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How to Calculate Vertical Center of Gravity?

Asked by: 13517 views ,
Light Sport Aircraft, Student Pilot

Here is a triangle, such as △ABC,∠ B as supporting an object's former vertices,

Its weight is:Gn,objects behind the fulcrum weight is: Gm,Is the total weight: Ge,As shown in figure.

The graph of the symbol definition is:d=AC,AB=c,AC=b,BC=a,AL=x,OL=y,

tanθ=BC/AC=a/b,Cosθ=AC/AB=b/a.

The problems need to be solved as follows:

1.When the object's centerline is horizontal reference line (green line) above,also is h1＞h2,θis positive angle(that is ＋3°、＋5°、＋7°、＋10°,etc),at this time B Angle (B point) is rising,so that the object front-end(B point) rising status.

2.1.When the object's centerline is horizontal reference line (green line) below,also is 

h1＜h2,θis negative angle,(that is -3°、-5°、-7°、-10°,etc),at this time B Angle (B point) is to drop,so that the object front-end(B point) declining state.

 ⑴ I write the formula is:Ge*(x－y*tanθ)*cosθGn*b,but I don't know whether it is right, please revise and work out y,

 ⑵ Calculate + 3°、+ 5°、+ 7°、+ 10°and - 3°、- 5°、- 10 °to correspond a (BC) in length.

Thanks for you.

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4 Answers



  1. Vance Cochrane on Nov 09, 2010

    What has this got to do with asking a flight instructor an aviation related question that is of use?

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  2. Brian on Nov 09, 2010

    Lee,
    Do you happen to have a picture?
    Without a picture there is only one thing I can tell you, this: 
    “tanθ=BC/AC=a/b,Cosθ=AC/AB=b/a.”
    Take a second look at how tangent and cosine are used. Regardless of the letter assignments used, if tan = x/y then cos cannot possibly be y/x. Hope that helps.
     
    ~Brian
     

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  3. Lee Bing on Nov 09, 2010

    Mr Brian,Hi,thanks for you,I have a picture, but this page I have no way to upload.If you have E-mail, please tell me, pictures sent to your mailbox.My E-Mail:aw_cgs@163.com.
    Mr B.Lee
     

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  4. Brian on Nov 10, 2010

    Lee, try this link to Image Shack. You can upload it there and then post a link to the picture here. 🙂

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