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.

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.

## Comment