Module 1: Mathematics

1. Overview

Module 1 of the EASA Part-66 basic knowledge syllabus establishes the fundamental mathematical principles required for the safe execution of aircraft maintenance tasks. This module is not about abstract mathematics; it is the practical toolkit used daily by certifying staff. From interpreting measurement tolerances and calculating engine parameters to converting units between imperial and SI systems, a solid grasp of these concepts is non-negotiable. The syllabus is divided into three main areas: Arithmetic (1.1), Algebra (1.2), and Geometry (1.3).

The knowledge levels for this module are defined in Appendix I of Part-66:

Level 1 (Overview): A general familiarisation with the principles. The candidate should recognise the concepts and their applications.

Level 2 (General Knowledge): A broader understanding of the theoretical concepts and the ability to apply them to typical maintenance scenarios.

Level 3 (Detailed Theory): A deep, detailed knowledge allowing the candidate to analyse, interpret, and solve complex problems, often involving multiple steps or unit conversions.

The exam questions provided cover all three levels, focusing on the practical application of arithmetic, algebra, and geometry.

2. Key Concepts Explained in Detail

2.1 Arithmetic (Syllabus Ref: 1.1)

This section deals with the manipulation of numbers and the interpretation of results, which is critical for making airworthiness decisions.

2.1.1 Averages and Their Limitations

Arithmetic Mean: The sum of a set of values divided by the number of values. It is used to find a representative value for a set of measurements.

Formula: Mean = (x₁ + x₂ + ... + xₙ) / n

Example (Q9): Mean thickness of a brake disc = (12.45 + 12.52 + 12.38 + 12.49) / 4 = 12.46 mm.

Critical Limitation: The mean can mask a localised defect. A single reading that exceeds a limit is cause for rejection, even if the average is within limits. The mean is a statistical tool, not a substitute for checking each individual measurement against the specified limit. The maximum measured value must always be compared directly to the manufacturer's absolute limit.

2.1.2 Tolerances and Deviations

Nominal Value: The specified, ideal, or target dimension (e.g., 23.00 mm).

Tolerance: The permissible variation from the nominal value, often expressed as a plus/minus (±) range.

Deviation: The difference between a measured value and the nominal value. It can be positive or negative.

Acceptance Criteria: A part is within tolerance only if all measured deviations fall within the specified ± limit.

Example (Q4): Nominal = 23.00 mm, Tolerance = ±0.05 mm. Measured values: 22.94 mm, 23.01 mm, 23.05 mm. Deviations are -0.06 mm, +0.01 mm, +0.05 mm. The maximum absolute deviation is 0.06 mm, which exceeds the 0.05 mm limit. The pin is out of tolerance.

Percentage Deviation: Expresses the deviation as a percentage of the nominal value.

Formula: Percentage Deviation = (Deviation / Nominal Value) × 100%

Example (Q10): Nominal = 2.00 mm, Tolerance = ±0.03 mm. Furthest measurement = 2.04 mm, deviation = +0.04 mm. Percentage deviation = (0.04 / 2.00) × 100% = 2.0%. This exceeds the tolerance, so the ring is unserviceable.

2.1.3 Ratios and Proportions

Ratio: A comparison of two or more quantities, showing their relative sizes (e.g., 1:4).

Proportion: An equation stating that two ratios are equal. Used to scale quantities up or down.

Application (Q11): For a 1:4 ratio (concentrate:water), the total number of parts is 5. The fraction of concentrate is 1/5. To make 2.5 litres of solution, the volume of concentrate required is (1/5) × 2.5 L = 0.5 L = 500 ml.

2.1.4 Unit Conversion and Dimensional Analysis

This is a fundamental skill, as maintenance documentation may use either SI (metric) or Imperial units. The key is to use conversion factors correctly.

Common Conversions:

Length: 1 inch = 25.4 mm

Volume: 1 US gallon = 3.785 litres; 1 Imperial gallon = 4.546 litres

Power: 1 hp = 745.7 W

Pressure: 1 bar = 14.5 psi; 1 atm = 14.7 psi

Torque: 1 N·m = 0.7376 lbf·ft

Method: Multiply the original value by a conversion factor expressed as a fraction equal to 1, ensuring the unwanted units cancel out.

Example (Q17): Convert 0.0035 inches to mm: 0.0035 in × (25.4 mm / 1 in) = 0.0889 mm.

Example (Q23): Convert 120 lbf·ft to N·m: 120 lbf·ft × (4.448 N / 1 lbf) × (0.3048 m / 1 ft) = 162.7 N·m.

2.1.5 Gauge vs. Absolute Pressure

Gauge Pressure: The pressure measured relative to ambient atmospheric pressure. A typical pressure gauge reads zero at atmospheric pressure.

Absolute Pressure: The total pressure, including atmospheric pressure. Absolute Pressure = Gauge Pressure + Atmospheric Pressure.

Application (Q5): A pressure gauge reading of 80 psi is a gauge pressure reading. To convert this to bar, use the conversion factor: 80 psi / 14.5 psi/bar = 5.52 bar. The ambient pressure is not added because the reading is already relative to it.

2.2 Algebra (Syllabus Ref: 1.2)

Algebra provides the framework for rearranging formulas and solving for unknown variables, which is essential for interpreting technical data.

2.2.1 Laws of Indices (Exponents)

These rules govern the manipulation of powers.

Multiplication: aᵐ × aⁿ = aᵐ⁺ⁿ

Division: aᵐ / aⁿ = aᵐ⁻ⁿ

Power of a Power: (aᵐ)ⁿ = aᵐⁿ

Negative Exponent: a⁻ⁿ = 1/aⁿ

Zero Exponent: a⁰ = 1

Example (Q22): 2⁵ × 2⁻³ = 2⁽⁵⁻³⁾ = 2² = 4.

2.2.2 Transposition of Formulae

This is the process of rearranging an equation to make a different variable the subject. The fundamental rule is that whatever you do to one side of the equation, you must do to the other.

Application (Q14): The torque formula is Torque (T) = Force (F) × Distance (d). To find the force, divide both sides by d: F = T / d. Therefore, F = 25 N·m / 0.3 m = 83.3 N.

2.3 Geometry (Syllabus Ref: 1.3)

Geometry provides the formulas for calculating areas and volumes of shapes commonly found on an aeroplane.

2.3.1 Area Calculations

Rectangle: Area = length × width

Example (Q19): An access panel of 30 cm × 45 cm has an area of 0.30 m × 0.45 m = 0.135 m². (Remember to convert units to be consistent).

Circle: Area = π × r² or Area = (π/4) × d², where r is the radius and d is the diameter.

Example (Q2): For a diameter of 15 mm, Area = 3.142 × (15/2)² = 3.142 × 56.25 = 176.7 mm².

2.3.2 Circumference of a Circle

Formula: Circumference = π × d or 2 × π × r.

Example (Q8): For a diameter of 20 cm, Circumference = 3.14 × 20 cm = 62.8 cm.

2.3.3 Volume Calculations

Cylinder (Swept Volume): This is the volume displaced by a piston moving from Top Dead Centre (TDC) to Bottom Dead Centre (BDC).

Formula: Swept Volume (V_s) = π × r² × stroke or V_s = (π/4) × bore² × stroke.

Example (Q15 & Q21): For a bore of 120 mm (12 cm) and a stroke of 140 mm (14 cm):

V_s = (3.142/4) × (12 cm)² × 14 cm = 0.7855 × 144 × 14 = 1583.36 cm³.

Note: It is often easier to use centimetres to get the answer directly in cm³.

2.3.4 The Pythagorean Theorem

This theorem applies to right-angled triangles and is used to find the length of one side when the other two are known.

Formula: a² + b² = c², where c is the hypotenuse (the longest side, opposite the right angle).

Example (Q13): If a = 60 mm and b = 80 mm, then c² = 60² + 80² = 3600 + 6400 = 10000. Therefore, c = √10000 = 100 mm.

3. Important Formulas and Procedures

The following table summarises the key formulas and concepts from this module.

Concept	Formula / Rule	Application Example
Arithmetic Mean	`(x₁ + x₂ + ... + xₙ) / n`	Calculating average thickness of a component.
Tolerance Check	`Nominal ± Tolerance`	Verifying if a measurement is within limits.
Percentage Deviation	`(Deviation / Nominal) × 100%`	Quantifying the error in a measurement.
Ratio/Proportion	`Part / Total = Fraction`	Mixing solutions, calculating fuel fractions.
Unit Conversion	Multiply by a conversion factor (e.g., `1 in = 25.4 mm`).	Converting imperial to metric dimensions.
Gauge Pressure	`P_gauge = P_absolute - P_atmospheric`	Interpreting pressure gauge readings.
Laws of Indices	`aᵐ × aⁿ = aᵐ⁺ⁿ`	Simplifying expressions with powers.
Transposition	Perform the same operation on both sides of the equation.	Solving for force in a torque equation.
Area (Rectangle)	`length × width`	Calculating the area of an access panel.
Area (Circle)	`π × r²` or `(π/4) × d²`	Calculating the cross-sectional area of a rod.
Circumference	`π × d`	Calculating the circumference of an inspection window.
Swept Volume	`(π/4) × bore² × stroke`	Calculating the swept volume of a cylinder.
Pythagoras	`a² + b² = c²`	Finding the length of a triangle's hypotenuse.

4. Common Relationships Between Concepts

Tolerances and Arithmetic: Tolerance checks are a direct application of arithmetic (subtraction for deviation) and are often used in conjunction with the mean. A mean can be within tolerance, but an individual reading may not be, highlighting the importance of checking all data points.

Geometry and Algebra: Geometric formulas (e.g., for swept volume) are used within algebraic equations to solve for other parameters, such as compression ratio.

Unit Conversion and Algebra: Unit conversion is a form of algebraic manipulation where the conversion factor is a multiplier. This is critical when combining data from different sources (e.g., an engine producing 180 hp needs to be converted to kW for comparison with SI-based data).

Ratios and Geometry: The compression ratio of an engine is a ratio that links the geometric volumes of the cylinder (total volume and clearance volume).

Arithmetic and Algebra: Arithmetic provides the basic operations, while algebra provides the framework for generalising these operations in formulas. For example, the formula for shim thickness (Shim = Measured - Desired) is an algebraic expression that uses arithmetic for its calculation.

5. Typical Exam Focus Points

Based on the syllabus and typical exam questions, candidates should focus on the following:

Interpretation of Tolerances: Be able to determine if a part is serviceable based on a set of measurements and a specified tolerance. Remember that the maximum deviation, not the average, is the deciding factor.

Unit Conversions: Be proficient in converting between imperial and SI units for length, volume, mass, force, pressure, and torque. Pay close attention to the conversion factors provided in the question.

Geometric Formulas: Know the formulas for the area and circumference of a circle, the area of a rectangle, and the volume of a cylinder. Be comfortable using π ≈ 3.142.

Algebraic Manipulation: Be able to transpose simple formulas to solve for an unknown variable (e.g., force from torque, or shim thickness).

Ratio and Proportion: Be able to calculate quantities based on given ratios, such as in mixing solutions or calculating fuel consumption.

Laws of Indices: Be able to simplify expressions involving powers, especially with multiplication and division.

Practical Application: The exam questions are scenario-based. You must be able to extract the relevant information from a maintenance scenario (e.g., "compression check", "piston pin measurement") and apply the correct mathematical principle to find the answer. The correct interpretation of the question is as important as the calculation itself.

Module 1: Mathematics