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by the total weight and multiply the answer by the
reduction factor. Figure 8-6 contains the specifications for
determining the CG using weights and moments indexes.
Determine the CG by using the data in Figure 8-6 and
following these steps:
1. Determine the total weight and record this number:
(830)(+)(836)(+)(340)(=) 2006
2. Determine the total moment index, divide this by the
total weight, and multiply it by the reduction factor of
100:
(1,062.4)(+)(1,070.1)(+)(170)(=)(2302.5)(÷)(2006)(=)
(1.148)(x)(100)(=)114.8
This airplane weighs 2,006 pounds and its CG is 114.8
inches form the datum.
Figure 8-6. Specifications for determining the CG of an airplane
using weights and moment indexes.
Determining CG in Percent of Mean Aerodynamic
Chord
• 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)(x) (100)(÷)(61.6)(=) 36.3
The CG of this airplane is located at 36.3% of the mean
aerodynamic chord.
Determining Lateral CG of a Helicopter
It is often necessary when working weight and balance of
a helicopter 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 8-7 contains the specifications for determining the
lateral CG of a typical helicopter.
Determine the lateral CG by using the data in Figure 8-7
and following these steps:
1. Add all of the weights:
(1545)(+)(170)(+)(200)(+)(288)(=) 2203
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.
(1545)(x)(.2)(=) 309
(170)(x)(13.5)(+/-)(=) -2295
(200)(x)(13.5)(=) 2700
(288)(x)(8.4)(+/-)(=) -2419
Figure 8-7. Specifications for determining the lateral CG of a
helicopter.
8–
3. Determine the algebraic sum of the lateral offset
moments.
(309)(+)(2295)(+/-)(+)(2700)(+)(2419)(+/-)(=) -1705
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 butt line zero.
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:
(50)(x)(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:
(246)(-)(118)(=) 128
2. Use this formula:
(2)(x)(4709)(÷)(128)(=) 73.6
Moving 73.6 pounds of baggage from the aft compartment
to forward compartment will shift the CG forward 2
inches.
Determining Distance Weight is Shifted to Move CG a
Specific Distance
How many inches aft will a 56-pound battery have 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:
(1.5)(x)(4026)(÷)(56)(=) 107.8
Moving the battery aft by 107.8 inches will shift the CG
aft 1.5 inches.
Determining Total Weight of an Aircraft That Will Have
a Specified ΔCG When Cargo is Moved
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:
(500)(x)(96)(÷)(2)(=) 24000
Moving 500 pounds of cargo 96 inches forward will cause
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:
(1876)(x)(.8)(÷)(195)(=) 7.7
Attaching 7.7 pounds of ballast to the bulkhead at station
　
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