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FAA ATPL test: CG shift in index arms

Old 18th Apr 2012, 11:46
  #1 (permalink)  
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Join Date: Aug 2002
Location: Switzerland, Singapore
Posts: 1,306
FAA ATPL test: CG shift in index arms

I'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:

Where is the new CG if the weight is shifted from the forward to the aft compartment under Loading Condition WS-5?

-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 is offline  
Old 21st Apr 2012, 13:30
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Join Date: Jun 2002
Location: GB
Posts: 22
easiest 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
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Old 29th Apr 2012, 17:23
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Clever solution, thank you very much!
Dani is offline  

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