Gear Train Ratio & Speed Simulator

why one of these gears does nothing to the ratio, and one of them does everything

Worked example — Gear Train Ratio & Speed Simulator

Overall Gear Ratio	3.00:1
Output Speed (RPM)	400.0 RPM
Output Torque	30.0 N·m
Gear 1 (N1)	12 teeth
Gear 2 (N2)	36 teeth
Gear 3 (N3)	24 teeth
Gear 4 (N4)	36 teeth
Input Speed (RPM)	1,200 RPM

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

The gears in the middle change the direction and nothing else 🖖

In a simple train the ratio is the last tooth count over the first, and every gear between them cancels. The idler preset runs 12 to 36 through gears of 36 and 24 and comes out at 3.00:1. Replace those two with 20 and 90, or 100 and 8, or 12 and 12, and the panel still reads 3.00:1 and 400 rpm. An idler earns its place by reversing the rotation or bridging a gap between shafts, which is a real job, but it is not a gear ratio.

Compounding is what actually multiplies 🖖

Fix a second gear to the same shaft and the stages multiply instead of cancelling. The compound preset uses 12, 36, 12 and 48 teeth and runs 3.00 then 4.00, landing on 12.000. Its gears matter as much as its staging: compound the simple train’s own 12, 36, 24 and 36 and the answer is 4.50, because the last two are 24 and 36 rather than 12 and 48. Compounding is how a gearbox reaches a large ratio in a small space. It is not the only way; a worm or a planetary set gets there differently.

Speed and torque are the same trade, and power is untouched 🖖

A ratio of 12 divides the speed by 12 and multiplies the torque by 12: 1,200 rpm and 10 N·m in, 100 rpm and 120 N·m out. The product is 12,000 either side, and it matches either side whatever tooth counts you set. Run it backwards with the overdrive preset and the same arithmetic gives 4,500 rpm at 11.1 N·m from 500 rpm at 100 N·m. Gears trade angular speed against torque; they never make any power.

Problems solved in full

The two gears in the middle that make no difference at all 7 steps
Load Simple Idler Train: 12, 36, 24 and 36 teeth, 1,200 rpm in at 10 N·m. The panel says a ratio of 3.00:1. Two of those four tooth counts are doing nothing to it. Prove which, and then switch the staging and see what those same four gears can actually give.
1. One rule covers every mesh: where two gears touch, the teeth pass the contact point at the same rate, so the smaller gear turns faster in exact proportion to the tooth counts.
  
  $\frac{ω _{1}}{ω _{2}} = \frac{N _{2}}{N _{1}} at every mesh$
2. A simple train is three meshes in a row, so multiply the three.
  
  $\frac{ω _{in}}{ω _{out}} = \frac{N _{2}}{N _{1}} \cdot \frac{N _{3}}{N _{2}} \cdot \frac{N _{4}}{N _{3}}$
3. Put the numbers in and each stage looks like it matters: a 3.00, then a 0.667, then a 1.50.
  
  $= \frac{36}{12} \cdot \frac{24}{36} \cdot \frac{36}{24} = 3 \times \frac{2}{3} \times \frac{3}{2} = 3.000$
4. Now cancel. Every middle count appears once on top and once underneath, so the whole product collapses to the last over the first, and the gears between them leave no trace in it.
  
  $\frac{N _{2}}{N _{1}} \cdot \frac{N _{3}}{N _{2}} \cdot \frac{N _{4}}{N _{3}} = \frac{N _{4}}{N _{1}} = \frac{36}{12} = 3$
5. The rest follows from the ratio alone. Speed divides, torque multiplies.
  
  $ω_{out} = \frac{1200}{3} = 400 rpm, τ_{out} = 10 \times 3 = 30 N m$
6. And the product is untouched. Gears move power between speed and force; they never make any, and the tool holds this at every setting.
  
  $ω τ (rpm \cdot N m) : 1200 \times 10 = 400 \times 30 = 12, 000$
7. To make the middle gears count, put two of them on one shaft so they are forced to turn together. Then the stages multiply instead of cancelling, and the same four tooth counts give 4.50 instead of 3.00.
  
  $compounded instead: \frac{36}{12} \cdot \frac{36}{24} = 3.00 \times 1.50 = 4.50$
Answer

The ratio is 36/12 = 3, printed as 3.00:1, and gears 2 and 3 contribute nothing to it. Swept over every idler pair the tool allows, the ratio never moves.

An idler is not useless; it is just not a gear ratio. It reverses the direction of the output, which the panel shows as the arrows alternating down the train, and it carries drive across a gap without a longer chain or a bigger pair. What it cannot do is change how fast the last shaft turns. Everything that needs a large ratio in a small space is compounded instead: switch the staging and 3.00 × 1.50 = 4.50 comes out of the identical tooth counts, 266.7 rpm and 45.0 N·m. That is the difference between a gearbox and a row of gears.
The shaft in the middle of a two-stage gearbox 6 steps
Load Compound Multiplier (12:1): 12, 36, 12 and 48 teeth, 1,200 rpm in at 10 N·m, 100 rpm and 120 N·m out. Between those two shafts sits a third one the panel never mentions. Work out what it is carrying, then decide which of the two stages should do the larger reduction.
1. The two stages multiply because gears 2 and 3 are keyed to one shaft and cannot turn independently of each other. Three times four is twelve, and the speed and torque cards follow from that number alone.
  
  $i = \frac{N _{2}}{N _{1}} \cdot \frac{N _{4}}{N _{3}} = \frac{36}{12} \cdot \frac{48}{12} = 3.00 \times 4.00 = 12.00 ⟹ 100.0 rpm, 120.0 N m$
2. That shaft is where the first stage ends and the second begins. It turns at the input speed divided by the first ratio and carries the input torque multiplied by it.
  
  $n_{mid} = \frac{n _{in}}{r _{1}} = \frac{1200}{3.00} = 400 rpm, T_{mid} = T_{in} r_{1} = 10 \times 3.00 = 30 N m$
3. Which leaves the product untouched a second time. The panel already shows it holding between input and output; it holds in the middle too, because a gear train has nowhere to store anything.
  
  $n T (rpm \cdot N m) : 1200 \times 10 = 400 \times 30 = 100 \times 120 = 12, 000$
4. Now swap the stages over — four first, three second. The overall ratio is still 12.00:1 and the output is still 100 rpm at 120 N·m, so nothing a reader can see on the page has changed. The middle shaft has: it is slower and carrying a third more torque.
  
  $stages swapped (12/48/12/36) : i = 4.00 \times 3.00 = 12.00, n_{mid} = 300 rpm, T_{mid} = 40 N m$
5. The general statement is short. The first ratio fixes the middle shaft entirely, and the two ratios have to multiply to whatever overall ratio you wanted, so the middle torque lands somewhere between the input’s and the output’s and never outside them.
  
  $T_{mid} = T_{in} r_{1}, n_{mid} = \frac{n _{in}}{r _{1}}, r_{1} r_{2} = i fixed ⟹ T_{in} \leq T_{mid} \leq T_{out}$
6. The choice is worth making because a shaft is sized by torque and not by speed. At one allowable shear stress the diameter goes as the cube root of the torque.
  
  $d \propto T^{1/3} (torsion, one allowable shear stress) : (\frac{40}{30})^{1/3} = 1.10$
Answer

400 rpm at 30 N·m, and 12,000 rpm·N·m — the same product as at both ends.

Put the small reduction first and the middle shaft runs fast and lightly loaded; put the large one first and it runs slow and heavy. This preset’s 3-then-4 arrangement gives 30 N·m, the 4-then-3 arrangement gives 40, and the cube-root rule turns that into about 10% on the shaft diameter and 21% on its weight. The fast shaft has costs of its own — its bearings run at 400 rpm rather than 300, and its gears mesh at the higher pitch-line speed — so the rule is a starting point rather than a verdict. What makes it easy to miss is that the overall ratio, the output speed, the output torque and the rotation direction are all identical either way. Every card on the page is blind to the decision.

$n_{mid} T_{mid} = n_{in} T_{in} always; min T_{mid} ⟺ min r_{1} (small stage first)$

Learning path

Simple machines: trading force for distance

Leads to screw-thread

By the 42Math team · Updated 2026-09-03

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Simple Idler Train - Twelve teeth to thirty-six through two idlers: ratio 3.00:1, and the two middle gears could be almost anything without changing it.
Speed Reducer (4:1) - A four-to-one reduction: 1,800 rpm in, 450 out, and torque from 15 to 60 N·m.
Compound Multiplier (12:1) - 12, 36, 12 and 48 teeth, compounded: 3.00 × 4.00 = 12.000, so 100 rpm out and 120 N·m.
Overdrive Train - Compounded the other way: 500 rpm becomes 4,500, and 100 N·m becomes 11.1. Power is 50,000 on both sides.