Module 1: Mathematics

Module 1: Mathematics — EASA Part-66 Category B13

1. Module Overview

Module 1 of the EASA Part-66 basic knowledge syllabus establishes the mathematical foundation required for all aircraft maintenance engineering disciplines, including helicopter-specific applications. For Category B13 (helicopter maintenance), this module covers arithmetic, algebra, geometry, trigonometry, and introductory statistics. The knowledge levels range from Level 1 (overview) for basic concepts to Level 3 (detailed theory) for topics directly applied in maintenance procedures such as weight and balance calculations, rotor tracking, gearbox ratios, and performance monitoring.

The module is divided into the following syllabus areas per Appendix I of Regulation (EU) No 1321/2014:

• 1.1 Arithmetic (Level 1–2)
• 1.2 Algebra (Level 1–2)
• 1.3 Geometry (Level 1–2)
• 1.4 Trigonometry (Level 1–2)
• 1.5 Graphs (Level 1–2)
• 1.6 Statistics (Level 2)

---

2. Key Concepts Explained in Detail

2.1 Arithmetic Fundamentals

2.1.1 SI Units and Conversions

The International System of Units (SI) is the standard for all aeronautical engineering. Maintenance personnel must be fluent in converting between SI units and imperial units, as many aircraft components are manufactured to imperial specifications while maintenance documentation may use SI.

Base SI units relevant to helicopter maintenance:

Common conversions for helicopter maintenance:

• 1 inch = 25.4 mm (exactly)
• 1 US gallon = 3.785 litres
• 1 kg = 2.20462 lb
• 1 N = 0.224809 lbf
• 1 hPa = 1 mbar = 100 Pa

Example application: A main rotor blade tip clearance is measured as 12 mm. The maintenance manual specifies a tolerance of ±0.5 inch. Converting: 12 mm ÷ 25.4 mm/inch = 0.472 inches. Since 0.472 < 0.5, the measurement is within tolerance.

2.1.2 Arithmetic Mean, Median, and Mode

These measures of central tendency are essential for interpreting maintenance data, particularly in rotor tracking and vibration analysis.

Arithmetic Mean: The sum of all values divided by the number of values.

$$\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$$

Example: Rotor blade tracking readings of 12.5, 12.7, 12.4, 12.6, and 12.8 mm give a mean of (12.5 + 12.7 + 12.4 + 12.6 + 12.8) ÷ 5 = 63.0 ÷ 5 = 12.6 mm.

Median: The middle value when data is arranged in ascending order. For an even number of values, the median is the average of the two middle values.

Example: Flight hours recorded over 6 days: 3.5, 4.2, 2.8, 5.1, 3.9, 4.5. Sorted: 2.8, 3.5, 3.9, 4.2, 4.5, 5.1. Median = (3.9 + 4.2) ÷ 2 = 4.05 hours.

Mode: The most frequently occurring value.

2.1.3 Standard Deviation

Standard deviation quantifies the spread of data around the mean. This is critical for assessing whether measurement variations in rotor tracking or engine parameters are within acceptable limits.

Population standard deviation (σ):

$$\sigma = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n}}$$

Sample standard deviation (s):

$$s = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n-1}}$$

Example: Blade track values of 12.4, 12.8, 12.6, 13.1, 12.9 mm.

Mean = 12.76 mm

Deviations: −0.36, 0.04, −0.16, 0.34, 0.14

Squared deviations: 0.1296, 0.0016, 0.0256, 0.1156, 0.0196

Sum of squared deviations = 0.292

Sample variance = 0.292 ÷ 4 = 0.073

Sample standard deviation = √0.073 = 0.27 mm

2.1.4 Ratios, Proportions, and Percentages

Gear ratios are fundamental to helicopter transmission systems. The relationship between input and output speeds is:

$$\frac{N_1}{N_2} = \frac{T_2}{T_1}$$

Where N = rotational speed (rpm) and T = number of teeth.

Example: Input pinion with 21 teeth at 6000 rpm driving a main gear with 98 teeth:

Output speed = 6000 × (21 ÷ 98) = 6000 × 0.2143 = 1285.7 rpm

Percentages express a ratio as a fraction of 100. Efficiency calculations use:

$$\text{Efficiency} = \frac{\text{Output power}}{\text{Input power}} \times 100\%$$

Example: Turbine engine producing 850 kW shaft power at 38% efficiency:

Input power = 850 ÷ 0.38 = 2237 kW

2.1.5 Rates

Rates express the change in one quantity relative to another, typically over time.

Example: A helicopter descending from 1500 m to 300 m in 4 minutes:

Rate of descent = (1500 − 300) ÷ 4 = 1200 ÷ 4 = 300 m/min

2.2 Algebra

2.2.1 Linear Equations

Linear equations take the form ax + b = 0. Solving requires isolating the variable using algebraic operations.

Example: 3(x − 4) = 2x + 6

Expand: 3x − 12 = 2x + 6

Subtract 2x: x − 12 = 6

Add 12: x = 18

Identities: Some equations are true for all values of the variable. For example, 3(x − 4) + 2x = 5x − 12 simplifies to 5x − 12 = 5x − 12, which is an identity. Any real value of x satisfies the equation.

2.2.2 Simultaneous Equations

Systems of linear equations with multiple unknowns are solved by substitution or elimination.

Substitution method:

Given: 3x + 2y = 17 and 2x − y = 6

From the second equation: y = 2x − 6

Substitute into the first: 3x + 2(2x − 6) = 17

3x + 4x − 12 = 17

7x = 29

x = 29/7 = 4.14

Elimination method:

Multiply the second equation by 2: 4x − 2y = 12

Add to the first: 7x = 29

x = 4.14

2.2.3 Quadratic Equations

Quadratic equations take the form ax² + bx + c = 0. Evaluation involves substituting values for the variable.

Example: For y = 3x² − 2x + 5, find y when x = −2:

y = 3(−2)² − 2(−2) + 5 = 3(4) + 4 + 5 = 12 + 4 + 5 = 21

The quadratic formula for solving ax² + bx + c = 0:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

2.2.4 Laws of Indices (Exponents)

The laws of indices govern operations with powers:

• aᵐ × aⁿ = aᵐ⁺ⁿ
• aᵐ ÷ aⁿ = aᵐ⁻ⁿ
• (aᵐ)ⁿ = aᵐⁿ
• a⁰ = 1
• a⁻ⁿ = 1/aⁿ

Example: Evaluate 2³ × 2⁻² ÷ 2⁴

2³ × 2⁻² = 2³⁻² = 2¹

2¹ ÷ 2⁴ = 2¹⁻⁴ = 2⁻³

2.2.5 Powers and Roots

Example: Calculate 7³ + √144

7³ = 7 × 7 × 7 = 343

√144 = 12

Sum = 343 + 12 = 355

2.2.6 Logarithms and Exponentials

The barometric altitude formula applies logarithms and exponentials to atmospheric physics:

$$P = P_0 \times (1 - 2.25577 \times 10^{-5} \times h)^{5.25588}$$

Where P = pressure at altitude h, P₀ = sea-level pressure (1013.25 hPa).

Example: Find altitude when pressure is 850 hPa:

(P/P₀)^(1/5.25588) = 1 − 2.25577 × 10⁻⁵ × h

(850/1013.25)^(0.1903) = 0.9657

1 − 0.9657 = 0.0343

h = 0.0343 ÷ (2.25577 × 10⁻⁵) = 1520 m

2.3 Geometry

2.3.1 Pythagoras' Theorem

For a right-angled triangle with perpendicular sides a and b and hypotenuse c:

$$c^2 = a^2 + b^2$$

Example: Sides of 30 mm and 40 mm:

c² = 30² + 40² = 900 + 1600 = 2500

c = √2500 = 50 mm

2.3.2 Circle Properties

Area of a circle: A = πr²

Circumference: C = 2πr = πd

Example: Rotor brake disc with diameter 400 mm:

Radius = 200 mm = 0.2 m

Area = 3.14 × (0.2)² = 3.14 × 0.04 = 0.1256 m²

2.3.3 Linear Expansion

Materials expand with temperature change according to:

$$\Delta L = L_0 \times \alpha \times \Delta T$$

Where ΔL = change in length, L₀ = original length, α = coefficient of linear expansion, ΔT = temperature change.

Example: Tail rotor drive shaft, 1.8 m long, α = 12 × 10⁻⁶/°C, temperature increases from 15°C to 45°C:

ΔL = 1.8 × 12 × 10⁻⁶ × (45 − 15) = 1.8 × 12 × 10⁻⁶ × 30 = 648 × 10⁻⁶ m = 0.648 mm

2.3.4 Angular Velocity and Linear Speed

For rotating components:

$$\omega = \frac{2\pi N}{60}$$

Where ω = angular velocity (rad/s), N = rotational speed (rpm).

Linear tip speed: v = ω × r

Example: Main rotor at 350 rpm, maximum tip speed 220 m/s:

ω = 2 × 3.1416 × 350 ÷ 60 = 36.65 rad/s

Maximum radius = 220 ÷ 36.65 = 6.00 m

2.3.5 Moments and Weight and Balance

The moment of a component about a reference point:

$$\text{Moment} = \text{Weight} \times \text{Arm}$$

Example: Moment = 2500 kg·m, arm = 2.5 m:

Weight = 2500 ÷ 2.5 = 1000 kg

2.4 Density and Fuel Calculations

Density relates mass to volume:

$$\rho = \frac{m}{V}$$

Example: Fuel flow of 420 kg/h with density 0.80 kg/L:

Volume flow = 420 ÷ 0.80 = 525 L/h

In US gallons: 525 ÷ 3.785 = 138.7 US gal/h

---

3. Important Formulas and Procedures

3.1 Essential Formula Summary

3.2 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 Mathematics syllabus with knowledge levels.
• AMC to Part-66: Provides acceptable means of compliance and guidance material for examination standards.

---

4. Common Relationships Between Concepts

4.1 Unit Conversions and Maintenance Limits

Maintenance manuals frequently mix metric and imperial units. The ability to convert accurately is essential for determining whether measurements fall within tolerance. For example, a blade track deviation of 12 mm must be compared against a limit of ±0.5 inch (12.7 mm). The conversion factor 1 inch = 25.4 mm is exact and must be memorised.

4.2 Statistics and Rotor Tracking

Rotor blade tracking produces multiple measurements that must be analysed statistically. The mean indicates the average blade tip position, while standard deviation reveals the consistency of the track. Large standard deviations may indicate mechanical problems such as worn pitch links or damaged bearings.

4.3 Gear Ratios and Transmission Systems

Helicopter transmissions use gear reductions between the engine and main rotor. Understanding gear ratios allows maintenance personnel to verify correct component selection and calculate expected output speeds during ground runs.

4.4 Algebra and Weight and Balance

Weight and balance calculations require rearranging formulas to solve for unknown quantities. The relationship Moment = Weight × Arm can be rearranged to find any of the three variables when the other two are known.

4.5 Exponentials and Atmospheric Physics

The barometric formula demonstrates the application of exponential functions to real-world phenomena. Understanding this relationship is important for interpreting altimeter readings and performance calculations at various altitudes.

---

5. Typical Exam Focus Points

5.1 Arithmetic (Module 1.1)

• Unit conversions between SI and imperial systems (metres to millimetres, litres to US gallons, millimetres to inches)
• Calculation of arithmetic mean from maintenance data sets
• Identification and calculation of median for both odd and even data sets
• Application of ratios and proportions to gear systems
• Percentage calculations for efficiency and tolerance assessments
• Rate calculations for descent rates, fuel consumption, and flow rates

5.2 Algebra (Module 1.2)

• Solving linear equations with one unknown
• Solving simultaneous equations using substitution and elimination methods
• Evaluating quadratic expressions at given values
• Application of laws of indices, including negative and zero exponents
• Calculation of powers and roots
• Recognition of algebraic identities
• Rearranging formulas to solve for different variables

5.3 Geometry (Module 1.3)

• Application of Pythagoras' theorem to right-angled triangles
• Calculation of circle area and circumference
• Linear expansion calculations for temperature effects on components
• Relationship between angular velocity and linear speed for rotating components
• Moment calculations for weight and balance

5.4 Statistics (Module 1.6)

• Calculation of arithmetic mean, median, and mode
• Calculation of standard deviation (both population and sample)
• Interpretation of statistical measures in maintenance contexts
• Understanding the difference between population and sample statistics

5.5 Examination Strategy

• Always check units before performing calculations
• Convert all values to consistent units before applying formulas
• Round answers appropriately based on the precision of the input data
• For standard deviation questions, note whether the question specifies population or sample
• Verify answers by substituting back into the original equation where possible
• Memorise key conversion factors: 1 inch = 25.4 mm, 1 US gallon = 3.785 L

---

6. Worked Examples for Examination Practice

Example 1: Fuel Consumption

A helicopter turbine engine consumes 210 US gallons per hour. Express this in litres per hour.

Solution: 210 × 3.785 = 794.85 L/h

Example 2: Component Diameter

A turbine engine component has a diameter of 0.75 m. Express in millimetres.

Solution: 0.75 × 1000 = 750 mm

Example 3: Rotor Blade Mean Displacement

Blade tip displacements measured as 2.5 cm, 2.7 cm, and 2.3 cm.

Solution: Mean = (2.5 + 2.7 + 2.3) ÷ 3 = 7.5 ÷ 3 = 2.5 cm

Example 4: Average Displacement with Unit Conversion

Blade tip displacements of 0.25 m, 0.30 m, and 0.28 m. Express the average in millimetres.

Solution: Average = (0.25 + 0.30 + 0.28) ÷ 3 = 0.2767 m = 276.7 mm

Example 5: Gearbox Output Speed

Input gear with 24 teeth at 3000 rpm driving an output gear with 72 teeth.

Solution: Gear ratio = 72 ÷ 24 = 3. Output speed = 3000 ÷ 3 = 1000 rpm

---

7. Summary

Module 1 Mathematics provides the quantitative foundation for all helicopter maintenance activities. Mastery of arithmetic operations, unit conversions, algebraic manipulation, geometric relationships, and basic statistics enables maintenance personnel to:

• Interpret maintenance manual data and tolerances
• Perform weight and balance calculations
• Analyse rotor tracking and vibration data
• Verify transmission system parameters
• Calculate fuel consumption and performance figures
• Apply engineering formulas to real-world maintenance scenarios

The knowledge levels required for Category B13 range from Level 1 (familiarity with basic concepts) to Level 2 (general knowledge with practical application). Candidates should focus on developing fluency in unit conversions and formula manipulation, as these skills are tested directly and appear throughout all other modules of the Part-66 syllabus.