4: Weight and Balance

Loading Definitions

Payload Computations

Definitions MAC/Datum

Center of Gravity (CG) Location

Weight × Arm = Moment

Total Moment ÷ Total Weight = CG

Index Units

Computing Aircraft CG and % MAC

Adding/Subtracting Weight and

Maximum Weight Addition

Shifting Weight

Computing Required Weight Shift

Loading Definitions

There are four kinds of weight and balance questions on the Flight Engineer test: (1) figuring the payload, (2) computing the center of gravity of three weights, or an aircraft, (3) computing the effect of adding or subtracting weight, and (4) computing the effect of shifting weight.

Empty Weight of an aircraft includes the weight of the airframe, engines, all permanently-installed equipment, unusable fuel and undrainable oil. Empty weight does not include usable fuel or oil, crew, passengers or other payload.

Basic Operating Weight (BOW) is applicable to transport category aircraft. BOW includes empty weight plus the required crew and other standard operating items such as meals and water.

Payload—the weight of the passengers, baggage and cargo.

Zero Fuel Weight—BOW plus the payload.

Ramp Weight (taxi weight)—the zero fuel weight plus all the usable fuel on board.

Takeoff Weight—the ramp weight less the fuel burned during start and taxi. Note: all calculations for the current flight engineer test assume zero fuel burn for taxi and use ramp weight and takeoff weight as the same value.

Landing Weight—the takeoff weight of the aircraft less the fuel burned and dumped en route.

There are maximum structural limits that must be observed for the zero fuel weight, ramp weight, takeoff weight and landing weight. See Figure 4-1.

Figure 4-1. Aircraft weight nomenclature (Transport category airplanes)

Payload Computations

There are several problems which require you to compute the maximum allowable payload. In order to account for all the limitations on weight, you should use a format similar to the one in Figure 4-2.

For example:

BOW...............................................101,500 lbs

Maximum zero fuel weight..............138,000 lbs

Maximum takeoff weight.................184,200 lbs

Maximum landing weight................142,500 lbs

What is the maximum payload which can be carried if the total fuel load is 52,000 pounds and 45,500 pounds of fuel will be burned en route?

1. Fill in the six known weights (See Figure 4-3). The heaviest possible landing weight will give you the greatest payload, so use 142,500 pounds, which is the aircraft's maximum landing weight. In this case, the computed takeoff weight is greater than the maximum takeoff weight.

2. Since all maximum limits must be observed, the maximum takeoff weight must be used for the next step. See Figure 4-4.

3. Compute the zero fuel weight by subtracting the fuel load from the takeoff weight. In this case the zero fuel weight is less than the maximum. The lower number must be used for the next step to avoid exceeding the maximum takeoff weight. See Figure 4-5.

4. Subtract the zero fuel BOW from the trip zero fuel weight to determine the maximum allowable pay-load for this trip.

132,200 - 101,500 = 30,700 lbs

Figure 4-2. Maximum payload problem

Figure 4-3. Maximum payload problem (step 1)

Figure 4-4. Maximum payload problem (step 2)

Figure 4-5. Maximum payload problem (step 3)

Definitions MAC/Datum

The Mean Aerodynamic Chord (MAC) is the average distance from the leading edge to the trailing edge of the wing. The MAC is specified for an aircraft by determining the average chord of an imaginary wing which has the same aerodynamic characteristics as the actual wing.

The Datum is an imaginary vertical plane or line from which all measuring of distance on the airplane is done. See Figure 4-6.

Arm is the horizontal distance in inches from the datum to the center of gravity on an item. For example, if the rear baggage compartment is 1,066 inches from the datum, the arm of any item placed there is 1,066 inches. Station is generally considered to be the same as the arm.

LEMAC stands for Leading Edge of Mean Aerodynamic Chord. Its position is expressed in inches aft of the datum.

TEMAC stands for Trailing Edge of the Mean Aerodynamic Chord. It is also defined in inches aft of the datum. Since TEMAC is always aft of LEMAC, it will always be a larger number than LEMAC.

Trim Station (TS) 0.0 is a point aft of the datum, which is a new zero reference. In some of the test questions, loading stations are referenced to the 0.0 TS. For example, a loading station at TS-229" is 229 inches ahead of the 0.0 trim station. The location of the 0.0 trim station is always given in the test question data when trim station is used.

Figure 4-6. CG terms

Center of Gravity (CG) Location

The center of gravity of an airplane is computed along its longitudinal axis. The CG location can be expressed in inches aft of the datum, inches aft of LEMAC or as a percentage of MAC. Allowable center of gravity locations always fall between LEMAC and TEMAC. The center of gravity position of large air-craft is normally expressed as a percent of the mean aerodynamic chord aft of LEMAC. If the CG were at LEMAC it would be at 0% of MAC. If it were at TEMAC it would be at 100% of MAC.

Weight × Arm = Moment

The basic formula used for all center of gravity computations is:

Weight × Arm = Moment

Moment, which is expressed in pound-inches, is the effect a weight has on the CG location. If cargo is moved from the aft cargo hold to the forward, the arm of the cargo would decrease and so the total moment would decrease. When the total moment decreases and the weight remains the same, the CG must move forward. This is true of even small weight shifts within the aircraft. If the landing gear moves forward when retracted, the CG will move forward a small amount because of the decrease in the total moment. If the landing gear retracts rearward, the CG will move rearward.

Total Moment ÷ Total Weight = CG

The location of the center of gravity in inches aft of the datum is found by using the formula:

Total moment ÷ Total weight = CG (in inches aft of datum)

For example, compute the CG of the following:

Weight X...................................1,330 pounds at 117" aft of datum

Weight Y...................................1,110 pounds at 110" aft of datum

Weight Z...................................750 pounds at 210" aft of datum

1. Compute the total weight and total moment.

Weight

×

Arm

=

Moment

X

1,330

×

117

=

155,610

Y

1,100

×

110

=

122,100

Z

+ 750

×

210

=

157,500

Total

3,190

435,210

2. Total moment ÷ Total weight = CG

435,210 ÷ 3,190 = 136.43"

Notice that the formula uses total moment. If you are given a moment index, you must multiply either the index or the CG times the reduction factor to get the correct CG.

Index Units

The distance from the datum to the CG is the arm of the entire aircraft. Suppose an aircraft weighs 150,000 pounds with the CG located 920" aft of the datum. The total moment of the aircraft is computed:

Weight × Arm = Moment

150,000 × 920 = 138,000,000 pound-inches

A Moment Index is moment divided by a constant such as 100, 1,000 or 10,000. The moment of 138,000,000 pound-inches could also be expressed as a moment index of 138,000 if a reduction factor of 1,000 is applied, or as 13,800 with a reduction factor of 10,000. The advantage of using an index is that it simplifies weight and balance computations for large aircraft where heavy weights and long arms would otherwise result in unmanageably large numbers.

Weight × Arm ÷ Reduction factor = Index units

Computing Aircraft CG and % MAC

The CG of a loaded aircraft can be computed using the Weight x Arm = Moment formula. The tables in FAA Figure 31 are available for the turbojet questions. The tables show the moment index for various weights at all the loading stations. The arm is given for each station. For the passengers, the arm is printed at the top of the moment index column. The cargo arms are above the weight columns. The fuel loading tables show the arm for each fuel load. If you multiply weight times the arm, be sure to divide the moment by the reduction factor (1000) to get the moment index. This is necessary when the weight in the question is not on the chart; in this case, the formula becomes:

Weight × Arm ÷ 1,000 = Moment Index (Moment/1,000)

In most computations you will be given a CG location in inches aft of the datum that you will express in percent of MAC. To do this, you need to know (1) the distance, in inches, from LEMAC to the CG and (2) the length of MAC in inches.

You can find the distance from LEMAC to the CG by finding the difference between the distance from datum to LEMAC and the distance from datum to the CG. The length of MAC is the difference between LEMAC and TEMAC.

The CG in percentage of MAC is computed by dividing the CG (in inches aft of LEMAC) by the length of MAC and multiplying by 100 to make it a percentage. Assume that the CG is at 924.2" aft of the datum and that MAC runs from station 860.2" (LEMAC) to 1040.9" (TEMAC).

1. Compute the length of MAC.

TEMAC - LEMAC = 1040.9 - 860.2 = 180.7"

2. Compute the distance of the CG aft of LEMAC.

CG - LEMAC = 924.2 - 860.2 = 64.0"

3. Compute the percent of MAC.

(CG aft of LEMAC ÷ MAC) × 100 = (64.0 ÷ 180.7) × 100 = 35.4%

For example:

Determine the CG in percent of MAC. Refer to FAA Figure 31.

Basic Operating Weight.......................................105,000 pounds

Basic Operating Index (Moment / 1000)................92,827.0

MAC......................................................................860.2 to 1040.9

Passenger Load:

Fwd Comp.......................................................Full

Aft Comp.........................................................Full

Fuel Load:

Tanks 1 and 3 (each)......................................10,500 pounds

Tank 2.............................................................26,000 pounds

Cargo Load:

Fwd Hold.........................................................2,500 pounds

Aft Hold...........................................................1,500 pounds

1. Determine the loaded weight and moment index using the tables. Interpolate as necessary, or multiply Weight × Arm and divide by 1,000, if the exact weight and moment index are not given.

Weight

Moment/1000

BOW

105,000

92,827

Fwd Pax

4,930

2,869

Aft Pax

22,610

22,243

Tank 1

10,500

10,451

Tank 3

10,500

10,451

Tank 2

26,000

23,767

Fwd Hold

2,500

1,700

Aft Hold

1,500

1,749

Total

183,450

167,057

2. Compute the CG in inches aft of datum using the formula:

Total Moment/Index × Reduction Factor ÷ Total Weight = CG

CG = 167,057 × 1,000 ÷ 183,540 = 910.2"

3. Determine the CG distance aft of LEMAC.

CG = 910.2 - 860.2 = 50.0"

4. Calculate the CG in percent of MAC.

Distance aft of LEMAC ÷ MAC × 100 = 50.0 ÷ 180.7 × 100 = 27.7%

If the number of passengers is not listed on FAA Figure 31, multiply by 170 to get the weight; then use Weight × Arm ÷ 1,000 = Moment/Index.

Adding/Subtracting Weight and

Computing CG in Inches From % MAC

Anytime weight is added or subtracted from an aircraft, the center of gravity will move. Consider the following problem:

What is the location of the CG if 1,460 pounds are removed from station 1500?

Aircraft Weight....................171,520 lbs

CG Location........................Station 820

1. Compute the current total moment for the loaded aircraft.

Weight × Arm = Moment

171,520 × 820 = 140,646,400

2. Compute the change in moment and determine the new weight and moment.

Weight × Arm = Moment

171,520 × 820 = 140,646,400

- 1,460 × 1,500 = - 2,190,000

170,060      = 138,456,400

3. Compute the new CG by dividing total moment by total weight.

138,456,400 ÷ 170,060 = 814.16

An alternate method of doing this computation is to use the formula:

Weight added or removed ÷ New weight = CG change ÷ Distance from weight station to old CG

or:

CG (change) = Weight added (or subtracted) × Distance to old CG ÷ New weight

1. Weight subtracted = 1,460 lbs

2. Distance to old CG = 1,500 - 820 = 680"

3. New weight = 171,520 - 1,460 = 170,060 lbs

4. CG change = -1,460 × 680 ÷ 170,060 = -5.84"

(weight removed aft of the CG moves the CG forward)

5. New CG = 820 - 5.84 = 814.16"

Either solution will yield the correct answer.

You may be given a problem in which the CG is in the percent of MAC. You need to convert that to a distance from LEMAC and to distance aft of datum. Use the following formulas:

(MAC × % MAC) ÷ 100 = Distance of CG from LEMAC

and

Distance of CG from LEMAC + LEMAC = Distance of CG aft of datum

For example, assume the following:

MAC = 180.7"

LEMAC = 860.2"

CG = 32.7% MAC

1. Compute distance of CG aft of LEMAC.

(180.7 × 32.7) ÷ 100 = 59.1" aft of LEMAC

2 Compute the distance of the CG aft of datum.

860.2 + 59.1 = 919.3" aft of datum

Maximum Weight Addition

Adding weight will change the CG position. At times it is necessary to compute the maximum weight that can be added without exceeding a CG limit. To do this, use a variation of the formula:

Weight added ÷ Old weight = CG change ÷ Distance from weight station to new CG

or:

Weight added = Old weight × CG change ÷ Distance from weight station to new CG

For example:

How much weight could be added at station 1600 without exceeding the aft CG limit?

Aircraft weight............................83,000 pounds

CG location .............................Station 900

Aft CG limit................................Station 905

1. Determine the known values of the formula assuming that the CG will move to the aft limit.

Old weight = 83,000 lbs

CG change = 905 - 900 = 5"

Distance from weight station to new CG = 1600 - 905 = 695"

2. Apply the formula.

Weight added = 83,000 × 5 ÷ 695 = 597.1 lbs

Shifting Weight

A problem involving a weight shift is essentially the same as one in which weight is added or subtracted. Weight is subtracted from one station and the same amount of weight is added in another. The total weight remains the same but the moment changes: increasing if the weight is moved rearward, or decreasing if it is moved forward.

For example:

What is the new CG location if 800 pounds of cargo are moved from the forward cargo hold to the aft cargo hold?

Airplane gross weight................150,000 pounds

CG prior to shift.........................998.0 inches aft of datum

Arm of forward hold...................667.0 inches aft of datum

Arm of aft hold..........................1,160 inches aft of datum

1. Compute the current total moment.

Weight × Arm = Moment

150,000 × 998.0 = 149,700,000

2. Compute the new moment.

Weight × Arm = Moment

150,000 × 998.0 = 149,700,000

- 800 x 667.0 = - 533,600

+ 800 x 1,160.0 = + 928,000

150,000       = 150,094,400

3. Calculate the new CG location.

New CG = 150,094,400 ÷ 150,000 = 1,000.6"

An alternate method of solving the problem is by using the formula:

CG change = Weight shifted × Distance shifted ÷ Total weight

In the previous problem:

Weight shifted = 800 pounds

Distance shifted = 1,160.0 - 667.0 = 493.0"

Total weight = 150,000 pounds

CG change = 800 × 493 ÷ 150,000 = 2.6"

New CG = 998.0" + 2.6" = 1,000.6"

When the original CG and the answers are in % MAC, convert the CG change to % MAC by dividing it by MAC and multiplying by 100. Then, add or subtract that percentage from the original CG in % MAC.

Computing Required Weight Shift

Occasionally it is necessary to compute the maximum amount of weight that can be shifted and still keep the CG within limits. At times, the CG is out of limits and it is necessary to compute the minimum amount of weight that has to be shifted to get back in. This can be done using the formula:

CG change = (Weight shifted × Distance shifted) ÷ Total weight

or

Weight shifted = (Total weight × CG change) ÷ Distance weight is shifted

For example:

The gross weight of the aircraft is 155,000 pounds. How much weight must be moved from station 1028.0 to station 582.0 to move the CG forward 1.2 inches?

Total weight....................................155,000 pounds

CG change.....................................1.2"

Distance weight is shifted...............1,028.0 - 582.0 = 446.0"

Weight shifted.................................155,000 × 1.2 ÷ 446.0 = 417 lbs

Another example:

A cargo aircraft loaded to maximum takeoff gross weight of 165,000 pounds is tail heavy. How many 50-pound boxes must be moved from the 1,200-inch station to the 710-inch station to move the CG forward 3.2 inches?

Total weight...............................165,000 pounds

CG change.................................3.2"

Distance weight is shifted.........1200 - 710 = 490"

Weight shifted...........................165,000 × 3.2 ÷ 490 = 1,077.55 lbs

Number of boxes......................1,077.55 ÷ 50 = 21.55 boxes

Since it is not possible to move a fraction of a box, and since 21 boxes would not move enough weight forward, the minimum number of boxes is 22.