Planetary Gear Train Calculator

All six configurations of a planetary set, solved from the Willis equation — with every ratio labelled input over output so it cannot be read upside down, the forced ring tooth count, the ideal torque split, the planet’s own speed, and both assembly conditions for two to eight planets.

Last updated 24 Sep 2026

Planetary gear train

Tooth counts and roles → ratio, speeds, torques, assembly

Which member is HELD, which drives, which is drivenThree separate choices, six combinations, and six different gearboxes out of one set of parts. The ratio changes by a factor of twelve across them and the sign changes twice. Every ratio this page prints is input speed DIVIDED BY output speed, so a number above 1 is a reduction and a number below 1 is an overdrive.

Sun teeth, SBelow about 17 teeth a 20° involute tooth undercuts unless it is profile-shifted, and the sun is the member that usually runs out of teeth first. The profile shift page has how far you can push it.

Planet teeth, PThis is the free choice, and it sets the ratio on its own: with the ring held, the reduction is 2 + 2P/S. The ring tooth count is then not a choice at all — R = S + 2P is forced by the fact that the three members are coaxial.

Number of planetsNot a free choice either. Two conditions have to pass and they are independent: (S + R) must divide by the planet count for the planets to space equally, and the planet tip circles have to clear each other. The series and the table below work through 2 to 8 for your gearset and say which ones are buildable.

ModuleOnly for the diameters and the tooth loads — the ratio and both assembly conditions are pure tooth counts and do not care. Series I of ISO 54. The module page converts to diametral pitch.

Input speed (rpm)Whatever turns the member the configuration above calls the input. Enter it positive; the page reports the other two members’ speeds with their own signs, and one of them is usually negative.

Input torque (N·m)The torque split printed below is the IDEAL one: no losses, and the planets sharing the load equally. Neither is quite true, and the second one is not true at all in a set built to ordinary tolerances.

Not a circuit: the planetary set in plan. The circles are drawn to scale for the page's DEFAULT gearset — sun 24, planet 18, ring 60, at module 2 — and they do not change when you change the tooth counts; the whole column of figures beside them does, and so do the three role markers and the number of planets. Two things are worth doing with it. First, read the role markers: which member is HELD, which is the INPUT and which is the OUTPUT are three separate choices, and getting them the wrong way round is the commonest error on planetary pages. Second, raise the planet count. At two, three, four and six planets the circles sit cleanly round the carrier circle; at seven and eight their TIP circles — the outer of the two circles drawn on each planet — overlap, and that overlap is the neighbour condition failing — the planets would foul one another before the set could be assembled. At seven planets the PITCH circles still just clear, which is why it is the tip circle the condition is written about. Five planets look perfectly fine and are rejected anyway, because 84 does not divide by five and they cannot be equally spaced; that failure is arithmetic and the drawing cannot show it, which is exactly why the page prints both conditions. The bar underneath is the magnitude of the ratio on a scale of nought to twelve.

3.5000:1Example

Sun 24, planet 18 and therefore ring 60, with the ring held, the sun driven at 1000 rpm with 50 N·m, three planets, module 2

One kinematic equation, one coaxial constraint, two assembly conditions

S·n_s + R·n_r = (S + R)·n_c · R = S + 2P · (S + R)/n ∈ ℤ · (S + P)·sin(180°/n) > P + 2

Willis: the first equation, and the only kinematics on the page. Everything else is algebra on it. It says that the sun, the ring and the carrier have TWO degrees of freedom between them, not one: fix any one member and the other two are locked together, which is why a configuration is three choices and not one. Written the way most textbooks write it, (n_s − n_c)/(n_r − n_c) = −R/S, it is the same statement: the gearset seen from the carrier is an ordinary internal pair
R = S + 2P: the coaxial condition. The sun, the ring and the carrier all turn about one axis, so the planet centre distance from the sun, m(S+P)/2, must equal its distance from the ring’s centre, m(R−P)/2. Rearranged, that is R = S + 2P, and inverted it is P = (R − S)/2. The ring tooth count is therefore not a design choice: pick the sun and the planet and the ring is decided. Note what this does to the ratio — 1 + R/S becomes 2 + 2P/S, so the ring size drops out of the ratio entirely
(S + R)/n: the equal-spacing, or homogeneity, condition: it must be a whole number for n planets to sit at equal angles. The published source that names the four planetary tooth-matching conditions also says which one gets forgotten, and it is this one. It is pure arithmetic, it has nothing to do with clearance, and it is the condition that decides whether the parts go together at all. Unequal spacing is possible and is sometimes done deliberately for noise, at the cost of a radial load on the sun and ring bearings that equal spacing cancels
(S+P)sin(180°/n) > P + 2: the neighbour condition: adjacent planets must not touch. Derived rather than quoted, because the published algebra for it that was read for this batch is dimensionally impossible as printed. The planet centres lie on a circle of radius m(S+P)/2, so the chord between adjacent centres is m(S+P)sin(180°/n); each planet’s tip circle is m(P + 2) across, with the 2 being twice ISO 53’s one-module addendum. The module cancels, so this is a tooth-count condition like the other one
S : R : (S+R): the ideal torque split, derived on this page from power balance rather than taken from a table. The three external torques sum to zero and the carrier always takes the biggest of them. The torque ratio between any two members is the inverse of their speed ratio, exactly, because the set is assumed lossless
n_p − n_c: the planet speed RELATIVE TO THE CARRIER, which is what the planet bearing actually turns at and the number a bearing life calculation needs. It is not the planet’s absolute speed, and on the default gearset the two are 666.7 and 952.4 rpm — a 43 per cent error if you use the wrong one. It comes from the sun-planet mesh seen from the carrier: n_p − n_c = −(S/P)(n_s − n_c), and the ring-planet mesh gives the identical answer, which is the check worth having

Worked example

Sun 24, planet 18 and therefore ring 60, with the ring held, the sun driven at 1000 rpm with 50 N·m, three planets, module 2
THE RING IS NOT A CHOICE. R = S + 2P = 24 + 36 = 60 teeth. The three members are coaxial, so the planet's centre distance from the sun axis has to be the same measured outwards to the sun and inwards from the ring, and that forces it. Turn it round and it is just as useful: a 60-tooth ring and a 24-tooth sun need P = (60 − 24)/2 = 18-tooth planets, and nothing else will mesh.
THE RATIO, AND WHICH WAY UP IT GOES. With the ring held the reduction is 1 + R/S = 1 + 60/24 = 3.5 : 1, and that is input speed DIVIDED BY output speed. So the 1000 rpm sun drives the carrier at 1000/3.5 = 285.71 rpm. The carrier is slower. Printing 3.5 as a multiplier — “output = input × 3.5” — is the single commonest error on planetary pages, and it is invisible because 3500 rpm is as plausible a number as 285.7. Drive the same gearset from the carrier instead and you get the overdrive: 1000 rpm into the carrier spins the sun to 3500.
AND THE SAME NUMBER WITHOUT THE RING. Because R = S + 2P, 1 + R/S is identically 2 + 2P/S = 2 + 36/24 = 3.5. The ring tooth count has cancelled out. That is worth carrying away: the ring-held reduction of a single stage is decided entirely by the planet-to-sun tooth ratio, it can never be 2 : 1 or less, and the ceiling is set by how few teeth you can put on the sun.
THE PLANETS GO ROUND AND ROUND THEMSELVES. The carrier turns at 285.71 rpm, and the planet turns at -666.67 rpm absolute — negative, because it rolls backwards inside the fixed ring. Its speed relative to the carrier, which is what its own bearing sees, is -952.38 rpm. Those two numbers are 43 per cent apart and a planet bearing calculation needs the second one. Both meshes agree on it: −(S/P)(n_s − n_c) from the sun side and +(R/P)(n_r − n_c) from the ring side give the same figure, and a wheel rolling inside a fixed circle gives it a third time.
THE TORQUE, AND THE REACTION NOBODY DESIGNS FOR. The split is S : R : (S+R) = 24 : 60 : 84. With 50 N·m into the sun, the ring takes 125.0 N·m and the carrier delivers 175.0 N·m — which is 50 × 3.5, as it has to be, because speed times torque is conserved in a lossless set. The three sum to zero. The number to notice is the ring's: 125.0 N·m is two and a half times the input torque, it is a static reaction rather than a transmitted load, and it is carried by whatever holds the annulus to the housing. That is the load that strips gearhead flange bolts.
CAN IT BE BUILT WITH THREE PLANETS? Two questions, both of which have to answer yes. Spacing: (S + R)/n = 84/3 = 28, a whole number, so three planets sit at exactly 120°. Clearance: the chord between adjacent planet centres is (S+P)sin 60° = 42 × 0.8660 = 36.373 modules against a planet tip diameter of P + 2 = 20, so they clear by 16.373 modules, or 32.75 mm at module 2. Both pass. Try five and the clearance is fine but 84/5 = 16.8 is not a whole number; try seven and 84/7 = 12 is perfect but the planets interfere by 1.777 modules. Six is the most this gearset will take, and neither condition on its own would have told you that.
WHAT THE EXTRA PLANETS BUY. The tangential load at the sun mesh is 2T_s/d_s = 2 × 50 000/48 = 2,083.3 N in total, which with three planets is 694.4 N each. That division is the whole reason planetary gearing exists: the same torque through a parallel-shaft pair would put all 2,083 N on one tooth. It is also the ideal, and the planets do not share equally — see the note below.

All six configurations of one gearset — sun 24, planet 18, ring 60

One set of parts, six gearboxes, and the ratio spans a factor of 12.25 from end to end. Read the fifth column as it is labelled: it is input speed DIVIDED BY output speed, so 3.5 means the output turns 3.5 times slower and 0.2857 means it turns 3.5 times faster. Every one of these six rows was solved from the Willis equation S·n_s + R·n_r = (S + R)·n_c with exact rational arithmetic, with the residual checked to be identically zero, and then compared with the closed form in column four — not the other way round. That matters because three of the six closed forms are reciprocals of the other three, and a page that writes them down from memory has a one-in-two chance of inverting each one. The two rows with the carrier held are not really planetary at all: with the carrier bolted down the planets are idlers on fixed centres, and the set is an ordinary internal gear train whose ratio is −R/S. That is exactly how a planetary automatic gets reverse. This is a first-pass calculation on an idealised part — uniform section, static load, no stress raisers beyond those stated, and room-temperature material properties. Real parts have fillets, keyways, surface finish and duty cycles that a closed-form answer cannot see.
Held	Input	Output	Closed form	Ratio, input ÷ output	Output at 1000 rpm in (rpm)	Direction	Effect	Where you meet it
Ring (annulus)	Sun	Carrier	1 + R/S	3.5000	285.71	same	reduction	the ordinary planetary reduction gearbox: a servo gearhead, a hub motor, the final drive of a hoist
Sun	Ring (annulus)	Carrier	1 + S/R	1.4000	714.29	same	reduction	the mild reduction of an automatic transmission’s upper gears, and the one case that can never reduce by more than 2
Carrier	Sun	Ring (annulus)	−R/S	-2.5000	-400.00	REVERSED	reduction	a reversing stage with the carrier bolted down — which is a plain internal gear pair with an idler, and is how reverse gear is got in a planetary automatic
Ring (annulus)	Carrier	Sun	1 / (1 + R/S)	0.2857	3,500.00	same	overdrive	the first case driven backwards: a speed increaser for a generator or a supercharger
Sun	Carrier	Ring (annulus)	1 / (1 + S/R)	0.7143	1,400.00	same	overdrive	an overdrive top gear, between 0.5 and 1
Carrier	Ring (annulus)	Sun	−S/R	-0.4000	-2,500.00	REVERSED	overdrive	a reversing speed increaser, the rarest of the six

How many planets the 24/18/60 set will take, and why

The two conditions in one table, and the reason both have to be printed. The spacing condition is arithmetic: the planets can only sit at equal angles if (S + R) divides by their number, because moving one planet round by one spacing rotates the sun by an exact number of teeth only when it does. The clearance condition is geometry: the planet centres sit on a circle of radius m(S+P)/2, so adjacent centres are m(S+P)sin(180°/n) apart, and each planet’s tip circle is m(P+2) across. Neither condition knows anything about the other, and on this gearset they disagree twice: five planets clear by 4.687 modules and will not space, seven planets space exactly and interfere by 1.777 modules. Six is the answer. Note also the last column, which is the real argument for more planets: it is the ideal tooth load per planet, and it is why a planetary set carries several times the torque of a parallel-shaft pair of the same size. The word ideal is load-bearing there — see the note under the calculator. The tip-diameter column assumes ISO 53’s basic rack, where the addendum is one module; a gearset cut with a different addendum coefficient needs that 2 changed. This is a first-pass calculation on an idealised part — uniform section, static load, no stress raisers beyond those stated, and room-temperature material properties. Real parts have fillets, keyways, surface finish and duty cycles that a closed-form answer cannot see.
Planets	(S+R) ÷ n	Spaces equally?	Centre chord (modules)	Tip diameter (modules)	Gap (modules)	Clearance	Verdict	Tooth load per planet (N)
2	42.000	yes	42.000	20.0	+22.000	clears	BUILDABLE	1,041.7
3	28.000	yes	36.373	20.0	+16.373	clears	BUILDABLE	694.4
4	21.000	yes	29.698	20.0	+9.698	clears	BUILDABLE	520.8
5	16.800	NO	24.687	20.0	+4.687	clears	no	416.7
6	14.000	yes	21.000	20.0	+1.000	clears	BUILDABLE	347.2
7	12.000	yes	18.223	20.0	-1.777	INTERFERES	no	297.6
8	10.500	NO	16.073	20.0	-3.927	INTERFERES	no	260.4

One ring size, nine gearsets: the reduction is 2 + 2P/S

The last two columns are the same number, and that is the point of the table. Because the three members are coaxial, R is forced to be S + 2P, and substituting that into 1 + R/S gives 2 + 2P/S identically. So the ring-held reduction of a single planetary stage depends on NOTHING but the planet-to-sun tooth ratio — the ring tooth count cancels out of it completely, even though the ring is the biggest and most expensive part in the set. Two consequences follow and both are useful. A single ring-held stage can never reduce by 2 : 1 or less, however the teeth are divided up, because 2P/S is strictly positive. And the practical ceiling is set by the sun running out of teeth. At 17 teeth a 20° involute sun is already at the undercut limit, and with a 60-tooth ring the closest integer set that respects it is an 18-tooth sun with 21-tooth planets, which gives 4.333 : 1. The 6 : 1 top row of this table needs a 12-tooth sun, which undercuts unless it is profile-shifted. To go further without shrinking the sun you have to grow the RING: a 10 : 1 single stage with an 18-tooth sun wants 72-tooth planets and a 162-tooth ring, and somewhere around there is where two stages or a compound planetary with stepped planets becomes the better answer. The nine rows here all use a 60-tooth ring and are therefore all the same size on the outside, which is what makes them comparable.
Sun S	Planet P	Ring R	P ÷ S	Reduction 1 + R/S (input ÷ output)	Output ÷ input	2 + 2P/S	Note
12	24	60	2.0000	6.0000	0.16667	6.0000	undercuts below 17 teeth without profile shift
16	22	60	1.3750	4.7500	0.21053	4.7500	undercuts below 17 teeth without profile shift
20	20	60	1.0000	4.0000	0.25000	4.0000
24	18	60	0.7500	3.5000	0.28571	3.5000
30	15	60	0.5000	3.0000	0.33333	3.0000
36	12	60	0.3333	2.6667	0.37500	2.6667
40	10	60	0.2500	2.5000	0.40000	2.5000
44	8	60	0.1818	2.3636	0.42308	2.3636
48	6	60	0.1250	2.2500	0.44444	2.2500

The ideal torque split, 24/18/60, with 50 N·m into the sun

Three torques, one constraint, and no choice about any of it. The three external torques must sum to zero, because nothing is accelerating; and power must be conserved at every admissible set of speeds. Put the Willis equation into the power balance and demand that it vanish for all independent sun and ring speeds, and the split falls out as T_sun : T_ring : −T_carrier = S : R : (S + R). It was derived that way for this page rather than quoted, because the one published statement of it that was read for this batch does not satisfy power conservation against its own kinematic constraint. The practical readings: the carrier always carries the largest torque of the three, which is why the carrier is the output of every reduction configuration; the ratio of the carrier’s torque to the sun’s is exactly 1 + R/S, the same number as the speed reduction and its inverse, which is the whole content of “speed times torque is conserved”; and the held member takes a real reaction torque that the housing and its fixings have to be designed for — 125 N·m here, two and a half times the input. That reaction is the load that pulls planetary gearhead flanges off. This is a first-pass calculation on an idealised part — uniform section, static load, no stress raisers beyond those stated, and room-temperature material properties. Real parts have fillets, keyways, surface finish and duty cycles that a closed-form answer cannot see. This page sizes a part; it does not certify one. Where the answer carries a consequence — a load path, a lifting duty, a pressure boundary, a fastener holding something that can fall — confirm it against the design code that governs the application, and against the manufacturer’s own rating, before relying on it.
Member	Share	Teeth	× the sun’s torque	Torque (N·m)
Sun	S	24	1.0000	50.000
Ring (annulus)	R	60	2.5000	125.000
Carrier	S + R	84	3.5000	175.000

Six gearboxes from one set of parts, a ring tooth count you do not get to choose, and two assembly conditions that disagree with each other

The ratio on this page is always input speed divided by output speed, and that is not a pedantic point. Take the commonest configuration: ring held, sun driven, carrier out. The reduction is 1 + R/S, which for a 24-tooth sun and a 60-tooth ring is 3.5. Drive the sun at 1000 rpm and the carrier turns at 285.7 rpm. It is slower. Printing that 3.5 as a multiplier — “the output runs at 3500 rpm” — is the commonest error on planetary pages anywhere, and it survives review because 3500 is exactly as plausible a number as 285.7 and because the formula is right either way. The same gearset driven from the carrier instead is a 3.5 times speed increase, and that is a different one of the six configurations, with a different closed form. Every ratio here carries its direction in the label.

A planetary set has two degrees of freedom, so a configuration is three choices and not one. The Willis equation, S·n_s + R·n_r = (S + R)·n_c, relates three speeds with one equation, which leaves two free. Hold one member and the other two are locked together; which one you hold, which one you drive and which one you take the output from are independent decisions, and the six combinations give six genuinely different gearboxes from one set of parts. The table above runs all six for the same 24/18/60 set: the ratios are 3.5, 1.4, −2.5, 0.2857, 0.7143 and −0.4, a span of more than twelve to one with two sign changes in it. An automatic transmission is built on exactly this: it gets several forward ratios and a reverse out of one or two planetary sets by changing which member a clutch or brake holds.

The ring tooth count is not a design choice, and the ratio does not depend on it. All three members turn about one axis, so the planet’s centre distance has to come out the same measured outwards from the sun and inwards from the ring, which forces R = S + 2P. Substitute that into 1 + R/S and the ring cancels: the ring-held reduction is identically 2 + 2P/S. It depends only on the planet-to-sun tooth ratio. Two things follow that are worth remembering. A single ring-held stage can never reduce by 2 : 1 or less, whatever you do with the teeth, because 2P/S is always positive. And the top of the range is set not by the ring but by how few teeth the sun can have. At 17 teeth a standard 20° tooth is at the undercut limit, so with a 60-tooth ring the span that respects it runs from 2.25 : 1 up to only 4.333 : 1, an 18-tooth sun with 21-tooth planets. Pushing past that means either a profile-shifted sun of 12 to 16 teeth, which gets you to 6 : 1 with the same ring, or a bigger ring: 10 : 1 with an 18-tooth sun wants 72-tooth planets and a 162-tooth ring. Beyond about ten to one, two stages or stepped planets win.

Two assembly conditions, independent of each other, and both have to pass. This is where a planetary design actually fails, and it is the part most pages leave out. The first is arithmetic: the planets can only be equally spaced if (S + R) divides exactly by their number, because stepping one planet round by one spacing has to advance the sun by a whole number of teeth. The second is geometry: the planet centres sit on a circle of radius m(S+P)/2, so adjacent centres are m(S+P)sin(180°/n) apart, and that has to exceed the planet tip diameter m(P+2) or the planets foul one another. Neither condition knows about the other. On the 24/18/60 set they disagree twice: five planets clear each other easily and will not space, because 84/5 is 16.8; seven planets space perfectly at twelve teeth each and interfere by nearly two modules. Six is the most the set will take, and you cannot get that from either condition alone. The chart and table above work both of them through 2 to 8 planets for whatever tooth counts you enter.

The torque split is forced, and the reaction on the held member is bigger than the input. Three external torques, summing to zero because nothing accelerates, with power conserved at every admissible speed: that is enough to fix the split completely as T_sun : T_ring : −T_carrier = S : R : (S + R). It was derived here rather than copied, because the one published statement of it read for this page does not satisfy power conservation against its own kinematic equation. With 50 N·m into a 24-tooth sun of the default set, the carrier delivers 175 N·m — 50 times the 3.5 reduction, as conservation demands — and the ring has to be held against 125 N·m. That last number is the one that gets forgotten. It is two and a half times the input torque, it is a static reaction rather than a transmitted load, and it is carried entirely by whatever fixes the annulus to the housing.

What the page does not know: the planets do not share equally. The load-per-planet figure here is the total tangential load divided by the number of planets, which is the ideal. Real sets do not achieve it. Pitch errors, carrier-bore position errors, ring rim flexibility and the fact that nothing is perfectly concentric all mean one planet takes more than its share, and the correction is a mesh load factor applied in a rating standard — ISO 6336 or ANSI/AGMA 2001 — rather than anything a kinematic calculation can see. Three planets are a special case worth knowing: three points define a plane, so a three-planet set with a floating sun or a floating carrier can self-centre and shares load far better than four or more can without deliberate compliance. That is why so many planetary gearheads have exactly three planets even when the arithmetic would allow six.

Frequently asked questions

What is the gear ratio of a planetary gear set?

It depends on which member you hold, and there are six answers rather than one. With the ring held and the sun driven, the reduction is 1 + R/S — input speed divided by output speed, so a sun of 24 teeth in a ring of 60 gives 3.5 : 1 and a 1000 rpm sun turns the carrier at 285.7 rpm. With the sun held and the ring driven it is 1 + S/R, which is always between 1 and 2. With the carrier held it is −R/S, reversing. The other three configurations are those three run backwards and are overdrives: 1/(1 + R/S), 1/(1 + S/R) and −S/R. The table on this page runs all six for one gearset and labels every ratio with its direction.

How many teeth should the ring gear have?

You do not get to choose. R = S + 2P is forced by the fact that the sun, the ring and the carrier are all on one axis — the planet’s centre distance has to be the same whether you measure it out from the sun or in from the ring. So a 24-tooth sun with 18-tooth planets needs a 60-tooth ring and nothing else will mesh. Turned round, a 60-tooth ring and a 24-tooth sun need (60 − 24)/2 = 18-tooth planets. If the sum comes out odd, there is no integer planet count that fits and one of the other two numbers has to move.

How many planets can I use?

Two conditions have to pass and they are independent. The planets space equally only if (S + R) divides exactly by their number; and they clear each other only if (S + P)·sin(180°/n) is greater than P + 2, which is the planet tip diameter in modules. On a 24/18/60 set the two conditions reject different counts: five planets clear comfortably but 84/5 is not a whole number, and seven planets divide 84 exactly but physically overlap. The answer is six, and the page prints the most your own gearset will take. Most commercial gearheads use three, not because the arithmetic stops there but because three planets can self-centre on a floating member and share load much better than more can.

Is 1 + R/S the reduction or the speed increase?

The reduction. It is input speed divided by output speed, so the output is slower by that factor: ring held, sun at 1000 rpm, 24-tooth sun in a 60-tooth ring gives a carrier speed of 1000/3.5 = 285.7 rpm. Printing it as a multiplier is the commonest error on planetary pages and it is hard to spot, because the wrong answer is as plausible as the right one. The speed increase from the same gearset is a different configuration — carrier driven, sun out — and its ratio is 1/(1 + R/S) = 0.2857, which the page also prints.

How much torque does the fixed ring have to hold?

R/S times the sun’s torque, which is more than the input. For the default 24/18/60 set with 50 N·m into the sun, the ring reaction is 125 N·m and the carrier output is 175 N·m; the three sum to zero because nothing is accelerating. The ring figure is the one designers forget, because it is a static reaction rather than a transmitted load — but it is carried by whatever fixes the annulus to the housing, and at two and a half times the input torque it is usually what sizes those fixings. The coupling page covers the other end of the same problem.

How fast do the planets turn?

Two different answers, and a bearing calculation needs the second one. On the default set with the ring held and the sun at 1000 rpm, the planet’s absolute speed is −666.7 rpm — negative, because it rolls backwards inside the fixed ring — while its speed relative to the carrier, which is what its own bearing or bush actually turns at, is −952.4 rpm. Those are 43 per cent apart. The page prints both and labels them. The relative figure comes from the sun-planet mesh seen from the carrier, and the ring-planet mesh gives the identical number, which is a useful check on any planetary arithmetic.

Why do planetary gearboxes carry more torque than ordinary ones?

Because the load is divided between the planets, and because the ring mesh is an internal one. The tangential load at the sun is 2T/d whatever you do, but with three planets each mesh carries a third of it, so the same torque needs a third of the face width or a smaller module. The internal ring mesh helps again: an internal tooth pair has a larger relative radius of curvature than an external one at the same size, so the contact stress is lower for the same load. The ideal split is what this page computes; the real one is worse, and the factor that corrects it belongs to a rating standard rather than to kinematics.

Can a single planetary stage give a 2 : 1 reduction?

With the ring held, no — not with any tooth counts at all. The reduction is 1 + R/S, and R = S + 2P turns that into 2 + 2P/S, which is strictly greater than 2 because the planets have to have teeth. The closest you can get is a very large sun with very small planets, and the planets then run out of teeth before the ratio reaches 2. Holding the sun instead gives 1 + S/R, which is always between 1 and 2, so that is where a ratio under 2 comes from; the page’s second configuration does it. This is the only hard limit in planetary gearing that comes out of the geometry rather than out of manufacturing.

Related calculators

References

tec-science, Transmission ratio of planetary gears (read 6 October 2026). The independent statement of the Willis equation used to check the solve on this page: n_rz_r = n_c(z_r + z_s) − z_sn_s, which rearranges to the form used here, z_sn_s + z_rn_r = (z_s + z_r)n_c. It is also the source for the three named closed forms and — the part that matters — for their DIRECTION: it gives i_r = 1 + z_r/z_s with the ring held, i_s = 1 + z_s/z_r with the sun held and i₀ = −z_r/z_s with the carrier held, all defined as input speed divided by output speed, and states that the first of those is above 2 and the second between 1 and 2.
Universidad de Almería, Teoría de Máquinas y Mecanismos course notes, lessons 8.1 (ordinary gear trains) and 8.2 (planetary gears) (read 6 October 2026). The source for the two rules the gear-train page is built on, stated in one place and in general form: for an ordinary train i = ω_out/ω_in = (−1)ⁿ·Z_in/Z_out with n the number of EXTERNAL meshes, and intermediate gears — idlers — “reverse the direction of rotation but do not affect the numerical value”; for a compound train i = (±1)·ΠZ_driving/ΠZ_driven. Its general Willis form, (ω_out − ω_carrier)/(ω_in − ω_carrier), is the one the planetary page solves.
MathWorks Simscape Driveline, Planetary Gear block reference (read 6 October 2026). Used for one thing only: its statement of the kinematic constraint in the compact form (1 + g_RS)ω_C = ω_S + g_RSω_R with g_RS = N_R/N_S, which is the Willis equation of this page divided through by the sun tooth count, and so an independent third statement of it. Its torque-balance line is NOT used: as the page prints it, the balance does not satisfy power conservation against its own kinematic constraint, so the torque split on this page is derived from power balance instead and proved rather than quoted.
Yang Jian-hua et al., Single Planetary Mechanism Teeth Matching Conditions, Proceedings of the 2015 International Conference on Advanced Manufacturing and Industrial Application (Atlantis Press, read 6 October 2026). The published source for treating the planetary tooth counts as a set of simultaneous CONDITIONS rather than a ratio calculation, and for naming the four of them: the concentric condition, the homogeneity (equal-spacing) condition (z_s + z_r) / n_p = integer, the neighbour condition, and the ratio-deviation condition. It also says which one gets forgotten: the equal-spacing condition is “most easily ignored” despite being the one that decides whether the parts go together at all. Its printed algebra for the concentric and neighbour conditions is garbled in the copy read here — the concentric condition appears as 2(z_s + z_p) = z_r, which is not dimensionally possible — so both are DERIVED from the geometry on this page rather than copied, and the derivation is shown.
ISO 53:1998, Cylindrical gears for general and heavy engineering — Standard basic rack tooth profile. Cited by number. One coefficient is taken from it and it is the one the planet-clearance condition on this page depends on: the basic rack addendum is one module, so a planet’s tip diameter is m(z_p + 2) and the neighbour condition reads (z_s + z_p)sin(180°/n) > z_p + 2. A gearset cut to a different addendum coefficient needs that 2 changed, and the page says so.
ISO 54:1996, Cylindrical gears for general engineering and for heavy engineering — Modules. Cited by number and not reproduced. The module series offered on these pages is series I, which was read from the standard’s own published preview during an earlier batch of this project and is reused rather than re-transcribed; series II’s module 6.5 is flagged there as one to avoid and is not offered.
ISO 6336, Calculation of load capacity of spur and helical gears, and ANSI/AGMA 2001-D04. Both cited by number only and neither reproduced nor used for a number on these pages. They are named because they are where a planetary load-sharing or tooth-rating question goes next: the load per planet computed here is the ideal equal split, and the mesh factor that corrects it for the fact that the planets never share equally is a rating-standard quantity, not a kinematic one.

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