Weight and Balance
Center of Gravity Computation
The first step in the solution of any weight and balance problem is the calculation of the total weight of the aircraft (gross weight) and the total moment. All weight and balance problems on the ATP-121 test use a moment index rather than the actual moment. The moment index is the actual moment divided by 1,000. Questions 8697 through 8711 require the calculation of the total weight and moment index for a Boeing 727-type aircraft. To determine the total weight and moment index, a separate weight and moment must be calculated for the Basic Operating Weight, the passenger loads in the forward and aft passenger compartments, the cargo loads in the forward and aft cargo compartments, and the fuel loads in fuel tanks 1, 2, and 3. The following example references Question 8697.
Basic Operating Weight (BOW) is defined as the empty weight of the aircraft plus the weight of the required crew, their baggage and other standard items such as meals and potable water. The BOW and the Basic Operating Index (Moment/1,000) are the same for all questions. The BOW is 105,500 pounds and the Basic Operating Index is 92,837. See FAA Figure 79.
The number of passengers is stated for each question. For example, Question 8697 refers to Load Condition WT-1. (See FAA Figure 76.) Load Condition WT-1 states that there are 18 passengers in the forward compartment and 95 passengers in the aft compartment. The weight of the passengers can be determined by use of the Passenger Loading Table in the upper left-hand corner of FAA Figure 80. Since neither 18 passengers for the forward compartment nor 95 passengers for the aft compartment is listed in the table, the weight must be calculated by multiplying the number of passengers times the average weight per passenger. A quick examination of the table reveals that the average passenger weight is 170 pounds. The weights are:
FWD Comp = 18 x 170 lbs = 3,060 lbs
AFT Comp = 95 x 170 lbs = 16,150 lbs
The Moment Index (MOM/1,000) is calculated by using the formula:
Weight x Arm/1,000 = MOM/1,000
The arms for the passenger compartments are listed at the top of each of the compartment loading tables after the words, "Forward Compartment Centroid" and "Aft Compartment Centroid." The arm for the forward compartment is 582.0 inches, and the aft compartment arm is 1028.0 inches. The easiest way to apply the 1,000 reduction factor is to move the decimal on the arm three places to the left (i.e., 582.0"/1,000 = .582). In the example used, the Moment/1,000 for the forward and aft passengers compartments (rounded to the nearest whole number) are:
FWD Comp Moment/1,000 = 3,060 x .582 = 1,781
AFT Comp Moment/1,000 = 16,150 x 1.028 = 16,602
The weights for the forward and aft cargo holds are stated for each question. For example, Load Condition WT-1 states that there is 1,500 pounds in the forward hold and 2,500 pounds in the aft hold. The Moment/1,000 can be determined from the tables in the upper right-hand corner of FAA Figure 80. For example, the Moment/1,000 for 1,500 pounds in the forward cargo hold is determined by adding the Moment/1,000 for 1,000 pounds (680) and the Moment/1,000 for 500 pounds (340). If necessary, the Moment/1,000 can also be determined by multiplying weight times arm (divided by 1,000). The Moment/1,000 for the cargo holds are:
FWD Hold = 1,020
AFT Hold = 2,915
Fuel tanks 1 and 3 are the wing tanks and are always loaded with the same weight of fuel. They will always have the Moment/1,000 as well. The number 2 tank is the center fuselage tank and will often have a fuel weight different from tanks 1 and 3. It will always have a different Moment/1,000. For example, Load Condition WT-1 states that the fuel load in tanks 1 and 3 is 10,500 pounds each and that the load in tank 2 is 28,000 pounds. The Moment/1,000 for each tank is determined from the table in the bottom portion of FAA Figure 80. The Moment/1,000 can be calculated, if necessary, by multiplying weight times arm (divided by 1,000). Notice that the arm varies with the fuel load in each tank. The Moment/1,000 for each tank is:
Tank 1 Moment/1,000 = 10,451
Tank 3 Moment/1,000 = 10,451
Tank 2 Moment/1,000 = 25,589
The total weight and total Moment/1,000 is the sum of all the items discussed above. The Total Weight and Moment/1,000 for Load Condition WT-1 is:
Weight Moment/1,000
BOW 105,500 92,837
18 PAX FWD 3,060 1,781
95 PAX AFT 16,150 16,602
FWD Cargo 1,500 1,020
AFT Cargo 2,500 2,915
Fuel Tank 1 10,500 10,451
Fuel Tank 3 10,500 10,451
Fuel Tank 2 + 28,000 + 25,589
Total 177,710 161,646
The Center of Gravity (CG) in inches aft of the Datum line can be determined by using the formula:
CG = Total Moment / Total Weight
Since these questions use a Moment Index instead of Moment, it is necessary to modify this formula by multiplying the (Total Moment/Total Weight) by the reduction factor (1,000). The formula then becomes:
CG = (Total Moment Index / Total Weight) x 1,000
Using the weight and Moment/1,000 we calculated above:
CG = (161,646/177,710) x 1,000 = 909.6 inches
The Center of Gravity of a properly loaded airplane must always fall somewhere along the Mean Aerodynamic Chord (MAC). The CG is often expressed as a percent of MAC. If the CG was at the Leading Edge of MAC (LEMAC), it would be at 0% of MAC. If it were at the Trailing Edge of MAC (TEMAC), it would be at 100% of MAC. The CG's percent of MAC is calculated by:
1. Determine the CG in inches aft of LEMAC by subtracting the distance Datum to LEMAC from the CG in inches aft of Datum. The distance from Datum to LEMAC is given in FAA Figure 79 as 860.5 inches. This is used for all calculations of percent of MAC for the 727. The CG in inches aft of Datum is calculated in the previous paragraph. Using those numbers:
CG (inches aft of LEMAC) = 909.6" - 860.5" = 49.1 inches
2. Determine the CG in percent of MAC by dividing the CG in inches aft of LEMAC by the length of MAC. The length of MAC is distance in inches from LEMAC to TEMAC. It is given in FAA Figure 79 and is 180.9 inches. The formula is:
CG (% of MAC) = (CG in inches aft of LEMAC ÷ MAC) x 100%
Using the numbers from above:
CG (% of MAC) = (49.1" ÷ 180.9") x 100% = 27.1%
Stabilizer Trim Setting
The correct horizontal stabilizer trim setting is very critical for proper takeoff performance of jet aircraft. The main determinants are the CG location and possibly the flap setting. Some aircraft, such as the DC-9, have their stabilizer trim indicators calibrated in percent of MAC, so it is necessary to calculate the CG to know the trim setting. Other aircraft (such as the B-737 and B-727) have their trim indicators marked off in units of nose up trim. In such cases it is necessary to refer to the trim table to determine the proper setting for a given CG. See FAA Figure 55.
The Stab Trim Setting Table at the bottom left side of FAA Figure 55 is used to determine the takeoff trim setting for a B-737. CG location in percent of MAC is used to determine the setting. For example, if the CG is at 8.0% of MAC, the stab trim setting is 7-3/4 units ANU (Airplane Nose Up).
The Stab Trim Setting Table at the left side of FAA Figure 83 is used to determine the takeoff trim setting for a B-727. Flap setting and CG location in percent of MAC are used to determine the setting. For example, if the CG is at 28% of MAC and the flaps are set at 15°, the stab trim setting is 4-1/2 units ANU.
Changing Loading Conditions
Anytime weight is either added to or subtracted from a loaded airplane, both the gross weight and the center of gravity location will change. The solution of such a problem is really a simplified loading problem. Instead of calculating a weight and moment for every section of the aircraft, it is only necessary to compute the original weight and moment and then the effect the change in weight had. Often in these problems the original CG is expressed in percent of MAC and it is necessary to convert this to an arm for the entire aircraft. The following example references Question 8578.
It is sometimes necessary to convert a CG position expressed in percent of MAC to the CG in inches aft of Datum. This is just the reverse of the process described above. This is done in two steps.
1. Convert the CG in percent of MAC to CG in inches aft of LEMAC. This is done by using the formula:
CG (inches aft of LEMAC) = (CG % of MAC ÷ 100%) x MAC.
Load Condition WS-1 (FAA Figure 44) gives a CG of 22.5% and a length of MAC of 141.5 inches. The formula is:
CG (inches aft of LEMAC) = (22.5% ÷ 100%) x 141.5" = 31.84 inches.
2. Add the CG in inches aft of LEMAC to the Distance from Datum to LEMAC. In FAA Figure 44, LEMAC is 549.13 inches aft of Datum.
CG (inches aft of Datum) = 549.13" + 31.84" = 580.97 inches.
Use Question 8578 and Conditions WS-1 (FAA Figure 44) for this example. Use the original weight and CG to calculate the original Moment/1,000. Next use the weight change and station to determine the Moment/1,000 change.
Weight Moment/1,000
Original Weight 90,000 52,287.08
Weight Change - 2,500 - 880.25
New Weight 87,500 51,406.83
Note: A reduction in weight results in a reduction in Moment/1,000. An increase in weight results in an increase in Moment/1,000.
Determine the new CG:
CG = (51,406.83 ÷ 87,500) x 1,000 = 587.51 inches
Convert CG to percent of MAC:
CG (inches aft of LEMAC) = 587.51" - 549.13" = 38.38"
CG (% of MAC) = (38.38" ÷ 141.5") x 100% = 27.1%
When a portion of an aircraft's load is shifted from one location to another, the CG of the loaded aircraft will change as well. Also, the CG will follow the weight. That is, if weight is shifted rearward, the CG will move rearward as well; and if weight is shifted forward, the CG will move forward. To calculate the effect of a weight shift on CG position, three numbers must be known: the weight shifted, the distance the weight was moved, and the total weight of the aircraft. The formula used is:
Change in CG = (Weight Shifted x Distance Shifted) ÷ Total Weight
Question 8573 asks what the effect on CG is if weight is shifted from the forward to aft cargo compartment under Load Condition WS-1 (See FAA Figure 44). Load Condition WS-1 gives the total weight as 90,000 pounds and the weight shifted as 2,500 pounds. The distance shifted is the difference between the forward compartment centroid (352.1 inches) and the aft compartment centroid (724.9 inches), which is 372.8 inches (724.9 - 352.1).
Note: These centroids are distances aft of the Datum line. The index arms are distances from the CG Index and will be discussed in a later example. Notice however, that the difference between the two index arms is also 372.8 inches (144.9 - (-227.9) = 372.8). The solution is:
Change in CG = (2,500 lbs x 372.8") ÷ 90,000 lbs = 10.4"
If weight is shifted forward, the CG will move forward as well. This is expressed by writing the distance shifted as a negative number. If weight is shifted from the aft to the forward cargo compartment, the distance shifted is -372.8 inches. For example, Question 8574 asks about such a shift of 1,800 pounds with an aircraft total weight of 85,000 pounds. The formula is:
Change in CG = (1,800 lbs x (-372.8)) ÷ 85,000 lbs = -7.89"
Questions 8573 and 8576 require an answer in percent of MAC. The change in CG can be converted to a percent of MAC by using the formula:
Change in CG (% of MAC) = (Change in CG/MAC) x 100%
Questions 8577 through 8581 express CG as an Index Arm. Index Arm is the distance, in inches, from an index set at a point close to the normal CG location. A positive Index Arm is a point aft of the index and a negative Index Arm is a point forward of the index. For example, FAA Figure 44 shows LEMAC as having an Index Arm of -30.87 inches or 30.87 inches forward of the index. The index point for all questions on this test is 580.0 inches. CG in Index Arm is calculated by the formula:
CG (Index Arm) = CG (inches aft of Datum) - 580 inches
Using the data from Load Condition WS-5, the formula is:
CG (Index Arm) = 585.21" - 580" = +5.21 inches
Questions 8429 through 8433 require calculation of the maximum weight that can be carried on a pallet. The limiting factor is the amount of weight that the aircraft floor can support per square foot. The calculation refers to FAA legends and involves the following steps:
1. Determine the area covered by the pallet. This is done by multiplying the pallet width by length. Since the pallet dimensions are in inches and the floor load limit is expressed in square feet, it is necessary to convert the pallet area from square inches to square feet. This is done by dividing by 144 (there are 144 square inches in a square foot). The formula is:
Pallet Area (square feet) = (Width x Length) ÷ 144
Using the example of Question 8429 which is a pallet 76" x 76", the area covered is:
Pallet Area = (76" x 76") ÷ 144 = 40.11 square feet
2. Determine the floor load limit by multiplying the pallet area in square feet times the floor load limit per square foot. Again using the example of Question 8429, if the floor load limit is 186 lbs/sq ft:
Floor Load Limit = 40.11 sq ft x 186 lbs/sq ft = 7,460.7 pounds
3. Determine the cargo weight which can be placed on the pallet by subtracting the weight of the pallet and tie-down devices. Since the floor has to support the pallet and tiedown weight, this reduces the total cargo which can be placed on the pallet. Once again, using the example of Question 8429, where the pallet weighs 93 pounds and the tiedown devices weigh 39 pounds (132 pounds total):
Allowable Weight = 7,460.7 lbs - 132 lbs = 7,328.7 pounds
Beech 1900 Weight and Balance
Note: By definition, "Basic Empty Weight" does not include crew weight, so you must include crew in the calculation. By definition, "Basic Operating Weight" includes crew weight so you do not include crew in the calculation.
Helicopter Weight and Balance: CG Shifts
These questions require a re-computation of CG based on a shift of weight only, i.e., CG will change but total weight does not change. AC 91-23A, Chapter 5 gives us a formula for working this type of problem.
Weight Shifted Change of CG
Total Weight = Distance of Shift
These problems may also be worked with a flight computer as shown in AC 91-23A, Chapter 5 in the following manner:
1. Set Weight Shifted (mile scale) over Total Weight (minute scale).
2. Find the Change in CG on the mile scale over the distance shifted on the minute scale.
Question 8518 is solved using both methods.
Helicopter Weight and Balance: Load Limits
In these questions, it will be necessary to compute both a takeoff and a landing weight and balance. Since the stations (CG) for fuel vary with weight, the most simple method of solving these problems is to compute the zero fuel weight for the given conditions, then perform a separate weight and balance for takeoff and landing. Some moments are given; others are not and therefore must be computed. Also, the fuel is stated in gallons, not pounds, which can be converted using the Jet A Table (FAA Figure 33).
Helicopter Weight and Balance: Lateral CG
These questions are answered by using the formula given in AC 91-23A.
1. For shifted weight:
Weight Shifted (WS) CG Shift (CS)
Total Weight (TW) = Distance Shifted (DS)
2. For added/removed weight (WA or WR):
(WA or WR) CG Shift (CS)
New Total Weight (NTW) = Distance shifted (DS)
Refer to answers to Questions 8523 through 8527 for total weights.
Floor Loading Limits
In addition to ensuring that an aircraft is loaded within its weight and balance limits, it is important to make sure that the floor of a cargo compartment is not overloaded. The load limit of a floor is stated in pounds per square foot. The questions on the test require you to determine the maximum load that can be placed on a pallet of certain dimensions.
For example: what is the maximum weight that may be carried on a pallet which has the dimensions of 37 x 39 inches, when the floor load limit is 115 pounds per square foot, the pallet weight is 37 pounds, and the weight of the tiedown devices is 21 pounds?
The first step is to determine the area of the floor (in square feet) covered by the pallet. This is done by multiplying the given dimensions (which calculates the area in square inches) and dividing by 144 (which converts the area to square feet).
37 inches x 39 inches ÷ 144 square inches = 10.02 square feet.
The next step is to determine the total weight that the floor under the pallet can support, by multiplying the area times the floor load limit given in the question.
10.02 square feet x 115 pounds per square foot = 1,152.39 pounds.
The final step is to determine the maximum weight which can be placed on the pallet by subtracting the weight of the pallet and the tiedown devices from the total load limit.
1,152.39 pounds - 58 pounds = 1,094.39 pounds.
The weight on the pallet must be equal to or less than this number (1,094.39, in this example). If it is more than this number, the combination of cargo, pallet, and tiedown weight would exceed the floor load limit. A review of the test questions reveals that the closest answer choice is always equal to or slightly less than the floor limit. All the calculations in this section were performed with a calculator carrying all digits to the right of the decimal point forward for the next step of the problem. The explanations show only two places to the right of the decimal.
A variation of the pallet loading problem is to determine the minimum floor load limit (in pounds per square foot) required to carry a particular loaded pallet. For example: what is the minimum floor load limit to carry a pallet of cargo with a pallet dimension of 78.9 inches x 98.7 inches, and a combination weight of pallet, cargo, and tiedown devices of 9,896.5 pounds?
The first step is to determine the floor area, multiplying the dimensions and dividing by 144 (78.9 x 98.7 ÷ 144 = 54.08 square feet). The second step is to determine the minimum required floor limit by dividing the total weight of the pallet, cargo, and tiedowns by the pallet area (9,896.5 ÷ 54.08 = 183.00 pounds). The correct answer must be at or above this weight (183.00 pounds, in this example).
