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Archive / FAA Aircraft Weight and Balance Handbook / Aircraft Weight and Balance Handbook: Chapter 10 — Use of Computer for Weight and Balance Computations

Chapter 10 — Use of Computer for Weight and Balance Computations, Part 2

Chapter 10 — Use of Computer for Weight and Balance Computations — Part 2

FAA-H-8083-1B (2025)

Figure 10-6. Specifications for determining the CG of an airplane using weight and arm.

Figure 10-7. Specifications for determining the CG of an airplane

using weights and moment indexes.

Weight Point Weight (lb) Arm (in) Moment (lb-in) CG

106,240

107,008

17,000

230,248 114.8

Right side

Left side

Nose

Total

830

836

340

2,006

128

128

50

Weighing Point Weight (lb) Moment/100 CG

114.8

Right side

Left side

Nose

Total

830

836

340

2,006

1,062.4

1,070.1

170

2,302.5

Determining CG, Given Weights, and Arms

Some weight and balance problems involve weights and arms

to determine the moments. Divide the total moment by the

total weight to determine the CG. Figure 10-6 contains the

specifications for determining the CG using weights and arms.

Determine the CG by using the data in Figure 10-6 and

following these steps:

1. Determine the total weight and record this number:

(830)(+)(836)(+)(340)(=) 2,006

2. Determine the moment of each weighing point and

record them:

(830)(×)(128)(=) 106,240

(836)(×)(128)(=) 107,008

(340)(×)(50)(=) 17,000

3. Determine the total moment and divide this by the

total weight:

(106240)(+)(107008)(+)(17000)(=)(÷)(2006)(=)

114.8

This airplane weighs 2,006 pounds and its CG is 114.8 inches

from the datum.

Determining CG, Given Weights, and Moment

Indexes

Other weight and balance problems involve weights and

moment indexes, such as moment/100 or moment/1,000. To

determine the CG, add all the weights and all the moment

indexes. Then, divide the total moment index by the total

weight and multiply the answer by the reduction factor.

Figure 10-7 contains the specifications for determining the

CG using weights and moments indexes.

Determine the CG by using the data in Figure 10-7 and

following these steps:

1. Determine the total weight and record this number:

(830)(+)(836)(+)(340)(=) 2,006

2. Determine the total moment index, divide this by the

total weight, and multiply it by the reduction factor

of 100:

(1062.4)(+)(1070.1)(+)(170)(=)(2302.5)(÷)(2006)(=)

(1.148)(×)(100)(=) 114.8

This airplane weighs 2,006 pounds and its CG is 114.8 inches

from the datum.

Determining CG in Percent Mean Aerodynamic

Chord (MAC)

• The loaded CG is 42.47 inches aft of the datum.

• MAC is 61.6 inches long.

• LEMAC is at station 20.1.

1. Determine the distance between the CG and LEMAC:

(42.47)(–)(20.1)(=) 22.37

2. Then, use this formula:

(22.37)(×)(100)(÷)(61.6)(=) 36.3

The CG of this airplane is located at 36.3 percent MAC.

Determining Lateral CG of a Helicopter

For a helicopter, it is often necessary to determine not only

the longitudinal CG, but the lateral CG as well. Lateral CG is

measured from butt line zero (BL 0). All items and moments

to the left of BL 0 are negative, and all those to the right of

BL 0 are positive. Figure 10-8 contains the specifications for

determining the lateral CG of a typical helicopter.

Figure 10-8. Specifications for determining the lateral CG of a helicopter.

Item Weight (lb) Lateral Arm (in) Lateral Offset Moment (lb-in) Lateral CG

309

−2,295

2,700

−2,419

−1,705 –0.77

Helicopter empty weight

Pilot

Passenger

Fuel (48 gal)

Total

1,545

170

200

288

2,203

+0.2

–13.5

+13.5

135.0

–8.4

x =

Determine the lateral CG by using the data in Figure 10-8 and

following these steps:

1. Add all of the weights:

(1545)(+)(170)(+)(200)(+)(288)(=) 2,203

2. Multiply the lateral arm (the distance between butt

line zero and the CG of each item) by its weight to

get the lateral offset moment of each item. Moments

to the right of BL 0 are positive and those to the left

are negative.

(1,545)(×)(.2)(=) 309

(170)(×)(13.5)(+/–)(=) –2,295

(200)(×)(13.5)(=) 2,700

(288)(×)(8.4)(+/–)(=) –2,419

3. Determine the algebraic sum of the lateral offset

moments.

(309)(+)(2295)(+/–)(+)(2700)(+)(2419)(+/–)(=)

–1,705

4. Divide the sum of the moments by the total weight to

determine the lateral CG.

(1705)(+/–)(÷)(2203)(=) –0.77

The lateral CG is 0.77 inch to the left of BL0.

Determining ΔCG Caused by Shifting Weights

Fifty pounds of baggage is shifted from the aft baggage

compartment at station 246 to the forward compartment at

station 118. The total airplane weight is 4,709 pounds. How

much does the CG shift?

1. Determine the number of inches the baggage is shifted:

(246)(–)(118)(=) 128

2. Use this formula:

Δ CG = Weight shifted × Distance weight is shifted

Total weight

(50)(×)(128)(÷)(4709)(=) 1.36

The CG is shifted forward 1.36 inches.

Determining Weight Shifted to Cause Specified

ΔCG

How much weight must be shifted from the aft baggage

compartment at station 246 to the forward compartment

at station 118 to move the CG forward 2 inches? The total

weight of the airplane is 4,709 pounds.

1. Determine the number of inches the baggage is shifted:

Weight shifted = Total weight shifted × Δ CG

Distance weight is shifted

(246)(–)(118)(=) 128

2. Use this formula:

(2)(×)(4709)(÷)(128)(=) 73.6

Moving 73.6 pounds of baggage from the aft compartment

to forward compartment shifts the CG forward 2 inches.

Determining Distance Weight Is Shifted to Move

CG a Specific Distanc

How many inches aft does a 56 pound battery need to be

moved to shift the CG aft by 1.5 inches? The total weight of

the airplane is 4,026 pounds.

Use this formula:

Distance weight is shifted = Total weight × Δ CG

Weight shifted

(1.5)(×)(4026)(÷)(56)(=) 107.8

Moving the battery aft by 107.8 inches shifts the CG aft 1.5

inches.

Determining Total Weight of an Aircraft With a

Specified ΔCG When Cargo Is Move

What is the total weight of an airplane if moving 500 pounds

of cargo 96 inches forward shifts the CG 2.0 inches?

Use this formula:

Total weight = Weight shifted × Distance weight is shifted

Δ CG

(500)(×)(96)(÷)(2)(=) 24,000

Moving 500 pounds of cargo 96 inches forward causes a

2.0-inch shift in CG of a 24,000-pound airplane.

Determining Amount of Ballast Needed to Move

CG to a Desired Location

How much ballast must be mounted at station 228 to move

the CG to its forward limit of +33? The airplane weighs

1,876 pounds and the CG is at +32.2, a distance of 0.8 inch

out of limit.

Use this formula:

Ballast weight = Aircraft empty weight × D i s t . o u t o f l i m i t s

Distance ballast to desired CG

(1876)(×)(.8)(÷)(195)(=) 7.7

Attaching 7.7 pounds of ballast to the bulkhead at station

228 moves the CG to +33.0.

Original source PDFPublished from pages 97–100 of the recorded source chapter.
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