Module 2: Physics

Module 2: Physics – B3 Licence Category

1. Module Overview

Module 2 of the EASA Part-66 basic knowledge syllabus provides the foundational physical principles required for the safe maintenance, inspection, and certification of aircraft. For the B3 (light aeroplane) category, this module covers the core areas of matter, mechanics, thermodynamics, optics, and basic electrical theory. A certifying engineer must apply these principles daily—from calculating centre of gravity for weight and balance, to understanding hydraulic system pressures, thermal expansion of components, and the behaviour of pitot-static instruments.

The syllabus is divided into the following key areas, each with defined knowledge levels (1 = overview, 2 = general knowledge, 3 = detailed theory):

This study material synthesises the knowledge required to answer typical exam questions and, more importantly, to apply these principles in a maintenance environment.

2. Key Concepts Explained in Detail

2.1 Units of Measurement and Conversions

The International System of Units (SI) is the standard for all aviation maintenance documentation and calculations. However, legacy tooling, American-manufactured aircraft, and certain maintenance manuals still use Imperial units. A certifying engineer must be fluent in converting between systems.

Example – Torque Conversion: A torque specification of 25 N·m is equivalent to 25 × 0.7376 = 18.44 lb-ft. This is critical when using Imperial-calibrated torque wrenches on components specified in SI units.

Example – Volume Conversion: A fuel tank of 200 litres has a volume of 200 × 0.001 = 0.2 m³. This conversion is used in fuel mass calculations (mass = density × volume).

2.2 Matter and Material Properties

Matter exists in three primary states relevant to aviation: solid, liquid, and gas. Each state has distinct physical properties that influence aircraft design and maintenance.

Material Properties in Maintenance: Understanding material properties is essential for selecting correct repair techniques. For example, ice is a crystalline solid with high compressive strength but low shear strength. Therefore, mechanical de-icing (scraping) applies a shear force to break the ice layer, rather than attempting to crush it.

2.3 Statics: Forces, Moments, and Equilibrium

Statics is the study of bodies at rest or moving at constant velocity. A body is in equilibrium when the vector sum of all forces and moments acting upon it is zero.

Example – Torque Application: A force of 100 N applied perpendicular to a torque wrench handle of 0.3 m length produces a torque of 100 N × 0.3 m = 30 N·m.

Centre of Gravity (CG): The point where the entire weight of an object is considered to act. For an aeroplane, the CG position is critical for stability and control.

Weight and Balance Calculation: The CG is calculated using the principle of moments:

Example – CG Shift: An aeroplane with a mass of 1,150 kg and a CG at 2.40 m aft of the datum has a moment of 1,150 × 2.40 = 2,760 kg·m. Loading 50 kg of cargo at a station 3.10 m aft of the datum adds a moment of 50 × 3.10 = 155 kg·m. The new total moment is 2,915 kg·m, and the new total mass is 1,200 kg. The new CG is 2,915 / 1,200 = 2.43 m aft of the datum. This must be checked against the CG envelope (forward and aft limits) to ensure the aeroplane remains within its certified limits.

Newton's Laws of Motion:

Example – Equilibrium: A helicopter hovering at constant altitude has zero net force acting on it. The upward lift (rotor thrust) exactly balances the downward weight. Similarly, a helicopter descending at a constant rate is in equilibrium; the upward forces (rotor thrust) equal the downward forces (weight).

Example – Resultant Force: An engine producing 2,000 N of thrust against a total drag of 1,500 N results in a net forward force of 500 N.

Example – Impact Force (Impulse-Momentum): A bird of mass 0.5 kg decelerated from 100 m/s to 0 m/s in 0.01 seconds experiences a force of F = (m × Δv) / t = (0.5 × 100) / 0.01 = 5,000 N. This principle is used to assess impact damage on components like compressor blades.

2.4 Kinematics: Linear and Rotational Motion

Kinematics describes motion without considering its causes.

Example – Rotor Tip Speed: A helicopter rotor blade of 5 m length rotating at 300 rpm. First, convert rpm to rad/s: ω = 300 × (2π / 60) = 31.42 rad/s. Then, tip speed v = 31.42 × 5 = 157.1 m/s. This can also be calculated by finding the circumference (2πr = 31.4 m) and multiplying by revolutions per second (5 rev/s), giving 157 m/s. Tip speed is a critical parameter for rotor performance and noise.

Rotational Kinetic Energy: A rotating body stores kinetic energy given by E = ½ I ω², where I is the moment of inertia (kg·m²). This energy is proportional to the square of the angular velocity.

Example – Autorotation Energy: If rotor speed decays from 400 rpm to 350 rpm, the ratio of final to initial energy is (350/400)² = 0.766. This represents a 23.4% decrease in stored rotational kinetic energy. This is a crucial consideration during autorotation, where the pilot manages this stored energy to control the rate of descent and flare.

2.5 Fluid Mechanics: Pressure, Hydraulics, and Aerodynamics

Pressure (P) is defined as force per unit area: P = F / A, measured in Pascals (Pa) where 1 Pa = 1 N/m². In aviation, pressure is often expressed in bar (1 bar = 100,000 Pa) or psi.

Example – Hydraulic Pressure: A force of 5,000 N applied to a piston of area 20 cm² (0.002 m²) produces a pressure of 5,000 / 0.002 = 2,500,000 Pa = 25 bar.

Pascal's Law: Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. This is the principle of hydraulic multiplication.

Example – Hydraulic Multiplication: A force of 50 N on a small piston of area 0.002 m² creates a pressure of 25,000 Pa. This pressure acts on a larger piston of area 0.1 m², producing a force of 25,000 × 0.1 = 2,500 N. This allows a small input force to lift a heavy load.

Hydraulic Power: The power delivered by a hydraulic pump is the product of pressure and flow rate: P = p × Q, where p is pressure in Pascals and Q is flow rate in cubic metres per second (m³/s).

Example – Pump Power: A pump delivering 3,000 psi (20,685,000 Pa) at a flow rate of 5 gallons per minute (0.0003154 m³/s) delivers approximately 20,685,000 × 0.0003154 = 6,524 W ≈ 6.5 kW.

Atmospheric Pressure and Pitot-Static Systems: The pitot-static system measures dynamic pressure for airspeed indication. A pitot tube measures total (pitot) pressure, while static ports measure ambient (static) pressure. The airspeed indicator (ASI) displays the difference (dynamic pressure).

Cabin Pressurisation: The pressure difference between the cabin and the outside atmosphere is critical for structural integrity. For example, a cabin pressure of 100 kPa at an outside pressure of 80 kPa results in a differential pressure of 20 kPa.

2.6 Thermodynamics: Temperature, Heat, and Expansion

Temperature is a measure of the average kinetic energy of the particles in a substance. It is measured in Kelvin (K), Celsius (°C), or Fahrenheit (°F). The Kelvin scale is an absolute scale, with 0 K being absolute zero.

Conversion: K = °C + 273.15.

Example: An ambient temperature of 15°C is equivalent to 15 + 273.15 = 288.15 K, which rounds to 288 K. This is essential for engine performance calculations, as gas turbine performance is referenced to International Standard Atmosphere (ISA) conditions.

Thermal Expansion: Most materials expand when heated and contract when cooled. The change in length (ΔL) of a solid is given by:

ΔL = α × L₀ × ΔT

Where:

Example – Aluminium Tail Boom: An aluminium tail boom (α = 23 × 10⁻⁶ /°C) of length 4.5 m, experiencing a temperature rise from 15°C to 45°C (ΔT = 30°C), will expand by ΔL = 23 × 10⁻⁶ × 4.5 × 30 = 0.003105 m = 3.1 mm. This expansion must be accommodated in control cable tensions and structural joints.

Example – Percentage Change: A rotor blade experiencing a temperature drop of 40 K will contract by a fractional amount ΔL/L = α × ΔT = 23 × 10⁻⁶ × 40 = 0.00092, or 0.092%.

Expansion of Liquids: Liquids also expand when heated. Hydraulic fluid reservoirs must have sufficient expansion space to accommodate the increased volume when the system is hot. A completely full cold reservoir can lead to excessive pressure and system failure when the fluid heats up.

Combustion and Gas Turbine Operation: In a gas turbine engine, the exhaust gas temperature (EGT) is a key performance indicator. A lean fuel/air mixture (excess air) burns hotter, leading to higher EGT. A rich mixture (excess fuel) has a cooling effect due to the latent heat of vaporisation of the excess fuel, resulting in lower EGT. A higher-than-normal EGT for a given power setting is a classic symptom of a lean condition.

2.7 Electrical Fundamentals

Basic electrical theory is part of Module 2. Key relationships are defined by Ohm's Law and the power equation.

Example – Navigation Light: A 28 V DC navigation light with a resistance of 4 ohms dissipates power P = V² / R = (28)² / 4 = 784 / 4 = 196 W. Alternatively, the current is I = V / R = 7 A, and power is P = V × I = 28 × 7 = 196 W.

3. Important Formulas and Procedures

4. Common Relationships Between Concepts

5. Typical Exam Focus Points

For the EASA Part-66 Module 2 exam (B3 category), candidates should focus on the following areas, which are frequently tested: