[color=blue]want to know the steps on how to go about it.[/color]

Pretty standard approach will fit the bill.

(a) stick the required load in and run the sums to see from where you are starting.

Put the occupants wherever you choose as you can move them after the first iteration. Put the cargo in the back as required and the fuel in the mains. Run the sums and see where that puts you. You should see that you have an aft CG consideration to address.

(b) then you need to make sure that you have the occupants 2-2-1 from the front which puts you as far forward as you can get prior to ballasting. You are after minimum ballast (which should always be a goal).

(c) play with the ballast - either trials if you are running longhand calculations or plot a suitable point on the graph if you want to work it that way.

The initial sums should make it pretty obvious what sort of ballast quantum you might need. Just be careful with the final answer as both the correct ballast answer and the pragmatic ballast answer are options.

Let us know how it turns out.

Hi John thank you for your explanation. Just also wondering when using the graphical method do we base our plotting points off our TOW or ZFW? ZFW is far aft of datum about 2505 kg at 2683 mm, when 120kg of fuel is added gives us 2625 kg at 2642mm which puts us in the CoG box. Just wanted to clarify when doing these. Thanks

[color=blue]when using the graphical method do we base our plotting points off our TOW or ZFW?[/color]

First, which of ZFW and TOW is causing you the greater problem, CG limit wise ? Let's get that one fixed first and then take stock of the situation ....

The ZFW CoG looks like is further aft then TOW, so from there we add as much ballast possible in the form of 5kg bags. Max we can add in the baggage compt is about 50 kg as we are trying to get as forward as possible also we can add weight in the 1st row ?

No need to rush. Sure, the aim is to get there reasonably quickly but, far more importantly, without making any mistakes. Make a significant mistake out in the aeroplane and you might kill yourself, in the exam, you might get the question wrong and fail the exam. Neither end result has much to recommend it.

Let's work it through, bit by bit, until we get to the final solution.

Let's have a think about your first observation.

[color=blue]The ZFW CoG looks like is further aft then TOW[/color]

Correct. But is that really important ? What we are really interested in is whether either of the ZFW and/or the TOW loading puts us outside the envelope. So, what is the story here ? Do we have either or both points outside the envelope ?

So we have the ZFW slightly outside the envelope and TOW inside the envelope. Sorry had to re plot ZFW just wanted to make sure.

Good man.

So, do we have a problem with the ZFW CG ?

Do we have a problem with the TOW CG ?

Looks like we have a problem with our ZFW CoG

That sounds to be a good call to me.

As an aside, we would be looking at the shape of the envelope and the slope of the individual load moments to make an assessment of where to go from here but let's just work it through and figure how it's going on the fly.

So, we have a ZFW problem. Pretty obviously, we don't have any options available to move things around once you have the occupants loaded 2 2 1 from the front so it's a case of needing some ballast. We wish to move the CG forward. The question says we want to use the minimum ballast feasible (that's always the case - unless the examiner makes some strange requirement for the questions - no point spending money (fuel) to carry weight which we have no commercial or other reason to carry). So, we plan to use the least amount of ballast by sticking it in the most forward location feasible. In practice, if that doesn't work by the time you have maxxed out in the most forward location, you might have to use some additional ballast in the next most forward location but that won't be necessary for this problem.

So, we can either play with some sums to zero in on the required ballast, or do it on the graph, plot a weight/CG delta and then interpolate from the graph to figure out the amount of ballast required.. Doesn't really matter which way you approach it in practice.

Have a go and perhaps you can walk us through your thought processes and working ?

With the graphic method I played around with the 00s, so I added 50 kg in the FWD locker which gave 2550kg, 2641mm 674.696IU. I then plotted this on the CoG box, which shows that it is now in the CoG box. I then drew a line connecting both of the points. Wherever the line intersected I then determined my probable weight of 2510Kg against my original ZFW which gives me about 5KG difference, if this sounds correct to you. Thanks for your help

Good man.

Now, you can see just how rough and ready using the graph is - absolutely dreadful.

How about you run the calculations longhand and tell me what the minimum ballast is to the nearer kilogram ?

Using the flow chart, I used the weight to add the formula which resulted in around 3 KG. At 2505 KG ZFW, there is a 3mm difference between the AFT limit of 2680 mm and the present CoG of 2683 mm. As we are trying to bring the CoG FWD as possible, we can determine that weight must be added in the Forward locker, as it is most available to us (2680-500 = 2180 mm diff). if we substitute all of this into the formula 2505 x 3mm / 2180 = 3.44 KG - nearest Kilogram gives us 3 KG. Correct me if I made an error.

We are now getting to the point where one needs to go to unrealistic precision to get a "correct" answer. Your working is fine for the question's requirement.

So, for the original question, which alternative will you select for your answer ?

I would go with the graphical method answer of 5KG. The reason being is by determining using trial an error to see what is the most accurate approach. What I have found in if we use 2505 + 5 kg our CoG will become 2679mm. However when using the calculative method, I found that 2505+3= 2508KG, CoG will become 2681 mm which is 1mm off.

3 x 500 / 10,000 = 0.15 IU + 672.196 = 672.346 / 2508KG = 2681 mm
5 x 500 / 10,000 = 0.25 IU + 672.196 = 672.446 / 2510 KG = 2679 mm

My best option is to go with the 5 KG which is our minimum ballast required to keep our Aircraft safe throughout flight. I hope this explanation make sense. Thanks