Machine Kinematics
solved, then drawn
Series

Transmission — the series

16 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. 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

    Every mechanism in this collection so far has had one input, and its output has been a function of that input. A differential does not. Its cage turns at the mean of two wheels, so knowing one of them tells you nothing at all about where the third shaft is going — and that is not a complication of the mechanism, it is a different kind of object.

    part 1 · transmission
  2. 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

    Write a gear train as a graph — bodies for vertices, meshes for edges, and on every edge the body the two axes are stationary in — and one formula covers a countershaft gearbox, a planetary, a harmonic drive and a car's differential. The ratio is the null space of a matrix whose entries are tooth counts, so it comes out as a fraction and not as a number that is nearly one.

    part 2 · transmission
  3. 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 lever diagram of an epicyclic is usually offered as a mnemonic. It is exact, and the reason is a fact about the null space: a gearset whose frame carries no teeth can turn as a block, and that one motion supplies the coordinate every member is plotted at. Where the line crosses the axis is the member standing still, and the ordering of the members on the lever settles which gears are reductions and which run backwards, without a formula anywhere.

    part 2 · transmission
  4. 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

    A gearset offers a plane of motions and a shift element is one linear condition, so a gear is a line in that plane. Enumerate every brake and every clutch and a Ravigneaux's twenty-seven combinations collapse to seven ratios — eighteen of them being the same gear, because locking any two members at all locks the whole gearset solid. Neutral is not one of the seven, because neutral is not a gear.

    part 3 · transmission
  5. 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

    A Ravigneaux gearset has five tooth counts and gives seven exact ratios. Two of the counts do not appear in any of them — the short planet is an idler and its size is free — and the remaining three enter only through two dimensionless numbers, so the whole shift ladder of a four-speed automatic is a function of ring-over-sun and ring-over-the-other-sun. A Simpson three-speed is a function of one number.

    part 3 · transmission
  6. 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

    A gearbox is supposed to have equal steps between its gears, so that the engine returns to the same speed after every shift. A gearset has one or two numbers to spend on three or four gears, so from the third one the steps are a consequence rather than a choice — and asking for them to be equal turns out to be a quadratic whose root is the golden ratio, realised in tooth counts by consecutive Fibonacci numbers.

    part 4 · transmission
  7. 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

    A differential with one wheel stopped turns the other at exactly twice the cage, and the relation it imposes is satisfied the whole time — nothing has failed, nothing is confused, and the reason the car does not move is not in this site. What is here is the other half: a locked axle is an overconstrained mechanism, and the sliding it produces is 2π times the track per circle driven, whatever the radius.

    part 4 · transmission
  8. 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

    Push a variable pulley's sheaves together and the belt rides further out. The other pulley's radius is then not a choice — the belt has a fixed length — so it is the root of an equation, solved rather than set. The rule of thumb that says the two radii add to a constant is true to first order and wrong by 7.4% of the ratio at full shift, and the departure has a closed form.

    part 5 · transmission
  9. 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

    A harmonic drive reduces by a hundred to one in a single stage with two gears in it, and the hundred is the flexspline's tooth count divided by the two teeth the circular spline has more than it. The same null space that answers a planetary answers it. What each of the three single-stage reductions pays for that arithmetic is different, and the compound epicyclic's price is a pair of meshes whose centre distances differ by half a tooth.

    part 5 · transmission
  10. 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

    A micrometer's differential screw, a chain hoist's differential pulley and a robot joint's compound epicyclic look nothing like each other and are the same device. Each takes two nearly equal quantities and returns their difference, each buys its enormous ratio with that difference, and each carries the same conditioning number — |a/(a−b)| — measured here by perturbing the mechanisms rather than by quoting the formula.

    part 5 · transmission
  11. 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

    One epicyclic offers six ratios, and the formula for each of them suggests the whole positive line is available. Sweep every design that can actually be cut and assembled and the reachable set has a hole in it running from 1.630 to 2.586 — the width of which has a closed form — and a reduction of exactly 2, the most ordinary thing anybody asks a gearbox for, sits in the middle of it.

    part 6 · transmission
  12. 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.

    part 6 · transmission
  13. 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

    The geometric half of force transmission is an angle in a triangle whose three sides scale together, so it is the same at every size — 54.31° at its worst on this machine, whatever units the drawing is in. Which means a measurement that cannot recover a single length recovers the whole of what this field computes.

    part 13 · transmission
  14. 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.

    part 14 · transmission
  15. 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.

    part 15 · transmission
  16. 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.

    part 16 · transmission

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