Module 1: Mathematics

Module 1: Mathematics — B3 Licence

1. Overview

Module 1 of the EASA Part-66 basic knowledge syllabus (Appendix I) provides the mathematical foundation required for aircraft maintenance certifying staff. For the B3 licence (rotorcraft), this module covers arithmetic, algebra, and geometry at knowledge levels 1–3, with an emphasis on practical applications in maintenance documentation, weight and balance, fuel calculations, and component sizing.

The module is divided into three main sections:

This textbook section synthesises the underlying knowledge tested in typical Part-66 Module 1 examination questions, focusing on the mathematical techniques themselves rather than memorised answers.

2. Key Concepts Explained in Detail

2.1 Arithmetic Fundamentals (Module 1.1)

2.1.1 Order of Operations

Aircraft maintenance calculations frequently require multiple operations in a single expression. The standard order of operations (BODMAS/BIDMAS) must be applied:

Example: In the resistance-temperature formula \( R = R_0(1 + \alpha T) \), the multiplication inside the brackets is performed before the addition, then the result is multiplied by \( R_0 \).

2.1.2 Fractions, Decimals, and Percentages

Converting between these forms is essential. For example, \( \frac{3}{4} = 0.75 = 75\% \).

Practical application: Fuel quantity indicators often display fractions of tank capacity. If a 180-litre tank reads \( \frac{3}{4} \) full, the volume is \( 180 \times \frac{3}{4} = 135 \) litres.

2.1.3 Ratios and Direct Proportion

A ratio expresses the relationship between two quantities. Direct proportion means that as one quantity increases, the other increases by the same factor.

Example: Fuel consumption is directly proportional to time. If an engine consumes 45 litres per hour, then in 2.5 hours it consumes:

\[ 45 \times 2.5 = 112.5 \text{ litres} \]

2.1.4 Unit Conversions

The International System of Units (SI) is the standard for aviation maintenance, but imperial units are still encountered in legacy documentation and US-manufactured components. Common conversions include:

Conversion method: To convert from one unit to another, multiply or divide by the appropriate conversion factor. When converting from a larger unit to a smaller unit, multiply; from a smaller unit to a larger unit, divide.

Worked examples:

Important note on torque conversions: When converting torque from N·m to lb-ft, always multiply by 0.73756. When converting from lb-ft to N·m, divide by 0.73756 (or multiply by 1.3558).

2.1.5 Significant Figures and Rounding

Maintenance documentation typically specifies the required precision. Common rounding rules:

When rounding, look at the digit immediately to the right of the desired precision. If it is 5 or greater, round up; otherwise, round down.

2.2 Algebra (Module 1.2)

2.2.1 Evaluating Formulae

A formula expresses a relationship between variables. To evaluate a formula, substitute known values and perform the arithmetic.

Example — Resistance-temperature relationship:

\[ R = R_0(1 + \alpha T) \]

Given \( R_0 = 100 \, \Omega \), \( \alpha = 0.004 \, /°C \), \( T = 25 \, °C \):

\[ R = 100 \times (1 + 0.004 \times 25) = 100 \times (1 + 0.1) = 100 \times 1.1 = 110 \, \Omega \]

2.2.2 Transposition of Formulae

Transposition (rearranging) allows solving for any variable in a formula. The rules are:

Example — Centre of gravity calculation:

The moment-arm relationship is:

\[ \text{Moment} = \text{Force} \times \text{Distance} \]

To find the arm (distance):

\[ \text{Distance} = \frac{\text{Moment}}{\text{Force}} \]

Given moment = 1250 N·m and force = 500 N:

\[ \text{Distance} = \frac{1250}{500} = 2.5 \, \text{m} \]

Example — Centre of gravity position from datum:

\[ \text{CG position} = \frac{\text{Total moment}}{\text{Total mass}} \]

Given total moment = 12,450 kg·m and total mass = 1,150 kg:

\[ \text{CG position} = \frac{12450}{1150} = 10.826 \, \text{m} \approx 10.83 \, \text{m} \]

2.2.3 Simple Equations

A linear equation has the form \( ax + b = c \). Solving involves isolating the unknown variable.

Example — Endurance calculation:

\[ \text{Endurance} = \frac{\text{Fuel volume}}{\text{Consumption rate}} = \frac{180 \, \text{L}}{42 \, \text{L/h}} = 4.2857 \, \text{h} \approx 4.3 \, \text{h} \]

Note: The fuel density (0.79 kg/L) is not required for this calculation. Density would only be needed to convert volume to mass.

2.2.4 Electrical Power (Mathematical Application)

While electrical theory is covered in Module 3, the mathematical relationship is used in Module 1:

\[ P = V \times I \]

Where:

Example: With \( V = 28 \, \text{V} \) and \( I = 4 \, \text{A} \):

\[ P = 28 \times 4 = 112 \, \text{W} \]

2.3 Geometry (Module 1.4)

2.3.1 Circles

The circle is fundamental to rotorcraft maintenance — rotor discs, inspection holes, and hydraulic pipes all involve circular geometry.

Circumference:

\[ C = \pi \times d = 2\pi r \]

Where:

Example: Inspection hole with diameter 40 mm:

\[ C = 3.14 \times 40 = 125.6 \, \text{mm} \]

Area:

\[ A = \pi r^2 = \frac{\pi d^2}{4} \]

Example — Rotor disc area: Helicopter rotor blade length 3.5 m (this is the radius of the swept disc):

\[ A = 3.14 \times (3.5)^2 = 3.14 \times 12.25 = 38.465 \, \text{m}^2 \]

2.3.2 Finding Diameter from Area

When the area of a circle is known, the diameter can be found by rearranging the area formula:

\[ A = \pi r^2 \]

\[ r = \sqrt{\frac{A}{\pi}} \]

\[ d = 2r = 2\sqrt{\frac{A}{\pi}} \]

Example — Hydraulic pipe: Area = 0.0007069 m², \( \pi = 3.1416 \):

\[ d = 2 \times \sqrt{\frac{0.0007069}{3.1416}} = 2 \times \sqrt{0.000225} = 2 \times 0.015 = 0.03 \, \text{m} = 30 \, \text{mm} \]

Note: The result is in metres; convert to millimetres by multiplying by 1000.

2.3.3 Volume

Volume calculations are used for fuel tanks, hydraulic reservoirs, and other containers.

\[ \text{Volume of a rectangular prism} = \text{length} \times \text{width} \times \text{height} \]

\[ \text{Volume of a cylinder} = \pi r^2 h \]

Volume unit conversions:

Example: 135 litres of fuel = 135 × 0.001 = 0.135 m³

3. Important Formulas and Relationships

3.1 Essential Formulae Summary

3.2 Key Relationships Between Concepts

4. Common Relationships Between Concepts

4.1 Weight and Balance Calculations

The weight and balance calculation is one of the most frequent mathematical tasks in maintenance. The fundamental relationship is:

\[ \text{Moment} = \text{Weight} \times \text{Arm} \]

Where:

The centre of gravity is then:

\[ \text{CG} = \frac{\sum \text{Moments}}{\sum \text{Weights}} \]

This relationship appears in two common forms:

4.2 Fuel Calculations

Fuel calculations involve several interconnected quantities:

The density is a distractor when calculating endurance from volume — it is only needed when converting between volume and mass.

4.3 Torque Conversions

Torque is a measure of rotational force. The conversion between metric and imperial units is a common real-world task:

\[ \text{Torque (lb-ft)} = \text{Torque (N·m)} \times 0.73756 \]

Examples:

5. Typical Exam Focus Points

Based on the pattern of Part-66 Module 1 examination questions for B3, candidates should focus on:

5.1 Arithmetic (Module 1.1)

5.2 Algebra (Module 1.2)

5.3 Geometry (Module 1.4)

5.4 Common Mistakes to Avoid

6. Regulatory References

The knowledge levels indicate:

For B3 licence, Module 1 is examined at levels 1–2, requiring proficiency in applying standard mathematical procedures to typical maintenance scenarios.