6.3 Center of Gravity Computations, Extreme Loading & Ballast

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:

• $W$ = Total net aircraft weight ($W = M + F$, where $M = \text{Left Main} + \text{Right Main}$ and $F = \text{Nose Reaction}$)
• $L$ = Wheelbase length (horizontal distance between nosewheel centerline and main wheel centerline)
• $D$ = 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 $D$, and the nose gear arm is $(D - L)$. Taking moments about the Reference Datum: $\text{Total Moment} = (M \times D) + [F \times (D - L)] = M D + F D - F L = (M + F)D - F L = W D - F L$ $\text{CG Station} = \frac{\text{Total Moment}}{W} = \frac{W D - F L}{W} = D - \left(\frac{F \times L}{W}\right)$

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

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

                     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:

• $W$ = Total net aircraft weight ($W = M + R$, where $M = \text{Left Main} + \text{Right Main}$ and $R = \text{Tail Reaction}$)
• $L$ = Wheelbase length (horizontal distance between main wheel centerline and tailwheel centerline)
• $D$ = 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 $D$, and the tailwheel arm is $(D + L)$. Taking moments about the Reference Datum: $\text{Total Moment} = (M \times D) + [R \times (D + L)] = M D + R D + R L = (M + R)D + R L = W D + R L$ $\text{CG Station} = \frac{\text{Total Moment}}{W} = \frac{W D + R L}{W} = D + \left(\frac{R \times L}{W}\right)$

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

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: $\text{TEMAC} = \text{LEMAC} + \text{MAC}$

                         % 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

1. Converting CG Station to % MAC: $\%\text{MAC} = \frac{\text{CG Station} - \text{LEMAC}}{\text{MAC}} \times 100\%$
2. Converting % MAC to CG Fuselage Station: $\text{CG Station} = \text{LEMAC} + \left(\frac{\%\text{MAC}}{100\%} \times \text{MAC}\right)$

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 $25%\text{ MAC}$ 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: $\text{Minimum Fuel (Pounds)} = \frac{\text{METO Horsepower}}{2}$ $\text{Minimum Fuel (Gallons)} = \frac{\text{METO Horsepower}}{12} \quad (\text{since Avgas} = 6.0\text{ lb/gal})$ Example: An engine with $240\text{ METO HP}$ requires a minimum fuel calculation weight of $240 / 2 = 120.0\text{ lb}$ ($20.0\text{ gallons}$).

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

• Flight Crew & Passengers: $170.0\text{ lb}$ per person (standard FAA calculation weight for general aviation).
• Baggage / Cargo: Maximum placard capacity for the respective compartment.
• Aviation Gasoline (100LL): $6.0\text{ lb/gal}$.
• Turbine Fuel (Jet A): $6.7\text{ lb/gal}$.
• Engine Oil: $7.5\text{ lb/gal}$ ($1.875\text{ lb/qt}$).

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

$\text{New CG} = \frac{\text{Original Total Moment} + \sum M_{\text{added}} - \sum M_{\text{removed}} + \sum \Delta M_{\text{relocated}}}{\text{Original Empty Weight} + \sum W_{\text{added}} - \sum W_{\text{removed}}}$

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: $\text{Ballast Weight} = \frac{\text{Aircraft Weight } (W) \times \Delta \text{CG}}{\text{Arm}_{\text{ballast}} - \text{CG}_{\text{desired}}} = \frac{W \times |\text{CG}_{\text{actual}} - \text{CG}_{\text{desired}}|}{|\text{Arm}_{\text{ballast}} - \text{CG}_{\text{desired}}|}$ Where:

• $W$ = Aircraft empty weight (or weight prior to ballast installation)
• $\Delta \text{CG}$ = Distance the center of gravity must be shifted ($|\text{CG}{\text{desired}} - \text{CG}{\text{actual}}|$)
• $\text{Arm}_{\text{ballast}}$ = Fuselage station where the ballast will be installed
• $\text{CG}_{\text{desired}}$ = Target center of gravity station (usually the forward or aft limit)

Ballast Installation and Placarding Standards

1. 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]
2. Structural Limits: The AMT must verify that the ballast weight does not exceed the maximum allowable localized floor loading ($\text{lb/ft}^2$) or bulkhead shear rating.
3. 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 = $965.0\text{ lb}$ at $\text{Station } +68.0\text{ in}$
• Right Main Wheel: Net reaction = $955.0\text{ lb}$ at $\text{Station } +68.0\text{ in}$
• Tail Wheel: Net reaction = $180.0\text{ lb}$ at $\text{Station } +284.0\text{ in}$
• Wheelbase ($L$) between main gear and tailwheel = $216.0\text{ in}$ ($284.0 - 68.0$)

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

Step-by-Step Mathematical Solution:

1. Method 1: Total Moment / Total Weight

   • Total Main Reaction $M = 965.0 + 955.0 = 1,920.0\text{ lb}$ at $\text{Station } +68.0\text{ in}$
   • Tail Reaction $R = 180.0\text{ lb}$ at $\text{Station } +284.0\text{ in}$
   • Total Weight $W = 1,920.0 + 180.0 = 2,100.0\text{ lb}$
   • Main Wheel Moment = $1,920.0\text{ lb} \times 68.0\text{ in} = 130,560.0\text{ in-lb}$
   • Tail Wheel Moment = $180.0\text{ lb} \times 284.0\text{ in} = 51,120.0\text{ in-lb}$
   • Total Moment = $130,560.0 + 51,120.0 = 181,680.0\text{ in-lb}$
   • $\text{EWCG} = \frac{181,680.0\text{ in-lb}}{2,100.0\text{ lb}} = +86.514\text{ inches} \approx +86.51\text{ in}$

2. Method 2: Tailwheel Formula ($CG = D + [R \times L / W]$)

   • $D = 68.0\text{ in}$, $R = 180.0\text{ lb}$, $L = 216.0\text{ in}$, $W = 2,100.0\text{ lb}$ $\text{EWCG} = 68.0 + \left(\frac{180.0 \times 216.0}{2,100.0}\right) = 68.0 + \left(\frac{38,880.0}{2,100.0}\right) = 68.0 + 18.514 = +86.514\text{ in}$ Conclusion: Both methods yield an identical EWCG of $\text{Station } +86.51\text{ in}$.

Example 2: Transport Category % MAC Computation

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

• Calculate the current Center of Gravity expressed as % MAC.
• If the certified allowable CG range is $18.0%\text{ to } 32.0%\text{ MAC}$, determine if the aircraft is within legal flight limits.

Step-by-Step Mathematical Solution:

1. Apply the % MAC formula: $\%\text{MAC} = \frac{\text{CG Station} - \text{LEMAC}}{\text{MAC}} \times 100\%$
2. Substitute the given values: $\%\text{MAC} = \frac{454.8\text{ in} - 420.0\text{ in}}{145.0\text{ in}} \times 100\% = \frac{34.8\text{ in}}{145.0\text{ in}} \times 100\% = 0.240 \times 100\% = 24.0\%\text{ MAC}$
3. Compliance Check: $18.0\% \le 24.0\% \le 32.0\%$ Conclusion: The aircraft CG is at $24.0%\text{ MAC}$, 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 $1,800.0\text{ lb}$ and an Empty Weight CG at $\text{Station } +42.5\text{ in}$. Following the installation of aft cabin equipment, an adverse loading check reveals that the loaded CG falls at $\text{Station } +44.2\text{ in}$, exceeding the certified aft CG limit of $\text{Station } +43.5\text{ in}$ by $0.7\text{ inches}$. The technician decides to install permanent lead ballast in the forward engine compartment at $\text{Station } -12.0\text{ in}$.

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

Step-by-Step Mathematical Solution:

1. Identify formula parameters:
   • Aircraft Weight ($W$) = $1,800.0\text{ lb}$
   • Current Actual CG ($\text{CG}_{\text{actual}}$) = $+44.2\text{ in}$
   • Desired Target CG ($\text{CG}_{\text{desired}}$) = $+43.5\text{ in}$ (the aft limit)
   • $\Delta \text{CG} = 44.2 - 43.5 = 0.7\text{ in}$
   • Ballast Arm ($\text{Arm}_{\text{ballast}}$) = $-12.0\text{ in}$
2. Apply the ballast formula: $\text{Ballast Weight} = \frac{W \times \Delta \text{CG}}{\text{CG}_{\text{desired}} - \text{Arm}_{\text{ballast}}} = \frac{1,800.0\text{ lb} \times 0.7\text{ in}}{43.5\text{ in} - (-12.0\text{ in})}$
3. Solve the denominator: $\text{Distance between Ballast and Target CG} = 43.5 - (-12.0) = 43.5 + 12.0 = 55.5\text{ in}$
4. Compute Ballast Weight: $\text{Ballast Weight} = \frac{1,260.0\text{ in-lb}}{55.5\text{ in}} = 22.7027\text{ lb} \approx 22.70\text{ lb}$
5. Proof Check:
   • Original Aircraft Moment = $1,800.0\text{ lb} \times 44.2\text{ in} = 79,560.0\text{ in-lb}$
   • Ballast Moment = $22.70\text{ lb} \times (-12.0\text{ in}) = -272.4\text{ in-lb}$
   • Total New Weight = $1,800.0 + 22.70 = 1,822.70\text{ lb}$
   • Total New Moment = $79,560.0 - 272.4 = 79,287.6\text{ in-lb}$
   • $\text{New CG} = \frac{79,287.6}{1,822.70} = +43.50\text{ in}$ (Matches aft limit exactly). Conclusion: Install $22.70\text{ lb}$ of permanent ballast at Station -12.0 inches.