Calculating Gearbox Efficiency

Introduction

Efficiency is one of the most critical parameters when selecting and sizing gearboxes. It directly determines how much of the motor power fed in is actually transmitted to the load — and how much is lost as unwanted heat generation. In practice this means: a gearbox with lower efficiency reduces the overall energy efficiency of the machine, increases operating costs, and can lead to thermal overloading.

This guide provides the necessary theory and practical calculation formulas for determining gearbox efficiency, highlights the most important influencing factors, and offers concrete optimization tips.

Fundamentals & Definition of Efficiency

Basic Formula

The efficiency of a gearbox is defined as the ratio of the output power to the input power:

Where:

• η = efficiency (dimensionless, value between 0 and 1)
• P_out = output power (power at the output shaft)
• P_in = input power (power at the input shaft)
• P_loss = power loss (converted to heat)

Percentage Representation

In practice, efficiency is often expressed as a percentage: η% = η × 100%

Example: A spur gearbox with η = 0.96 has an efficiency of 96%. This means that 96% of the input power is transmitted to the load, and 4% is dissipated as heat loss.

Conversion between Power and Torque

For practical use it is often useful to also express efficiency in terms of torques. Since P = M × ω (power = torque × angular velocity), at constant speed this simplifies to:

With speed change (gearbox with ratio i ≠ 1): n_out = n_in / i.

Efficiency by Gearbox Type

The following table provides an overview of typical efficiencies for various gearbox types:

Loss Types and Their Causes

The power loss (P_loss) is composed of several components:

1. Gear Mesh Losses (Tooth Friction)

This is the largest loss in gear transmissions. Causes include sliding friction between tooth flanks, surface irregularities, and deformation under load. Gear mesh losses are particularly dominant in worm gearboxes (the sliding component can make up 100% of relative motion, while in spur gearboxes typically 5–20% is sliding).

2. Bearing Losses (Rolling Bearing Friction)

Every shaft is supported in rolling bearings (ball, roller, or needle bearings). These generate friction, especially at higher speeds. In typical gearboxes this component accounts for 2–5% of total loss.

3. Seal Losses (Leakage Flow)

Oil can leak through seals or be displaced through gaps. This creates pressure build-up and thus friction in the seals. This component is normally small (1–2%), but can become significant with poor seal design.

4. Churning Losses (Oil Splash Friction)

At higher speeds, lubricating oil is carried along by rotating gears and "splashed" inside the housing. This creates friction in the oil mass. Churning losses are speed-dependent and at very high speeds can account for 10–15% of total losses. Particularly relevant in planetary gearboxes with low viscosity (ISO VG 32).

Magnitude of Losses

For a typical spur gearbox with η = 0.96 (4% total loss), the breakdown is approximately as follows:

• Gear mesh losses: ~2.5%
• Bearing losses: ~1.0%
• Churning losses: ~0.4%
• Seal losses: ~0.1%

Efficiency of Multi-Stage Gearboxes

For multi-stage gearboxes (e.g., two-stage planetary gearboxes, cascades of spur gearboxes), the overall efficiency is determined by multiplying the individual stage efficiencies:

Practical Example: Two-Stage Planetary Gearbox

Given two planetary stages with efficiencies η1 = 0.96 and η2 = 0.95, the overall efficiency is:

η_total = 0.96 × 0.95 = 0.912 = 91.2%

This shows: although each stage has a high efficiency of 95–96%, the overall efficiency becomes significantly lower. Multi-stage gearboxes should therefore only be used when the higher gear ratios justify it.

Comparison: One vs. Two Stages

Suppose you need an overall ratio of 25:1. Two options:

• Option 1: A single-stage worm gearbox with i=25:1, η≈0.40 (very poor!)
• Option 2: Two planetary stages with i1=5:1, i2=5:1, η_total = 0.96 × 0.96 = 0.922 (92.2%, much better!)

This example shows why planetary gearboxes are often the preferred choice despite higher cost.

Effect of Temperature and Lubrication

Temperature Dependence

The viscosity of lubricating oil decreases as temperature rises. This has two opposing effects:

• Positive: Lower viscosity reduces churning and bearing losses → efficiency increases
• Negative: Thinner lubricating film increases gear tooth friction → efficiency decreases

In practice there is an optimal temperature window (typically 60–80°C for mineral oils). Below 40°C churning losses are high; above 90°C the load-bearing capacity of the lubricating film decreases.

Selecting the Lubricating Oil

Oil viscosity according to ISO classification is decisive:

• ISO VG 32: Low viscosity, for high speeds and planetary gearboxes, lower wear through reduced churning
• ISO VG 100: Standard for bevel and spur gearboxes, good compromise
• ISO VG 220: High viscosity, for low speeds and heavy loads, better lubricating film

Practical Example: Complete Calculation

Task: A 7.5 kW electric motor drives a screw conveyor via a two-stage planetary gearbox (ratio 20:1). Calculate the output power and power loss.

Given Data:

• P_in = 7.5 kW (motor power)
• i_total = 20:1 (ratio)
• η1 = 0.96 (stage 1, e.g. i=4:1)
• η2 = 0.95 (stage 2, e.g. i=5:1)

Calculation:

Step 1: Overall efficiency

η_total = η1 × η2 = 0.96 × 0.95 = 0.912 (91.2%)

Step 2: Output power

P_out = P_in × η_total = 7.5 kW × 0.912 = 6.84 kW

Step 3: Power loss

P_loss = P_in - P_out = 7.5 kW - 6.84 kW = 0.66 kW = 660 W

Result:

The screw conveyor receives 6.84 kW of power. 660 W is converted to heat and must be dissipated by the gearbox housing. This requires a sufficiently large housing and, if necessary, cooling fins for heat dissipation.

TEA Recommendation

Efficiency is not merely a technical specification — it is a significant economic factor. A gearbox that loses 10% of power instead of 5% will cost you considerably more over its service life in energy costs and thermal management infrastructure. Our engineers can help you find the optimal balance between capital cost, efficiency, and heat balance for your application.

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