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  • Q?????

    Hi All,

    I trust you're well

    I'm having one of the blond moments with following M&B question:

    "Length of the Mean Aerodynamic Chord: 1 m; Moment Arm of the forward cargo: -0,50 m; Moment Arm of the aft cargo: +2,50 m; Aircraft mass: 2.200 kg; Centre of gravity: 25% MAC; To move the centre of gravity to 40%, which mass has to be transferred from the forward to the aft cargo hold?"

    So my thought process is as follows:

    1) I use MAC% formula which is MAC% = (A-B)/C x 100%, where A- distance from DATUM to CG, B- distance from DATUM to MAC Leading Edge, C- length of MAC, so:

    25% = (A - B)/1 -> A - B = 0.25 -> B = A - 0.25. So far so good.

    2) We know that new CG is placed 40%MAC, therefore:

    40% = (A1 - B)/1 by substituting B -> 40% = A1 - A - 0.25 -> A1 = A + 0.65.

    3) using the formula for shifting the cargo m x d = M x cc, where m - mass to be moved, d - distance, M - total mass, cc - movement of CG.

    From the 2) we know that cc = 0.65 so let's use this knowledge m x d = 2200 x 0.65 -> m x d = 1430 -> m = 1430/d So now the missing bit is to "find the d".

    4) Now the fun part. Moment arms (defined as Force x arm) are given in the question so:

    -0.5 = 2200 x d1, where d1 is from cg to end of arm
    2.5 = 2200 x d2, where we can substitute d2 as d - d1 which brings us to 2,5 = 2200 x (d - d1)

    Transforming 1st for d1 = -0.5 / 2200 and substituting 2nd for d1 we end up with:

    2.5 = 2200 x d - 2200 x ( - 0.5 / 2200) -> d = 1 / 1100

    But it can't be OK as if we go back to m = 1430/d and put d = 1/1100, we'll end up with staggering mass of 1573000 which seems bit odd

    Sorry for stupid question but I'm banging my head against the wall for last 45 minutes and can't figure what's wrong with 4) onward.

    Many thanks in advance for your help

    Cheers,
    M.


    Last edited by Maciej; 16-05-2017, 22:21.

  • #2
    Hah! I'm answering my own questions

    OK so there were two issues:

    1) I mixed up the signs

    Instead of B -> 40% = A1 - A - 0.25 -> A1 = A + 0.65, it should be B -> 40% = A1 - (A - 0.25) -> A1 = A + 0.15

    So this brings us to m = 330 / d.

    Also the 4) onward is obviously wrong. the d length is simply 2.5 PLUS 0.5 which is 3

    so m = 330 / 3 = 110kg.

    Hope that this will help someone

    Comment


    • #3
      25% of a MAC that is 1m long means the CG is 25cm back from the wing leading edge, LEMAC.

      The question asks us to move the CG back to 40% MAC, this is 40cm back from LEMAC, a movement of 15cm or 0.15m.
      The distance between the holds is 3m

      Use the formula:
      Mass Change / Total Mass = Change of CG / Distance Moved

      Transpose the formula and substitute known values.

      Mass Change =(Total Mass x Change of CG) / Distance Moved
      Mass Change = (2200 x 0.15) / 3
      Mass Change = 110 kg

      Comment


      • #4
        OMG. How did I miss that MAC = 1m ?
        I guess time to go back to primary school...
        Many thanks Col

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