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:

• 1.1 Arithmetic (levels 1–2): basic operations, fractions, decimals, ratios, percentages, and unit conversions
• 1.2 Algebra (levels 1–2): evaluating simple formulae, transposition of formulae, and solving linear equations
• 1.4 Geometry (levels 1–2): area, circumference, volume, and simple geometric relationships

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.

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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:

1. Brackets (parentheses)
2. Orders (powers, roots)
3. Division and Multiplication (left to right)
4. Addition and Subtraction (left to right)

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

• A fraction represents a part of a whole (e.g., \( \frac{3}{4} \) full fuel tank)
• A decimal is a fraction expressed in base-10 notation (e.g., 0.75)
• A percentage is a fraction with denominator 100 (e.g., 75%)

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:

• Length: 0.75 m = 0.75 × 100 = 75 cm
• Mass: 180 lb ÷ 2.2046 = 81.65 kg ≈ 81.6 kg (or 82 kg to nearest whole number)
• Torque: 25 N·m × 0.73756 = 18.44 lb-ft ≈ 18.4 lb-ft
• Volume: 135 L × 0.001 = 0.135 m³

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:

• Round to the nearest whole number (e.g., 81.65 kg → 82 kg)
• Round to one decimal place (e.g., 4.2857 h → 4.3 h)
• Round to two decimal places (e.g., 10.826 m → 10.83 m)

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.

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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:

• Whatever operation is performed on one side of the equation must be performed on the other
• To isolate a variable, apply inverse operations (addition ↔ subtraction, multiplication ↔ division)

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:

• \( P \) = power (watts, W)
• \( V \) = voltage (volts, V)
• \( I \) = current (amperes, A)

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

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

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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:

• \( C \) = circumference
• \( d \) = diameter
• \( r \) = radius
• \( \pi \approx 3.14 \) or \( 3.1416 \)

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:

• 1 L = 0.001 m³
• 1 m³ = 1000 L

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

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3. Important Formulas and Relationships

3.1 Essential Formulae Summary

3.2 Key Relationships Between Concepts

• Unit conversions underpin all other calculations — always check that units are consistent before substituting into a formula
• Algebraic transposition is the tool for solving any formula for an unknown variable
• Geometry provides the area and volume relationships needed for component sizing and fluid capacity
• Ratios and proportions connect fuel volume to endurance, and tank fraction to actual volume
• Weight and balance links force (weight), distance (arm), and moment — the same mathematical structure appears in both N·m and kg·m systems

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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:

• Moment is expressed in N·m (or kg·m when using mass)
• Weight is expressed in N (or mass in kg)
• Arm is the distance from the datum in metres

The centre of gravity is then:

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

This relationship appears in two common forms:

• Using force (N): \( \text{Arm} = \frac{\text{Moment (N·m)}}{\text{Weight (N)}} \)
• Using mass (kg): \( \text{CG} = \frac{\text{Moment (kg·m)}}{\text{Mass (kg)}} \)

4.2 Fuel Calculations

Fuel calculations involve several interconnected quantities:

• Volume (litres) — what the tank holds
• Mass (kg) — volume × density
• Endurance (hours) — volume ÷ consumption rate

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:

• 25 N·m = 25 × 0.73756 = 18.44 lb-ft ≈ 18.4 lb-ft
• 35 N·m = 35 × 0.73756 = 25.82 lb-ft ≈ 25.8 lb-ft

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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)

• Unit conversions — particularly metric↔imperial for mass (lb↔kg) and torque (N·m↔lb-ft)
• Metric conversions — metres↔centimetres↔millimetres, litres↔cubic metres
• Time calculations — converting hours and minutes to decimal hours (e.g., 2 h 30 min = 2.5 h)
• Ratios and fractions — applying fractions of tank capacity to actual volumes
• Rounding — to specified precision (nearest whole number, one decimal place, etc.)

5.2 Algebra (Module 1.2)

• Substituting values into given formulae (e.g., resistance-temperature)
• Transposing formulae to solve for the unknown (e.g., finding arm from moment and weight)
• Recognising irrelevant data — exam questions often include extra information (distractors) that is not needed for the calculation

5.3 Geometry (Module 1.4)

• Circumference of circles from diameter
• Area of circles from radius
• Finding diameter from area (requires rearranging the formula and using square roots)
• Unit consistency — ensure all dimensions are in the same units before calculating

5.4 Common Mistakes to Avoid

1. Using the wrong conversion direction — when converting N·m to lb-ft, multiply; when converting lb to kg, divide by 2.2046
2. Forgetting to convert units — e.g., leaving a diameter in metres when millimetres are required
3. Using irrelevant data — e.g., fuel density when calculating endurance from volume
4. Incorrect rounding — always check the required precision in the question
5. Mixing force and mass — moments can be expressed in N·m (force) or kg·m (mass); do not mix systems

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6. Regulatory References

• Regulation (EU) No 1321/2014, Annex III (Part-66) — establishes the basic knowledge requirements for aircraft maintenance licences
• Appendix I to Part-66 — defines the Module 1 syllabus content and knowledge levels:
• Module 1.1 Arithmetic — levels 1–2
• Module 1.2 Algebra — levels 1–2
• Module 1.4 Geometry — levels 1–2
• AMC to Part-66 — provides guidance on examination standards and question formats

The knowledge levels indicate:

• Level 1: Overview — recall of basic facts and simple procedures
• Level 2: General knowledge — understanding of principles and ability to apply standard procedures
• Level 3: Detailed theory — comprehensive understanding with ability to analyse and synthesise

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