**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:- Determine the total weight and record this number:

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

- Determine the total weight and record this number:

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

- Determine the distance between the CG and LEMAC:

(42.47)(–)(20.1)(=) 22.37 - 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.

Determine the lateral CG by using the data in Figure 10-8 and following these steps:- Add all of the weights:

(1545)(+)(170)(+)(200)(+)(288)(=) 2,203 - 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 - Determine the algebraic sum of the lateral offset moments.

(309)(+)(2295)(+/–)(+)(2700)(+)(2419)(+/–)(=) –1,705 - 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?

- Determine the number of inches the baggage is shifted:

(246)(–)(118)(=) 128 - Use this formula:

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.

- Determine the number of inches the baggage is shifted:

(246)(–)(118)(=) 128 - 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 Distance**

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:

(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 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)(×)(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:

(1876)(×)(.8)(÷)(195)(=) 7.7

Attaching 7.7 pounds of ballast to the bulkhead at station 228 moves the CG to +33.0.