Module 1: Mathematics

Module 1: Mathematics — B1.1 Category Study Material

1. Overview of Module 1

Module 1 of the EASA Part-66 basic knowledge syllabus provides the mathematical foundation required for aircraft maintenance certifying staff. It covers arithmetic, algebra, geometry, and trigonometry as applied to typical maintenance tasks such as weight and balance calculations, unit conversions, tolerance assessments, and mechanical ratio computations. The module is divided into four main sections: Arithmetic (1.1), Algebra (1.2), Geometry (1.3–1.4), and Trigonometry (1.5), with knowledge levels ranging from overview (level 1) to detailed theory (level 3) depending on the licence category.

For B1.1 (aeroplane turbine) certifying staff, the mathematics syllabus is examined at level 2 for most topics, meaning candidates must have a general knowledge of the material with a practical understanding of its application in maintenance contexts. This study material synthesises the core concepts and their applications.

2. Arithmetic (Module 1.1)

2.1 The Number System and Notation

Aircraft maintenance documentation uses a variety of numerical notations. Certifying staff must be proficient in:

2.2 Unit Conversions

The aviation industry uses both SI (metric) and imperial units. Legacy documentation, particularly from North American manufacturers, frequently uses imperial units. Certifying staff must convert between systems accurately.

Common conversion factors:

Worked example — fuel volume conversion:

A fuel quantity of 2,500 US gallons must be recorded in litres:

2,500 × 3.78541 = 9,463.525 litres ≈ 9,463.5 litres

Worked example — torque conversion:

A maintenance manual specifies 25 N·m. Convert to lbf·ft:

25 × 0.73756 = 18.439 lbf·ft ≈ 18.4 lbf·ft

Conversely, converting 150 ft-lb to N·m:

150 × 1.356 = 203.4 N·m

Temperature conversions:

For example, 15°C = (9/5 × 15) + 32 = 27 + 32 = 59°F.

2.3 Ratios and Proportions

A ratio expresses the relationship between two quantities. Ratios are used extensively in gear calculations, fuel-oil mixtures, and weight and balance.

Gear ratio calculation:

For a gear train, the gear ratio is determined by the number of teeth on each gear:

This means the input gear rotates three times for every single rotation of the output gear — a speed reduction of 3:1.

Ratio simplification:

A ratio of 150 ml of oil to 1 litre of fuel must be expressed as a fraction in simplest form:

2.4 Percentages

Percentages express a proportion relative to 100. Converting between decimals, fractions, and percentages is a fundamental skill.

Worked example — wear assessment:

A turbine blade has an original thickness of 12.5 mm and now measures 11.8 mm.

Percentage reduction = (12.5 − 11.8) / 12.5 × 100 = 0.7 / 12.5 × 100 = 5.6%

2.5 Tolerances and Limits

Engineering drawings specify nominal dimensions with tolerance limits. Each individual measurement must fall within the specified range; averaging is not an acceptable acceptance criterion unless explicitly stated in the maintenance manual.

Worked example — shaft diameter assessment:

Nominal diameter: 50.00 mm

Tolerance: +0.05 / −0.02 mm

Upper limit = 50.00 + 0.05 = 50.05 mm

Lower limit = 50.00 − 0.02 = 49.98 mm

Measurements of 50.01, 50.02, and 50.03 mm all fall within the acceptable range (49.98 to 50.05 mm) and are therefore within tolerance.

2.6 Linear Interpolation

Linear interpolation estimates values between two known data points, assuming a constant rate of change. This is commonly applied to wear predictions.

Worked example — brake disc wear:

Original thickness: 24.50 mm

Thickness after 1,200 landings: 23.90 mm

Wear per landing = (24.50 − 23.90) / 1200 = 0.60 / 1200 = 0.0005 mm/landing

After a further 600 landings:

Additional wear = 0.0005 × 600 = 0.30 mm

New thickness = 23.90 − 0.30 = 23.60 mm

3. Algebra (Module 1.2)

3.1 Solving Linear Equations

Linear equations take the form ax + b = c. Solving involves isolating the unknown variable using inverse operations.

Worked example:

3x + 6 = 18

Subtract 6 from both sides: 3x = 12

Divide both sides by 3: x = 4

3.2 Transposition of Formulae

Certifying staff must be able to rearrange formulae to solve for any variable. This is essential for weight and balance, electrical, and mechanical calculations.

Worked example — moment balance:

1200 kg × 2.5 m = 800 kg × x

x = (1200 × 2.5) / 800 = 3000 / 800 = 3.75 m

Worked example — Ohm's law:

R = V/I

Given V = 28 V and I = 4 A:

R = 28 / 4 = 7 Ω

3.3 Weight, Mass, and Force

The relationship between mass and weight is a common source of confusion. Mass is a measure of the amount of matter (in kilograms), while weight is the force exerted by gravity on that mass (in newtons).

Weight (N) = mass (kg) × gravitational acceleration (m/s²)

Worked example:

A component has a mass of 2.5 kg.

Weight = 2.5 × 9.81 = 24.525 N

Standard gravity is taken as 9.81 m/s² for aviation calculations.

4. Geometry (Module 1.3–1.4)

4.1 Area Calculations

Rectangles:

Area = length × width

Worked example — access panel:

A panel measures 30 cm × 20 cm.

Convert to metres: 0.3 m × 0.2 m

Area = 0.3 × 0.2 = 0.06 m²

Triangles:

Area = ½ × base × height

Worked example — triangular access panel:

Base = 300 mm, height = 200 mm

Area = ½ × 300 × 200 = 30,000 mm²

Convert to m²: 30,000 × (0.001)² = 30,000 × 1×10⁻⁶ = 0.03 m²

Circles:

Area = πr²

Worked example — hydraulic pipe cross-section:

Internal radius = 8 mm, π = 3.1416

Area = 3.1416 × 8² = 3.1416 × 64 = 201.0624 mm² ≈ 201 mm²

4.2 Volume Calculations

Rectangular prisms (fuel tanks, cargo compartments):

Volume = length × width × height

Worked example — rectangular fuel tank:

Length = 1.2 m, width = 0.8 m, height = 0.5 m

Volume = 1.2 × 0.8 × 0.5 = 0.48 m³

Convert to litres: 0.48 × 1000 = 480 litres

Cylinders (fuel tanks, hydraulic accumulators, pipes):

Volume = πr²h

Worked example — cylindrical fuel tank:

Diameter = 0.6 m, length = 2.0 m

Radius = 0.3 m

Volume = 3.1416 × 0.3² × 2.0 = 3.1416 × 0.09 × 2 = 0.565488 m³

Convert to litres: 0.565488 × 1000 = 565.488 litres ≈ 566 litres

4.3 Triangles and Pythagoras' Theorem

Pythagoras' theorem states that in a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides: a² + b² = c².

Worked example — triangle classification:

A triangle has sides of 3 cm, 4 cm, and 5 cm.

3² + 4² = 9 + 16 = 25 = 5²

Since the relationship satisfies Pythagoras' theorem, the triangle is right-angled.

5. Trigonometry (Module 1.5)

Trigonometry deals with the relationships between angles and sides of triangles. It is used in aircraft for calculating climb gradients, component angles, and structural measurements.

5.1 Basic Trigonometric Ratios

For a right-angled triangle with angle θ:

5.2 Standard Trigonometric Values

Certifying staff should know the standard values:

Worked example:

sin 30° = 0.5

This value is used in calculations involving inclined components, such as determining the vertical component of a force or the height of a triangular structure.

6. Common Relationships Between Concepts

The mathematical concepts in Module 1 are interconnected and frequently combined in maintenance tasks:

7. Typical Exam Focus Points

Candidates should pay particular attention to:

8. Regulatory References

For B1.1, Module 1 is examined at level 2 for most topics, requiring a general knowledge of the theoretical concepts with a practical understanding of their application in aircraft maintenance.