CPL Performance

[john_tullamarine]

The original poster has emailed me the current Echo docs etc. I'll put together some background explanation etc which might make it easier for you to see why you are doing what you are doing to come up with the answers. Main thing with weight control work is good housekeeping otherwise it is very easy to go off at a tangent ... unless you get involved with the more complex trimsheets, it doesn't get any more complicated than step a, step b, step c.

I'll potter about with this post progressively to add commentary on the files provided which include what appears to be a schedule of formulae and how-to-do-it flowcharts ...

Any specific questions .. please post and I'll endeavour to give you a sensible answer. Also, should you see a typo, please advise and I'll correct it.

A. Moment Index

The file gives the following

moment index = (weight x arm)/index unit

index unit = 100, 1000, etc

I have some reservation with terminology aspects of this.

Weight and balance calculations necessarily involve simple arithmetic moment calculations or an easy housekeeping record of clockwise and anticlockwise moment sums.

Moment is just the product (arithmetic jargon for multiplication) of a distance (what we usually call the "loading arm" or, more simply, the "arm") and a weight (strictly mass but that's not too important a distinction for pilot work).

The concept called "centre of gravity" is the point in a loading arrangement where there is no residual moment left over trying to rotate the system about the CG position (imagine a fulcrum position located at the CG position) ie we refer to the CG as being the point where the sum (jargon for the total addition) of the moments (with the calculations done for the CG position) is zero. Strictly this will be true for any point (datum) for which the calculations are done .. as seen later when we consider the significance of the datum position.

[I'll belabour the moment concept for a bit as, if this is not understood, then understanding the whole lot falls down in a heap.]

The moment sum idea results in the usual CG calculation table with which pilots have to contend. Initially, I'll draw this in the usual pilot way .. which is not strictly correct ..



Clearly this system is "balanced" (like a see-saw in the local park) and there is nothing there tending to rotate the support beam. If we calculate a simple moment equation as shown in the sketch, we can see that the total (the "sum") of the moments tending to rotate the beam clockwise is the same in magnitude as the total of the moments tending to rotate the beam anticlockwise.

Note that the moment is the product of a length and a weight and MUST have units of length x weight. In this calculation the units are metres and kilograms, generally written as kg.m

Clearly, you can generate more complicated systems to emphasise the point.

Conventionally, we set clockwise moments to be positive (plus values) and anticlockwise moments to be negative (minus values). This results in

(a) arms to the left of the datum being negative

(b) arms to the right of the datum being positive

(c) loads added being positive

(d) loads removed being negative

This sign (plus and minus) convention allows us to run the calculations arithmetically rather than keep tabs on clockwise and anticlockwise moments.

If we do that in this first simple case we get

clockwise moments = (+2) kg x (+3) m = (+6) kg.m = 6 kg.m

anticlockwise moments = (+2) kg x (-3) m = (-6) kg.m

and, if we add the two together (sum the calculations) we get

total moment = 6 kg.m - 6 kg.m = 0 kg.m

Note that, if you don't like pluses and minuses, you can run all the calculations by keeping track of clockwise and anticlockwise moments.

We say that the present datum position is the CG of the system because there is no tendency for any rotation .. ie the clockwise and anticlockwise moments balance out .. or, the sum of the moments is zero.

I suggested before that the picture was not quite correct. If we were to pick up the whole thing and put it onto some bathroom scales, the scales would indicate 4 kg (assuming that the weight of the plank and fulcrum were negligible). This 4 kg load doesn't affect the moments in this calculation (as it is in line with the datum and has a zero arm) but will for any other datum.

The next consideration is whether the datum location is significant ? Answer = no. In particular, the datum has no greater significance than being somewhere convenient to hang the end of one's tape measure on .. the ONLY important thing is that, once chosen, you MUST do all your sums with arms measured from the same datum .. otherwise it all turns into a wonderful mess ...

Look at the case where we take the datum for calculation to be moved to the left, say .. Has this changed the see-saw balance situation ? Of course it hasn't, but let's have a look at the sums to see what happens to the numbers. (This graphic insists on posting too big .. eventually I'll work out how to fix that).



If we use the arithmetic approach, the clockwise total moment is 24 kg.m and the anticlockwise moment - 24 kg.m and the total moment remains zero. You can play with moving the datum here and there and the situation stays the same .. the numbers change but the significance (and the balance consideration) doesn't.

Index Units

If we venture into the real world, especially with large aeroplanes, the moments can become VERY big numbers, especially if arms are measured in millimetres (mm). Considering the accuracy required and manipulation convenience, we really don't want to play with the sort of big numbers which can arise .. so we need a technique to simplify things .. and we end up with index units, by convention.

Index Unit (IU) is defined (ie we say that this is what an IU is)

IU = moment/a convenient constant moment

eg = 134,534,058 kg.mm / 1,000,000 kg.mm

= 134.534058

which we would round off to something like 134.53, say.

Note that the quantity we divide by is a "convenient constant moment". This means it is

(a) a moment (ie it has moment units .. kg.m, kg.mm, ton.yards .. whatever)

(b) a constant and is used for ALL calculations in a given problem or weight control situation

(c) chosen for convenience .. ie there is nothing intrinsically important about the number .. you might use 10, 10000, 12345, whatever .. the ONLY important thing is that, once chosen, you stick with the same convenient constant

and that the resulting IU has no dimensions (units) .. ie the IU is a plain, old, ordinary number.

Turning to the how-to-do-it document, the terminology chosen is very confusing in that it is not compatible with the normal conventions. I suggest that you think more conventionally and replace it with

IU = weight x arm/constant moment

where "constant moment" is any convenient value, usually powers of 10 for arithmetic simplicity.

The only other thing to keep in mind is whether you are working in moments or IU as the equations needed to work things out are a little different to take into account the difference between moment and IU as defined in the equation above. So long as you remember that the sums must involve moments and, if you are working in IU you have to convert to/from moment for the sums, you won't go too far wrong.

B. Calculating the CG

Using the example above, the CG calculation falls out pretty simply. We are trying to figure where to put a "balance point" so that the moments all cancel out. A very simple way to do this is to consider all the aircraft loads first and then work out a balancing load. The IU column is not necessary in this case but makes the numbers a bit easier to work with generally when we are looking at big numbers .. For this case, let's use an IU constant of, say, 10 to see what is going on.



Notice that

(a) moment = weight * arm

(b) IU = moment / constant

In practice we would choose either to include the moment or the IU column rather than both.

Notice that we have

(a) added the weights to get a total weight .. this gives us the weight on the scales as before

(b) added the moments (or IU) to give a total moment (or IU)

Please don't do anything with the "Arm" column and note very clearly that moments are NOT the same as IU .. related, yes, but not the same thing at all.

Now, to figure the CG, we recall that we need to achieve a zero total moment. That is, we need to do a sum to work out what CG value would make the following equation correct

total weight x CG = total moment

rearranging we might just as easily say we need to figure out

CG = total moment / total weight.

Plugging in the numbers gives

CG = 24/4 = 6

which we know to be the correct answer from the earlier calculations.

If we had chosen to do the sums in IU, rather than moments, we just need to make sure that we incorporate the IU to moment conversions in the CG sum.

CG = total moment / total weight

= (total IU *constant) / total weight

= (2.4 * 10) / 6

= 24 /6

= 6

which is what we expected to see.

... and, conceptually, that's about as hard as it gets .. the problem is housekeeping to make sure that we don't lose numbers along the way.

Conventionally, we "park" the CG answer in the table between the total weight and total moment values but it is very important to remember that the CG value is just parked there and doesn't follow the calculation pattern of the other line items in the table.

C. Mean Aerodynamic Chord (MAC)

Why do we get involved with MAC when, for a pilot (other than test pilots), it is absolutely of no interest to us ?

Solely because, conventionally, the AFM/POH loading information is prepared by the aerodynamicists in the design team (good chaps like djpil). The aerodynamicist is very interested in MAC for a host of non-W&B reasons but, being busy and not having time to rework the sums back to station values, generally presents all the limitations in MAC terms, rather than arms. It doesn't make any difference, other than for making the pilot's job a tad more involved arithmetically.

Relationship between Station Arms and Percentage MAC

For weight and balance, we are interested only in being able to convert percent (or fraction) MAC to/from fuselage station (arm). The two are only different ways of writing the same thing .. how far is something from the loading datum .. no more, no less.



This is all we need to know about %MAC.


Converting from Fuselage Station (arm) to % MAC


First we need to work out the distance back to the station from the MAC leading edge. Clearly this is

b = arm - a

where a is the distance from the datum to the LE MAC. Then we plug the value into the percentage MAC equation

%MAC = b/MAC * 100

= (arm - distance to LE MAC)/length MAC * 100


Converting from % MAC to Fuselage Station (arm)


Now, we can get b from the MAC equation

b = %MAC * MAC/100

so

arm = a + b

= distance to LE MAC + %MAC * MAC/100

It is important to be clear that %MAC is just another (inconvenient for pilots) way of saying loading arm.


For the Echo POH, the equations simplify to

arm = 2190 + 19 * %MAC

%MAC = (arm - 2190)/19


The formula schedule provided contains the following for MAC



I suggest that you don't try to use this .. it is nonsense as it is presented .. ie plain wrong. However, we can make some sense of it with a bit of tweaking and changing this and that.

The expressions above appear to be an attempt to describe the equation of the upper forward CG envelope limit (ie the slopey bit). If we reorder the terms and correct the odd error in the first expression we end up with

upper forward limit = 2400 + 0.2711864 * (GW - 2360)

where GW is gross weight within the range 2360 - 2950 kg and 0.2711864 is the slope (gradient) of the limit in mm/kg. What this is doing is saying "add to 2400 mm the prorated arm increment as weight increases above 2360 kg to get the forward limit at the higher weight".

A slightly simpler equation which does the same task is

upper forward limit = 1760 + 0.2711864 * GW

What it doesn't do is calculate the %MAC which is what it appears to be trying to do. If we look at the second expression in the formula schedule, we can see that for 2765 kg, the forward limit is 2510 mm and an arm of 2510 mm works out to be the same as 16.8 %MAC.


D. Floor Loading

The how-to-do-it document is a bit lax in that it doesn't keep track of the units.

Main point to keep in mind is that floor loading limits are similar to pressure or stress limits with units of force/area. If the actual value is less than the limitation (for the Echo 450 kg/m2), then you are fine (for that limitation, anyway).

Looking at the example given, with a 20kg box of base dimensions 200mm x 200mm (remember to convert mm to metres)

load = 20 kg

and

area = 200/1000 * 200/1000

= 0.2 * 0.2

= 0.04 m2

resulting in a floor loading intensity

FL = force / area

= 20 kg/0.04 m2

= 500 kg/m2

which is too high so you will need some shoring to spread the load over a greater floor area.




(Continued at post 19 to avoid the attention of the post length limit police ...)