Aft centre of gravity in a loaded PA-28-161
You are loading a Piper PA-28-161 for a private day VFR trip. The AFM uses pounds and inches. Basic empty weight is 1,350 lb with an empty CG of 84.0 in aft of datum. Maximum takeoff weight is 2,325 lb. CG limits are 80.5 in to 91.5 in aft of datum. Station arms are: front seats 80.5 in, rear seats 118.0 in, baggage 142.0 in, fuel 95.0 in. Fuel load is 235 lb (about 39 US gal usable). On board: pilot 170 lb (front), front passenger 160 lb, rear passenger 150 lb, rear passenger 160 lb, baggage 100 lb. The baggage compartment limit is 100 lb. The aircraft has a current maintenance release. Your destination is a 1.5-hour flight with no refuelling available; you plan to land with at least 45 minutes reserve. You need to decide whether the loading is acceptable and, if not, what to change.
1. Compute total weight and CG; determine compliance.
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Use the moment method. Weights x arms:
- BEW: 1,350 x 84.0 = 113,400 lb-in
- Pilot: 170 x 80.5 = 13,685
- Front passenger: 160 x 80.5 = 12,880
- Rear passenger 1: 150 x 118.0 = 17,700
- Rear passenger 2: 160 x 118.0 = 18,880
- Baggage: 100 x 142.0 = 14,200
- Fuel: 235 x 95.0 = 22,325 Total weight = 2,325 lb. Total moment = 213,070 lb-in. CG = 213,070 / 2,325 = 91.64 in.
The weight is exactly at the 2,325 lb maximum, so it is not below max. The CG limit is 91.5 in aft; 91.64 in is 0.14 in aft of the limit. The loading is therefore not acceptable: it is both at max weight and outside the aft CG limit.
2. What is the smallest practical change to bring CG within limits?
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The smallest practical change is to remove baggage. Let x be the baggage removed from the 142.0 in station. Solve for CG = 91.5 in: (213,070 - 142x) / (2,325 - x) = 91.5 213,070 - 142x = 212,737.5 - 91.5x 332.5 = 50.5x x = 6.58 lb.
So removing about 7 lb of baggage brings the CG to the aft limit, and removing 10 lb gives a small margin. New weight with 10 lb removed = 2,315 lb; new moment = 213,070 - 1,420 = 211,650 lb-in; new CG = 91.43 in, within limits. This is better than offloading fuel because it preserves range and reserve. The pilot should also recheck that the baggage compartment remains within its 100 lb limit and secure the load.
3. If the pilot refuses to remove baggage, how much fuel must be offloaded?
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Offload fuel at the 95.0 in station. Solve for x pounds of fuel removed: (213,070 - 95x) / (2,325 - x) = 91.5 213,070 - 95x = 212,737.5 - 91.5x 332.5 = 3.5x x = 95 lb.
So about 95 lb of fuel must be removed, leaving 235 - 95 = 140 lb, roughly 23 US gal. That may still satisfy the 1.5-hour flight and 45-minute reserve, but it is a large reduction and leaves the aircraft at the aft CG limit with no margin. The better operational decision is to remove baggage or redistribute load, not to trade away fuel unless the resulting fuel plan is checked carefully. If the pilot removes fuel, recalculate weight, CG, and reserve before departure.