Exam Question about Weight, Balance and MAC

[Pugachev Cobra]

Hi all.

Recently I did an exam, and came upon this question which I would like to share, and want to know what do you think is the proper answer, since I can't remember the options.

The question went on like this:

The BOW for an aircraft is 4,200 lbs and is located at STA 90, at 20% of the MAC. The fuel weighs 500 lbs and is at STA 100, at 30% of the MAC. Where on the MAC is located the CG in % (for the fuel)?

I've put the (for the fuel) in parenthesis because I think the question was for the fuel CG, but maybe you can help me calculating both the fuel CG and the total CG.

Thanks for anyone that helps!

[Ex Cargo Clown]

I think I understand the question, although it is not the clearest !! I'm guessing they are looking for the MAC% with fuel loaded.

Answer = 21.4%

Standing by to get shot down on this though (I will show you the calculation if this does turn out to be correct !!!

[john_tullamarine]

One of those not-terribly-useful questions which examiners come up with from time to time. You have to tolerate it .. keeping in mind that these poor folk are kept in dark dungeons out in the back of the main building and deserve our empathy.

What is the question on about ?

(a) looking to see if you have some very basic algebraic/arithmetic understanding

(b) if you get past (a) the answer falls out very easily .. if not, then you are up for some heavy simultaneous equations .. all good fun, but quite unnecessary

First off, figure the CG for the load (ignore the asterisks - I still have trouble with formatting in this medium) -

BOW***4200****90*****378000
FUEL****500***100******50000
LOAD***4700****91.1***428000

(if you want to do the calc to 40 decimal places, that's OK, just a bit pointless).

Now, the MAC bit.

Two straightforward ways to approach this problem.

First, a bit more complicated, but relates the solution back to the usual MAC equation -

What you don't have given is the MAC length and LEMAC, so we need to figure a simple way to work them out.

The trick is to be aware of the following -

(a) both the %MAC and FS scales are linear so you can use proportions

(b) for the BOW, 20% = 90 units on the FS scale (doesn't matter what scale they are)
for the fuel load, 30% = 100 units
therefore it follows that 10% = 10 units

without too much strain you can figure out that

length of the MAC (ie 100%MAC) = 100 units (10% = 10 units .. multiple both by 10)
LEMAC = 70 units (if 90 units = 20% and 10 units is equal to 10%, then LEMAC = 90-20)

this gives you the MAC equation at

%MAC = ((FS-LEMAC)/MAC)*100

%MAC = ((FS - 70)/100)*100

which, in this case, comes out to be

%MAC = FS-70 (aren't round numbers great ?)

So the final answer is %MAC = 91.1-70 = 21.1%


The slightly simpler way to get the answer relies on the fact that the scales are linear and that you can use simple proportion (normal back of the prayer wheel sort of stuff)

(a) figure out, as before, that 10% = 10 units

(b) taking the measures from the BOW position as a temporary datum

delta CG = 1.1
delta %MAC = 1.1% (as 10 units = 10%)

So, from the original datum, %MAC = 20 + 1.1 = 21.1%

Not too sure if you would ever find a real world use for this particular exercise but that's the nature of exams, I guess.

For this sort of question, do it, get it right, collect the mark, and move on ....

If the explanation is too rambling and convoluted, ask again for clarification and I'll keep it to the point.

[ft]

Assuming they are looking for the fuelled up CG as %MAC:

Ignore the stations. All you need is a distance from a datum (i e a fulcrum) and the mass. There is no requirement to convert between STA and %MAC, nor do you need to know the actual position of the datum.

Just use %MAC as your fulcrums. A %MAC is a given length from the %MAC datum, and that's all you need. Just calculate the moments as usual in the rather odd unit pound-%MAC, rather than the pound-inches, kg-inches or kg-meters which you are probably more accustomed to.

BOW: 4200 lbs * 20 %MAC = 84.000 pound-%MAC
Fuel: 500 lbs * 30 %MAC = 15.000 pound-%MAC

Sum up the mass and the moments, again as per standard procedure.

TOW: 4700 lbs, moment 99.000 pound-%MAC.

And, as always, calculate the CG in %MAC by dividing the moment by the mass:

TOW: 4700 lbs @ 99.000/4700 = 21.06 %MAC.

[john_tullamarine]

ft - oh, indeed, elegance and simplicity, good sir - salut ! I shall away and beat myself into a pulp for not seeing it earlier.

However, lest a couple of unintended (I'm sure) typos in your post confuse our colleague, Pugachev Cobra (http://www.pprune.org/members/305898-pugachev-cobra), I suspect that you intended, for your solution

4200****20****84000
*500****30****15000
4700****21.1***99000

(or, 21.06 if one were to desire two decimals)

I presume that you did the calc and then retyped it with the first typo resulting in the second but the answer transcribed correctly from your original calculation ? However, 85500/4700 just doesn't quite fit the bill.

No matter .. I am more than impressed by your insight in the technique used .. well done, good sir.

[Harry Burns]

Was the exam for future pilots or engineers? :bored:

HB

[ft]

JT,
right you are and a confusing typo it was! My post is now edited, thank you very much for pointing it out.

[NoJoke]

You people are far too smart. Pull back ........... houses ..............

Well done boys.

[Pugachev Cobra]

Thank you all for the responses!

Harry Burns: The exam is for a Commercial Pilot rating.

john_tullamarine: In your first explanation, I understood everything but the "FS scale" part. I'm not familiar with this "FS" acronym. What does it stand for?

The answer seems just about right, there was the 21% answer.

Now, on a side question, forgive my physics knowledge, but I can't quite grasp the understanding of finding the CG by dividing the total moment with the total mass.

What is the logic or reasoning behind it? I know it's standard CG calculation, but in my books I can't find the explanation for the calculations.

Again thank you all, it's amazing how many people are willing to help here!

I only hope that I can contribute too.

[ft]

For weight and balance purposes, you find the moment around a given point created by a given object by multiplying the distance between the given point and the object and the mass of the object.

Similarly, the moment about that same given point created by the aircraft as a whole is given by the distance between the center of gravity and that given point multiplied by the total mass of the aircraft.

If the given point is the CoG datum, the distance between the center of gravity and the given point is the figure you are looking for when doing your weight and balance calculations.

Hence, if you calculate the moments created by all the objects you wish to include about the CoG datum and sum them up, the summed moments will equal the moment created by the entire aircraft around the CoG datum.

You can also easily calculate the mass of the aircraft as a whole by summing up the mass of all the objects you wish to include.

Now, let's put this in algebra.

The moment M is the product of the CoG distance d and the mass of the entire aircraft m.

M = d * m

The CoG distance d is what we are looking for. Rearranging the above by dividing both sides by the total mass m we get

M/m = d * m / m

Or, as m/m is equal to one

d = M / m * 1 = M / m

That's your division, right there.

The moment M1 created by object 1, say, the unfuelled aircraft, is similarly given as the product of the distance from the CoG datum to the object d1 and the mass of the object m1.

M1 = d1 * m1

Similarly, for object 2 (e g the fuel added)

M2 = d2 * m2

Now, by the above the moment created by the fuel AND the unfuelled aircraft will be the same as the moment created by the entire aircraft.

M1 + M2 = M

The total mass is m1 added to m2.

m = m1 + m2

Now, go back to

d = M / m

insert M = M1 + M2 and m = m1 + m2

d = (M1 + M2) / (m1 + m2)

The CoG distance is the sum of the moments divided by the sum of the masses.

[john_tullamarine]

I'm not familiar with this "FS" acronym. What does it stand for?

For W&B work, the easiest way to think about arms is to consider a datum being some place where you can hang your tape measure on a nail tapped into the side of the aircraft. The scale reading on the tape measure gives you the arm of the location (fore and aft of the datum in all cases - and sideways from a separate lateral datum for rotorcraft).

Fuselage Station is the usual longitudinal scale used by the OEM to define locations along the aircraft and, usually, is just a linear scale along the aircraft starting at the OEM's declared datum.

Note that the datum is just some declared position and has no intrinsic significance ie you can use a datum different to the OEM's if you chose and this is quite the typical circumstance with trimsheet design. Likewise, in ft's solution earlier, he chose to nominate a datum position at the LEMAC.

I can't quite grasp the understanding of finding the CG by dividing the total moment with the total mass.

ft's explanation might be a tad involved for your question unless you are comfortable with algebra. The following says much the same but puts a slightly different emphasis on it.

Basics for working out a centre of gravity (centre of mass, for the purists) are -

(I'm presuming that you are reasonably comfortable with (a) through (d) but it is convenient to review the material for completeness)

(a) we are looking for some position which can be thought of as a balance point for the whole aircraft

(b) if we could (and most times that is impracticable) put a pivot at this point, the aircraft would balance, ie not tip one way or the other. In physical terms this means that there is no leftover tipping tendency. The buzzword we use for "tipping tendency" is "moment" but it doesn't matter which term you use.

(c) it follows that (b) requires the total tipping tendency one way to be the same (size-wise) as the total tipping tendency the other way.

Working this out is the purpose of the normal moment sum we do when calculating W&B longhand. It would be just as valid to replace the pluses and minuses with two columns labelled something like "nosedown tipping" and "noseup tipping" - indeed, on some standard calculation forms one sees two separate columns for + and - which harks back to the olden days when we used to do it all longhand.

For the purpose of thinking about tipping tendency, we might consider the datum to be the point at which we put a pivot point for the calculation ...

(d) now, unless you happen to be very lucky, and guess a position for the datum which coincides with the CG (when you would cancel all the moments out), you are most likely to end up with some residual, or left over, moment.

(e) we now need to figure an arm from the datum to a position where a load equal to the total aircraft weight would give a moment sufficient to balance the residual moment. Why is this important ? .. because, if we then moved the datum to this position and redid the sums, then we would end up with no residual moment .. ie we would be at the CG.

(f) we could figure this position out by trial and error along the lines of -

total aircraft weight x trial and error arm = residual moment

and eventually end up with the answer ... or,

using a basic algebraic technique (transposition) we can figure the arm directly by reworking the equation to a similar form

arm = residual moment/total aircraft weight

and this arm is the CG.

[Matt101]

Top tip for Mass & Balance - realise it is all basic maths - ignore the fluff the examiners add in and get to the numbers - 100% shall be yours. :ok: (There are a few other facts to remember but even most of that is in the CAP)

[Pugilistic Animus]

Just to add a little one EXCELLENT reference [even for maintenance personnel] is the FAA 'weight and balance handbook'

especially since all FAA books are written at an assumed seventh grade level---perfect for me:}

[Pugachev Cobra]

Thank you John Tullamarine, ft and everyone else for the excellent explanations.

I come here for this topic every now and then to refresh my knowledge. I also have the FAA Weight and Balance handbook, and it's pretty neat.

Still, it's good to know the name of the technique, transposition, for calculating the arm from the datum.

I need to know where do these calculations come from in order to fully understand them, I don't like just swallowing in formulas without knowing the meaning of them.

Again, thank you all folks!

[arba]

You people are far too smart.

you guys are great, I know only how to fill trim_sheet and check computerized load_sheet ! may be they are right , I'm overpaid!