1. Add all of the weights: 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. 3. Determine the algebraic sum of the lateral offset  moments. 4. Divide the sum of the moments by the total weight to  determine the lateral CG. 1. Determine the number of inches the baggage is shifted: (246)(–)(118)(=) 128 2. Use this formula: 1. Determine the number of inches the baggage is shifted: 2. Use this formula: 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. Moving 73.6 pounds of baggage from the aft compartment  to forward compartment shifts the CG forward 2 inches. 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. The lateral CG is 0.77 inch to the left of BL0. Use this formula: Moving the battery aft by 107.8 inches shifts the CG aft 1.5  inches. 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? The CG is shifted forward 1.36 inches. 10-6 Determine the lateral CG by using the data in  Figure 10-8  and  following these steps: Determining Weight Shifted to Cause Specified ΔCG Determining Distance Weight Is Shifted to Move  CG a Specific Distanc Determining ΔCG Caused by Shifting Weights Figure 10-8.  Specifications for determining the lateral CG of a helicopter. Item Weight (lb) Lateral Arm (in) Lateral Offset Moment (lb-in) Lateral CG 309 −2,295 2,700 −2,419 −1,705 –0.77 Helicopter empty weight Pilot Passenger Fuel (48 gal) Total 1,545 170 200 288 2,203 +0.2  –13.5 +13.5 135.0 –8.4 x = (2)(×)(4709)(÷)(128)(=) 73.6 (246)(–)(118)(=) 128 Weight shifted =  Total weight shifted × Δ CG Distance weight is shifted (309)(+)(2295)(+/–)(+)(2700)(+)(2419)(+/–)(=)  –1,705 (1705)(+/–)(÷)(2203)(=) –0.77 Distance weight is shifted =  Total weight × Δ CG Weight shifted (1.5)(×)(4026)(÷)(56)(=) 107.8 (50)(×)(128)(÷)(4709)(=) 1.36 Δ CG  =  Weight shifted × Distance weight is shifted Total weight (1,545)(×)(.2)(=) 309 (170)(×)(13.5)(+/–)(=) –2,295 (200)(×)(13.5)(=) 2,700 (288)(×)(8.4)(+/–)(=) –2,419 (1545)(+)(170)(+)(200)(+)(288)(=) 2,203