Hi Guys
I'm struggling with this question for exam prep...hoping someone can help me with how to work it out. Thanks in advance!

Question 12 of 20 (Marks: 4)
Refer VFR (Day) Booklet - Loading System ECHO, Pages 18 to 21. An Echo aeroplane is loaded as follows:
Aeroplane weight with row 3 seats removed ...... 1970 kg
Aeroplane moment with row 3 seats removed .... 475.66 index units
Row 1 .................................................................. pilot and passenger
Row 2 .................................................................. 2 passengers
Row 3 .................................................................. 4 crates @ 40 kg each
Rear compartment ............................................... 5 packages @ 30 kg each
Ballast available .................................................. 5 kg sand bags

The minimum amount of fuel required as ballast in the main tanks is closest to -


A - 80 kg
B - 40 kg
C - 135 kg
D - nil

You need at least 35kg of ballast fuel. So the only answer you can pick is 40kg.

Thank you so much Bob, stupidly forgot to take ZFW into account and therefore added 55kg to the nose and as a result didn't 'need' any fuel ballast

Hi Bob

What is the best way to approach the following question? I can get to the correct answer 180kg by adding weights by trial and error and calculating how the different answers compare to their FWD CoG Limit. By doing this, I get +-160kg which is obviously closest to 180kg...this method is just quite time consuming considering the time pressure in this exam.. thanks!

Question 15 of 20
Refer VFR (Day) Booklet - Loading System ECHO, Pages 18 to 21.
An Echo aeroplane is loaded with the main tanks full and the auxiliary tanks empty, giving a gross weight and moment index of 2655 kg and 666 index units respectively.
The maximum fuel that may now be added to the auxiliary tanks is -

180 kg
215 kg
100 kg
295 kg

This question involves the forward limit of the Echo loading graph. If you are asked to find the weight that must be added or subtracted to move the C of G to the forward limit you cannot do it by a formula. The only way is to use the method you have described. Either by trial and error using each of the given answers, or by adding or subtracting any arbitrary amount of weight to establish a second point on the chart.

Join the two points and note where the line joining them crosses the forward limit line.

Note that if you are SHIFTING weight to get to the forward limit, then you can always use the formula or flow chart.

Thank you Bob!