Machine Kinematics
solved, then drawn
Teeth

Epicyclic ratios, two ways

An epicyclic train has three shafts and one equation relating them, so fixing any one gives a different ratio from the same gears. The sign errors are notorious, so this site computes every ratio by Willis's equation and by the tabular method and requires them to agree.
17 min read 6 figures Two routes to the same numberThe constant that is not constant

An ordinary gear train has fixed shaft centres, and its ratio is the product of its stages. An epicyclic does not: its planet gears are carried on an arm that itself rotates, so a planet both spins about its own axis and orbits the sun.

That gives three shafts — sun, ring and carrier — with one equation between them. Two must be specified before the third is determined, and which two is a design decision that changes the ratio completely.

Sun 24, ring 72, planet 24An epicyclic train has three shafts and one equation relating them, so fixing any one gives a different ratio from the same gears. With the ring held, a sun input turns the carrier at 1 + 72/24 = 4.000 to one. Willis's equation and the tabular superposition method are independent derivations and both are computed here; the site requires them to agree, because planetary ratios are the most commonly mis-stated numbers in mechanism work and the sign errors are notorious.ringsunhold the ringsun in, carrier out4.000 : 1same directionhold the carriersun in, ring out−3.000 : 1output reverseshold the sunring in, carrier out1.333 : 1same directionWillis: (ω_s − ω_c)/(ω_r − ω_c) = −72/24both derivations agree, and the build requires it
Fig. 1 Sun 24, ring 72, planets 24. Three different ratios from the same gears, depending on which shaft is held.

Willis’s equation

The trick is to look at the train from the carrier’s point of view. Sit on the carrier and the shaft centres are fixed again, so it becomes an ordinary train.

Relative to the carrier, the sun drives the ring through the planet with ratio −Nr/Ns-N_r/N_s — negative because the sun and ring turn opposite ways when the carrier is held. Written out:

ωs−ωcωr−ωc=−NrNs\frac{\omega_s - \omega_c}{\omega_r - \omega_c} = -\frac{N_r}{N_s}

That is the whole relation. Every epicyclic ratio is this equation with two of the three speeds substituted.

The tabular method

The second route superposes two motions rather than changing frame.

First lock the entire train and rotate it as a rigid body by some amount: every member turns equally. Then hold the carrier and rotate the gears against each other by whatever is needed to bring one member back to its required speed. Add the two motions.

It is a completely different derivation — no change of reference frame, no relative-velocity argument — and it must produce the same answer.

This site computes both for every epicyclic and asserts agreement. Planetary ratios are the most commonly mis-stated numbers in mechanism work, and the errors are almost always signs: a minus dropped in the relative-velocity step gives a plausible ratio that is simply wrong. Two derivations that share no algebra will not make the same sign error.

Why the relative-velocity trick works

Willis’s equation looks like an algebraic manipulation and is a change of reference frame, which is worth seeing clearly because the same move solves every epicyclic problem including the ones with compound planets.

The difficulty with an epicyclic is that the planet’s axis moves. Ordinary gear train arithmetic — ratios multiply, external meshes reverse — assumes fixed shaft centres, and it fails here for that reason alone.

Sit on the carrier and the centres stop moving. From there the train is ordinary: sun meshes with planet, planet meshes with ring, and the ratio is the product with one sign reversal, because sun-to-planet is external and planet-to-ring is internal. Internal meshes do not reverse, which is why an epicyclic’s sign behaviour surprises people who have only met external ones.

Having got the relation in the carrier’s frame, add the carrier’s own rotation back to every term. That is the whole derivation, and every ω − ω_c in Willis’s equation is a velocity measured from the carrier.

What the planet count does not change, and what it does

Three planets or one, the ratio is identical: the planets are idlers between sun and ring, and idlers cancel.

Three things do change, and none is kinematic.

Load sharing. Three planets divide the torque three ways, so each mesh carries a third of it. That is why epicyclic gearheads have the torque density they do.

Radial balance. With planets equally spaced, the radial forces on the sun cancel, so the sun shaft carries no net side load and needs no bearing to resist one. A single-planet epicyclic works kinematically and pushes its sun sideways hard.

An assembly condition. The planets can only be equally spaced if the tooth counts allow it: (N_sun + N_ring) must be divisible by the number of planets. Otherwise the planets cannot all mesh at equal spacing and the thing will not go together. That is a real constraint on gearset design and it is arithmetic rather than geometry.

The extra planets are redundant in exactly the sense the mobility audit measures — they add constraint equations that the existing ones already imply — and they are deliberate.

Three ratios, one gearset

With sun 24 and ring 72:

Ring held, sun in, carrier out. 1+Nr/Ns=1+3=41 + N_r/N_s = 1 + 3 = 4. A four-to-one reduction, same direction, and the standard arrangement in an automatic transmission or a planetary gearhead.

Carrier held, sun in, ring out. −Nr/Ns=−3-N_r/N_s = -3. A three-to-one reduction with the output reversed — which is an ordinary train, since holding the carrier fixes the centres.

Sun held, ring in, carrier out. 1+Ns/Nr=1.3331 + N_s/N_r = 1.333. A gentle reduction, same direction.

Same gears in all three cases. The ratio is a property of the arrangement rather than of the gearset, which is what makes epicyclics useful for multi-speed transmissions: engaging different brakes and clutches selects different ratios from one set of teeth.

Why the planet count does not appear

Three planets or one, the ratio is identical. The planets are idlers — they transmit motion between sun and ring without affecting the ratio, exactly as an idler does in a simple train.

What multiple planets buy is load sharing and balance. Three planets divide the torque three ways and cancel the radial forces on the sun, so the sun shaft carries no net side load. That is a structural benefit and a kinematic redundancy — the extra planets constrain nothing new, and they only work if the tooth counts allow them to be spaced equally.

That last condition is real: (Ns+Nr)(N_s + N_r) must be divisible by the number of planets. Otherwise the planets cannot all mesh at equal spacing and the assembly will not go together.

Compound epicyclics

Very large reductions come from compounding: a planet with two different tooth counts on one shaft, meshing with two different rings.

Ratios of thousands to one are available from a single stage, which no ordinary train manages without absurdly large wheels, and the arithmetic is the same equation with the compound planet’s two counts entering separately. What also arrives is very low efficiency — the sliding velocities in the mesh grow with the reduction — so the extreme ratios are useful where the load is small and the reduction matters more than the power.

Ratios, and the gear that does not change oneThe ratio of a train is the product of its stages, so the middle gear of a simple three-gear train cancels: 30/20 × 40/30 = 40/20, exactly as if it were not there. What it does change is the direction, since every external mesh reverses — which is the entire reason idlers exist. A compound train, where two gears share a shaft, does not cancel, and that is how large reductions are built without absurdly large wheels.reduction ratio20 → 402.00 : 1reversed20 → 30 → 40 (idler)2.00 : 1same direction20 → 40, 15 → 45 compound6.00 : 1same direction20 → 60, 20 → 60, 20 → 6027.00 : 1reversedproduct of the stages · sign flips at every external meshthe idler cancels exactly
Fig. 2 Ordinary trains for comparison, where the ratio is a product and an idler cancels. An epicyclic’s ratio does not factor that way, which is why it needs its own equation.
20 teeth driving 32Both flanks generated from the involute, not approximated. The orange line is the line of action — tangent to both base circles, and the only place contact happens. Its length between the two tip circles divided by the base pitch is the contact ratio, 1.612 here, which means that for 61% of the cycle two tooth pairs are carrying the load and for the rest just one. The velocity ratio is 0.6250, and it is constant because the common normal never moves.pitch pointline of actionmodule 1, 20° pressure angle, centre distance 26contact ratio 1.612
Fig. 3 The mesh underneath all of it. Every engagement in an epicyclic is an ordinary involute mesh with a constant ratio; what makes the train interesting is that one of the shaft centres is moving.
Undercutting, either side of 17.10 teethFive gears, drawn whole and scaled to a common pitch circle so the tooth counts can be compared by counting them. The dedendum sits a fixed 1.25 modules below the pitch circle and the base circle sits at r·cos α, so as the tooth count falls the base circle rises relative to the root. Below N = 2/sin²α = 17.097 it rises above it, and the part of the flank between them lies where no involute exists — the cutter removes it. The familiar rule says seventeen; the exact figure is 17.10, so seventeen undercuts slightly and eighteen is the smallest count that does not. The teeth are not to a common scale, because at a common module the small gears would be unreadable.10 teethundercut14 teethundercut17 teethundercut18 teethclean24 teethcleanred: the root circle has risen above the base circlethreshold N = 2/sin²α = 17.097
Fig. 4 A constraint on the sun. Epicyclic reductions want a small sun and a large ring, and the undercut threshold sets how small the sun can be before its teeth are cut away.
Mobility, counted and measuredGrübler's criterion counts links and joints and knows nothing about the dimensions; the rank of the constraint Jacobian measures the dimensions and knows nothing about the topology. They agree for four of these five. The parallelogram with a redundant third bar is the exception: the formula declares it a structure with zero degrees of freedom, and it moves. The formula is the one that is wrong, because it cannot see that the third bar's constraint equations are already implied by the other two.GrüblerJacobiantriangulated frame00agreefour-bar11agreeslider-crank11agreePeaucellier cell11agreeparallelogram + third bar01they disagree — the mechanism moves3(n−1) − 2j₁ − j₂ · free coordinates − rank(J)one row where the formula loses
Fig. 5 And a constraint on the planets. Three planets instead of one add no kinematic freedom — they are redundant in exactly the sense the audit measures, and they are there for load sharing.

Where the ratios get used

Three ratios from one gearset is not a curiosity; it is why epicyclics are in things that need more than one ratio without more than one gearset.

Automatic transmissions are stacks of epicyclic sets with brakes and clutches that hold different members. Engaging a brake on the ring gives one ratio; releasing it and locking two members together gives direct drive; holding the sun gives another. The gears never move — only which member is held changes.

Gearheads on motors use the ring-held case for its compactness: the reduction of 1 + N_ring/N_sun comes from a package no longer than the gears are wide, because the load path goes through the middle rather than sideways to a second shaft.

Differentials are the case run the other way, with power going in at the carrier and out at two sun gears — so the two outputs can turn at different speeds as long as their average is fixed. That is why a car’s driven wheels can round a corner, and why a car with an open differential and one wheel on ice goes nowhere: the equation fixes the average, and nothing fixes the split.

Bicycle hub gears are epicyclics with the selection done by pawls, which is the same idea with the brakes replaced by ratchets.

In every case the useful property is the one Willis’s equation states: three shafts and one relation, so fixing any one is a design choice rather than a fixed consequence of the teeth.

Ratios, and the gear that does not change oneThe ratio of a train is the product of its stages, so the middle gear of a simple three-gear train cancels: 30/20 × 40/30 = 40/20, exactly as if it were not there. What it does change is the direction, since every external mesh reverses — which is the entire reason idlers exist. A compound train, where two gears share a shaft, does not cancel, and that is how large reductions are built without absurdly large wheels.reduction ratio20 → 402.00 : 1reversed20 → 30 → 40 (idler)2.00 : 1same direction20 → 40, 15 → 45 compound6.00 : 1same direction20 → 60, 20 → 60, 20 → 6027.00 : 1reversedproduct of the stages · sign flips at every external meshthe idler cancels exactly
Fig. 6 The ordinary case for comparison, where ratios simply multiply because the shaft centres do not move. An epicyclic’s arithmetic looks harder only because one of its centres is orbiting, and the fix is to sit on it.

The differential, which is an epicyclic run backwards

Every epicyclic in this essay has one input and one output with a member held. Let nothing be held and the gearset becomes a differential: two outputs whose speeds must average to the input, in whatever proportion the geometry sets.

That is the same Willis relation with no member fixed. One equation, three unknown speeds, so two of them may be chosen freely and the third follows. In an automotive differential the two are the road wheels and the constraint is the road: whichever wheel meets less resistance turns faster, and their average is what the driveshaft delivered.

The property that makes it useful is exactly the property that makes it fail. On a corner, the outer wheel must turn faster than the inner or something must slip, and the differential supplies the difference without anyone specifying it. On ice, the wheel with no grip takes all the speed and the one with grip takes none, and the vehicle stays where it is — because torque through a simple differential is split equally regardless of speed, and equal to nearly nothing is nothing.

Every fix is a way of breaking the differential’s own equation: a lock, a clutch pack, a viscous coupling, or brake intervention on the spinning wheel. All of them reintroduce the constraint that the open differential removed.

The same arrangement predates the car by a long way. The South-pointing chariot is a differential driven by its own road wheels, arranged so that a pointer keeps its absolute heading as the vehicle turns — an integrator built from gears, and one whose error accumulates exactly as the wheel-slip does. It is the earliest device that computes with a gear ratio rather than merely transmitting one.

Efficiency, and the recirculating-power trap

A gear train’s ratio is exact and its efficiency is not, and epicyclics have an efficiency behaviour that is genuinely surprising.

The loss in a mesh comes from sliding between the flanks, and the sliding depends on the relative speeds of the gears, not their absolute ones. In an epicyclic the meshes are between members that are all moving, so the relative speeds — and hence the losses — do not follow from the overall ratio in any simple way.

The consequence is that two epicyclic arrangements with the same overall ratio can have very different efficiencies. A high-ratio arrangement in which the members move slowly relative to each other is efficient. One in which a large ratio is produced by taking a small difference between two large numbers is not: internally, much more power flows around the loop than passes through it, and every watt of that circulating power pays the mesh loss.

That is recirculating power, and it is what makes some high-ratio compound epicyclics — the Wolfrom arrangement, for instance — capable of ratios in the hundreds at efficiencies that fall into the tens of percent. The ratio is exactly what the tooth counts say. The power that arrives is another matter, and at the extreme the train is not back-drivable at all: the losses exceed what the output can supply.

This site computes the ratios and does not compute the losses, which is the usual boundary here — the geometry is exact and the tribology is absent. The point worth keeping is that an exact ratio is not a claim about power, and epicyclics are the family where the two diverge most.

Why two routes, on a subject with a formula

Willis’s equation is a formula, it is correct, and it is a single line. Computing every ratio a second way by tabular superposition doubles the work for no new answers, which is a fair objection and worth answering directly.

The answer is that the two routes fail differently.

Willis’s equation fails on sign conventions. Which member is the ring, whether an internal mesh reverses direction, whether the ratio is written as sun-over-ring or the reverse — every one of these is a place to be off by a minus sign, and the resulting number is plausible, wrong, and often out by exactly a factor whose form does not announce itself.

Tabular superposition fails on bookkeeping. It works by locking the train, rotating everything, then holding the carrier and unwinding, and adding the two — and a row added to the wrong column gives an answer that is wrong in a completely different way.

An error that produces the same wrong number through both routes has to be an error in the tooth counts themselves, which is checkable by inspection. Anything else shows up as a disagreement.

That is the entire argument for redundancy, and this site has enough evidence for it to state it as a working rule. A gear test that parameterised both flanks the same way agreed with itself and measured nothing. A velocity solve that double-negated its right-hand side drew perfectly and reversed every velocity, and was caught only by a finite-difference comparison. A mobility routine that deleted a link instead of substituting a bar reported a plausible integer for the wrong mechanism.

None of those was found by inspection, and every one was found by a second route. The cost of the second route is a few lines; the cost of not having it is a published figure that is wrong in a way nothing in the figure reveals.

Why epicyclics are everywhere

An epicyclic gearset is more expensive than a pair of gears on parallel shafts: it has more parts, it needs a carrier, and the planets must be positioned accurately. It is used anyway in an enormous range of machinery, and the reasons are worth listing.

Coaxial input and output. The sun and ring share an axis with the carrier, so the gearset introduces no offset. That is decisive in a wheel hub, a winch, a turbofan reduction stage and any drive that must fit in line.

Load sharing. Three or more planets divide the torque, so each mesh carries a fraction of it. A parallel-shaft pair carries all of it at one mesh, which for the same torque means much larger gears.

High ratio in one stage. Ratios of 3:1 to 10:1 are routine in a single epicyclic and higher in compound arrangements, where a parallel-shaft stage above about 6:1 becomes awkward.

Two inputs or two outputs. Leaving one member free turns the gearset into a differential, which is a function a parallel-shaft pair simply does not have.

Ratio changing without disengaging. Holding a different member gives a different ratio from the same gears, which is the basis of every automatic transmission built before the belt-CVT: brakes and clutches select which member is held, and the ratio changes without any gear ever leaving mesh.

That last property is why the tabular method survives as a teaching tool. Working out what a compound gearset does in each of its selected states is exactly the bookkeeping the method is for, and doing it twice — once by Willis and once by superposition — is how the answer becomes trustworthy without a test rig.

Where this goes next

pitch pointline of actionmodule 1, 20° pressure angle, centre distance 26contact ratio 1.612 What happens in a mesh Contact between two gear teeth happens only along one straight line, and only over part of it. How much of that line lies between the two tip circles, divided by the base pitch, is the contact ratio — and if it drops below one the drive periodically stops being driven. the traced pointtangencybase circle r = 9.4020 teeth, module 1, 20° pressure anglenormal meets the base circle to 2.3e-5 Why a tooth is an involute A gear tooth is not a shape anybody chose for its looks. It is what one requirement forces — that the ratio of the two shafts' speeds stays exactly constant while the contact point slides along the flank. Impose that and the curve is essentially determined. -0.25000.2500.5000100200300crank angle (degrees)output ÷ input angular velocitya 20:40 gear pair, 0.500mean 0.000four-bar 4/1/3.5/3-0.40 to 0.33 through one turn The ratio that is not a number A four-bar's output-to-input speed ratio runs from −0.29 to 0.51 through one turn and changes sign on the way. Quoting a single figure for it quotes the average of that curve, which the mechanism never exhibits. A gear pair is the case where the same phrase is honest, and it is honest by construction.