Two-Place Loading With Tail Ballast
Your club's two-place glass trainer has an empty weight of 600 lb at 10.0 in aft of datum, maximum gross weight 1,000 lb, and approved CG range 9.0–15.0 in. You weigh 200 lb and your passenger weighs 150 lb; each parachute is 20 lb and both are loaded at 5.0 in. To keep the CG aft for a weak-thermal day, the previous pilot left 30 lb of lead tail ballast at 70.0 in. The glider has no water ballast; the lead is in 10-lb plates. Density altitude is 5,500 ft and the runway is 2,200 ft. The passenger is a friend who has driven two hours, and the forecast is for 1–2 kt thermals. You have a calculator and the weight and balance sheet.
1. Compute gross weight and CG with the 30 lb of tail ballast, and state whether the loading is within limits.
Read the worked answer
Use moment = weight × arm.
| Item | Weight (lb) | Arm (in) | Moment (lb-in) |
|---|---|---|---|
| Empty glider | 600 | 10.0 | 6,000 |
| Pilot | 200 | 5.0 | 1,000 |
| Passenger | 150 | 5.0 | 750 |
| Parachutes | 40 | 5.0 | 200 |
| Tail ballast | 30 | 70.0 | 2,100 |
| Total | 1,020 | 10,050 |
CG = 10,050 ÷ 1,020 = 9.85 in aft of datum. The CG is inside the 9.0–15.0 in range, but gross weight is 1,020 lb, which is 20 lb over the 1,000 lb maximum. The glider is not legal to fly as loaded.
2. Compute the loading with the ballast removed, and explain why simply dumping the ballast is not a valid fix.
Read the worked answer
Without ballast: weight = 600 + 200 + 150 + 40 = 990 lb. Moment = 6,000 + 1,000 + 750 + 200 = 7,950 lb-in. CG = 7,950 ÷ 990 = 8.03 in, which is 0.97 in forward of the 9.0 in forward limit. So removing ballast fixes the overweight condition but creates a forward-CG condition. The minimum ballast b for CG 9.0 solves (7,950 + 70b)/(990 + b) = 9.0, giving 61b = 960, so b ≈ 15.7 lb. At 16 lb the weight is 1,006 lb, still over gross. No ballast amount between 16 and 30 lb satisfies both limits: at 30 lb the CG is legal but weight is over; at 16 lb the weight is still over; at 0–15 lb the CG is forward of limit. The loading is unworkable as proposed.
3. Recommend a specific loading or operational decision that respects both weight and CG limits.
Read the worked answer
The flight should not go as loaded. A practical fix is to fly solo or with a lighter passenger. For example, a 140-lb passenger with 16 lb of ballast gives 600 + 200 + 140 + 40 + 16 = 996 lb; moment = 6,000 + 1,000 + 700 + 200 + 1,120 = 9,020; CG = 9.06 in. That is within both limits. Flying solo with about 10 lb of ballast also works: 600 + 200 + 20 + 10 = 830 lb; moment = 6,000 + 1,000 + 100 + 700 = 7,800; CG = 9.40 in. Do not simply remove all ballast, and do not add nose weight to fix an aft-CG problem. At 5,500 ft density altitude and a 2,200-ft runway, the lighter loading also improves climb and margin. Explain the arithmetic to the passenger and offer a later flight rather than accepting an illegal loading.