A derivative aircraft differs from a clean-sheet design in that it
Answer: A
Stretching a fuselage or re-engining an airframe reaches a market far more cheaply than starting again. The penalty is that the original design point constrains the result, and successive derivatives drift further from optimal.
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The field length requirement affects the design principally through
Answer: B
A short field demands a low wing loading for lift and a high thrust-to-weight ratio for acceleration, both of which penalise cruise efficiency. The balance between field and cruise requirements is one of the first trades made.
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A design specification differs from a set of requirements in that the specification
Answer: C
Requirements state what must be achieved; the specification records what the chosen design actually is and does. Confusing the two leads to requirements that quietly embed a solution and close off alternatives before they are examined.
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Certification requirements are treated differently from customer requirements because they
Answer: D
Airworthiness requirements are a condition of being permitted to fly at all, so they enter the design as absolute constraints rather than as criteria to be weighed. Customer requirements, by contrast, can and frequently are negotiated.
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Market analysis contributes to the requirements set by
Answer: A
An aircraft that meets no identified need will not sell however elegant it is, so route structures, fleet replacement cycles and competitor capability all feed into the top-level requirements. Several technically excellent aircraft have failed on this point alone.
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An aircraft has an operating empty weight of \( 45 \) t, payload of \( 20 \) t and fuel of \( 25 \) t. Its take-off weight is
Answer: B
Correct answer: 90 t
The take-off weight is the sum of the three:
\[ 45 + 20 + 25 = 90 \text{ t} \]Operating empty weight covers the structure, systems, engines and crew; payload is what is being carried for revenue; and fuel is the third component. This simple addition is the constraint that generates the payload-range diagram, since the total is capped.
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If that aircraft is limited to a \( 90 \) t maximum and its tanks are enlarged to hold \( 30 \) t, the payload it can carry becomes
Answer: C
Correct answer: 15 t
With the take-off weight capped, extra fuel can only come out of payload:
\[ 90 - 45 - 30 = 15 \text{ t} \]The extra 5 t of fuel has cost 5 t of payload, one for one. That exchange is exactly the sloping portion of the payload-range diagram, along which range is bought at the price of revenue load.
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Using the Breguet equation with \( V = 250 \) m/s, \( c = 1.7 \times 10^{-4} \) per second, \( L/D = 18 \) and a weight ratio of \( 1.4 \), the range is nearest to
Answer: D
The Breguet range equation for a jet is \[ R \;=\; \frac{V}{c}\,\frac{L}{D}\,\ln\!\frac{W_1}{W_2}. \] Taking the logarithm first, \( \ln 1.4 = 0.336 \). Then \[ \frac{V}{c} \;=\; \frac{250}{1.7 \times 10^{-4}} \;=\; 1.47 \times 10^{6}\ \text{m}, \] and multiplying by the remaining two factors, \[ R \;=\; \left(1.47 \times 10^{6}\right)\left(18\right)\left(0.336\right) \;=\; 8.9 \times 10^{6}\ \text{m} \;=\; 8900\ \text{km}. \] Note the units: \( V/c \) has the dimensions of a length, so the whole expression comes out in metres and must be divided by a thousand. A specific fuel consumption of \( 1.7 \times 10^{-4} \) per second is about 0.6 lb per lbf per hour, which is typical of a cruising turbofan. The logarithm means that doubling the fuel fraction does far less than double the range.
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Improving \( L/D \) from \( 18 \) to \( 20 \) at constant everything else changes the range by
Answer: A
Correct answer: About 11 per cent
The Breguet range equation makes range directly proportional to lift-to-drag ratio at constant weight fraction and specific fuel consumption, so
\[ \frac{20}{18} = 1.111 \]an improvement of about 11 per cent. Two points of \( L/D \) is a substantial aerodynamic achievement, and the fact that it converts one for one into range is why so much design effort goes into drag reduction.
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An aircraft of \( 90 \) t on a wing of \( 120 \) m\( ^2 \) has a wing loading of
Answer: B
Converting to force, \( 90\,000 \times 9.81 = 883 \) kN, and dividing by the area gives \( 883\,000/120 = 7360 \) N/m\( ^2 \). Wing loading is among the first numbers chosen in conceptual design, since it drives field length and cruise efficiency in opposite directions.
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