Gear Ratios
Part of Simple Machines
Understanding the speed-torque tradeoff in gear systems and how to calculate the ratios you need.
Why This Matters
Every power-driven machine has an ideal operating speed. A grinding mill needs the millstone to spin at 100-150 rpm for efficient grinding. A water wheel turns at 8-15 rpm. Without the right gear ratio connecting them, either the stone grinds too slowly and produces poor flour, or it runs too fast and overheats or vibrates apart. The gear ratio is the translation layer between the speed your power source provides and the speed your machine requires.
Gear ratios also encode the fundamental tradeoff of all simple machines: you cannot get something for nothing. If gears multiply speed by 10:1, they divide torque by 10:1. If they multiply torque by 5:1, they divide speed by 5:1. Total power (torque × speed) is conserved (minus friction losses). Understanding this tradeoff allows you to diagnose why a machine isn’t working and design machines that match your power source to your load correctly.
The Fundamental Formula
The gear ratio relates the number of teeth on the two gears:
Gear ratio = Teeth on driven gear / Teeth on driver gear
The effects cascade from this single ratio:
Output speed = Input speed / Gear ratio
Output torque = Input torque × Gear ratio × Efficiency
Example:
- Water wheel: 12 rpm, 500 kg⋅m torque
- Gear pair: 60-tooth driver on wheel shaft, 12-tooth driven on millstone shaft
- Gear ratio = 12/60 = 0.2 (or expressed as 5:1 speed increase)
- Millstone speed = 12 rpm / 0.2 = 60 rpm
- Millstone torque = 500 kg⋅m × 0.2 × 0.85 (efficiency) = 85 kg⋅m
Speed vs. Torque: The Tradeoff
The most important thing to understand about gear ratios is the tradeoff they represent:
To get more speed: Use a large driver and small driven gear. You gain speed but lose torque in the same proportion.
To get more torque (force): Use a small driver and large driven gear. You gain torque but lose speed in the same proportion.
You cannot get both. The power (torque × speed) delivered to the driven shaft always equals the power at the driver shaft, minus friction losses. Gears are not amplifiers of power — they are translators between speed and torque.
Practical implication: When a machine is stalling (not enough torque), you need either more power at the input, or you need a higher gear ratio (more torque, less speed). When a machine is running too slowly, you need a lower gear ratio (more speed, less torque) — but check first that you have enough torque to handle the higher speed.
Calculating Required Gear Ratio
Step 1: Determine your power source speed and torque.
Typical power sources:
| Source | Approximate Speed | Approximate Torque |
|---|---|---|
| Undershot water wheel (2 m diameter) | 8-15 rpm | 200-800 kg⋅m |
| Overshot water wheel (3 m diameter) | 5-10 rpm | 400-1,500 kg⋅m |
| Windmill (small, 4 sails) | 15-40 rpm | 100-400 kg⋅m |
| Horse capstan | 3-6 rpm | 200-400 kg⋅m |
| Ox capstan | 2-4 rpm | 300-600 kg⋅m |
| Two-person treadwheel | 10-20 rpm | 30-80 kg⋅m |
Step 2: Determine your machine’s required speed and torque.
Typical machines:
| Machine | Required Speed | Required Torque |
|---|---|---|
| Grain millstone (60 cm) | 80-150 rpm | 30-100 kg⋅m |
| Grain millstone (120 cm) | 50-100 rpm | 100-300 kg⋅m |
| Oil press | 5-20 rpm | Very high (500+ kg⋅m) |
| Lathe (woodworking) | 200-800 rpm | Low (5-20 kg⋅m) |
| Water pump (reciprocating) | 30-80 rpm | 20-100 kg⋅m |
| Hammer mill | 200-400 rpm | Moderate (30-80 kg⋅m) |
Step 3: Calculate required ratio.
Required gear ratio = Required output speed / Input speed
Example — water wheel driving a millstone:
- Water wheel: 10 rpm
- Millstone required: 100 rpm
- Required ratio = 100/10 = 10:1 speed increase
Verify the torque also works:
- Water wheel torque: 400 kg⋅m
- Torque available at millstone: 400 × (1/10) × 0.80 efficiency = 32 kg⋅m
- Millstone requires: 50 kg⋅m
The torque is insufficient! Either use a bigger water wheel, reduce friction in the millstones, or reduce the grinding load. In practice, you would reduce the grain feed rate to keep the torque requirement within available power.
Multi-Stage Gear Trains
When the required ratio is large, use multiple gear stages.
Why not use a single large ratio? A single 10:1 gear pair means the driver gear has 10 times as many teeth as the driven gear. If the driven pinion has 12 teeth (smallest practical), the driver needs 120 teeth — a very large wheel. Two stages of 3:1 each give 9:1 total, with moderate gear sizes.
Compound gear train calculation:
Total ratio = Stage 1 ratio × Stage 2 ratio × … × Stage n ratio
Example: Water wheel to millstone requiring 12:1 ratio:
- Stage 1: 48-tooth gear driving 12-tooth pinion = 4:1 speed increase
- Stage 2: 36-tooth gear (on same shaft as Stage 1 pinion) driving 9-tooth pinion = 4:1 speed increase
- Total: 4 × 4 = 16:1 speed increase (slightly more than needed — adjust tooth counts to fine-tune)
Shaft speeds at each stage:
- Water wheel: 10 rpm
- After Stage 1: 10 × 4 = 40 rpm
- After Stage 2: 40 × 4 = 160 rpm (at millstone)
Selecting Tooth Counts
You cannot always use fractional numbers of teeth — you need whole numbers. This constrains what ratios you can achieve.
Finding tooth counts for a target ratio:
Strategy: Express the ratio as a fraction, multiply to get reasonable tooth counts.
Example: Need 3.5:1 ratio:
- Express as fraction: 3.5/1 = 7/2
- Multiply both by 6: 42 teeth / 12 teeth → ratio = 42/12 = 3.5:1 ✓
Example: Need exactly 4.33:1 ratio:
- Cannot achieve exactly with integers easily
- Closest approximation: 39/9 = 4.333… ✓ (good enough for mill applications)
- Or: 52/12 = 4.333… ✓
Hunting tooth ratios: When the two meshing gears have no common factor in their tooth counts, every tooth on the driver meshes with every tooth on the driven gear over time. This distributes wear evenly.
- 48 and 12: GCD = 12, each driver tooth only contacts 4 of the 12 driven teeth
- 47 and 12: GCD = 1, each driver tooth contacts all 12 driven teeth over time
For durability, choose tooth counts with no common factor (or small GCD).
Variable Ratios: Speed Shifting
Some applications benefit from adjustable gear ratios — lower ratio for starting under load, higher ratio for running light.
Sliding pinion (simple speed shift): Mount a pinion that can slide along its shaft. Cut two sets of teeth on the pinion (or two separate pinions side by side on the same shaft). Slide the pinion to engage either a large or small driven gear.
This requires a momentary disconnect (stop the power, shift, restart) but is mechanically simple.
Step pulley (for belt drives): A stepped pulley has two or three different-diameter flanges on the same shaft. Shift the belt from one step to another to change the effective diameter and therefore the ratio.
Practical design: For most mill and workshop applications, a fixed single ratio is adequate. Design it correctly for the primary operating condition (full load, typical speed) and the machine will function well for its intended purpose.
Common Ratio Mistakes
Too high a ratio for available torque: The driven shaft stalls under load because not enough torque is available at the high speed. Solution: reduce the ratio, reduce the load, or increase the power input.
Too low a ratio for required speed: The driven shaft doesn’t reach operating speed even at maximum power source speed. Solution: increase the ratio or increase the power source speed.
Ratio based on unloaded speed: Power sources (water wheels, windmills) run faster with no load than under full load. Always calculate ratio based on the loaded speed of the power source, not the unloaded (free-running) speed.
Ignoring friction: Multi-stage gear trains lose 5-15% of power per stage to friction. A three-stage gear train with 85% efficiency per stage transmits: 0.85 × 0.85 × 0.85 = 61% of input power. Calculate this loss and verify you still have enough torque at the output.