Gear Ratio, Speed and Torque Calculator

Two to four gears, in a line or in pairs — with the idler shown to cancel out of the ratio exactly, compound mesh ratios shown to multiply, the direction taken from the external mesh count, and an efficiency per mesh so the output torque is honest rather than ideal.

Last updated 24 Sep 2026

Gear ratio, speed and torque

Two to four gears → ratio, direction, speed, torque

How the gears are arranged — and this changes the answer more than the tooth counts doThe same four gears give 2.7 : 1 in a line and 9 : 1 in two pairs. A simple train’s overall ratio is the last tooth count over the first and the middle gears do nothing to it; a compound train multiplies its mesh ratios. Everything else on this page follows from which of the two you have.

How many gearsOnly for a simple train. A compound train as this page builds it is four gears in two pairs, so this field is locked to 4 in that mode.

Gear 1 teeth — the DRIVERThe input gear. In a simple train the overall ratio is gear-n over gear-1, so this tooth count and the last one are the only two that matter to the ratio.

Gear 2 teethIn a simple train of three or four gears this is an idler and the ratio does not depend on it at all — the chart below sweeps it from 15 to 90 teeth to show the line stay flat. In a compound train it is the first driven gear and it matters a great deal.

Gear 3 teethIn a compound train this gear shares a shaft with gear 2 and is the second mesh’s driver. In a four-gear simple train it is the second idler.

Gear 4 teeth — the final DRIVEN gearThe output gear in both arrangements, and in a simple train one of only two tooth counts the overall ratio depends on.

How many of the meshes are INTERNAL (a gear inside a ring)Direction is (−1) to the power of the number of EXTERNAL meshes. An internal mesh — a pinion running inside an annulus — does not reverse, so each one you declare here flips the predicted output direction back. Leave it at nought for an ordinary train of external gears. It does not change the ratio.

Input speed (rpm)The speed of gear 1. The page reports the output speed unsigned and the direction separately, and also gives the signed figure for anyone who wants one number.

Input torque (N·m)Torque at gear 1. The ideal output torque is this times the ratio; the honest one is that times the efficiency, which is the next field.

Efficiency per meshNot per gearbox: PER MESH, and they multiply. A measured spur mesh is better than 98 per cent and peaks between 99.3 and 99.9; a published spur GEARBOX is 94 to 98, and the difference is bearings, seals and oil churning rather than teeth. Use 0.98 to 0.99 for spur or helical, and the worm page for a worm, where the figure can be anywhere from 0.2 to 0.98 and depends on the lead angle.

ModuleOnly for the centre distances. The ratio, the direction and the efficiency are all independent of it. Series I of ISO 54.

Not a circuit: two gear trains drawn to scale at the same module, so the sizes are comparable. The upper one is a simple train on four shafts — 20, 35, 60 and 40 teeth — and its overall ratio is 40 ÷ 20 = 2.000. The 35 and the 60 are idlers: each contributes a mesh ratio that is cancelled by the next, so they change the direction and the span and nothing else. The lower one is a compound train on three shafts, where the 60 and the 18 are keyed together and therefore turn at the same speed. That one shared shaft breaks the cancellation, the two mesh ratios multiply, and 3 × 3 gives 9.000 out of gears no larger than the ones above. Count the shafts rather than the gears and you can tell which arrangement a gearbox has from the outside. The circles are drawn at the tooth counts printed on them, not at the numbers you enter; the three ratios underneath are live, and the middle one is a useful check — it is what your own tooth counts would give arranged in a line.

2.0000:1Example

A simple train of four gears, 20 → 35 → 60 → 40, at 1450 rpm and 50 N·m with 98 per cent per mesh, module 2

Two rules for two arrangements, and one for the direction

SIMPLE: i = z_last / z_first · COMPOUND: i = Π(z_driven / z_driver) · direction = (−1)^{external meshes} · n_out = n_in/i · T_out = T_in·i·η^m

i: the overall ratio, written throughout this page as INPUT SPEED DIVIDED BY OUTPUT SPEED. Above 1 is a reduction and below 1 is a step-up. Stating that is not pedantry: the same number read the other way round is the commonest error on gear-ratio pages, and both readings are plausible, so nothing in the answer catches it
z_last/z_first: the simple train. Each mesh contributes z_n+1/z_n and the product telescopes: every intermediate tooth count appears once in a numerator and once in a denominator and cancels. So a simple train’s ratio depends on exactly two of its gears, however many it has
idler: any gear in a simple train that is neither the first nor the last. It reverses the direction and it sets the centre distance, and it has no effect on the ratio at all. That is what it is FOR: an idler lets a train span a distance the ratio does not want to span, and lets the output turn the way you need it to
Π: the compound train. Two gears keyed to the same shaft turn at the same speed, which breaks the telescoping — the shared count appears twice in the same place instead of once each side — so the mesh ratios multiply. This is how a large reduction is got in a small box, and it is also how a large step-up is got by accident if a pair is the wrong way round
m: the number of meshes: one fewer than the gear count in a simple train, and half the gear count in a compound one. The efficiency enters as η to this power, so losses multiply rather than add
external: a mesh between two gears whose teeth face each other. Each one reverses the direction. An internal mesh — a pinion inside an annulus — does not, so it is excluded from the count. This is why a planetary with its carrier held runs backwards and why an idler puts an output back the right way round
η: efficiency PER MESH, and the least certain number on the page. A measured spur mesh is better than 98 per cent and peaks between 99.3 and 99.9; a published spur gearbox is 94 to 98 per cent, and the gap is bearings, seals and churning. It is an input rather than a constant because over four meshes the difference between 0.995 and 0.98 is a factor of four in the losses

Worked example

A simple train of four gears, 20 → 35 → 60 → 40, at 1450 rpm and 50 N·m with 98 per cent per mesh, module 2
THE RATIO IS TWO GEARS' BUSINESS. Every gear is on its own shaft and each meshes the next, so the three mesh ratios are 35/20, 60/35 and 40/60, and their product is (35 × 60 × 40)/(20 × 35 × 60) = 40/20 = 2.0000. The 35 and the 60 cancelled. Replace them with any tooth counts you like and the overall ratio is still 2.0000 — the chart on this page sweeps the first of them from 15 to 90 teeth to make that visible, and the check behind the page walks every integer in that range.
SO WHAT ARE THE MIDDLE GEARS FOR? Two things, both real. They set the CENTRE DISTANCE: at module 2 the three meshes sit 55, 95 and 100 mm apart, so the train spans 250 mm from the first shaft to the last, where a single 20:40 pair would span only 60. An idler train is how a drive reaches across a gap that the ratio alone would not reach. And they set the DIRECTION, which is the next step.
DIRECTION IS A MESH COUNT, NOT A GEAR COUNT. Three meshes, all external, so three reversals, so the output turns the opposite way to the input: (−1)³ = −1. Drop one idler and you have two meshes and the output runs the same way as the input. That is the one thing a single idler is most often fitted to do. An internal mesh does not reverse and does not count, which is why the field for it is separate.
SPEED. The output shaft turns at 1450/2 = 725 rpm. The idlers turn at whatever their own tooth counts give — 1450 × 20/35 = 828.6 rpm for the 35, and 828.6 × 35/60 = 483.3 rpm for the 60 — and those speeds matter for their bearings and for pitch line velocity, but not for the answer.
TORQUE, IDEALLY AND HONESTLY. Ideally the output torque is 50 × 2 = 100 N·m, because speed times torque is conserved. Three meshes at 98 per cent each give 0.98³ = 94.1192 per cent overall, so the real figure is 94.1192 N·m — nearly 6 N·m of the ideal 100 gone as heat in the teeth. Note that the losses MULTIPLY: 0.98³ is 0.9412 and not 0.94. Over three meshes the difference is small; over six it is not.
AND THE POWER BOOKS BALANCE. In: 50 × 1450/9549.2966 = 7.5922 kW. Out: 94.1192 × 725/9549.2966 = 7.1457 kW. The ratio of the two is exactly 0.98³, which is the check worth doing on any drive-train calculation — if the power ratio is not the efficiency you put in, something in the torque or the speed has been inverted. The 446 W of difference has to go somewhere, and in a closed box it goes into the oil.
WHAT COMPOUNDING THE SAME GEARS WOULD DO. Put gears 2 and 3 on one shaft instead and the ratio becomes (35/20) × (40/60) = 1.1667 — worse than the simple train, because the second pair is a step-up. Compounding multiplies whatever you give it. Arrange the same idea with 20:60 and 18:54 and you get 3 × 3 = 9.000 from four gears of the same size class, where a simple train of those four counts would give 54/20 = 2.7. That is the whole argument for a compound train, and it is a factor of 3.33 for nothing but a different shaft layout.

Six trains, and what the arrangement does to the ratio

Rows one to three are the finding this page exists for: three different trains, three different gear counts, and exactly the same ratio of 2.000. In a simple train — every gear on its own shaft, each meshing the next — the overall ratio is the last tooth count divided by the first, and every gear in between cancels. It is easy to see why once stated: each mesh contributes z_n+1/z_n, and in the product every intermediate count appears once on top and once underneath. The chart on this page sweeps the middle gear from 15 to 90 teeth to make the point visible, and the verification behind the page walks every integer value in that range and finds the ratio constant to one part in 10¹². Rows four and five are the other half: the same four tooth counts give 9.000 as two compound pairs and 2.7 in a line, because in a compound train the mesh ratios multiply instead of cancelling. Row six is there to stop the lesson being over-learned — compounding this page’s own default gears would give 1.167 rather than 2.000, because the second pair is a step-up. Compounding multiplies what you give it, including fractions. 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.
Train	Arrangement	Meshes	Overall ratio is	Ratio, input ÷ output	Output direction	What to notice
20 → 40	Simple, 2 gears	1	40 ÷ 20	2.0000	REVERSED	one external mesh, so the output turns the other way
20 → 35 → 40	Simple, 3 gears	2	40 ÷ 20	2.0000	same	the 35 is an idler: it cancels out of the ratio and puts the direction back
20 → 35 → 60 → 40	Simple, 4 gears	3	40 ÷ 20	2.0000	REVERSED	two idlers, both cancelling. Identical ratio to the two-gear pair
20 → 60 ‖ 18 → 54	Compound, 2 pairs	2	(60 ÷ 20) × (54 ÷ 18)	9.0000	same	the mesh ratios multiply: 3 × 3
20 → 60 → 18 → 54	Simple, 4 gears	3	54 ÷ 20	2.7000	REVERSED	the SAME four gears in a line give 2.7, not 9 — a factor of 3.33 thrown away by the arrangement
20 → 35 ‖ 60 → 40	Compound, 2 pairs	2	(35 ÷ 20) × (40 ÷ 60)	1.1667	same	this page’s own default four gears, compounded instead — and compounding is not always a gain

Which way the output turns

One rule, and the only subtlety in it is which meshes count. Each EXTERNAL mesh — two gears with their teeth facing each other — reverses the direction, so the output runs the same way as the input when the number of external meshes is even. An INTERNAL mesh, a pinion running inside an annulus, does not reverse, because the two gears turn the same way; so it drops out of the count entirely. That is why a planetary set with its carrier held runs backwards — one external mesh at the sun and one internal at the ring, so one reversal — and it is the whole reason an idler exists as a part. An idler costs a gear, a shaft and a bearing, adds a mesh’s worth of loss, and buys exactly two things: a direction change, and a centre distance you choose rather than one the ratio dictates. The planetary page works through the internal-mesh case.
Meshes	of which internal	External meshes	(−1) to that power	Output direction	Typical case
1	0	1	−1	reversed	a single external pair
1	1	0	+1	same as the input	a pinion running inside an annulus
2	0	2	+1	same as the input	one idler, or a compound pair of pairs
2	1	1	−1	reversed	a planetary set with its carrier held — one external mesh at the sun and one internal at the ring
2	2	0	+1	same as the input	two internal meshes in series, and nothing external
3	0	3	−1	reversed	two idlers, or a three-stage box
3	1	2	+1	same as the input	2 external and 1 internal mesh in series
3	2	1	−1	reversed	1 external and 2 internal meshes in series
4	0	4	+1	same as the input	three idlers, or a four-stage box
4	1	3	−1	reversed	3 external and 1 internal mesh in series
4	2	2	+1	same as the input	2 external and 2 internal meshes in series
5	0	5	−1	reversed	four idlers
5	1	4	+1	same as the input	4 external and 1 internal mesh in series
5	2	3	−1	reversed	3 external and 2 internal meshes in series

Efficiency per mesh, compounded through the train

Mesh losses multiply, they do not add, and over a short train the difference between the two is small enough that adding them is a fair approximation — four meshes at 98 per cent is 92.24 and not 92.00. The row to look at is 0.995 against 0.98. Both are defensible per-mesh figures for a spur mesh depending on where you read, and over four meshes they are 98.0 and 92.2 per cent — a factor of nearly four in the heat you have to get rid of, from a difference of one and a half per cent per mesh. That is why the efficiency figure on this page is an input and not a constant: it is the least certain number in the calculation and it compounds. The 0.90 row is not a spur mesh at all; it is there because it is roughly where a single-stage worm sits, and because it shows what happens when a lossy mesh is put in series with anything. A drive is rated for the duty it sees, not for the power it nominally transmits. The service factor used here is stated; the manufacturer’s own factor for your machine class and daily running hours takes precedence over any general table.
η per mesh	1 mesh (%)	2 (%)	3 (%)	4 (%)	6 (%)	Lost over 4 meshes (%)
0.995	99.500	99.002	98.507	98.015	97.037	1.99
0.990	99.000	98.010	97.030	96.060	94.148	3.94
0.980	98.000	96.040	94.119	92.237	88.584	7.76
0.970	97.000	94.090	91.267	88.529	83.297	11.47
0.950	95.000	90.250	85.737	81.451	73.509	18.55
0.900	90.000	81.000	72.900	65.610	53.144	34.39

Published efficiency, and the two things it can mean

Two columns, two sources, and for years they looked like a contradiction. The mesh column is RoyMech’s table: spur, helical, double helical and bevel all 98 to 99 per cent. The gearbox column is Cotta’s: spur and helical boxes 94 to 98, planetary 95 to 98, worm 49 to 90, with a rule of thumb that each loaded mesh costs about 2 per cent. The two do not disagree, and the thing that reconciles them is in the NASA report cited below, which measured spur gear system power loss directly: mesh efficiency is “generally greater than 98 percent except at very low torque levels” and peaks between 99.3 and 99.9 per cent, while support bearing losses alone account for more than 35 per cent of full-load system losses above 20 m/s of pitch line velocity. A mesh is 99 per cent. A gearbox is 94 to 98, and the missing few per cent are bearings, seals and oil churning, not teeth. So use the mesh column in the calculator above, where the losses are being counted per mesh, and expect the complete unit to come out below what it predicts. The worm row is the exception to everything: its efficiency depends on the lead angle and the sliding velocity rather than on the quality of the gears, which is why its range is 20 to 98 rather than one per cent wide. The worm gear page computes it from the geometry. A drive is rated for the duty it sees, not for the power it nominally transmits. The service factor used here is stated; the manufacturer’s own factor for your machine class and daily running hours takes precedence over any general table.
Gear type	Per MESH (%)	Whole GEARBOX (%)	η low	η high
Spur	98 to 99	94 to 98	0.980	0.990
Helical	98 to 99	94 to 98	0.980	0.990
Double helical (herringbone)	98 to 99	—	0.980	0.990
Straight bevel	98 to 99	—	0.980	0.990
Spiral bevel	98 to 99	95 to 99	0.980	0.990
Planetary (one stage)	—	95 to 98	—	—
Crossed helical	70 to 98	—	0.700	0.980
Worm	20 to 98	49 to 90	0.200	0.980

Why the gears in the middle do not matter, and why two on one shaft matter enormously

A simple gear train’s ratio depends on exactly two of its gears, however many it has. If every gear sits on its own shaft and each meshes the next, the overall ratio is the last tooth count divided by the first. The reason is one line of algebra: each mesh contributes z_n+1/z_n, so in the product every intermediate count appears once on top and once underneath and cancels. [20, 35, 40] gives 2.000. [20, 35, 60, 40] gives 2.000. [20, 40] gives 2.000. The gears in between are idlers, and an idler does not change the ratio at all — not approximately, exactly. The chart on this page sweeps a middle gear from 15 to 90 teeth so you can watch the line stay flat, and the verification behind the page walks every integer in that range and finds the ratio constant to one part in 10¹².

So why fit an idler at all? Two reasons, and both are practical rather than theoretical. The first is direction: each external mesh reverses it, so an odd number of meshes turns the output backwards and an even number does not. One idler converts a reversing pair into a non-reversing train without touching the ratio, and that is the commonest reason there is one. The second is centre distance. A 20:40 pair at module 2 puts its shafts 60 mm apart and there is nothing to be done about it; put two idlers in and the same 2 : 1 ratio spans 250 mm. An idler is how a drive reaches somewhere the ratio does not want to reach. Neither reason is free: each idler is a gear, a shaft and a bearing, and it costs one mesh’s worth of efficiency.

A compound train multiplies instead of cancelling, and that is the whole difference. Key two gears to the same shaft and they turn at the same speed, which breaks the telescoping — the shared tooth count now appears twice in the same position rather than once on each side — so the mesh ratios multiply. Take 20:60 and 18:54: as a compound pair-of-pairs that is 3 × 3 = 9.000, and as a simple train of the same four gears it is 54/20 = 2.7. Same parts, same size class, 3.33 times the reduction, and the only difference is which shaft two of the gears are on. That is why every multi-stage gearbox is compound. But the multiplication is blind: compound this page’s own default gears, 20:35 and 60:40, and you get 1.75 × 0.667 = 1.167, which is worse than the simple train, because the second pair is a step-up. Compounding multiplies whatever you hand it, fractions included.

Direction is a count of external meshes, not of gears. (−1) to the power of the number of external meshes, and the only subtlety is the word external. Two gears with their teeth facing each other turn opposite ways. A pinion running inside an annulus turns the same way as the annulus, so an internal mesh does not reverse and does not count. That one fact explains two things that otherwise look arbitrary: why a planetary set with its carrier held runs backwards — one external mesh at the sun, one internal at the ring, so one reversal — and why gearbox drawings so often have a gear in them that appears to do nothing.

Mesh efficiency and gearbox efficiency are different numbers, and most of the confusion about gear losses is the two being compared. The tabulated per-mesh figures are 98 to 99 per cent for spur, helical and bevel. The tabulated whole-gearbox figures for the same gear types are 94 to 98. For years that looked like a disagreement between sources. It is not: NASA measured spur gear system power loss directly and found mesh efficiency generally above 98 per cent with peaks between 99.3 and 99.9, while reporting that support bearing losses alone account for more than 35 per cent of full-load system losses above 20 m/s of pitch line velocity. The teeth are not where a gearbox loses its power. So the efficiency input on this page is per MESH and the losses multiply across them, and the complete unit will always come out below what this page predicts, by whatever the bearings, seals and oil take. One more thing about that input: it is the least certain number in the calculation and it compounds. Over four meshes, 0.995 and 0.98 give 98.0 and 92.2 per cent — nearly four times the heat from a difference of one and a half per cent per mesh.

Torque, and the check that catches an inversion. The ideal output torque is the input torque times the ratio, because speed times torque is conserved; the honest one is that times η to the power of the mesh count. Always check the power: input power is T·n/9549.2966 in kW with T in newton metres and n in rpm, and the ratio of output power to input power must come out as exactly the efficiency you put in. If it does not, a ratio has been inverted somewhere, and that is worth knowing because an inverted ratio produces a perfectly plausible wrong answer. The default train here takes 7.5922 kW in and delivers 7.1457 out, a ratio of 0.98³ to twelve decimal places, and the 446 W of difference goes into the oil.

Frequently asked questions

Does an idler gear change the gear ratio?

No — not approximately, exactly not. In a simple train where every gear is on its own shaft, the overall ratio is the last tooth count divided by the first, and every gear in between cancels out of the product algebraically. [20, 35, 40] and [20, 35, 60, 40] both give exactly 2.000, the same as a plain 20:40 pair. What an idler does change is the direction — one reversal each — and the centre distance, which is usually why it is there.

How do I calculate the ratio of a train of three or four gears?

It depends on the arrangement, and that matters more than the tooth counts. If each gear is on its own shaft and meshes the next, divide the last tooth count by the first and ignore everything in between. If pairs of gears share shafts — a compound train — work out each mesh’s driven-over-driver ratio and multiply them. The same four gears can give 2.7 one way and 9.0 the other, so the first thing to establish about a gearbox you are measuring is how many shafts it has, not how many gears.

Which way does the output gear turn?

Count the external meshes and raise −1 to that power. An odd number of external meshes reverses the output, an even number does not. So a single pair reverses, one idler puts it back, two idlers reverse it again. The word external is load-bearing: a pinion running inside an internal gear or annulus turns the same way as the annulus, so an internal mesh does not reverse and is left out of the count. That is why a planetary set with the carrier held gives a negative ratio.

Why is a compound gear train better for a large reduction?

Because the ratios multiply rather than cancelling, so the reduction grows without any single pair having to be extreme. Two stages of 3 : 1 give 9 : 1 with no gear more than three times another; one stage of 9 : 1 needs a wheel nine times the pinion’s diameter, which is a large, expensive and heavy part with a small pinion meshing it. The practical threshold is around 7 : 1 per external stage. Compounding is not automatically a gain, though: it multiplies whatever you give it, so a step-up pair in the chain multiplies the reduction down.

What efficiency should I use per mesh?

0.98 to 0.99 for a spur or helical mesh, and expect the complete gearbox to be worse than the multiplication suggests. The published mesh figures are 98 to 99 per cent for spur, helical and bevel; NASA’s direct measurements on spur gears found better than 98 per cent except at very low torque, peaking between 99.3 and 99.9. Published whole-gearbox figures for the same types are 94 to 98 per cent, and the gap is bearings, seals and oil churning rather than teeth — bearing losses alone are more than a third of full-load system losses at high pitch line velocity. A worm is a different animal entirely, from about 20 to 98 per cent depending on the lead angle.

How do I get the output torque?

Input torque times the ratio, times the efficiency to the power of the number of meshes. The first factor is just conservation — speed times torque is constant in a lossless train — and the second is the honest part. Then check the power: T·n/9549.2966 gives kilowatts from newton metres and rpm, and output power divided by input power must come out as exactly the efficiency you assumed. If it does not, a ratio has been inverted. That check is worth building into any drive-train spreadsheet, because an inverted ratio gives a wrong answer that looks entirely reasonable.

What is the difference between a simple and a compound gear train?

The number of shafts. In a simple train every gear has its own shaft and each meshes the next, so the ratio telescopes down to last-over-first and the intermediate gears are idlers. In a compound train at least one shaft carries two gears, which forces them to turn together, breaks the cancellation, and makes the mesh ratios multiply. You can tell them apart from the outside by counting shafts rather than gears: four gears on four shafts is simple, four gears on three shafts is compound.

Related calculators

References

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.
N. E. Anderson and S. H. Loewenthal, Effect of Geometry and Operating Conditions on Spur Gear System Power Loss, NASA TM-81426 / AVRADCOM TR-80-C-2 (read 6 October 2026). The one measured source in this batch for gear mesh efficiency, and the one that resolves the apparent disagreement between the two tabulated sources below. It reports that “gear efficiencies were generally greater than 98 percent except at very low torque levels”, with peak efficiencies “approximately 99.3 to 99.9 percent”, and — the reconciling fact — that support bearing losses account for more than 35 per cent of full-load system losses at pitch line velocities above 20 m/s. A mesh is 99 per cent; a gearbox is not, and the difference is not in the teeth.
RoyMech, Gear Efficiency (read 6 October 2026). The tabulated per-mesh figures used on this page: spur, helical, double helical and bevel all 98 to 99 per cent, worm 20 to 98, crossed helical 70 to 98. It states the rule this page implements — that the efficiencies of each mesh in a train multiply — and attributes its own equations mainly to the SDP/SI technical library.
Cotta Transmission, Gearbox Efficiency: Complete Guide to Performance Optimization (read 6 October 2026). The second tabulated source, and the pessimistic one: whole-gearbox efficiencies of 94 to 98 per cent for spur and helical, 95 to 98 for planetary, 95 to 99 for spiral bevel and 49 to 90 for worm, with the rule of thumb that “each loaded gear mesh reduces efficiency by approximately 2%”. Its worm examples are worth the read on their own: a single-stage 5:1 worm box at 90 per cent against a 300:1 double reduction at 49.
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.
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.

CalcEngines mechanical calculators are for education and design estimates. Results depend on the values you enter and the assumptions stated on the page, and they do not replace a qualified engineer's check. Standards, material grades and manufacturer data change — verify against the current standard and the supplier's own catalogue before you specify, machine or build anything.