Torque Multiplication Calculator
Solve the gear reduction relation τout = τin · N for any of its three unknowns — output torque, input torque, or gear ratio — with step-by-step math, a live gear schematic, and companion transmission metrics.
Gears trade speed for torque
A gearbox with ratio N multiplies torque by N and divides speed by N. Power is conserved (minus losses): P = τ·ω is the same on both sides. This is why a small, fast electric motor can drive a heavy, slow robot joint: the motor spins quickly with modest torque, and the gearbox converts that into slow rotation with large torque. The gear ratio is the fundamental design parameter linking motor characteristics to load requirements.
Higher ratio means more torque but less speed
Increasing the gear ratio N increases the output torque proportionally but reduces the output speed by the same factor. There is no free lunch: the product τ·ω is fixed by the input power. This is why a vehicle in low gear can climb a steep hill (high torque, low speed) but cannot go fast, while in high gear it cruises efficiently (low torque, high speed) but struggles on hills. The transmission selects the ratio that best matches the engine's torque curve to the driving condition.
Inertia is reflected by N²
When a load with inertia Iload is connected through a gearbox of ratio N, the inertia seen by the motor is Ireflected = Iload / N². This is a powerful result: a high gear ratio makes a heavy load appear light to the motor, so the motor can accelerate it quickly. It also means that adding a gearbox is often more effective than choosing a larger motor, because the inertia reduction scales with N² while the torque multiplication scales with N. This is why almost all robot joints use high-ratio gearboxes.
Real gearboxes lose 1–10% per stage
The ideal relation τout = τin·N assumes no losses. Real gearboxes have friction, oil churning, and windage losses, so the actual output torque is τout = η·τin·N, where η is the efficiency (typically 0.95–0.99 per stage for spur gears, 0.70–0.90 for worm gears). The output speed is also reduced slightly less than the ideal ratio predicts. For precision design, the efficiency must be included; for rough estimates, the ideal relation is adequate.
How to Use This Calculator
The torque multiplication relation links three quantities: τout, τin, and N. Pick which one you want the tool to solve, then fill in the other two.
1. Choose what to solve for
Pick τout, τin, or N from the toggle. The calculator rearranges τout = τin · N automatically for whichever you choose.
2. Use realistic presets
Presets cover typical values: output torque from 1 N·m (small joint) to 5,000 N·m (heavy machinery), input torque from 0.1 N·m (small motor) to 100 N·m (engine), and gear ratio from 1 (direct) to 500 (high reduction).
3. Match units carefully
Use N·m for both torques — the ratio N is then dimensionless. If you have torque in ft·lbf or kgf·m, convert first — or use the unit converter further down the page. The hint under each field shows the value in alternative units.
4. Read the result and verdict
The result card shows the solved value plus equivalent units. A colour-coded verdict classifies the torque from low (small motor) through moderate (gearmotor) to high (heavy machinery).
5. Check the schematic and charts
The schematic shows a small input gear meshing with a large output gear, with the input and output shafts, speed, and torque labelled. The chart plots output torque against gear ratio for three input torques.
6. Switch to Advanced for depth
Advanced mode adds a chart of τout vs. τin, a sensitivity check on a +1% input change, and companion metrics including output speed, power, efficiency, and reflected inertia.
What This Tool Offers
Built around the gear reduction relation τout = τin · N, the fundamental link between input torque, gear ratio, and output torque.
Three-way solving
Solve for output torque, input torque, or gear ratio. The calculator rearranges the formula for whichever variable you choose.
Clean gear schematic
A small input gear meshing with a large output gear, with shafts, teeth, speeds, and torques all clearly labelled.
Torque-class verdict
Each result is classified from low (small motor) through moderate (gearmotor) to high (heavy machinery), with a coloured badge.
Multi-unit hints
Each field shows the value converted to related units — ft·lbf and kgf·m for torque; ratio notation for gear ratio.
Dual comparison charts
τout vs. N for three input torques, plus an advanced τout vs. τin chart for three gear ratios.
Reference tables
Typical gear ratios by application, torque unit conversions, and a full variable reference table.
Reference Tables & Conversions
A torque unit converter, a τout lookup grid, and typical gear ratios.
Torque unit converter
Output Torque Lookup at τin = 1 N·m
Computed from τout = τin · N. Values in N·m.
| N | τ_in = 0.1 N·m | τ_in = 0.5 N·m | τ_in = 1 N·m | τ_in = 5 N·m | τ_in = 10 N·m |
|---|
Typical Gear Ratios by Application
Representative values for gearboxes and transmissions.
| Application | Typical N | Output speed (rpm) | Notes |
|---|---|---|---|
| Direct drive (no gearbox) | 1 | Same as motor | Used in some torque motors and direct-drive robots |
| Small servo with gearhead | 10 – 100 | 30 – 300 | Robot joints, CNC axes |
| Robot harmonic drive | 50 – 160 | 10 – 60 | Zero-backlash, compact |
| Industrial planetary gearbox | 3 – 1000 | 0.5 – 500 | High torque density |
| Worm gearbox | 5 – 100 | 10 – 300 | High ratio in one stage, self-locking |
| Car manual transmission (1st) | 3 – 4 | — | Low gear for acceleration |
| Car manual transmission (5th) | 0.7 – 1.0 | — | Overdrive for cruising |
| Car differential (final drive) | 3 – 4.5 | — | Wheel speed reduction |
| Truck transmission (1st) | 6 – 10 | — | Heavy hauling |
| Wind turbine gearbox | 60 – 120 | 1,200 – 1,800 (gen) | Rotor 10–20 rpm |
| Clock gear train | 1,000 – 100,000 | — | Multi-stage |
Variable Reference
| Symbol | Meaning | Typical range | Units |
|---|---|---|---|
| τout | Output torque | 0.01 – 10⁶ | N·m |
| τin | Input torque | 0.001 – 10⁵ | N·m |
| N | Gear ratio (τout/τin) | 0.1 – 10⁵ | — |
| ωin | Input angular velocity | — | rad/s |
| ωout | Output angular velocity | — | rad/s; ωout = ωin/N |
| η | Gearbox efficiency | 0.70 – 0.99 | — |
| P | Power through gearbox | — | W; P = τ·ω |
| Iref | Reflected inertia | — | kg·m²; Iload/N² |
What Does τout = τin · N Measure?
The torque multiplication relation gives the output torque of a gearbox in terms of the input torque and the gear ratio. It is the fundamental relation of mechanical power transmission and the basis for designing gear trains, selecting motors, and understanding how a transmission changes the torque–speed trade-off. The relation assumes an ideal, lossless gearbox; for real gearboxes, an efficiency factor is applied.
Torque Multiplication Formula
The defining relationship is:
Rearranged to solve for the other two unknowns:
N = τout / τin
The gear ratio also relates the input and output speeds, and the number of teeth:
N = Nteeth,out / Nteeth,in
Power is conserved (minus losses):
Pout = τout · ωout = η · Pin
The inertia of a load reflected to the input side is:
Gear Reduction Variables
| Symbol | Meaning | Notes |
|---|---|---|
| τout | Output torque | Torque delivered by the gearbox output shaft |
| τin | Input torque | Torque supplied to the gearbox input shaft |
| N | Gear ratio | N = τout/τin = ωin/ωout; dimensionless |
| ωin | Input angular velocity | Speed of the input shaft |
| ωout | Output angular velocity | ωout = ωin / N |
| η | Gearbox efficiency | Accounts for friction and other losses |
| P | Power | P = τ·ω; conserved across the gearbox (minus η) |
How Each Variable Affects Output Torque
Output torque is directly proportional to both input torque and gear ratio. Doubling the input torque doubles the output torque; doubling the gear ratio also doubles the output torque. This is why gearboxes are so useful: they allow a small, fast, low-torque motor to drive a large, slow, high-torque load. The same physical principle is behind bicycle gears, car transmissions, and industrial gearmotors.
The trade-off is speed: increasing the gear ratio reduces the output speed by the same factor, so the product τ·ω (the power) is unchanged (minus losses). This is why a vehicle in low gear accelerates quickly but has a low top speed, while in high gear it cruises efficiently but accelerates slowly. In a complete design, the gear ratio is chosen to match the motor's torque–speed curve to the load's torque–speed requirement, usually with the goal of maximizing acceleration, efficiency, or some weighted combination of the two.
Torque Multiplication Calculation Examples
τout = 2 × 50 = 100 N·m
τin = 100 / 50 = 2 N·m
N = 100 / 2 = 50
τout = 0.5 × 100 = 50 N·m
N = 500 / 5 = 100
ωout = 1,500 / 20 = 75 rpm
Frequently Asked Questions
Because gears work by the principle of the lever: a small force acting at a large radius produces a larger force at a smaller radius (or vice versa). In a gear pair, the input gear's teeth push on the output gear's teeth at the pitch radius. A small input gear driving a large output gear means the input force acts at a small radius and the output force acts at a large radius, so the output torque is larger by the ratio of the radii (which equals the ratio of the teeth, which equals N). The same physics applies to belt drives, chain drives, and even levers. Power is conserved because the output force moves through a proportionally smaller distance per unit time (lower speed).
"Gear ratio" is the general term for the ratio of output torque to input torque (equivalently, input speed to output speed). "Gear reduction" specifically refers to the case where the output speed is lower than the input speed (N > 1), which is the common case for motor-to-load transmissions. If N < 1, the gearbox is an overdrive — the output speed is higher and the output torque is lower, as in the top gear of a car. Most gearboxes used in robotics and industry are reduction gearboxes, because motors typically run fast with low torque while loads typically need slow, high torque. The terms are often used interchangeably when the context is clearly reduction.
Choosing a gear ratio is a trade-off between torque and speed. First, determine the required output torque (from the load and acceleration requirements). Second, determine the required output speed (from the motion profile). Third, find a ratio that lets the motor operate within its torque and speed limits at that operating point — the motor's continuous torque must exceed τout/(η·N), and its no-load speed must exceed N·ωout. Fourth, consider inertia matching: the reflected load inertia Iload/N² should be comparable to the motor's own inertia Imotor for best dynamic performance (a ratio between 1:1 and 10:1 is often targeted). Finally, check the gearbox's rated torque, backlash, and efficiency against the requirements. In practice, a few candidate ratios are evaluated against all these criteria, and the best compromise is selected.
Reflected inertia is the effective inertia that a load presents to the motor when viewed through a gearbox. It is Ireflected = Iload / N². This is important because the motor must accelerate both its own inertia and the reflected load inertia. A high gear ratio reduces the reflected load inertia dramatically: a load of 1 kg·m² seen through a 100:1 gearbox appears to the motor as 0.0001 kg·m² — a factor of 10,000 reduction. This makes the load easy to accelerate, allowing the motor to respond quickly and use its full torque capability. The reflected inertia also determines the system's natural frequency and stability: if the reflected load inertia is much larger than the motor inertia, the system may oscillate or be difficult to control. A common rule of thumb is to keep Iload/N² between 1× and 10× the motor inertia.
Real gearboxes lose energy to friction, oil churning, and windage, so the output torque is less than the ideal value: τout = η · τin · N, where η is the gearbox efficiency. For a single spur gear stage, η is typically 0.97–0.99. For a worm gear, it can be as low as 0.30–0.70, depending on the lead angle. For a multi-stage gearbox, the efficiencies multiply: ηtotal = η₁ · η₂ · ... · ηk. A three-stage planetary gearbox with 0.97 efficiency per stage has an overall efficiency of about 0.91, so the output torque is about 9% less than the ideal calculation. When sizing a motor, the efficiency must be included, or the motor will be undersized and the system will not meet its performance targets.
No. A gearbox cannot increase both torque and speed simultaneously, because that would violate the conservation of energy: the output power (τ·ω) cannot exceed the input power. If the output torque is multiplied by N, the output speed is divided by N, so τout·ωout = τin·ωin (minus losses). This is the fundamental trade-off of all transmissions. A gearbox can multiply torque (at the cost of speed), multiply speed (at the cost of torque), or change the direction of rotation, but it cannot increase the product τ·ω. The only way to increase both is to increase the input power, which means a larger motor, a higher supply pressure, or a different energy source.
For an ideal gearbox, the torque ratio equals the gear ratio: N = τout/τin. But for a real gearbox with efficiency η, the torque ratio is η·N, while the speed ratio is exactly N (in the ideal case) or N/η in some formulations. The distinction matters when efficiency is not 1: the gear ratio N is a purely geometric quantity (ratios of teeth or radii) and is the same for torque and speed, while the actual torque multiplication is reduced by efficiency losses. In most engineering contexts, "gear ratio" refers to the geometric ratio N, and the efficiency is applied separately in the torque calculation.
Variables
| Symbol | Quantity | Unit |
|---|---|---|
| tau_out | Output torque after the gear reduction | N.m |
| tau_in | Input torque from the motor | N.m |
| N | Gear reduction ratio (output speed / input speed) |
What it means
Gear reduction increases the output torque by the gear ratio N (assuming ideal efficiency), while decreasing output speed by the same factor. This is used in robotics to match motor characteristics to the load: a high‑speed, low‑torque motor can be geared to provide high torque at low speed. This formula is used in actuator selection and gearbox design. Understanding torque multiplication is essential for choosing gear ratios to achieve required joint torques and speeds.
Worked example
Torque Multiplication (Gear Reduction) – Two Detailed Examples
Real‑World| Parameter | Value |
|---|---|
| τ_in (N·m) | 5 |
| N | 3 |
| Parameter | Value |
|---|---|
| τ_in | 2 |
| N | 10 |
Common mistakes
- Gear ratio: τ_out = τ_in × N – where N is the gear ratio (output teeth / input teeth).
- Speed reduction: If N > 1, output torque increases, speed decreases (reduction gear).
- Ideal: Assumes 100% efficiency – real gears have losses (use efficiency factor).
- Sign: Both input and output torque directions are related; if gear reverses, sign changes.
- Inertia reflection: The output inertia is reflected to the input multiplied by N² – not covered by this formula.
Applications
Torque multiplication through gear reduction, τ_out = τ_in·N, describes how a gearbox amplifies torque by a factor N (the gear ratio), while reducing speed by the same factor. This is widely used in robotics and automotive drivetrains to increase the torque available from a motor. Engineers use it to match motor characteristics to load requirements, to reduce motor size and cost, and to improve motion control precision. By selecting appropriate gear ratios, they can achieve the desired balance between speed and torque. This formula is essential for designing efficient power transmission systems in robots and vehicles.
- Gearbox design for robotics and automation
- Motor selection and optimisation for torque requirements
- Transmission design in automotive and industrial machinery
- Power train efficiency and speed‑torque trade‑offs
- Simulation of geared mechanical systems
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