Weight Estimation Methods Questions and Answers

Answer: A

Correct answer: 0.164 m

Adding weight increases the total, so it appears in the denominator: \[ \Delta x = \frac{w \times d}{W + w} = \frac{3 \times 4}{70 + 3} = \frac{12}{73} = 0.164\ \text{m aft.} \] Using 70 rather than 73 gives 0.171 m, which is offered as a distractor. The rule is the same in both directions: divide the moment by the weight the aircraft finishes with.

Answer: B

Both are true and the second explains the first. An aircraft can therefore be within its take-off weight limit and still overload the wing root, if too much of that weight sits in the fuselage as payload rather than in the wings as fuel.

Answer: C

Both statements are true and the second is the reason for the first. A return shortly after departure therefore requires fuel jettison or an overweight landing inspection, since the aircraft has not yet burned down to the landing limit.

Answer: D

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason is the justification. A single aircraft-level target tells you only that the aircraft is heavy, and only once everything has been added up - by which time the design is frozen. Allocating a share to each group means every increment is compared against a budget as it happens, so growth is visible while there is still time and freedom to recover it.

Answer: A

Both statements are true and the second is the reason for the first. The regression describes aircraft as built, including their weight growth, but applied at aircraft level it hides where the mass sits and so understates awkward details.

Answer: B

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason gives the purpose. Expressing centre of gravity as a percentage of the mean aerodynamic chord non-dimensionalises it, so 25 per cent means the same aerodynamic thing on a regional jet as on a widebody. Absolute distances from a datum would be meaningless between types, and the certified envelope limits could not be compared at all.

Answer: C

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason gives the mechanism. With the centre of gravity far forward the aircraft is very stable, which is not the problem; the problem is that the elevator must generate a large nose-up moment against a long arm to rotate at take-off and to flare at landing. When the required deflection exceeds what is available, the limit has been reached - so it is control authority, not stability, that sets it.

Answer: D

Both are true and the second explains the first. Shifting the centre of gravity aft within the envelope reduces the tail load needed to trim, and the fuel saving over a long sector justifies the system's weight and complexity.

Answer: A

Both statements are true and the second is the reason for the first. Quoting one definition against another flatters whichever aircraft is described by the narrower one, so the basis must be established before any comparison is made.

Answer: B

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason explains why a single statement would not do. Weight growth arrives as a stream of small, individually reasonable increments - a bracket here, a thicker skin there - and no one of them looks alarming. Only by plotting the same statement at intervals does the accumulation become a trend, and only a trend can be acted on before the design is committed.