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- aalim21
- Topic Author

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

Last edit: 1 year 3 months ago by aalim21.

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- John.Heddles
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- ATPL/consulting aero engineer

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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 ?

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 ?

Engineering specialist in aircraft performance and weight control.

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- aalim21
- Topic Author

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.

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- John.Heddles
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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 ?

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

Engineering specialist in aircraft performance and weight control.

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- aalim21
- Topic Author

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

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

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- John.Heddles
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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.

Actually, had you run the sums for 4 kg, you would have been just about spot on. Not really worth worrying about in the real world as the numbers you start with for the sums are nowhere near that accuracy. So, 3 or 5 kg is fine. The answer to select for the question has to be 5 kg as that's the mass of the ballast items you have available to you. I doubt that the examiner would be interested in splitting hairs to the nearer kilogram so all should be fine for the exam as well.

It is obvious that you have run the sums, either longhand or in a spreadsheet, so I am a little surprised that you didn't pick up on the 4 kg - not to worry - it's a trivial point.

So, well done. I might have given you the odd clue along the way but you have successfully run the analysis and arrived at the correct final solution by yourself. Do you think this sort of problem will give you any further problems ? I shouldn't think so.

Perhaps run a few more examples to reinforce the approach and you will have this sort of question well and truly sorted out.

Actually, had you run the sums for 4 kg, you would have been just about spot on. Not really worth worrying about in the real world as the numbers you start with for the sums are nowhere near that accuracy. So, 3 or 5 kg is fine. The answer to select for the question has to be 5 kg as that's the mass of the ballast items you have available to you. I doubt that the examiner would be interested in splitting hairs to the nearer kilogram so all should be fine for the exam as well.

It is obvious that you have run the sums, either longhand or in a spreadsheet, so I am a little surprised that you didn't pick up on the 4 kg - not to worry - it's a trivial point.

So, well done. I might have given you the odd clue along the way but you have successfully run the analysis and arrived at the correct final solution by yourself. Do you think this sort of problem will give you any further problems ? I shouldn't think so.

Perhaps run a few more examples to reinforce the approach and you will have this sort of question well and truly sorted out.

Engineering specialist in aircraft performance and weight control.

Last edit: 1 year 3 months ago by John.Heddles.

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- aalim21
- Topic Author

Thanks a lot for your help and explanation. Much appreciated! Is there a possibility whereby the examiner gives us data that both TOW and ZFW fall withing the CoG range? If so would I pick the one with the worst case scenario, like a ZFW more aft then TOW but still in range?

Last edit: 1 year 3 months ago by aalim21.

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- John.Heddles
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Is there a possibility whereby the examiner gives us data that both TOW and ZFW fall withing the CoG range?

Absolutely the case.

Add, say, another 10 kg to the previous 5 kg ballast. This puts the CG values to around 2670.4 (ZFW) and 2630 (TOW).

Simple enough then to pose a problem like the following -

"The aircraft will be expected to perform optimally for fuel burn if the CG is as aft as can be arranged. What is the minimum weight you need to leave behind for your flight to achieve this if you don't relocate the passengers ?"

Absolutely the case.

Add, say, another 10 kg to the previous 5 kg ballast. This puts the CG values to around 2670.4 (ZFW) and 2630 (TOW).

Simple enough then to pose a problem like the following -

"The aircraft will be expected to perform optimally for fuel burn if the CG is as aft as can be arranged. What is the minimum weight you need to leave behind for your flight to achieve this if you don't relocate the passengers ?"

Engineering specialist in aircraft performance and weight control.

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- aalim21
- Topic Author

No worries will have a look at problems like those as well. Thank you for all your help! You're a life saver

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- John.Heddles
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Not at all. You did the work, not I.

Engineering specialist in aircraft performance and weight control.

The following user(s) said Thank You: Stuart Tait, aalim21

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