Solution
2.3.10 Aircraft Weight and Balance
2.3.10.3 Weight and Balance Computation
We now discuss the determination of the aircraft weight and the location of the longitudinal center of gravity, commonly called the computation of aircraft weight and balance. The weight and CG limits, as described in the previous sections, are usually depicted on a weight versus station number chart, called a center of gravity chart, as shown in Figure 2.38. The computed aircraft weight and CG location are plotted on this chart to determine if the aircraft weight and balance are within the specified limits.
The aircraft gross weight is the sum of the basic empty weight and the weight of all items that are loaded into or onto the aircraft. Loaded items include the people (aircrew and passengers), baggage, usable fuel, and external stores, if attached. For safe operation of the aircraft, the computed aircraft gross weight must be less than the specified maximum gross weight.
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2,750
2,500
2,250
78 80 82 84 86
Station (inches) Forward
CG limit
Maximum gross weight
Aft CG limit
88 90 92
Aircraft weight (lb)
2,000
1,750
1,500
Figure 2.38 Aircraft center of gravity chart.
Datum Forward CG limit
Aft CG limit
FS 0 hCG
hi
Wi CG range
Figure 2.39 Aircraft center of gravity limits.
The aircraft center of gravity location is computed as follows. Each item that is loaded in the aircraft is located a certain distance from the datum, hi, as shown in Figure 2.39. The product of the item weight and this distance is a moment about the datum (FS 0). The distance from the datum is called a moment arm, or simply, an arm. The aircraft center of gravity location, hCG, is obtained by dividing the sum of the moments created by each item weight,∑
iMi, by the sum of the item weights,∑
iWi, as given by
hCG=
∑
iMi
∑
iWi (2.52)
The weight of each item is known or is obtained through weighing. The value of each item’s arm is obtained by measuring the distance of the item’s location from the datum.
These distances are usually specified by the aircraft manufacturer to the aircraft operator in the operating manuals, such as the distances to the seats, fuel tanks, baggage compartments, etc. The
k k basic empty weight of the aircraft has an arm and a moment associated with it, and these values
are also supplied by the aircraft manufacturer or are obtained through weighing. The computation of an aircraft weight and balance is illustrated in the example below.
Example 2.10 Calculation of Aircraft Weight and Balance A Beechcraft A36 Bonanza, six-place, high-performance, general aviation aircraft (Figure 2.23) has a basic empty weight of 2230 lb (1012 kg), a maximum takeoff weight of 3600 lb (1632.9 kg) and a maximum landing weight of 3100 lb (1406.1 kg). The moment and arm corresponding to the empty weight is supplied by the aircraft manufacturer as 171,130.2 in-lb (19,335.1 N-m) and 76.74 in (194.9 cm), respectively. At the maximum takeoff weight, the forward and aft center of gravity limits are 81.0 in (205.7 cm) and 87.7 in (222.8 cm), respectively, aft of the datum. At the maximum landing weight, the forward and aft center of gravity limits are 74.0 in (188.0 cm) and 87.7 in (222.8 cm), respectively, aft of the datum.
A 165 lb (74.8 kg) pilot and a 182 lb (82.6 kg) co-pilot are seated in the cockpit at FS 79. Two 180 lb (81.6 kg) passengers are seated in the cabin at FS 117.5. The aircraft has 74 gal (280 liters) of usable fuel in the wing tanks at FS 75. (See Figure 2.40 for fuselage station numbers.)
If the Bonanza takes off, flies for 3 hours, then lands, calculate the zero fuel weight, the takeoff weight and center of gravity location, and the landing weight and center of gravity location. Assume that the aviation gasoline has a weight of 6.0 lb (2.72 kg) per gallon with a fuel burn rate of 15.1 gallons per hour (57.2 liters/h). Also, determine whether the takeoff and landing weights and center of gravity locations are within the allowable limits. Draw a weight versus center of gravity diagram showing all of the limits and the takeoff and landing conditions.
Solution
The following calculations are made in the order given below.
1. The zero fuel weight, Wzero fuel, is the sum of the basic empty weight and the weights of the pilot, co-pilot, and passengers.
Wzero fuel= Wempty+ Wpilot+ Wco−pilot+ Wpassengers
FS 75
FS 79
FS 117.5
Figure 2.40 Beechcraft A-36 Bonanza fuselage station numbers (not to scale).
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Inserting numerical values, the zero fuel weight is
Wzero fuel= 2230 lb + 165 lb + 182 lb + 2 × 180 lb = 2937 lb
2. The fuel load is converted from gallons to pounds using the conversion of 6.0 lb per gallon.
Wfuel= 74 gal × 6.0 lb
gal = 444 lb
3. The moments, M, corresponding to the pilot, co-pilot, passengers, and fuel are the product of their respective weight, W, and arm, h (fuselage station).
Mpilot= (Wh)pilot= (165 lb)(79.00in) = 13,035 in ⋅ lb Mco−pilot= (Wh)co−pilot= (182 lb)(79.00in) = 14,378 in ⋅ lb Mpassengers= (Wh)passengers= (2 × 180 lb)(117.50in) = 42,300in ⋅ lb
Mfuel= (Wh)fuel= (444 lb)(75.0in) = 33,300in ⋅ lb 4. The takeoff weight is the sum of the zero fuel weight and the fuel load.
Wtakeoff= Wzero fuel+ Wfuel= 2937 lb + 444.0lb = 3381lb
5. The takeoff moment is the sum of the moments due to the empty weight, pilot, co-pilot, passen-gers, and fuel load.
Mtakeoff= Mempty+ Mpilot+ Mco−pilot+ Mpassengers+ Mfuel
Mtakeoff= 171, 130.2 + 13,035 + 14,378 + 42,300 + 33,300 = 274,143in ⋅ lb 6. Using Equation (2.52), the takeoff center of gravity is the takeoff moment divided by the takeoff
weight.
hCG, takeoff= Mtakeoff
Wtakeoff = 274,143in ⋅ lb
3381 lb = 81.08in
7. The fuel burn, in gallons, is the flight time (3 hours) multiplied by the fuel burn rate (15.1 gallons/h). The fuel burn weight is then converted from gallons to pounds.
Wfuel burn= (3 h × 15.1 gal) × 6.0 lb
gal = 271.8lb 8. The fuel burn moment is the fuel burn weight multiplied by the fuel arm.
Mfuel burn = (Wh)fuel burn= (271.8lb)(75.0in) = 20,385in ⋅ lb 9. The landing weight is the takeoff weight minus the fuel burn weight.
Wland = Wtakeoff− Wfuel burn= 3381 lb − 271.8lb = 3109.2lb 10. The landing moment is the takeoff moment minus the fuel burn moment.
Mland= Mtakeoff− Mfuel burn
Mland= 274,143in ⋅ lb − 20,385in ⋅ lb = 253,758in ⋅ lb
k k 11. Using Equation (2.52), the landing center of gravity is the landing moment divided by the
landing weight.
hCG,land= Mland
Wland = 253,758in ⋅ lb
3109.2lb = 81.62in
The results of these calculations, along with the data supplied for the empty weight, are used to fill in Table 2.8. The zero fuel weight is 2937.0 lb (1332.2 kg). The takeoff weight and center of gravity location are 3381.0 lb (1553.6 kg) and 81.08 inches (205.9 cm), respectively. The land-ing weight and center of gravity location are 3109.2 lb (1410.3 kg) and 81.62 inches (207.3 cm), respectively.
The weight and center of gravity limits provided in the problem statement are used to develop the center of gravity chart shown in Figure 2.41. The takeoff and landing conditions are placed on the chart. Both conditions are within the acceptable weight and CG limits for safe flight.
Table 2.8 Beechcraft A-36 Bonanza weight and balance data.
Item Weight, W (lb) Arm, h (in) Moment/100, M∕100 (in-lb)
Empty weight 2230.0 76.74 1711.30
Pilot 165.0 79.00 130.35
Co-pilot 182.0 79.00 143.78
Passengers (2) 360.0 117.5 423.00
Zero fuel weight 2937.0 − −
Fuel load 444.0 75.00 333.00
Takeoff weight 3381.0 81.08 2741.43
Fuel Burn 271.8 75.00 203.85
Landing weight 3109.2 81.62 2537.58
3,600
Takeoff
Land 3,400
3,200
3,000 2,800
Aircraft weight, lb
2,600 2,400
2,200 2,000
70 75 80
CG location, inches aft of datum
85 90
Figure 2.41 Beechcraft A-36 Bonanza center of gravity chart.
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