FAA General Written TestChapter 8 · 40 practice questions

Chapter 8: Mathematics

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Chapter: Mathematics for Aircraft Maintenance Engineers

Overview

Mathematics is the foundational language of aircraft maintenance engineering. Every task an AME performs—from measuring fuel lines and calculating wire resistance to determining aircraft center of gravity and torque specifications—requires precise mathematical computation. This chapter covers the essential mathematical concepts and applications that appear throughout aviation maintenance, including unit conversions, geometric calculations, statistical analysis, percentage calculations, and the fundamental formulas governing electrical and mechanical systems.

The ability to perform accurate calculations is not merely an academic exercise; it directly impacts flight safety. An incorrectly calculated cable stretch, an improperly sized drill bit, or a miscalculated center of gravity can lead to catastrophic failure. This chapter equips you with the mathematical tools and understanding necessary to perform maintenance tasks correctly and document results in accordance with FAA regulations.

Unit Conversion and Dimensional Analysis

The Metric-Imperial Conversion System

Aircraft maintenance documentation frequently requires converting between metric and imperial units. The aviation industry predominantly uses the imperial system (inches, feet, pounds) in the United States, but many components and international aircraft use metric dimensions. The fundamental conversion factors every AME must know include:

  • 1 inch = 2.54 centimeters (exact conversion)
  • 1 foot = 12 inches
  • 1 meter = 100 centimeters
  • 1 pound = 16 ounces
  • 1 kilogram ≈ 2.2046 pounds

Converting Between Systems

Converting Between Systems: 2.5 Meters to Inches UNIT CONVERSION CHAIN — METERS TO INCHES START 2.5 meters Given value ×100 STEP 1 250 cm meters → centimeters ÷2.54 RESULT 98.4 in centimeters → inches 2.5 m 98.4 in FUEL LINE SAMPLE 98.4 inches DIMENSIONAL ANALYSIS — UNITS CANCEL 2.5 m × 100 cm × 1 in = 98.4 in 1 m 2.54 cm m cm CONVERSION FACTORS 1 m = 100 cm 1 in = 2.54 cm FAA A&P Exam Prep — Mathematics: Converting Between Systems (AC 43.13-1B)

When converting measurements, always work systematically through the units. For example, converting 2.5 meters to inches requires a two-step process:

  1. Convert meters to centimeters: 2.5 m × 100 cm/m = 250 cm
  2. Convert centimeters to inches: 250 cm ÷ 2.54 cm/in = 98.4 inches

Critical Application: Fluid line fabrication and routing require exact dimensional accuracy. A fuel line cut even slightly too short or too long can result in improper fit, inadequate clearance, or stress on fittings. Per AC 43.13-1B Chapter 8, all dimensional conversions must be verified before cutting or bending any fluid line.

Fraction-Decimal Equivalents

Fraction-Decimal Equivalents: 3/16" = 0.1875" Drill Bit Selection FRACTION-DECIMAL EQUIVALENTS — 3/16" = 0.1875" 0 1 2 3 RULER (inches) 3/16 inch mark on precision ruler 0.1875 MICROMETER (0.0001" resolution) Micrometer reads 0.1875 inches #11 DRILL 3/16" = 0.1875" Correct drill bit for structural repair (AC 43.13-1B) EQUIVALENT VALUES 3/16 0.1875 CONVERSION RULE Numerator ÷ Denominator 3 ÷ 16 = 0.1875 AC 43.13-1B Ch. 4 DRILL BIT EQUIVALENTS FRACTION DECIMAL 5/32 0.1562 3/16 0.1875 7/32 0.2188 FAA A&P Exam Prep — Mathematics: Fraction-Decimal Equivalents | 14 CFR Part 65, AC 43.13-1B

Aircraft maintenance requires fluency in converting between fractions and decimals, particularly when selecting drill sizes, fasteners, and measuring components. Common equivalents include:

FractionDecimal
1/160.0625
1/80.125
3/160.1875
1/40.25
5/160.3125
3/80.375
7/160.4375
1/20.5

Critical Application: Drill bit selection for structural repairs demands precision. A 0.1875-inch hole requires a 3/16-inch drill bit. Using an incorrect drill size creates oversized holes that reduce joint strength, promote fastener loosening, and can lead to crack formation in sheet metal structures. AC 43.13-1B Chapter 4 provides detailed guidance on proper hole preparation.

Geometric Calculations

Area of a Circle

Area of a Circle — Hydraulic Line Sizing (FAA A&P Mathematics) Area of a Circle — A = π × r² | Hydraulic Line Sizing Circle Area Visualization r = 0.75 in d = 1.5 in A = π × r² r² = 0.75² = 0.5625 in² A = 3.1416 × 0.5625 A = 1.767 in² Hydraulic Line Sizing — AC 43.13-1B Fluid Line Cross-Section (0.75 in ID) Flow Rate Relationship Q = A × V (Continuity Equation) Larger area → Higher flow at same velocity A = 1.767 in² at 0.75 in line ID Pressure Drop Consideration ΔP ∝ 1/r⁴ — Small radius, large ΔP Reference: AC 43.13-1B Chapter 6 ⚠ Undersized Lines Restrict Flow Minimum area per FAA guidance: 0.75 in line → A ≥ 0.44 in² required FAA A&P General Mathematics — Area of a Circle | 14 CFR Part 65 | AC 43.13-1B

The cross-sectional area of circular components—fluid lines, wire conductors, and structural members—is calculated using:

A = π × r²

Where:

  • A = area (square inches or square centimeters)
  • π ≈ 3.14 (or 3.1416 for greater precision)
  • r = radius (half the diameter)

For a fluid line with an inside diameter of 0.75 inches:

  • Radius = 0.75 ÷ 2 = 0.375 inches
  • Area = 3.14 × (0.375)² = 3.14 × 0.1406 = 0.442 square inches

Critical Application: Hydraulic system performance depends on accurate cross-sectional area calculations. The area determines fluid flow rate, pressure drop, and system response time. An undersized line restricts flow and causes excessive pressure drops; an oversized line adds unnecessary weight and cost. AC 43.13-1B Chapter 8 specifies the relationship between line size and system requirements.

Other Geometric Formulas

  • Circumference of a circle: C = π × d (where d = diameter)
  • Area of a rectangle: A = length × width
  • Volume of a cylinder: V = π × r² × h (where h = height or length)
  • Area of a triangle: A = ½ × base × height

Statistical Analysis in Maintenance

Calculating Averages

Calculating Averages - Brake Disc Wear Measurement Calculating Averages — Brake Disc Wear Measurement P1 P2 P3 P4 MICROMETER Brake Disc (rotor) SUM 0.825 in AVERAGE (÷ 4) 0.206 in MIN SPEC (AC 43.13-1B) 0.200 in 0.206 ≥ 0.200 ✓ WITHIN LIMITS UNEVEN WEAR CHECK ⚠ MAX VARIANCE: 0.015 in Individual Readings P1: 0.208 P2: 0.205 P3: 0.207 P4: 0.205 Average of multiple measurements evaluates overall wear per AC 43.13-1B §5-3 Measurement point Within limits

The arithmetic mean (average) is used extensively in maintenance to evaluate wear patterns, measurement consistency, and system performance. The formula is:

Average = Sum of all measurements ÷ Number of measurements

For example, measuring brake disc thickness at four points:

  • Measurements: 0.250, 0.248, 0.251, 0.249 inches
  • Sum: 0.250 + 0.248 + 0.251 + 0.249 = 0.998 inches
  • Average: 0.998 ÷ 4 = 0.2495 inches (rounds to 0.250 inches)

Critical Application: Brake disc wear evaluation requires measuring thickness at multiple points around the disc to detect uneven wear. The average thickness is compared against manufacturer minimum specifications. Per AC 43.13-1B Chapter 12, brake components must be replaced when wear exceeds limits, regardless of whether the average or any individual measurement exceeds the threshold.

Rounding Conventions

When performing maintenance calculations, follow these rounding rules:

  • Round to the precision specified in the maintenance manual
  • For measurements, round to the nearest thousandth of an inch (0.001)
  • For percentages, round to one decimal place unless otherwise specified
  • Never round intermediate values—only round the final answer

Percentage Calculations

Basic Percentage Formula

Percentage = (Part ÷ Whole) × 100

This fundamental formula applies to numerous maintenance scenarios, including compliance tracking, wear measurement, and performance evaluation.

Compliance Tracking

When calculating fleet compliance with airworthiness directives (ADs) or service bulletins:

Compliance % = (Aircraft completed ÷ Total aircraft) × 100

Example: 18 of 24 aircraft modified

  • Percentage = (18 ÷ 24) × 100 = 75%

Critical Application: Per 14 CFR 91.417, accurate AD status records must be maintained for every aircraft. Misreporting compliance percentages can result in operating aircraft with unresolved airworthiness issues, creating serious safety and legal consequences.

Elongation and Stretch Calculations

Control cable stretch is a critical safety parameter. The percentage of elongation is calculated as:

Elongation % = [(Final length - Original length) ÷ Original length] × 100

Example 1: Cable stretched from 10 feet to 10 feet 1 inch

  • Stretch = 1 inch = 1/12 foot = 0.0833 feet
  • Elongation % = (0.0833 ÷ 10) × 100 = 0.833%

Example 2: Cable stretched from 120 inches to 121.2 inches

  • Elongation % = (121.2 - 120) ÷ 120 × 100 = 1.0%

Critical Application: Per AC 43.13-1B Chapter 7, excessive cable stretch affects control rigging and can compromise flight control integrity. Cables that exceed manufacturer-specified elongation limits must be replaced. The difference between "stretch" (permanent deformation) and "tension" (elastic deformation under load) is important—tensioning procedures must account for both.

Electrical Calculations

Ohm's Law

Ohm's Law is the cornerstone of all electrical troubleshooting:

I = V ÷ R

Where:

  • I = Current in amperes (amps)
  • V = Voltage in volts
  • R = Resistance in ohms

Inverse Relationship: With constant voltage, current is inversely proportional to resistance. If resistance is halved, current doubles; if resistance is doubled, current halves.

Example: Wire resistance reduced from 0.5 ohms to 0.25 ohms at constant voltage

  • Current doubles (2× increase)

Critical Application: Understanding this relationship is essential when replacing wiring. A lower-resistance wire carries more current, which may overload downstream components. Conversely, a higher-resistance wire reduces current, potentially causing system malfunction. Per AC 43.13-1B Chapter 11, replacement wiring must match the original specifications for resistance, current-carrying capacity, and voltage drop.

Resistance of Wire

Resistance of Wire - Length and Cross-Sectional Area Effects Resistance of Wire — Length vs. Cross-Sectional Area LENGTH: Longer Wire → Higher Resistance 50 ft (short) 100 ft (long) OHMS 0.05 CROSS-SECTION: Thicker Wire → Lower Resistance AWG #22 (thin) AWG #10 (thick) Resistance 16.5 Ω 1.0 Ω RESISTANCE PER 1000 FEET — 50-FOOT WIRE CALCULATION R = (ρ × L) / A Where: R = Resistance (Ω), ρ = Resistivity (Ω·cmil/ft), L = Length (ft), A = Cross-sectional area (cmil) Copper (Cu): ρ = 10.4 Ω·cmil/ft Aluminum (Al): ρ = 17.0 Ω·cmil/ft AWG #10 copper: A = 10,380 cmil EXAMPLE: 50 ft of AWG #10 Copper Wire R/1000ft = (10.4 × 1000) / 10,380 R/1000ft = 10,400 / 10,380 = 1.002 Ω For 50 ft: R = 1.002 × (50/1000) R = 1.002 × 0.05 = 0.0501 Ω R ≈ 0.05 Ω for 50 ft of #10 AWG Cu Per AC 43.13-1B Chapter 11, Table 11-1 14 CFR §43.13 / AC 43.13-1B — Acceptable methods for aircraft electrical wire sizing

Total wire resistance is calculated as:

R_total = Resistance per foot × Length in feet

Example: 14 AWG copper wire, 50 feet long, 0.00162 ohms per foot

  • R_total = 0.00162 × 50 = 0.081 ohms

Critical Application: Excessive resistance causes voltage drops and overheating. During 100-hour inspections, measuring wire resistance helps identify corrosion, damaged conductors, or loose connections. The resistance per foot varies with wire gauge (AWG), material (copper vs. aluminum), and temperature.

Electrical Power

P = V × I

Where:

  • P = Power in watts
  • V = Voltage in volts
  • I = Current in amps

Example: 24-volt circuit drawing 8 amps

  • P = 24 × 8 = 192 watts

Critical Application: Power calculations determine whether a circuit is overloaded. Exceeding the rated power of a component or circuit can cause overheating, insulation failure, and fire. Per FAA-H-8083-30 and AC 43.13-1B Chapter 11, all electrical components must be derated appropriately and protected with correctly sized circuit breakers or fuses.

Combined Electrical Formulas

Additional relationships derived from Ohm's Law and power formulas:

  • P = I² × R (power in terms of current and resistance)
  • P = V² ÷ R (power in terms of voltage and resistance)
  • V = I × R (voltage in terms of current and resistance)

Weight and Balance Calculations

Center of Gravity Determination

The center of gravity (CG) is calculated using the moment method:

CG = Total Moment ÷ Total Weight

Where:

  • Moment = Weight × Arm (distance from datum)
  • Total Moment = Sum of all individual moments
  • Total Weight = Sum of all individual weights

Example: Aircraft with nose wheel and two main wheels

  • Nose wheel: 550 lb at 20 inches from datum
  • Each main wheel: 1,000 lb at 78 inches from datum
  • Total weight = 550 + 1,000 + 1,000 = 2,550 lb
  • Total moment = (550 × 20) + (1,000 × 78) + (1,000 × 78) = 11,000 + 78,000 + 78,000 = 167,000 lb-in
  • CG = 167,000 ÷ 2,550 = 65.5 inches from datum

Critical Application: Accurate CG determination is essential for aircraft stability and control. Per 14 CFR Part 43.9 and FAA Form 337 requirements, weight and balance records must be updated after any major repair or alteration. A misplaced decimal point or arithmetic error can produce an incorrect CG that places the aircraft outside its certified envelope.

Moment Calculations

Moment = Weight × Arm

Example: 25-pound component at 60.0 inches arm

  • Moment = 25 × 60.0 = 1,500 pound-inches

Critical Application: When recording weight and balance data, always verify units and arithmetic. A single misplaced decimal point can shift the CG significantly. Double-check all calculations and have them verified by a second person when possible.

Mechanical Calculations

Torque Calculations

Torque Calculations - Force × Distance Relationship TORQUE CALCULATIONS T = F × d | FAA A&P General Mathematics (14 CFR §65.75, AC 43.13-1B) PIVOT d = d = 12 in F = 50 lb T = 600 in-lb TORQUE FORMULA T = F × d T = torque (in-lb / ft-lb) F = force (lb) d = distance (in/ft) UNITS (AC 43.13-1B) 1 ft-lb = 12 in-lb 1 N·m = 8.851 in-lb Always verify spec units first KEY POINTS • Torque = force × lever arm • Longer arm = more torque • Calibrate wrench regularly EXAMPLE: Cylinder head bolt requires 240 in-lb (20 ft-lb). Using a 12-in wrench: F = T ÷ d = 240 ÷ 12 = 20 lb | With 6-in extension: F = 240 ÷ 18 = 13.3 lb Common torque specs (14 CFR §43.13): Spark plugs: 30 ft-lb (360 in-lb) Oil drain plug: 18-22 ft-lb Propeller bolts: 25-30 ft-lb F = 50 lb Never exceed spec 90° ± 5° Interactive torque visualization — adjust force or distance to see torque change | FAA A&P General Mathematics

The torque required for a fastener is calculated using:

T = K × D × P

Where:

  • T = Torque in inch-pounds
  • K = Torque coefficient (friction factor, typically 0.15–0.25)
  • D = Nominal diameter in inches
  • P = Clamp load (preload) in pounds

Example: 0.50-inch bolt, K = 0.20, clamp load = 4,000 lb

  • T = 0.20 × 0.50 × 4,000 = 400 inch-pounds
  • Convert to foot-pounds: 400 ÷ 12 = 33.3 ft-lb

Critical Application: Per AC 43.13-1B Chapter 7, accurate torque application prevents fastener failure from undertorquing (loosening) or overtorquing (stripping or breaking). The torque coefficient varies with lubrication, surface finish, and thread condition. Always consult the maintenance manual for the correct K value and torque specification.

Speed and Velocity

Speed = Distance ÷ Time

Example: Actuator stroke of 12 inches in 8 seconds

  • Speed = 12 ÷ 8 = 1.5 inches per second

Critical Application: Actuator extension speed must match specifications for proper system operation. Per AC 43.13-1B Chapter 10, hydraulic actuators that extend too quickly or too slowly indicate internal leakage, incorrect orifice sizing, or system pressure problems.

Documentation and Regulatory Requirements

FAA Form 337

Major repairs and alterations require documentation on FAA Form 337. This form must include:

  • Accurate dimensional measurements before and after repair
  • Weight and balance data
  • Materials used
  • Compliance with applicable regulations

Per 14 CFR Part 43, Appendix A, major repairs include structural repairs that affect the aircraft's strength, flight characteristics, or airworthiness. The difference between original and repaired dimensions must be precisely recorded.

Maintenance Record Requirements

Per 14 CFR 91.417, maintenance records must include:

  • AD compliance status
  • Total time in service
  • Current status of life-limited parts
  • Major alterations and repairs

Measurement Precision

The precision required for measurements varies by application:

  • Structural dimensions: ±0.01 inch or better
  • Brake disc thickness: ±0.001 inch
  • Cable tension: ±5% of specified value
  • Torque: ±5% of specified value

Common Relationships and Connections

Interconnected Concepts

Understanding how mathematical concepts relate to each other enhances troubleshooting ability:

  1. Geometry ↔ Fluid Systems: Cross-sectional area determines flow rate, which affects actuator speed
  2. Electrical ↔ Mechanical: Power (watts) relates to mechanical work; torque and RPM determine horsepower
  3. Statistics ↔ Wear Monitoring: Average measurements track wear trends, predicting component life
  4. Percentages ↔ Compliance: AD compliance percentages guide maintenance scheduling

Unit Consistency

Always maintain consistent units throughout calculations:

  • Convert all measurements to the same unit system before calculating
  • Verify that torque is in the correct units (inch-pounds vs. foot-pounds)
  • Check that moment calculations use consistent units (pound-inches vs. pound-feet)

Verification Procedures

Professional AMEs always verify calculations:

  1. Perform the calculation twice using different methods when possible
  2. Check that the answer is reasonable (order-of-magnitude check)
  3. Have critical calculations verified by a second technician
  4. Document all calculations in maintenance records

Summary

Mathematics in aircraft maintenance is not abstract theory—it is a practical tool applied daily to ensure airworthiness. Mastery of unit conversions, geometric formulas, statistical analysis, percentage calculations, and electrical/mechanical formulas is essential for:

  • Fabricating and installing components correctly
  • Troubleshooting system malfunctions
  • Maintaining accurate maintenance records
  • Ensuring compliance with FAA regulations
  • Protecting the safety of passengers and crew

The formulas and concepts in this chapter appear throughout the FAA General curriculum and in daily maintenance practice. Practice these calculations until they become second nature, and always verify your work before signing off any maintenance task.

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