6.3 Center of Gravity Computations, Extreme Loading & Ballast

FAA Aviation Mechanic — General AMG

Track your progress as you read

Finished sections get a check mark.

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

6.3 Center of Gravity Computations, Extreme Loading & Ballast

Key Takeaways Explain This Topic AI

Center of gravity from scale reactions is derived using static moment equilibrium: for tricycle gear with datum ahead of the nose, CG = D − ( F × L / W ) \text{CG} = D - (F \times L / W) CG = D − ( F × L / W ) ; for tailwheel gear with datum ahead of main wheels, CG = D + ( R × L / W ) \text{CG} = D + (R \times L / W) CG = D + ( R × L / W ) .

Percent Mean Aerodynamic Chord (% MAC) standardizes longitudinal balance on transport category airframes: % MAC = CG Station − LEMAC MAC × 100 % \%\text{MAC} = \frac{\text{CG Station} - \text{LEMAC}}{\text{MAC}} \times 100\% % MAC = MAC CG Station − LEMAC ​ × 100% .

Adverse-loaded CG checks verify that the aircraft will not exceed certified forward or aft CG limits under worst-case loading configurations, utilizing minimum fuel formulas ( Minimum Fuel = METO HP / 2 lb \text{Minimum Fuel} = \text{METO HP} / 2\text{ lb} Minimum Fuel = METO HP /2 lb ) for forward adverse checks.

Equipment additions, removals, or relocations modify empty weight and total moment; the revised CG is calculated as New CG = Original Moment ± Δ Moments Original Weight ± Δ Weights \text{New CG} = \frac{\text{Original Moment} \pm \Delta \text{Moments}}{\text{Original Weight} \pm \Delta \text{Weights}} New CG = Original Weight ± Δ Weights Original Moment ± Δ Moments ​ .

Permanent ballast installed to correct out-of-limit CG conditions must be calculated using Ballast Weight = Empty Weight × Δ CG Arm ballast − CG desired \text{Ballast Weight} = \frac{\text{Empty Weight} \times \Delta \text{CG}}{\text{Arm}_{\text{ballast}} - \text{CG}_{\text{desired}}} Ballast Weight = Arm ballast ​ − CG desired ​ Empty Weight × Δ CG ​ , bolted to primary structure, painted in red with a warning placard, and recorded on FAA Form 337.

Last updated: August 2026

6.3 Center of Gravity Computations, Extreme Loading & Ballast

Once an aircraft has been weighed and net scale reactions established, the Aviation Maintenance Technician must calculate the precise Center of Gravity (CG) location, verify adherence to certified forward and aft limits, evaluate extreme loading conditions, and compute equipment alteration impacts. For transport category aircraft, balance is expressed as a percentage of the Mean Aerodynamic Chord (% MAC) . When alterations shift the CG beyond certified limits, technicians must calculate and install ballast and document all revisions in accordance with 14 CFR Part 43 .

1. Center of Gravity Mathematical Computations from Scale Data

Scale reaction geometry depends on whether the aircraft utilizes a tricycle landing gear (nosewheel) or a tailwheel (conventional) landing gear configuration. TRICYCLE GEAR WEIGHING GEOMETRY

• Datum (Ahead of Nose)
• │
• │◄──────────────── D (Distance to Main Wheels) ────────────────►│
• │
• │◄──── (D - L) ────►│ │
• │ │◄───────────── L (Wheelbase) ─────────────►│
• │ │ │
• ─┴───────────────────▼───────────────────────────────────────────▼─
• Nose Wheel (F) Main Wheels (M)
• │◄─── CG ───►│
• Tricycle Landing Gear CG Derivations

Let:

W W W = Total net aircraft weight ( W = M + F W = M + F W = M + F , where M = Left Main + Right Main M = \text{Left Main} + \text{Right Main} M = Left Main + Right Main and F = Nose Reaction F = \text{Nose Reaction} F = Nose Reaction )

• L L L = Wheelbase length (horizontal distance between nosewheel centerline and main wheel centerline)
• D D D = Horizontal distance from the Reference Datum to the main wheel centerline

Case 1: Datum Located Forward of the Aircraft Nose (All Arms Positive) The main gear arm is D D D , and the nose gear arm is ( D − L ) (D - L) ( D − L ) .

Taking moments about the Reference Datum: Total Moment = ( M × D ) + [ F × ( D − L ) ] = M D + F D − F L = ( M + F ) D − F L = W D − F L \text{Total Moment} = (M \times D) + [F \times (D - L)] = M D + F D - F L = (M + F)D - F L = W D - F L Total Moment = ( M × D ) + [ F × ( D − L )] = M D + F D − F L = ( M + F ) D − F L = W D − F L CG Station = Total Moment W = W D − F L W = D − ( F × L W ) \text{CG Station} = \frac{\text{Total Moment}}{W} = \frac{W D - F L}{W} = D - \left(\frac{F \times L}{W}\right) CG Station = W Total Moment ​ = W W D − F L ​ = D − ( W F × L ​ )

Case 2: Datum Located at the Nosewheel Centerline ( D nose = 0 D_{\text{nose}} = 0 D nose ​ = 0 , D main = L D_{\text{main}} = L D main ​ = L ) CG Station = M × L W \text{CG Station} = \frac{M \times L}{W} CG Station = W M × L ​

Case 3: Datum Located at the Main Wheel Centerline ( D main = 0 D_{\text{main}} = 0 D main ​ = 0 , D nose = − L D_{\text{nose}} = -L D nose ​ = − L ) CG Station = − ( F × L W ) (Inches forward of main gear) \text{CG Station} = -\left(\frac{F \times L}{W}\right) \quad \text{(Inches forward of main gear)} CG Station = − ( W F × L ​ ) (Inches forward of main gear) TAILWHEEL GEAR WEIGHING GEOMETRY

• Datum (Ahead of Main Wheels)
• │
• │◄────────── D ──────────►│
• │ │◄───────────── L (Wheelbase) ─────────────►│
• │ │ │
• ─┴─────────────────────────▼───────────────────────────────────────────▼─
• Main Wheels (M) Tail Wheel (R)
• │◄─── CG ───►│
• Tailwheel (Conventional) Landing Gear CG Derivations

Let:

W W W = Total net aircraft weight ( W = M + R W = M + R W = M + R , where M = Left Main + Right Main M = \text{Left Main} + \text{Right Main} M = Left Main + Right Main and R = Tail Reaction R = \text{Tail Reaction} R = Tail Reaction )

• L L L = Wheelbase length (horizontal distance between main wheel centerline and tailwheel centerline)
• D D D = Horizontal distance from Reference Datum to main wheel centerline

Case 1: Datum Located Forward of the Main Wheels (All Arms Positive) The main gear arm is D D D , and the tailwheel arm is ( D + L ) (D + L) ( D + L ) .

Taking moments about the Reference Datum: Total Moment = ( M × D ) + [ R × ( D + L ) ] = M D + R D + R L = ( M + R ) D + R L = W D + R L \text{Total Moment} = (M \times D) + [R \times (D + L)] = M D + R D + R L = (M + R)D + R L = W D + R L Total Moment = ( M × D ) + [ R × ( D + L )] = M D + R D + R L = ( M + R ) D + R L = W D + R L CG Station = Total Moment W = W D + R L W = D + ( R × L W ) \text{CG Station} = \frac{\text{Total Moment}}{W} = \frac{W D + R L}{W} = D + \left(\frac{R \times L}{W}\right) CG Station = W Total Moment ​ = W W D + R L ​ = D + ( W R × L ​ )

Case 2: Datum Located at the Main Wheel Centerline ( D main = 0 D_{\text{main}} = 0 D main ​ = 0 , D tail = + L D_{\text{tail}} = +L D tail ​ = + L ) CG Station = + ( R × L W ) (Inches aft of main gear) \text{CG Station} = +\left(\frac{R \times L}{W}\right) \quad \text{(Inches aft of main gear)} CG Station = + ( W R × L ​ ) (Inches aft of main gear)

Loading diagram... Mean Aerodynamic Chord (MAC) Geometry and % MAC Calculation

2. Percent Mean Aerodynamic Chord (% MAC) Computations

On large multi-engine, swept-wing, and commercial transport aircraft, the center of gravity is universally expressed as a percentage of the Mean Aerodynamic Chord (% MAC) rather than in raw fuselage station inches.

Concept of Mean Aerodynamic Chord (MAC)

Mean Aerodynamic Chord (MAC): The chord length of an imaginary rectangular airfoil that has the exact same pitching moment, aerodynamic center, and lift characteristics as the actual complex, tapered, or swept wing planform.

LEMAC (Leading Edge of MAC): The fuselage station representing the leading edge of the Mean Aerodynamic Chord.

TEMAC (Trailing Edge of MAC): The fuselage station representing the trailing edge of the Mean Aerodynamic Chord: TEMAC = LEMAC + MAC \text{TEMAC} = \text{LEMAC} + \text{MAC} TEMAC = LEMAC + MAC % MAC RELATIONSHIP DIAGRAM

Fuselage Station Scale:

• STA 0.0 STA 500.0 (LEMAC) STA 540.0 (CG) STA 700.0 (TEMAC)
• │ │ │ │
• │◄──── LEMAC Arm ────►│ │ │
• │ │◄──── CG - LEMAC ───►│ │
• │ │◄──────────────────── MAC (200 in) ───────────►│
• │ │ │ │
• ▼ ▼ ▼ ▼
• Datum 0% MAC 20.0% MAC 100% MAC
• Mathematical Formulas for % MAC

Converting CG Station to % MAC: % MAC = CG Station − LEMAC MAC × 100 % \%\text{MAC} = \frac{\text{CG Station} - \text{LEMAC}}{\text{MAC}} \times 100\% % MAC = MAC CG Station − LEMAC ​ × 100%

Converting % MAC to CG Fuselage Station: CG Station = LEMAC + ( % MAC 100 % × MAC ) \text{CG Station} = \text{LEMAC} + \left(\frac{\%\text{MAC}}{100\%} \times \text{MAC}\right) CG Station = LEMAC + ( 100% % MAC ​ × MAC )

Why % MAC is Essential: Expressing CG as % MAC allows pilots and dispatchers to instantly assess aerodynamic balance regardless of aircraft gross weight. A CG at 25 % MAC 25\%\text{ MAC} 25% MAC represents the exact same relative stability margin whether the aircraft is at minimum operating weight or maximum takeoff weight.

3. Adverse-Loaded Center of Gravity Checks (Extreme Condition Checks)

An Adverse-Loaded CG Check (also termed an extreme condition check ) is a mathematical verification performed by an AMT following a major repair or alteration to prove that the aircraft cannot be loaded in any plausible operational configuration that would exceed the forward or aft CG limits. ADVERSE LOADING STRATEGY MATRIX

• ┌────────────────────────────────────────────────────────────────────────┐
• │ FORWARD ADVERSE-LOADED CG CHECK │
• │ Goal: Drive CG as far FORWARD as physically possible │
• │ • Maximum load in all stations FORWARD of Forward CG Limit │
• │ • Minimum load in all stations AFT of Forward CG Limit │
• │ • Minimum Fuel (METO Fuel Formula) unless tank is ahead of limit │
• ├────────────────────────────────────────────────────────────────────────┤
• │ AFT ADVERSE-LOADED CG CHECK │
• │ Goal: Drive CG as far AFT as physically possible │
• │ • Maximum load in all stations AFT of Aft CG Limit (Baggage/Pax) │
• │ • Minimum load in all stations FORWARD of Aft CG Limit (Single Pilot) │
• │ • Minimum Fuel unless fuel tank is located aft of Aft CG Limit │
• └────────────────────────────────────────────────────────────────────────┘
• Minimum Fuel Formula for Forward Adverse Check

When conducting a forward adverse check on reciprocating engine aircraft, fuel weight must be reduced to the certified Minimum Fuel (also known as Maximum Except Takeoff / METO Fuel) to simulate the critical low-fuel condition: Minimum Fuel (Pounds) = METO Horsepower 2 \text{Minimum Fuel (Pounds)} = \frac{\text{METO Horsepower}}{2} Minimum Fuel (Pounds) = 2 METO Horsepower ​ Minimum Fuel (Gallons) = METO Horsepower 12 ( since Avgas = 6.0 lb/gal ) \text{Minimum Fuel (Gallons)} = \frac{\text{METO Horsepower}}{12} \quad (\text{since Avgas} = 6.0\text{ lb/gal}) Minimum Fuel (Gallons) = 12 METO Horsepower ​ ( since Avgas = 6.0 lb/gal )

Example: An engine with 240 METO HP 240\text{ METO HP} 240 METO HP requires a minimum fuel calculation weight of 240 / 2 = 120.0 lb 240 / 2 = 120.0\text{ lb} 240/2 = 120.0 lb ( 20.0 gallons 20.0\text{ gallons} 20.0 gallons ).

Standard Weights for Adverse Computations (FAA AC 43.13-1B)

Flight Crew & Passengers: 170.0 lb 170.0\text{ lb} 170.0 lb per person (standard FAA calculation weight for general aviation).

Baggage / Cargo: Maximum placard capacity for the respective compartment.

Aviation Gasoline (100LL): 6.0 lb/gal 6.0\text{ lb/gal} 6.0 lb/gal .

Turbine Fuel (Jet A): 6.7 lb/gal 6.7\text{ lb/gal} 6.7 lb/gal .

Engine Oil: 7.5 lb/gal 7.5\text{ lb/gal} 7.5 lb/gal ( 1.875 lb/qt 1.875\text{ lb/qt} 1.875 lb/qt ).

4. Equipment Addition, Removal, and Relocation Mathematics

Whenever equipment is installed, removed, or repositioned on an aircraft, the AMT must calculate the new Basic Empty Weight and EWCG. EQUIPMENT CHANGE ALGEBRAIC RULES

• ┌────────────────────────────────────────────────────────────────────────┐
• │ 1. ADDING Equipment: │
• │ + Weight, + Moment (Weight × Arm) │
• │ │
• │ 2. REMOVING Equipment: │
• │ - Weight, - Moment (Weight × Arm) │
• │ │
• │ 3. RELOCATING Equipment: │
• │ Weight Delta = 0 │
• │ Moment Shift (ΔM) = Weight × (New Arm - Old Arm) │
• └────────────────────────────────────────────────────────────────────────┘

Master Revision Formula New CG = Original Total Moment + ∑ M added − ∑ M removed + ∑ Δ M relocated Original Empty Weight + ∑ W added − ∑ W removed \text{New CG} = \frac{\text{Original Total Moment} + \sum M_{\text{added}} - \sum M_{\text{removed}} + \sum \Delta M_{\text{relocated}}}{\text{Original Empty Weight} + \sum W_{\text{added}} - \sum W_{\text{removed}}} New CG = Original Empty Weight + ∑ W added ​ − ∑ W removed ​ Original Total Moment + ∑ M added ​ − ∑ M removed ​ + ∑ Δ M relocated ​ ​

5. Permanent and Temporary Ballast Calculations

When structural alterations or heavy equipment installations move the Empty Weight CG outside allowable limits, ballast must be installed to restore proper balance.

Sizing Ballast Weight

The required ballast weight is derived from the principle of static moments: Ballast Weight = Aircraft Weight ( W ) × Δ CG Arm ballast − CG desired = W × ∣ CG actual − CG desired ∣ ∣ Arm ballast − CG desired ∣ \text{Ballast Weight} = \frac{\text{Aircraft Weight } (W) \times \Delta \text{CG}}{\text{Arm}_{\text{ballast}} - \text{CG}_{\text{desired}}} = \frac{W \times |\text{CG}_{\text{actual}} - \text{CG}_{\text{desired}}|}{|\text{Arm}_{\text{ballast}} - \text{CG}_{\text{desired}}|} Ballast Weight = Arm ballast ​ − CG desired ​ Aircraft Weight ( W ) × Δ CG ​ = ∣ Arm ballast ​ − CG desired ​ ∣ W × ∣ CG actual ​ − CG desired ​ ∣ ​

Where:

W W W = Aircraft empty weight (or weight prior to ballast installation)

Δ CG \Delta \text{CG} Δ CG = Distance the center of gravity must be shifted ( ∣ CG desired − CG actual ∣ |\text{CG}_{\text{desired}} - \text{CG}_{\text{actual}}| ∣ CG desired ​ − CG actual ​ ∣ )

Arm ballast \text{Arm}_{\text{ballast}} Arm ballast ​ = Fuselage station where the ballast will be installed

CG desired \text{CG}_{\text{desired}} CG desired ​ = Target center of gravity station (usually the forward or aft limit)

Ballast Installation and Placarding Standards

Permanent Ballast: Typically fabricated from lead bar stock or high-density brass plates. Must be bolted securely to primary structural members, safety-wired, and protected against shifting. It must be painted in bright red enamel and permanently stenciled with white lettering: PERMANENT BALLAST — DO NOT REMOVE — [XX] LBS AT STATION [YY]

Structural Limits: The AMT must verify that the ballast weight does not exceed the maximum allowable localized floor loading ( lb/ft 2 \text{lb/ft}^2 lb/ft 2 ) or bulkhead shear rating.

Temporary Ballast: Carried in baggage compartments during special ferry flights. Must be secured with cargo nets and placarded with its exact weight and arm.

6. Weight & Balance Revision Logbook Entry & 14 CFR Documentation

Under 14 CFR § 43.9 and 14 CFR § 91.417 , any change in aircraft equipment that affects weight and balance requires an immediate record entry: WEIGHT & BALANCE REVISION DOCUMENTATION

• ┌────────────────────────────────────────────────────────────────────────┐
• │ 1. Aircraft Maintenance Record Entry (§ 43.9): │
• │ • Detailed description of work performed │
• │ • Date of completion │
• │ • Name, signature, and certificate number of person approving │
• │ │
• │ 2. Weight and Balance Revision Sheet (In AFM / POH): │
• │ • New Basic Empty Weight (lbs) │
• │ • New Empty Weight Center of Gravity (inches / station) │
• │ • New Useful Load (lbs) │
• │ • Updated Moment / Index │
• │ │
• │ 3. Equipment List Update: │
• │ • Red-line removed items; add newly installed items with weight, │
• │ arm, part number, and serial number │
• │ │
• │ 4. FAA Form 337 (Major Repair and Alteration): │
• │ • Required if alteration is classified as Major under Part 43 App A│
• └────────────────────────────────────────────────────────────────────────┘

• 7. Worked Numerical Examples
• Example 1: Scale Reaction CG Determination (Tailwheel Aircraft)

Scenario: A vintage tailwheel aircraft with the reference datum at the tip of the propeller spinner is weighed on wheel scales in a level attitude. The following net scale readings and dimensions are recorded:

Left Main Wheel: Net reaction = 965.0 lb 965.0\text{ lb} 965.0 lb at Station + 68.0 in \text{Station } +68.0\text{ in} Station + 68.0 in

Right Main Wheel: Net reaction = 955.0 lb 955.0\text{ lb} 955.0 lb at Station + 68.0 in \text{Station } +68.0\text{ in} Station + 68.0 in

Tail Wheel: Net reaction = 180.0 lb 180.0\text{ lb} 180.0 lb at Station + 284.0 in \text{Station } +284.0\text{ in} Station + 284.0 in

Wheelbase ( L L L ) between main gear and tailwheel = 216.0 in 216.0\text{ in} 216.0 in ( 284.0 − 68.0 284.0 - 68.0 284.0 − 68.0 )

Calculate the Empty Weight Center of Gravity (EWCG) using both the Total Moment method and the Tailwheel Formula.

Step-by-Step Mathematical Solution:

Method 1: Total Moment / Total Weight

Total Main Reaction M = 965.0 + 955.0 = 1 , 920.0 lb M = 965.0 + 955.0 = 1,920.0\text{ lb} M = 965.0 + 955.0 = 1 , 920.0 lb at Station + 68.0 in \text{Station } +68.0\text{ in} Station + 68.0 in

Tail Reaction R = 180.0 lb R = 180.0\text{ lb} R = 180.0 lb at Station + 284.0 in \text{Station } +284.0\text{ in} Station + 284.0 in

Total Weight W = 1 , 920.0 + 180.0 = 2 , 100.0 lb W = 1,920.0 + 180.0 = 2,100.0\text{ lb} W = 1 , 920.0 + 180.0 = 2 , 100.0 lb

Main Wheel Moment = 1 , 920.0 lb × 68.0 in = 130 , 560.0 in-lb 1,920.0\text{ lb} \times 68.0\text{ in} = 130,560.0\text{ in-lb} 1 , 920.0 lb × 68.0 in = 130 , 560.0 in-lb

Tail Wheel Moment = 180.0 lb × 284.0 in = 51 , 120.0 in-lb 180.0\text{ lb} \times 284.0\text{ in} = 51,120.0\text{ in-lb} 180.0 lb × 284.0 in = 51 , 120.0 in-lb

Total Moment = 130 , 560.0 + 51 , 120.0 = 181 , 680.0 in-lb 130,560.0 + 51,120.0 = 181,680.0\text{ in-lb} 130 , 560.0 + 51 , 120.0 = 181 , 680.0 in-lb

EWCG = 181 , 680.0 in-lb 2 , 100.0 lb = + 86.514 inches ≈ + 86.51 in \text{EWCG} = \frac{181,680.0\text{ in-lb}}{2,100.0\text{ lb}} = +86.514\text{ inches} \approx +86.51\text{ in} EWCG = 2 , 100.0 lb 181 , 680.0 in-lb ​ = + 86.514 inches ≈ + 86.51 in

Method 2: Tailwheel Formula ( C G = D + [ R × L / W ] CG = D + [R \times L / W] CG = D + [ R × L / W ] )

D = 68.0 in D = 68.0\text{ in} D = 68.0 in , R = 180.0 lb R = 180.0\text{ lb} R = 180.0 lb , L = 216.0 in L = 216.0\text{ in} L = 216.0 in , W = 2 , 100.0 lb W = 2,100.0\text{ lb} W = 2 , 100.0 lb EWCG = 68.0 + ( 180.0 × 216.0 2 , 100.0 ) = 68.0 + ( 38 , 880.0 2 , 100.0 ) = 68.0 + 18.514 = + 86.514 in \text{EWCG} = 68.0 + \left(\frac{180.0 \times 216.0}{2,100.0}\right) = 68.0 + \left(\frac{38,880.0}{2,100.0}\right) = 68.0 + 18.514 = +86.514\text{ in} EWCG = 68.0 + ( 2 , 100.0 180.0 × 216.0 ​ ) = 68.0 + ( 2 , 100.0 38 , 880.0 ​ ) = 68.0 + 18.514 = + 86.514 in

Conclusion: Both methods yield an identical EWCG of Station + 86.51 in \text{Station } +86.51\text{ in} Station + 86.51 in .

Example 2: Transport Category % MAC Computation

Scenario: A twin-turboprop regional airliner has a certified Mean Aerodynamic Chord (MAC) of 145.0 inches 145.0\text{ inches} 145.0 inches . The leading edge of the MAC (LEMAC) is located at fuselage Station 420.0 in \text{Station } 420.0\text{ in} Station 420.0 in . After passenger and cargo loading, dispatch calculates the loaded aircraft CG to be at Station 454.8 in \text{Station } 454.8\text{ in} Station 454.8 in .

Calculate the current Center of Gravity expressed as % MAC.

If the certified allowable CG range is 18.0 % to 32.0 % MAC 18.0\%\text{ to } 32.0\%\text{ MAC} 18.0% to 32.0% MAC , determine if the aircraft is within legal flight limits.

Step-by-Step Mathematical Solution:

Apply the % MAC formula: % MAC = CG Station − LEMAC MAC × 100 % \%\text{MAC} = \frac{\text{CG Station} - \text{LEMAC}}{\text{MAC}} \times 100\% % MAC = MAC CG Station − LEMAC ​ × 100%

Substitute the given values: % MAC = 454.8 in − 420.0 in 145.0 in × 100 % = 34.8 in 145.0 in × 100 % = 0.240 × 100 % = 24.0 % MAC \%\text{MAC} = \frac{454.8\text{ in} - 420.0\text{ in}}{145.0\text{ in}} \times 100\% = \frac{34.8\text{ in}}{145.0\text{ in}} \times 100\% = 0.240 \times 100\% = 24.0\%\text{ MAC} % MAC = 145.0 in 454.8 in − 420.0 in ​ × 100% = 145.0 in 34.8 in ​ × 100% = 0.240 × 100% = 24.0% MAC

Compliance Check: 18.0 % ≤ 24.0 % ≤ 32.0 % 18.0\% \le 24.0\% \le 32.0\% 18.0% ≤ 24.0% ≤ 32.0%

Conclusion: The aircraft CG is at 24.0 % MAC 24.0\%\text{ MAC} 24.0% MAC , which is safely inside the certified allowable range.

Example 3: Forward Ballast Sizing Calculation

Scenario: A four-seat utility aircraft has an Empty Weight of 1 , 800.0 lb 1,800.0\text{ lb} 1 , 800.0 lb and an Empty Weight CG at Station + 42.5 in \text{Station } +42.5\text{ in} Station + 42.5 in . Following the installation of aft cabin equipment, an adverse loading check reveals that the loaded CG falls at Station + 44.2 in \text{Station } +44.2\text{ in} Station + 44.2 in , exceeding the certified aft CG limit of Station + 43.5 in \text{Station } +43.5\text{ in} Station + 43.5 in by 0.7 inches 0.7\text{ inches} 0.7 inches . The technician decides to install permanent lead ballast in the forward engine compartment at Station − 12.0 in \text{Station } -12.0\text{ in} Station − 12.0 in .

Calculate the minimum weight of permanent ballast required to bring the CG to the aft limit of + 43.5 in +43.5\text{ in} + 43.5 in .

Step-by-Step Mathematical Solution:

Identify formula parameters:

• Aircraft Weight ( W W W ) = 1 , 800.0 lb 1,800.0\text{ lb} 1 , 800.0 lb
• Current Actual CG ( CG actual \text{CG}_{\text{actual}} CG actual ​ ) = + 44.2 in +44.2\text{ in} + 44.2 in

Desired Target CG ( CG desired \text{CG}_{\text{desired}} CG desired ​ ) = + 43.5 in +43.5\text{ in} + 43.5 in (the aft limit)

• Δ CG = 44.2 − 43.5 = 0.7 in \Delta \text{CG} = 44.2 - 43.5 = 0.7\text{ in} Δ CG = 44.2 − 43.5 = 0.7 in
• Ballast Arm ( Arm ballast \text{Arm}_{\text{ballast}} Arm ballast ​ ) = − 12.0 in -12.0\text{ in} − 12.0 in

Apply the ballast formula: Ballast Weight = W × Δ CG CG desired − Arm ballast = 1 , 800.0 lb × 0.7 in 43.5 in − ( − 12.0 in ) \text{Ballast Weight} = \frac{W \times \Delta \text{CG}}{\text{CG}_{\text{desired}} - \text{Arm}_{\text{ballast}}} = \frac{1,800.0\text{ lb} \times 0.7\text{ in}}{43.5\text{ in} - (-12.0\text{ in})} Ballast Weight = CG desired ​ − Arm ballast ​ W × Δ CG ​ = 43.5 in − ( − 12.0 in ) 1 , 800.0 lb × 0.7 in ​

Solve the denominator: Distance between Ballast and Target CG = 43.5 − ( − 12.0 ) = 43.5 + 12.0 = 55.5 in \text{Distance between Ballast and Target CG} = 43.5 - (-12.0) = 43.5 + 12.0 = 55.5\text{ in} Distance between Ballast and Target CG = 43.5 − ( − 12.0 ) = 43.5 + 12.0 = 55.5 in

Compute Ballast Weight: Ballast Weight = 1 , 260.0 in-lb 55.5 in = 22.7027 lb ≈ 22.70 lb \text{Ballast Weight} = \frac{1,260.0\text{ in-lb}}{55.5\text{ in}} = 22.7027\text{ lb} \approx 22.70\text{ lb} Ballast Weight = 55.5 in 1 , 260.0 in-lb ​ = 22.7027 lb ≈ 22.70 lb

Proof Check:

Original Aircraft Moment = 1 , 800.0 lb × 44.2 in = 79 , 560.0 in-lb 1,800.0\text{ lb} \times 44.2\text{ in} = 79,560.0\text{ in-lb} 1 , 800.0 lb × 44.2 in = 79 , 560.0 in-lb

Ballast Moment = 22.70 lb × ( − 12.0 in ) = − 272.4 in-lb 22.70\text{ lb} \times (-12.0\text{ in}) = -272.4\text{ in-lb} 22.70 lb × ( − 12.0 in ) = − 272.4 in-lb

Total New Weight = 1 , 800.0 + 22.70 = 1 , 822.70 lb 1,800.0 + 22.70 = 1,822.70\text{ lb} 1 , 800.0 + 22.70 = 1 , 822.70 lb

Total New Moment = 79 , 560.0 − 272.4 = 79 , 287.6 in-lb 79,560.0 - 272.4 = 79,287.6\text{ in-lb} 79 , 560.0 − 272.4 = 79 , 287.6 in-lb

New CG = 79 , 287.6 1 , 822.70 = + 43.50 in \text{New CG} = \frac{79,287.6}{1,822.70} = +43.50\text{ in} New CG = 1 , 822.70 79 , 287.6 ​ = + 43.50 in (Matches aft limit exactly). Conclusion: Install 22.70 lb 22.70\text{ lb} 22.70 lb of permanent ballast at Station -12.0 inches.