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A M+B question

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Old 14th Jan 2004, 17:46
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TightYorksherMan
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A M+B question

QU26 PP1

An aircraft has two holds. the arm for hold 1 is +94" and the arm for hold 2 is +210". The CG limits for the aircraft are forward +156" and aft +165".
The aircraft is loaded to an AUM of 10650lbs, and the CG position is 172" aft of the datum.

HOw much cargo must be moved between the holds to shift the CG to the aft limit?

172" - 165" x 10650lbs
------------------------------
210 - 94

answer is 826lbs from hold2 to hold 1

IS this correct? reason I am asking this muppetry question is I can find the answers to the question paper it came from.
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Old 14th Jan 2004, 19:27
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PPRuNe Knight in Shining Armour
 
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Nope, 642.7 lbs from 2 to 1

10650/(210-94)*((210-165)-(210-172))


Try it from first principals.

Lets say x = weight in hold 1 and y = weight in hold 2

You know that 94x + 210y = 10650*172
and
x + y = 10650

So you can work out x and y

Then work out x and y when 94x + 210y = 10650*165

take one x from the other and bob's your uncle.

A bit of algebraeic manipulation and you get the equation at the top.

Hope that helps
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Old 14th Jan 2004, 21:18
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I agree with snigs' answer.

Formula is :-

Weight to be moved = (total mass) times (distance CofG is to move) divided by (distance weight is to be moved)

w = (10650) * (172 - 165) / ( 210 - 94)

w= 642.6724lb

Jinkster, looking back at your mail you have the correct formula, but the wrong answer. Time to change the batteries in your calculator

D
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Old 15th Jan 2004, 03:27
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To understand how to answer these load shift questions it is best to start with a little bit of common sense logic.

Shifting a load from one point to another in an aircraft will not change the total mass of the aircraft, but will change the total moment. This change in total moment will cause the C of G to move in the direction of the load shift.

To solve this problem use

M to represent the total mass of the aircraft
D to represent the distance (+ or -) that the C of G moves.
m to represent the mass of the load that is shifted.
d to represent the distance (+ or -) that the load is shifted.

Then the change in total moment is the total mass x the distance the C of G moves (which is M x D)

And the moment change caused by shifting the load is the mass of the load x the distance it is shifted (m x d)

The total moment change (M x D) will be equal to the moment change caused by the load shift (m x d).

So we have the simple equation MD = md

We want to know the mass (m) so we divide both sides of the equation by d to give

(MD)/d = m

The question states that

Holds are at +94" and +210"
The CG limits +156" (forward limit) and aft +165" (aft limit)
The total mass is 10650lbs
The CG position is at +172"

To calculate D and d we must use the equation

Distance moved = new location - initial location

We want to get the C of G from +172" to +165" so the required distance D = +165 - (+172) which is -7".

The load is to be moved from +210" to +94" so the distance
d = +94 - (+210) which is d = -116"

Inserting these values into our equation gives

m = (10650 lb x (-7")) / -116"

This gives m = 642.672 lb.

The important thing in this kind of problem is to ensure that you include the signs. It made no difference in this particular question, but it would have done had you been calculating distances forward or aft.

I also find that things work best if you start with a diagram and a simple statement of what is happening, before jumping into the equations.
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Old 15th Jan 2004, 03:46
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Ah Keith (the man who taught me all I know, but wait.... wasn't that POF...?).

You can tell he's an experienced ground instructor, Jinkster, so take note (and I had to pay for his experience, tut!!!!)

For the exam you need to be quick, hence know the equation, but understanding why is the important thing!
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