10.2 Geometric Computations (Area, Volume & Cylinder Displacement)

FAA Aviation Mechanic — General AMG

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1 Chapter 1: Introduction & AMG Exam Overview

1.1 About the FAA Aviation Mechanic General Knowledge Test 1.2 How to Study for the AMG Exam

2 Chapter 2: Maintenance Forms, Records, Publications & Regulations

2.1 14 CFR Parts 43, 65, and 91 Regulatory Framework 2.2 Aviation Mechanic Privileges, Limitations & Recent Experience 2.3 Maintenance Recordkeeping & Return-to-Service Entries (14 CFR 43.9 & 43.11) 2.4 FAA Form 337 Major Repairs and Major Alterations 2.5 Airworthiness Directives (ADs) & Mandatory Compliance 2.6 Type Certificate Data Sheets (TCDS), STCs, ACs & Maintenance Manuals 2.7 FAA Forms, MMEL/MEL, Inoperative Equipment & Parts Effectivity

3 Chapter 3: Basic Electricity

3.1 DC Electrical Circuits, Ohm's Law & Power 3.2 AC Electrical Fundamentals, Reactance & Impedance 3.3 Aircraft Lead-Acid & Nickel-Cadmium Storage Batteries 3.4 DC Generators, Alternators & Electric Motors 3.5 Aircraft Wiring, Bonding & Circuit Protection Devices 3.6 Resistor Color Codes, Semiconductors, Digital Logic & Electrostatic Discharge

4 Chapter 4: Materials & Processes

4.1 Aircraft Ferrous and Nonferrous Metals & Heat Treatment 4.2 Metal Hardness Testing (Rockwell, Brinell, Vickers) 4.3 Nondestructive Testing (NDT/NDI) Inspection Methods 4.4 Aircraft Hardware, Fasteners & Safetying Methods 4.5 Composite Materials, Plastics & Control Cables 4.6 Soldering, Welding & Weld Quality Inspection

5 Chapter 5: Cleaning & Corrosion Control

5.1 Corrosion Types, Chemistry & Identification 5.2 Corrosion Removal, Chemical Conversion Coatings & Surface Treatments 5.3 Aircraft Cleaning Agents, Precautions & Preventative Maintenance 5.4 Aircraft Finishing Systems, Registration Markings & Placards

6 Chapter 6: Weight & Balance

6.1 Weight and Balance Terminology, Datum, Arm & Moment Principles 6.2 Aircraft Weighing Procedures, Scale Reactions & Tare Weight 6.3 Center of Gravity Computations, Extreme Loading & Ballast

7 Chapter 7: Ground Operation, Servicing & Safety

7.1 Engine Starting Procedures, Danger Areas & Ground Taxi Safety 7.2 Aircraft Servicing (Oxygen, Hydraulic, Pneumatic & De-Icing Fluids) 7.3 Aircraft Fueling Safety, Static Grounding & Hangar Fire Protection 7.4 Airport Movement Areas, Hazardous Materials, Tool Control & FOD Prevention

8 Chapter 8: Aircraft Drawings

8.1 Aircraft Drawing Types, Title Blocks & Orthographic Projections 8.2 Line Conventions, Dimensions, Tolerances & Material Symbols 8.3 Electrical Wiring Diagrams, Logic Flowcharts & Fluid Schematics

9 Chapter 9: Fluid Lines & Fittings

9.1 Rigid Metal Tubing Fabrication, Bending & 37° AN Flaring 9.2 Flexible Hose Construction, Pressure Ratings & Lay Lines 9.3 AN/MS/Quick-Disconnect Fittings, Identification & Installation

10 Chapter 10: Mathematics for Aviation Maintenance

10.1 Fractions, Decimals, Percentages & Ratio/Proportion 10.2 Geometric Computations (Area, Volume & Cylinder Displacement) 10.3 Algebraic Equations, Powers, Roots & Maintenance Conversions

11 Chapter 11: Physics for Aviation Maintenance

11.1 Matter, Mass, Weight, Density & Specific Gravity 11.2 Work, Energy, Power & Simple Machines (Mechanical Advantage) 11.3 Fluid Mechanics, Pascal's Law & Bernoulli's Principle 11.4 Thermodynamics, Heat Transfer, Gas Laws, Sound & Aerodynamics 11.5 Standard Atmosphere, Density & Pressure Altitude, Flight Controls & Aerodynamic Devices

12 Chapter 12: Inspection Concepts, Techniques & Precision Measurement

12.1 Aircraft Inspection Programs & Inspection Concepts 12.2 Precision Measuring Tools & Tool Calibration

13 Chapter 13: Human Factors in Aviation Maintenance

13.1 Human Error, the Dirty Dozen & Maintenance Safety Culture 13.2 Human Performance, Limitations, Fatigue & the Work Environment 13.3 Communication, Shift Turnover, Event Investigation & Professionalism

10.2 Geometric Computations (Area, Volume & Cylinder Displacement)

Key Takeaways Explain This Topic AI

Plane geometric area calculations form the basis for aerodynamic wing loading, hydraulic piston force generation, and structural sheet metal repair patch sizing.

The area of a circle is computed as A = π r 2 = π 4 D 2 ≈ 0.7854 D 2 A = \pi r^2 = \frac{\pi}{4} D^2 \approx 0.7854 D^2 A = π r 2 = 4 π ​ D 2 ≈ 0.7854 D 2 ; the constant 0.7854 0.7854 0.7854 represents π / 4 \pi / 4 π /4 and is the standard FAA formulation for piston cross-sectional area and cylinder bore computations.

Total piston displacement of a reciprocating engine is the volume swept by all pistons during one stroke: V total = 0.7854 × Bore 2 × Stroke × N cylinders V_{\text{total}} = 0.7854 \times \text{Bore}^2 \times \text{Stroke} \times N_{\text{cylinders}} V total ​ = 0.7854 × Bore 2 × Stroke × N cylinders ​ , expressed in cubic inches ( in 3 \text{in}^3 in 3 ) or cubic centimeters ( cm 3 \text{cm}^3 cm 3 / liters).

The compression ratio (CR) of an internal combustion engine is the ratio of total cylinder volume with the piston at Bottom Dead Center (BDC) to the clearance volume with the piston at Top Dead Center (TDC): CR = V D + V C L V C L = V total cylinder V clearance \text{CR} = \frac{V_D + V_{CL}}{V_{CL}} = \frac{V_{\text{total cylinder}}}{V_{\text{clearance}}} CR = V C L ​ V D ​ + V C L ​ ​ = V clearance ​ V total cylinder ​ ​ .

Solid geometric volume formulas determine fuel tank capacities ( 1 U.S. gal = 231 in 3 1\text{ U.S. gal} = 231\text{ in}^3 1 U.S. gal = 231 in 3 ), cargo bay hold volumes, accumulator pre-charge chambers, and structural fluid reservoirs across rectangular prisms ( V = L × W × H V = L \times W \times H V = L × W × H ), cylinders ( V = 0.7854 D 2 H V = 0.7854 D^2 H V = 0.7854 D 2 H ), and spheres ( V = 4 3 π r 3 V = \frac{4}{3} \pi r^3 V = 3 4 ​ π r 3 ).

Last updated: August 2026

10.2 Geometric Computations (Area, Volume & Cylinder Displacement)

Geometric mathematics enables aviation maintenance technicians to quantify two-dimensional surface areas and three-dimensional volumetric spaces. Technicians apply geometric equations daily when calculating the required surface area of a sheet metal doubler patch, determining the wing planform loading of an aircraft, sizing hydraulic actuator pistons under Pascal's principle, and computing the swept displacement and compression ratio of reciprocating aircraft engines. The FAA exam emphasizes precise understanding of both plane and solid geometric formulas as detailed in FAA-H-8083-30B (Aviation Maintenance Technician Handbook — General) .

1. Plane Geometry: 2D Surface Area Computations

Area is the measure of two-dimensional surface space enclosed within a boundary, expressed in square units (square inches in 2 \text{in}^2 in 2 , square feet ft 2 \text{ft}^2 ft 2 , or square millimeters mm 2 \text{mm}^2 mm 2 ). PRIMARY TWO-DIMENSIONAL GEOMETRIC SHAPES

• Rectangle / Square Triangle Trapezoid
• ┌──────────────────┐ /\ ┌──────────┐ (b)
• │ │ / \ / \
• (W)│ A = L × W │ (h)/ \ (h) (h)/ A = (a+b)h \
• │ │ / A = ½bh / 2 \
• └──────────────────┘ /________\ /──────────────────\
• (L) (b) (a)

Circle

. --- .

• / │ \
• | (r)│ | A = π·r²
• | o──────| A = (π/4)·D² = 0.7854·D²
• \ /
• ' --- '
• (D)
• Area Formulas for Plane Shapes
• Rectangle: A = L × W A = L \times W A = L × W

where L L L is length and W W W is width.

2. Square: A = s 2 A = s^2 A = s 2

where s s s is the side length.

3. Triangle: A = 1 2 b h = b × h 2 A = \frac{1}{2} b h = \frac{b \times h}{2} A = 2 1 ​ bh = 2 b × h ​

where b b b is the base length and h h h is the true vertical height (altitude) perpendicular to the base.

4. Trapezoid: A four-sided polygon possessing exactly one pair of parallel sides (bases a a a and b b b ): A = ( a + b 2 ) h = ( a + b ) × h 2 A = \left(\frac{a + b}{2}\right) h = \frac{(a + b) \times h}{2} A = ( 2 a + b ​ ) h = 2 ( a + b ) × h ​

where a a a and b b b are the lengths of the parallel top and bottom edges, and h h h is the perpendicular distance between them. Tapered wing planforms and sheet metal doubler plates are calculated using the trapezoid formula.

5. Circle: A = π r 2 = π ( D 2 ) 2 = π 4 D 2 A = \pi r^2 = \pi \left(\frac{D}{2}\right)^2 = \frac{\pi}{4} D^2 A = π r 2 = π ( 2 D ​ ) 2 = 4 π ​ D 2 Since π 4 = 3.14159265 4 ≈ 0.7854 : A = 0.7854 D 2 \text{Since } \frac{\pi}{4} = \frac{3.14159265}{4} \approx 0.7854: \quad A = 0.7854 D^2 Since 4 π ​ = 4 3.14159265 ​ ≈ 0.7854 : A = 0.7854 D 2

Radius ( r r r ): The distance from the center point to the outer perimeter.

Diameter ( D D D ): The total distance across the circle through the center ( D = 2 r D = 2r D = 2 r ).

Circumference ( C C C ): The linear distance around the perimeter of the circle: C = π D = 2 π r C = \pi D = 2\pi r C = π D = 2 π r .

The Aviation Constant 0.7854 0.7854 0.7854 : FAA maintenance examinations standardly express circular cross-sectional areas using diameter rather than radius: A = 0.7854 D 2 A = 0.7854 D^2 A = 0.7854 D 2 . This formulation eliminates the step of halving the diameter into a radius, directly utilizing the measured cylinder bore or piston diameter.

Practical Aviation Surface Area Applications

Wing Planform Area ( S S S ): The projected gross surface area of an aircraft wing including the area covered by the fuselage. Wing Loading ( W L WL W L ) equals total aircraft weight divided by wing area: Wing Loading = Gross Weight ( W ) Wing Area ( S ) ( expressed in lb/ft 2 ) \text{Wing Loading} = \frac{\text{Gross Weight } (W)}{\text{Wing Area } (S)} \quad \left(\text{expressed in } \text{lb/ft}^2\right) Wing Loading = Wing Area ( S ) Gross Weight ( W ) ​ ( expressed in lb/ft 2 )

Hydraulic Actuator Piston Force (Pascal's Law): Force ( F ) = Pressure ( P ) × Piston Area ( A ) \text{Force } (F) = \text{Pressure } (P) \times \text{Piston Area } (A) Force ( F ) = Pressure ( P ) × Piston Area ( A ) F = P × ( 0.7854 D 2 ) F = P \times (0.7854 D^2) F = P × ( 0.7854 D 2 )

Sheet Metal Repair Patch Sizing: AC 43.13-1B requires replacement doublers to possess equal or greater cross-sectional area and skin contact area than the damaged material removed.

Loading diagram... Geometric Area and Solid Volume Relationships in Aviation

2. Solid Geometry: 3D Volume Computations

Volume is the measure of three-dimensional space occupied by a solid body or enclosed fluid container, expressed in cubic units (cubic inches in 3 \text{in}^3 in 3 , cubic feet ft 3 \text{ft}^3 ft 3 , or cubic centimeters cm 3 \text{cm}^3 cm 3 ). PRIMARY THREE-DIMENSIONAL SOLID BODIES

Rectangular Prism Cylinder Sphere

┌──────────────────┐ . --- .

• / /│ / │ \
• ┌──────────────────┐ │ (D)| │(H) | / \
• │ │ │ | │ | | o────| (r)
• │ V = L × W × H │ │ \ │ / \ /
• │ │/ ' --- '
• └──────────────────┘
• (L) V = 0.7854 · D² · H V = (4/3) · π · r³
• Volume Formulas for Solid Shapes
• Rectangular Solid (Prism): V = L × W × H V = L \times W \times H V = L × W × H

where L L L is length, W W W is width, and H H H is height/depth. Used for cargo hold volumes, avionics rack cooling volumes, and rectangular fuel cells.

2. Cylinder: V = Base Area × Height = π r 2 H = π 4 D 2 H = 0.7854 D 2 H V = \text{Base Area} \times \text{Height} = \pi r^2 H = \frac{\pi}{4} D^2 H = 0.7854 D^2 H V = Base Area × Height = π r 2 H = 4 π ​ D 2 H = 0.7854 D 2 H

where D D D is internal barrel diameter and H H H is axial height/stroke. Used for engine cylinders, hydraulic accumulators, and landing gear shock struts.

3. Sphere: V = 4 3 π r 3 = 4 3 π ( D 2 ) 3 = π 6 D 3 ≈ 0.5236 D 3 V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \left(\frac{D}{2}\right)^3 = \frac{\pi}{6} D^3 \approx 0.5236 D^3 V = 3 4 ​ π r 3 = 3 4 ​ π ( 2 D ​ ) 3 = 6 π ​ D 3 ≈ 0.5236 D 3

where r r r is radius and D D D is diameter. Used for spherical fire extinguishing bottles (Halon/Novec), high-pressure gaseous oxygen spheres, and spherical hydraulic accumulator pressure vessels.

Liquid Volumetric Conversion Constants

Standard U.S. Gallon: Exactly 231.0 cubic inches 231.0\text{ cubic inches} 231.0 cubic inches ( 231 in 3 231\text{ in}^3 231 in 3 ). Volume in U.S. Gallons = Volume in Cubic Inches ( in 3 ) 231 \text{Volume in U.S. Gallons} = \frac{\text{Volume in Cubic Inches } (\text{in}^3)}{231} Volume in U.S. Gallons = 231 Volume in Cubic Inches ( in 3 ) ​ Volume in Cubic Inches = Volume in U.S. Gallons × 231 \text{Volume in Cubic Inches} = \text{Volume in U.S. Gallons} \times 231 Volume in Cubic Inches = Volume in U.S. Gallons × 231

Cubic Feet to Gallons: 1 ft 3 = 1 , 728 in 3 = 7.4805 U.S. Gallons 1\text{ ft}^3 = 1,728\text{ in}^3 = 7.4805\text{ U.S. Gallons} 1 ft 3 = 1 , 728 in 3 = 7.4805 U.S. Gallons .

Metric Volume Equivalence: 1 Liter = 1 , 000 cm 3 = 61.024 in 3 1\text{ Liter} = 1,000\text{ cm}^3 = 61.024\text{ in}^3 1 Liter = 1 , 000 cm 3 = 61.024 in 3 ; 1 U.S. Gallon = 3.7854 Liters 1\text{ U.S. Gallon} = 3.7854\text{ Liters} 1 U.S. Gallon = 3.7854 Liters .

3. Reciprocating Engine Piston Displacement

Piston Displacement (or swept volume) is the total volume of air-fuel mixture displaced by an engine piston as it moves from its lowest position, Bottom Dead Center (BDC) , to its highest position, Top Dead Center (TDC) , during a single stroke. ENGINE CYLINDER PISTON DISPLACEMENT GEOMETRY

• ┌────────────────────────────────┐
• │ Clearance Volume (V_CL) │ <--- Space at TDC
• Top Dead ──────┼───┬────────────────────────┬───┼────── (Spark Plug / Valves)
• Center (TDC) │ │ Piston Crown │ │
• │ └────────────────────────┘ │ ▲
• │ │ │ │
• │ │ (Connecting │ │ Piston Stroke (L)
• │ │ Rod) │ │ (BDC to TDC Distance)
• │ ▼ │ │
• │ ┌────────────────────────┐ │ ▼
• Bottom Dead ───┼───┼───── Piston Crown ─────┼───┼──────
• Center (BDC) │ └────────────────────────┘ │
• │ │
• └────────────────────────────────┘
• ◄──────── Cylinder Bore (D) ─────►
• Definitions of Engine Parameters

Cylinder Bore ( D D D ): The inside diameter of the cylinder barrel, measured in inches (or millimeters).

Stroke ( L L L ): The linear distance traveled by the piston from TDC to BDC. The stroke is mechanically determined by the crankshaft throw geometry: Stroke = 2 × Crankshaft Throw Radius \text{Stroke} = 2 \times \text{Crankshaft Throw Radius} Stroke = 2 × Crankshaft Throw Radius .

Number of Cylinders ( N N N ): The total count of functioning power cylinders in the engine block ( 4 , 6 , 8 , 9 , 12 , 14 , 18 , 28 4, 6, 8, 9, 12, 14, 18, 28 4 , 6 , 8 , 9 , 12 , 14 , 18 , 28 ).

Piston Displacement Formulas

Displacement of a Single Cylinder ( V D V_D V D ​ ): V D = Cross-Sectional Area × Stroke = 0.7854 × Bore 2 × Stroke V_D = \text{Cross-Sectional Area} \times \text{Stroke} = 0.7854 \times \text{Bore}^2 \times \text{Stroke} V D ​ = Cross-Sectional Area × Stroke = 0.7854 × Bore 2 × Stroke

Total Engine Piston Displacement ( V total V_{\text{total}} V total ​ ): V total = V D × N cylinders = 0.7854 × Bore 2 × Stroke × N V_{\text{total}} = V_D \times N_{\text{cylinders}} = 0.7854 \times \text{Bore}^2 \times \text{Stroke} \times N V total ​ = V D ​ × N cylinders ​ = 0.7854 × Bore 2 × Stroke × N

where:

• V total V_{\text{total}} V total ​ is total engine displacement in cubic inches ( in 3 \text{in}^3 in 3 )
• Bore \text{Bore} Bore is cylinder inside diameter in inches ( D D D )
• Stroke \text{Stroke} Stroke is piston travel length in inches ( L L L )
• N N N is total number of cylinders

Aviation Engine Designations: Aircraft engine model numbers standardly incorporate their total cubic inch displacement rounded to the nearest integer. For example:

Lycoming O-320: Opposed, 320 in 3 320\text{ in}^3 320 in 3 displacement ( 4 cylinders 4\text{ cylinders} 4 cylinders , 5.125 in 5.125\text{ in} 5.125 in bore, 3.875 in 3.875\text{ in} 3.875 in stroke).

Continental IO-520: Fuel-injected opposed, 520 in 3 520\text{ in}^3 520 in 3 displacement ( 6 cylinders 6\text{ cylinders} 6 cylinders , 5.25 in 5.25\text{ in} 5.25 in bore, 4.00 in 4.00\text{ in} 4.00 in stroke).

Pratt & Whitney R-2800 Double Wasp: Radial, 2 , 800 in 3 2,800\text{ in}^3 2 , 800 in 3 displacement ( 18 cylinders 18\text{ cylinders} 18 cylinders , 5.75 in 5.75\text{ in} 5.75 in bore, 6.00 in 6.00\text{ in} 6.00 in stroke).

4. Engine Compression Ratio (CR) Mathematics

The Compression Ratio (CR) is the ratio comparing the maximum total internal volume of a cylinder when the piston is at the bottom of its stroke (BDC) to the minimum volume remaining when the piston reaches the top of its stroke (TDC). COMPRESSION RATIO VOLUME BREAKDOWN

• [ Piston at BDC: Maximum Volume ] [ Piston at TDC: Minimum Volume ]
• ┌───────────────────────────────┐ ┌───────────────────────────────┐
• │ Clearance Volume (V_CL) │ │ Clearance Volume (V_CL) │
• ├───────────────────────────────┤ └───────────────────────────────┘
• │ │
• │ │ CR = (V_D + V_CL) / V_CL
• │ Piston Displacement (V_D) │ = (Total Volume at BDC) / (Volume at TDC)
• │ │
• │ │
• └───────────────────────────────┘

Compression Ratio Formulas Compression Ratio ( C R ) = Total Cylinder Volume at BDC Clearance Volume at TDC = V D + V C L V C L \text{Compression Ratio } (CR) = \frac{\text{Total Cylinder Volume at BDC}}{\text{Clearance Volume at TDC}} = \frac{V_D + V_{CL}}{V_{CL}} Compression Ratio ( CR ) = Clearance Volume at TDC Total Cylinder Volume at BDC ​ = V C L ​ V D ​ + V C L ​ ​ CR = V D V C L + 1 \text{CR} = \frac{V_D}{V_{CL}} + 1 CR = V C L ​ V D ​ ​ + 1

where:

V D V_D V D ​ is piston swept displacement volume ( 0.7854 × Bore 2 × Stroke 0.7854 \times \text{Bore}^2 \times \text{Stroke} 0.7854 × Bore 2 × Stroke )

V C L V_{CL} V C L ​ is clearance volume (combustion chamber cavity plus cylinder head spark plug recess volume above piston at TDC)

Solving for Clearance Volume ( V C L V_{CL} V C L ​ )

When engine designers or overhaul technicians know the desired compression ratio and measured displacement, clearance volume is derived algebraically: CR × V C L = V D + V C L ⟹ CR × V C L − V C L = V D \text{CR} \times V_{CL} = V_D + V_{CL} \implies \text{CR} \times V_{CL} - V_{CL} = V_D CR × V C L ​ = V D ​ + V C L ​ ⟹ CR × V C L ​ − V C L ​ = V D ​ V C L ( CR − 1 ) = V D ⟹ V C L = V D CR − 1 V_{CL} (\text{CR} - 1) = V_D \implies V_{CL} = \frac{V_D}{\text{CR} - 1} V C L ​ ( CR − 1 ) = V D ​ ⟹ V C L ​ = CR − 1 V D ​ ​

Practical Engineering Effects of Compression Ratio

Thermal Efficiency: Higher compression ratios ( 8.5 : 1 8.5:1 8.5 : 1 vs. 6.5 : 1 6.5:1 6.5 : 1 ) compress the air-fuel charge more tightly, producing higher peak combustion pressures, greater mechanical power output, and lower specific fuel consumption.

Detonation Risk & Octane Requirements: Excessively high compression ratios increase combustion chamber temperatures and pressures, necessitating high-octane anti-detonation aviation fuels (100LL Avgas) to prevent pre-ignition and damaging engine detonation.

• 5. Worked Step-by-Step Calculation Examples
• Example 1: Total Engine Displacement of a 6-Cylinder Aircraft Powerplant

Scenario: An AMT is overhauling a Continental IO-520 six-cylinder aircraft engine. The manufacturer specification sheet lists the cylinder barrel inside diameter (bore) as 5.250 inches 5.250\text{ inches} 5.250 inches and the piston travel distance (stroke) as 4.000 inches 4.000\text{ inches} 4.000 inches .

Calculate the piston cross-sectional area ( A A A ).

Calculate the swept displacement volume of a single cylinder ( V D V_D V D ​ ).

Calculate the total piston displacement of the complete six-cylinder engine ( V total V_{\text{total}} V total ​ ).

Step-by-Step Mathematical Solution:

Calculate the cross-sectional area of one cylinder using the FAA circular area formula: A = 0.7854 × Bore 2 A = 0.7854 \times \text{Bore}^2 A = 0.7854 × Bore 2 A = 0.7854 × ( 5.250 in ) 2 = 0.7854 × 27.5625 sq in = 21.6476 in 2 A = 0.7854 \times (5.250\text{ in})^2 = 0.7854 \times 27.5625\text{ sq in} = 21.6476\text{ in}^2 A = 0.7854 × ( 5.250 in ) 2 = 0.7854 × 27.5625 sq in = 21.6476 in 2

Calculate the swept displacement volume of a single cylinder ( V D V_D V D ​ ): V D = A × Stroke = 21.6476 in 2 × 4.000 in = 86.5904 cubic inches V_D = A \times \text{Stroke} = 21.6476\text{ in}^2 \times 4.000\text{ in} = 86.5904\text{ cubic inches} V D ​ = A × Stroke = 21.6476 in 2 × 4.000 in = 86.5904 cubic inches

Calculate total engine displacement ( V total V_{\text{total}} V total ​ ) across all 6 6 6 cylinders: V total = V D × N cylinders = 86.5904 in 3 × 6 = 519.5424 cubic inches V_{\text{total}} = V_D \times N_{\text{cylinders}} = 86.5904\text{ in}^3 \times 6 = 519.5424\text{ cubic inches} V total ​ = V D ​ × N cylinders ​ = 86.5904 in 3 × 6 = 519.5424 cubic inches Rounding to nearest integer: V total ≈ 520 in 3 \text{Rounding to nearest integer: } V_{\text{total}} \approx 520\text{ in}^3 Rounding to nearest integer: V total ​ ≈ 520 in 3

Conclusion: Total engine piston displacement is 519.54 in 3 519.54\text{ in}^3 519.54 in 3 (commercially designated as the Continental 520 520 520 series).

Example 2: Engine Compression Ratio and Clearance Volume Determination

Scenario: A four-cylinder aircraft engine has a single cylinder displacement ( V D V_D V D ​ ) of 90.0 cubic inches 90.0\text{ cubic inches} 90.0 cubic inches and a measured clearance volume ( V C L V_{CL} V C L ​ ) of 12.0 cubic inches 12.0\text{ cubic inches} 12.0 cubic inches .

Part A: Calculate the compression ratio ( C R CR CR ) of this engine.

Part B: If an engine modifier alters the piston crown to achieve a target compression ratio of 9.0 : 1 9.0:1 9.0 : 1 on an engine with a displacement of 80.0 in 3 80.0\text{ in}^3 80.0 in 3 per cylinder, calculate the required clearance volume ( V C L V_{CL} V C L ​ ).

Step-by-Step Mathematical Solution: Part A Solution:

Apply the compression ratio formula: CR = V D + V C L V C L \text{CR} = \frac{V_D + V_{CL}}{V_{CL}} CR = V C L ​ V D ​ + V C L ​ ​ CR = 90.0 in 3 + 12.0 in 3 12.0 in 3 = 102.0 in 3 12.0 in 3 = 8.5 \text{CR} = \frac{90.0\text{ in}^3 + 12.0\text{ in}^3}{12.0\text{ in}^3} = \frac{102.0\text{ in}^3}{12.0\text{ in}^3} = 8.5 CR = 12.0 in 3 90.0 in 3 + 12.0 in 3 ​ = 12.0 in 3 102.0 in 3 ​ = 8.5

Conclusion: The compression ratio is 8.5 : 1 8.5 : 1 8.5 : 1 .

Part B Solution:

Apply the derived clearance volume formula: V C L = V D CR − 1 V_{CL} = \frac{V_D}{\text{CR} - 1} V C L ​ = CR − 1 V D ​ ​ V C L = 80.0 in 3 9.0 − 1 = 80.0 in 3 8.0 = 10.0 cubic inches V_{CL} = \frac{80.0\text{ in}^3}{9.0 - 1} = \frac{80.0\text{ in}^3}{8.0} = 10.0\text{ cubic inches} V C L ​ = 9.0 − 1 80.0 in 3 ​ = 8.0 80.0 in 3 ​ = 10.0 cubic inches

Conclusion: To obtain a 9.0 : 1 9.0:1 9.0 : 1 compression ratio on an 80.0 in 3 80.0\text{ in}^3 80.0 in 3 cylinder, the clearance volume must measure exactly 10.0 cubic inches 10.0\text{ cubic inches} 10.0 cubic inches .

Example 3: Rectangular Fuel Tank Volume & Fuel Capacity Calculation

Scenario: A rectangular aluminum auxiliary fuel tank fabricated for a transport category aircraft cargo floor has the following internal dimensions:

• Length ( L L L ) = 36.0 inches 36.0\text{ inches} 36.0 inches
• Width ( W W W ) = 14.0 inches 14.0\text{ inches} 14.0 inches
• Depth ( H H H ) = 11.0 inches 11.0\text{ inches} 11.0 inches

The aircraft maintenance manual specifies that 1.5 U.S. gallons 1.5\text{ U.S. gallons} 1.5 U.S. gallons of fuel in this tank is unusable (residual fuel trapped beneath the fuel pickup standpipe). Standard 100LL aviation gasoline weighs 6.0 lb/gal 6.0\text{ lb/gal} 6.0 lb/gal .

Calculate the internal volume of the tank in cubic inches ( in 3 \text{in}^3 in 3 ).

Calculate the total fuel capacity of the tank in U.S. gallons ( 1 gal = 231 in 3 1\text{ gal} = 231\text{ in}^3 1 gal = 231 in 3 ).

Calculate the usable fuel capacity in U.S. gallons.

Calculate the weight of the usable fuel in pounds.

Step-by-Step Mathematical Solution:

Calculate total internal volume in cubic inches: V = L × W × H = 36.0 in × 14.0 in × 11.0 in = 5 , 544.0 cubic inches V = L \times W \times H = 36.0\text{ in} \times 14.0\text{ in} \times 11.0\text{ in} = 5,544.0\text{ cubic inches} V = L × W × H = 36.0 in × 14.0 in × 11.0 in = 5 , 544.0 cubic inches

Convert cubic inches to total U.S. gallons ( 231 in 3 / gal 231\text{ in}^3/\text{gal} 231 in 3 / gal ): Total Capacity = 5 , 544.0 in 3 231.0 in 3 / gal = 24.0 U.S. Gallons \text{Total Capacity} = \frac{5,544.0\text{ in}^3}{231.0\text{ in}^3/\text{gal}} = 24.0\text{ U.S. Gallons} Total Capacity = 231.0 in 3 / gal 5 , 544.0 in 3 ​ = 24.0 U.S. Gallons

Calculate usable fuel volume by subtracting unusable fuel allowance: Usable Fuel Volume = Total Capacity − Unusable Fuel = 24.0 gal − 1.5 gal = 22.5 U.S. Gallons \text{Usable Fuel Volume} = \text{Total Capacity} - \text{Unusable Fuel} = 24.0\text{ gal} - 1.5\text{ gal} = 22.5\text{ U.S. Gallons} Usable Fuel Volume = Total Capacity − Unusable Fuel = 24.0 gal − 1.5 gal = 22.5 U.S. Gallons

Calculate the weight of usable Avgas ( 6.0 lb/gal 6.0\text{ lb/gal} 6.0 lb/gal ): Usable Fuel Weight = 22.5 gal × 6.0 lb/gal = 135.0 pounds \text{Usable Fuel Weight} = 22.5\text{ gal} \times 6.0\text{ lb/gal} = 135.0\text{ pounds} Usable Fuel Weight = 22.5 gal × 6.0 lb/gal = 135.0 pounds

Conclusion: Total tank volume is 5 , 544.0 in 3 5,544.0\text{ in}^3 5 , 544.0 in 3 ( 24.0 gallons 24.0\text{ gallons} 24.0 gallons ), providing 22.5 gallons 22.5\text{ gallons} 22.5 gallons of usable fuel weighing 135.0 lbs 135.0\text{ lbs} 135.0 lbs .