Introduction
Weight and balance in aircraft is based on the law of the lever.
This chapter discusses the application of the law of the lever
and its applications relative to locating the balance point of a
beam or lever on which various weights are located or shifted.
The chapter also discusses the documentation pertaining to
weight and balance that is furnished by the Federal Aviation
Administration (FAA) and aircraft manufacturers.
Weight and Balance Theory
Chapter 2
Figure 2-1. Balance lever.
B = 200
A = 100
Arm B (+25")Arm A (−50")
Fulcrum
Datum− Moment + Moment
−Forces +Forces
Figure 2-2. Balance point locations.
Item Weight (lb) Arm (in) Moment (lb-in)
−5,000
+5,000
0
Weight A
Weight B
100
200
300
−50
+25
Figure 2-3. Balance lever datum located off the lever.
C = 200
A = 100
B = 100
+50
+90
+110
+150
CG
Datum
Weight and Balance Theory
Two elements are vital in the weight and balance considerations
of an aircraft.
• The total weight of the aircraft must be no greater
than the maximum weight allowed by the FAA for
the make and model of the aircraft.
• The center of gravity (CG), or the point at which
all of the weight of the aircraft is considered to be
concentrated, must be maintained within the allowable
range for the operational weight of the aircraft.
Arm
The arm is usually measured and expressed in inches and
refers to the horizontal distance between the CG of an item
or object and the datum, a point from where all measurements
are taken. Arms to the left of the datum are negative (–) and
those to the right of the datum are positive (+). The datum is an
imaginary vertical plane from which all horizontal distances
are measured for balance purposes. The position of the
reference datum varies by aircraft design and manufacturer.
When the datum is located off of the lever and to the left, all of
the arms are positive and computational errors are minimized.
Note: When the datum is established ahead of the aircraft, for
example at the aircraft nose, all of the arms are positive and
computational errors are minimized.
Moment
A moment is a force that tries to cause rotation and is the
product of the arm, in inches, and the weight, in pounds.
Moments are generally expressed in pound-inches (lb-in)
and may be either positive or negative.
The Law of the Lever
Weight and balance problems are based on the physical law
of the lever. This law states that a lever is balanced when the
weight on one side of the fulcrum (a pivot point for the lever)
multiplied by its arm is equal to the weight on the opposite
side multiplied by its arm. In other words, the lever is balanced
when the sum of the moments about the fulcrum is zero. This
is the condition in which the positive moments (those that
try to rotate the lever clockwise) are equal to the negative
moments (those that try to rotate it counterclockwise). In an
aircraft, the balance point is referred to as the CG.
One of the easiest ways to understand weight and balance is
to consider a lever with weights placed at various locations.
The balance point or CG of the lever can be changed by either
moving the weights closer or farther from the fulcrum or by
increasing or decreasing the weights. The balance point or
CG of a lever may be determined by using these four steps:
1. Measure the arm of each weight in inches from the
datum.
2. Multiply each arm by its weight in pounds to determine
the moment in pound-inches of each weight.
3. Determine the total of all weights and of all the
moments. (Disregard the weight of the lever).
4. Divide the total moment by the total weight to
determine the balance point.
Consider these facts about the lever in Figure 2-1 . The
100-pound weight A is located 50 inches to the left of the
fulcrum (the datum, in this instance), and it has a moment of
100 × –50 = –5,000 lb-in. The 200-pound weight B is located
25 inches to the right of the fulcrum, and its moment is
200 × +25 = +5,000 lb-in. In Figure 2-2, the sum of the
moments is –5,000 + 5,000 = 0, and the lever is balanced. The
forces that try to rotate it clockwise have the same magnitude
as those that try to rotate it counterclockwise. If either weight
is moved or changed, the balance point or CG changes and
the lever becomes unbalanced.
In Figure 2-3, the datum is located off the lever to the left
of weight A. Using the information provided in Figure 2-3,
Figure 2-4. Finding balance point with datum located off the lever.
Item Weight (lb) Arm (in) Moment CG
5,000
9,000
30,000
44,000 110
Weight A
Weight B
Weight C
100
100
200
400
50
90
150
Figure 2-5. Locating balance point.
C = 200
A = 100
B = 100
−20
−60
110
+40
CG
Original datum New datum
Figure 2-6. Proving balance point with three weights is correct.
Item Weight (lb) Arm (in) Moment (lb-in)
−6,000
−2,000
+8,000
0
Weight A
Weight B
Weight C
100
100
200
−60
−20
+40
Figure 2-7. Locating balance point with datum at C.
C = 200
A = 100
B = 100
−60
−100
Datum
Figure 2-8. Determining new balance point.
Item Weight (lb) Arm (in) Moment CG
−10,000
−6,000
0
−16,000 −40
Weight A
Weight B
Weight C
100
100
200
400
−100
−60
Figure 2-9. Locating balance point with datum left of original.
C = 200
A = 100
B = 100
−40
Datum
−20
−60
+40
determine the balance point by making a chart like the one
in Figure 2-4.
As noted in Figure 2-4, A weighs 100 pounds and is 50 inches
from the datum; B weighs 100 pounds and is 90 inches from
the datum; C weighs 200 pounds and is 150 inches from the
datum. The total of the weights is 400 pounds, and the total
moment is 44,000 lb-in.
Determine the balance point by dividing the total moment
by the total weight. A balance point is equal to the CG and
can be mathematically written as:
CG = total moment
total weight
To prove this is the correct balance, move the datum to a
location 110 inches to the right of the original datum and
determine the arm of each weight from this new datum.
[Figure 2-5] Then, make a new chart similar to the one in
Figure 2-6. If the balance point is correct, the sum of the
moments is zero.
The new arm of weight A is 60 inches (the difference between
110 and 50), and since this weight is to the left of the datum,
its arm is negative or –60 inches. The new arm of weight B is
20 inches (110 – 90), and it is also to the left of the datum, so
it is –20; the new arm of weight C is 40 inches (150 – 110).
It is to the right of the datum and is therefore positive.
The lever is balanced when the sum of the moments is zero.
The location of the datum used for determining the arms of
the weights is not important; it may be in various locations,
but all of the measurements must be made from the same
datum location.
The procedure for finding the balance point is the same
anywhere the datum is located. In Figure 2-7, the datum is
located at C. Weight A has an arm of –100 inches (negative
because it is to the left) of the datum and weight B has an
arm of –60 inches from the datum. The table in Figure 2-8
is used to determine the new balance point.
To verify that this is the correct balance point, move the datum
40 inches to the left of the original datum and determine the
arm of each weight from this new datum as in Figure 2-9.
Figure 2-10. Proving the new balance point is correct.
Item Weight (lb) Arm (in) Moment (lb-in)
−6,000
−2,000
+8,000
0
Weight A
Weight B
Weight C
100
100
200
−60
−20
40
Figure 2-11. Locating balance point with three weights.
C = 200
B = 200
A = 100
80
72
100
CG
Datum
Before Weight Shift
Figure 2-12. Proving the new balance point is correct.
Item Weight (lb) Arm (in) Moment (lb-in)
−5,000
+10,000
+5,000
Weight A
Weight B
Weight C
100
200
−50
+50
Figure 2-13. Weight distribution to balance lever.
B = 200
A = 100
C = 100
−50
50
−25
CG
Datum
After Weight Shift
Figure 2-14. Weight shift provides correct CG.
Item Weight (lb) Arm (in) Moment (lb-in)
−5,000
−5,000
+10,000
0
Weight A
Weight B
Weight C
100
200
200
−50
−25
+50
The new arm for weight A would be –100 + 40 = –60; for
weight B, –60 + 40 = –20; and point C, is +40. The lever is
balanced and the balance point is correct when the sum of
the moments is zero. [Figure 2-10]
Shifting the Balance Point or CG
One common weight and balance problem involves moving
or shifting weight from one point to another in order to move
the balance point or CG to a desired location. This can be
demonstrated by using a lever with three weights to work
out the problem.
Solution by Chart
As the lever is loaded in Figure 2-11, it balances at a point
72 inches from the CG of weight A.
To shift weight B so the lever balances about its center, 50
inches from the CG of weight A, first determine the arm of
weight B that produces a moment that causes the total moment
of all three weights around this desired balance point to be
zero. The combined moment of weights A and C around this
new balance point is 5,000 lb-in, so the moment of weight B
must be –5,000 lb-in for the lever to balance. [Figure 2-12]
Determine the arm of weight B by dividing its moment,
–5,000 lb-in, by its weight of 200 pounds. The arm is –25
inches. To balance the lever at its center, weight B must be
placed so its CG is 25 inches to the left of the center of the
lever. [Figure 2-13]
Figure 2-14 indicates that the shift in weight depicted in
Figure 2-13 allows the lever to balance as the sum of the
moments is zero.
Basic Weight and Balance Equation
The following formulas can be used to determine the distance
weight must be shifted to obtain a desired change in the CG
location. The equation can also be rearranged to fi d the
amount of weight required to be shifted to move the CG
to a desired location, to find the distance the CG is moved
when a specified amount of weight is shifted, or to find the
total weight that would allow shifting a specified amount of
weight to move the CG a given distance.
Weight to be shifted = Δ CG
Total weight Distance weight is shifted
Total weight = Weight shifted × Distance weight is shifted
Δ CG
Weight shifted = Total weight shifted × Δ CG
Distance weight is shifted
Δ CG = Weight shifted × Distance weight is shifted
Total weight
Distance weight is shifted = Total weight × Δ CG
Weight shifted