Discussion
Sea level static thrust-to-weight ratios required by the cruise constraint at four altitudes.
| Altitude (m) | Thrust lapse factor | \( T/W \) at altitude | Sea level static \( T/W \) |
|---|---|---|---|
| 9000 | 0.36 | 0.058 | 0.161 |
| 10000 | 0.32 | 0.059 | 0.184 |
| 11000 | 0.28 | 0.062 | 0.221 |
| 12000 | 0.24 | 0.068 | 0.283 |
What does the table demonstrate about cruise altitude and engine size?
Answer: C
Correct answer: A higher cruise altitude demands more static thrust
The sea level static column rises steadily with altitude - \( 0.161 \) at 9000 m to \( 0.283 \) at 12,000 m - even though the thrust actually required at altitude barely changes, from \( 0.058 \) to \( 0.068 \).
The cause is the lapse factor, which falls from \( 0.36 \) to \( 0.24 \) across the same range. Since the sea level figure is the requirement at altitude divided by the lapse, a shrinking denominator inflates it. Higher cruise altitudes therefore cost engine size, which is the counterweight to the aerodynamic and fuel-burn advantages of flying high.
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