PPRuNe Forums > Ground & Other Ops Forums > Flight/Ground Ops, Crewing and Dispatch > FAA ATPL test: CG shift in index arms PDA View Full Version : FAA ATPL test: CG shift in index arms Dani18th Apr 2012, 10:46I'm stuck here with these W&B questions. I figured out the MAC part, but on the index arms I'm completly confused, since there is no real explanation in the theory book: Q: Where is the new CG if the weight is shifted from the forward to the aft compartment under Loading Condition WS-5? A: -97.92, +19.15, +13.93 index arm. The theory book only gives you the explanation: CG (index arm) = CG (inches of the Datum) - 580 The table is the good old one from DC-9: Loaded weight 88 300 lbs Loaded MAC 25.5% Weight Change 3300 lbs Fwd Comp centroid -227.9 Index arms Aft Comp centroid +144.9 Index arms Lemac -30.87 Index arms Does anyone have a how to do? Many thanks Dani pavil21st Apr 2012, 12:30easiest way to solve it is using the equation; weight to be shifted / total weight = Δcg / distance the weight is shifted you are trying to work out the difference, which is the Δcg, Δcg = ( weight to be shifted * distance the weight is shifted ) / total weight we know the weight to be shifted is 3300 lbs, the distance to be shifted is from fwd hold (-227.9) to aft hold (144.9), therefore the distance between these two holds is 144.9 plus 227.9, which gives us 372.8, and the total weight is 88300 lbs. so now Δcg = ( 3300 * 372.8 ) / 88300 Δcg = 13.93 Dani29th Apr 2012, 16:23Clever solution, thank you very much!