Weight Estimation Methods Questions and Answers

Answer: C

Correct answer: 76.5 t

\[ 0.85 \times 90 = 76.5\ \text{t.} \] The landing limit is below the take-off limit because the undercarriage and its backup structure are sized for touchdown at a given sink rate, and designing them for the full take-off weight would cost weight the aircraft carries on every flight for a case that arises only in an emergency - which is why an overweight landing needs fuel jettison or an inspection afterwards.

Answer: D

Correct answer: 10.6 per cent

\[ \frac{9.5}{90} = 0.1056 \approx 10.6\ \text{per cent.} \] Ten to twelve per cent of take-off weight is the usual range for the wing group of a transport, so the answer is also a sanity check on the estimate. The distractor 9.5 per cent simply repeats the mass in tonnes as a percentage, which is a coincidence of the numbers rather than a calculation.

Answer: A

Correct answer: 4 per cent

\[ \frac{3.6}{90} = 0.04 = 4\ \text{per cent.} \] Undercarriage groups typically fall between three and five per cent of take-off weight, so this is a normal figure. It is worth remembering against the distractor that offers a fifth of take-off weight - no aircraft carries an undercarriage anything like that heavy.

Answer: B

Correct answer: 0.53

\[ \frac{48}{90} = 0.533. \] The empty weight fraction is one of the two numbers a Class I estimate turns on, and for a transport it usually lies between about 0.45 and 0.55. A value far outside that range is a signal that the estimate is wrong rather than that the design is remarkable.

Answer: C

Correct answer: 6.8 per cent

The increase is expressed as a fraction of the earlier figure: \[ \frac{94 - 88}{88} = \frac{6}{88} = 0.068 = 6.8\ \text{per cent.} \] Dividing by 94 instead gives 6.4 per cent and is the usual slip. A Class II estimate coming out above Class I is the normal direction of travel, because the group formulae capture attachments and secondary structure that aircraft-level fractions smooth over.

Answer: D

Correct answer: 0.077 m

Burning fuel removes weight, so both the total and the moment change: \[ \Delta x = \frac{w \times d}{W - w} = \frac{2 \times 3}{80 - 2} = \frac{6}{78} = 0.077\ \text{m forward.} \] The denominator is the weight remaining, not the weight before - using 80 gives 0.075 m, close enough here to look plausible, which is why the distinction is worth being deliberate about.

Answer: A

Correct answer: Outside the aft limit

The envelope runs from 15 to 35 per cent MAC, and 38 per cent lies beyond its aft edge. An aft centre of gravity reduces the static margin, so the aircraft becomes less stable and the pitch control lighter than certified - which is why the loading must be corrected before flight rather than treated as a small exceedance.

Answer: B

Correct answer: 7240 N/m^2

Wing loading is weight, not mass, divided by area: \[ \frac{90\,000 \times 9.81}{122} = \frac{882\,900}{122} = 7237\ \text{N/m}^2. \] Forgetting the factor \( g \) gives 738 N/m², which is one of the distractors. A figure of around 7000 N/m² is typical for a transport, so the magnitude is itself a check.

Answer: C

Compounding, \( 48 \times 1.015^4 = 48 \times 1.0614 = 50.9 \) t. Growth of one and a half per cent a quarter looks negligible in any single review and amounts to nearly three tonnes in a year, which is exactly why the trend is tracked rather than the snapshot.

Answer: D

Correct answer: 6.7 per cent

The overrun is measured against the allocation: \[ \frac{9.6 - 9.0}{9.0} = \frac{0.6}{9.0} = 0.0667 = 6.7\ \text{per cent.} \] Dividing by the reported weight instead gives 6.25 per cent. The point of allocating targets by group is precisely that a number like this can be computed and acted on while the design can still be changed.