Loading the Archer: Weight, Balance, and a Hot High Departure
A client has loaded a PA-28-181 for a business trip and asks you to sign off the flight. Empty weight is 1,650 lb with an empty moment of 147,015 lb·in. Moment arms: pilot and front passenger 85.5 in, each rear seat 118.1 in, baggage 142.8 in, fuel 95.0 in. Load: pilot 185 lb, front passenger 175 lb, two rear passengers at 165 lb each, 40 lb of baggage, and 48 gal of usable fuel at 6 lb/gal. Maximum takeoff weight is 2,550 lb; the CG range is 82.0 to 95.0 in. The departure airport is at 3,800 ft MSL, altimeter 29.92, OAT 30 degrees C. The runway is 2,900 ft long with rising terrain off the departure end.
1. Compute takeoff weight and CG. May the airplane legally depart as loaded?
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Weights: 1,650 + 185 + 175 + 330 + 40 + 288 (48 gal x 6) = 2,668 lb.
Moments (lb·in):
- Empty: 147,015
- Pilot: 185 x 85.5 = 15,817.5
- Front pax: 175 x 85.5 = 14,962.5
- Rear pax: 330 x 118.1 = 38,973
- Baggage: 40 x 142.8 = 5,712
- Fuel: 288 x 95.0 = 27,360
- Total: 249,840
CG = 249,840 / 2,668 = 93.6 in.
That CG is inside 82.0-95.0, so balance is fine. Weight is not: 2,668 - 2,550 = 118 lb over maximum gross. The airplane may not depart this way. Note also how close 93.6 in is to the 95.0 aft limit - any substitution of a heavier rear passenger, or a bag thrown in at the last minute, pushes you out the back of the envelope where stall and spin recovery degrade.
2. The pilot proposes to offload 20 gallons of fuel, keeping 28 gallons aboard. Evaluate that plan: does it solve the problem, and what does it cost in CG and range?
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28 gal = 168 lb, so weight becomes 1,650 + 185 + 175 + 330 + 40 + 168 = 2,548 lb. Technically legal - by 2 lb.
A 2-lb margin is not a plan. Any ramp CHECK addition (a chart bag, a coat, a passenger's actual weight versus their stated weight) puts you back over gross, and you will be departing at a density altitude near 6,500 ft. CG barely moves: 249,840 - (120 x 95.0) = 238,440; / 2,548 = 93.6 in, unchanged to the nearest tenth.
Range cost: 168 lb / 6 = 28 gal. At roughly 10 gph that is 2.8 hr total, and reserving 45 min (7.5 gal) leaves about 2.0 hr of useful endurance - on the order of 230 nm at 115 kt. If the trip is longer than that, the plan fails twice.
The better fix is to remove the 118 lb from the back, not the tanks: offloading one 165-lb rear passenger plus the 40 lb of baggage gives 1,650 + 185 + 175 + 165 + 288 = 2,463 lb, and CG falls to about 91 in - a genuine margin in both directions.
3. Compute density altitude for the departure and explain what it means for the 2,900-ft runway and the rising terrain.
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Pressure altitude = 3,800 ft (altimeter 29.92). ISA temperature at 3,800 ft is about 15 - (2 x 3.8) = 7.4 degrees C. The spread is 30 - 7.4 = 22.6 degrees C, and each degree above ISA adds roughly 120 ft:
DA = 3,800 + (120 x 22.6) = approximately 6,510 ft.
Operating implications:
- Normally aspirated takeoff and climb performance at 6,500 ft DA is a fraction of the sea-level book numbers; ground roll and the 50-ft obstacle distance both grow sharply. A 2,900-ft runway that is comfortable at ISA becomes a decision, not a formality.
- True airspeed is higher for the same indicated speed, so the airplane covers more ground while accelerating and while climbing - the obstacle ahead arrives sooner.
- Rising terrain off the departure end removes the option of trading altitude for airspeed after liftoff.
Mitigations to brief: depart in the cool of the morning, reduce to well under gross, lean for maximum power before brake release, plan Vx (best angle) to the obstacle rather than Vy, and identify a specific abort point on the runway. If no combination gives a comfortable margin over the terrain, the correct answer is to wait.