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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 1

Chapter 10 — Use of Computer for Weight and Balance Computations — Part 1

FAA-H-8083-1B (2025)

Introduction

Almost all weight and balance problems involve only simple

math. This allows slide rules and hand-held electronic

calculators to relieve much of the tedium involved with these

problems. This chapter compares the methods of determining

the center of gravity (CG) of an airplane while it is being

weighed. First, it shows how to determine the CG using a

simple electronic calculator, then solves the same problem

using an E6-B flight computer. Finally, it shows how to solve

it using a dedicated electronic flight computer

Use of Computer for Weight

and Balance Computations

Chapter 10

Figure 10-2. Weight and balance data of a typical nosewheel

airplane.

Weighing Point Weight (lb) Arm (in)

0

0

–78

830

836

340

2,006

Right side

Left side

Nose

Total

Figure 10-1. A typical electronic calculator is useful for solving

most types of weight and balance problems.

Examples of typical weight and balance problems (solved

with an electronic calculator) that pilots and airframe and

powerplant (A&P) maintenance technicians encounter

throughout their aviation endeavors are shown later in this

chapter.

Electronic Calculator

Determining the CG of an airplane in inches for the main-

wheel weighing points can be done with any simple electronic

calculator that has addition (+), subtraction (–), multiplication

(×), and division (÷) functions. [Figure 10-1] Scientifi

calculators with such additional functions as memory (M),

parentheses (( )), plus or minus (+/–), exponential (y x),

reciprocal (1/x), and percentage (%) functions allow more

complex problems to be solved or simple problems to be

solved using fewer steps.

According to Figure 10-2, the weight of the nosewheel (F) is

340 pounds, the distance between main wheels and nosewheel

(L) is –78 inches, and the total weight (W) of the airplane is

2,006 pounds. L is negative because the nosewheel is ahead

of the main wheels.

To determine the CG, use this formula:

CG = F × L

W

= 340 × –78

2,006

Key the data into the calculator and press the equal (=) key.

The answer should read as follows:

(340)(×)(78)(+/–)(÷)(2006)(=) –13.2

The arm of the nosewheel is negative, so the CG is –13.2 or

13.2 inches ahead of the main wheel weighing points.

E6-B Flight Computer

The E6-B uses a special kind of slide rule. Instead of its scales

going from 1 to 10, as on a normal slide rule, both scales

go from 10 to 100. The E6-B cannot be used for addition or

subtraction, but it is useful for making calculations involving

multiplication and division. Its accuracy is limited, but it is

sufficiently accurate for most weight and balance problems.

The same problem that was just solved with the electronic

calculator can be solved on an E6-B by following these steps:

CG = F × L

W

= 340 × –78

2,006

First, multiply 340 by 78 (disregard the minus sign)

[Figure 10-3]:

• Step 1—place 10 on the inner scale (this is the index

opposite 34 on the outer scale that represents 340).

• Step 2—opposite 78 on the inner scale, read 26.5 on

the outer scale.

Determine the value of these digits by estimating:

300 × 80 = 24,000, so 340 × 78 = 26,500.

Then, divide 26,500 by 2,006 [Figure 10-4]:

Figure 10-3. E6-B computer set up to multiply 340 by 78.

Figure 10-4. E6-B computer set up to divide 26,500 by 2,006.

Step 2

Step 1

Step 4

Step 3

Figure 10-5. Dedicated electronic flight computers are programmed

to solve weight and balance problems, as well as flight problems.

• Step 3—on the inner scale, place 20, which represents

2,006 opposite 26.5 on the outer scale (26.5 represents

26,500).

• Step 4—opposite the index, 10, on the inner scale read

13.2 on the outer scale.

Determine the value of 13.2 by estimating:

20,000 ÷ 2000 = 10, so 26,500 ÷ 2,006 = 13.2.

The arm (–78) is negative, so the CG is also negative.

The CG is –13.2 inches or 13.2 inches ahead of the datum.

Dedicated Electronic Flight Computer

Dedicated electronic flight computers, like the one in

Figure 10-5, are programmed to solve many flight problems

such as wind correction, heading and ground speed,

endurance, and true airspeed (TAS), as well as weight and

balance problems.

The problem just solved with an electronic calculator and

an E6-B can also be solved with a dedicated flight computer

using the information shown in Figure 10-2 . Each fligh

computer handles the problems in a slightly different way,

but all are programmed with prompts that solicit the required

data to be inputted so there is no need to memorize any

formulas. Weight and arms are inputted as called for, and a

running total of the weight, moment, and CG are displayed.

Typical Weight and Balance Problems

A hand-held electronic calculator like the one in Figure 10-5

is a valuable tool for solving weight and balance problems. It

can be used for a variety of problems and has a high degree

of accuracy. The examples given here are solved with a

calculator using only the (×),(÷),(+),( – ), and (+/–) functions.

If other functions are available on your calculator, some of

the steps may be simplified

Determining CG in Inches From the Datum

This type of problem is solved by firs determining the

location of the CG in inches from the main wheel weighing

points, then measuring this location in inches from the datum.

There are four types of problems involving the location of

the CG relative to the datum.

Nosewheel Airplane With Datum Ahead of the

Main Wheels

The datum (D) is 128 inches ahead of the main wheel

weighing points; the weight of the nosewheel (F) is 340

pounds, and the distance between main wheels and nosewheel

(L) is 78 inches. The total weight (W) of the airplane is 2,006

pounds. Refer to Figure 3-5 on page 3-5.

Use this formula:

CG = D – F × L

W

1. Determine the CG in inches from the main wheel:

(340)(×)(78)(÷)(2006)(=) 13.2

2. Determine the CG in inches form the datum:

(128)(–)(13.2)(=) 114.8

The CG is 114.8 inches behind the datum.

Nosewheel Airplane With Datum Behind the Main

Wheels

The datum (D) is 75 inches behind the main wheel weighing

points, the weight of the nosewheel (F) is 340 pounds, and

the distance between main wheels and nosewheel (L) is 78

inches. The total weight (W) of the airplane is 2,006 pounds.

Refer to Figure 3-6 on page 3-5.

Use this formula:

CG = – D + F × L

W

1. Determine the CG in inches from the main wheels:

(340)(×)(78)(÷)(2006)(=) 13.2

2. Determine the CG in inches from the datum:

(75)(+)(13.2)(=) 88.2

The minus sign before the parenthesis in the formula

means the answer is negative. The CG is 88.2 inches

ahead of the datum (–88.2).

Tailwheel Airplane With Datum Ahead of the Main

Wheels

The datum (D) is 7.5 inches ahead of the main wheel

weighing points, the weight of the tailwheel (R) is 67 pounds,

and the distance between main wheels and tailwheel (L) is

222 inches. The total weight (W) of the airplane is 1,218

pounds. Refer to Figure 3-7 on page 3-6.

Use this formula:

CG = D + R × L

W

1. Determine the CG in inches from the main wheels.

(67)(×)(222)(÷)(1218)(=) 12.2

2. Determine the CG in inches from the datum:

(7.5)(+)(12.2)(=) 19.7

The CG is 19.7 inches behind the datum.

Tailwheel Airplane With Datum Behind the Main

Wheels

The datum (D) is 80 inches behind the main wheel weighing

points, the weight of the tailwheel (R) is 67 pounds, and the

distance between main wheels and tailwheel (L) is 222 inches.

The total weight (W) of the airplane is 1,218 pounds. Refer

to Figure 3-8 on page 3-6.

Use this formula:

CG = –D + R × L

W

1. Determine the CG in inches from the main wheels:

(67)(×)(222)(÷)(1218)(=) 12.2

2. Determine the CG in inches from the datum:

(80)(+/–)(+)(12.2)(=) –67.8

The CG is 67.8 inches ahead of the datum.

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Original source PDFPublished from pages 93–96 of the recorded source chapter.
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