Figure 9-9. Stabilizer trim setting in ANU units.
Stabilizer Trim Setting—Units Airplane Nose Up
6
8
10
12
14
16
18
20
22
24
26
28
30
32
8
73/4
71/2
7
63/4
61/4
53/4
51/2
5
41/2
4
31/2
3
21/2
Flaps (all)CG
Figure 9-10. Determining the distance from CG to the LEMAC.
Distance CG to LEMAC = Datum to CG – Datum to LEMAC
= 10.7 inches
= 635.7 – 625
Figure 9-11. Determining the location of CG in percent MAC.
( )CG in % MAC = × 100Distance CG to LEMAC
MAC
( )= × 10010.7
134.0
= 8.0 % MAC
Figure 9-12. Determining the location of CG in inches before cargo
is removed.
( )CG (inches aft of LEMAC) = × MACCG in % MAC
100
( )= × 141.522.5
100
= 31.84 inches
Determining the Correct Stabilizer Trim Setting
It is important before takeoff to set the stabilizer trim for the
existing CG location. There are two ways the stabilizer trim
setting systems may be calibrated: in percent MAC and in
units airplane nose up (ANU).
If the stabilizer trim is calibrated in percent MAC, determine
the CG location in percent MAC as has just been described,
then set the stabilizer trim on the percentage figure thus
determined. Some aircraft give the stabilizer trim setting in
units of ANU that correspond with the location of the CG
in percent MAC. When preparing for takeoff in an aircraft
equipped with this system, first determine the CG in percent
MAC in the way described above, then refer to the stabilizer
trim setting chart on the takeoff performance page of the
pertinent AFM. Figure 9-9 is an excerpt from the AFM chart
on the takeoff performance of a Boeing 737.
Consider an airplane with these specifications
CG location ................................................ station 635.7
LEMAC ........................................................ station 625
MAC ..................................................................134.0 in
1. Determine the distance from the CG to the LEMAC
by using the formula in Figure 9-10.
2. Determine the location of the CG in percent MAC by
using the formula found in Figure 9-11.
Refer to Figure 9-9 for all flap settings and a CG located
at 8 percent MAC; the stabilizer setting is 73⁄4 units ANU.
Determining CG Changes Caused by
Modifying the Cargo
Since large aircraft can carry substantial cargo, adding,
subtracting, or moving any of the cargo from one hold to
another can cause large shifts in the CG.
Effects of Loading or Offloading Cargo
Both the weight and CG of an aircraft are changed when
cargo is loaded or offloaded. In the following example, the
new weight and CG are calculated after 2,500 pounds of
cargo is offloaded from the forward cargo hold
Aircraft specifications are
Loaded weight .................................................90,000 lb
Loaded CG ........................................ 22.5 percent MAC
Weight change ...................................................2,500 lb
Forward cargo hold centroid ...................... station 352.1
MAC ..................................................................141.5 in
LEMAC ................................................... station 549.13
1. Determine the CG location in inches from the
datum before the cargo is removed. Do this by first
determining the distance of the CG aft of the LEMAC.
[Figure 9-12]
Figure 9-13. Determining the distance between CG and the datum.
CG (inches from datum) = CG inches aft of LEMAC
+ Datum to LEMAC
= 580.97 inches
= 31.84 + 549.13
Figure 9-14. Determining the moment/1,000 for the original weight.
Moment/1,000 = Weight × Arm
1,000
90,000 × 580.97
1,000
= 52,287.3
=
Figure 9-15. Determining the moment/1,000 of the removed weight.
Moment/1,000 = Weight × Arm
1,000
2,500 × 352.1
1,000
= 880.25
=
Figure 9-16. New weights and CG.
Weight (lb) Moment/1,000 CG (inches from datum) CG (percent MAC)
580.97
587.51
22.5
27.12
Original data
Changes
New data
90,000
– 2,500
87,500
52,287.3
– 880.3
51,407.0
Figure 9-17. Determining the location of new CG.
CG = × 1,000Total moment/1,000
Total weight
51,407.0
87,500= × 1,000
= 587.51 inches behind the datum
Figure 9-18. Determining the distance between the CG and LEMAC.
CG (inches aft of LEMAC) =
CG (inches from datum) – LEMAC
= 38.38 inches
= 587.51 – 549.13
Figure 9-19. Determining the new CG in percent MAC.
( )CG % MAC = × 100Distance CG to LEMA
MAC
( )= × 10038.38
141.5
= 27.12% MAC
2. Determine the distance between the CG and the datum
by adding the CG in inches aft of LEMAC to the
distance from the datum to LEMAC. [Figure 9-13]
3. Determine the moment/1,000 for the original weight.
[Figure 9-14]
4. Determine the new weight and new CG by first
determining the moment/1,000 of the removed weight.
Multiply the weight removed (2,500 pounds) by the
centroid of the forward cargo hold (352.1 inches), and
then divide the result by 1,000. [Figure 9-15]
5. Subtract the removed weight from the original weight
and subtract the moment/1,000 of the removed weight
from the original moment/1,000. [Figure 9-16]
6. Determine the location of the new CG by dividing
the total moment/1,000 by the total weight and
multiplying this by the reduction factor of 1,000.
[Figure 9-17]
7. Convert the new CG location to percent MAC. First,
determine the distance between the CG location and
LEMAC. [Figure 9-18]
8. Then, determine the new CG in percent MAC.
[Figure 9-19]
Loading 3,000 pounds of cargo into the forward cargo hold
moves the CG forward 5.51 inches, from 27.12 percent MAC
to 21.59 percent MAC.
Figure 9-24. Determining the change in CG caused by shifting
2,500 pounds of cargo.
CG = × 100Weight shifted × Distance shifted
Total weight
= 2,500 × (227.9 + 144.9)
90,000
= 2,500 × 372.8
90,000
= 10.36 inches
Figure 9-21. Determining the new CG after shifting cargo weight.
New CG = Old CG ± CG
= 591.33 inches
= 580.97 + 10.36
Figure 9-22. Converting the location of CG to percent MAC.
( )CG % MAC = × 100CG inches
MAC
( )= × 10010.36
141.5
= 7.32% MAC
Figure 9-23. Determining the new CG in percent MAC.
New CG % MAC = Old CG ± CG
= 29.82% MAC
= 22.5% + 7.32%
Figure 9-20. Calculating the change in CG, using index arms.
CG = Weight shifted × Distance shifted
Total weight
= 2,500 × (724.9 – 352)
90,000
= 2,500 × 372.9
90,000
= 10.36 inches
Effects of Shifting Cargo From One Hold to
Another
When cargo is shifted from one cargo hold to another, the CG
changes, but the total weight of the aircraft remains the same.
For example, use the following data:
Loaded weight ................................................ 90,000 lb
Loaded CG ............. station 580.97 (22.5 percent MAC)
Forward cargo hold centroid ........................ station 352
Aft cargo hold centroid ............................. station 724.9
MAC ..................................................................141.5 in
LEMAC ....................................................... station 549
To determine the change in CG ( ΔCG) caused by shifting
2,500 pounds of cargo from the forward cargo hold to the
aft cargo hold, use the formula in Figure 9-20.
Since the weight was shifted aft, the CG moved aft and the
CG change is positive. If the shift were forward, the CG
change would be negative.
Before the cargo was shifted, the CG was located at station
580.97, which is 22.5 percent of MAC. The CG moved aft
10.36 inches, so the new CG is found using the formula from
Figure 9-21.
Convert the location of the CG in inches aft of the datum to
percent MAC by using the formula in Figure 9-22.
The new CG in percent MAC caused by shifting the cargo is
the sum of the old CG plus the change in CG. [Figure 9-23]
Some AFMs locate the CG relative to an index point rather
than the datum or the MAC. An index point is a location
specified by the aircraft manufacturer from which arms used
in weight and balance computations are measured. Arms
measured from the index point are called index arms, and
objects ahead of the index point have negative index arms,
while those behind the index point have positive index arms.
Use the same data as in the previous example, except for
these changes:
Loaded CG .......................... index arm of 0.97, which is
22.5 percent of MAC
Index point ................................... fuselage station 580.0
Forward cargo hold centroid ...............–227.9 index arm
Aft cargo hold centroid .......................+144.9 index arm
MAC ..................................................................141.5 in
LEMAC ..............................................–30.87 index arm
The weight was shifted 372.8 inches (–227.9 + Δ = +144.9,
Δ =372.8).
The change in CG can be calculated by using this formula
found in Figure 9-24.
Figure 9-25. Determining the new CG, moved aft 10.36 inches.
New CG = Old CG ± CG
= 11.33 index arm
= 0.97 + 10.36
Figure 9-26. The change in the CG in percent MAC.
New CG % MAC = Old CG ± CG
= 29.82% MAC
= 22.5% + 7.32%
Figure 9-27. The new CG in percent MAC.
( )CG % MAC = × 100CG inches
MAC
( )= × 10010.36
141.5
= 7.32% MAC
Figure 9-28. Determining pallet area in square feet.
Area (sq. ft.) = Length (inches) × Width (inches)
144 square inches/square foot
= 48.5 × 33.5
144
= 1,624.7
144
= 11.28 square feet
Since the weight was shifted aft, the CG moved aft, and the
CG change is positive. If the shift were forward, the CG
change would be negative. Before the cargo was shifted,
the CG was located at 0.97 index arm, which is 22.5 percent
MAC. The CG moved aft 10.36 inches, and the new CG is
shown using the formula in Figure 9-25.
The change in the CG in percent MAC is determined by using
the formula in Figure 9-26.
The new CG in percent MAC is the sum of the old CG plus
the change in CG. [Figure 9-27]
Notice that the new CG is in the same location whether the
distances are measured from the datum or from the index
point.
Determining Cargo Pallet Loads and Floor
Loading Limits
Each cargo hold has a structural floor loading limit based on
the weight of the load and the area over which this weight is
distributed. To determine the maximum weight of a loaded
cargo pallet that can be carried in a cargo hold, divide its total
weight, which includes the weight of the empty pallet and
its tie down devices, by its area in square feet. This load per
square foot must be equal to or less than the floor load limit.
In this example, determine the maximum load that can be
placed on this pallet without exceeding the floor loading limit.
Pallet dimensions ..........................................36 by 48 in
Empty pallet weight ................................................47 lb
Tie down devices ....................................................33 lb
Floor load limit ............................169 lb per square foot
The pallet has an area of 36 inches (3 feet) by 48 inches
(4 feet), which equals 12 square feet, and the floor has a
load limit of 169 pounds per square foot. Therefore, the total
weight of the loaded pallet can be 169 × 12 = 2,028 pounds.
Subtracting the weight of the pallet and the tie down devices
gives an allowable load of 1,948 pounds (2,028 – [47 + 33]).
Determine the floor loading limit that is needed to carry a
loaded cargo pallet having the following dimensions and
weights:
Pallet dimensions ...................................48.5 by 33.5 in
Pallet weight ..........................................................44 lb
Tiedown devices ....................................................27 lb
Cargo weight .....................................................786.5 lb
First, determine the number of square feet of pallet area as
shown in Figure 9-28.
Then, determine the total weight of the loaded pallet:
Pallet ....................................................................44.0 lb
Tiedown devices .................................................27.0 lb
Cargo .................................................................786.5 lb
Total ...................................................................857.5 lb
Determine the load imposed on the floor by the loaded pallet.
[Figure 9-29] The floor must have a minimum loading limit
of 76 pounds per square foot.
Figure 9-29. Determining the load imposed on the floor by the
loaded pallet.
Floor Load = Loaded weight
Pallet area
857.5
11.28
= 76.0 pounds/square foot
=
Figure 9-30. Finding the maximum takeoff weight.
Max limit
142,000
184,200
Landing weight
+ trip fuel
Takeoff weight
Trip limit
142,000
+ 40,000
182,000
Figure 9-31. Determining zero fuel weight with lower trip limits.
Max limit
184,200
138,000
Landing weight
– fuel load
Zero fuel weight
Trip limit
182,000
– 54,000
128,000
Figure 9-32. Finding maximum payload with lower trip limits.
Max limit
138,000 Zero fuel weight
– BOW
Payload (pounds)
Trip limit
128,000
–100,500
27,500
Determining the Maximum Amount of Payload
That Can Be Carried
The primary function of a transport or cargo aircraft is to carry
payload, which is the portion of the useful load, passengers,
or cargo that produces revenue. To determine the maximum
amount of payload that can be carried, both the maximum
limits for the aircraft and the trip limits imposed by the
particular trip must be considered. In each of the following
steps, the trip limit must be less than the maximum limit. If
it is not, the maximum limit must be used.
These are the specifications for the aircraft in this example
Basic operating weight (BOW) .....................100,500 lb
Maximum zero fuel weight ............................138,000 lb
Maximum landing weight ..............................142,000 lb
Maximum takeoff weight ..............................184,200 lb
Fuel tank load .................................................54,000 lb
Estimated fuel burn en route ............................40,000 lb
1. Compute the maximum takeoff weight for this trip.
This is the maximum landing weight plus the trip fuel.
[Figure 9-30]
2. The trip limit is lower than the maximum takeoff
weight, so it is used to determine the zero fuel weight.
[Figure 9-31]
3. The trip limit is again lower than the maximum takeoff
weight, so use it to compute the maximum payload
for this trip. [Figure 9-32]
Under these conditions, 27,500 pounds of payload may be
carried.
Determining the Landing Weight
It is important to know the landing weight of the aircraft in
order to set up the landing parameters and to be certain the
aircraft is able to land safely at the intended destination.
In this example of a four-engine turboprop airplane,
determine the airplane weight at the end of 4.0 hours of cruise
under these conditions:
Takeoff weight ...............................................140,000 lb
Pressure altitude during cruise ..........................16,000 ft
Ambient temperature during cruise .....................–32 °C
Fuel burned during descent and landing ............1,350 lb
Refer to the U.S. Standard Atmosphere Table in
Figure 9-33 and the gross weight table in Figure 9-34 when
completing the following steps:
1. Use the U.S. Standard Atmosphere Table to
determine the standard temperature for 16,000 feet
(–16.7 °C).
2. The ambient temperature is –32 °C, which is a
deviation from standard of 15.3 °C. (–32° – (–16.7°)
= –15.3°). It is below standard.
3. In the gross weight table, follow the vertical line
representing 140,000 pounds gross weight upward
until it intersects the diagonal line for 16,000 feet
pressure altitude.
4. From this intersection, draw a horizontal line to the left
to the temperature deviation index (0 °C deviation).
5. Draw a diagonal line parallel to the dashed lines for
Below Standard from the intersection of the horizontal
line and the Temperature Deviation Index.
6. Draw a vertical line upward from the 15.3 °C
Temperature Deviation From Standard.