Hi all,
Just finishing up my revision for my PERF exam this Tues and have a couple of questions about some of the content in Bob's book.

Firstly to do with finding ISA TEP. I understand that the formula to use is 15 - 2 X (thousands ft). However in EX 1.5 Q1 the formula 15 + 2 X (thousands ft) is used. Is this due to the P.H being a negative or is it a typo, noting my book is issue 3 datin back to Jan 08?

The next question is to do with finding MTOW using the beetle diagram. When attempting EX 5.10 I noticed that you only take into consideration the TKOF limit and LDG limit + burn. I thought that you only dropped considering the 2630 + fuel at TKOF if you were trying to find Maximum fuel? The answer in the back said that the ZFW limit can't be a prob due to the ZFW being under 2630KG. Why don't we take it into consideration for this senario???

The last one is to do with Flow chart v Plotting on the graph when adding/subtracting wt in ECHO. I am aware that you only need to plot on the graph when the C of G falls outside the FWD limit and when the option of shifting is not available. However, when attempting Q6 in EX 5.13 ("An ECHO is loaded at ZFW with a otal wt of 2400KG and total moment of 590. Find the greatest amoun of fuel that may now be added to the man tanks only")
I started by finding the C of G (2458mm) then FWD limit (2411mm). As it fell inside the FWD limit I continued with the flow chart. However, I put the required C of G in as 2680mm. (Now getting of track a bit, but can someone please correct me here if im wrong, but shouldn't it go the other way? Since adding fuel to the mains would cause the C of G to move FWD... Therefore, should 2411mm (FWD limit) have been used as the Req C of G instead?)
So I continued with the flow chart still using the 2680mm as my C of G required, and came up with a figure that even two ECHO's combined could never carry!
So my question is, why can't this question be solved using the flow chart??? It has me confused

I think that is it for now. Hope these questions arn't too silly, but I just can't get my head around them at the moment.

Cheers Mick!

Hi Mick,

Here's some comments to your questions:

ISA Temp:
Yes, the formula is slightly different because you are dealing with a negative number.
The working: 15 -( 2 x -0.5) is mathematically equivalent to: 15 + (2 x 0.5). You are just subtracting a negative number which is the same as adding.

MTOW, Beetle Diagram Ex 5.10:
Think about what this particular exercise is asking: you are adding fuel as ballast and you are being given the ZFW. Any extra weight you add will be fuel and so will not change the ZFW of the aircraft in the question. You are already given a valid ZFW so you don't need to check it: it is already within limits.

On the other hand, if the question told you to add ballast you WOULD need to check the ZFW limit. You might end up adding so many bricks that the ZFW of the aircraft increases above the 2630kg ZFW limit.

Fuel as ballast, Forward CofG Ex 5.13 Question 6:
You can't use the flow chart in this case because you are adding weight (in the form of fuel) and if you change the weight, you change the forward C of G limit and the flowchart technique doesn't work. The forward limit for 2400 kg will not be the forward limit after you have added the fuel. The new weight will have a new forward limit.

You must use the 2 point method on the chart in this case. In fact, you must do this for any forward limit problems where you are required to add (not shift) weight.

By the way, you used 2680 but that is the aft limit for the Echo and has nothing to do with the forward limit. No wonder you were getting some odd answers

Just as a tip for the exam: if you have a question where you must add ballast to get you to the forward limit, you can use the chart to work it out but you can also simply check the answer options by calculating the CofG for the new weight (after adding the amount of ballast specified in the answer) and that CofG position should be the same as the forward limit that you can calculate for that new weight. If it is, you know you have the right answer.

Does that clear it up a bit for you? Let me know if it doesn't.

> Hope these questions arn't too silly
There's no such thing as silly questions...

Cheers,

Rich

Hi Richard,
Thanks for the reply mate. It has cleared a few things up for me

Now understand the change in the ISA TEMP formula...

Still trying to get my head around the ZFW limits and when to consider it. But after reading your answer, it is helping me figure it out a lot easier.

Would it be possible for you to send an example of the EXAM TIP you wrote about in your reply. I think I get what you are saying, however if I see some numbers, it may help me put everything into perspective!

Thanks mate!

Mick

Here's an example using the question you mentioned:

An Echo is loaded at ZFW with a total weight of 2400 kg and a total moment index of 590. What is the greatest amount of fuel that may now be added?
a) 90 kg
b) 120 kg
c) 70 kg
d) 65 kg

Obviously the best way to deal with this is to plot it on the graph and use the two point method that is described in the book. However you can check your answer quite simply (if you have time).

Let's assume in the heat of the moment, you came up with option (c) as the answer. We can check if this is right by looking at the new C of G position after we have added the 70 kg of fuel that option (c) suggests.

ITEM	WEIGHT	ARM	MOMENT
ZFW	2400		590
Fuel	70	1780	12.5
TOTAL	2470	--> 2439	602.5

From this you see that the CofG position after adding 70 kg of fuel is 2439mm aft of the datum. What's the forward limit for the new weight of 2470kg? If we are right, the CofG position should be exactly on the forward limit for that weight.

First off, you'll need the forward limit formula for the Echo out of the textbook:



So, the forward limit for 2470kg is:
= ( 2470 - 2360 ) x 0.27 + 2400
= 2430mm

Oops. 2430 ??? But we got 2439 for our CofG position. We're actually 9mm aft of the forward limit for this weight. That means we must still be able to add a bit more fuel.

So, now we can either go back to the graph and try again or we can simply choose another answer. The correct answer has to be one of the three remaining options. We need the CofG to move forward just a little so let's try the option with just a bit more fuel - let's try (a) and add 90kg instead:

ITEM	WEIGHT	ARM	MOMENT
ZFW	2400		590
Fuel	90	1780	16.0
TOTAL	2490	--> 2434	606.0

The forward limit at this new weight is:
= ( 2490 - 2360 ) x 0.27 + 2400
= 2435

Now 2435 is close enough to 2434 which was the CofG we calculated using the fuel load from option (a). It's sure as heck closer than 120kg or 65kg will be that's for sure!

So, with this trick, we have not only checked our answer but we have also been able to correct it without going back to the graph and starting all over again. This way you have the answer in front of you in black and white: you will be able to add 90kg of fuel.

It might save some time in the exam but still make sure you know how to use the graph method. This is really just a handy back-up check.

Cheers,

Rich

Thanks Rich!

That little 'double check' formula is a good one! And one that I have used a few times now to check an answer I got.
Completed my first practise exam this morning and got 90% Not too bad and am pretty happy with that. Few questions stumped me.. However I did suprise myself, as I managed to bluff my way through some that I thought I had no idea about! I may just be learning this subject after all!!!

Mick

Excellent backup method Richard. Very handy. Thank you.

Did you mean 'graph' when you wrote 'flowchart'?

rubber

Hi rubber,

Yes, I meant the two-point method on the "chart". Thanks for pointing that out. It must have been a slip of the typing fingers. I've changed it in the original post now

Cheers,

Rich