Welcome Guest. Sign in or Signup

1 Answers

CG changes with removal of weight

Asked by: 8311 views Aircraft Systems, General Aviation

Can someone please help me understand how they got the answer here? When I add, subtract, divide or do anything I cannot come up with the answer. Maybe somone can break it down a bit further. 

Thanks

Aircraft Total Weight

 6,860 lb

CG Station

 80.0

 

Determine the location of the CG if 140 pounds of baggage is added to station 150.

 

Solution:

Added Weight/ New Total Weight

=

ΔCG/ Distance between weight and old CG

140/(6,860 + 140)

=

ΔCG/(150–80)

140/ 7,000

=

ΔCG/70

CG

=

1.4 in aft

 

Ace Any FAA Written Test!
Actual FAA Questions / Free Lifetime Updates
The best explanations in the business
Fast, efficient study.
Pass Your Checkride With Confidence!
FAA Practical Test prep that reflects actual checkrides.
Any checkride: Airplane, Helicopter, Glider, etc.
Written and maintained by actual pilot examiners and master CFIs.
The World's Most Trusted eLogbook
Be Organized, Current, Professional, and Safe.
Highly customizable - for student pilots through pros.
Free Transition Service for users of other eLogs.
Our sincere thanks to pilots such as yourself who support AskACFI while helping themselves by using the awesome PC, Mac, iPhone/iPad, and Android aviation apps of our sponsors.

1 Answers



  1. John D. Collins on Dec 12, 2011

    Doing it the long way, the total weight and CG station are the weight and CG location before adding the 140 pound baggage load at station 150.  Multiply the weight of the additional load times its station to obtain the moment or the baggage (140 x 150 = 21,000). Convert the original weight and CG location to a moment by the same process (6860 x 80 = 548,800).  Add the two weights and the two moments. The new total weight is (6860+140) 7000 pounds.  The new total moment is (21,000 + 588,800) 569,800 inch-pounds. Divide the total moment by the total weight (569,800/7000) to get the adjusted CG of 81.4 inches. The difference is 1.4 inches aft.

     

    The method you are using is similar.  As long as one is consistent, the location of the datum is arbitrary. In effect, the equation works because the datum is being redefined to be at the current CG location before the baggage is added. Assuming D1 is the original CG location, D2 is the location of the baggage, and D3 is the new CG location. Also assume W1 is the original weight and W2 is the baggage. The basic equation for the CG location is:

     

    D3 = [(W1 x D1) + (W2 x D2)]/(W1+W2) .

     

    If we redefine the moment arm to be at the location of D1, then the equation becomes

     

    D3-D1 = [(W1 x (D1 – D1)) + (W2 x (D2 – D1))] / (W1+W2)

     

    The first term on the numerator of the right side of the equation becomes zero, so the equation can be rewritten as:

     

    D3-D1 = W2 x (D2 – D1) / (W1+W2)

     

    Divide both sides by the term (D2 – D1), yeilds the form of the equation you started with:

     

    (D3-D1) / (D2 – D1) = W2 / (W1+W2), where D3 – D1 is the delta CG location.

    0 Votes Thumb up 0 Votes Thumb down 0 Votes


Answer Question

Our sincere thanks to all who contribute constructively to this forum in answering flight training questions. If you are a flight instructor or represent a flight school / FBO offering flight instruction, you are welcome to include links to your site and related contact information as it pertains to offering local flight instruction in a specific geographic area. Additionally, direct links to FAA and related official government sources of information are welcome. However we thank you for your understanding that links to other sites or text that may be construed as explicit or implicit advertising of other business, sites, or goods/services are not permitted even if such links nominally are relevant to the question asked.