PPLA exam number 2 …question number 53
Can this question be done on the Charlie system loading chart ?
After 2 hrs and total confused I looked at the answer and explanation seen there was a formula . For weight shift .

The Charlie loading system is the only one (except for Echo) that publishes the location of the aft limit. If you know the location of the aft limit and the location of the current CofG, you can always use the weight to shift formula.

You cannot use the formula (flow chart), for Alpha or Bravo systems because the actual position of the aft limit has not be provided.

Bob I can get to w = 982. A=2950. Iu 28770 is this my first plot on the graph ??
Are you able to explain where the new zfw comes from 995 and the 3004 please

Bryan, if you like to post the question, I can offer a few more comments to amplify Bob's observations.

After doing the calculations again I came up with 133 ,, this is Mtow - tow 1115-982 …aft limit 3004 on graph .. I must be missing something . This is all done on the graph not using the formula .

Bryan, Appears you have repeated the one page. Any chance we might have the second page, please ?

While we are waiting for Bryan to post the second page, I can make a few comments which might assist folks.

As Bob observes, the preferred way to approach this sort of problem is to use the ballast formula. Keep in mind that there are two different formulae which are used. One is if you are rearranging load to alter CG (you really don't need to remember the formula, as all you are doing is figuring out the required moment change and then figuring the equivalent weight change to achieve the moment change). The other is the ballast formula which covers the case where you are adding (or removing) weight to (or from) a location to achieve a CG change. Easiest to use the formula for that one.

Some observations.

The CG Envelope

The certification has to present the CG limits as CG limits so this is what we see in the Limitations Section of the POH.

However, when we run calculations (as pilots) CG values are of not much use as we prefer to use moments (IU, if you like) to figure out what the mathematical mechanicists refer to as moment balances. So, when you go to the Weight and Balance Section of the POH (Section 6) you see the CG data usually recast as moment data. Now, be very sure that these two presentations provide precisely the same data, just recast into a different format style. I have heard a variety of nonsensical fairy tales about how and why they differ. They do NOT differ at all in the data information provided.

When you have a weight by CG envelope, such as in the Charlie (which is a Beech Sundowner - with a few Examiner fiddles in the data presentation format to keep students on their toes - as Bob observes, on occasion, who, in their right mind, would plot graphical divisions at 8 mm ?) you need to be a tad careful. If you plot a couple of points and join them with a straight line to get an intersection with the CG limit, you are creating a problem as the actual line is not a straight line but a curve on this CG presentation format. If you plot a few points in between the end points you will see this quite clearly for Charlie. So, if you plot a straight line, you will get a "wrong" answer for the intersection with the limit but not incredibly wrong as the curve is relatively gentle. If you can't figure out any other way to get to the solution, by all means plot some end points but then you have to follow up with several bracketing calculations to figure the correct answer, having started with an approximate answer from the initial straight line plotting exercise.

However, when you have a weight by moment (or IU) format CG envelope then the line joining the two end points IS a straight line (providing that the arm is constant with varying weight - this can cause difficulties, say, with fuel loads in some aircraft) and the intersection with the envelope limit line will be fine. (One caveat - the usual sloping forward upper limit CG line often is drawn as a straight line but should be drawn as a curve to be correct. You can get yourself into minor trouble out in the real world if you forget this).

Envelope Data Values

When Bob suggests we should only use the data explicitly provided by the examiner in the loading systems, he is doing his best to keep things reasonably simple for you as students. However, be quite comfortable that there is no reason why you can't, or shouldn't, read data from the charts and, where necessary, run calculations on the data which you have read off the charts. So, for example, using Charlie, we can read the lower weight forward limit to be 2744 mm.

Bryan's main problem, at the moment, I suggest, is that he is using the envelope and getting approximate answers. It is essential, then, that the approximate answer is used to run some bracketing calculations to find an acceptably correct answer.

When Bryan has time to post the second sheet, there may be some further comments which I can add appropriately.

Thanks, Stuart, that fits OK. (As an aside, for those who have gone via a longhand calculation, there is a minor round off error accumulation discrepancy between the formula and the longhand calculation. The actual answer is a bit over 113 but a bit below 114. Not something one would worry about too much in practice as the approach would be to load not more than 113 in the exam to avoid any round off error accumulation problems).

It's a bit difficult to work out just what Bryan has done as he has only given us glimpses of his working rather than the whole solution - any time you have a problem to resolve it is always preferable to include a scan of your whole solution as that makes it very much easier to find the error(s) - a LOT easier as we don't have to second guess what is going on.

It appears that he has run the initial load calculation at post #3 (with a mistake in copying the fuel arm instead of the initial TOW arm). In respect of his question regarding the ZFW of 995/3004, that is the ZFW load with the required row 2 loading included (113/114 kg).

At post #5 , he observes 133, this is Mtow - tow 1115-982 …aft limit 3004 on graph 133 kg in row 2 brings the TOW load up to 1115 kg but with the aft CG breached for both ZFW (the more critical) and TOW cases. But then he provides no information as to where he has proceeded so we can't provide any further assistance at this point. If he were to plot the loaded case he would have found that the intersection with the aft envelope limit (on a weight by CG chart) would give him an incorrect answer due to the load line's being plotted as a straight line when it should have been plotted as a curve.

The only way to proceed from this point would be to run some longhand calculations to eventually get to the answer. Certainly not the ideal way to go but, if all else failed and you had a mental blank in the exam, it would allow you to get to the answer in a rather long-winded way. Better to have done enough practice examples so that the process is burnt into the memory banks, I suggest.

Perhaps Bryan can give us some more information as to how he has proceeded and we can offer some further assistance to him ? Overall, though, the way to approach this sort of problem is to use the ballast formula.