FAA Private Pilot Airplane (PAR)Performance, Weight and BalanceEasy
An airplane's empty weight is 1,350 lb at an arm of 78.0 in. The pilot and passenger weigh a total of 340 lb at an arm of 85.0 in. Fuel is loaded at 30 gallons (6 lb/gal) at an arm of 95.0 in, and baggage of 40 lb is loaded at an arm of 140.0 in. What is the airplane's center of gravity?
- A85.0 in
- B90.5 in
- C78.0 in
- D82.1 in
Show answer & explanationAnswer & explanation
Correct answer: D. 82.1 in
Total weight = 1,350 + 340 + 180 + 40 = 1,910 lb. Total moment = (1,350×78.0)+(340×85.0)+(180×95.0)+(40×140.0) = 105,300+28,900+17,100+5,600 = 156,900 lb-in. CG = 156,900 ÷ 1,910 = 82.1 in.
Why the other options are wrong
- A. This is only the arm of the occupants, not the overall CG.
- B. This value is too far aft and does not match the calculated total moment.
- C. This is only the empty weight's arm, not the loaded CG.
Weight and Balance CG Calculation
CG location is found by dividing total moment (sum of each item's weight × arm) by total weight.
- Moment = weight × arm
- Total moment ÷ total weight = CG
- Each item (empty weight, occupants, fuel, baggage) contributes its own moment
Memory trick: Weigh it, arm it, sum it, divide it.