedoms.

A ratio is a null space

The gear train as a graph, Willis's relation as one row per mesh, mobility as the nullity of an integer matrix, and gear ratios computed exactly in rational arithmetic.

 The lever of a simple planetary. Each member sits at a position on the lever fixed by the tooth counts alone, and its speed is the height of one straight line over that position. The line here is drawn through sun and ring; every other member is plotted where the train's null space puts it, and lands on the line exactly — the residual is zero in rationals, and 2.2e-16 once the coordinates have been rounded to doubles for the drawing. Where the line crosses the axis is the member that is standing still, and that is what a brake does: it pins the line to the axis at one position and leaves it free to pivot there. The planets are on the lever too, off the end of it, which is where they belong — they are members of the train and are not shafts anybody can reach.

The lever that is the gearset

The epicyclic lever analogy computed rather than remembered — the members' positions as coordinates in the null space, the exactness proved in rational arithmetic, and the trains that have no lever at all.

 What a Simpson gearset can be made to do. Every distinct ratio the gearset offers, as a reduction. Each one is the same mechanism with one more constraint imposed — a brake holding a member to the case or a clutch locking two members together — and the input is a choice as well, which is why a four-speed needs an input clutch rather than four brakes. The shaded rows are the 4 a real transmission on this gearset is sold with; the rest are ratios the mechanism has and the gearbox does not buy the elements to reach. Every value is exact: the reductions are ratios of integers and are printed as such.

Holding a member chooses the ratio

Every brake and clutch a gearset offers, enumerated: the three outcomes, why all six clutches give the same direct drive, and why a gearbox in neutral is the mechanism in its own natural state.

 What a Ravigneaux gearset can be made to do. Every distinct ratio the gearset offers, as a reduction. Each one is the same mechanism with one more constraint imposed — a brake holding a member to the case or a clutch locking two members together — and the input is a choice as well, which is why a four-speed needs an input clutch rather than four brakes. The shaded rows are the 5 a real transmission on this gearset is sold with; the rest are ratios the mechanism has and the gearbox does not buy the elements to reach. Every value is exact: the reductions are ratios of integers and are printed as such.

Four speeds from two numbers

The Ravigneaux and Simpson gearsets' ratio sequences derived as closed forms in one and two dimensionless ratios, checked against an exact null space and against a published transmission's gears.

 The steps, against the ones the rule asks for. The design rule everybody quotes is that the steps between gears should be equal in ratio, so that the engine returns to the same speed after every shift. That makes the sequence geometric, and the ideal step for this spread over this many gears is 1.5324, marked. The steps a gearset actually gives are not free: the whole sequence is a function of the tooth counts, so once the top and bottom are chosen there is nothing left to spend on the middle. The worst step here is off the ideal by 10.8%.

The steps are not free

Why a gearset's ratio steps cannot all be chosen, the equal-step condition solved as a quadratic, and the Fibonacci gearsets whose ratio sequences are geometric to a hundredth of a per cent.

 A bevel differential, turning. cage in, hold left, with every member's speed taken from the train's null space and every angular position that speed integrated. The teeth are marked at the pitch points rather than cut as involutes — the flank is the teeth field's subject — but the count is the tooth count and the positions are the solved ones, so what turns and how fast is real. Drag it and watch which way each member goes: left 0.000 · right 2.000 · cage 1.000.

One wheel on ice

The differential's speed relation as a kinematic statement, what it can and cannot explain about a wheel spinning on ice, and the locked axle as an overconstrained mechanism measured in scrub.

 A ratio with no steps in it. Two pulleys whose sheaves slide on their shafts, and one belt. Pushing the primary's sheaves together makes the belt ride further out; the secondary's radius is then not a choice, because the belt is a fixed length and its length over two pulleys at a fixed centre distance is a function of both radii. So the secondary's radius here is the root of that equation, solved rather than assumed, and the drawn belt is the length it is supposed to be to 0.0e+0 mm. Ratio 1.000, with the two radii adding to 110.00 — a number the received rule of thumb says should not change and which changes by 2.6% across the travel.

A ratio with no steps in it

A continuously variable transmission's running radii solved from a fixed belt length, the ratio range that follows, and why the two radii do not move by equal and opposite amounts.

 One planet shaft, two centre distances. A compound epicyclic gets its enormous reduction from two meshes whose tooth counts are nearly in the same proportion. The two planet gears are on one shaft, so their axes are at one radius — and at a common module the two rings ask for radii that differ by 0.50 of a tooth. The exactness the reduction is famous for is bought with a pair of meshes running away from the centre distance they were cut at, and the difference is made up by profile shift — the same correction the teeth field applies for a different reason. It is not a rounding: it is the mechanism's own condition, and it is the reason a catalogue reduction of this kind comes in a short list of tooth counts rather than in any combination.

A hundred to one from a difference of one

Harmonic and cycloidal drives as epicyclics with one mesh, the tooth-difference ratio derived from the same graph as a planetary, and the geometric price a compound epicyclic pays for its exactness.

 Everything one planetary can do, and the gap in the middle. A single epicyclic has three shafts, so there are six ways of choosing which is held, which is driven and which comes out. Each gives a band of reductions as the tooth counts run over every design that can be cut, assembled with three planets and kept clear of undercutting. The bands above 1 are drawn; between them is a gap running from 1.6304 to 2.5862 that no single planetary reaches in any configuration — and a reduction of exactly 2, which is the most ordinary thing anybody asks a gearbox for, is inside it. The gap's width as a factor is exactly the smallest achievable ring-over-sun ratio, 1.5862, which is a statement about how small a planet may be and how large a sun may be.

Three mechanisms, one subtraction

Three unrelated-looking reduction mechanisms shown to compute one subtraction, with the conditioning number derived, measured by perturbation, and found to be the same expression in all three.

 Everything one planetary can do, and the gap in the middle. A single epicyclic has three shafts, so there are six ways of choosing which is held, which is driven and which comes out. Each gives a band of reductions as the tooth counts run over every design that can be cut, assembled with three planets and kept clear of undercutting. The bands above 1 are drawn; between them is a gap running from 1.6304 to 2.5862 that no single planetary reaches in any configuration — and a reduction of exactly 2, which is the most ordinary thing anybody asks a gearbox for, is inside it. The gap's width as a factor is exactly the smallest achievable ring-over-sun ratio, 1.5862, which is a statement about how small a planet may be and how large a sun may be.

The reductions a planetary cannot give

The complete set of reductions a single planetary can reach, the gap in the middle, and why the bound comes from the assembly conditions rather than from anything about ratios.

 The transmission angle through one turn. μ is the angle at B between coupler and rocker, computed from each solved position rather than from a formula. It runs from 54.3° to 100.3° for these lengths. The shaded band is the usual design rule — keep μ between 40° and 140° — and this linkage stays inside it throughout. The rule is about geometry alone: nothing here knows about friction, and a mechanism with a comfortable μ can still be a poor machine.

The transmission angle has no size

Why the transmission angle, the pressure angle and every quantity in the transmission field survive a scaling, and what that says about measuring one.

 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

Reverted gear trains as a Diophantine problem, enumerated exhaustively: which ratios admit equal stages, which need larger wheels than their neighbours, and what a coaxial shaft costs.

 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

A power-split transmission measured as arithmetic: the hyperbola the output ratio traces against the variator's, the neutral setting shown to belong to the gearset over four planetaries and three belt lengths, the (1 + K)/(K − v)² sensitivity checked two ways, and the variator's own tolerance stack carried through to an output uncertainty that exceeds its own ratio.

 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

The fixed gear ratio between a variator and a power-split planetary treated as the design variable; the usable forward and reverse spans after trimming wherever the output ratio is uncertain by more than a stated share; the closed-form optimum 1 + τ(1 − r)/p for a constant tolerance, checked by sweeping at three trims and three planetaries; the variator's tolerance measured to rise from 0.98% to 2.24% along its travel; and the flat optimum that rise produces.

 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

The branch powers of a power split from the planetary's own lossless torque split, 1 : K : −(1+K): the variator carrying v/(K−v) of the engine and the straight path K/(K−v) flowing the other way, the crossing at v = K/2 across nine gearsets, and a usable span of 1.70 where the kinematic arithmetic promises 5.375.

 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

Power splits on a simple planetary: the six attachments of engine, variator and output to sun, ring and carrier, with the variator coupled at the input or the output; the identity span = (1 + R a)/(1 + a) from the planetary's lossless torque split; the three branches of span against the variator's largest share (no circulation, the variator's branch running backwards, the variator carrying more than the engine); the previous essay's 1.70 as 2 − 1/R; the best span for a given variator rating; and multi-mode machines as products of spans without circulation.

 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

Power splits on a four-shaft gearset: every attachment on two Simpson and two Ravigneaux gearsets as exact levers; splits with a free fourth shaft on the three single-planetary curves; the compound split's output ratio C(1 + Av)/(1 + Bv) and its variator share as a difference; the frontier span [(√R + k)/(1 + k√R)]² on a lever with the variator's shafts at (1 ∓ S)/2; the shares along one frontier travel; and what four real gearsets reach.

 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

Which tooth counts can build the compound split's lever: a gearset's four shafts on its lever with the variator's pair marked, for six gearsets; the largest share a compound split beats one planetary at, against the cross-ratio of its shafts, as a law and as a search; every buildable Simpson and Ravigneaux up to a ring of 100 by their ring-to-sun ratios; the census at two planet minima; and the best of each family solved on its lever against one planetary and the frontier.

 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

Every way to join two simple planetaries into a four-shaft gearset, measured for a compound power split: two planetaries' levers laid on the four-shaft lever they make; each join's largest cross-ratio over buildable teeth and with ordinary proportions; the largest member power each pays to build the half-share lever; the cross-ratio over both ring-to-sun ratios for one join; the law of share against cross-ratio; and a ledger of all twelve.

 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 a 26/52 and a 41/71 planetary built as a gearbox: its lever with each fixed gear's speed line through the braked shaft; the ratios of four gearsets with their splits' ranges and fixed gears; the mesh losses along each split; the loss in each fixed gear for the winner and a Ravigneaux; and a ledger of the four as gearboxes.

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